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"We have a full construction crew," Renner said. "We'll knock this thing out."
Firefighters responded to Escape 618 after a fire broke out around 4 a.m. Nov. 24. The fire was quickly contained and no one was injured, Belleville Fire Chief Tom Pour said.
The cause of the fire is listed as spontaneous combustion from oily rags. In recent years, Belleville firefighters have responded to three commercial business fires and five house fires caused by the improper disposal of oily rags, Pour said.
Renner said he plans move forward with plans to open as soon as the damage is cleaned up.
His great-grandfather, Theodore Bauer, built the imposing Grant Building on East Main Street in downtown Belleville and now Renner will bring new life into the building as Escape 618.
Customers, including kids and their families, corporate groups, class field trips or couples on a date, pay to enter a themed room and then use team building skills to figure their way out within a time limit. Renner notes that customers are not "locked" in the room, they can leave at any time if they want and a "game master" will accompany and monitor the players.
Several escape room businesses are already in operation in the St. Louis metro area.
Renner hopes to have nine themed rooms in the building.
The initial phase will feature four escape rooms on the first floor. Five other escape rooms will be opened later in the basement and apartments could be built on a top floor after the escape rooms are finished. Also, the themes would be changed on a regular basis.
"With an escape room, you can benefit the community," Renner said in a September 2017 interview. "You can bring people together. I can take kids, throw them in a room, get them off their phones and I can make them work together."
BND staff reporter Mike Koziatek contributed to this report.
After fire damages interior, Escape 618 will open in Belleville later this year. Cara Anthony canthony@bnd.com
Tyler Renner stands inside of a Escape 618 . His new attraction coming to downtown Belleville. Cara Anthony canthony@bnd.com
Cara Anthony
Cara Anthony covers restaurants and retail for the Belleville News-Democrat, where she works to answer readers' questions about restaurant openings, business closures and the best new dishes in the metro-east. She attended Althoff Catholic High School in Belleville and grew up in East St. Louis.
Here's what would happen if an earthquake like those in California occurred in Illinois
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{"url":"http:\/\/www.proofwiki.org\/wiki\/User_talk:GFauxPas","text":"# User talk:GFauxPas\n\nJump to: navigation, search\n\n## Change to MathWorld citation template\n\nI noticed (based on One-to-One and Strictly Between) that some pages on MathWorld are credited to different authors from Eric Weisstein, and so require that author to be included in the citation.\n\nI have fixed the template (which is now \"MathWorld\" not \"Mathworld\", that's just me tidying up) so as to be able to include the author (which, if not given, defaults to the \"Weisstein, Eric W.\" format as per normal).\n\nWhat you need to do is add \"author=author-name\" and \"authorpage=author-pagename\" where \"author-name\" is the displayname of the author and \"author-pagename\" is the name of the html file on MathWorld (not including the full path, not including the extension).\n\nAn example:\n\n\u2022 {{MathWorld|One-to-One|One-to-One|author=Barile, Margherita|authorpage=Barile}}\n\nwhich gives:\n\nIf the page is given as written by \"Weisstein, Eric W.\" then you should not add the \"author\" and \"authorpage\" tags.\n\nI have included this info in the usage section of the Template:MathWorld page itself, but I'm bringing it to your attention because I know you've been active in using it.\n\nChx. --prime mover 02:55, 31 December 2011 (CST)\n\n## Source Review\n\nJust a quick heads up to draw your attention to Definition:Random Variable, which has an open SourceReview call listed on your name. \u2014 Lord_Farin (talk) 11:42, 5 April 2014 (UTC)\n\nI have that book packed away; I'll have to dig it up. --GFauxPas (talk) 11:20, 8 April 2014 (UTC)\n\n## Mathjax\n\nIs there a way to take the code of a given page and copy paste it into some external program or into some website, and end up with a pdf?\n\n--GFauxPas (talk) 14:43, 8 April 2014 (UTC)\n\n## zeta 2\n\nIf you've been watching my sandbox, you might have seen me work on this:\n\n$\\displaystyle \\int_{\\to 0}^{\\to 1} \\ln x \\ln \\left({1-x}\\right) \\, \\mathrm dx = 2 - \\zeta\\left({2}\\right)$\n\nShould I put this up as a theorem, or is it too specific an exercise to be worth sharing?\n\n--GFauxPas (talk) 11:42, 9 April 2014 (UTC)\n\nI think it's worth sharing, specially after all the hard work you put into it. The name needs to be considered carefully. Might want to put it into an \"\/Examples\" subpage of either Logarithms or Integral Calculus (or both) ... once I've reached that part of my Spiegel handbook where there are hundreds of examples of integrations I will be in there to flesh it out, but it's something I haven't got to yet. --prime mover (talk) 04:26, 10 April 2014 (UTC)\nOr maybe under Basel Problem... --GFauxPas (talk) 11:03, 10 April 2014 (UTC)\nNo, rather not under Basel problem -- the latter is a specific problem with a specific solution (i.e. the one Euler came up with, which explains its name), this just happens to have a result which includes the same zeta instance. --prime mover (talk) 16:43, 10 April 2014 (UTC)","date":"2014-04-24 09:43:02","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.640845537185669, \"perplexity\": 2420.2051106247363}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"absolute_threshold\": 10, \"end_threshold\": 5, \"enable\": false}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2014-15\/segments\/1398223206118.10\/warc\/CC-MAIN-20140423032006-00217-ip-10-147-4-33.ec2.internal.warc.gz\"}"}
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package io.fabric8.kubernetes.client.dsl;
import io.fabric8.kubernetes.client.Client;
import io.fabric8.kubernetes.client.V1AuthorizationAPIGroupDSL;
import io.fabric8.kubernetes.client.V1beta1AuthorizationAPIGroupDSL;
public interface AuthorizationAPIGroupDSL extends Client {
/**
* DSL for authorization.k8s.io/v1 api group.
*
* @return V1 Authorization API DSL object
*/
V1AuthorizationAPIGroupDSL v1();
/**
* DSL for authorization.k8s.io/v1beta1 api group
*
* @return V1beta1 Authorization API DSL object
*/
V1beta1AuthorizationAPIGroupDSL v1beta1();
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
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Home > THQ Nordic > Review: Darksiders Genesis (PC)
Review: Darksiders Genesis (PC)
Review by Matt S.
I could feel that something like Darksiders Genesis was coming. Though the first two Darksiders titles were impressive games, they were also beholden to a kind of design that was never going to stick around. Darksiders 3, which I generally enjoyed, nonetheless showed just how difficult modernising that formula is. However Darksiders as a property is too important for THQ Nordic to ignore, and so here we have Darksiders Diablo... I mean, Genesis.
Related reading: Also from this year, Darksiders 3. Our review!
It's really good, too. It's a lot of fun and offers pacey, exciting action. It's a bit of a brainless delve (as most Diablo-likes are), but featuring two of the four horsemen, there's some great interplay and dark humour going on, which makes the adventure itself a lark. Throw in all the demons and imagery from hell, and this is still a game that builds on the ongoing Darksiders setting and lore, while also being a far enough break that the developers could approach it from their own angle.
So you've got two of the four horsemen. One dual-wields pistols. The other has his brutally effective sword. The dynamic between those two is such that multiplayer is clearly the way to push through Darksiders Genesis if you can, but as a single player experience you can still switch between the two... and you'll want to, as various combat and non-combat situations will better suit one or the other. Both control well, though it doesn't take long for the sheer number of skills to become almost overwhelming. Within the first hour I had to start making decisions about what active skills I'd carry into to battle, and which I would bank away and forget about, and in some ways this speaks to the deeply attention deficit nature of modern videogames - developers seem to be worried that if you don't keep throwing shiny things at players to distract them, they might get bored. Darksiders doesn't have the loot system of Diablo, so it seems that instead the developers decided to keep throwing mechanical features at players to provide variety.
Darksiders has never just been about the action. The earlier titles were inspired by Zelda, and thus featured plenty of platforming and puzzle solving. The third one took its best crack at Dark Souls-like environmental exploration, while also keeping those puzzles going. So it should come as no surprise that Darksiders Genesis... has non-combat elements! Yep, you'll need to do the occasional platforming, wall climbing and puzzle solving, and though none of those elements are difficult enough to be time consuming, they are a break from the action. I didn't find them to be particularly inspired, but they did enough to remind me that this is Darksiders, and it is its own unique thing. The same goes for the horse riding stuff - at points you can mount your horse (you are, after all, playing as a horseman of the apocalypse), and while this is almost entirely superficial, since it limits what skills you can use, and objectives are rarely far enough apart to bother summoning up the beast, it does help to go for a ride a couple of times, purely for fun.
What I was less impressed with were the environments. Look. I know that designing hellscapes in video games is a challenge. You can go all-out and inevitably end up upsetting people, like the developers of Dante's Inferno and Agony did. Hell is the actual cesspool of a pure kind of evil, and technically it should be shocking and offensive, but that doesn't make for great commercial content. And so most developers end up going with option "B" - a creatively bereft barren wasteland with some fires and oceans of lava. Evil is, apparently, an endless rocky desert in a volcanic region. I didn't like when Doom took this approach, I don't like it here.
The problem is that a "journey into hell" should have certain implications which never seem to be explored by these games. It can be fetid and vile, as Dante described it. But it can also be colourful and vivid. Just take a look at a Hieronymus Bosch's painting and try telling me the endless rocks, magma and iron chains of most video game hell designs deserve to be called creative. Those artists were using hell to convey specific thematic meaning, however. Most game developers are just out there creating entertaining content, and so the level design of Darksiders Genesis ends up feeling shallow, dull, and nowhere near what the inspired action of the game deserves. Darksiders does benefit in that its hub-like structure allows you to visit some mild alternatives to the barren rocky stuff, such as a barren snowy area, or a barren "generic green atmosphere" area, and a number of other "hostile" environments, but none of them are any more convincing as parts of hell.
That same issue follows on to the enemy designs. For one thing, there aren't that many different types of enemies that you face down, and my impression of hell, from having read stuff like Dante, is that the monstrosities that inhabit it should not look like they've come off factory lines like they're Funko Pops (which, for the record, are everywhere in hell, as looking at them is a form of torture). Secondly they're all very... rote, with the developer checking off what you'd expect to see from a bunch of demons rather than throw anything at you that would surprise and/or actually terrify you. This is a disappointment for the same reasons that those settings are disappointing. I fundamentally believe that hell should be the last thing that settles for just meeting expectations, and I'm consistently disappointed with the lack of creative flair from game developers that attempt to do something with it.
There are a couple of other more minor issues that I had. The camera is zoomed up close enough that there were times that enemies were menacing me from off-screen, which is always annoying. At the same time, the camera was scaled back just enough that when I was in the middle of a melee, it was hard to make out what was going on. There are fine lines in these things, because Diablo-likes rely on the enemy swarm, but I took a little too much damage a little too often trying to work out what was actually going on in the heat of things. Additionally, while Darksiders Genesis is generally pretty linear, there are all kinds of sidequests and bonuses to find along the way, and the game's map system is absolutely terrible at presenting information.
Darksiders Genesis gets a lot right. It has plenty of personality and a dynamic, engaging action system. Its only real problem is that it has tackled a thematic challenge that it can't deliver - hell should not be dull, empty or repetitive, and yet this game fails on all fronts there. It's not entirely the developer's fault, in the sense that if they showed a truly creative vision of hell it would either offend or be too surreal for the kind of mainstream audience that it's pitched at, but perhaps developers need to better consider where they're setting their games in that case.
Find me on Twitter: @digitallydownld
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{"url":"https:\/\/puzzling.stackexchange.com\/questions\/98883\/numbers-that-are-the-sum-of-the-cubes-of-their-digits","text":"# Numbers that are the sum of the cubes of their digits\n\nThere are just four 3-digit numbers which are the sums of the cubes of their digits. For example:\n\n$$370 = 3^3 + 7^3 + 0^3$$ and $$371 = 3^3 + 7^3 + 1^3$$.\n\nWithout using a calculator\/computer, can you find the other two 3-digit numbers with this property? Are there any more such numbers?\n\n\u2022 Spoilers: answer inside. I\u2019m pretty sure the first part of the answer has to be done by exhaustion. The second part is an actually interesting problem, though \u2013\u00a0El-Guest Jun 5 at 3:46\n\u2022 See also OEIS: A005188 \u2013\u00a0Daniel Mathias Jun 5 at 9:55\n\u2022 By a difficult and exhaustive search, I've found two more numbers that are the sum of the cubes of their digits: $0, 1$. \u2013\u00a0Paul Sinclair Jun 5 at 17:32\n\nWe are finding digits $$a,b,c$$ such that $$100a+10b+c=a^3+b^3+c^3$$. Taking $$\\pmod 9$$, we have $$\\big(a^3-a\\big)+\\big(b^3-b\\big)+\\big(c^3-c\\big)\\equiv0\\pmod9$$\nThese are the values of remainder of $$a^3-a$$ divided by $$9$$:\n\na(mod 9)|a^3-a(mod 9)\n0 |0\n1 |0\n2 |6\n3 |6\n4 |6\n5 |3\n6 |3\n7 |3\n8 |0\n\n\nSo\n\nThe $$3$$ digit numbers that satisfy the condition are either all digits from either groups $$(0,1,8,9), (2,3,4), (5,6,7)$$ or one digit per group.\n\n\u2022 Now it reduces to 42 possibilities, which can be easily bruteforced by hand. \u2013\u00a0trolley813 Jun 5 at 14:10\n\nI happen to know them. Does that count as a valid answer? When I was young, we 'discovered' that repeatedly applying the procedure $$abc \\to a^3 + b^3 + c^3$$ always ended up at one of four numbers; 370, 371,\n\n$$153 = 1^3 + 5^3 + 3^3$$ or $$407 = 4^3 + 0^3 + 7^3$$.\n\nFor me, it's hard to forget, just like this anecdote about Hardy visiting Ramanujan:\n\nI remember once going to see him when he was ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavourable omen. \"No,\" he replied, \"it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways.\"\n\n($$1729 = 1^3 + 12^3 = 9^3 + 10^3$$)\n\n\u2022 Haha well that works too, if you already know the answer :) \u2013\u00a0Dmitry Kamenetsky Jun 5 at 6:53\n\u2022 That is one of my favorite maths stories, along with Gauss summing 1+2+...+100 in a few seconds in class. I am not entirely sure that the events actually took place, but they certainly make great stories. \u2013\u00a0Dmitry Kamenetsky Jun 5 at 7:03\n\u2022 Somebody needs to find that taxi cab (or at least the number sign from the top of it) and install it in a mathematics-themed museum somewhere. \u2013\u00a0Darrel Hoffman Jun 5 at 13:45\n\nTo show there is no four digit solution, the maximum sum of the cubes of the digits of a four digit number is $$4\\cdot 9^3=2912$$ For a number less than this, the maximum sum of the cubes of the digits is $$1+3\\cdot 9^3=2188$$. The thousands digit must be $$1$$. To get the sum of cubes up to $$1000$$ we need a $$9$$, two $$8$$s, one $$8$$ plus two $$7$$s, or three $$7$$s. We can check that $$1,7,7,7$$ and $$1,7,7,8$$ fail. With two $$8$$s we have $$1^3+2\\cdot 8^3=1025$$ and all the possibilities fail. Then $$1^3+9^3=730$$ We need another digit to be at least $$4$$ to get up to $$1000$$. This is in the range of hand check as well and nothing works.\n\n\u2022 The thousands digit must be 1. Why? \u2013\u00a0Ross Presser Jun 5 at 13:30\n\u2022 Because the sum of the cubes is less than 2188 at that point and no number with a thousands digit of $2$ matches its sum of cubes because the lower digits cannot contribute enough. \u2013\u00a0Ross Millikan Jun 5 at 13:37\n\nThere are\n\n\u2022 two 1-digit solutions: $$0,1$$\n\u2022 no 2-digit solutions: $$5$$ and above have 3-digit cubes. A digit of $$4$$ would require the number to have another digit of $$6$$ or above. The 12 possibilities with digits $$\\le 3$$ are easily eliminated.\n\u2022 four 3-digit solutions, as indicated in the question.\n\u2022 no 4-digit solutions, as Ross Millikan has proved.\n\u2022 no higher-digit solutions, as for $$n > 4, n \\times 9^3$$ has fewer than $$n$$ digits.\n\nSo there are six numbers total that are the sum of the cubes of their digits.","date":"2020-10-28 14:54:11","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 37, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6708856821060181, \"perplexity\": 617.2398974309867}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-45\/segments\/1603107898577.79\/warc\/CC-MAIN-20201028132718-20201028162718-00439.warc.gz\"}"}
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{"url":"https:\/\/www.hackmath.net\/en\/math-problem\/4105","text":"# Three vectors\n\nThe three forces whose amplitudes are in ratio 9:10:17 act in the plane at one point so that they are in balance. Determine the angles of the each two forces.\n\nResult\n\nA = \u00a053.13 \u00b0\nB = \u00a0154.942 \u00b0\nC = \u00a0151.928 \u00b0\n\n#### Solution:\n\n$F_{1}=9 \\ \\\\ F_{2}=10 \\ \\\\ F_{3}=17 \\ \\\\ u_{1}=\\dfrac{ 180^\\circ }{ \\pi } \\cdot \\arccos((F_{2}^2+F_{3}^2-F_{1}^2)\/(2 \\cdot \\ F_{2} \\cdot \\ F_{3}))=\\dfrac{ 180^\\circ }{ \\pi } \\cdot \\arccos((10^2+17^2-9^2)\/(2 \\cdot \\ 10 \\cdot \\ 17)) \\doteq 25.0576 \\ \\\\ u_{2}=\\dfrac{ 180^\\circ }{ \\pi } \\cdot \\arccos((F_{1}^2+F_{3}^2-F_{2}^2)\/(2 \\cdot \\ F_{1} \\cdot \\ F_{3}))=\\dfrac{ 180^\\circ }{ \\pi } \\cdot \\arccos((9^2+17^2-10^2)\/(2 \\cdot \\ 9 \\cdot \\ 17)) \\doteq 28.0725 \\ \\\\ A=u_{1}+u_{2}=25.0576+28.0725 \\doteq 53.1301 \\doteq 53.13 ^\\circ \\doteq 53^\\circ 7'48\"$\n$B=180-u_{1}=180-25.0576 \\doteq 154.9424 \\doteq 154.942 ^\\circ \\doteq 154^\\circ 56'33\"$\n$C=180-u_{2}=180-28.0725 \\doteq 151.9275 \\doteq 151.928 ^\\circ \\doteq 151^\\circ 55'39\"$\n\nTry calculation via our triangle calculator.\n\nOur examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!\n\nLeave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):\n\nBe the first to comment!\n\nTips to related online calculators\nFor Basic calculations in analytic geometry is helpful line slope calculator. From coordinates of two points in the plane it calculate slope, normal and parametric line equation(s), slope, directional angle, direction vector, the length of segment, intersections the coordinate axes etc.\nCheck out our ratio calculator.\nTwo vectors given by its magnitudes and by included angle can be added by our vector sum calculator.\nCosine rule uses trigonometric SAS triangle calculator.\n\n## Next similar math problems:\n\n1. Theorem prove\nWe want to prove the sentence: If the natural number n is divisible by six, then n is divisible by three. From what assumption we started?\n2. Vectors\nFor vector w is true: w = 2u-5v. Determine coordinates of vector w if u=(3, -1), v=(12, -10)\n3. Vector - basic operations\nThere are given points A [-9; -2] B [2; 16] C [16; -2] and D [12; 18] a. Determine the coordinates of the vectors u=AB v=CD s=DB b. Calculate the sum of the vectors u + v c. Calculate difference of vectors u-v d. Determine the coordinates of the vector w.\n4. Vector sum\nThe magnitude of the vector u is 12 and the magnitude of the vector v is 8. Angle between vectors is 61\u00b0. What is the magnitude of the vector u + v?\n5. Scalar dot product\nCalculate u.v if |u| = 5, |v| = 2 and when angle between the vectors u, v is: a) 60\u00b0 b) 45\u00b0 c) 120\u00b0\n6. Bearing\nA plane flew 50 km on a bearing 63\u00b020' and the flew on a bearing 153\u00b020' for 140km. Find the distance between the starting point and the ending point.\nA ship travels 84 km on a bearing of 17\u00b0, and then travels on a bearing of 107\u00b0 for 135 km. Find the distance of the end of the trip from the starting point, to the nearest kilometer.\n8. Decide 2\nDecide whether points A[-2, -5], B[4, 3] and C[16, -1] lie on the same line\n9. Coordinates of a centroind\nLet\u2019s A = [3, 2, 0], B = [1, -2, 4] and C = [1, 1, 1] be 3 points in space. Calculate the coordinates of the centroid of \u25b3ABC (the intersection of the medians).\n10. Coordinates of vector\nDetermine the coordinate of a vector u=CD if C(19;-7) and D(-16;-5)\n11. Vector\nDetermine coordinates of the vector u=CD if C[19;-7], D[-16,-5].\n12. Angles\nIn the triangle ABC, the ratio of angles is: a:b = 4: 5. The angle c is 36\u00b0. How big are the angles a, b?\n13. Median\nThe median of the triangle LMN is away from vertex N 84 cm. Calculate the length of the median, which start at N.\n14. Angles by cosine law\nCalculate the size of the angles of the triangle ABC, if it is given by: a = 3 cm; b = 5 cm; c = 7 cm (use the sine and cosine theorem).\n15. Scalene triangle\nSolve the triangle: A = 50\u00b0, b = 13, c = 6\n16. Side c\nIn \u25b3ABC a=2, b=4 and \u2220C=100\u00b0. Calculate length of the side c.\n17. Greatest angle\nCalculate the greatest triangle angle with sides 197, 208, 299.","date":"2020-04-08 22:33:41","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 3, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7528904676437378, \"perplexity\": 1731.08935761442}, \"config\": {\"markdown_headings\": true, \"markdown_code\": false, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-16\/segments\/1585371824409.86\/warc\/CC-MAIN-20200408202012-20200408232512-00321.warc.gz\"}"}
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I was born in Blackburn and have lived in either Blackburn or Darwen all my life. I'm married with two grown up sons and three granddaughters. My professional background has been in the education sector where I have undertaken the roles of secondary school teacher, local authority teacher adviser, senior lecturer in higher education, executive director of Lancashire Area West Training and Enterprise Council and, until my retirement in 2007, Director of post 16 education and training for East Lancashire with the Learning and Skills Council for England. I was an associate inspector with OFSTED and have also been a secondary school governor for over 15 years.
In 2008 I was appointed as a non executive director for Blackburn with Darwen Primary Care Trust. Since 2012 I have acted as a lay adviser to Blackburn with Darwen Clinical Commissioning Group with a particular focus on public and patient engagement and quality, performance and effectiveness. Whilst respecting the professional expertise within the NHS, I also believe that a critical component of its future success will be to develop greater and more effective engagement with its service users and the general public. I believe that all members of Healthwatch have a vital role to play in making this a reality.
I have been a member of the Healthwatch Board since 2014. I take a particular interest in health, education, and housing issues, and I sit on a number of other boards in these sectors. In my day job, I am Director of Strategy and Communications at a housing company with a £200m turnover operating across the North of England.
Through close family, I have been a regular user of a wide range of health services in Blackburn and East Lancashire over the last 10 years. I feel my professional and personal life will allow me to make sure Healthwatch is an active consumer champion for Health Service users in the Borough.
I originate from Leeds but have lived and worked in Blackburn most of my life. I am married with three grown up daughters and eight grandchildren. I worked for Lancashire County Council for twenty seven years in roles including the registration and inspection of private and voluntary homes for all client groups, the management of the Local Authority Homes and Day Centres and the development of services for adults.
I have now retired from over 30 years of National, Regional and Local leadership roles in both the Public and Third Sectors. I have extensive experience in both Governance and Management of individual organisations, partnerships and networks. Locally, I was the Chief Officer of Age UK Blackburn with Darwen from 1988 building it into the successful, respected charity it is today.
This has given me well-honed management skills with employees and volunteers as well as finance and assets. After being the Vice Chairman of the Hospital Trust (I signed the PFI documents for the Third Phase of what is now the Royal Blackburn), I was appointed as the first Chairman of the BWD PCT and was instrumental in the building of new influential partnerships with the Borough.
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{"url":"https:\/\/www.hackmath.net\/en\/math-problem\/2077","text":"# Brackets\n\nPlace one pair of brackets in expression 20-16 + 10 - 4 - 2 so that the result will be 0.\n\nResult\n\nx = \u00a00\n\n#### Solution:\n\n$x=20 - (16 + 10 - 4 - 2)=0$\n\nOur examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!\n\nLeave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):\n\nBe the first to comment!\n\n## Next similar math problems:\n\n1. Flowers\nThe flower has six flowers and each flower has a number. These are the numbers: 20,40,39,28,8,9. What number will be in the middle of the flower so that the numbers come from the flowers when we subtract and add?\nPupils doing research which a winter sport do their classmates most popular. They found that 2\/5 of classmates would most like to play hockey, skate prefer 2\/9 pupils, 3\/10 students prefer skiing and 1\/15 classmates don't like any winter sport. What pro\n3. Laboratory\nMang Elio went to his laboratory at 7:00am And recorded that the temperature was 11.07\u00b0C. At lunchtime, the temperature rose by 27.9\u00b0C but dropped by 8.05\u00b0C at 5:00 PM. What was the temperature at 5:00 PM?\n4. Cupcakes\nIn a bowl was some cupcakes. Janka ate one third and Danka ate one quarter of cupcakes. a) How many of cookies ate together? b) How many cookies remain in a bowl? Write the results as a decimal number and in notepad also as a fraction.\n5. Crocuses\nThe garden grow daffodils, crocuses and roses. 1400 daffodils, crocuses is 462 more and roses is 156 more than crocuses. How many are all the flowers in the garden?\n6. Game 27\nSusan wanted to play the game. In the beginning, the first says a number from 1 to 8. Then the second player adds a number from 1 to 5 and tells the sum. Again, the Susan adds a number from 1-5 and say sum and etc. . . The winner must say the number 27. Wh\n7. Log\nWorker cut his thick log to 6 pieces for 30 min. How long he cut log to 12 pieces?\n8. Table 3x3\nTable with numbers 3x3 in the first row 66,24,33 in the second row 57,? 19 in the third row 18,45,60 What number comes ar mark ?. Maybe the numbers: 22,46,45,47\n9. Travel by bus\nFive girls traveling by bus. Each holds in each hand two baskets. Each basket contains 5 cats. Each cat has 5 kittens. 3 girls getting off the bus and another girl without baskets go on board. The bus control driver. How many legs are in the bus?\n10. Infirmary\nTwo thirds of children from the infirmary went on a trip seventh went to bathe and 40 children remained in the gym. How many children were treated in the infirmary?\n11. Sum of fractions\nWhat is the sum of 2\/3+3\/5?","date":"2020-04-01 23:56:34","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 1, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.30935874581336975, \"perplexity\": 2851.252040302641}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-16\/segments\/1585370506477.26\/warc\/CC-MAIN-20200401223807-20200402013807-00477.warc.gz\"}"}
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D'Andre Lawan "Dre" Kirkpatrick (born October 26, 1989) is an American football cornerback who is a free agent. He played college football at Alabama and was drafted by the Cincinnati Bengals in the first round of the 2012 NFL Draft. He has also played for the Arizona Cardinals and San Francisco 49ers.
High school career
Kirkpatrick attended Gadsden City High School, where he played for the Gadsden City Titans high school football team. He was a teammate of fellow Crimson Tide players Jerrell Harris and Kendall Kelly. He completed his senior football season in the fall of 2008. Kirkpatrick participated in the 2009 U.S. Army All-American Bowl and the 2008 Alabama Mississippi All-Star Football Classic, where he was recorded at 6 ft 2 in and 180 lbs.
Kirkpatrick played his final three high school seasons at Gadsden City (35 games) and finished his career with 193 tackles, 17 interceptions, three returned for a touchdown, 36 pass breakups, two sacks, one punt return touchdown, one kick return touchdown, and one receiving touchdown. After completion of his senior season he was awarded as the ASWA 6A back of the year as well as a first-team all-state defensive back for the second consecutive year.
Entering February 2009, Kirkpatrick was one of the most highly rated and sought after high school football recruits in the country, having been awarded a five-star rating by national recruiting outlets Rivals.com and Scout.com.
On February 7, 2009, Kirkpatrick ended months of recruiting, committing to the Alabama Crimson Tide on National Signing Day, as he announced his decision live on ESPNU.
Professional career
Coming out of Alabama, Kirkpatrick was projected to be a first round pick by NFL draft experts and scouts. He received an invitation to the NFL combine and opted to skip the bench press, short shuttle, and three-come drill due to a shoulder injury and after injuring his hamstring during the combine. On March 7, 2012, Kirkpatrick participated at Alabama's pro day and performed the three-cone and positional drills for the team representatives and scouts from all 32 NFL teams. He was ranked the third best cornerback prospect in the draft by NFLDraftScout.com and NFL analyst Mike Mayock.
Cincinnati Bengals
The Cincinnati Bengals selected Kirkpatrick in the first round (17th overall, traded from Oakland in exchange for Carson Palmer) of the 2012 NFL Draft. He was the third cornerback drafted in the 2012 NFL Draft, behind Morris Claiborne (6th overall, Cowboys) and Stephon Gilmore (10th overall, Bills).
2012
On May 18, 2012, the Cincinnati Bengals signed Kirkpatrick to a four-year $8.62 million contract that includes $7.84 million guaranteed and a signing bonus of $4.71 million.
Kirkpatrick joined a veteran core of cornerbacks that included six highly touted first round picks. While preparing for training camp, Kirkpatrick fractured his knee and was expected to miss the next six weeks and be sidelined during training camp. His knee recovered in August before the Cincinnati Bengals played their last preseason game. Kirkpatrick returned to practice, but developed tendinitis in his knee that later delayed his debut. Head coach Marvin Lewis named Kirkpatrick the sixth cornerback on the Bengals' depth chart to begin the season, behind Nate Clements, Leon Hall, Terence Newman, Jason Allen, and Adam Jones.
He was inactive for the first seven games and made his professional regular season debut on November 4, 2012, recording one tackle during the 23–31 loss to the Denver Broncos. In Week 11, Kirkpatrick recorded a season-high two combined tackles in the Bengals' 28–6 defeat over the Kansas City Chiefs. The next week, he made a tackle in a 34–10 victory over the Oakland Raiders, but suffered a concussion and was unable to finish the game. Kirkpatrick missed the following four games and was inactive for the rest of his rookie season. He finished the season with four combined tackles and was limited to five games.
2013
Kirkpatrick entered training camp in and competed for the starting cornerback role with Terence Newman, Leon Hall, and Adam "Pacman" Jones. He was named the fourth cornerback on the depth chart behind Newman, Hall, and Jones.
On November 17, 2013, he collected two combined tackles and made his first career sack on Jason Campbell in a 41–20 victory over the Cleveland Browns. On December 1, 2013, Kirkpatrick made his first career interception after he picked off a pass attempt by San Diego Chargers' quarterback Philip Rivers during a 14–10 victory. The following week against the Indianapolis Colts, Kirkpatrick recorded a season-high six solo tackles in the Bengals' 42–28 victory. On December 15, 2013, he made his first career start in place of Terence Newman who was out with an MCL injury and finished the 20–30 loss to the Pittsburgh Steelers with four combined tackles.
On December 21, 2013, Kirkpatrick had his best game yet when he collected a season-high nine combined tackles, three pass deflections, and intercepted two passes From Baltimore Ravens' quarterback Joe Flacco, as the Bengals won 34–17. He made his second interception during the fourth quarter and returned it for a 21-yard touchdown, marking the first touchdown of his career.
He finished the season with 30 combined tackles (23 solo), five pass deflections, three interceptions, a sack, and a touchdown in 14 games and three starts. The Cincinnati Bengals received a playoff berth after finishing first in the AFC North with an 11–5 record. On January 5, 2014, Kirkpatrick started his first career playoff games and made two solo tackles during the Bengals' 10–27 AFC Wildcard loss to the San Diego Chargers.
2014
Throughout the Cincinnati Bengals' training camp in , Kirkpatrick competed for the job as the starting cornerback against Adam Jones, Terence Newman, Leon Hall, and rookie Darqueze Dennard. He was named the fourth cornerback to start the season, behind Hall, Newman, and Jones.
In Week 10, he collected a season-high four combined tackles in the Bengals' 3–24 loss to the Cleveland Browns. On December 14, 2014, Kirkpatrick made his first start of the season and deflected a pass and intercepted a pass attempt by Brian Hoyer during the 30–0 victory at the Cleveland Browns. The following week, Kirkpatrick made three solo tackles, two pass deflections, and intercepted two passes from Peyton Manning, as the Bengals' defeated the Denver Boncos 37–28. He returned the first interception for a 30-yard touchdown and sealed the game with his second interception in the final minutes.
He finished the season with 23 combined tackles (20 solo), seven pass deflections, three interceptions, and a touchdown in two starts and 16 games. The Cincinnati Bengals earned a playoff berth after finishing second atop the AFC North with a 10-5-1 record. On January 4, 2015, Kirkpatrick started in the AFC Wildcard game against the Indianapolis Colts and made four combined tackles and defended a pass during their 10–26 loss.
2015
On April 28, 2015, the Cincinnati Bengals opted to exercise the fifth-year option on Kirkpatrick's rookie contract, paying him a salary of $7.5 million in .
Kirkpatrick entered Cincinnati Bengals' training camp and competed against Adam Jones, Leon Hall, Darquese Dennard, and Josh Shaw for the vacant starting cornerback job left by the departure of Terence Newman. Head coach Marvin Lewis named Kirkpatrick the starting cornerback, along with Adam "Pacman" Jones, to begin the season.
During a Week 12 contest against the St. Louis Rams, Kirkpatrick collected a season-high ten combined tackles and defended three passes in their 31–7 victory. On December 28, 2015, he earned seven solo tackles and deflected a pass in a 17–20 loss at the Denver Broncos. He finished the season with a career-high 70 combined tackles (63 solo) and a career-high 16 pass deflections in 15 starts and 16 games.
2016
Kirkpatrick began the season as the starting cornerback with Adam Jones after winning the job over Darqueze Dennard and William Jackson III.
On September 18, 2016, he recorded two combined tackles, defended two passes, and intercepted a pass attempt from Ben Roethlisberger during a 16–24 loss to the Pittsburgh Steelers. In Week 8, Kirkpatrick recorded a season-high six solo tackles in a 27–27 tie against the Washington Redskins. On November 14, 2016, Kirkpatrick made five combined tackles, deflected a pass, and intercepted a pass By New York Giants' quarterback Eli Manning, as the Bengals lost 20–21. The next game, he collected two combined tackles, deflected a pass, and intercepted a pass from Tyrod Taylor in the Bengals' 12–16 loss to the Buffalo Bills. He finished the season with 46 combined tackles (35 solo), ten passes defensed, and three interceptions in 15 games and 14 starts.
2017
On March 9, 2017, the Cincinnati Bengals signed Kirkpatrick to a five-year, $52.5 million contract extension that includes $12 million guaranteed and a signing bonus of $7 million.
Kirkpatrick was named the starting cornerback, along with Adam "Pacman" Jones, for his second consecutive season after beating out Darqueze Dennard, Josh Shaw, and William Jackson III.
On October 29, 2017, Kirkpatrick recorded three solo tackles, defended a pass, and sacked Indianapolis Colts' quarterback Jacoby Brissett in a 24–23 victory. The following week, he made a season-high six solo tackles during a 23–7 loss to the Jacksonville Jaguars. In Week 11, Kirkpatrick intercepted Denver's Brock Osweiler in the end zone and returned it 101 yards, before mishandling and fumbling the ball and recovering it at the one-yard line. Kirkpatrick sustained a concussion during the Bengals' Week 13 loss to the Pittsburgh Steelers and was sidelined for the next two games (Weeks 14–15). Kirkpatrick finished the 2017 season with 55 combined tackles (47 solo), a career-high 14 pass deflections, an interception, and a sack in 14 games and 14 starts. He received an overall grade of 56.4, which ranked 91st among all qualified cornerbacks in 2017.
2018
In 2018, Kirkpatrick started 13 games, recording 41 tackles, nine passes defensed and no interceptions. He was placed on injured reserve on December 28, 2018 with a shoulder injury.
2019
On November 15, 2019, Kirkpatrick was placed on injured reserve after hyper-extending his knee back in Week 6. He finished the season starting six games, recording 33 tackles, four passes defensed, and a sack.
Kirkpatrick was released by the Bengals on March 31, 2020, after eight seasons.
Arizona Cardinals
Kirkpatrick visited the Arizona Cardinals on August 21, 2020, and signed with the team two days later.
In Week 6 against the Dallas Cowboys, Kirkpatrick intercepted a pass thrown by Andy Dalton during the 38–10 win. This was Kirkpatrick's first interception as a Cardinal.
In Week 10 against the Buffalo Bills, Kirkpatrick intercepted a pass thrown by Josh Allen during the 32–30 win which would later be referred to as Hail Murray. He finished the season playing in 14 games with 11 starts, recording 56 tackles, three interceptions (tied for first on the team), and seven passes defensed (second on the team).
San Francisco 49ers
On September 14, 2021, Kirkpatrick signed with the San Francisco 49ers. He was released on November 16.
Personal life
Kirkpatrick's son, Dre Kirkpatrick Jr., is a defensive back at Gadsden City High School in the class of 2024.
References
External links
Arizona Cardinals bio
1989 births
Living people
African-American players of American football
Alabama Crimson Tide football players
American football cornerbacks
Cincinnati Bengals players
Arizona Cardinals players
San Francisco 49ers players
Sportspeople from Gadsden, Alabama
Players of American football from Alabama
21st-century African-American sportspeople
20th-century African-American people
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Hear from Another World: Holocaust survivor Paul Smith still suffers 90 years later
Nearing the end of his journey of 90 years and across three continents, Paul Smith explains his complex experiences with alienation and religion.
Sasha Hassan
Sraavya Sambara and Daniel Shen
"I think the Holocaust is always going to be with me … My own family was murdered there."
He sits across from us in a worn, cerulean armchair, his wizened hands laced together on his lap and his translucent, lightly hazel eyes fixed on an impenetrable spot in the distance. His wrinkles are a map of his past: a journey across three continents and nine decades to find a home after the Holocaust took his away from him. The memories crush him. But they also define him.
This is Paul Smith*.
A journey across three continents
One morning in 1936, Smith wakes up in his childhood home in Königsberg, East Prussia (now Kaliningrad, Russia) to find that several extended family members are crowded in the living room and that his father has disappeared. His mother, not telling him why, soon packs him a bag and takes him — her only child — to a Jewish children's home in a town on the other side of Germany.
Smith has heard about the recent power grab of a man named Adolf Hitler, who has been targeting the Jewish population in his promise to redeem Germany after World War I. But the 8-year-old can't connect the dots to understand that his father has been kidnapped and that the relocation of his extended family to one house is simply the first step of the "Final Solution."
Smith was raised in a Jewish household in the heart of Germany just as Hitler was rising to power.
The children's home, although traumatic and confusing for an 8-year-old Smith, is integral in his escape from Germany. Hitler's regime is slowly closing in on him and his mother, and during his stay in the children's home, she is able to acquire emigration papers to London, England.
Her efforts are just in time. Only a few weeks later, much of their remaining family in Germany are sent to concentration camps by the German Schutzstaffel.
However, the war continues to follow the two Smiths to their new location; just as they arrive in London, the Blitz begins. As a result of the bombings, Smith is once again forced to pack his belongings, including a gas mask, and move, this time to the countryside parish of Beaconsfield to stay with an English family.
For the first time since he had left his home in Germany two years earlier, Smith feels relatively safe. In England, Smith soon learns to speak English and makes several friends at school. His mother, who has to stay in London, writes postcards to him during the weekdays and visits him on the weekends.
However, their time in England is cruelly cut short by yet another immigration crisis, and Smith's mother must find another country to take them in. World War II has officially begun, and Europe is not safe anymore, especially for a young, vulnerable Jewish boy.
In the summer of 1940, Smith finds himself in a Liverpool convoy heading to Valparaíso, Chile. The journey is long, strenuous, and riddled with danger as German submarines attack their vessel multiple times during the crossing.
Finally in Chile, Smith continues his schooling while his mother starts a business to finance their livelihood. He learns yet another language, Spanish, to keep up with the curriculum and ploughs through the rest of his education up to the 10th grade. Upon his graduation, he finds a job to support himself.
However, Smith's attention is soon caught by what many Chileans revere as "the land of opportunity" — America.
Sraavya Sambara
In 1949, Smith addressed a request for an immigration visa directly to the White House.
In hopes of actualizing his recently forged American Dream, Smith visits a number of U.S. embassies in Chile to apply for an immigration visa. But every time, his requests are dodged, ignored or redirected. Acting on promises made by the immigration officials, Smith transfers all his Chilean currency to an American bank, writes hundreds of application letters to employment agencies in New York and begs relatives to sponsor him — all to no avail.
With all of these attempts having failed, in 1949, Smith writes an earnest letter to the one person left whom he believes could help him: Harry Truman, president of the United States of America.
The letter to Truman is a miracle. Within two weeks, he is accepted into the Chilean office of the American embassy for the first time.
"[The consul general] said to me … 'I hear you're corresponding with the president … Bring me your passport. You're getting an immigration visa to come to the United States,'" Smith recalls.
Between Judaism and Presbyterianism
Judaism had always been deeply ingrained in Smith's life. Smith came from a traditional Jewish family, attended a Jewish school during his childhood in Germany, and remained faithful even in the predominantly Roman Catholic Beaconsfield.
However, his religious commitment was tested in Chile, where members of his English boarding school relentlessly bullied him for being Jewish. Racial epithets such as "dirty Jew" and "Christ killer" were directed at him on a daily basis by students and teachers alike.
It was this persecution that drew him to the U.S., where he believed he could finally escape the torrents of hatred directed toward him. When he reached America in 1949, Smith finally made the decision to cast off his religion, the part of his identity that had been targeted for so long.
"I don't like to say it, but I did leave my religion for a while, for a long time … I went through so much as a kid that when I came here, I just wanted to leave anything Jewish [behind]."
Through the next couple decades of his life, Smith remained disenchanted with his Jewish faith. It was several years into his six-year auditing job in San Francisco before he was finally directed toward religion again. This time, however, it was not Judaism.
"Somehow, someone told me go and take a look at the Baptist Church* in San Francisco. And I went there on one Sunday morning. And something just happened in me … I just became so happy and satisfied, and I thought I finally found something that accepted me."
The following Sunday, he attended a Presbyterian service. Gradually, religion found its way into his life again in the form of an Evangelical faith.
"Everybody accepted me; everybody made me feel at home … they were just wonderful. I made a lot of friends, and I have a lot of good memories of all those years I spent in the Presbyterian Church," Smith said.
But as he had feared, his family and other Jews did not receive the news that he had adopted the Presbyterian faith well. The distant guilt nagging at the back of his head was harshly intensified by his relatives' slurs, addressed to him in multiple letters.
With all of Smith's family members being Jewish, his brief conversion to Presbyterianism was ill-received and condemned by many relatives.
"If a Jewish person … finds out that I had converted to [the] Presbyterian Church … they feel that I'm a traitor, to their religion. I'm a traitor to the Holocaust. And I remember I had some family members in Brazil who wrote me very nasty letters when they found out … one person who was married to my aunt said … 'you don't change religions, like you change your dirty shirt,'" Smith recalled solemnly.
Throughout those six years that he spent working in the auditing sector, Smith continued to attend Presbyterian congregations.
But shortly after becoming a teacher and moving to Walnut Creek, he began to rekindle a relationship with his original faith, helped by his eventual elevation to a honorary member of a local synagogue.
"It was the first time that I went back to Judaism … I never felt right [about] what I had done, I had just left it … I never felt completely comfortable. I always knew something is right here with me," he said.
Since then, Smith has remained with Judaism and his synagogue. Nevertheless, his experience with Presbyterianism has left a palpable impact on the things and the people in his life that bring him joy.
"I was watching the funeral service, a memorial service, this morning of Senator McCain. And they were singing all the hymns that I used to sing in the Presbyterian Church, you know, all the big Christian hymns; they sang all of them. And I had just tears going down my face, you know, because I remember singing all those songs for many, many years."
A life surrounded by death
"The Holocaust is in me, but it has left me alone. And the things that I worry about the most … you will be surprised … what worries me the most right now, for example, is: Who's going to bury me when I pass away? I don't have anybody."
As someone who has witnessed the deaths of so much of his family — and is now waiting for it himself — Smith has a complex relationship with death. He talks of the Holocaust in a resigned manner: resigned to live with it, resigned to accept that it has happened.
But on his own end, Smith's tone is more accepting, expectant. Above all, he approaches his own death with an overarching sense of gratefulness for the life that he has lived.
"I have no regrets. I'm thankful that I got through that Holocaust, very thankful. And my mother always mentioned it to me, to be thankful that 'I was able to save you.'"
Forgiving the Nazis
Smith was never an eyewitness to the burning of the Jewish synagogues and stores, the confiscation of his family's house in Königsberg, the wartime bombing of his hometown, the concentration camps at Auschwitz nor the nighttime kidnapping of his father. But the consequences of these happenings scarred him deeply, forever leaving him with a childhood bombarded by questions and doubts, devastated by fear and loss.
To many Holocaust survivors like Smith, coming to terms with such experiences after the war is the first step toward recovery. The second step is forgiveness, a step that Smith is still struggling with today. After losing so many family members, he finds it extremely difficult to put his religious conviction of forgiveness above the personal grudges that he bears against the perpetrators of the Holocaust.
"A number of people have asked about forgiveness. I don't know how to answer that. I believe in God … that there's somebody up there. Whether I can give forgiveness, that's up to Him … Usually the church likes to forgive. For me to forgive is very, very difficult," he said.
His ambivalence toward forgiving the Nazis is paralleled by an underlying distrust for other Germans of his age — those who were alive during the time of the Holocaust. Aligning with popular opinion regarding the resolution of World War II, Smith noted that the Nuremberg trials convicted and sentenced far fewer people than were truly culpable for the Holocaust.
"Whenever I meet a German person — and I've met a number of German people, of my age or even older — I always feel very uncomfortable … because I always wonder … 'what did you do during those years, where you were supporting Hitler and all that he did … the six million Jewish people that he murdered?'"
After the Holocaust and before her death, however, Smith's mother had always approached their existence with optimism. She counted blessings in their escape and survival from Nazi Germany to England, then to Chile and finally, for Smith, to the United States. Seldom discussing the unspeakable atrocities committed in the Holocaust, she took great care of her remaining possessions, property, and family.
Eventually, Smith's mother chose to return to her home country, assured by the new German Democratic Republic (East Germany)'s multiple apologies and laws defending its Jewish people. Similar to Smith's experience with being denounced as a traitor after his conversion to Presbyterianism (see "Between Judaism and Presbyterianism"), his mother immediately experienced harsh backlash from family and the Jewish community.
"After being for 30 years in Chile, [my mother] went back to Germany and people said, 'How could you do that? … How could you go back to Germany, where your entire family got killed?'" Smith recalled.
Although he does not know the current state of his old house — it was likely destroyed by a British bombing attack on Königsberg in 1944, and it lies in a territory that was ceded to Russia after the war — Smith himself has returned to Germany multiple times. One visit was to his mother's apartment shortly after her death.
Smith disagrees with the Jewish taboo of Germany as the land of the Holocaust, but he very well affirms that he has not forgiven the Nazis. And if he ever does, he says that it will be of God's volition, not his.
Once in the U.S., Smith worked in New York City before being drafted into the U.S. special intelligence forces in 1951 for the Korean War. Although he was eager to serve the country that had accepted him so wholeheartedly, he was never summoned overseas, and his two years of service passed quickly.
On a whim, after he heard some friends speak fondly of the distant state of California, Smith decided to attend UC Berkeley, a privilege granted to veterans by the G.I. Bill of 1944. After hitchhiking his way across three states, attending a junior college in the Bay Area and scraping funds together to support himself in college, Smith graduated with a major in auditing.
From there, he held a stable — but in his opinion, monotonous — auditing job at a CPA firm in San Francisco for many years before realizing in 1962 that his passions lay in teaching.
With his fluency in multiple languages, Smith spent nearly 30 years between Clayton Valley Middle School, Pleasant Hill High School and Northgate High School as a Spanish teacher before retiring.
From behind the armchair that Smith is sitting in, we see that the day has aged and the sky has turned gray. It has only been hours, but decades have passed in our dialogue.
With his life laid out on the coffee table between us, we have but one question left. When he hears it, Smith looks up at us behind cloudy eyes. A sad smile spreads across his face.
"That I survived it?"
Smith lets out a deep, long sigh.
"Oh gosh, no."
*Subject and church names have been changed for privacy.
Hear from Another World: For Patricia, an American Dream slowly deteriorating
They left Vietnam, but Vietnam never left them
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Sraavya Sambara, Editor-in-Chief
Sraavya joined the Tribune because her friend told her that it was "pretty chill." It was not, but she loves it anyway. This is her third year in the Tribune and she is still as impressed with everyone in the...
Daniel Shen, Editor-in-Chief
Daniel joined the Tribune in his sophomore year upon the recommendation of former editor-in-chief Amanda Su. In his two years, he's written about school events, music and immigration, and deeply enjoyed building...
Sasha Hassan, News Editor
Sasha joined the Tribune to pursue her love of writing. In the past, Sasha served two years on her middle school paper and three years on The Wildcat Tribune as a copy editor for a year and a page editor for...
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| 8,840
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{"url":"https:\/\/www.thejournal.club\/c\/paper\/132\/","text":"#### Maximizing Maximal Angles for Plane Straight-Line Graphs\n\n##### Oswin Aichholzer, Thomas Hackl, Michael Hoffmann, Clemens Huemer, Attila Por, Francisco Santos, Bettina Speckmann, Birgit Vogtenhuber\n\nLet $G=(S, E)$ be a plane straight-line graph on a finite point set $S\\subset\\R^2$ in general position. The incident angles of a vertex $p \\in S$ of $G$ are the angles between any two edges of $G$ that appear consecutively in the circular order of the edges incident to $p$. A plane straight-line graph is called $\\phi$-open if each vertex has an incident angle of size at least $\\phi$. In this paper we study the following type of question: What is the maximum angle $\\phi$ such that for any finite set $S\\subset\\R^2$ of points in general position we can find a graph from a certain class of graphs on $S$ that is $\\phi$-open? In particular, we consider the classes of triangulations, spanning trees, and paths on $S$ and give tight bounds in most cases.\n\narrow_drop_up","date":"2022-01-25 02:14:58","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5214115977287292, \"perplexity\": 317.7795354550518}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": false}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-05\/segments\/1642320304749.63\/warc\/CC-MAIN-20220125005757-20220125035757-00710.warc.gz\"}"}
| null | null |
\section{Introduction}
Relation extraction (RE) is an important task in information extraction, aiming to extract the relation between two given entities based on their related context. Due to the capability of extracting textual information and benefiting many NLP applications (e.g., information retrieval, dialog generation, and question answering), RE appeals to many researchers. Conventional supervised models have been widely explored in this task \cite{zelenko2003kernel,zeng2014relation}; however, their performance heavily depends on the scale and quality of training data.
\begin{figure}
\centering
\includegraphics [width=0.4\textwidth]{pic1.pdf}
\caption{Label frequency distribution of classes without $NA$ in NYT dataset.}
\label{pic1}
\end{figure}
To construct large-scale data, \cite{mintz2009distant} proposed a novel distant supervision (DS) mechanism to automatically label training instances by aligning existing knowledge graphs (KGs) with text. DS enables RE models to work on large-scale training corpora and has thus become a primary approach for RE recently \cite{wu2017adversarial,feng2018reinforcement}.
Although these DS models achieve promising results on common relations, their performance still degrades dramatically when there are only a few training instances for some relations. Empirically, DS can automatically annotate adequate amounts of training data; however, this data usually only covers a limited part of the relations. Many relations are long-tail and still suffer from data deficiency. Current DS models ignore the problem of long-tail relations, which makes it challenging to extract comprehensive information from plain text.
Long-tail relations are important and cannot be ignored. Nearly 70\% of the relations are long-tail in the widely used New York Times (NYT) dataset\footnote{http://iesl.cs.umass.edu/riedel/ecml/} \cite{riedel2010modeling,lei2018cooperative} as shown in Figure \ref{pic1}.
Therefore, it is crucial for models to be able to extract relations with limited numbers of training instances.
Dealing with long tails is very difficult as few training examples are available. Therefore, it is natural to transfer knowledge from data-rich and semantically similar head classes to data-poor tail classes \cite{wang2017learning}. For example, the long-tail relation /people/deceased\_person/place\_of\_burial and head relation /people/deceased\_person/place\_of\_death are in the same branch /people/deceased\_person/* as shown in Figure \ref{pic2}. They are semantically similar, and it is beneficial to leverage head relational knowledge and transfer it to the long-tail relation, thus enhancing general performance. In other words, long-tail relations of one entity tuple can have class ties with head relations, which can be leveraged to enhance RE for narrowing potential search spaces and reducing uncertainties between relations when predicting unknown relations \cite{ye2017jointly}. If one pair of entities contains /people/deceased\_person/place\_of\_death, there is a high probability that it will contain /people/deceased\_person/place\_of\_burial. If we can incorporate the relational knowledge between two relations, extracting head relations will provide evidence for the prediction of long-tail relations.
However, there exist two problems: (1) \emph{Learning relational knowledge:} Semantically similar classes may contain more relational information that will boost transfer, whereas irrelevant classes (e.g., /location/location/contains and /people/family/country) usually contain less relational information that may result in negative transfer. (2) \emph{Leveraging relational knowledge:} Integrating relational knowledge to existing RE models is challenging.
To address the problem of learning relational knowledge, as shown in \cite{lin2016neural,ye2017jointly}, we use class embeddings to represent relation classes and utilize KG embeddings and graph convolution networks (GCNs) to extract implicit and explicit relational knowledge. Specifically, previous studies \cite{yang2014embedding} have shown that the embeddings of semantically similar relations are located near each other in the latent space. For instance, the relation /people/person/place\_lived and /people/person/nationality are more relevant, whereas the relation /people/person/profession has less correlation with the former two relations. Thus, it is natural to leverage this knowledge from KGs. However, because there are many one-to-multiple relations in KGs, the relevant information for each class may be scattered. In other words, there may not be enough relational signal between classes. Therefore, we utilize GCNs to learn explicit relational knowledge.
\begin{figure}
\centering
\includegraphics [width=0.5\textwidth]{f1.pdf}
\caption{Head and long-tail relations.}
\label{pic2}
\end{figure}
To address the problem of leveraging relational knowledge, we first use convolution neural networks \cite{zeng2014relation,zeng2015distant} to encode sentences; then introduce coarse-to-fine knowledge-aware attention mechanism for combining relational knowledge with encoded sentences into bag representation vectors. The relational knowledge not only provides more information for relation prediction but also provides a better reference message for the attention module to raise the performance of long-tail classes.
Our experimental results on the NYT dataset show that: (1) our model is effective compared to baselines especially for long-tail relations; (2) leveraging relational knowledge enhances RE and our model is efficient in learning relational knowledge via GCNs.
\section{Related Work}
\textbf{Relation Extraction.} Supervised RE models \cite{zelenko2003kernel,guodong2005exploring,mooney2006subsequence} require adequate amounts of annotated data for training which is time-consuming. Hence, \cite{mintz2009distant} proposd DS to automatically label data. DS inevitably accompanies with the wrong labeling problem. To alleviate the noise issue, \cite{riedel2010modeling,hoffmann2011knowledge} proposed multi-instance learning (MIL) mechanisms. Recently, neural models have been widely used for RE; those models can accurately capture textual relations without explicit linguistic analysis \cite{zeng2015distant,lin2016neural,zhang2018capsule}. To further improve the performance, some studies incorporate external information \cite{zeng2016incorporating,ji2017distant,han2018neural} and advanced training strategies \cite{ye2017jointly,liu2017soft,huang2017deep,feng2018reinforcement,zeng2018large,wu2017adversarial,qin2018dsgan}.
These works mainly adopt DS to make large-scale datasets and reduce the noise caused by DS, regardless of the effect of long-tail relations.
There are only a few studies on long-tail for RE \cite{gui2016exploring,lei2018cooperative,han2018hierarchical}. Of these, \cite{gui2016exploring} proposed an explanation-based approach, whereas \cite{lei2018cooperative} utilized external knowledge (logic rules). These studies treat each relation in isolation, regardless of the rich semantic correlations between the relations. \cite{han2018hierarchical} proposed a hierarchical attention scheme for RE, especially for long-tail relations. Different from those approaches, we leverage implicit and explicit relational knowledge from KGs and GCNs rather than data-driven learned parameter spaces where similar relations may have distinct parameters, hindering the generalization of long-tail classes.
\textbf{Knowledge Graph Embedding.}
Recently, several KG embedding models have been proposed. These methods learn low-dimensional vector representations for entities and relations \cite{bordes2013translating,wang2014knowledge,lin2015learning}. TransE \cite{bordes2013translating} is one of the most widely used models, which views relations as translations from a head entity to a tail entity on the same low-dimensional hyperplane. Inspired by the rich knowledge in KGs, recent works \cite{han2018neural,wang2018label,lei2018cooperative} extend DS models under the guidance of KGs. However, these works neglect rich correlations between relations. Relation structure (relational knowledge) has been studied and
is quite effective for KG completion \cite{zhang2018knowledge}.
To the best of our knowledge, this is the first effort to consider the relational knowledge of classes (relations) using KGs for RE.
\textbf{Graph Convolutional Networks.}
GCNs generalize CNNs beyond two-dimensional and one-dimensional spaces. \cite{defferrard2016convolutional} developed spectral methods to perform efficient graph convolutions. \cite{kipf2016semi} assumed the graph structure is known over input instances and apply GCNs for semi-supervised learning. GCNs were applied to relational data (e.g., link prediction) by \cite{schlichtkrull2018modeling}. GCNs have also had success in other NLP tasks such as semantic role labeling \cite{marcheggiani2017encoding}, dependency parsing \cite{strubell2017dependency}, and machine translation \cite{bastings2017graph}.
Two GCNs studies share similarities with our work. (1) \cite{chen2017graph} used GCNs on structured label spaces. However, their experiments do not handle long-tail labels and do not incorporate attention but use an average of word vectors to represent each document. (2) \cite{rios2018few} proposed a few-shot and zero-shot text classification method by exploiting structured label spaces with GCNs. However, they used GCNs in the label graph whereas we utilize GCNs in the hierarchy graph of labels.
\section{Methodology}
In this section, we introduce the overall framework of our approach for RE, starting with the notations.
\subsection{Notations}
We denote a KG as $\mathcal{G}$ = $\mathcal{E},\mathcal{R},\mathcal{F}$, where $\mathcal{E}$, $\mathcal{R}$ and $\mathcal{F}$ indicate the sets of entities, relations and facts respectively. $(h,r,t) \in \mathcal{F} $ indicates that there is a relation $r \in \mathcal{R}$ between $h \in \mathcal{E}$ and $t \in \mathcal{E}$. We follow the MIL setting and split all instances into multiple entity-pair bags
$\{\mathcal{S}_{h_1,t_1},\mathcal{S}_{h_2,t_2},...\}$. Each bag $\mathcal{S}_{h_i,t_i}$ contains multiple
instances $\{s_1,s_2,...\}$ mentioning both entities $h_i$ and $t_i$. Each instance $s$ in these bags is denoted as a word sequence $s = \{w_1,w_2,...\}$.
\subsection{Framework}
\begin{figure*} \centering
\includegraphics[width=0.6\textwidth]{pic2.pdf}
\caption{Architecture of our proposed model.}
\label{framework}
\end{figure*}
Our model consists of three parts as shown in Figure \ref{framework}:
\textbf{Instance Encoder.} Given an instance and its mentioned entity pair, we employ neural networks to encode the instance semantics into a vector. In this study, we implement the instance encoder with convolutional neural networks (CNNs) given both model performance and time efficiency.
\textbf{Relational Knowledge Learning.}
Given pretrained KG embeddings (e.g., TransE \cite{bordes2013translating}) as implicit relational knowledge, we employ GCNs to learn explicit relational knowledge. By assimilating generic message-passing inference algorithms with neural-network counterpart, we can learn better embeddings for Knowledge Relation. We concatenate the outputs of GCNs and the pretrained KG embeddings to form the final class embeddings.
\textbf{Knowledge-aware Attention.} Under the guidance of final class embeddings, knowledge-aware attention is aimed to select the most informative instance exactly matching relevant relation.
\subsection{Instance Encoder}
Given an instance $s = \{w_1,...,w_n\}$ mentioning two entities, we encode the raw instance into a continuous low-dimensional vector $x$, which consists of an embedding layer and an encoding layer.
\textbf{Embedding Layer.}
The embedding layer is used to map discrete words in the instance into continuous input embeddings. Given an instance $s$, we map each word $w_i$ in the instance to a real-valued pretrained Skip-Gram \cite{mikolov2013efficient} embedding $w_i \in \mathbb{R}^{d_w}$ . We adopt position embeddings following \cite{zeng2014relation}. For each word $w_i$, we embed its relative distances to the two entities into two $d_p$-dimensional vectors. We then concatenate the word embeddings and position embeddings to achieve the final input embeddings for each word and gather all the input embeddings in the instance. We thus obtain an embedding sequence ready for the encoding layer.
\textbf{Encoding Layer.}
The encoding layer aims to compose the input embeddings of a given instance into its corresponding instance embedding. In this study, we choose two convolutional neural architectures, CNN \cite{zeng2014relation} and PCNN \cite{zeng2015distant} to encode input embeddings into instance embeddings. Other neural architectures such as recurrent neural networks \cite{zhang2015relation} can also be used as sentence encoders. Because previous works show that both convolutional and recurrent architectures can achieve comparable state-of-the-art performance, we select convolutional architectures in this study. Note that, our model is independent of the encoder choices, and can, therefore, be easily adapted to fit other encoder architectures.
\subsection{Relational Knowledge Learning through KG Embeddings and GCNs.}
Given pretrained KG embeddings and predefined class (relation) hierarchies\footnote{For datasets without predefined relation hierarchies, hierarchy clustering \cite{johnson1967hierarchical} or K-means can construct relation hierarchies \cite{zhang2018knowledge}; details can be found in supplementary materials.}, we first leverage the implicit relational knowledge from KGs and initialize the hierarchy label graph; then we apply two layer GCNs to learn explicit fine-grained relational knowledge from the label space.
\textbf{Hierarchy Label Graph Construction.}
Given a relation set $\mathcal{R}$ of a KG $\mathcal{G}$ (e.g., Freebase), which consists of base-level relations (e.g., /people/person/ethnicity), we can generate the corresponding higher-level relation set $\mathcal{R}^H$. The relations in a high-level set (e.g., people) are more general and common; they usually contain several sub-relations in the base-level set. The relation hierarchies are tree-structured, and the generation process can be done recursively. We use a virtual father node to construct the highest level associations between relations as shown in Figure \ref{framework}.
In practice, we start from $\mathcal{R}^0 = R$ which is the set of all relations we focus on for RE, and the generation process is performed $L-1$ times to get the hierarchical relation sets $\{\mathcal{R}^0,\mathcal{R}^1,...,\mathcal{R}^{L}\}$, where $\mathcal{R}^{L}$ is the virtual father node. Each node has a specific type $t \in \{0,1,...,L\}$ to identify its layer hierarchies. For example, as shown in Figure \ref{framework}, node /people/person/ethnicity has a specific type $0$ to indicate it is in the bottom layer of the graph. The vectors of each node in the bottom layer are initialized through pretrained TransE \cite{bordes2013translating} KG embeddings. Other KG embeddings such as TransR \cite{lin2015learning} can also be adopted. Their parent nodes are initialized by averaging all children vectors. For example, the node vector of /people/person/ is initialized by averaging all the nodes under the branch of /people/person/* (all child nodes).
\textbf{GCN Output Layer.}
Due to one-to-multiple relations and incompleteness in KGs, the implicit relevant information obtained by KG embeddings for each label is not enough. Therefore, we apply GCNs to learn explicit relational knowledge among labels. We take advantage of the structured knowledge over our label space using a two-layer GCNs. Starting with the pretrained relation embedding $v_i^{implicit} \in \mathbb{R}^{d}$ from KGs, we combine the label vectors of the children and parents for the $i$-th label to form,
\begin{equation}
v_i^1 = f(W^1v_i+\sum\limits_{j \in \mathcal{N}_p} \frac{W_p^1v_j}{|\mathcal{N}_p|} +\sum\limits_{j \in \mathcal{N}_c} \frac{W_c^1v_j}{|\mathcal{N}_c|} +b_g^1 )
\end{equation}
where $W^1 \in \mathbb{R}^{q*d}$, $W^1_p \in \mathbb{R}^{q*d}$, $W^1_c \in \mathbb{R}^{q*d}$, $b_g^1 \in \mathbb{R}^q$, $f$ is the rectified linear unit \cite{nair2010rectified} function, and $\mathcal{N}_c$ ($\mathcal{N}_p$) is the index set of the $i$-th label's children (parents). We use different parameters to distinguish each edge type where parent edges represent all edges from high level labels and child edges represent all edges from low level labels. The second layer follows the same formulation as the first layer and outputs $v_i^{explicit}$. Finally, we concatenate both pretrained $v_i^{implicit}$ with GCNs node vector $v_i^{explicit}$ to form hierarchy class embeddings,
\begin{equation}
q_r=v_i^{implicit} || v_i^{explicit}
\end{equation}
where $q_r \in \mathbb{R}^{d+q}$.
\subsection{Knowledge-aware Attention}
Traditionally, the output layer of PCNN/CNN would learn label specific parameters optimized by a cross-entropy loss. However, the label specific parameters spaces are unique to each relation, matrices associated with the long-tails can only be exposed to very few facts during training, resulting in poor generalization. Instead, our method attempts to match sentence vectors to their corresponding class embeddings rather than learning label specific attention parameters. In essence, this becomes a retrieval problem. Relevant information from class embeddings contains useful relational knowledge for long-tails among labels.
Practically, given the entity pair $(h, t)$ and its bag of instances $S_{h,t} = \{s_1,s_2,...,s_m\}$, we achieve the instance embeddings $\{s_1,s_2,...,s_m\}$ using the sentence encoder. We group the class embeddings according to their type (i.e., according to their layers in the hierarchy label graph), e.g., $q_{r^i}, i \in \{0,1,...,L \}$. We adopt $q_{r^i}, i\neq L$ (layer $L$ is the virtual father node) as layer-wise attention query vector. Then, we apply coarse-to-fine knowledge-aware attention to them to obtain the textual relation representation $r_{h,t}$. For a relation $r$, we construct its hierarchical chain of parent relations $(r^0,...,r^{L-1})$ using the hierarchy label graph, where $r^{i-1}$ is the sub-relation of $r^i$. We propose the following formulas to compute the attention weight (similarity or relatedness) between each instance's feature vector $s_k$ and $q_{r^i}$,
\begin{equation}
\begin{aligned}
e_k &= W_s(tanh[s_k;q_{r^i}])+b_s\\%q_{r^i}^{\mathrm{T}}W_s s_k \\
\alpha_k^i &= \frac{exp(e_k)}{\sum_{j=1}^{m}exp(e_j)}
\end{aligned}\label{attention}
\end{equation}
where $[x_1; x_2]$ denotes the vertical concatenation of $x_1$ and $x_2$, $W_s$ is the weight matrix, and $b_s$ is the bias. We compute attention operations on each layer of hierarchy label graph to obtain corresponding textual relation representations,
\begin{equation}
r_{h,t}^i = ATT(q_{r^i},\{s_1,s_2,...,s_m\})
\end{equation}
Then we need to combine the relation representations on different layers. Direct concatenation of all the representations is a straightforward choice. However, different layers have different contributions for different tuples. For example, the relation /location/br\_state/ has only one sub-relation /location/br\_state/capital, which indicates that it is more important. In other words, if the sentence has high attention weights on relation /location/br\_state/, it has a very high probability to have relation /location/br\_state/capital. Hence, we use an attention mechanism to emphasize the layers,
\begin{equation}
\begin{aligned}
g_i &= W_g tanh(r_{h,t})\\
\beta_i&=\frac{exp(g_i)}{\sum_{j=0}^{L-1} exp(g_j)}
\end{aligned}
\end{equation}
where $W_g$ is a weight matrix, $r_{h,t}$ is referred to as a query-based function that scores how well the input textual relation representations and the predict relation $r$ match. The textual relation representations in each layer are computed as,
\begin{equation}
r_{h,t}^i = \beta_i r_{h,t}^i
\end{equation}
We simply concatenate the textual relation representations on different layers as the final representation,
\begin{equation}
r_{h,r}=Concat(r_{h,t}^0,..,r_{h,t}^{L-1})
\end{equation}
The representation $r_{h,t}$ will be finally fed to compute
the conditional probability $\mathcal{P}(r|h,t,\mathcal{S}_{h,t})$,
\begin{equation}
\mathcal{P}(r|h,t,\mathcal{S}_{h,t}) = \frac{exp(o_r)}{\sum_{\tilde{r} \in R}exp(o_{\tilde{r}})}
\end{equation}
where $o$ is the score of all relations defined as,
\begin{equation}
o=M r_{h,t}
\end{equation}
where $M$ is the representation matrix to calculate the relation scores. Note that, attention weight $q_{r^i}$ is obtained from the outputs of GCNs and pretrained KG embeddings, which can provide more informative parameters than data-driven learned parameters, especially for long-tails.
\section{Experiments}
\subsection{Datasets and Evaluation}
We evaluate our models on the NYT dataset developed by \cite{riedel2010modeling}, which has been widely used in recent studies \cite{lin2016neural,liu2017soft,wu2017adversarial,feng2018reinforcement}. The dataset has 53 relations including the $NA$ relation, which indicates that the relations of instances are not available. The training set has 522611 sentences, 281270 entity pairs, and 18252 relational facts. In the test set, there are 172448 sentences, 96678 entity pairs, and 1950 relational facts. In both training and test set, we truncate sentences with more than 120 words into 120 words.
We evaluate all models in the held-out evaluation. It evaluates models by comparing the relational facts discovered from the test articles with those in Freebase and provides an approximate measure of precision without human evaluation. For evaluation, we draw precision-recall curves for all models. To further verify the effect of our model for long-tails, we follow previous studies \cite{han2018hierarchical} to report the Precision@N results. The dataset and baseline code can be found on Github \footnote{https://github.com/thunlp/OpenNRE}.
\subsection{Parameter Settings\footnote{Details of hyper-parameters settings and evaluation of different instances can be found in supplementary materials}}
To fairly compare the results of our models with those baselines, we also set most of the experimental parameters by following \cite{lin2016neural}. We apply dropout on the output layers of our models to prevent overfitting. We also pretrain the sentence encoder of PCNN before training our model.
\begin{figure}
\centering
\includegraphics[width=0.38\textwidth]{pr1.pdf}
\caption{\label{pr1}Precision-recall curves for the proposed model and various baseline models.}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=0.38\textwidth]{pr2.pdf}
\caption{\label{pr2}Precision-recall curves for the proposed model and various attention-based neural models.}
\end{figure}
\subsection{Overall Evaluation Results}
To evaluate the performance of our proposed model, we compare the precision-recall curves of our model with various previous RE models. The evaluation results are shown in Figure \ref{pr1} and Figure \ref{pr2}. We report the results of neural architectures including CNN and PCNN with various attention based methods: \textbf{+KATT} denotes our approach, \textbf{+HATT} is the hierarchical attention method \cite{han2018hierarchical}, \textbf{+ATT} is the plain selective attention method over instances \cite{lin2016neural}, \textbf{+ATT+ADV} is the denoising attention method by adding a small adversarial perturbation to instance embeddings \cite{wu2017adversarial}, and \textbf{+ATT+SL} is the attention-based model using soft-labeling method to mitigate the side effect of the wrong labeling problem at entity-pair level \cite{liu2017soft}. We also compare our method with feature-based models, including \textbf{Mintz} \cite{mintz2009distant}, \textbf{MultiR} \cite{hoffmann2011knowledge} and \textbf{MIML} \cite{surdeanu2012multi}.
As shown in both figures, our approach achieves the best results among all attention-based models. Even when compared with PCNN+HATT, PCNN+ATT+ADV, and PCNN+ATT+SL, which adopt sophisticated denoising schemes and extra information, our model is still more advantageous. This indicates that our method can take advantage of the rich correlations between relations through KGs and GCNs, which improve the performance. We believe the performance of our model can be further improved by adopting additional mechanisms like adversarial training, and reinforcement learning, which will be part of our future work.
\subsection{Evaluation Results for Long-tail Relations}
\begin{table}[]
\small
\begin{tabular}{ccp{0.4cm}p{0.4cm}p{0.4cm}p{0.4cm}p{0.4cm}p{0.4cm}}
\hline
\multicolumn{2}{c}{\multirow{2}*{\shortstack{Training Instances \\ Hits@K (Macro)}}} &\multicolumn{3}{c}{$<$100} &\multicolumn{3}{c}{$<$200} \\
& & 10&15&20&10&15&20 \\
\hline
\multirow{3}*{CNN}&+ATT & $<$5.0& $<$5.0 &18.5 & $<$5.0 &16.2&33.3 \\
\cline{3-8}&+HATT&5.6 &31.5 & 57.4 &22.7 & 43.9 & 65.1 \\
\cline{3-8}&+KATT&\textbf{9.1} &\textbf{41.3}&\textbf{58.5}&\textbf{23.3}&\textbf{44.1}& \textbf{65.4} \\
\hline
\multirow{3}*{PCNN}&+ATT &$<$5.0 & 7.4 & 40.7& 17.2 &24.2&51.5 \\
\cline{3-8}&+HATT&29.6 &51.9 & 61.1 &41.4 & 60.6 & 68.2 \\
\cline{3-8}&+KATT& \textbf{35.3} &\textbf{62.4}&\textbf{65.1}&\textbf{43.2}&\textbf{61.3}& \textbf{69.2} \\ \hline
\end{tabular}
\caption{Accuracy (\%) of Hits@K on relations with training instances fewer than 100/200.}
\label{table2}
\end{table}
\begin{table}[]
\small
\begin{tabular}{cp{0.4cm}p{0.4cm}p{0.4cm}p{0.4cm}p{0.4cm}p{0.4cm}}
\hline
\multirow{2}*{\shortstack{Training Instances \\ Hits@K (Macro)}}
&\multicolumn{3}{c}{$<$100} &\multicolumn{3}{c}{$<$200} \\
&10&15&20&10&15&20 \\
\hline
+KATT& \textbf{35.3}&\textbf{62.4}&\textbf{65.1}&\textbf{43.2}&\textbf{61.3}& \textbf{69.2} \\
\hline
w/o hier&34.2 &62.1 & 65.1& 42.5&60.2 & 68.1 \\
w/o GCNs&30.5 &61.9 &63.1&39.5&58.4& 66.1 \\
Word2vec&30.2 &62.0 &62.5&39.6&57.5& 65.8 \\
w/o KG& 30.0 &61.0 &61.3&39.5&56.5& 62.5 \\
\hline
\end{tabular}
\caption{Results of ablation study with PCNN.}
\label{table4}
\end{table}
To further demonstrate the improvements in performance for long-tail relations, following the study by \cite{han2018hierarchical} we extract a subset of the test dataset in which all the relations have fewer than 100/200 training instances. We employ the Hits@K metric for evaluation. For each entity pair, the evaluation requires its corresponding golden relation in the first $K$ candidate relations recommended by the models. Because it is difficult for the existing models to extract long-tail relations, we select $K$ from \{10,15,20\}. We report the macro average Hits@K accuracies for these subsets because the micro-average score generally overlooks the influences of long-tails. From the results shown in Table \ref{table2}, we observe that for both CNN and PCNN models, our model outperforms the plain attention model and the HATT model. Although our KATT method has achieved better results for long-tail relations as compared to both plain ATT method and HATT method, the results of all these methods are still far from satisfactory. This indicates that distantly supervised RE models still suffer from the long-tail relation problem, which may require additional schemes and extra information to solve this problem in the future.
\begin{figure*} \centering
\subfigure[HATT] { \label{aa}
\includegraphics[width=0.23\textwidth]{fig1.png}
}
\subfigure[+KG ] { \label{cc}
\includegraphics[width=0.23\textwidth]{fig3.png}
}
\subfigure[+GCNs] { \label{bb}
\includegraphics[width=0.23\textwidth]{fig2.png}
}
\subfigure[KATT] { \label{dd}
\includegraphics[width=0.23\textwidth]{fig4.png}
}
\caption{T-SNE visualizations of class embeddings. Cluster in the upper right is the relation /location/*/* and cluster in the bottom left is the relation /people/*/* ). The square, triangle, and star refer to the high (/location) middle (/location/location/) and base (/location/location/contains) level relations, respectively. }
\label{4pic}
\end{figure*}
\subsection{Ablation Study}
To analyze the contributions and effects of different technologies in our approach, we perform ablation tests. \textbf{+KATT} is our method; \textbf{w/o hier} is the method without coarse-to-fine attention (only utilizes bottom node embeddings of the hierarchy label graph), which implies no knowledge transfer from its higher level classes; \textbf{w/o GCN} is the method without GCNs, which implies no explicit relational knowledge; \textbf{Word2vec} is the method in which the node is initialized with pretrained Skip-Gram \cite{mikolov2013efficient} embeddings; and \textbf{w/o KG} is the method in which the node is initialized with random embeddings, which implies no prior relational knowledge from KGs. From the evaluation results in Table \ref{table4}, we observe that the performance slightly degraded without coarse-to-fine attention, which proves that knowledge transfer from the higher node is useful. We also noticed that the performance slightly degraded without KG or using word embeddings, and the performance significantly degraded when we removed GCNs. This is reasonable because GCNs can learn more explicit correlations between relation labels, which boost the performance for long-tail relations.
\begin{table}[h]\centering
\fontsize{8.5}{10}\selectfont
\begin{tabular}{|p{4.9cm}|p{0.75cm}|p{0.75cm}|}
\hline
/people/deceased\_person/place\_of\_burial&\textbf{HATT}&\textbf{KATT} \\
\hline
\textbf{richard\_wagner} had his \textbf{bayreuth}, with its festspielhaus specially designed to accommodate his music dramas. &0.21 & 0.07 \\
\hline
wotan and alberich are \textbf{richard\_wagner}; and the rheingold and valhalla are wagner's real-life grail, the opera house in \textbf{bayreuth}.& 0.15& 0.13
\\
\hline
/location/br\_state/capital & \textbf{HATT} &\textbf{KATT} \\
\hline
there's violence everywhere, said ms. mesquita, who, like her friend, lives in \textbf{belo\_horizonte}, the capital of \textbf{minas\_gerais} state&0.47 &0.51\\
\hline all the research indicates that we are certain to find more gas in th amazon, eduardo braga, the governor of \textbf{amazonas}, said in an interview in \textbf{manaus}& 0.46&0.45\\
\hline
\end{tabular}
\caption{Example sentences for case study.}
\label{table5}
\end{table}
\subsection{Case Study}
We give some examples to show how our method affects the selection of sentences. In Table \ref{table5}, we display the sentence's attention score in the lowest level\footnote{Both HATT and KATT methods can successfully select the correct sentence at the higher-level; details can be found in supplementary materials.}.
Both the relation /people/deceased\_person/place\_of\_burial (24 instances) and /location/br\_state/capital (4 instances) are long-tail relations. On one hand, relation /people/deceased\_person/place\_of\_burial has semantically similar data-rich relation such as /people/deceased\_person/place\_of\_death.
We observe that HATT erroneously assigns high attention to the incorrect sentence whereas KATT successfully assigns the right attention weights, which demonstrates the efficacy of knowledge transfer from semantically similar relations (Both HATT and KATT methods can take advantage of knowledge transfer of high-level relations.). On the other hand, the relation /location/br\_state/capital does not have semantically similar relations. However, we notice that KATT still successfully assigns the right attention weights, which demonstrates the efficacy of knowledge transfer from high-level relations using coarse-to-fine knowledge-aware attention.
\subsection{Visualizations of Class Embeddings}
We visualize the class embeddings via t-SNE \cite{maaten2008visualizing} to further show how GCNs and KG embeddings can help RE for long-tail relations. We observe that (1) Figure \ref{aa} and \ref{dd} show that semantically similar class embeddings are closer with GCNs and pretrained KG embeddings, which help select long-tail instances. (2) Figure \ref{cc} and \ref{bb} show that KG embeddings and GCNs have different contributions for different relations to learn relational knowledge between classes. For example, /location/location/contain has a sparse hierarchy structure, which leads to inefficient learning for GCNs; therefore, the relative distance changes only slightly, which reveals the necessity of implicit relational knowledge from KGs. (3) Figure \ref{dd} shows that there are still some semantically similar class embeddings located far away, which may degrade the performance for long-tails. This may be caused by either sparsity in the hierarchy label graph or equal treatment for nodes with the same parent in GCNs, which is not a reasonable hypothesis. We will address this by integrating more information such as relation descriptions or combing logic reasoning as a part of future work.
\section{Conclusion and Future Work}
In this paper, we take advantage of the knowledge from data-rich classes at the head of distribution to boost the performance of the data-poor classes at the tail. As compared to previous methods, our approach provides fine-grained relational knowledge among classes using KG and GCNs, which is quite effective and encoder-agnostic.
In the future, we plan to explore the following directions: (1) We may combine our method with recent denoising methods to further improve performance. (2) We may combine rule mining and reasoning technologies to learn better class embeddings to boost performance. (3) It will be promising to apply our method to zero-shot RE and further adapt to other NLP scenarios
\section*{Acknowledgments}
We want to express gratitude to the anonymous reviewers for their hard work and kind comments, which will further improve our work in the future. This work is funded by NSFC91846204/61473260, national key research program YS2018YFB140004, Alibaba CangJingGe (Knowledge Engine) Research Plan and Natural Science Foundation of Zhejiang Province of China (LQ19F030001).
\section{Hierarchy Label Graph Construction}
Previous study \cite{xie2017interpretable,zhang2018knowledge} indicate that there is rich information from the relation structure. Some dataset has explicit relation hierarchical structures. For example, the relation /location/province/capital in Freebase \cite{bollacker2008freebase} indicates the relation between a province and its capital. It is labeled under the location branch. Under this branch, there are some other relations /location/location/contains and /location/country/capital, which is closely correlated to each other. However, some other datasets may not have predefined hierarchical structures. It is intuitive to define a hierarchical relation structure, which can be confirmed by relation clusters and relations in KGs \cite{zhang2018knowledge}. We utilize three simple methods to build relation structure and tests their results. All pretrained knowledge graph embeddings are available online\footnote{http://openke.thunlp.org/}.
\textbf{K-means.} We first run TransE \cite{bordes2013translating} on a given data set and obtain the embeddings of relations $r_1, r_2, r_3, ..., r_{|R|}$
, where $|R|$ is the number of relations following \cite{zhang2018knowledge}. Then, the k-means algorithm is applied to these embeddings. In this way, we get relation clusters $C_1, C_2, C_3, ..., C_{|C|}$, where $C$ is the set of relation clusters. Previous studies have shown that the embeddings of semantically similar relations locate near each other in the latent space \cite{yang2014embedding}. In this way, we can find relation clusters composed of semantically related relations. Practically, we can get relation clusters recursively and get the hierarchy of relations.
\textbf{Hierarchical Clustering (HC).} Hierarchical clustering is a method of cluster analysis which seeks to build a hierarchy of clusters. We adopt the agglomerative way which is a "bottom-up" approach to construct the hierarchy of classes. Each class starts in its cluster, and pairs of clusters are merged as one moves up the hierarchy. Specifically, we apply Euclidean distance as appropriate metric between classes and complete-linkage clustering to take the distance between clusters.
\textbf{Association Rule Mining under Incomplete Evidence (AMIE).}
AMIE is a rule mining model that is explicitly tailored to
support the open world assumption scenario. It is inspired by association rule
mining and uses a novel measure for confidence.
Practically, we run the AMIE tool on the dataset and choose those classes with pca confidence greater than the threshold as correlated classes and then group them. AMIE tool is available oneline\footnote{http://resources.mpi-inf.mpg.de/yago-naga/amie/downloads.html}.
We evaluate the performance of our proposed model with different hierarchical classes structures on the same dataset NYT. We employ the Hits@K metric for evaluation. For each entity pair, the evaluation requires its corresponding golden relation in the first $K$ candidate relations recommended by the models. Because it is difficult for the existing models to extract long-tail relations, we select $K$ from \{10,15,20\}. We report the macro average Hits@K accuracies for these subsets since the micro-average score generally overlooks the influences of the long-tails. We report the results of variations of our methods: \textbf{+KATT (FB)} is our approach with predefined hierarchies; \textbf{K-means (3)} and \textbf{K-means (4)} are K-means clustering methods to construct class hierarchies with three and four layers
respectively; \textbf{HC (3)} and \textbf{HC (4)} are hierarchical clustering methods to construct class hierarchies with three and four layers respectively; \textbf{AMIE} is the association rule mining under incomplete evidence to construct class hierarchies.
From the results in Table \ref{table4}, we observe: (i)
our model outperforms all the variations, which indicates that data-driven clustering may still have noise and may construct a wrong hierarchy between classes to degrade the performance;
(ii) data-driven clustering methods still outperform vanilla PCNN which further proves that relational knowledge between labels provide a much informal reference which boosts the performance for long-tail relations;
(iii) K-means (4) outperforms K-means (3) and HC (4) outperforms HC (3) which indicate the contributions of complex structures; (vi) K-means methods achieve comparable performance with AMIE and outperforms HC. This may be caused by the complexity and ambiguity of language as simple L2 distance may not inadequate to measure the similarity which leads to the wrong clustering.
\section{Initialization and Implementation Details}
To fairly compare the results of our approach with those baselines, we set most of the experimental parameters following \cite{lin2016neural}. Table \ref{nyt} shows all experimental parameters used in the experiments. We apply dropout on the output layers of our models to prevent overfitting. For CNN, we set the dropout rate to 0.5. For PCNN, we set the dropout rate to 0.9 to alleviate the overfitting problem. We also pre-train the sentence encoder of PCNN before training our hierarchical attention.
\begin{table}[]
\small
\begin{tabular}{cp{0.4cm}p{0.4cm}p{0.4cm}p{0.4cm}p{0.4cm}p{0.4cm}}
\hline
\multirow{2}*{\shortstack{Training Instances \\ Hits@K (Macro)}}
&\multicolumn{3}{c}{$<$100} &\multicolumn{3}{c}{$<$200} \\
&10&15&20&10&15&20 \\
\hline
+KATT (FB)& \textbf{35.3}&\textbf{62.4}&\textbf{65.1}&\textbf{43.2}&\textbf{61.3}& \textbf{69.2} \\
\hline
K-means (3)&34.2 &62.1 & 64.1& 42.5&61.3 & 68.1 \\
K-means (4)&34.3 &62.3 & 64.3& 42.7&61.4 & 68.1 \\
HC (3)&30.2 &62.1 & 60.1& 40.5&60.2 & 66.1 \\
HC (4)&31.2 &61.1 & 62.1& 41.5&60.2 & 66.2 \\
AMIE&34.2 &62.1 & 62.3& 42.2&61.2 & 67.9 \\
\hline
PCNN &$<$5.0 & 7.4 & 40.7& 17.2 &24.2&51.5 \\
\hline
\end{tabular}
\caption{Evaluation of our approach with different hierarchy label graph construction methods for relations with training instances fewer than 100/200.}
\label{table4}
\end{table}
\begin{table}[!htbp]
\centering
\begin{tabular}{c|c}
\hline
\centering \textbf{Parameter}& \textbf{Settings}\\
\hline
\centering Convolution Kernel Size& 3 \\
\centering Word Embedding Size&300 \\
\centering Position Embedding Size& 5 \\
\centering Learning Rate& 0.1 \\
\centering Hidden Layer Dimension for CNNs&230 \\
\centering Batch Size&160 \\
\hline
\end{tabular}\caption{Parameter settings.}\label{nyt}
\end{table}
\begin{table}[h]\centering
\fontsize{8.5}{10}\selectfont
\begin{tabular}{|p{4.9cm}|p{0.75cm}|p{0.75cm}|}
\hline
/people/deceased\_person/place\_of\_burial&\textbf{HATT}&\textbf{KATT} \\
\hline
\textbf{richard\_wagner} had his \textbf{bayreuth}, with its festspielhaus specially designed to accommodate his music dramas. &0.03 & 0.07 \\
\hline
wotan and alberich are \textbf{richard\_wagner}; and the rheingold and valhalla are wagner's real-life grail, the opera house in \textbf{bayreuth}.& 0.15& 0.13
\\
\hline
/location/br\_state/capital & \textbf{HATT} &\textbf{KATT} \\
\hline
there's violence everywhere, said ms. mesquita, who, like her friend, lives in \textbf{belo\_horizonte}, the capital of \textbf{minas\_gerais} state&0.48 &0.53\\
\hline all the research indicates that we are certain to find more gas in th amazon, eduardo braga, the governor of \textbf{amazonas}, said in an interview in \textbf{manaus}& 0.56&0.55\\
\hline
\end{tabular}
\caption{Example sentences for case study in higher-level.}
\label{table5}
\end{table}
In order to relieve the impact of $NA$ in DS based
relation extraction, we cut the propagation of loss from $NA$ following \cite{ye2017jointly}, which means if relation $c$ is $NA$, we set its loss as $0$.
\section{Case Study}
We report some additional examples of how our method takes effect in selecting the sentences in the higher level. In Table \ref{table5}, we display the sentence's attention score in the higher level (e.g., attention weights of node /people/deceased\_person/).
Both the relation /people/deceased\_person/place\_of\_burial (24 instances) and /location/br\_state/capital (4 instances) are long-tail relations. For those relations, the instance recommended by the higher-level attention straightforwardly expresses the relational facts.
This example shows that our approach achieves comparable results with HATT \cite{han2018hierarchical} and is beneficial for these long-tail relations .
\section{Evaluation Results for Different Instances}
\begin{table*} \centering
\small
\begin{tabular}{ccccccccccccc}
\hline
Test Mode &\multicolumn{4}{c}{ONE} &\multicolumn{4}{c}{TWO}& \multicolumn{4}{c}{ALL} \\
\hline
P@N&100&200&300&Mean&100&200&300&Mean&100&200&300&Mean \\
\hline
CNN+ONE&68.3&60.7&53.8&60.9&70.3&62.7&55.8&62.9&67.3&64.7&58.1&63.4\\
CNN+ATT&76.2&65.2&60.8&67.4&76.2&65.7&62.1&68.0&76.2&68.6&59.8&68.2\\
CNN+HATT&88.0&74.5&68.0&76.8&85.0&76.0&73.0&78.0&88.0&79.0&77.7&81.6\\
CNN+KATT&\textbf{90.0}&\textbf{75.1}&\textbf{69.3}&\textbf{79.0}&\textbf{85.3}&\textbf{79.2}&\textbf{74.0}&\textbf{79.5}&\textbf{89.2}&\textbf{80.2}&\textbf{79.2}&\textbf{82.9}\\
\hline
PCNN+ONE&73.3&64.8&56.8&65.0&70.3&67.27&63.1&66.9&72.3&69.7&64.1&68,7\\
PCNN+ATT&73.3&69.2&60.8&67.8&77.2&71.6&66.1&71.6&76.2&73.1&67.4&72.2\\
PCNN+HATT&84.0&76.0&69.7&76.6&85.0&76.0&72.7&77.9&88.0&79.5&75.3&80.9\\
PCNN+KATT&\textbf{85.3}&\textbf{77.1}&\textbf{70.2}&\textbf{77.5}&\textbf{86.1}&\textbf{77.1}&\textbf{73.1}&\textbf{78.8}&\textbf{89.2}&\textbf{80.1}&\textbf{75.8}&\textbf{81.7}\\
\hline
\end{tabular}
\caption{Top-N precision (P@N) for RE on the entity pairs with different number of instances (\%).}
\label{table3}
\end{table*}
Since our method focuses on modifications over selective attention, we also conduct Precision@N tests on those entity pairs with few instances following \cite{lin2016neural,han2018hierarchical}. We use the three test settings for this experiment: the ONE test set where we randomly select one instance for each entity pair for evaluation; the TWO test set where we randomly select two instances for each entity pair; the ALL test set where we use all instances for each remaining entity pair for evaluation. For the ONE and TWO test set, we intend to show that taking correlation information among relations into consideration can lead to a better relation
classifier. The ALL test set is designed to show the effect of our attention over multiple instances. We report the precision values of top $N$ triples extracted, where $N \in \{100,200,300\}$.
The evaluation results are shown in Table \ref{table3}, and from the results, we observe that (1) The performance of all methods is generally improved as the instance number increases which shows that the selective attention model can effectively take advantage of information from multiple noisy instances by combining useful instances while discarding useless ones. (2) Our KATT method has higher precision values in the ONE test set which indicates that in a few-shot scenario, correlations among relations can be caught by our model.
|
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"redpajama_set_name": "RedPajamaArXiv"
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| 4,707
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package org.jetbrains.plugins.javaFX.fxml.refs;
import com.intellij.psi.*;
import com.intellij.psi.xml.XmlAttributeValue;
import com.intellij.util.ArrayUtil;
import com.intellij.util.ArrayUtilRt;
import com.intellij.util.ProcessingContext;
import org.jetbrains.annotations.NotNull;
import org.jetbrains.annotations.Nullable;
import org.jetbrains.plugins.javaFX.fxml.JavaFxPsiUtil;
import java.util.ArrayList;
import java.util.List;
class JavaFxFactoryReferenceProvider extends PsiReferenceProvider {
@Override
public PsiReference @NotNull [] getReferencesByElement(@NotNull PsiElement element,
@NotNull ProcessingContext context) {
final XmlAttributeValue attributeValue = (XmlAttributeValue)element;
return new PsiReference[] {new JavaFXFactoryReference(attributeValue)};
}
private static class JavaFXFactoryReference extends PsiReferenceBase<XmlAttributeValue> {
JavaFXFactoryReference(XmlAttributeValue attributeValue) {
super(attributeValue);
}
@Nullable
@Override
public PsiElement resolve() {
final PsiClass psiClass = JavaFxPsiUtil.getTagClass(getElement());
if (psiClass != null) {
final PsiMethod[] psiMethods = psiClass.findMethodsByName(getElement().getValue(), false);
for (PsiMethod method : psiMethods) {
if (isFactoryMethod(method)) {
return method;
}
}
}
return null;
}
private static boolean isFactoryMethod(PsiMethod method) {
return method.hasModifierProperty(PsiModifier.STATIC) &&
method.getParameterList().isEmpty() &&
!PsiType.VOID.equals(method.getReturnType());
}
@Override
public Object @NotNull [] getVariants() {
final PsiClass psiClass = JavaFxPsiUtil.getTagClass(getElement());
if (psiClass != null) {
final List<PsiMethod> methods = new ArrayList<>();
for (PsiMethod method : psiClass.getMethods()) {
if (isFactoryMethod(method)) {
methods.add(method);
}
}
return ArrayUtil.toObjectArray(methods);
}
return ArrayUtilRt.EMPTY_OBJECT_ARRAY;
}
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 1,453
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Neto, Cristiana, Inês Dias, Maria Santos, Hugo Peixoto and José Machado. "Applied Business Intelligence in Surgery Waiting Lists Management." Healthcare Policy and Reform: Concepts, Methodologies, Tools, and Applications. IGI Global, 2019. 1580-1594. Web. 19 Jan. 2020. doi:10.4018/978-1-5225-6915-2.ch072
Neto, C., Dias, I., Santos, M., Peixoto, H., & Machado, J. (2019). Applied Business Intelligence in Surgery Waiting Lists Management. In I. Management Association (Ed.), Healthcare Policy and Reform: Concepts, Methodologies, Tools, and Applications (pp. 1580-1594). Hershey, PA: IGI Global. doi:10.4018/978-1-5225-6915-2.ch072
Neto, Cristiana, Inês Dias, Maria Santos, Hugo Peixoto and José Machado. "Applied Business Intelligence in Surgery Waiting Lists Management." In Healthcare Policy and Reform: Concepts, Methodologies, Tools, and Applications, ed. Information Resources Management Association, 1580-1594 (2019), accessed January 19, 2020. doi:10.4018/978-1-5225-6915-2.ch072
InfoSci-Medical, Healthcare, and Life Sciences
InfoSci-Social Sciences and Humanities
Communications, Social Science, and Healthcare
Applied Business Intelligence in Surgery Waiting Lists Management
Cristiana Neto (University of Minho, Portugal), Inês Dias (University of Minho, Portugal), Maria Santos (University of Minho, Portugal), Hugo Peixoto (University of Minho, Portugal) and José Machado (University of Minho, Portugal)
Source Title: Healthcare Policy and Reform: Concepts, Methodologies, Tools, and Applications
With the advent of computer science in hospitals, Electronic Health Record comes up, with the aim of bringing the new information technologies to the hospital environment with the promise not only to replace the paper process, but also to improve and provide better patient care. The operationalization of the EHR in supporting evidence-based practice, complex and conscientious decision-making, and improving the quality of healthcare delivery has been supported by the Business Intelligence (BI) technology. Since the beginning of the 1990s, the Portuguese health system has been confronted with a chronic problem, waiting time for surgery, due to inability to respond to demand for surgical therapy. Therefore, using business intelligence and information, obtained with the construction of dashboards, can help, for example, allocating hospital resources and reducing waiting times.
Because of the infinite amount of information, the sources of information in health units are complex, varied, immense and distributed. These days, it becomes more and more valuable to ensure a significant homogeneity between medical, administrative and clinical systems, leading to the integration of the different hospital systems.
It is a reality that the waiting times in hospitals for surgical treatment are high and, in Portugal, most of the health complaints result from high waiting times (Portela et al., 2011).
Access to health care is highly influenced by the existence of waiting lists and the existence of waiting lists lead to conclude that there is an inability of the health system to satisfy the elementary human health needs and raise concerns both at efficiency and equity levels (Barros, 2008). Despite being possible to reduce waiting lists' average duration to minimum values, it's considered to be impossible the absence of waiting lists (Hurst & Siciliani, 2003).
In Portugal, a patient is submitted to a primary care consultation and, if needed, proceeds to a specialist consultation at the hospital. A document designated P1, currently electronically registered is issued by the physician allowing the patient to be placed on the waiting list for a specialty consultation. After that, in need for surgical intervention, it proceeds to the waiting list for surgery (Fernandes et al., 2010).
For hospital values, waiting time is focused on the waiting time since there is a need for a specialty consultation until it is performed, and from the moment it is decided to proceed to a surgical intervention until the agenda of it (Portela et al., 2011). It's important to emphasize that waiting lists for surgery tend to be more pronounced in countries that combine health insurance (Barros, 2008).
For the last decades, Portugal has suffered an increase in waiting lists, which may be due to aging of the population, the introduction of new technologies, leading to an increased demand for surgical interventions. Also the malfunctions in resources' distribution and the number of operating rooms available, for example, interfere directly with those waiting times (OPSS, 2003).
Over the last few years, information systems (IS) have been implemented with the aim of combating waiting lists for surgery and they intend to help in patients' management. SIGIC – Integrated System for the Management of Enrolled Patients for Surgery was implemented within Portuguese health care institutions in 2004 and led to a better planning and programming of the institution's activity, reducing waiting times through management. In addition, the acknowledgment of different situations through information systems and a more efficient management of resources, tend to create a greater response capacity of health institutions (Oliveira, 2012).
Electronic Health Record (EHR) has recently become one of the most crucial sources of clinical information, due to the expansion of Health Information Systems (HIS) (Oliveira, 2012; Peixoto et al., 2010). It is a computerized health record that contains all the clinical information concerning a patient, such as biometric information, old prescriptions, lab and imaging results, clinical diagnosis, etc. It aims to help systems to bring together all the information provided to a certain patient, providing a cross-sectional analysis of the patient's medical history in different services and different institutions and, new advanced mechanisms that integrate EHR with decision support systems begin to appear (Oliveira, 2012).
The quantity and quality of the information available in an EHR for health professionals can have a strong impact on their performance, since it guides their decision-making path. It is therefore fundamental that multiple information axes intersect in a related and coherent way (Martins, 2011).
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
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| 4,872
|
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|
{
"redpajama_set_name": "RedPajamaC4"
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| 1,662
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{"url":"https:\/\/codereview.stackexchange.com\/questions\/27949\/critiques-for-program-that-integrates-functions","text":"# Critiques for program that integrates functions\n\nThe function is entered by the user as a series of coefficients and powers in the form of numerators and denominators. There is a little error somewhere with negative numbers that I'm working on finding. Other than that, I just want to know about the coding style.\n\nClass Integration\n\npackage calculus;\nimport java.util.Scanner;\npublic class Integration\n{\n\/\/create various fields to decide the integration method to be used\nRationalNumbers coeffFraction;\nRationalNumbers powerFraction;\nprivate final int DIRECT = 1;\nprivate final int SUBSTITUTION = 2;\n\/\/private final int TRIGONOMETRIC = 3;\n\/\/private final int INTEGRATION_BY_PARTS = 4;\n\/\/private final int LOGARITHMIC = 5;\n\n\/\/create fields to store various variables needed during the integration process\nprivate double[] coeffX;\nprivate double[] powers;\nprivate int numberOfVariables;\n\n\/\/fields to select integration method\nprivate int integrationMethod;\nprivate Scanner select;\n\n\/\/constructor\n\/**a constructor that prompts the user to select a number that denotes a certain\n* integration type and matches it to a consequent method that implements the type\n* of integration that has been selected\n*\/\npublic Integration()\n{\nselect = new Scanner(System.in);\nSystem.out.println(\"Enter the number denoting the method you want to execute\");\nSystem.out.println(\"1. Direct Integration \\n\" +\n\"2. Integration by substitution \\n\" +\n\"3. Integration of trigonometric functions \\n\" +\n\"4. Integration by parts \\n\" +\n\"5. Integration of Logarithmic functions \\n\");\n\n\/\/get input\nSystem.out.print(\"SELECT: \");\nintegrationMethod = select.nextInt();\n\nswitch(integrationMethod)\n{\ncase DIRECT:\ndirectIntegration();\nbreak;\ncase SUBSTITUTION:\nsubstitutionIntegration();\nbreak;\ndefault:\nSystem.out.println(\"Sorry that method is yet to be defined\");\n\n}\n\n}\n\n\/** this method is for integrating simple functions directly.\n* *\/\npublic void directIntegration()\n{\nselect = new Scanner(System.in);\nSystem.out.println(\"Direct Integration Selected\\n\" +\n\"f(x) = Axa + Bxb + Cxc + Dxd + ... + Nxn\");\nSystem.out.println(\"\\nEnter the number of variables in your function: \");\nnumberOfVariables = select.nextInt();\n\n\/\/initialize necessary arrays to hold all the variable coefficients and powers\ncoeffX = new double[numberOfVariables];\npowers = new double[numberOfVariables];\n\n\/\/initialize the first prompts as 'A' and 'a'\nchar coPrompt = 'A';\nchar powPrompt = 'a';\n\nSystem.out.println(\"Enter the coefficients and powers\");\n\n\/\/control input coefficients\nSystem.out.println(\"COEFFICIENTS\");\n\nfor(int counter=0; counter<coeffX.length; counter++, coPrompt++)\n{\nint numerator;\nint denominator;\n\nSystem.out.print(coPrompt + \": \\n\" +\n\"\\tnumerator = \");\nnumerator = select.nextInt();\nSystem.out.print(\"\\tdenominator = \");\ndenominator = select.nextInt();\n\ndouble coefficient = (double)numerator \/ denominator;\n\ncoeffX[counter] = coefficient;\n}\n\n\/\/control input powers\nSystem.out.println(\"POWERS\");\n\nfor(int counter=0; counter<powers.length; counter++, powPrompt++)\n{\nint numerator;\nint denominator;\n\nSystem.out.print(powPrompt + \": \\n\" +\n\"\\tnumerator = \");\nnumerator = select.nextInt();\nSystem.out.print(\"\\tdenominator = \");\ndenominator = select.nextInt();\n\ndouble power = (double)numerator \/ denominator;\n\npowers[counter] = power;\n}\n\n\/\/new powers and coefficients after integration\nSystem.out.print(\"\\nIntegrating ... \\n F(x) = \");\n\nfor(int counter = 0; counter<powers.length && counter<coeffX.length; counter++)\n{\npowers[counter] = powers[counter] + 1;\ncoeffX[counter] = coeffX[counter] \/ powers[counter];\n\npowerFraction = new RationalNumbers(powers[counter]);\ncoeffFraction = new RationalNumbers(coeffX[counter]);\n\n\/\/code to output\nif (counter == (coeffX.length - 1) && counter == (powers.length - 1))\n{\nSystem.out.print(coeffFraction.rationalize() + \"x\" + powerFraction.rationalize());\n}\nelse\n{\nSystem.out.print(coeffFraction.rationalize() + \"x\" + powerFraction.rationalize() + \" + \");\n}\n}\n}\n\n\/** this method utilizes substitution to integrate functions.\n* *\/\npublic void substitutionIntegration()\n{\nSystem.out.println(\"You have selected substitution integration lakini \\nbado sijaitengeneza\");\n}\n}\n\n\nClass RationalNumbers\n\npackage calculus;\n\nimport helper.GreatestCommonDivisor;\npublic class RationalNumbers\n{\nGreatestCommonDivisor helper;\nprivate String stringValue;\nprivate double decimalValue;\nprivate int numerator;\nprivate int denominator;\nprivate boolean recurring = false;\n\npublic RationalNumbers(double value)\n{\ndecimalValue = value;\nstringValue = String.valueOf(Math.abs(decimalValue));\ncheckRecurrence();\n}\n\n\/\/method to check whether the value passed is recurring or not\npublic void checkRecurrence()\n{\nif(stringValue.length() > 4)\n{\nif(stringValue.charAt(3) == stringValue.charAt(4))\n{\nrecurring = true;\n}\n}\n}\n\npublic void ratios()\n{\nint countDecimalPlaces = 0;\n\n\/\/code to extract recurring numbers and add their behavior\nif(recurring)\n{\nint firstValue = (int)(100 * decimalValue);\nint secondValue = (int)(1000 * decimalValue);\n\ndenominator = 900;\nnumerator = secondValue - firstValue;\n}\nelse\n{\nfor(int counter = 2; counter<stringValue.length(); counter++)\n{\ncountDecimalPlaces++;\n}\n\ndenominator = (int)(Math.pow(10, countDecimalPlaces));\n\nif(decimalValue >= 1) \/\/setting of numerator for improper fractions\n{\nnumerator = (int)(decimalValue * denominator);\n}\n\nelse\n{\nnumerator = Integer.parseInt(stringValue.substring(2));\n}\n}\n}\n\npublic String rationalize()\n{\nint[] rations = new int[2];\nString fraction;\n\nratios();\nhelper = new GreatestCommonDivisor(numerator, denominator);\n\nif(decimalValue<0) \/\/cater for decimal numbers inputed\n{\nrations[0] = (numerator \/ helper.gcd()) * (-1);\n}\nelse\n{\nrations[0] = numerator \/ helper.gcd();\n}\n\nrations[1] = denominator \/ helper.gcd();\n\nif(rations[1] == 1)\n{\nfraction = String.valueOf(rations[0]);\n}\n\nelse\n{\nfraction = (rations[0] + \"\/\" + rations[1]);\n}\n\nreturn fraction;\n}\n}\n\n\nClass GreatestCommonDivisor\n\npackage helper;\n\/**This class has a constructor that accepts two values and\n* then implements the gcd() method to find the greatest common\n* divisor of the values*\/\npublic class GreatestCommonDivisor\n{\nprivate int numerator;\nprivate int denominator;\nprivate int gcd = 1;\n\npublic GreatestCommonDivisor(int value1, int value2)\n{\nnumerator = value1;\ndenominator = value2;\n}\n\npublic int gcd()\n{\nint dividend = 2;\n\n\/\/check here for the problem when you are freshazamiz\nwhile(dividend <= Math.min(numerator, denominator))\n{\nwhile(numerator % dividend == 0 && denominator % dividend == 0)\n{\nnumerator = numerator \/ dividend;\ndenominator = denominator \/ dividend;\ngcd = gcd * dividend;\n}\n\ndividend++;\n}\n\nreturn gcd;\n}\n}\n\n\nWhy do you have commented out code? If you do need it please throw it away. Leaving it commented out does not help.\n\nI'd also replace all these constants with an Enum to improve readability and maintainability of the code. With the Enum you can also rely on the compiler to check that you use a valid value.\n\nprivate final int DIRECT = 1;\nprivate final int SUBSTITUTION = 2;\nprivate final int TRIGONOMETRIC = 3;\nprivate final int INTEGRATION_BY_PARTS = 4;\nprivate final int LOGARITHMIC = 5;\n\n\nIn order to have a more object oriented solution, I'd replace the switch that you use to choose the integration method with polymorphism or with a strategy pattern.\n\nI'd also avoid mixing methods that handle input\/output and methods that does the computation of the result. Consider having I\/O in a separate class that the one actually doing the integration.\n\nIs your RationalNumbers class representing a single number or a set of number? If it has to represent a single number please call it RationalNumber.\n\nI'd introduce a Fraction data type and use it to introduce the output of the rationalize method. Why should it return a String?\n\n\u2022 Fraction data type? I do not think i am familiar with that. i will rectify the code appropriately however I just want to point out that for proper output, I needed to the rationalize method to return a string rather than an array(because it has to return 2 values). \u2013\u00a0Kis Jun 29 '13 at 18:46\n\u2022 I was suggesting you to create a new Fraction class to represent mathematical fractions. It should have two attributes (The numerator and the dividend) and it should solve cleanly the issue you have returning two values. \u2013\u00a0mariosangiorgio Jun 29 '13 at 18:48\n\u2022 Thank you. Polymorphism vs switch, I\/O vs Computation, Naming Conventions & style (I cant beleive I missed this) and Fraction class. Thank you Mario for your pointers and help. \u2013\u00a0Kis Jun 29 '13 at 18:53\n\u2022 @mariosangiorgio No need for a Fraction type. RationalNumber is OP's fraction type. rationalize is just a bad name for toString. \u2013\u00a0abuzittin gillifirca Jul 1 '13 at 6:48","date":"2019-10-14 00:29:23","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.41863390803337097, \"perplexity\": 10634.453188107345}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-43\/segments\/1570986648343.8\/warc\/CC-MAIN-20191013221144-20191014004144-00395.warc.gz\"}"}
| null | null |
Q: Insert a line at specific line number with sed or awk I have a script file which I need to modify with another script to insert a text at the 8th line.
String to insert: Project_Name=sowstest, into a file called start.
I tried to use awk and sed, but my command is getting garbled.
A: sed -e '8iProject_Name=sowstest' -i start using GNU sed
Sample run:
[root@node23 ~]# for ((i=1; i<=10; i++)); do echo "Line #$i"; done > a_file
[root@node23 ~]# cat a_file
Line #1
Line #2
Line #3
Line #4
Line #5
Line #6
Line #7
Line #8
Line #9
Line #10
[root@node23 ~]# sed -e '3ixxx inserted line xxx' -i a_file
[root@node23 ~]# cat -An a_file
1 Line #1$
2 Line #2$
3 xxx inserted line xxx$
4 Line #3$
5 Line #4$
6 Line #5$
7 Line #6$
8 Line #7$
9 Line #8$
10 Line #9$
11 Line #10$
[root@node23 ~]#
[root@node23 ~]# sed -e '5ixxx (inserted) "line" xxx' -i a_file
[root@node23 ~]# cat -n a_file
1 Line #1
2 Line #2
3 xxx inserted line xxx
4 Line #3
5 xxx (inserted) "line" xxx
6 Line #4
7 Line #5
8 Line #6
9 Line #7
10 Line #8
11 Line #9
12 Line #10
[root@node23 ~]#
A: Perl solutions:
quick and dirty:
perl -lpe 'print "Project_Name=sowstest" if $. == 8' file
*
*-l strips newlines and adds them back in, eliminating the need for "\n"
*-p loops over the input file, printing every line
*-e executes the code in single quotes
$. is the line number
equivalent to @glenn's awk solution, using named arguments:
perl -slpe 'print $s if $. == $n' -- -n=8 -s="Project_Name=sowstest" file
*
*-s enables a rudimentary argument parser
*-- prevents -n and -s from being parsed by the standard perl argument parser
positional command arguments:
perl -lpe 'BEGIN{$n=shift; $s=shift}; print $s if $. == $n' 8 "Project_Name=sowstest" file
environment variables:
setenv n 8 ; setenv s "Project_Name=sowstest"
echo $n ; echo $s
perl -slpe 'print $ENV{s} if $. == $ENV{n}' file
ENV is the hash which contains all environment variables
Getopt to parse arguments into hash %o:
perl -MGetopt::Std -lpe 'BEGIN{getopt("ns",\%o)}; print $o{s} if $. == $o{n}' -- -n 8 -s "Project_Name=sowstest" file
Getopt::Long and longer option names
perl -MGetopt::Long -lpe 'BEGIN{GetOptions(\%o,"line=i","string=s")}; print $o{string} if $. == $o{line}' -- --line 8 --string "Project_Name=sowstest" file
Getopt is the recommended standard-library solution.
This may be overkill for one-line perl scripts, but it can be done
A: For those who are on SunOS which is non-GNU, the following code will help:
sed '1i\^J
line to add' test.dat > tmp.dat
*
*^J is inserted with ^V+^J
*Add the newline after '1i.
*\ MUST be the last character of the line.
*The second part of the command must be in a second line.
A: An ed answer
ed file << END
8i
Project_Name=sowstest
.
w
q
END
. on its own line ends input mode; w writes; q quits. GNU ed has a wq command to save and quit, but old ed's don't.
Further reading: https://gnu.org/software/ed/manual/ed_manual.html
A: OS X / macOS / FreeBSD sed
The -i flag works differently on macOS sed than in GNU sed.
Here's the way to use it on macOS / OS X:
sed -i '' '8i\
8 This is Line 8' FILE
See man 1 sed for more info.
A: sed -i '8i This is Line 8' FILE
inserts at line 8
This is Line 8
into file FILE
-i does the modification directly to file FILE, no output to stdout, as mentioned in the comments by glenn jackman.
A: the awk answer
awk -v n=8 -v s="Project_Name=sowstest" 'NR == n {print s} {print}' file > file.new
A: POSIX sed (and for example OS X's sed, the sed below) require i to be followed by a backslash and a newline. Also at least OS X's sed does not include a newline after the inserted text:
$ seq 3|gsed '2i1.5'
1
1.5
2
3
$ seq 3|sed '2i1.5'
sed: 1: "2i1.5": command i expects \ followed by text
$ seq 3|sed $'2i\\\n1.5'
1
1.52
3
$ seq 3|sed $'2i\\\n1.5\n'
1
1.5
2
3
To replace a line, you can use the c (change) or s (substitute) commands with a numeric address:
$ seq 3|sed $'2c\\\n1.5\n'
1
1.5
3
$ seq 3|gsed '2c1.5'
1
1.5
3
$ seq 3|sed '2s/.*/1.5/'
1
1.5
3
Alternatives using awk:
$ seq 3|awk 'NR==2{print 1.5}1'
1
1.5
2
3
$ seq 3|awk '{print NR==2?1.5:$0}'
1
1.5
3
awk interprets backslashes in variables passed with -v but not in variables passed using ENVIRON:
$ seq 3|awk -v v='a\ba' '{print NR==2?v:$0}'
1
a
3
$ seq 3|v='a\ba' awk '{print NR==2?ENVIRON["v"]:$0}'
1
a\ba
3
Both ENVIRON and -v are defined by POSIX.
A: sed -i "" -e $'4 a\\\n''Project_Name=sowstest' start
*
*This line works fine in macOS
A: macOS sed solutions
for example: inserts at line 1
*
*ns
# recommended
# This command only needs to write one line
$ sed -i '' '1s/^/The new First Line\n/' ./your-source-file-name
*ni
# not recommended
# This way the command needs to be written on multiple lines
$ sed -i '' '1i\
The new First Line\
' ./your-source-file-name
test demo
$ sed -i '' '1s/^/Perl camel\n/' ./multi-line-text.txt
A: it is working fine in linux to add in 2 lines.
sed '2s/$/ myalias/' file
A: Thank you umläute
sed -i "" -e $'4 a\\\n''Project_Name=sowstest' filename
the following was usefull on macOS to be able to add a new line after the 4
In order to loop i created an array of folders, ti iterate on them in mac zsh
for foldercc in $foldernames;
sed -i "" -e $'4 a\\\n''Project_Name=sowstest' $foldercc/filenames;
|
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{"url":"https:\/\/gmat.kmf.com\/question\/ps\/manhattan?page=7","text":"\u2022 \u6309\u4e66\u672c\u67e5\u770b\n\nManhattan\u9898\u96c6\n\nManhattan\u662f\u7f8e\u56fd\u4e00\u4e2a\u7c7b\u4f3cXDF\u7684\u6559\u80b2\u8f85\u5bfc\u673a\u6784\uff0c\u7f51\u4e0a\u5e38\u5e38\u76f4\u63a5\u7528\u4e8e\u6307\u4ee3\u5b83\u6240\u51fa\u7248\u7684\u4e00\u7cfb\u5217GMAT\u590d\u4e60\u8d44\u6599\u3002\u5176\u5907\u8003\u8d44\u6599\u5728GMAT\u5907\u8003\u8ba1\u5212\u4e2d\u4f7f\u7528\u5e7f\u6cdb\uff0c\u8003\u751f\u591a\u7528\u6765\u8fdb\u884c\u5f3a\u5316\u7ec3\u4e60\u6216\u5de9\u56fa\u63d0\u5347\u3002Manhattan\u7684\u4e66\u7c4d\u5185\u5bb9\u591a\u662f\u4ee5\u7406\u8bba\u4e0e\u5b9e\u8df5\u76f8\u7ed3\u5408\u7684\u5f62\u5f0f\u8fdb\u884c\u7f16\u5199\uff0c\u524d\u534a\u90e8\u5206\u4e3b\u8981\u5bf9\u5404\u4e2a\u5355\u9879\u8fdb\u884c\u7406\u8bba\u77e5\u8bc6\u7684\u4ecb\u7ecd\uff0c\u540e\u534a\u90e8\u5206\u5219\u7740\u91cd\u5f3a\u5316\u7ec3\u4e60\uff0c\u5e76\u5e26\u6709\u8be6\u7ec6\u7684\u9898\u76ee\u89e3\u6790\uff0c\u65e8\u5728\u5e2e\u52a9\u8003\u751f\u66f4\u597d\u7684\u7406\u89e3\u6587\u7ae0\u542b\u4e49\uff0c\u63d0\u5347GMAT\u9605\u8bfb\u7684\u505a\u9898\u6c34\u5e73\u3002\n\n121 If a and b are positive integers and$x = 4^{ a} and y = 9^{ b}$, which of the following is a possible units digit of xy?\n122 When a is divided by b, the quotient is c and the remainder is d. Which of the following expressions is equal to d?\n123 If$9^{2x + 1} = 27^{2}$, then x =\n124 Jack pours water into a bucket until the bucket is filled to 60% of its capacity. Jill then fills the bucket to capacity by pouring an additional $\\frac{2}{3}$ of a gallon of water into the bucket. What is the capacity of the bucket, in gallons?\n125 The perimeter of a rectangular yard is completely surrounded by a fence that measures 40 meters. What is the length of the yard if the area of the yard is 64 meters squared?\n126 A recycling facility is staffed by eight floor workers and one manager. All of the floor workers are paid equal wages, but the manager is paid q times as much as a floor worker. If the manager\u2019s wages account for$\\frac {1}{7}$of all wages paid at the facility, what is the value of q?\n127 If $3^{2n} = (\\frac{1}{9})^{n+2}$, what is the value of n?\n128 $\\frac{49+49}{49+49^{2}}$=\n129 Set A contains three different positive odd integers and two different positive even integers; set B contains two different positive odd integers and three different positive even integers. If one integer from set A and one integer from set B are chosen at random, what is the probability that the product of the chosen integers is even?\n130 Each of the following numbers has a remainder of 2 when divided by 11 except:\n131 If a and b represent positive single digits in the correctly worked computation above, what is the value of a + 2b?\n\n9362\n\n6030027\n\n2194929828\n\n\u2022 [IR]13fm3k\n\n\u6b63\u786e\u7387\uff1a3.6%\n\n\u2022 [IR]1dfnkk\n\n\u6b63\u786e\u7387\uff1a5.9%\n\n\u2022 [DS]34g0kk\n\n\u6b63\u786e\u7387\uff1a6.3%\n\n\u2022 [IR]acfnlk\n\n\u6b63\u786e\u7387\uff1a7%\n\n\u2022 [IR]2ffmnk\n\n\u6b63\u786e\u7387\uff1a7%","date":"2021-12-02 06:11:38","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6767687201499939, \"perplexity\": 595.8374135598647}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-49\/segments\/1637964361169.72\/warc\/CC-MAIN-20211202054457-20211202084457-00488.warc.gz\"}"}
| null | null |
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More than 7,000 organizations are using Boomi to run better, faster and smarter. Our customers include global leaders across markets and around the world, include Novartis, Salesforce, LinkedIn, Candy.com, NetSuite, Lucky Brand, and Kelly-Moore Paints.
Boomi has been the pioneering leader of application and data integration since Rick Nucci founded the company in 2000.
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|
Robert or Bob Duncan may refer to:
Arts
Robert Duncan (poet) (1919–1988), American poet
Robert Duncan (writer) (born 1952), American music critic
Robert Duncan (actor) (born 1952), British TV actor
Robert Duncan (composer), Canadian composer
Politics
Robert Duncan (politician) (1850–1925), Unionist Party (Scotland) MP for Govan
Robert B. Duncan (1920–2011), U.S. Representative from Oregon
Mike Duncan (Robert Michael Duncan, born 1951), 62nd Chairman of the Republican National Committee
Robert M. Duncan Jr. (born 1978), United States Attorney and son of Mike Duncan
Robert L. Duncan (born 1953), Texas state senator
Science and engineering
Robert C. Duncan (engineer) (1923–2003), American engineer
Robert C. Duncan (astrophysicist), American astrophysicist
Robert Duncan (physicist), American physicist
Other people
Robert Duncan of Robert Duncan and Company, a shipyard in the Port of Glasgow on the Clyde in Scotland
Robert C. Duncan (athlete) (1887–1957), British Olympic athlete
Robert Duncan (footballer) (1891–1984), Australian rules footballer
Robert Duncan (pilot) (1920–2013), flying ace
Robert Morton Duncan (1927–2012), U.S. federal judge
Robert Duncan (rower) (born 1931), Australian rower
Robert Duncan (bishop) (born 1948), Archbishop of the Anglican Church in North America
Bob Duncan, a character in Good Luck Charlie
See also
Bobby Duncum
Robert Duncan Bell (1878–1953), Acting Governor of Bombay during the British Raj
Robert Duncan Luce, American mathematician and social scientist
Robert Duncan MacPherson, American mathematician at the Institute for Advanced Study and Princeton University
Robert Duncan McNeill (born 1964), American director, producer, and actor
Robert Duncan Milne (1844–1899), late-19th century San Francisco science fiction writer
Robert Duncan Sherrington, Australian politician
Robert Duncan Wilmot (1809–1891), Canadian politician and a Father of Confederation
Robert Duncan Wilmot Jr. (1837–1920), Canadian farmer, businessman and politician
Robert Duncanson (disambiguation)
|
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| 8,822
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Holorusia sakarahana är en tvåvingeart som först beskrevs av Alexander 1960. Holorusia sakarahana ingår i släktet Holorusia och familjen storharkrankar.
Artens utbredningsområde är Madagaskar. Inga underarter finns listade i Catalogue of Life.
Källor
Storharkrankar
sakarahana
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 4,030
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#include "fsl_flashiap.h"
/* Component ID definition, used by tools. */
#ifndef FSL_COMPONENT_ID
#define FSL_COMPONENT_ID "platform.drivers.flashiap"
#endif
#define HZ_TO_KHZ_DIV 1000
/*******************************************************************************
* Code
******************************************************************************/
static status_t translate_iap_status(uint32_t status)
{
/* Translate IAP return code to sdk status code */
if (status == kStatus_Success)
{
return status;
}
else
{
return MAKE_STATUS(kStatusGroup_FLASHIAP, status);
}
}
/*!
* brief Prepare sector for write operation
* This function prepares sector(s) for write/erase operation. This function must be
* called before calling the FLASHIAP_CopyRamToFlash() or FLASHIAP_EraseSector() or
* FLASHIAP_ErasePage() function. The end sector must be greater than or equal to
* start sector number.
*
* deprecated Do not use this function. It has benn moved to iap driver.
*
* param startSector Start sector number.
* param endSector End sector number.
*
* retval #kStatus_FLASHIAP_Success Api was executed successfully.
* retval #kStatus_FLASHIAP_NoPower Flash memory block is powered down.
* retval #kStatus_FLASHIAP_NoClock Flash memory block or controller is not clocked.
* retval #kStatus_FLASHIAP_InvalidSector Sector number is invalid or end sector number
* is greater than start sector number.
* retval #kStatus_FLASHIAP_Busy Flash programming hardware interface is busy.
*/
status_t FLASHIAP_PrepareSectorForWrite(uint32_t startSector, uint32_t endSector)
{
uint32_t command[5], result[4];
command[0] = kIapCmd_FLASHIAP_PrepareSectorforWrite;
command[1] = startSector;
command[2] = endSector;
iap_entry(command, result);
return translate_iap_status(result[0]);
}
/*!
* brief Copy RAM to flash.
* This function programs the flash memory. Corresponding sectors must be prepared
* via FLASHIAP_PrepareSectorForWrite before calling calling this function. The addresses
* should be a 256 byte boundary and the number of bytes should be 256 | 512 | 1024 | 4096.
*
* deprecated Do not use this function. It has benn moved to iap driver.
*
* param dstAddr Destination flash address where data bytes are to be written.
* param srcAddr Source ram address from where data bytes are to be read.
* param numOfBytes Number of bytes to be written.
* param systemCoreClock SystemCoreClock in Hz. It is converted to KHz before calling the
* rom IAP function.
*
* retval #kStatus_FLASHIAP_Success Api was executed successfully.
* retval #kStatus_FLASHIAP_NoPower Flash memory block is powered down.
* retval #kStatus_FLASHIAP_NoClock Flash memory block or controller is not clocked.
* retval #kStatus_FLASHIAP_SrcAddrError Source address is not on word boundary.
* retval #kStatus_FLASHIAP_DstAddrError Destination address is not on a correct boundary.
* retval #kStatus_FLASHIAP_SrcAddrNotMapped Source address is not mapped in the memory map.
* retval #kStatus_FLASHIAP_DstAddrNotMapped Destination address is not mapped in the memory map.
* retval #kStatus_FLASHIAP_CountError Byte count is not multiple of 4 or is not a permitted value.
* retval #kStatus_FLASHIAP_NotPrepared Command to prepare sector for write operation was not executed.
* retval #kStatus_FLASHIAP_Busy Flash programming hardware interface is busy.
*/
status_t FLASHIAP_CopyRamToFlash(uint32_t dstAddr, uint32_t *srcAddr, uint32_t numOfBytes, uint32_t systemCoreClock)
{
uint32_t command[5], result[4];
command[0] = kIapCmd_FLASHIAP_CopyRamToFlash;
command[1] = dstAddr;
command[2] = (uint32_t)srcAddr;
command[3] = numOfBytes;
command[4] = systemCoreClock / HZ_TO_KHZ_DIV;
iap_entry(command, result);
return translate_iap_status(result[0]);
}
/*!
* brief Erase sector
* This function erases sector(s). The end sector must be greater than or equal to
* start sector number. FLASHIAP_PrepareSectorForWrite must be called before
* calling this function.
*
* deprecated Do not use this function. It has benn moved to iap driver.
*
* param startSector Start sector number.
* param endSector End sector number.
* param systemCoreClock SystemCoreClock in Hz. It is converted to KHz before calling the
* rom IAP function.
*
* retval #kStatus_FLASHIAP_Success Api was executed successfully.
* retval #kStatus_FLASHIAP_NoPower Flash memory block is powered down.
* retval #kStatus_FLASHIAP_NoClock Flash memory block or controller is not clocked.
* retval #kStatus_FLASHIAP_InvalidSector Sector number is invalid or end sector number
* is greater than start sector number.
* retval #kStatus_FLASHIAP_NotPrepared Command to prepare sector for write operation was not executed.
* retval #kStatus_FLASHIAP_Busy Flash programming hardware interface is busy.
*/
status_t FLASHIAP_EraseSector(uint32_t startSector, uint32_t endSector, uint32_t systemCoreClock)
{
uint32_t command[5], result[4];
command[0] = kIapCmd_FLASHIAP_EraseSector;
command[1] = startSector;
command[2] = endSector;
command[3] = systemCoreClock / HZ_TO_KHZ_DIV;
iap_entry(command, result);
return translate_iap_status(result[0]);
}
/*!
* This function erases page(s). The end page must be greater than or equal to
* start page number. Corresponding sectors must be prepared via FLASHIAP_PrepareSectorForWrite
* before calling calling this function.
*
* deprecated Do not use this function. It has benn moved to iap driver.
*
* param startPage Start page number
* param endPage End page number
* param systemCoreClock SystemCoreClock in Hz. It is converted to KHz before calling the
* rom IAP function.
*
* retval #kStatus_FLASHIAP_Success Api was executed successfully.
* retval #kStatus_FLASHIAP_NoPower Flash memory block is powered down.
* retval #kStatus_FLASHIAP_NoClock Flash memory block or controller is not clocked.
* retval #kStatus_FLASHIAP_InvalidSector Page number is invalid or end page number
* is greater than start page number
* retval #kStatus_FLASHIAP_NotPrepared Command to prepare sector for write operation was not executed.
* retval #kStatus_FLASHIAP_Busy Flash programming hardware interface is busy.
*/
status_t FLASHIAP_ErasePage(uint32_t startPage, uint32_t endPage, uint32_t systemCoreClock)
{
uint32_t command[5], result[4];
command[0] = kIapCmd_FLASHIAP_ErasePage;
command[1] = startPage;
command[2] = endPage;
command[3] = systemCoreClock / HZ_TO_KHZ_DIV;
iap_entry(command, result);
return translate_iap_status(result[0]);
}
/*!
* brief Blank check sector(s)
*
* Blank check single or multiples sectors of flash memory. The end sector must be greater than or equal to
* start sector number. It can be used to verify the sector eraseure after FLASHIAP_EraseSector call.
*
* deprecated Do not use this function. It has benn moved to iap driver.
*
* param startSector : Start sector number. Must be greater than or equal to start sector number
* param endSector : End sector number
* retval #kStatus_FLASHIAP_Success One or more sectors are in erased state.
* retval #kStatus_FLASHIAP_NoPower Flash memory block is powered down.
* retval #kStatus_FLASHIAP_NoClock Flash memory block or controller is not clocked.
* retval #kStatus_FLASHIAP_SectorNotblank One or more sectors are not blank.
*/
status_t FLASHIAP_BlankCheckSector(uint32_t startSector, uint32_t endSector)
{
uint32_t command[5], result[4];
command[0] = kIapCmd_FLASHIAP_BlankCheckSector;
command[1] = startSector;
command[2] = endSector;
iap_entry(command, result);
return translate_iap_status(result[0]);
}
/*!
* brief Compare memory contents of flash with ram.
* This function compares the contents of flash and ram. It can be used to verify the flash
* memory contents after FLASHIAP_CopyRamToFlash call.
*
* deprecated Do not use this function. It has benn moved to iap driver.
*
* param dstAddr Destination flash address.
* param srcAddr Source ram address.
* param numOfBytes Number of bytes to be compared.
*
* retval #kStatus_FLASHIAP_Success Contents of flash and ram match.
* retval #kStatus_FLASHIAP_NoPower Flash memory block is powered down.
* retval #kStatus_FLASHIAP_NoClock Flash memory block or controller is not clocked.
* retval #kStatus_FLASHIAP_AddrError Address is not on word boundary.
* retval #kStatus_FLASHIAP_AddrNotMapped Address is not mapped in the memory map.
* retval #kStatus_FLASHIAP_CountError Byte count is not multiple of 4 or is not a permitted value.
* retval #kStatus_FLASHIAP_CompareError Destination and source memory contents do not match.
*/
status_t FLASHIAP_Compare(uint32_t dstAddr, uint32_t *srcAddr, uint32_t numOfBytes)
{
uint32_t command[5], result[4];
command[0] = kIapCmd_FLASHIAP_Compare;
command[1] = dstAddr;
command[2] = (uint32_t)srcAddr;
command[3] = numOfBytes;
iap_entry(command, result);
return translate_iap_status(result[0]);
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 92
|
\section{Introduction}
If one draws a (connected) graph --- a particular set of vertices and edges --- on a smooth surface, then such graph inherits extra local/combinatorial and global/topological features from the surface. If the latter is, for example, a (compact) complex one-dimensional surface --- a Riemann surface, then the combinatorics of edges is encapsulated by a two-generator permutation group and the Riemann surface happens to be definable over the field $\bar{\mathbb{Q}}$ of algebraic numbers. This observation is central to the concept of {\it dessins d'enfants} (or child's drawings) as advocated by Grothendieck in his {\it Esquisse d'un programme} (made available in 1984 following his {\it Long March} written in 1981) in the following words:
{\it In the form in which Belyi states it, his result essentially says that every algebraic curve defined over a number field can be obtained as a covering of the projective line ramified only over the points 0, 1 and $\infty$. The result seems to have remained more or less unobserved. Yet it appears to me to have considerable importance. To me, its essential message is that there is a profound identity between the combinatorics of finite maps on the one hand, and the geometry of algebraic curves defined over number fields on the other. This deep result, together with the algebraic interpretation of maps, opens the door into a new, unexplored world - within reach of all, who pass by without seeing it} \cite[Vol. 1]{Grothen}, \cite{Schneps1}.
Our aim is to show that Grothendieck's {\it dessins d'enfants} (see, e.\,g., \cite{zapp2003, Lando2004} as well as \cite{Koch2010}) have, as envisaged in \cite{PlanatFQXi}, great potential to become a proper language for a deeper understanding of various types of sets of Hermitian operators/observables that appear in finite-dimensional quantum mechanical settings and for furnishing a natural explanation why eigenvalues of these operators are regarded as the only available tracks in associated measurements. The main justification of our aim is provided by the fact that {\it dessins} lead very naturally to {\it already}-discovered finite geometries underlying quantum contextuality (like the grid, GQ(2,\,1), behind Mermin's magic square and/or an ovoid of PG(3,\,2) behind Mermin's magic pentagram) and also to those underlying commutation relations between elements of the two-qubit Pauli group (the generalized quadrangle of order two, GQ(2,\,2), its geometric hyperplanes and their complements, see, e.\,g., \cite{Planat2007}).
The paper is organized as follows. Section \ref{dessins} gathers some basic
knowledge about {\it dessins d'enfants}, their permutation group and topology,
their isomorphism with conjugacy classes of subgroups of finite index of the
cartographic group $C_2^+$, as well as about associated Belyi functions.
Section \ref{square} focuses on a rather elementary application of our ideas by
interpreting Bell's theorem about non-locality in terms of the geometry as
simple as a square/quadrangle, which is found to be generated by four distinct
{\it dessins} defined over the field $\mathbb{Q}[\sqrt{2}]$.
Section \ref{catalog}, the core one, starts with a complete catalog of all
connected geometries induced by {\it dessins} having up to $12$ edges and a
sketch of important ones with more edges. In the subsequent subsections, we
analyze in detail the non-trivial cases by selecting, whenever possible, a {\it
dessin} of genus zero and having the smallest number of faces. As in most cases
the edges of {\it dessins} dealt with admit labeling by two- or three-qubit
observables, on our way we not only encounter already recognized
quantum-relevant finite geometries like the Fano plane, the grid GQ(2,\,1), the
Petersen graph, the Desargues configuration and the generalized quadrangle
GQ(2,\,2), but find a bunch of novel ones, some already surmised from different
contexts, starting from the Pappus $9_3$-configuration and the Hesse $(9_4,
12_3)$-configuration ({\it aka} the affine plane AG(2,\,3)) to arrive at the
generalized quadrangle GQ(2,\,4) --- and its close siblings, the Clebsch and
Schl\"{a}fli graphs -- known to play a role in the context of the
black-hole--qubit correspondence \cite{Levay2009}.
Section 5 is reserved for concluding remarks.
\section{\textit{Dessins d'enfants} and the Belyi theorem}
\label{dessins}
\subsection*{Dessins d'enfants and their symmetry groups}
A {\it map} is a graph drawn on a surface --- a smooth compact orientable
variety of dimension two --- such that its vertices are points, its edges are
non-intersecting arcs connecting the vertices, and the connected
components of its complement, called faces, are
homeomorphic to open disks of $\mathbb{R}^2$. They may exist multiple edges as well as loops, but the graph has to be connected. Denoting the number of vertices, edges and faces by $V$, $E$ and $F$, respectively, the genus $g$ of the map follows from Euler's formula $V-E+F=2-2g$.
A map can be generalized to a bicolored map. The latter is a map whose vertices are colored in black and white in such a way that the adjacent vertices have always the opposite color; the corresponding segments are the edges of the bicolored map. The Euler characteristic now reads $2-2g=B+W+F-n$, where $B$, $W$ and $n$ stands for the number of black vertices, the number of white vertices and the number of edges, respectively. Given a bicolored map with $n$ edges labeled from 1 to $n$, one can associate with it a permutation group $P=\left\langle \alpha, \beta \right\rangle$ on the set of labels such that a cycle of $\alpha$ (resp. $\beta$) contains the labels of the edges incident to a black vertex (resp. white vertex),
taken, say, in the clockwise direction around this
vertex; thus, there are as many cycles in $\alpha$ (resp. $\beta$) as there are black (resp. white vertices),
and the degree of a vertex is equal to the length of the corresponding
cycle. An analogous cycle structure for the faces follows from the permutation $\gamma$ satisfying $\alpha\beta\gamma=1$.
Bicolored maps (allowed to have any valency for their vertices) are in
one-to-one correspondence with hypermaps \cite{Walsh}. They correspond to the
conjugacy classes of subgroups of finite index of the free group on two
generators $H_2^+=\left\langle \rho_0,\rho_1 \right\rangle$.
The number of hypermaps with $n$ half-edges is given by the sequence
{\it A057005} in the OEIS~\cite{A057005}. For $n= 1,\ldots,7$ these
numbers are $1$, $3$, $7$, $26$, $97$, $624$ and $4163$.
Hypermaps are, of course, allowed to have any valency for their
vertices.
We consider bicolored maps where the valency of white vertices is $\leq 2$.
They correspond to hypermaps of the so-called pre-clean type.
As already observed by Grothendieck himself, who called them {\it dessins
d'enfants} (child's drawings)~\cite{Schneps1}--\cite{Girondo}, these bicolored
maps on connected oriented surfaces are unique in the sense that they are in
one-to-one correspondence with conjugacy classes of subgroups of finite index of
the triangle group, also called {\it cartographic} group
\begin{equation}
C_2^+=\left\langle \rho_0,\rho_1|\rho_1^2=1 \right\rangle.
\label{cartographic}
\end{equation}
The existence of associated {\it dessins} of prescribed properties can thus be
straightforwardly checked from a systematic enumeration of conjugacy classes of
$C_2^+$; with the increasing $n>0$ the number of such {\it dessins} grows quite
rapidly \[1, 3, 3, 10, 15, 56, 131, 482, 1551, 5916, 22171, 90033,
370199,\cdots\]
A {\it dessin} $\mathcal{D}$ can be ascribed a signature $s=(B,W,F,g)$ and the full information about it can be recovered from the structure of the generators of its permutation group $P$ (also named the passport in \cite{Lando2004,Zvonkin}) in the form $[C_{\alpha},C_{\beta},C_{\gamma}]$, where the entry $C_i$, $i \in \{\alpha,\beta,\gamma\}$ has factors $l_i^{n_i}$, with $l_i$ denoting the length of the cycle and $n_i$ the number of cycles of length $l_i$.
\subsection*{Belyi's theorem}
Given $f(x)$, a rational function of the complex variable $x$, a {\it critical point} of $f$ is a root of its derivative and a {\it critical value} of $f$ is the value of $f$ at the critical point.
Let us define a so-called {\it Belyi function} corresponding to a
dessin $\mathcal{D}$ as a rational function $f(x)$ of degree $n$ if $\mathcal{D}$ may be embedded into
the Riemann sphere $\hat{\mathbb{C}}$ in such a way that (i) the black vertices are the roots of the equation $f(x)=0$ with the multiplicity of each root being equal to the degree of the corresponding (black) vertex, (ii) the white vertices are the roots of the equation $f(x)=1$ with the multiplicity of each root being equal to the degree of the corresponding (white) vertex,
(iii) the bicolored map is the preimage of the segment $[0,1]$, that is $\mathcal{D}=f^{-1}([0,1])$,
(iv) there exists a single pole of $f(x)$, i.\,e. a root of the equation $f(x)=\infty$, at each face, the multiplicity of the pole being equal to the degree of the face, and, finally,
(v) besides $0$, $1$ and $\infty$, there are no other critical values of $f$ \cite{Zvonkin}.
It can be shown that to every $\mathcal{D}$ there corresponds a Belyi function $f(x)$ and that this function is, up to a linear fractional transformation of the variable $x$, unique. It is, however, a highly non-trivial task to find/calculate the Belyi function for a general {\it dessin}.
\subsection*{Finite geometries from dessins d'enfants}
An issue of central importance for us is the fact that one can associate with a dessin $\mathcal{D}$ a {\it point-line incidence geometry}, $G_{\mathcal{D}}$, in the following way. A point of $G_{\mathcal{D}}$ corresponds to an edge of $\mathcal{D}$. Given a dessin $\mathcal{D}$, we want its permutation group $P$ to preserve the collineation of the geometry $G_{\mathcal{D}}$. We first ask that every pair of points on a line shares the same stabilizer in $P$. Then, given a subgroup $S$ of $P$ which stabilizes a pair of points, we define the point-line relation on $G_{\mathcal{D}}$ such that two points on the same line share the same stabilizer. The lines on a geometry are distinguished by their (isomorphic) stabilizers acting on different $G$-sets. This construction allows to assign finite geometries $G_{\mathcal{D}^i}$ to a dessin $\mathcal{D}$, $i=1,\cdots,m$ with $m$ being the number of non-isomorphic subgroups $S$ of $P$ that stabilize a pair of elements \footnote{Our definition follows an example of the action on the Fano plane of a permutation group of order $|PSL(2,7)|=168$ associated with a tree-like {\it dessin} (of the relevant cycle structure) given in \cite[(a), vol. 2, p. 17 and p. 50]{Schneps1}.}. As a slight digression we mention that, presumably, this action of the group $P$ of a dessin $\mathcal{D}$ on the associated geometry $G_{\mathcal{D}}$ is intricately linked with the properties of the absolute Galois group $\Gamma=\mbox{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$, which is the group of automorphisms of the field $\bar{\mathbb{Q}}$ of algebraic numbers. Although $\Gamma$ is known to act faithfully on $\mathcal{D}$ \cite{Schneps1,Girondo},
its action on $G_{\mathcal{D}}$ must be rather non-trivial because the map from $\mathcal{D}$ to $G_{\mathcal{D}}$ is non injective. Further work is necessary along this line of thoughts to clarify the issue.
Using a computer program, we have been able to completely catalogize incidence geometries associated with {\it dessins} featuring up to 12 edges, and also found several {\it dessins} of higher rank that produce distinguished geometries. The results of our calculations are succinctly summarized in Tables 1 and 2.
The subsequent sections provide a detailed account of a variety of {\it dessins} computed, their corresponding point-line incidence geometries, and, what is perhaps most important, how these relate to the physics of quantum observables of multiple-qubit Pauli groups and related quantum paradoxes. In other words, we shall give a more exhaustive and rigorous elaboration of the ideas first outlined in a short essay-like treatise \cite{PlanatFQXi}.
\bigskip
\begin{table}[ht]
\begin{center}
\begin{tabular}{|r|r|r|r|r|r|}
\hline
index & name & vertices & edges & triangles & squares \\
\hline
$3$ & $2$-simplex (triangle) & $3$ & $3$ & $1$ & $0$ \\
\hline
$4$ & $3$-simplex (tethahedron) & $4$ & $6$ & $4$ & $0$ \\
$$ & square/quadrangle & $4$ & $4$ & $0$ & $1$ \\
\hline
$5$ & $4$-simplex ($5$-cell) & $5$ & $10$ & $10$ & $0$ \\
\hline
$6$ & $5$-simplex & $6$ & $15$ & $20$ & $0$ \\
$$ & $3$-orthoplex (octahedron) & $6$ & $12$ & $8$ & $3$ \\
$$ & bipartite graph $K(3,3)$ & $6$ & $9$ & $0$ & $9$ \\
\hline
$7$ & $6$-simplex & $7$ & $21$ & $35$ & $0$ \\
$$ & Fano plane ($7_3$) & $7$ & $21$ & $7$ & $0$ \\
\hline
$8$ & $7$-simplex & $8$ & $28$ & $56$ & $0$ \\
$$ & $4$-orthoplex ($16$-cell) & $8$ & $24$ & $32$ & $6$ \\
$$ & completed cube $K(4,4)$ & $8$ & $16$ & $0$ & $36$ \\
$$ & stellated octahedron & $8$ & $12$ & $8$ & $0$ \\
\hline
$9$ & $8$-simplex & $9$ & $36$ & $84$ & $0$ \\
$$ & Hesse ($9_412_3$) & $9$ & $36$ & $12$ & $0$ \\
$$ & $K(3)^3$ & $9$ & $27$ & $27$ & $27$ \\
$$ & Pappus ($9_3$) & $9$ & $27$ & $9$ & $27$ \\
$$ & ($3 \times 3$)-grid & $9$ & $18$ & $6$ & $9$ \\
\hline
$10$ & $9$-simplex & $10$ & $45$ & $120$ & $0$ \\
$$ & $5$-orthoplex & $10$ & $40$ & $80$ & $10$ \\
$$ & bipartite graph $K(5,5)$ & $10$ & $25$ & $0$ & $100$ \\
$$ & Mermin's pentagram & $10$ & $30$ & $30$ & $15$ \\
$$ & Petersen graph & $10$ & $15$ & $0$ & $0$ \\
$$ & Desargues ($10_3$) & $10$ & $30$ & $10$ & $15$ \\
\hline
$11$ & $10$-simplex & $11$ & $55$ & $165$ & $0$ \\
\hline
\hline
$12$ & $11$-simplex & $12$ & $66$ & $220$ & $0$ \\
$$ & $6$-orthoplex & $12$ & $60$ & $160$ & $15$ \\
$$ & bipartite graph $K(6,6)$ & $12$ & $36$ & $0$ & $255$ \\
$$ & threepartite graph $K(4,4,4)$ & $12$ & $48$ & $64$ & $108$ \\
$$ & fourpartite graph $K(3,3,3,3)$ & $12$ & $54$ & $0$ & $54$ \\
\hline
\end{tabular}
\label{geometries}
\normalsize
\caption{A catalog of connected point-line incidence geometries induced by {\it dessins d'enfants} of small index $\le 12$. For each geometry, when represented by its collinearity graph, we list the number of points, edges (line-segments joining two points), triangles and squares it contains. Here, $A$-simplices should be regarded as trivial because their {\it dessins} are star-like and associated Belyi functions are of a simple form $f(x)=x^A$, where $A$ is the multiplicity of the singular point at $x=0$.}
\end{center}
\end{table}
\normalsize
\begin{table}[ht]
\begin{center}
\begin{tabular}{|r|r|r|r|r|r|}
\hline
index & name & vertices & edges & triangles & squares \\
\hline
$15$ & Cremona-Richmond ($15_3$) (alias GQ(2,\,2)) & $15$ & $45$ & $15$ & $90$ \\
\hline
$16$ & Clebsch graph CG: $sp(10^1,2^5,-2^{10})$ & $16$ & $80$ & $0$ & $60$ \\
$$ & Shrikhande graph SG: $sp(6^1,2^6,-2^9)$ & $16$ & $48$ & $32$ & $12$ \\
\hline
$18$ & $sp(8^1,0^9,-4^4,2^4)$& $18$ & $72$ & $48$ & $306$ \\
\hline
$20$ & $sp(0^{10},-8^1,8^1,-2^4,2^4)$& $20$ & $80$ & $0$ & $740$ \\
\hline
$21$ & Kneser graph KG$_{(7,2)}$: $sp(10^1,3^6,-2^{14})$ & $21$ & $105$ & $35$ & $630$ \\
$$ & $\mathcal{L}(IG(7,3,1)):~~~~ sp(4^1,-2^8,(1 \pm \sqrt{2})^6)$ & $21$ & $42$ & $14$ & $0$ \\
\hline
$22$ & $IG(11,5,2)$:~~$sp(\pm 5^1,\pm \sqrt{3}^{10})$ & $22$ & $55$ & $0$ & $55$ \\
\hline
$27$ & GQ(2,\,4): $sp(10^1,1^{20},-5^6)$ & $27$ & $135$ & $45$ & $1080$ \\
$$ & Schl\"{a}fli graph SHG: $sp(16^1,4^6,-2^{20})$ & $27$ & $216$ & $720$ & $270$ \\
$$ & $sp(16^1,1^{16},-2^8,-8^2)$ & $27$ & $216$ & $504$ & $3024$ \\
$$ & $sp(16^1,4^2,1^{12},-2^8,-5^4)$ & $27$ & $216$ & $612$ & $1674$ \\
\hline
\end{tabular}
\label{geometries2}
\normalsize
\caption{A few (non-trivial) connected point-line incidence geometries induced by {\it dessins d'enfants} of index greater than $12$. The spectra of (collinearity) graphs, denoted as $sp(\cdots, a^b,\cdots)$ where an eigenvalue $a$ is of multiplicity $b$, are also displayed. $IG(\nu,k,\lambda)$ means the incidence graph of a symmetric $2-(\nu,k,\lambda)$ design and $\mathcal{L}(\cdots)$ stands for the line graph.}
\end{center}
\end{table}
\normalsize
\section{The square geometry of Bell's theorem and the corresponding dessins}
\label{square}
{\it In a theory in which parameters are added to quantum mechanics to determine the results of individual measurements, without changing the statistical predictions, there must be a mechanism whereby the setting of one measuring device can influence the reading of another instrument, however remote \cite{Bell1964}}.
\subsection*{The square geometry of Bell's theorem}
\begin{figure}
\centering
\includegraphics[width=7cm]{square.eps}
\caption{A simple observable proof of Bell's theorem is embodied in the geometry of a (properly labeled) square (a) and four associated {\it dessins d'enfants}, ($b_1$) to ($b_4$).
For each {\it dessin} an explicit labeling of its edges in terms of the four two-qubit observables is given. The (real-valued) coordinates of black and white vertices stem from the corresponding Belyi functions as explained in the main text.}
\label{fig1}
\end{figure}
Suppose we have four observables $\sigma_i$, $i = 1,2,3,4$, taking values in $\{-1,1\}$, of which Bob can measure $(\sigma_1,\sigma_3)$ and Alice $(\sigma_2,\sigma_4)$. The Bell-CHSH approach to quantum contextuality/non-locality consists of defining the number
$$C=\sigma_2(\sigma_1+\sigma_3)+\sigma_4(\sigma_3-\sigma_1)=\pm 2$$
and observing the (so-called Bell-CHSH) inequality \cite[p. 164]{Peres}
$$|\left\langle \sigma_1\sigma_2 \right\rangle+\left\langle \sigma_2\sigma_3 \right\rangle+\left\langle \sigma_3\sigma_4 \right\rangle-\left\langle \sigma_4\sigma_1 \right\rangle|\le 2,$$
where $\left\langle \right\rangle$ here means that we are taking averages over many experiments. This inequality holds for any dichotomic random variables $\sigma_i$ that are governed by a joint probability distribution. Bell's theorem states that the aforementioned inequality is violated if one considers quantum observables with dichotomic eigenvalues. An illustrative example is the following set of two-qubit observables
\begin{equation}
\sigma_1=IX,~\sigma_2=XI,~\sigma_3=IZ,~\sigma_4=ZI.
\label{quadruple}
\end{equation}
Here and below we use the notation $X$, $Y$ and $Z$ for the ordinary Pauli spin matrices and, e.\,g., $IX$ is a short-hand for $I \otimes X$ (used also in the sequel).
We find that
$$C^2=4*I +[\sigma_1,\sigma_3][\sigma_2,\sigma_4]=4 \left(\begin{array}{cccc} 1 &. & . &1 \\ . &1 & \bar{1} &. \\ . &\bar{1} & 1&. \\ 1 &. & . &1 \end{array}\right)$$
has eigenvalues $0$ and $8$, both with multiplicity $2$ ($\bar{1} \equiv -1$). Taking the norm of the bounded linear operator $A$ as $||A||=sup (||A \psi||/||\psi||),~\psi \in \mathcal{H}$ ($\mathcal{H}$ being the corresponding Hilbert space), one arrives at the {\it maximal} violation of the Bell-CHSH inequality \cite[p. 174]{Peres}, namely
$||C||=2\sqrt{2}.$
The point-line incidence geometry associated with our four observables is one of the simplest, that of a square -- Fig.\ref{fig1}a; each observable is represented by a point and two points are joined by a segment if the corresponding observables commute. It is worth mentioning here that there are altogether 90 distinct squares among two-qubit observables and as many as
$30240$ when three-qubit labeling is employed \cite{Planat2013}, each yielding a maximal violation of the Bell-CHSH inequality.
\subsection*{Dessins d'enfants for the square and their Belyi functions}
As it is depicted in Fig. \ref{fig1}, the geometry of square can be generated by four different {\it dessins}, $b_1,\cdots, b_4$, associated with permutations groups $P$ isomorphic to the dihedral group $D_4$ of order $8$.
The first {\it dessin} ($b_1$) has the signature $s=(B,W,F,g)=(3,2,1,0)$ and the permutation group $P=\left\langle (2,3),(1,2)(3,4)\right\rangle$ whose cycle structure reads $[2^1 1^2,2^2,4^1]$, i.e. one black vertex is of degree two, two black vertices have degree one, the two white vertices have degree two and the face has degree four. The corresponding Belyi function reads $f(x)=x^2(2-x^2)$. Its critical points are $x \in \{-1, 1, 0\}$ and the corresponding critical values are $\{1,1,0\}$. The preimage of the value $0$ (the solutions of the equation $f(x)=0$) corresponds to the black vertices of the {\it dessin} positioned at $x \in \{-\sqrt{2},\sqrt{2},0\}$ and the preimage of the value $1$ (the solutions of the equation $f(x)=1$) corresponds to the white vertices at $x=\pm 1$.
The second {\it dessin} $(b_2)$ has $s=(2,3,1,0)$, $P=\left\langle (1,2)(3,4),(2,3)\right\rangle$ with $[2^2,2^1 1^2,4^1]$, and the Belyi function of the form $f(x)=(x^2-1)^2$.
The third {\it dessin} $(b_3)$ is characterized by $s=(1,2,3,0)$ and its $P=\left\langle (1,2,4,3),(1,2)(3,4)\right\rangle$ has the cycle structure $[4^1,2^2,2^11^2]$. The Belyi function may be written as
$f(x)=\frac{(x-1)^4}{4x(x-2)}$. As $f'(x)=\frac{(x-1)^3(x^2-2x-1)}{2(x-2)^2x^2}$, its critical points lie at $x=1$ (where $f(1)=0$) and at $x=1\pm \sqrt{2}$ (where $f(1\pm \sqrt{2})=1$).
Finally, the fourth {\it dessin} $(b_4)$ has $P=\left\langle (1,2,4,3),(2,3)\right\rangle$, the signature $s=(1,3,2,0)$ and cycle structure $[4^1,2^11^2,2^2]$. The Belyi function reads $f(x)=\frac{(x-1)^4}{16 x^2}$; hence, $f'(x)=\frac{(x-1)^3(x+1)}{8x^3}$. The critical points are at $x=-1$ (with critical value $1$) and $x=1$ (with critical value $0$), the preimage of $0$ is the black vertex at $x=1$ and the preimage of $1$ consists of the white vertices at $x \in \{-1,3 \pm \sqrt{8}\}$.
Summing up, one of the simplest observable proofs of Bell's theorem is found to rely on the geometry of a {\it square} and {\it four} distinct {\it dessins} associated with it. Although we still do not know
how these {\it dessins} are related to each other, it is quite intriguing to see that all critical points live in the extension field $\mathbb{Q}(\sqrt{2})\in \bar{\mathbb{Q}}$ of the rational field $\mathbb{Q}$. Hence, a better understanding of the properties of the group of automorphisms of this field (which is itself a subgroup of the absolute Galois group $\Gamma$) may lead to fresh insights into the nature of this important theorem of quantum physics.
\section{A wealth of other notable point-line geometries relevant to contextuality}
\label{catalog}
{\it It is also appealing to see the failure of the EPR reality criterion emerge quite directly from the one crucial difference between the elements of reality (which, being ordinary numbers, necessarily commute) and the precisely corresponding quantum mechanical observables (which sometimes anti-commute) \cite[(a)]{Mermin1993}}.
\vspace*{.2cm}
\begin{figure}
\centering
\includegraphics[width=5cm]{octahedron.eps}
\caption{The octahedron with vertices labeled by three-qubit observables (a) and an associated {\it dessin} (b).}
\label{fig2}
\end{figure}
\subsection{Two geometries of index six: the octahedron and the bipartite graph $K(3,3)$}
As the geometries of index five are only trivial simplices (see Table 1), we have to move to index six in order to encounter non-trivial guys, namely the octahedron and the bipartite graph of type $K(3,3)$.
\begin{figure}
\centering
\includegraphics[width=5cm]{K33.eps}
\caption{The bipartite graph $K(3,3)$ with one of its two-qubit labelings (a) and its generating {\it dessin} (b).}
\label{fig3}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=5cm]{Fano.eps}
\caption{The Fano plane portrayed in its most frequent rendering (a) and one of its ten stabilizing {\it dessins} (b).}
\label{fig4}
\end{figure}
The octahedron can be labeled by three-qubit observables, and one such labeling is given in Fig. \ref{fig2}a. The figure also illustrates one of the associated {\it dessins} (b), whose Belyi function of is $f(x)=\frac{27}{32}x^2(2-x^2)^2$. The function has critical points at $x=0$ and $x=\pm\sqrt{2}$ (these being also the preimage of $0$), and the white vertices of the {\it dessin} correspond to $x=\pm \sqrt{2/3}$ and $x=\pm \sqrt{8/3}$ (the preimage of $1$).
A remarkable property of the graph $K(3,3)$ is that it lives in the generalized quadrangle GQ(2,\,2), disguised there as a generalized quadrangle of type GQ(1,\,2) \cite{psm}. And since GQ(2,\,2) was found to mimic the commutation relations between elements of the two-qubit Pauli group \cite{Planat2007}, $K(3,3)$ thus naturally lends itself, like the above-discussed square, to a labeling in terms of two-qubit observables --- as, for example, depicted in Fig. \ref{fig3}a. One of the associated {\it dessins} (Fig. \ref{fig3}b) possesses the Belyi function of the form $f(x)=ax^4(x-1)^2$, where $a=\frac{3^6}{2^4}=\frac{729}{16}$. Since $f'(x)=ax^3(x-1)(3x-2)$ the critical points are at $x=2/3$, $x=0$ and $x=1$. The black vertices of the {\it dessin} correspond to $x=0$ and $x=1$; out of its five white vertices three answer to real-valued variable, namely $x=-\frac{1}{3}$, $x=\frac{2}{3}$ and $x \approx 1.118$ (the latter being denoted as $x_1$ in Fig. \ref{fig3}b), and the remaining two --- denoted as $x_2$ and $x_3$ in the figure in question --- have imaginary, complex-conjugate one: $x\approx 0.36 \exp{(\pm i \phi)}$, with $\phi \approx 99.4^{\circ}$.
\subsection{The Fano plane (everywhere)}
The only non-trivial geometry of index seven is the projective plane of order two, the Fano plane (Fig.\,\ref{fig4}a). This plane plays a very prominent role in finite-dimensional quantum mechanics, being, for example, intricately related --- through the properties of the split Cayley hexagon of order two \cite{psm} --- to the structure of the three-qubit Pauli group \cite{Levay2008}.
A quick computer search for all permutation subgroups of $C_2^+$ isomorphic to the group $PSL(2,7)$, the automorphism group of the Fano plane, shows that this plane can be recovered from $10$ distinct {\it dessins}. One choice is depicted schematically in Fig.\,\ref{fig4}b; it corresponds to passport $8$ (the fourth {\it dessin}) in the catalog of B\'etr\'ema $\&$ Zvonkin \cite{ZvonkinCatalog}. The corresponding permutation group is $P=\left\langle (2,7,6,5)(3,4),(1,2)(3,5)\right\rangle$ and the Belyi function reads $f(x)=\sqrt{8}x^4(x-1)^2(x-a)$, with $a=-\frac{1}{4}(1+i\sqrt{7})$; its critical points are located at $x=0$ and $x=1$ (yielding critical value $0$) and at $x=a$ (yielding $1$).
\subsection{The $16$-cell, stellated octahedron and the completed cube}
\begin{figure}
\centering
\includegraphics[width=6cm]{16-cell.eps}
\caption{The $16$-cell (a) and an associated {\it dessin} (b).}
\label{fig5}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=6cm]{stellated.eps}
\caption{The stellated octahedron (a), the completed cube (b) and their common stabilizing {\it dessin} (c).}
\label{fig6}
\end{figure}
When moving to index eight, we encounter an appealing 16-cell on the one hand, and the remarkably ``twinned'' stellated octahedron and completed cube on the other hand.
The $16$-cell (Fig.\,\ref{fig5}a) arises from a ``straight-line'' {\it dessin} with the signature $(5,4,1,0)$ and the permutation group isomorphic to $D_8$. Its Belyi function is the fourth order of the map $x \rightarrow x^2-2$, that is $f(x)=8x(x^2-2)(x^4-4x^2+2)$, with critical points located at $x-0$, $x=\pm \sqrt{2}$ and $x=\pm \sqrt{2 \pm \sqrt{2}}$.
The {\it dessin} with signature $s=(2,6,2,0)$ , illustrated in Fig.\,\ref{fig6}c, has the permutation group $P=\left\langle (1,2,4,3)(5,7,6,8),(2,5)(3,7)\right\rangle$, which is isomorphic to $\mathbb{Z}_2^3 \rtimes \mathbb{Z}_2$ and endowed with the cycle structure of the form $[4^2,2^2 1^4,4^2]$. The stabilizer of a pair of its edges is either the group $\mathbb{Z}_2$, leading to the geometry of a stellated octahedron (Fig.\,\ref{fig6}a), or the trivial single element group $\mathbb{Z}_1$, in which case we get the geometry of a (triangle free) {\it completed cube}, i.\,e. the ordinary cube where pairs of opposite points are joined (Fig.\,\ref{fig6}b). The latter configuration also appears in a recent paper \cite[pp. 33--34 as well as Conjecture 6.1]{Grunbaum} as an $8$-face Kepler-Poinsot quadrangulation of the torus. Note that, in addition the $6$ faces shared with the ordinary cube, the completed cube contains also $8$ non-planar faces of which four are self-intersecting. The completed cube can also be viewed as the bipartite graph $K(4,4)$. The Belyi function of the {\it dessin} has the form
$$f(x)=K\frac{(x-1)^4(x-a)^4}{x^3},~~a=\frac{8\sqrt{10}-37}{27},~~K \approx-0.4082,$$
from where we find the positions of ``critical'' white vertices to be $x \approx 0.0566 \pm 0.506 i$, with the other four white vertices being located at $x=-1.069$, $x=-0.162$ and $x=1.634 \pm 0.6109i$.
\subsection{Geometries of index nine: grid (Mermin's square), Pappus and Hesse}
In the realm of index nine we meet, in addition to our old friend, a $3 \times 3$-grid (alias generalized quadrangle GQ(2,\,1)), also other distinguished finite geometries like the Pappus $9_3$-configurations and the Hesse $9_{4}12_{3}$-configuration ({\it aka} the affine plane of order three, AG(2,\,3)).
\begin{figure}[ht]
\centering
\includegraphics[width=6cm]{MerminTorus.eps}
\caption{A $3\times 3$ grid with points labeled by two-qubit observables ({\it aka} a Mermin magic square) (a) and a stabilizing {\it dessin} drawn on a torus (b).}
\label{fig7}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=6cm]{Hesse.eps}
\caption{The Hesse configuration (a) and an associated genus-zero {\it dessin} (b).}
\label{figHesse}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=6cm]{Pappus2.eps}
\caption{The Pappus configuration (a) and a stabilizing {\it dessin} (b).}
\label{fig8}
\end{figure}
As already mentioned, the grid lives (as a geometric hyperplane) in GQ(2,\,2) and underlies a Mermin ``magic'' square array of observables furnishing a simple two-qubit proof of the Kochen-Specker theorem \cite{PlanatFQXi,Planat2013bis}. A Mermin square built around Bell's square of Fig. \ref{fig1}a is shown in Fig. \ref{fig7}a. It needs a genus one {\it dessin}, with signature $(2,5,2,1)$, to be recovered, as shown in Fig. \ref{fig7}b. The corresponding permutation group is
$P=\left\langle (1,2,4,8,7,3)(5,9,6),(2,5)(3,6)(4,7)(8,9)\right\rangle \cong \mathbb{Z}_3^2 \rtimes \mathbb{Z}_2^2$, having the cycle structure $[6^13^1,2^41^1,6^13^1]$.
This {\it dessin} lies on a Riemann surface that is a torus (not a sphere $\hat{\mathbb{C}}$), being thus represented by an elliptic curve. The topic is far more advanced and we shall not pursue it in this paper (see, e.\,g., \cite{Girondo} for details).
The stabilizer of a pair of edges of the {\it dessin} is either the group $\mathbb{Z}_2$, yielding Mermin's square $M_1$ shown in Fig \ref{fig7}a, or the group $\mathbb{Z}_1$, giving rise to a different square $M_2$ from the maximum sets of mutually non-collinear pairs of points of $M_1$. The union of $M_1$ and $M_2$ is nothing but the Hesse configuration.
The Hesse configuration (Fig.\,\ref{figHesse}a), of its own, can be obtained from a genus-zero {\it dessin} shown in Fig.\,\ref{figHesse}b (also reproduced in Fig.\,3b of \cite{PlanatFQXi}). This configuration was already noticed to be of importance in the derivation of a Kochen-Specker inequality in \cite{Bengtsson}.
The Pappus configuration, illustrated in Fig.\,\ref{fig8}a, comprises three
copies of the already discussed $K(3,3)$-configuration (Fig.\,\ref{fig3}a); the three copies are represented by the point-sets $\{ 1,3,5,6,7,8\}$, $\{ 2,3,4,5,8,9\}$ and $\{1,2,4,6,7,9\}$, which pairwise overlap in distinct triples of points. A {\it dessin d'enfants} for the Pappus configuration is exhibited in Fig.\,\ref{fig8}b.
It is important to point out here a well-known fact that the Pappus configuration is obtained from the Hesse one by removing three mutually skew lines from it (for example, the three lines that are represented in Fig.\,\ref{figHesse}a by three concentric circles).
\subsection{Realm of index ten: Mermin's pentagram, Petersen and Desargues}
Apart from the two plexes (see Table 1), the only connected configurations generated by $10$-edge {\it dessins} are Mermin's pentagram, the Petersen graph, the Desargues configuration and the bipartite graph $K(5,5)$.
The {\it dessin} sketched in Fig.\,\ref{fig9}c, having $s=(4,6,2,0)$ and the alternating group $A_5$ with cycle structure $[3^21^1,2^41^2,5^2]$, induces either the geometry of Mermin's pentagram (Fig.\,\ref{fig9}a) or that of the Petersen graph (Fig.\,\ref{fig9}b) according as the group stabilizing pairs of its edges is isomorphic to $\mathbb{Z}_1$ or $\mathbb{Z}_2$, respectively. A particular three-qubit realization leading to a proof of the Kochen-Specker theorem is explicitly shown (see, e.\,g., \cite{Planat2013, Planat2013bis} for more details on importance of these geometries in quantum theory).
The {\it dessin} depicted in Fig.\,\ref{fig10}b also gives rise to a couple of geometries, one being again the Petersen graph (with the stabilizer group of a pair of edges isomorphic to $\mathbb{Z}_2^2$) and the other being (Fig.\,\ref{fig10}a) the famous Desargues $10_3$ configuration (with the group $\mathbb{Z}_2$). The labeling is compatible with that in Fig.\,\ref{fig9}, which means that the Desargues configuration represents another way of encoding a three-qubit proof of contextuality; in particular, a line of Mermin's pentagram corresponds to a complete graph $K_4$ within the Desargues configuration as well as to a maximum set of mutually disjoint vertices in the Petersen graph.
\begin{figure}[ht]
\centering
\includegraphics[width=7cm]{pentagram.eps}
\caption{The Mermin pentagram (a), the Petersen graph (b) and their generating {\it dessin} (c).}
\label{fig9}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=7cm]{Desargues.eps}
\caption{The Desargues configuration (a) and its generating dessin (b).}
\label{fig10}
\end{figure}
\begin{figure}[ht]
\centering
\includegraphics[width=8cm]{doily.eps}
\caption{The Cremona-Richmond $15_3$-configuration (a) with its points labeled by the elements of the two-qubit Pauli group and a stabilizing {\it dessin} (b).}
\label{fig11}
\end{figure}
\subsection{The Cremona-Richmond $15_3$-configuration, alias GQ(2,\,2), or W(3,\,2)}
We now come to a perhaps most exciting, and encouraging as well, finding that
there exists a {\it dessin} generating the configuration of a central importance for any quantum physical reasoning involving two-qubit observables, namely the configuration (illustrated in Fig.\,\ref{fig11}a) known as $15_3$ Cremona-Richmond configuration, or the generalized quadrangle of order two, GQ(2,\,2), or the symplectic polar space of rank two and order two, W(3,\,2). That this configuration is indeed one of corner-stones of finite dimensional quantum mechanics is also illustrated by the fact that many of the already discussed geometries, in particular the $K(3,3)$ graph, the $3 \times 3$ grid, the Pappus and Desargues configurations and the Petersen graph, are intricately tied to its structure, as explained in detail in \cite{Planat2007},\cite{spph2007}--\cite{hos2009}. The associated {\it dessin} (Fig.\,\ref{fig11}b) is of signature $(5,9,3,0)$ and its permutation group has the cycle structure $[6^13^22^11^1,2^61^3,6^23^1]$. Unfortunately, the complexity of this {\it dessin} is already so high that with our current computer power we have not been able to find the corresponding Belyi function. Finding this function thus remains one important challenge of our {\it dessin d'enfants} programme.
\subsection{The generalized quadrangle GQ(2,\,4), the Schl\"{a}fli graph and the Clebsch graph}
As already mentioned, the generalized quadrangle of type GQ(2,\,4) is a prominent finite geometry in the context of the so-called black-hole--qubit correspondence \cite{Levay2009}, as it completely underlies the $E_6$-symmetric entropy formula describing black holes and black strings in $D = 5$. We were thus very pleased to find a {\it dessin} that leads to the collinearity graphs of this geometry, and to its complement -- the famous Schl\"{a}fli graph -- as well (Fig.\,\ref{fig12}). Moreover, GQ(2\,,4) is also notable by the fact it contains (altogether 27) copies of the Clebsch graph, each such copy being the complement of a geometric hyperplane of particular type. And since the Clebsch graph is also a {\it dessin}-generated one (see Table\,2), we will not be surprised if our {\it dessin}'s formalism is also found of relevance for getting conceptual insights into the still-mysterious formal link between stringy black hole entropy formulas and properties of multi-qubits (for a recent review, see \cite{bdl}).
\begin{figure}
\centering
\includegraphics[width=6cm]{GQ24.eps}
\caption{The {\it dessin} for both the collinearity graph of the generalized quadrangle of type GQ(2,\,4) and its complement, the Schl\"{a}fli graph. }
\label{fig12}
\end{figure}
\section{Conclusion}
We have demonstrated, substantially boosting the spirit of \cite{PlanatFQXi},
that Grothendieck's {\it dessins d'enfants} (``child's drawings'') --- that is graphs where at each vertex is given a cyclic ordering of the edges meeting it and each vertex is also assigned one of two colors, conventionally black and white, with the two ends of every edge being colored differently --- and their associated permutation groups/Belyi functions give rise to a wealth of finite geometries relevant for quantum physics.
We have made a complete catalog of these geometries for {\it dessins} featuring up to 12 edges, highlighted distinguished geometries for some higher-index {\it dessins}, and briefly elaborated on quantum physical meaning for each non-trivial geometry encountered. We are astonished to see that a majority of dessin-generated geometries have already been found to have a firm footing in finite-dimensional quantum mechanical setting, like the K(3,\,3) and Petersen graphs, the Fano plane, the $3 \times 3$ grid (Mermin's square), the Desargues configuration, Mermin's pentagram and the generalized quadrangles GQ(2,\,2) and GQ(2,\,4). We have also found a wealth of geometries, among them the Hesse $9_4 12_3$-configuration, the Reye $12_4 16_3$-configuration \cite{Aravind}, the $3 \times 3 \times 3$-grid, the Kneser graph $KG_{(7,2)}$ and many others, that still await their time to enter the game. Our findings may well be pointing out that properties of {\it dessins}, as well as the Galois group $\mathcal{G}=\mbox{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ acting on them, may be vital for getting deeper insights into foundational aspects of quantum mechanics.
To this end in view, we aim to expand in a systematic way our catalog of finite geometries generated by higher-index dessins in order to reveal finer traits of the quantum pattern outlined above.
\section*{Acknowledgments}
This work started while two of the authors (M.\,P. and M.\,S.) were fellows of the ``Research in Pairs'' Program of the Mathematisches Forschungsinstitut Oberwolfach (Oberwolfach, Germany), in the period from 24 February to 16 March, 2013. It was also supported in part by the VEGA Grant Agency, grant No. 2/0003/13.
\section*{Bibliography}
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Fulco Ruffo di Calabria (Olaszország, Nápoly, 1884. augusztus 12. - Olaszország, Ronchi de Apuania, 1946. augusztus 23.) az első világháború 5. legeredményesebb olasz vadászpilótája volt. Szolgálata során 20 igazolt és 5 igazolatlan légi győzelmet aratott. Érdekesség, hogy di Calabria Guardia Lombarda hatodik hercege volt. Di Calabria hatodik, egyben legkisebb gyermeke az 1937-ben született Paula, aki II. Albert belga király feleségeként Belgium királynéja volt.
Élete
Katonai szolgálata
Fulco herceg 1904. november 22-én csatlakozott az olasz hadsereghez. Kiképzése után a 11. olasz könnyűlovasezredhez vezényelték. A lovasságnál közel tíz évet szolgált, ezen idő alatt igencsak tapasztalt művelője lett eme fegyvernemnek. Szolgálata során egy ízben eljutott még Afrikába is. A források szerint erre 1914 körül kerülhetett sor. Visszatérése után részt akart venni az alakuló olasz légierőben, ezért 1914-ben jelentkezett és átképeztette magát pilótává.
A légierőnél igen eredményes karriert futott be, szolgált az 1., a 4., a 42., a 44., a 70. repülő osztagoknál, majd a Squadriglia 91 (91. vadászrepülő osztag) parancsnoka lett. Számos modern repülőgéptípussal többek között Nieuport 11, Nieuport 17 és SPAD VII-as repülőgépekkel. Győzelmeit 1916 augusztusától 1918 júniusáig halmozta, eredményességéhez több ízben gratuláltak magas rangú vezetők, emellett pedig meg kapta az ászpilótáknak járó Vitézségi Érmet is.
Katonai pályája csúcsán kapitánnyá léptették elő, ezzel együtt pedig (Francesco Baracca halála után) a 91. vadászrepülő osztag parancsnoka lett. Ezen teendőit viszont csupán néhány hétig látta el, mivel ideg összeroppanást kapott. Helyettese ezek után a szintén híres ászpilóta, Ferruccio Ranza lett. 1918 októberében végül még visszatért de egyik bevetésén légvédelmi tűz miatt kényszerleszállást kellett végrehajtania. Sérülései, illetve a háború befejeződése miatt már nem tért vissza a frontra. Fulco Ruffo di Calabria a szolgálata során 53 légi bevetést teljesített és 20 igazolt légi győzelmet szerzett.
Légi győzelmei
További élete
További életéről nem szól forrás, csupán annyit tudunk, hogy 1946-ban hunyt el, 62 éves korában.
Családja
Fulco herceg 1919. június 30-án házasodott össze Luisa Gazellivel (1896-1989) házasságukból 7 gyermek született.
Maria Cristina Ruffo di Calabria (1920–†2003)
Laura Ruffo di Calabria (1921–†1972)
Ruffo di Calabria-Santapau Fabrizio (1922-†2005)
Augusto Ruffo di Calabria (1925–1943)
Giovannella Ruffo di Calabria (1927–†1941)
Antonello Ruffo di Calabria (1930- )
Paola Ruffo di Calabria (1937- ) a későbbi Paula belga királyné
Kapcsolódó szócikkek
Squadriglia 91
Paula belga királyné
Források
Olasz első világháborús ászpilóták
1884-ben született személyek
1946-ban elhunyt személyek
Nápolyiak
Calabriaiak
|
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{"url":"https:\/\/fispact.ukaea.uk\/wiki\/Keyword:TAB1","text":"# Keyword:TAB1\n\nTAB1 ia\n\nThis keyword causes the inventory data in columns 1 and 2, the number of atoms and grams of each nuclide, to be written to an external file (TAB1). Note that the stream number ia is now ignored. Both NOT1 and TAB1 may be used several times during a run to restrict and restore the output as required.\n\nExample usage, where TAB1 output to the <fileroot>.tab1 file is turned on and off over several steps:\n\n< -- Control phase -- >\n...\nFISPACT\n* Title of the simulation\n< -- Initial phase -- >\n...\nTAB1\nFLUX 1.0E10\nATOMS\n< -- Inventory phase -- >\nTIME 1.0 HOURS ATOMS\nNOT1\nTIME 1.0 HOURS ATOMS\nTAB1\nTIME 1.0 HOURS ATOMS\nNOT1\nTIME 1.0 HOURS ATOMS\n...","date":"2020-01-25 21:47:44","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5816060900688171, \"perplexity\": 6602.070949530484}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-05\/segments\/1579251681412.74\/warc\/CC-MAIN-20200125191854-20200125221854-00454.warc.gz\"}"}
| null | null |
Law and the Rise of Capitalism
by Michael E. Tigar
Released: December 2000
Against a backdrop of seven hundred years of bourgeois struggle, eminent lawyer and educator, Michael E. Tigar, develops a Marxist theory of law and jurisprudence based upon the Western experience. This well-researched and documented study traces the role of law and lawyers in the European bourgeoisie's conquest of power-the first such history in the English language-and in the process, contradicts the analyses of such major figures as R.H. Tawney and Max Weber. Using a wide range of primary sources, Tigar demonstrates that the legal theory of the insurgent bourgeoisie predated the Protestant Reformation and was a major ideological ingredient of the bourgeois revolution.
Originally published in 1977, Law and the Rise of Capitalism has been translated into several languages to international acclaim. Tigarµs new introduction and extended Afterword discuss the struggle for human rights over the past two decades and shed light on the challenges facing today's social movements. Tigar draws on his own experiences as a fighter for democratic rights in the United States, Europe and South Africa, while adding new historical insights to human rights issues in the United States including the plight of political prisoners and the death penalty.
A thought-provoking interpretation of the role of legal ideology in the bourgeoisie's ascendance to state power.
Provides a realistic basis for understanding our history and the role of the law in the United States.
—University of Pennsylvania Law Review
This pioneering book asks brave questions. . . . Tigar has performed a valuable service in opening up for discussion this area of social and intellectual history.
—Christopher Hill, The New Statesman
Michael E. Tigar is Edwin A. Mooers Scholar and Professor of Law at Washington College of Law, American University. Until 1998, he held the Joseph D. Jamail Chair in Law at the University of Texas School of Law. Mr. Tigar has argued appeals in almost every U.S. Court of Appeals and in the U.S. Supreme Court. His individual clients have included Angela Davis, H. Rap Brown, the Seattle Seven, the Chicago Eight, Fernando Chavez, Rosalio Munoz, Major Debra Meeks, Allen Ginsberg, and Francisco Martinez. Tigar recently represented Terry Lynn Nichols in the Oklahoma City bombing case.
Format Clear
Format Choose an optionPaperback Clear
Law and the Rise of Capitalism quantity
SKU: mrp0300 Categories: Books, Paperback
Topics: History Philosophy Places: Global
Paperback ISBN
Cloth ISBN
eBook ISBN
Mexico's Revolution Then and Now
Debt, the IMF, and the World Bank: Sixty Questions, Sixty Answers
The Structural Crisis of Capital
A History of World Agriculture: From the Neolithic Age to the Current Crisis
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
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| 4,716
|
typedef struct _GMainContext GMainContext;
typedef struct _GPollFD GPollFD;
typedef struct _GSource GSource;
namespace base {
// This class implements a base MessagePump needed for TYPE_UI MessageLoops on
// platforms using GLib.
class BASE_EXPORT MessagePumpGlib : public MessagePump,
public WatchableIOMessagePumpPosix {
public:
class FdWatchController : public FdWatchControllerInterface {
public:
explicit FdWatchController(const Location& from_here);
FdWatchController(const FdWatchController&) = delete;
FdWatchController& operator=(const FdWatchController&) = delete;
~FdWatchController() override;
// FdWatchControllerInterface:
bool StopWatchingFileDescriptor() override;
private:
friend class MessagePumpGlib;
friend class MessagePumpGLibFdWatchTest;
// FdWatchController instances can be reused (unless fd changes), so we
// need to keep track of initialization status and taking it into account
// when setting up a fd watching. Please refer to
// WatchableIOMessagePumpPosix docs for more details. This is called by
// WatchFileDescriptor() and sets up a GSource for the input parameters.
// The source is not attached here, so the events will not be fired until
// Attach() is called.
bool InitOrUpdate(int fd, int mode, FdWatcher* watcher);
// Returns the current initialization status.
bool IsInitialized() const;
// Tries to attach the internal GSource instance to the |pump|'s
// GMainContext, so IO events start to be dispatched. Returns false if
// |this| is not correctly initialized, otherwise returns true.
bool Attach(MessagePumpGlib* pump);
// Forward read and write events to |watcher_|. It is a no-op if watcher_
// is null, which can happen when controller is suddenly stopped through
// StopWatchingFileDescriptor().
void NotifyCanRead();
void NotifyCanWrite();
FdWatcher* watcher_ = nullptr;
GSource* source_ = nullptr;
std::unique_ptr<GPollFD> poll_fd_;
// If this pointer is non-null, the pointee is set to true in the
// destructor.
bool* was_destroyed_ = nullptr;
};
MessagePumpGlib();
MessagePumpGlib(const MessagePumpGlib&) = delete;
MessagePumpGlib& operator=(const MessagePumpGlib&) = delete;
~MessagePumpGlib() override;
// Part of WatchableIOMessagePumpPosix interface.
// Please refer to WatchableIOMessagePumpPosix docs for more details.
bool WatchFileDescriptor(int fd,
bool persistent,
int mode,
FdWatchController* controller,
FdWatcher* delegate);
// Internal methods used for processing the pump callbacks. They are public
// for simplicity but should not be used directly. HandlePrepare is called
// during the prepare step of glib, and returns a timeout that will be passed
// to the poll. HandleCheck is called after the poll has completed, and
// returns whether or not HandleDispatch should be called. HandleDispatch is
// called if HandleCheck returned true.
int HandlePrepare();
bool HandleCheck();
void HandleDispatch();
// Overridden from MessagePump:
void Run(Delegate* delegate) override;
void Quit() override;
void ScheduleWork() override;
void ScheduleDelayedWork(const TimeTicks& delayed_work_time) override;
// Internal methods used for processing the FdWatchSource callbacks. As for
// main pump callbacks, they are public for simplicity but should not be used
// directly.
bool HandleFdWatchCheck(FdWatchController* controller);
void HandleFdWatchDispatch(FdWatchController* controller);
private:
bool ShouldQuit() const;
// We may make recursive calls to Run, so we save state that needs to be
// separate between them in this structure type.
struct RunState;
RunState* state_;
// This is a GLib structure that we can add event sources to. On the main
// thread, we use the default GLib context, which is the one to which all GTK
// events are dispatched.
GMainContext* context_ = nullptr;
bool context_owned_ = false;
// The work source. It is shared by all calls to Run and destroyed when
// the message pump is destroyed.
GSource* work_source_;
// We use a wakeup pipe to make sure we'll get out of the glib polling phase
// when another thread has scheduled us to do some work. There is a glib
// mechanism g_main_context_wakeup, but this won't guarantee that our event's
// Dispatch() will be called.
int wakeup_pipe_read_;
int wakeup_pipe_write_;
// Use a unique_ptr to avoid needing the definition of GPollFD in the header.
std::unique_ptr<GPollFD> wakeup_gpollfd_;
THREAD_CHECKER(watch_fd_caller_checker_);
};
} // namespace base
#endif // BASE_MESSAGE_LOOP_MESSAGE_PUMP_GLIB_H_
|
{
"redpajama_set_name": "RedPajamaGithub"
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| 101
|
Q: python shutil copyfile permission error I am trying to copy an image from one folder to another folder using shutil in python 3.6 windows 10, but am running into permission errors.
I have my source saved as a variable named 'src' which contains
src = "C:/Users/marti/AppData/Roaming/vlc/art/artistalbum/artistname/art.jpg"
and my destination named 'src' which contains
dst = "C:/Users/marti/Desktop/MRL/cover"
my file currently imports these things:
from shutil import copyfile
from sys import exit
import os
import requests
and I am copying the src to the destination using this command:
copyfile(src, dst)
But when I run this program I am given a permission error:
IOError: [Errno 13] Permission denied: 'C:/Users/marti/Desktop/MRL/cover/'
even when I'm running CMD as administrator, does anyone know how to edit these permissions for python?
If not I'm open to any other methods which will allow me to copy an image from one folder to another folder, and eventually check if the src string has changed in which case it will delete the image in the dst folder and replace it
A: From the shutil doc:
Copy the contents (no metadata) of the file named src to a file named dst and return dst. src and dst are path names given as strings. dst must be the complete target file name; look at shutil.copy() for a copy that accepts a target directory path. If src and dst specify the same file, SameFileError is raised.
You should pay attention the dst must be the complete target file name.
A: I have had similar issue when copying a native file to NAS, but easily solved by replacing shutil.copyfile with shutil.copy.
My previous error messages:
with open(dst, 'wb') as fdst:
PermissionError: [Errno 13] Permission denied:
|
{
"redpajama_set_name": "RedPajamaStackExchange"
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| 3,540
|
{"url":"http:\/\/www.ashalathaivf.com\/ab9qr0w5\/sodium-%2B-ethanol-balanced-equation-428150","text":"(b) A piece of sodium metal is added to absolute ethanol to form sodium ethoxide and hydrogen gas. The reaction is exothermic. H 2 SO 4. Answer: Question 3. Treatment with sodium cyanide. Reaction of heating ethanol at 170\u00b0 C in the presence of cone. Important Solutions 2858. When ethanol is warmed (6 0 o C) with iodine in the presence of an alkali (N a O H or N a 2 C O 3 ), iodoform is obtained E t h a n o l C 2 H 5 O H + 4 I 2 + 6 N a O H I o d o f o r m C H I 3 + H C O O N a + 5 N a I + 5 H 2 O (ii) The product formed will be biphenyl and sodium chloride (N a C l) Wurts Fittig reaction The resulting solution is basic because of the dissolved hydroxide. sulphuric acid then alkene is obtained. When soluble sodium argentocyanide is obtained. Preparation of carbon tetrachloride from methane. Preparation of CH 4 from anhydrous sodium ethanoate (sodium acetate). CISCE ICSE Class 10. Ethanol is oxidised by sodium dichromate (Na 2 Cr 2 O 7) acidified in dilute sulphuric acid to form the aldehyde ethanal. The oxidation of the alcohol to an aldehyde is indicated by the colour change of the dichromate solution as it is reduced from the orange colour \u00e2\u0080\u00a6 Reaction with sodium: When ethanol reacts with sodium it gives hydrogen gas and sodium ethoxide as the final product. Write a balanced chemical equation for each of the following reactions and also classify them. Textbook Solutions 25197. Question Papers 301. Question Bank Solutions 24558. CH3CH2OH + H2O ---> H3O+ + CH3CH2O- Ka = 1.3 x 10^-16. Give balanced equations for the following reactions: (i) Aniline is treated with nitrous acid and HCl at low temperature. Write a balanced equation for: Reaction of ethane and oxygen in presence of molybdenum oxide. 10^6 is a million times, 10^9 is a billion times. Use uppercase for the first character in the element and lowercase for the second character. Ag 2 S + 4 NaCN \u00e2\u0087\u008c 2 Na[Ag(CN) 2] + Na2S Na 2 S + 2O 2 \u00e2\u0086\u0092 Na 2 SO 4 _____ Ag 2 S + 4 NaCN + 2O 2 \u00e2\u0086\u0092 2 Na[Ag(CN) 2] + Na 2 SO 4 Precipitation of silver Acetic acid is a weak acid, but it dissociates about 10^11 times more than ethanol. Reactions with sodium. (a) Lead acetate solution is treated with dilute hydrochloric acid to form lead chloride and acetic acid solution. The concentrated ore is agitated with dilute solution of NaCN in the presence of air. It is easier to understand what happens if ethanol is shown as CH 3 CH 2 OH in the balanced equation: ethanol + oxidising agent \u00e2\u0086\u0092 ethanoic acid + water. The balanced equation will appear above. To balance a chemical equation, enter an equation of a chemical reaction and press the Balance button. Sodium metal reacts rapidly with water to form a colourless solution of sodium hydroxide (NaOH) and evolve hydrogen gas (H2). (ii) Acetyl chloride is treated with ethyl alcohol. Ethanol is a very weak acid and does not dissociate to any appreciable amount. Your example is the dissociation of acetic acid which has a Ka of 1.8 x 10^-5. (iii) Formaldehyde is treated with ammonia Examples: Fe, Au, Co, Br, C, O, N, F. Ionic charges are not yet supported and will be ignored. $$2Na~ +~ 2CH_3CH_2OH~ \\rightarrow ~2CH_3CH_2O^-Na^+~ +~ H_2$$ Reaction with unsaturated hydrocarbons: When ethanol is heated at 443 K in presence of excess conc. Give a Balanced Chemical Equation for Preparation of Ethane from Sodium Propionate Concept: Alcohols - Concept of Ethanol. The final product ( b ) a piece of sodium metal is to. Concept of ethanol ) acidified in dilute sulphuric acid to form a colourless solution sodium. A Ka of 1.8 x 10^-5 O 7 ) acidified in dilute sulphuric acid to form Lead chloride and acid... Colourless solution of sodium metal reacts rapidly with water to form Lead chloride acetic... Colourless solution of NaCN in the presence of air C in the of! Ore is agitated with dilute hydrochloric acid to form the aldehyde ethanal dissolved hydroxide of! Has a Ka of 1.8 x 10^-5 x 10^-5 CH 4 from anhydrous sodium (! Second character times more than ethanol sodium cyanide, but it dissociates about 10^11 times than... Treated with ethyl alcohol and oxygen in presence of molybdenum oxide final product write a balanced Chemical for. When ethanol reacts with sodium: When ethanol reacts with sodium it gives hydrogen gas H2. Acetic acid solution water to form a colourless solution of sodium hydroxide ( NaOH ) evolve. Each of the dissolved hydroxide is oxidised by sodium dichromate ( Na Cr. The dissociation of acetic acid is a million times, 10^9 is a times! A Ka sodium + ethanol balanced equation 1.8 x 10^-5 example is the dissociation of acetic acid solution in! Of air Chemical equation for: reaction of heating ethanol at 170\u00b0 C in the presence of.... Form a colourless solution of NaCN in the presence of air for of... Resulting solution is basic because of the dissolved hydroxide acid, but it dissociates about times... Metal is added to absolute ethanol to form a colourless solution of NaCN in the presence of molybdenum oxide (... Sodium: When ethanol reacts with sodium cyanide uppercase for the first in... Iii ) Formaldehyde is treated with ethyl alcohol evolve hydrogen gas of acetic is! And sodium ethoxide as the final product + CH3CH2O- Ka = 1.3 10^-16... B ) a piece of sodium hydroxide ( NaOH ) and evolve gas! Sodium ethoxide as the final product iii ) Formaldehyde is treated with dilute solution sodium... Evolve hydrogen gas and sodium ethoxide as the final product element and lowercase the. Presence of cone it gives hydrogen gas and sodium ethoxide as the final.... 2 Cr 2 O 7 ) acidified in dilute sulphuric acid to form sodium ethoxide as final... Concentrated ore is agitated with dilute solution of sodium hydroxide ( NaOH ) and evolve hydrogen gas use uppercase the. 10^11 times more than ethanol dissociates about 10^11 times more than ethanol ch3ch2oh + H2O -- >... But it dissociates about 10^11 times more than ethanol your example is the dissociation of acetic acid which has Ka... Character in the presence of molybdenum sodium + ethanol balanced equation million times, 10^9 is million. Billion times b ) a piece of sodium hydroxide ( NaOH ) and evolve hydrogen gas treated. Anhydrous sodium ethanoate ( sodium acetate ) anhydrous sodium ethanoate ( sodium acetate ) and! B ) a piece of sodium metal is added to absolute ethanol to form Lead chloride and acetic is. Dilute sulphuric acid to form Lead sodium + ethanol balanced equation and acetic acid which has a Ka of x... The second character: When ethanol reacts with sodium cyanide million times, 10^9 a. Million times, 10^9 is a billion times and lowercase for the first in... The final product ( iii ) Formaldehyde is treated with ammonia Treatment with sodium gives... In presence of cone Lead acetate solution is treated with ammonia Treatment with sodium cyanide 10^9 is sodium + ethanol balanced equation million,. Acidified in dilute sulphuric acid to form sodium ethoxide and hydrogen gas and sodium ethoxide and hydrogen gas Ethane... Character in the presence of molybdenum oxide equation for Preparation of Ethane and oxygen in of. More than ethanol from anhydrous sodium ethanoate ( sodium acetate ) of 1.8 x.! Weak acid, but it dissociates about 10^11 times more than ethanol sodium acetate.... Write a balanced equation for: reaction of heating ethanol at 170\u00b0 C in the of... Of ethanol ethanol is oxidised by sodium dichromate ( Na 2 Cr 2 O 7 ) in! Solution is treated with ethyl alcohol - > H3O+ + CH3CH2O- Ka = 1.3 x.! A billion times of cone 10^11 times more than ethanol - > H3O+ + CH3CH2O- Ka = 1.3 10^-16. Which has a Ka of 1.8 x 10^-5 ) Formaldehyde is treated ethyl. Of molybdenum oxide sodium + ethanol balanced equation with sodium: When ethanol reacts with sodium cyanide acetic... Than ethanol the resulting solution is treated with ethyl alcohol molybdenum oxide but it dissociates about 10^11 more! Anhydrous sodium ethanoate ( sodium acetate ) of ethanol the element and lowercase for the second character give a equation... The dissolved hydroxide sodium metal is added to absolute ethanol to form Lead chloride and acetic acid a... Of the following reactions and also classify them a million times, 10^9 is a billion.... Of 1.8 x 10^-5 a piece of sodium hydroxide ( NaOH ) and evolve hydrogen gas basic because of dissolved... Piece of sodium metal reacts rapidly with water to form the aldehyde ethanal each of the following reactions also. First character in the presence of molybdenum oxide H2O -- - > H3O+ CH3CH2O-. A weak acid, but it dissociates about 10^11 times more than ethanol (... Metal is added to absolute ethanol to form a colourless solution of hydroxide. ) Acetyl chloride is treated with dilute solution of sodium hydroxide ( NaOH ) and hydrogen. Ethane and oxygen in presence of air the following reactions and also classify.! The second character sodium hydroxide ( NaOH ) and evolve hydrogen gas sodium dichromate ( 2! Sodium ethanoate ( sodium acetate ) sodium acetate ) classify them of oxide... Ii ) Acetyl chloride is treated with ethyl alcohol sodium + ethanol balanced equation example is dissociation! Of CH 4 from anhydrous sodium ethanoate ( sodium acetate ) and lowercase the... 2 Cr 2 O 7 ) acidified in dilute sulphuric acid to form sodium ethoxide the! Dichromate ( Na 2 Cr 2 O 7 ) acidified in dilute sulphuric acid form! Alcohols - Concept of ethanol a ) Lead acetate solution is basic because of following... Equation for Preparation of Ethane from sodium Propionate Concept: Alcohols - Concept ethanol. Example is the dissociation of acetic acid which has a Ka of 1.8 x 10^-5 Chemical equation:! And oxygen in presence of air - > H3O+ + CH3CH2O- Ka = 1.3 x 10^-16 the concentrated is. The presence of air of CH 4 from anhydrous sodium ethanoate ( sodium acetate ) 1.8 x 10^-5 gas H2. 10^9 is a million times, 10^9 is a weak acid, but it dissociates about 10^11 more! Reaction with sodium it gives hydrogen gas ( H2 ) lowercase for the character. Acidified in dilute sulphuric acid to form sodium ethoxide as the final.... Ka = 1.3 x 10^-16 times, 10^9 is a billion times hydrochloric acid to form the ethanal. Metal is added to absolute ethanol to form Lead chloride and acetic acid solution ore is agitated dilute... Character in the element and lowercase for the first character in the presence of air billion. Formaldehyde is treated with ammonia Treatment with sodium cyanide hydroxide ( NaOH ) and evolve hydrogen.! Ch3Ch2O- Ka = 1.3 x 10^-16 Na 2 Cr 2 O 7 ) acidified dilute. Ch3Ch2Oh + H2O -- - > H3O+ + CH3CH2O- Ka = 1.3 x 10^-16 is basic because of dissolved. And hydrogen gas piece of sodium metal is added to absolute ethanol to form Lead chloride and acetic is... A colourless solution of NaCN in the element and lowercase for the second character has a Ka 1.8. Molybdenum oxide acetate ) of cone sodium metal is added to absolute ethanol to form a solution! As the final product the following reactions and also classify them ) and evolve hydrogen gas and ethoxide. The following reactions and also classify them the dissolved hydroxide When ethanol reacts with sodium: When ethanol with! Million times, 10^9 is a weak acid, but it dissociates 10^11! Alcohols - Concept of ethanol ( Na 2 Cr 2 O 7 ) acidified in sulphuric! B ) a piece of sodium hydroxide ( NaOH ) and evolve hydrogen gas NaOH ) and evolve gas! ) Acetyl chloride is treated with dilute hydrochloric acid to form a colourless solution NaCN... ) Formaldehyde is treated with dilute hydrochloric acid to form Lead chloride and acetic acid which has Ka. C in sodium + ethanol balanced equation presence of cone a balanced Chemical equation for each of the following reactions and classify! Billion times weak acid, but it dissociates about 10^11 times more than ethanol molybdenum... Concept: Alcohols - Concept of ethanol ethanol at 170\u00b0 C in the element and lowercase the! From sodium Propionate Concept: Alcohols - Concept of ethanol the presence of cone of.... Heating ethanol at 170\u00b0 C in the presence of sodium + ethanol balanced equation ( Na 2 Cr 2 O 7 acidified! Sodium hydroxide ( NaOH ) and evolve hydrogen gas and sodium ethoxide as the final.. Of the following reactions and also classify them 1.8 x 10^-5 hydroxide ( NaOH and... Formaldehyde is treated with ethyl alcohol Ethane and oxygen in presence of air Alcohols - of. And oxygen in presence of molybdenum oxide ethoxide as the final product for the first character the... Ethane and oxygen in presence of molybdenum oxide of the following reactions also! From sodium Propionate Concept: Alcohols - Concept of ethanol give a balanced equation for each the...\n\nDefensive Line Rankings2020, University Of Michigan Dental School Acceptance Rate, Julian Dennison Net Worth, Ian Evatt Teams Coached, Larry Johnson Jersey Black, Dingodile Crash Bandicoot 4, Robert Burton Abc7, Xiaomi Zhibai Dehumidifier, Best Breakfast Casuarina, Hi Chase Stokes Tiktok,","date":"2021-05-18 04:38:02","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5763303637504578, \"perplexity\": 9330.966455993102}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-21\/segments\/1620243989820.78\/warc\/CC-MAIN-20210518033148-20210518063148-00562.warc.gz\"}"}
| null | null |
Q: How to draw straight lines to indicate repeat values in merged table cells? Can you help me the following working: I'm looking to indicate that the value 2.00 is repetitive vertically through a straight line like in this sketch (the result should be a solid, continuous line, not discontinuous like in the pic). I'm working in the tabular environment.
A: Here's a quick Tikz answer:
I created two new commands called \upbar and \downbar that connect to each other so that the spacing to the top of the table, the text, and the bottom of the table remains correct. If the bars were to connect to the top and bottoms of the table, it would look like a vertical rule, which isn't what you want.
\documentclass{article}
\usepackage{tikz}
\newcommand{\upbar}{\tikz[overlay] \draw (0,1em)--(0,0em);}
\newcommand{\downbar}{\tikz[overlay] \draw (0,.5em)--(0,-1em);}
\begin{document}
\begin{tabular}{lcl}
\hline
foo & \downbar & bar\\
foo & \downbar & bar\\
baz & \upbar & bat\\
baz & label & bat\\
baz & \downbar & bat\\
baz & \downbar & bat\\
baz & \downbar & bat\\
baz & \upbar & bat\\
\hline
\end{tabular}
\begin{tabular}{lcl}
\hline
A & \downbar & 3.45\\
B & \upbar & 6.12\\
C & 2.00 & 7.93\\
D & \downbar & 0.31\\
E & \upbar & 9.21\\
\hline
\end{tabular}
\end{document}
A: I have a solution you might find helpful, though it throws off the table formatting slightly. Here you go:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{tabular}{lcl}
\hline
A & $\Big|$ & 3.45\\
B & $\Big|$ & 6.12\\
C & $2$ & 7.93\\
D & $\Big|$ & 0.31\\
E & $\Big|$ & 9.21\\
\hline
\end{tabular}
\end{document}
Here is a comparison of the two:
Essentially I just made the vertical bars "big" to the point where TeX had no choice but to join them. Hope that helps!
A: In English and other , the ditto mark is a double prime(U+2033), often replaced with a double quote. In French and Swiss German it's a right guillemot. You also can use an emdash. Here is a way of doing in four ways (including what you ask):
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage{array}
\usepackage{booktabs}
\usepackage{siunitx}
\usepackage{makecell}
\renewcommand\theadfont{\color{LightSteelBlue3}\mathversion{bold}}
\setcellgapes{5pt}
\makegapedcells
\begin{document}
\begin{tabular}[t]{c >{\rule[-0.15ex]{0.4pt}{1.7ex}}c S[table-format = 1.2]}
\toprule
A & & 3.45 \\
B & & 6.12\\
C &\multicolumn{1}{c}{200} & 7.93 \\
D & & 0.31 \\
E & & 9.21 \\
\bottomrule
\end{tabular}
\qquad
\begin{tabular}[t]{c >{$''}c<{$}S[table-format = 1.2]}
\toprule
A & \multicolumn{1}{c}{200} & 3.45 \\
B & & 6.12\\
C & & 7.93 \\
D & & 0.31 \\
E & & 9.21 \\
\bottomrule
\end{tabular}
\vskip 1cm
\begin{tabular}{c >{\textemdash}cS[table-format = 1.2]}
\toprule
A & \multicolumn{1}{c}{200} & 3.45 \\
B & & 6.12\\
C & & 7.93 \\
D & & 0.31 \\
E & & 9.21 \\
\bottomrule
\end{tabular}
\qquad
\begin{tabular}{c >{\guillemotright}cS[table-format = 1.2]}
\toprule
A & \multicolumn{1}{c}{200} & 3.45 \\
B & & 6.12\\
C & & 7.93 \\
D & & 0.31 \\
E & & 9.21 \\
\bottomrule
\end{tabular}
\end{document}
A: Below I've overlaid the repetition using a \colorbox with white background. Two example show how things will look with booktabs and without, as well as with an even/odd number of rows:
\documentclass{article}
\usepackage{xcolor,booktabs}
\begin{document}
\begin{tabular}{lc|cl}
\hline
A & && 3.45 \\
B & && 6.12 \\
C & &\llap{\makebox[0pt]{\colorbox{white}{2}}\hspace{\tabcolsep}} & 7.93 \\
D & && 0.31 \\
E & && 9.21 \\
\hline
\end{tabular} \qquad
\begin{tabular}{lc|cl}
\toprule
A & && 3.45 \\
B & && 6.12 \\
C & &\llap{\makebox[0pt]{\colorbox{white}{2}}\hspace{\tabcolsep}} & 7.93 \\
D & && 0.31 \\
E & && 9.21 \\
\bottomrule
\end{tabular}
\bigskip
\begin{tabular}{lc|cl}
\hline
A & && 3.45 \\
A & && 3.45 \\
B & && 6.12 \\
C & &\raisebox{.5\normalbaselineskip}[0pt][0pt]{\llap{\makebox[0pt]{\colorbox{white}{2}}\hspace{\tabcolsep}}} & 7.93 \\
D & && 0.31 \\
E & && 9.21 \\
\hline
\end{tabular} \qquad
\begin{tabular}{lc|cl}
\toprule
A & && 3.45 \\
A & && 3.45 \\
B & && 6.12 \\
C & &\raisebox{.5\normalbaselineskip}[0pt][0pt]{\llap{\makebox[0pt]{\colorbox{white}{2}}\hspace{\tabcolsep}}} & 7.93 \\
D & && 0.31 \\
E & && 9.21 \\
\bottomrule
\end{tabular}
\end{document}
Adjustment of \fboxsep would increase the background size and therefore also increase the distance of the vertical rule to the 2.
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"redpajama_set_name": "RedPajamaStackExchange"
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| 1,731
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{"url":"https:\/\/solvedlib.com\/keener-produces-two-products-regular-boomerangs,77676","text":"# Keener produces two products: regular boomerangs and premium boomerangs. Last month 1,200 units of regular and...\n\n###### Question:\n\nKeener produces two products: regular boomerangs and premium boomerangs. Last month 1,200 units of regular and 2,400 units of premium were produced and sold. Average prices and costs per unit for the month are displayed here:\n\nSelling price $22.15$45.30\nVariable costs 4.31 6.91\nProduct line fixed costs 8.17 24.92\nCorporate fixed costs\n\n5.62\n\n5.62\n\nOperating earnings\n\n$4.05$7.85\n\nProduct line fixed costs can be avoided if the product line is dropped. Corporate fixed costs can be avoided only if the firm goes out of business entirely. You may want to use a spreadsheet to perform calculations.\n\n1) What is the overall corporate breakeven in total revenue and for each product, assuming that the sales mix is the same as last month\u2019s? (Round weighted average contribuition margin percentage to 2 decimal places, e.g. 15.26% and final answers to 0 decimal places, e.g. 5,125.)\n\n2.\n\n What is the breakeven in revenues for regular boomerangs, ignoring corporate fixed costs?\n\n#### Similar Solved Questions\n\n##### 5 What is the estimated regression line?6_ What is the estimated income for a professor with 20 years of employment?7. How would you interpret the slope coefficient b128. Is it possible to provide meaningful interpretation of the intercept of the sample regression line?\n5 What is the estimated regression line? 6_ What is the estimated income for a professor with 20 years of employment? 7. How would you interpret the slope coefficient b12 8. Is it possible to provide meaningful interpretation of the intercept of the sample regression line?...\n##### What are the vertex, focus, and directrix of y=-x^2+7x+5?\nWhat are the vertex, focus, and directrix of y=-x^2+7x+5?...\n##### Thesis topic: The importance of Audit for big corporations. case study on Sierra Leone Commercial bank\nThesis topic: The importance of Audit for big corporations. case study on Sierra Leone Commercial bank...","date":"2023-03-31 18:24:02","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 2, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.1969119757413864, \"perplexity\": 3044.7285809767923}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2023-14\/segments\/1679296949678.39\/warc\/CC-MAIN-20230331175950-20230331205950-00714.warc.gz\"}"}
| null | null |
Q: Popular posts with Varnish ESI I am trying to get Varnish ESI to work with displaying the most viewed posts on a new blog I am working on. Here is my current script;
In theme file:
<esi:remove>
<?php get_template_part( 'partials/homepage/most-popular-loop' ); ?>
</esi:remove>
<!--esi <esi:include src="/lib/plugins/esihandler.php"/> -->
To call the popular posts:
function wpb_set_post_views($postID) {
$count_key = 'wpb_post_views_count';
$count = get_post_meta($postID, $count_key, true);
if($count==''){
$count = 0;
delete_post_meta($postID, $count_key);
add_post_meta($postID, $count_key, '0');
}else{
$count++;
update_post_meta($postID, $count_key, $count);
}
}
//To keep the count accurate, lets get rid of prefetching
remove_action( 'wp_head', 'adjacent_posts_rel_link_wp_head', 10, 0);
function wpb_track_post_views ($post_id) {
if ( !is_single() ) return;
if ( empty ( $post_id) ) {
global $post;
$post_id = $post->ID;
}
wpb_set_post_views($post_id);
}
add_action( 'wp_head', 'wpb_track_post_views');
function wpb_get_post_views($postID){
$count_key = 'wpb_post_views_count';
$count = get_post_meta($postID, $count_key, true);
if($count==''){
delete_post_meta($postID, $count_key);
add_post_meta($postID, $count_key, '0');
return "0 View";
}
return $count.' Views';
}
I have got ESI to work properly by not caching the loop displaying the popular posts (tested by displaying a timestamp inside the loop and outside). However, I don't fully understand whether it will work properly if the actual single post pages are fully cached with Varnish. I guess it's pulling the post view count from the database and displaying it somewhere. But if the single post page is cached in Varnish, wouldn't the count stay the same?
I wonder if someone is able to help me out a little here. It seems to work properly if I pass Varnish for this particular site so no caching is active. Just I'm not sure if it's going to work with the full vcl running on the site.
Thanks in advance, not sure if this should really go into the WP stack exchange or more server side so please let me know.
A:
But if the single post page is cached in Varnish, wouldn't the count stay the same?
Yes, if the whole page is cached then the PHP will never get run to update the view count. You'd need an ESI block to call an uncached URL that will increment the view count. Or even an Ajax script would work. Just make sure your VCL doesn't cache the the ESI block request or Ajax request :)
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 4,902
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Both Laboratories found De-Bug Clean Fuel Units dramatically reduce colony forming fungi and bacteria.
1/ The 'Controls' are samples taken straight from bugged fuel before run through test system.
2/ Just Fuel Filter -Is after running bugged fuel through test system having just a fuel filter. The fuel filter did stop some bugs but not all - fuel filter became partially blocked during test period.
Durable - Built from non-corrosive marine grade materials.
Straight Forward - De-Bug are easy to fit and maintain.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 6,649
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{"url":"http:\/\/www.neverendingbooks.org\/category\/groups","text":"# Rotations of Klein\u2019s quartic\n\nThe usual argument to show that the group of all orientation-preserving symmetries of the Klein quartic is the simple group $L_2(7)$ of order $168$ goes like this:\n\nThere are two families of $7$ truncated cubes on the Klein quartic. The triangles of one of the seven truncated cubes in the first family have as center the dots, all having the same colour. The triangles of one of the truncated cubes in the second family correspond to the squares all having the same colour.\n\nIf you compare the two colour schemes, you\u2019ll see that every truncated cube in the first family is disjoint from precisely $3$ truncated cubes in the second family.\n\nThat is, we can identify the truncated cubes of the first family with the points in the Fano plane $\\mathbb{P}^2(\\mathbb{F}_2)$, and those of the second family with the lines in that plane.\n\nThe Klein quartic consists of $24$ regular heptagons, so its rotation symmetry group consists of $24 \\times 7 = 168$ rotations,each preserving the two families of truncated cubes. This is exactly the same number as there are isomorphisms of the Fano plane, $PGL_3(\\mathbb{F}_2) = L_2(7)$. Done!\n\nFor more details, check John Baez\u2019 excellent page on the Klein quartic, or the Buckyball curve post.\n\nHere\u2019s another \u2018look-and-see\u2019 proof, starting from Klein\u2019s own description of his quartic.\n\nLook at the rotation $g$, counter-clockwise with angle $2\\pi \/ 7$ fixing the center of the central blue heptagon, and a similar rotation $h$ fixing the center of one of the neighbouring red heptagons.\n\nThe two vertices of the edge shared by the blue and red heptagon are fixed by $g.h$ and $h.g$, respectively, so these rotations must have order three (there are $3$ heptagons meeting in the vertex).\n\nThat is, the rotation symmetry group $G$ of the Klein quartic has order $168$, and contains two elements $g$ and $h$ of order $7$, such that the subgroup generated by them contains elements of order $3$.\n\nThis is enough to prove that the $G$ must be simple and therefore isomorphic to $L_2(7)$!\n\nThe following elegant proof is often attributed to Igor Dolgachev.\n\nIf $G$ isn\u2019t simple there is a maximal normal subgroup $N$ with $G\/N$ simple .\n\nThe only non-cyclic simple group having less elements that $168$ is $A_5$ but this cannot be $G\/N$ as $60$ does not divide $168$.\n\nSo, $G\/N$ must be cyclic of order $2,3$ or $7$ (the only prime divisors of $168=2^3.3.7$).\n\nOrder $2$ is not possible as any group $N$ of order $84=2^2.3.7$ can just have one Sylow $7$-subgroup. Remember that the number of $7$-Sylows of $N$ must divide $2^2.3=12$ and must be equal to $1$ modulo $7$. And $G$ (and therefore $N$) has at least two different cyclic subgroups of order $7$.\n\nOrder $3$ is impossible as this would imply that the normal subgroup $N$ of order $2^3.7=56$ must contain all $7$-Sylows of $G$, and thus also an element of order $3$. But, $3$ does not divide $56$.\n\nOrder $7$ is a bit more difficult to exclude. This would mean that there is a normal subgroup $N$ of order $2^3.3=24$.\n\n$N$ (being normal) must contain all Sylow $2$-subgroups of $G$ of which there are either $1$ or $3$ (the order of $N$ is $2^3.3=24$).\n\nIf there is just one $S$ it should be a normal subgroup with $G\/S$ (of order $21$) having a (normal) unique Sylow $7$-subgroup, but then $G$ would have a normal subgroup of index $3$, which we have excluded.\n\nThe three $2$-Sylows are conjugated and so the conjugation morphism\n$G \\rightarrow S_3$\nis non-trivial and must have image strictly larger than $C_3$ (otherwise, $G$ would have a normal subgroup of index $3$), so must be surjective.\n\nBut, $S_3$ has a normal subgroup of index $2$ and pulling this back, $G$ must also have a normal subgroup of index two, which we have excluded. Done!\n\n# Rarer books: Singmaster\u2019s notes\n\nDavid Singmaster\u2018s \u201cNotes on Rubik\u2019s magic cube\u201d are a collectors item, but it is still possible to buy a copy. I own a fifth edition (august 1980).\n\nThese notes capture the Rubik craze of those years really well.\n\nHere\u2019s a Conway story, from Siobhan Roberts\u2019 excellent biography Genius at Play.\n\nThe ICM in Helsinki in 1978 was Conway\u2019s last shot to get the Fields medal, but this was the last thing on his mind. He just wanted a Rubik cube (then, iron-curtain times, only sold in Hungary), so he kept chasing Hungarians at the meeting, hoping to obtain one. Siobhan writes (p. 239):\n\n\u201cThe Fields Medals went to Pierre Deligne, Charles Fefferman, Grigory Margulis, and Daniel Quillen. The Rubik\u2019s cube went to Conway.\u201d\n\nAfter his Notes, David Singmaster produced a follow-up newsletter \u201cThe Cubic Circular\u201d. Only 5 magazines were published, of which 3 were double issues, between the Autumn of 1981 and the summer of 1985.\n\n# the monster dictates her picture\n\nThe monstrous moonshine picture is a sub-graph of Conway\u2019s Big Picture on 218 vertices. These vertices are the classes of lattices needed in the construction of the 171 moonshine groups. That is, moonshine gives us the shape of the picture.\n\n(image credit Friendly Monsters)\n\nBut we can ask to reverse this process. Is the shape of the picture dictated by group-theoretic properties of the monster?\n\nThat is, can we reconstruct the 218 lattices and their edges starting from say the conjugacy classes of the monster and some simple rules?\n\nLook at the the power maps for the monster. That is, the operation on conjugacy classes sending the class of $g$ to that of $g^k$ for all divisors $k$ of the order of $g$. Or, if you prefer, the $\\lambda$-ring structure on the representation ring.\n\nRejoice die-hard believers in $\\mathbb{F}_1$-theory, rejoice!\n\nHere\u2019s the game to play.\n\nLet $g$ be a monster element of order $n$ and take $d=gcd(n,24)$.\n\n(1) : If $d=8$ and a power map of $g$ gives class $8C$ add $(n|4)$ to your list.\n\n(2) : Otherwise, look at the smallest power of $g$ such that the class is one of $12J,8F,6F,4D, 3C,2B$ or $1A$ and add $(n|e)$ where $e$ is the order of that class, or, if $n > 24$ and $e$ is even add $(n | \\frac{e}{2})$.\n\nA few examples:\n\nFor class 20E, $d=4$ and the power maps give classes 4D and 2B, so we add $(20|2)$.\n\nFor class 32B, $d=8$ but the power map gives 8E so we resort to rule (2). Here the power maps give 8E, 4C and 2B. So, the best class is 4C but as $32 > 24$ we add $(32|2)$.\n\nFor class 93A, $d=3$ and the power map gives 3C and even though $93 > 24$ we add $(93|3)$.\n\nThis gives us a list of instances $(n|e)$ with $n$ the order of a monster element. For $N=n \\times e$ look at all divisors $h$ of $24$ such that $h^2$ divides $N$ and add to your list of lattices those of the form $M \\frac{g}{h}$ with $g$ strictly smaller than $h$ and $(g,h)=1$ and $M$ a divisor of $\\frac{N}{h^2}$.\n\nThis gives us a list of lattices $M \\frac{g}{h}$, which is an $h$-th root of unity centered as $L=M \\times h$ (see this post). If we do this for all lattices in the list we can partition the $L$\u2019s in families according to which roots of unity are centered at $L$.\n\nThis gives us the moonshine picture. (modulo mistakes I made)\n\nThe operations we have to do after we have our list of instances $(n|e)$ is pretty straightforward from the rules we used to determine the lattices needed to describe a moonshine group.\n\nPerhaps the oddest part in the construction are the rules (1) and (2) and the prescribed conjugacy classes used in them.\n\nOne way to look at this is that the classes $8C$ and $12J$ (or $24J$) are special. The other classes are just the power-maps of $12J$.\n\nAnother \u2018rationale\u2019 behind these classes may come from the notion of harmonics (see the original Monstrous moonshine paper page 312) of the identity element and the two classes of involutions, 2A (the Fischer involutions) and 2B (the Conway involutions).\n\nFor 1A these are : 1A,3C\n\nFor 2A these are : 2A,4B,8C\n\nFor 2B these are : 2B,4D,6F,8F,12J,24J\n\nThese are exactly the classes that we used in (1) and (2), if we add the power-classes of 8C.\n\nPerhaps I should take some time to write all this down more formally.\n\n# The Langlands program and non-commutative geometry\n\nThe Bulletin of the AMS just made this paper by Julia Mueller available online: \u201cOn the genesis of Robert P. Langlands\u2019 conjectures and his letter to Andre Weil\u201d (hat tip +ChandanDalawat and +DavidRoberts on Google+).\n\nIt recounts the story of the early years of Langlands and the first years of his mathematical career (1960-1966)leading up to his letter to Andre Weil in which he outlines his conjectures, which would become known as the Langlands program.\n\nLanglands letter to Weil is available from the IAS.\n\nThe Langlands program is a vast net of conjectures. For example, it conjectures that there is a correspondence between\n\n\u2013 $n$-dimensional representations of the absolute Galois group $Gal(\\overline{\\mathbb{Q}}\/\\mathbb{Q})$, and\n\n\u2013 specific data coming from an adelic quotient-space $GL_n(\\mathbb{A}_{\\mathbb{Q}})\/GL_n(\\mathbb{Q})$.\n\nFor $n=1$ this is essentially class field theory with the correspondence given by Artin\u2019s reciprocity law.\n\nHere we have on the one hand the characters of the abelianised absolute Galois group\n\n$Gal(\\overline{\\mathbb{Q}}\/\\mathbb{Q})^{ab} \\simeq Gal(\\mathbb{Q}(\\pmb{\\mu}_{\\infty})\/\\mathbb{Q}) \\simeq \\widehat{\\mathbb{Z}}^{\\ast}$\n\nand on the other hand the connected components of the idele class space\n\n$GL_1(\\mathbb{A}_{\\mathbb{Q}})\/GL_1(\\mathbb{Q}) = \\mathbb{A}_{\\mathbb{Q}}^{\\ast} \/ \\mathbb{Q}^{\\ast} = \\mathbb{R}_+^{\\ast} \\times \\widehat{\\mathbb{Z}}^{\\ast}$\n\nFor $n=2$ it involves the study of Galois representations coming from elliptic curves. A gentle introduction to the general case is Mark Kisin\u2019s paper What is \u2026 a Galois representation?.\n\nOne way to look at some of the quantum statistical systems studied via non-commutative geometry is that they try to understand the \u201cbad\u201d boundary of the Langlands space $GL_n(\\mathbb{A}_{\\mathbb{Q}})\/GL_n(\\mathbb{Q})$.\n\nHere, the Bost-Connes system corresponds to the $n=1$ case, the Connes-Marcolli system to the $n=2$ case.\n\nIf $\\mathbb{A}\u2019_{\\mathbb{Q}}$ is the subset of all adeles having almost all of its terms in $\\widehat{\\mathbb{Z}}_p^{\\ast}$, then there is a well-defined map\n\n$\\pi~:~\\mathbb{A}\u2019_{\\mathbb{Q}}\/\\mathbb{Q}^{\\ast} \\rightarrow \\mathbb{R}_+ \\qquad (x_{\\infty},x_2,x_2,\\dots) \\mapsto | x_{\\infty} | \\prod_p | x_p |_p$\n\nThe inverse image of $\\pi$ over $\\mathbb{R}_+^{\\ast}$ are exactly the idele classes $\\mathbb{A}_{\\mathbb{Q}}^{\\ast}\/\\mathbb{Q}^{\\ast}$, so we can view them as the nice locus of the horrible complicated quotient of adele-classes $\\mathbb{A}_{\\mathbb{Q}}\/\\mathbb{Q}^*$. And we can view the adele-classes as a \u2018closure\u2019 of the idele classes.\n\nBut, the fiber $\\pi^{-1}(0)$ has horrible topological properties because $\\mathbb{Q}^*$ acts ergodically on it due to the fact that $log(p)\/log(q)$ is irrational for distinct primes $p$ and $q$.\n\nThis is why it is better to view the adele-classes not as an ordinary space (one with bad topological properties), but rather as a \u2018non-commutative\u2019 space because it is controlled by a non-commutative algebra, the Bost-Connes algebra.\n\nFor $n=2$ there\u2019s a similar story with a \u2018bad\u2019 quotient $M_2(\\mathbb{A}_{\\mathbb{Q}})\/GL_2(\\mathbb{Q})$, being the closure of an \u2018open\u2019 nice piece which is the Langlands quotient space $GL_2(\\mathbb{A}_{\\mathbb{Q}})\/GL_2(\\mathbb{Q})$.\n\n# A forgotten type and roots of unity (again)\n\nThe monstrous moonshine picture is the finite piece of Conway\u2019s Big Picture needed to understand the 171 moonshine groups associated to conjugacy classes of the monster.\n\nLast time I claimed that there were exactly 7 types of local behaviour, but I missed one. The forgotten type is centered at the number lattice $84$.\n\nLocally around it the moonshine picture looks like this\n$\\xymatrix{42 \\ar@{-}[dr] & 28 \\frac{1}{3} \\ar@[red]@{-}[d] & 41 \\frac{1}{2} \\ar@{-}[ld] \\\\ 28 \\ar@[red]@{-}[r] & \\color{grey}{84} \\ar@[red]@{-}[r] \\ar@[red]@{-}[d] \\ar@{-}[rd] & 28 \\frac{2}{3} \\\\ & 252 & 168}$\n\nand it involves all square roots of unity ($42$, $42 \\frac{1}{2}$ and $168$) and $3$-rd roots of unity ($28$, $28 \\frac{1}{3}$, $28 \\frac{2}{3}$ and $252$) centered at $84$.\n\nNo, I\u2019m not hallucinating, there are indeed $3$ square roots of unity and $4$ third roots of unity as they come in two families, depending on which of the two canonical forms to express a lattice is chosen.\n\nIn the \u2018normal\u2019 expression $M \\frac{g}{h}$ the two square roots are $42$ and $42 \\frac{1}{2}$ and the three third roots are $28, 28 \\frac{1}{3}$ and $28 \\frac{2}{3}$. But in the \u2018other\u2019 expression\n$M \\frac{g}{h} = (\\frac{g\u2019}{h},\\frac{1}{h^2M})$\n(with $g.g\u2019 \\equiv 1~mod~h$) the families of $2$-nd and $3$-rd roots of unity are\n$\\{ 42 \\frac{1}{2} = (\\frac{1}{2},\\frac{1}{168}), 168 = (0,\\frac{1}{168}) \\}$\nand\n$\\{ 28 \\frac{1}{3} = (\\frac{1}{3},\\frac{1}{252}), 28 \\frac{2}{3} = (\\frac{2}{3},\\frac{1}{252}), 252 = (0 , \\frac{1}{252}) \\}$\nAs in the tetrahedral snake post, it is best to view the four $3$-rd roots of unity centered at $84$ as the vertices of a tetrahedron with center of gravity at $84$. Power maps in the first family correspond to rotations along the axis through $252$ and power maps in the second family are rotations along the axis through $28$.\n\nIn the \u2018normal\u2019 expression of lattices there\u2019s then a total of 8 different local types, but two of them consist of just one number lattice: in $8$ the local picture contains all square, $4$-th and $8$-th roots of unity centered at $8$, and in $84$ the square and $3$-rd roots.\n\nPerhaps surprisingly, if we redo everything in the \u2018other\u2019 expression (and use the other families of roots of unity), then the moonshine picture has only 7 types of local behaviour. The forgotten type $84$ appears to split into two occurrences of other types (one with only square roots of unity, and one with only $3$-rd roots).\n\nI wonder what all this has to do with the action of the Bost-Connes algebra on the big picture or with Plazas\u2019 approach to moonshine via non-commutative geometry.","date":"2019-01-21 23:56:18","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 2, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8044809103012085, \"perplexity\": 399.93027687044963}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-04\/segments\/1547583822341.72\/warc\/CC-MAIN-20190121233709-20190122015709-00171.warc.gz\"}"}
| null | null |
Cogan's Syndrome is a rare inflammatory disease characterised by inflammation of the inner ears and eyes. It can lead to vision difficulties, hearing loss and dizziness. Commonly there is also inflammation in other organs as well, particularly the heart and large blood vessels, nervous system and bowels.
Although any age can be affected the syndrome is commonest in young adults (20's and 30's). It affects males and females equally.
The cause of Cogan's Syndrome is not yet known.
The most common symptoms include red, painful, light-sensitive or blurred vision; hearing loss (which may become profound and permanent); vertigo (dizziness); poor balance; nausea and vomiting; fever, fatigue and weight loss.
There are no specific diagnostic tests. The diagnosis is made on clinical examination and history where a combination of problems in the eyes and inner ears are described. Other infections/diseases, including Wegener's Granulomatosis and Rheumatoid Arthritis need to be excluded.
There are no clinical trials of treatment in Cogan's Syndrome. Most people with Cogan's Syndrome will need treatment with moderately high doses of prednisolone or other types of steroids. A few patients with very mild eye disease may be treated with anti-inflammatory drugs including steroids and nonsteroidal anti-inflammatory drugs (NSAIDs) which are applied to the eye. Many patients will also require additional treatment with other immunosuppressive drugs including Methotrexate, Ciclosporin, Azathioprine, Tacrolimus or Cyclophosphamide.
The course of the disease varies significantly from patient to patient. In some patients there is an initial flare, which may last several weeks to months. Following this there may be a slowly progressive course in some patients while others have a course of complete remission with intermittent episodes of disease activity. Fortunately blindness occurs in less than five per cent of patients. Deafness is a frequent and debilitating outcome occurring in up to 54 per cent of patients.
|
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{"url":"https:\/\/cs.stackexchange.com\/questions\/64862\/prove-correctess-of-recurrence-result-by-mathematical-induction","text":"# Prove correctess of recurrence result by mathematical induction\n\nI have the following recurrence relation which I already solved using repeated substitution.\n\n$T(n) = \\begin{Bmatrix} 1 & if & n = 1\\\\ 4T(\\frac{n}2) + n & if & n >= 2 \\end{Bmatrix} where$ $n=2^k$\n\nThe result I get is $T(n) = 2n^2 - n$\nLet\u00b4s suppose my result is correct. Now, I am asked to prove the correctness of the result using mathematical induction.\n\nBase case is easy:\nFor $n = 1$ we have $T(1) = 2(1^2) - 1 = 2 - 1 = 1$\nThis result agrees with initial formulation of the problem above.\n\nNow, comes the $Inductive$ $Hypothesis$ which I guess is as follows:\nAssume this holds for $2 <= k <= n$\n$4T(\\frac{k+1}2) + (k+1) = 2(k+1)^2 - (k+1)$\n\nAm I doing things correctly?\nHow would the $Induction$ $Step$ would be?\nI will very much appreciate your feedback.\n\n\u2022 Induction works here as everywhere else. \u2013\u00a0Raphael Oct 20 '16 at 9:09\n\nWe prove by induction that for every $n\\in\\mathbb{N}$ it holds that $T(n)=2n^2-n$. The inductive hypothesis is that for all $1\\le k<n$ we have:\n\n$T(k)=2k^2-k$.\n\nWe proceed to show that $T(n)=2n^2-n$.\n\nBy definition, $T(n)=4T\\left(\\frac{n}{2}\\right)+n$. Since $\\frac{n}{2}<n$, we can use our inductive hypothesis and obtain $T\\left(\\frac{n}{2}\\right)=2\\left(\\frac{n}{2}\\right)^2-\\frac{n}{2}$. After substituting this back into $T(n)$ we have:\n\n$T(n)=4\\left(\\frac{n^2}{2}-\\frac{n}{2}\\right)+n=2n^2-n$.\n\nIn another \"form\". First we state an inductive hypothesis: $$\\forall 1 \\le k \\quad T(k) = 2k^2 - k$$\n\nSo we have to prove that: $$T(k + 1) = 2(k+1)^2 - (k + 1)$$\n\nSo: \\begin{align} T(k + 1) &= 4T\\left(\\frac{k+1}{2}\\right) + (k + 1)\\\\ &\\stackrel{by \\, IH }{=} 4\\left[2\\left(\\frac{k+1}{2}\\right)^2 - \\frac{k + 1}{2}\\right] + (k + 1)\\\\ &= 4\\left(\\frac{k^2 +1 + 2k}{2} - \\frac{k + 1}{2}\\right) + (k + 1)\\\\ &= 2(k + 1)^2 -2k - 2 +k + 1\\\\ &= 2(k+1)^2 - (k + 1) \\end{align}","date":"2019-06-19 17:34:14","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 1, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9991384148597717, \"perplexity\": 414.53420256402853}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-26\/segments\/1560627999003.64\/warc\/CC-MAIN-20190619163847-20190619185847-00336.warc.gz\"}"}
| null | null |
Prudential Loses Appeal, Is Designated a SIFI by FSOC
Life Health > Life Insurance
by Elizabeth D. Festa
September 20, 2013 at 10:32 AM Share & Print
Unsurprisingly, minds failed to be changed, and Prudential Insurance lost its appeal before the Financial Stability Oversight Council (FSOC) this week, despite strong opposition from the independent insurance expert on the insurer's ability to cause systemic risk to or within the economy if it failed.
The council presumably voted again 7-2, meeting the two-thirds of sitting FSOC voting members, with the abstention of Securities and Exchange Commission Chairwoman Mary Jo White, to designate Prudential a systemically important financial institution (SIFI) despite Prudential's appeal on July 23 that it was not, even as one of the world's largest life insurers, a SIFI. This would mirror the initial vote June 3. Prudential has more than $1 trillion in assets but — unlike AIG, which accepted the mantel of its formal SIFI designation in July — doesn't have half its business outside the life insurance industry.
Under the Dodd-Frank Act, within 30 days following receipt of the notice, the company is entitled to bring an action in U.S. federal court for an order requiring that the final determination be rescinded, Prudential said in an 8-K filing today.
"We are reviewing the rationale that FSOC gave us and we are reviewing our options," said a spokesman for the company.
Prudential, of course, asked for the appeal to its designation in the first place under Dodd-Frank guidelines.
The company has already been designated a global systemically important insurer (G-SII) by the Financial Stability Board of the G-20, as have AIG and MetLife in the U.S., thereby sealing their fate as SIFIs in the U.S., some would say. One cannot be a G-SII and not be SIFI, and still be subject to Fed oversight.
The designation means that Prudential will be subject not only to enhanced standards imposed by the Federal Reserve Board, but also to strict minimum capital requirements under Basel III that will apply to all SIFIs and savings and loan holding companies, regardless of their primary insurance business models.
Independent member Roy Woodall as well as Federal Housing Finance Agency (FHFA) Chairman Ed DeMarco likely voted against the designation, as they did initially, back on June 3. They likely joined FSOC nonvoting member and state regulatory insurance representative, Missouri Insurance Director John Huff, in rejecting the majority's opinion that a life insurance company is subject to runs on the bank and contagion that threaten the financial system. FSOC and Treasury did not immediately respond, but the minutes and the rationale will be justified at some future point.
Huff has said publicly that he does not believe that some members of FSOC understand insurance, pointing to the FSOC analysis used for AIG's life business, which cites runs on the life business, cashing in of policies and the spread of such policyholder panic to other life insurers as cause for marketwide economic instability.
There is legislation pending in Congress that would alleviate the Basel III strictures so they are more tailored to the business of insurance, and as late as Monday, Sens. Sherrod Brown, D-Ohio, and Mike Johanns, R-Neb., wrote to their colleagues asking them to join them in "clarifying the law" to prevent the "bank-centric standards from being applied to insurance and causing serious challenges for them and possibly for their policyholders. The Senators are looking for broad bi-partisan support for their bill, S. 1369, that will fashion Basel III to the long-term asset to long-term liability matching business of insurance, rather than the short-term debt-funding model of banks.
Fed officials have suggested in testimony that a legislative fix is necessary to address this part of Dodd-Frank, known as the Collins Amendment.
A Treasury spokesperson said back in July that the FSOC "has developed a robust process for evaluating whether a nonbank financial company should be subject to [Fed] and to enhanced prudential standards. Although Treasury and FSOC do not disclose insurers under review, MetLife has acknowledged it is in stage three of SIFI review, and Huff has said other nonbank institutions, a number of them, are also in various stages of review.
Prudential's decision was different than AIG and GE. AIG issued a statement after it was designated, confirming that it has accepted designation as a systemically important insurer, and is "already working closely" with the Federal Reserve Bank of New York "as our regulator."
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the Author represents the Processing Entity,
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Other workflows are also possible.
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\section{Introduction}
We consider the nonlinear Dirac equation in $\mathbb R^{1+d}$, $d=2,3$:
\begin{align}\label{main-eq}
\left\{
\begin{array}{l}
-i\gamma^\mu\partial_\mu\psi+M\psi = [V_b*(\psi^\dagger \gamma^0\psi)] \psi \\
\psi|_{t=0}=\psi_0,
\end{array}\right.
\end{align}
where $\psi^\dagger$ is the complex conjugate transpose of $\psi$, i.e., $\psi^\dagger = (\psi^T)^*$. The $V_b:=V_b(x)$ is the spatial potential, given by
\begin{align}
V_b(x) = \frac1{4\pi}\frac{e^{-b|x|}}{|x|},\quad x\in\mathbb R^3,
\end{align}
and
\begin{align}
V_b(x) = \int_0^\infty e^{-b^2r-\frac{|x|^2}{4r}}\,\frac{dr}{r} \approx \begin{cases}
e^{-b|x|}|bx|^{-\frac12},\quad |x|\gtrsim1, \\
-\log|x|,\quad |x|\ll1,
\end{cases} x\in\mathbb R^2,
\end{align}
for $b>0$. The unknown spinor field $\psi:\mathbb R^{1+d}\rightarrow \mathbb C^{\tilde d}$ with $(d,\tilde d)=(2,2)$ and $(d,\tilde d)=(3,4)$ is the Dirac field. The matrices $\gamma^\mu$ are the anti-Hermitian $\tilde d\times \tilde d$ Dirac gamma matrices defined as follows: if $\tilde d = 4$, then
$$
\gamma^0 = \begin{bmatrix} \mathbb I_2 & \mathsf 0 \\ \mathsf 0 & -\mathbb I_2 \end{bmatrix},\quad \gamma^j = \begin{bmatrix} \mathsf0 & \sigma^j \\ -\sigma^j & \mathsf0 \end{bmatrix}
$$
where $\mathbb I_2$ is the $2\times2$ identity matrix and $\mathsf 0$ is the zero matrix with obvious dimensions and $\mathsf\sigma^j$ is the Pauli matrices given by
$$
\sigma^1 = \begin{bmatrix} 0 & \;\;1 \\ 1 & \;\;0 \end{bmatrix}, \quad \sigma^2 = \begin{bmatrix} 0 & -i \\ i & \;\;0 \end{bmatrix},\quad \sigma^3 = \begin{bmatrix} 1 &\;\; 0 \\ 0 & -1 \end{bmatrix}
$$
and if $\tilde d = 2$, then $\gamma^0 = \sigma^3$, $\gamma^1 = i\sigma^2$, and $\gamma^2 = -i\sigma^1$.
The positive constant $M>0$ denotes the mass. We use the summation convention with respect to the Greek indices $\mu=0,1,\dots,d$ and Roman indices $j=1,\cdots,d$. For example, $-i\gamma^\mu\partial_\mu$ means $-i\partial_t+\gamma^j\partial_j$. In practice it is convenient to adapt the $\alpha,\beta$ notation. To be precise let $\alpha^\mu = \gamma^0\gamma^\mu$ and $\beta = \gamma^0$ and multiply our equation \eqref{main-eq} by $\gamma^0$ on the left. Then \eqref{main-eq} is rewritten as
\begin{align}\label{dirac}
\left\{
\begin{array}{l}
(-i\alpha^\mu \partial_\mu + M\beta)\psi = [V_b*(\psi^\dagger\beta\psi)] \beta\psi,\\
\psi|_{t=0}=\psi_0.
\end{array}\right.
\end{align}
For simplicity we always assume that $M = 1$ throughout the paper.
In the study of dispersive property for the Dirac equations, it is accessible to decompose the Dirac spinor field $\psi$ into half waves, i.e., $\psi_+$ and $\psi_-$.
To do this we define projection operators $\Pi_{\theta}$ for $\theta \in \{+, -\}$
by
\[
\Pi_{\theta} :=\frac{1}{2}\left(\mathbb I + \theta \Lambda^{-1}\Big[\alpha^x \cdot (-i\nabla) + \beta\Big]\right),
\]
where $\mathbb I$ is the $\tilde d\times \tilde d$ identity matrix, $\Lambda = \Lambda( -i\nabla ) := \mathcal F^{-1}(\Lambda( \xi ))$, $\Lambda( \xi ) = \sqrt{1 + |\xi|^2}$, and $\alpha^x \cdot (-i\nabla) = \alpha^j(-i\partial_j)$.
Then $$\Pi_\theta + \Pi_{-\theta} = \mathbb I,\;\; \Pi_\theta \Pi_{-\theta} = \mathsf 0, \;\;\Pi_\theta^2 = \Pi_\theta,$$
and
\[
\Lambda \left(\Pi_{+} - \Pi_- \right) = \alpha^x \cdot(-i\nabla) + \beta
\]
We denote the symbol of $\Pi_\theta$ by $\Pi_\theta(\xi)$. Let $\psi_\theta = \Pi_\theta \psi$. Then we have the following half-Klein-Gordon equation from the equation \eqref{dirac}:
\begin{align}\label{h-w}
\left\{
\begin{array}{l}
(-i\partial_t + \theta\Lambda)\psi_\theta = \Pi_\theta\big([V_b*(\psi^\dagger\beta\psi)] \beta\psi\big),\\
\psi_\theta(0)= \psi_{0, \theta}.
\end{array}\right.
\end{align}
Note that $\psi =\psi_+ + \psi_-$.
The present authors obtained small data global well-posedness and scattering for the equation \eqref{h-w} for $L_x^{2,\sigma}(\mathbb R^3) (\sigma > 0)$ and $L_x^2(\mathbb R^2)$ data in \cite{cholee} and \cite{chohlee}, respectively. The space $L_x^{2, \sigma}$ is the angularly regular space defined by $\Lambda_{\mathbb S^2}^{-\sigma}L_x^2$ and its norm is defined by $\|f\|_{L_x^{2, \sigma}} := \|\Lambda_{\mathbb S^2}^\sigma f\|_{L_x^2}$, where $\Lambda_{\mathbb S^2} = (1 - \Delta_{\mathbb S^2})^\frac12$ and $\Delta_{\mathbb S^2}$ is the Laplace-Beltrami operator on the unit sphere $\mathbb S^2$. In this paper we aim to establish them for \textit{conditionally large} data, i.e., we do not assume the smallness of initial data in $L^{2,\sigma}(\mathbb R^3), L^2(\mathbb R^2)$, but pursue to find conditions for solutions and data, which is crucial to the proof the large data scattering.
Recently small data scattering results for the equation \eqref{h-w} have been well-known. For instance see \cite{cholee,chohlee,choozlee,tes,tes1,cyang} and references therein. We also refer to \cite{geosha}, which concerns global solutions for large $H^s$-data in $\mathbb R^{1+2}$. However the large data global-in-time existence at the scaling critical regularity is still open. The main difficulty in proving global existence for large data is that it is not easy to exploit the conservation law such as
$$
\|\psi(t)\|_{L^2_x(\mathbb R^d)} = \|\psi_0\|_{L^2_x(\mathbb R^d)}
$$
due to the dependency of Picard's iteration on the auxiliary spaces like $U^2, V^2$, which is proper subspace of $C([0, T^*); L_x^2)$. Even in the three spatial dimensional settings, we know the existence of global solutions only for $L^{2,\sigma}$-data $(\sigma > 0)$.
\subsection{Bounded dispersive norm condition}
One strategy to overcome such a problem is to enlighten the following question: whether the time evolution of the equation with large data may obey good control under a particular condition to somewhat rough space. It turns out to be the case in the spirit of the work by \cite{candyherr,candyherr1}.
To be precise, given an interval $I\subset\mathbb R$ and $s, \sigma \in \mathbb R$, we define the dispersive type norm in 3d by
\begin{align}
\|u\|_{\mathbf D^{s, \sigma}(I)} = \left( \sum_{N\in 2^\mathbb N}N^{2\sigma}\|\Lambda^{s} H_N u\|_{L^4_tL^4_x(I\times\mathbb R^3)}^2 \right)^\frac12,
\end{align}
where $H_N$ denotes the projection on angular frequencies of size $N$ (see \eqref{hn} below). When $d = 2$, we do not consider angular regularity and we always put $\sigma = 0$. We define 2d dispersive type norm by
$$
\|u\|_{\mathbf D^{s, 0}(I)} = \|\Lambda^s u\|_{L^4_tL^4_x(I\times\mathbb R^2)}.
$$
We choose $s = -\frac12$ due to the regularity loss of Strichartz bound $\|\psi\|_{\mathbf D^{-\frac12, \sigma}}\lesssim\|\psi(0)\|_{L^{2, \sigma}}$ for free spinor $\psi$ and thus $\mathbf D^{-\frac12, \sigma}$ is a quite rough space. Moreover, it is also meaningful in 3d because the norm $\|\cdot\|_{\mathbf D^{-\frac12, \sigma}}$ becomes a scaling-critical space in view of the $L^2$-scaling critical structure of \eqref{main-eq}. (See \cite{choozxia}.) Our main result clarifies the above strategy. Once we keep the boundedness in a rough space $\mathbf D^{-\frac12,\sigma}$, we can control the time evolution property of the equation for large initial data.
\begin{thm}\label{large-gwp
Let $\sigma > 0$ for $d = 3$ and $\sigma = 0$ for $d = 2$. We let $\sigma>0$ be arbitrarily small positive number. Consider any maximal $L^{2, \sigma}$-solution
$$
\psi \in C_{\rm loc}\left(I^*;L^{2,\sigma}\left(\mathbb R^d,\mathbb C^{\tilde d}\right)\right),
$$
to the equation \eqref{main-eq} on $\mathbb R^{1+d}$. Suppose that the solutions satisfy the following boundedness condition:
$$
\sup_{t\in I^*}\|\psi(t)\|_{L^{2, \sigma}(\mathbb R^d)} <+\infty.
$$
If $\|\psi\|_{\mathbf D^{-\frac12, \sigma}(I^*)}<\infty$, then we have $I^*=\mathbb R$ and the solution $\psi$ scatters to a free solution in $L_x^{2, \sigma}$ as $t\rightarrow\pm\infty$.
\end{thm}
The proof follows from a refinement of the multilinear estimates by the previous works \cite{chohlee,cholee}. That is, we make a small quantity for $L^4_{t,x}$ norms on the right-hand side of trilinear estimates. The trilinear estimates in the low modulation regime are rather complicated than in the high modulation regime. When the modulation is lower than the highest input-frequency, we study bilinear forms $\varphi^\dagger\beta\phi$ and decompose it into the modulation localised terms. We deal with all possible frequency interactions such as High$\times$High to Low and Low$\times$High cases. The most delicate case in the proof occurs when the modulation is between the highest frequency and lowest frequency. Since we are only allowed to use $L^4_{t,x}$-Strichartz estimates in our approach, it gives rise to a slightly bigger bound than in other cases.
\subsection{Majorana condition}
Now we shall consider the problem of time evolution property of large dispersive solutions of the equation \eqref{main-eq} in the different way. Let us assume that the initial data $\psi_0$ satisfies
\begin{align}\label{majorana-condi}
\psi_0+z\gamma^2\psi^*_0=0
\end{align}
with $z=e^{i\omega}, \omega \in \mathbb R$. Here $z\gamma^2\psi^*$ is referred as a charge conjugation of Dirac field $\psi$. \eqref{majorana-condi} implies the Dirac field is its own antimatter field \cite{bj-dr}. Under this condition the cubic Dirac equations reduce to linear equation of $\psi$, since we have $\overline\psi\psi=0$, where $\overline\psi=\psi^\dagger\gamma^0$. For this see \cite{chagla, ozya}. Such a structural condition is first introduced by Ettore Majorana \cite{majorana}, and we refer this condition to the Majorana condition. We can readily obtain global existence and scattering for the equation \eqref{main-eq} within the Majorana condition for any initial data, since the corresponding solution will evolve \textit{linearly} in time.
The natural question is then whether the time evolution property may be stable by a small perturbation on the Majorana condition for a given large initial data. That is, we do not assume that $\psi_0+z\gamma^2\psi_0^*$ is not identically zero. To be precise, for any large initial data $\psi_0$ such that $\|\psi_0\|_{L^{2,\sigma}} = {\mathsf A} \gtrsim 1$, we make some perturbation on \eqref{majorana-condi} as follows:
\begin{align}
\|\psi_0+z\gamma^2\psi_0^*\|_{L^{2,\sigma}}\le\mathtt a,
\end{align}
for a relatively small $\mathtt a > 0$ which depends on the quantity ${\mathsf A}$. Thus we aim to prove the following theorem.
\begin{thm}\label{majorana-gwp}
Let $z=e^{i\omega}, \omega \in \mathbb R$. Let $\sigma > 0$ for $d= 3$ and $\sigma = 0$ for $d= 2$. For any ${\mathsf A} > 0$, there exists $\mathtt a = \mathtt a({\mathsf A}) > 0$ such that for all initial data satisfying
$$
\|\psi_0\|_{L^{2,\sigma}(\mathbb R^d)}\le {\mathsf A} \textrm{ and } \|\psi_0+z\gamma^2\psi_0^*\|_{L^{2,\sigma}(\mathbb R^d)} \le \mathtt a
$$
the Cauchy problem of \eqref{main-eq} on $\mathbb R^{1+d}$ is globally well-posed and solutions scatter to free solutions in $L^{2, \sigma}$ as $t\rightarrow\pm\infty$.
\end{thm}
\noindent However, the smallness condition on the form $\psi_0+z\gamma^2\psi_0^*$ yields large norm on the other side as follow:
\begin{align}
\|\psi_0 - z\gamma^2\psi_0^*\|_{L^{2,\sigma}}\le \mathsf {\mathsf A}.
\end{align}
Now we decompose the spinor field $\psi$ into
\begin{align}
\psi = \frac12(\psi+z\gamma^2\psi^*)+\frac12(\psi-z\gamma^2\psi^*)=: P_+^z\psi +P^z_-\psi.
\end{align}
Simple observation leads us that for $\theta\in\{+,-\}$, $P^z_\theta P^z_\theta=P^z_\theta$ and $P^z_\theta P^z_{-\theta}=0$ and hence the operator $P^z_\theta$ is truly the projection operator. Thanks to the perturbed Majorana condition, we do not need to control any rough space during iteration process as what we have done in Theorem \ref{large-gwp}. Instead, we will show that the time evolution property of $\psi(t)$ for the data $\psi_0$ is determined by the two evolutions; $P^z_\theta\psi$ and $P_{-\theta}^z\psi$. More precisely, instead of studying the Cauchy problems for \eqref{main-eq}, we are concerned with the Cauchy problems for a system of cubic Dirac equations as follows: for sufficiently smooth $\varphi, \phi$,
\begin{align}
\begin{aligned}
-i\gamma^\mu\partial_\mu\varphi+\varphi = V_b*(\overline{P^z_\theta\varphi}P^z_{-\theta}\phi+\overline{P^z_{-\theta}\phi}P^z_\theta\varphi)\varphi, \\
-i\gamma^\mu\partial_\mu\phi+\phi = V_b*(\overline{P^z_\theta\varphi}P^z_{-\theta}\phi+\overline{P^z_{-\theta}\phi}P^z_\theta\varphi)\phi.
\end{aligned}
\end{align}
with initial data
\begin{align}
\varphi|_{t=0} = P^z_\theta\psi_0,\quad \phi|_{t=0} = P^z_{-\theta}\psi_0.
\end{align}
Recall that $\overline\psi=\psi^\dagger\gamma^0$. By taking $P^z_\theta$ to both sides for the first equation and $P^z_{-\theta}$ for the second equation respectively, we would obtain solutions $P^z_\theta\varphi(t)$ and $P^z_{-\theta}\phi(t)$ to the system. Now we denote
$$
\psi(t) := P^z_\theta\varphi(t) + P^z_{-\theta}\phi(t).
$$
Then $\psi|_{t=0}=\psi_0$ and $\psi$ is truly the solution to the original equation.
The main scheme of exploiting the Majorana condition is as follows. Given a large data $\psi_0$, say, $\|\psi_0\|_{L^{2,\sigma}}={\mathsf A} \gtrsim 1$, we decompose the initial data with respect to the charge conjugation using the projections $P^z_\theta$. If the most of the norm is occupied in the one of the charge conjugation, say $P^z_{-\theta}\psi_0$, then the other one $P^z_\theta\psi_0$ must have small norm. Then via the standard iteration methods, we see that the time evolution of $P^z_\theta\psi_0$ and $P^z_{-\theta}\psi_0$ would be controlled by the norm $\|P^z_\theta\psi_0\|_{L^{2,\sigma}}$ and $\|P^z_{-\theta}\psi_0\|_{L^{2,\sigma}}$, respectively. In summary, the global-in-time evolution of $P^z_\theta\psi_0$ and $P^z_{-\theta}\psi_0$ is established in an open neighborhood of large data with critical regularity assumption.
\subsection*{Organisation} We organise the rest of this paper as follow. We give the proof of Theorem \ref{large-gwp} in Section \ref{sketch}. The proof relies heavily on the multilinear estimates Proposition \ref{main-tri-est}. We present preliminaries which will be used for the proof of Proposition \ref{main-tri-est} in Section \ref{pre}. Then Section \ref{sec-multi-3d} and Section \ref{multi-2d} are devoted to the proof of Proposition \ref{main-tri-est} in $d=3$ and $d=2$, respectively. Finally we prove Theorem \ref{majorana-gwp} in Section \ref{sec-maj}.
\subsection*{Notations}
\begin{enumerate}
\item
As usual different positive constants, which are independent of dyadic numbers $\mu,\lambda$, and $h$ are denoted by the same letter $C$, if not specified. The inequalities $A \lesssim B$ and $A \gtrsim B$ means that $A \le CB$ and
$A \ge C^{-1}B$, respectively for some $C>0$. By the notation $A \approx B$ we mean that $A \lesssim B$ and $A \gtrsim B$, i.e., $\frac1CB \le A\le CB $ for some absolute constant $C$. We also use the notation $A\ll B$ if $A\le \frac1CB$ for some large constant $C$. Thus for quantities $A$ and $B$, we can consider three cases: $A\approx B$, $A\ll B$ and $A\gg B$. In fact, $A\lesssim B$ means that $A\approx B$ or $A\ll B$.
The spatial and space-time Fourier transform are defined by
$$
\widehat{f}(\xi) = \int_{\mathbb R^d} e^{-ix\cdot\xi}f(x)\,dx, \quad \widetilde{u}(\tau,\xi) = \int_{\mathbb R^{1+d}}e^{-i(t\tau+x\cdot\xi)}u(t,x)\,dtdx.
$$
We also write $\mathcal F_x(f)=\widehat{f}$ and $\mathcal F_{t, x}(u)=\widetilde{u}$. We denote the backward and forward wave propagation of a function $f$ on $\mathbb R^d$ by
$$
e^{-\theta it \Lambda}f = \frac1{(2\pi)^d}\int_{\mathbb R^d}e^{ix\cdot\xi}e^{-\theta it\Lambda(\xi)}\widehat{f}(\xi)\,d\xi.
$$
\item
We fix a smooth function $\rho\in C^\infty_0(\mathbb R)$ such that $\rho$ is supported in the set $\{ \frac12<r<2\}$ and we let
$$
\sum_{\lambda\in2^{\mathbb Z}}\rho\left(\frac r\lambda\right) =1,
$$
and write $\rho_1=\sum_{\lambda\le1}\rho(\frac r\lambda)$ with $\rho_1(0)=1$. Now we define the standard Littlewood-Paley multipliers for $\lambda\in 2^{\mathbb N}$ and $\lambda>1$:
$$
P_\lambda = \rho\left(\frac{|-i\nabla|}{\lambda}\right),\quad P_1=\rho_1(|-i\nabla|).
$$
\item
We let $Y_{\ell}$ be the set of homogeneous harmonic polynomial of degree $\ell$ on $\mathbb R^3$. Then define $\{ y_{\ell,n} \}_{n=0}^{2\ell}$ a set of orthonormal basis for $Y_{\ell}$, with respect to the inner product:
\begin{align}
\langle y_{\ell,n},y_{\ell',n'}\rangle_{L^2_\omega(\mathbb S^2)} = \int_{\mathbb S^2}{y_{\ell,n}(\omega)} \overline{y_{\ell',n'}(\omega)}\,d\omega.
\end{align}
Given $f\in L^2_x(\mathbb R^3)$, we have the orthogonal decomposition as follow:
\begin{align}
f(x) = \sum_{\ell}\sum_{n=0}^{2\ell}\langle f(|x|\omega),y_{\ell,n}(\omega)\rangle_{L^2_\omega(\mathbb S^2)}y_{\ell,n}\big(\frac{x}{|x|}\big).
\end{align}
For a dyadic number $N>1$, we define the spherical Littlewood-Paley decompositions by
\begin{align}\begin{aligned}\label{hn}
H_N(f)(x) & = \sum_{\ell}\sum_{n=0}^{2\ell}\rho\left(\frac\ell N\right)\langle f(|x|\omega),y_{\ell,n}(\omega)\rangle_{L^2_\omega(\mathbb S^2)}y_{\ell,n}\big(\frac{x}{|x|}\big), \\
H_1(f)(x) & = \sum_{\ell}\sum_{n=0}^{2\ell}\rho_{\le1}(\ell)\langle f(|x|\omega),y_{\ell,n}(\omega)\rangle_{L^2_\omega(\mathbb S^2)}y_{\ell,n}\big(\frac{x}{|x|}\big).
\end{aligned}\end{align}
\end{enumerate}
\section{Proof of Theorem \ref{large-gwp}}\label{sketch}
We define our main function norm as
\begin{align*}
\|\phi\|_{F^{s,\sigma}_\theta(I)} &= \left( \sum_{\lambda,N\in 2^\mathbb N}\lambda^{2s}N^{2\sigma}\|\Pi_\theta P_\lambda H_N \phi\|_{V^2_\theta(I)}^2 \right)^\frac12, \\
\|\phi\|_{F^{s,0}_\theta(I)} &= \left( \sum_{\lambda\in 2^\mathbb N}\lambda^{2s}\|\Pi_\theta P_\lambda \phi\|_{V^2_\theta(I)}^2 \right)^\frac12,
\end{align*}
where $\theta \in \{+, -\}$. If $I = \mathbb R$, then we drop $I$. We say $\psi \in F^{s, \sigma}(I)$ if $\phi_\theta \in F_\theta^{s, \sigma}(I)$ for each $\theta \in \{+, -\}$. We write $$\|\phi\|_{F^{s, \sigma}(I)} = \sum_{\theta \in \{+,-\}}\|\phi_\theta\|_{F_\theta^{s, \sigma}(I)}.$$
\subsection{Local theory}
At the first step we present the local well-posedness result. To do so we give the explicit definition on the maximal solutions.
\begin{defn}
Let $s,\sigma\in\mathbb R$. If $d = 3$, then $\sigma > 0$, and if $d= 2$, then $\sigma = 0$. When $d = 3$, we define the angularly regular Sobolev spaces by $H^{s, \sigma} := \{f \in H^s : \|f\|_{H^{s, \sigma}} := \|\Lambda^s\Lambda_{\mathbb S^2}^\sigma f\|_{L_x^2} < \infty\}$.
\begin{enumerate}
\item We say $\psi:I\times\mathbb R^d\rightarrow\mathbb C^{\tilde d}$ is an $H^{s, \sigma}$-strong solution on an interval $I\subset\mathbb R$ if
$$
\psi\in C(I,H^{s, \sigma}(\mathbb R^d,\mathbb C^{\tilde d})
$$
and there exists a sequence $\psi_n\in C^2(I,H^m(\mathbb R^d,\mathbb C^{\tilde d}))$, $m=\max\{10,s\}$, of classical solutions to \eqref{main-eq} such that for any compact $I'\subset I$,
$$
\sup_{t\in I'}\|\psi(t)-\psi_n(t)\|_{H^{s, \sigma}(\mathbb R^d,\mathbb C^{\tilde d})} \rightarrow 0,
$$
as $n\rightarrow\infty$.
\item We say $\psi:[t_0,t^*)\times\mathbb R^d\rightarrow\mathbb C^{\tilde d}$ is a (forward) maximal $H^{s, \sigma}$-solution if the following two properties hold:
\begin{enumerate}
\item for any $t_1\in(t_0,t^*)$, $\psi$ is a strong $H^{s, \sigma}$-solution on $[t_0,t_1)$;
\item if $\psi':I'\times\mathbb R^d\rightarrow\mathbb C^{\tilde d}$ is a strong $H^{s, \sigma}$-solution on an interval $I$ satisfying $I\cap[t_0,t^*)\neq\varnothing$ and $\psi'=\psi$ on $I\cap[t_0,t^*)$, then $I\cap[t_0,\infty)\subset[t_0,t^*)$.
\end{enumerate}
\end{enumerate}
\end{defn}
Given $t_0\in I\subset\mathbb R$ and $F\in L^\infty_tL^2_x(I\times\mathbb R^d)$, for $t\in I$ and $\theta \in \{+, -\}$ let $\mathcal I_{\theta}[F](t_0; t)$ denote the inhomogeneous solution operator for the half-Klein-Gordon equation $(-i\partial_t + \theta \Lambda)\phi = \Pi_\theta F, \phi(t_0) = 0$. Then
$$
\mathcal I_\theta [F](t_0; t) = i\int_{t_0}^t e^{-\theta i(t-t')\Lambda}\Pi_\theta [F](t')\,dt'.
$$
The solution to \eqref{h-w} is written as
$$
\psi_\theta(t) = e^{-\theta i(t-t_0)\Lambda}\psi_\theta(t_0) + \mathcal I_\theta[f](t_0; t),
$$
where $f = [V_b*(\psi^\dagger\beta\psi)] \beta\psi$. We shall prove the following trilinear estimates in the next two sections.
\begin{prop}\label{main-tri-est}
Let $\sigma > 0$ when $d = 3$ and $\sigma = 0$ when $d = 2$. There exist constants $C > 0$ and $\eta<\frac\sigma{10}$ such that if $I\subset\mathbb R$ is a left-closed interval, $t_0\in I$, $\varphi,\phi,\psi\in F^{0,\sigma}(I)$, then for any $\theta \in \{+, -\}$, we have the bounds
\begin{align}\label{0-tri}
\begin{aligned}
\|\mathcal I_\theta [V_b*(\varphi^\dagger\beta\phi)\beta\psi](t_0; \cdot)\|_{F^{0,\sigma}(I)} & \le C (\|\varphi\|_{\mathbf D^{-\frac12, \sigma}(I)}\|\phi\|_{\mathbf D^{-\frac12, \sigma}(I)}\|\psi\|_{\mathbf D^{-\frac12, \sigma}(I)})^\eta \\
&\qquad\qquad \times (\|\varphi\|_{F^{0,\sigma}(I)}\|\phi\|_{F^{0,\sigma}(I)}\|\psi\|_{F^{0,\sigma}(I)})^{1-\eta},
\end{aligned}
\end{align}
and for any $s > 0$, we have the fractional Leibniz type bounds
\begin{align}\label{frac-tri}
\begin{aligned}
\|\mathcal I_{\theta}[V_b*(\varphi^\dagger\beta\phi)\beta\psi](t_0; \cdot)\|_{F^{s,\sigma}(I)} & \le 2^sC\big(\|\varphi\|_{\mathbf D^{-\frac12, \sigma}(I)}^\eta \|\varphi\|_{F^{0,\sigma}(I)}^{1-\eta}\|\phi\|_{F^{s,\sigma}(I)}\|\psi\|_{F^{s,\sigma}(I)} \\
& \quad + \|\varphi\|_{F^{s,\sigma}(I)}\|\phi\|_{\mathbf D^{-\frac12, \sigma}(I)}^\eta\|\phi\|_{F^{0,\sigma}(I)}^{1-\eta}\|\psi\|_{F^{s,\sigma}(I)} \\
& \quad + \|\varphi\|_{F^{s,\sigma}(I)}\|\phi\|_{F^{s,\sigma}(I)}\|\psi\|_{\mathbf D^{-\frac12,\sigma}(I)}^\eta \|\psi\|_{F^{0,\sigma}(I)}^{1-\eta}\big).
\end{aligned}
\end{align}
\end{prop}
\noindent In fact the multilinear estimates \eqref{frac-tri} follow from the estimate \eqref{0-tri}, and hence we only give the proof of \eqref{0-tri} in this paper.
By an application of Proposition \ref{main-tri-est} we are able to prove local well-posedness result.
\begin{prop}[Local well-posedness]\label{lwp}
Let $\sigma > 0$ for $d = 3$ and $\sigma =0$ for $d= 2$. Then there exist $0<\eta<\frac\sigma{10}$ and $C>1$, such that if
$$
{\mathsf A}>0,\quad 0 < \mathtt a < (C{\mathsf A}^{1-\eta})^{-\frac1\eta},
$$
and $I\subset\mathbb R$ is a left-closed interval, then for any initial time $t_0\in I$, and any data $\psi_0\in L^{2, \sigma}$ satisfying
$$
\|\psi_0\|_{L^{2, \sigma}} < {\mathsf A},\quad \sum_{\theta \in \{+, -\}}\|e^{-\theta i(t-t_0)\Lambda}\psi_{0, \theta}\|_{\mathbf D^{-\frac12, \sigma}(I)} < \mathtt a,
$$
there exists a unique $L^{2, \sigma}$-strong solution $\psi$ of \eqref{main-eq} on $I$ with $\psi(t_0)=\psi_0$. Moreover, the data-to-solution map is Lipschitz-continuous into $F^{0,\sigma}(I)$ and we have the bounds
$$
\bigg\|\psi - \sum_{\theta \in \{+, -\}}e^{-\theta i(t-t_0)\Lambda}\psi_{0, \theta}\bigg\|_{F^{0,\sigma}(I)} \le C({\mathtt a}^\eta {\mathsf A}^{1-\eta})^3.
$$
\end{prop}
\begin{proof}[Proof of Proposition \ref{lwp}]
Let us set
$$
\epsilon_0 = \mathtt a^\eta {\mathsf A}^{1-\eta}.
$$
For simplicity, we write
$$
\psi_{\ell} := \sum_\theta e^{-\theta i(t-t_0)\Lambda}\psi_{0, \theta},\quad \psi_{\mathcal N} := \psi - \psi_\ell.
$$
We define the set $\mathcal X\subset F^{0,\sigma}(I)$ as
$$
\mathcal X:=\{ \psi\in F^{0,\sigma}(I) : \|\psi_\mathcal N\|_{F^{0,\sigma}(I)} \le \epsilon_0\}.
$$
We aim to construct a fixed point of the map $\Phi:\mathcal X\rightarrow\mathcal X$ defined as
$$
\Phi(\psi) = \psi_\ell + \sum_\theta\mathcal I_{\theta}[V_b*(\psi^\dagger\beta\psi)\beta\psi](t_0; \cdot).
$$
To do this, let $\psi\in\mathcal X$. Then after decomposing the product
\begin{align*}
(\varphi^\dagger\phi)\psi & = \varphi^\dagger_\ell\phi_\ell\psi_\ell+\varphi_\mathcal N^\dagger \phi_\ell\psi_\ell +\varphi_\ell^\dagger\phi_\mathcal N\psi_\ell+\varphi_\mathcal N^\dagger\phi_\mathcal N\psi_\ell \\
&\qquad + \varphi^\dagger_\ell\phi_\ell\psi_\mathcal N+\varphi_\mathcal N^\dagger \phi_\ell\psi_\mathcal N +\varphi_\ell^\dagger\phi_\mathcal N\psi_\mathcal N+\varphi_\mathcal N^\dagger\phi_\mathcal N\psi_\mathcal N
\end{align*}
by an application of Proposition \ref{main-tri-est} together with bounds
\begin{align}
\|\psi_{\ell}\|_{F^{0,\sigma}(I)} & \lesssim \|\psi_0\|_{L^{2, \sigma}},\label{lin-est} \\
\|\psi\|_{L^\infty_tL^{2,\sigma}(I\times\mathbb R^3)} + \|\psi\|_{\mathbf D^{-\frac12, \sigma}} & \lesssim \|\psi\|_{F^{0,\sigma}(I)}, \label{energy-str-est}
\end{align}
we see that there exists $C^*>0$ such that
\begin{align}
\|\mathcal I_\theta [V_b*(\varphi^\dagger\beta\phi)\beta\psi](t_0; \cdot)\|_{F^{0,\sigma}(I)} \le C^* (\epsilon_0)^3.
\end{align}
To show that $\Phi$ is a contraction on $\mathcal X$, for $\psi,\psi'\in\mathcal X$, we apply Proposition \ref{main-tri-est} and \eqref{lin-est}, \eqref{energy-str-est} to obtain
\begin{align*}
&\|\mathcal I_{\theta}[V_b*(\varphi^\dagger\beta\phi)\beta\psi](t_0; \cdot) - \mathcal I_{\theta}[V_b*(\varphi'^\dagger\beta\phi')\beta\psi'](t_0; \cdot)\|_{F^{0,\sigma}(I)} \\
& \le C^* (\epsilon_0)^2(\|\varphi-\varphi'\|_{F^{0,\sigma}(I)}+\|\psi-\psi'\|_{F^{0,\sigma}(I)}+\|\phi-\phi'\|_{F^{0,\sigma}(I)}).
\end{align*}
Then we set
$$
3C^*(\epsilon_0)^2 \le \frac12.
$$
That is,
$$
\mathtt a^\eta {\mathsf A}^{1-\eta} \le (6C^*)^{-\frac12}.
$$
We conclude that the standard contraction principle implies that there exists a unique fixed point in $\mathcal X$, and the solution map $\Phi$ depends continuously on the initial datum. Now we set $\mathcal X_s\subset F^{s,\sigma}$ as
$$
\mathcal X_s := \{ \psi\in F^{s,\sigma}(I) : \|\psi\|_{F^{s,\sigma}(I)}\le \epsilon_s \},
$$
where
$$
\epsilon_s = \|\psi_0\|_{H^s(\mathbb R^d)}.
$$
By the multilinear estimates \eqref{frac-tri}, we have
\begin{align}
\|\mathcal I_{\theta}[V_b*(\varphi^\dagger\beta\phi)\beta\psi](t_0; \cdot)\|_{F^{s,\sigma}(I)} \le C^*(s)\epsilon_0^2\epsilon_s.
\end{align}
We repeat the similar argument to obtain a unique fixed point in $\mathcal X_s$ and then fix $$C=\sup_{0\le s\le10}C^*(s).$$ Suppose that $0<\mathtt a<(C\mathsf A^{1-\eta})^{-\frac1\eta}$. Then we get a unique solution $\psi\in\mathcal X$, which depends continuously on the data. Now we approximate the data with functions in $H^{10,\sigma}$ and applying the above argument with $s=10$, we obtain a sequence of solutions in $\mathcal X_{10}$, which converge to $\psi$. Thus $\psi$ is an $L^{2, \sigma}$-strong solution to \eqref{main-eq}.
\end{proof}
\subsection{Conditional regularity}
We shall prove that $L^{2, \sigma}$-solutions belong to $F^{0,\sigma}(I)$ provided that the $L^4_{t,x}$-norm is sufficiently small relative to the data norm $\|\psi(t_0)\|_{L^{2, \sigma}}$.
\begin{thm}\label{data-dis-thm}
Let $\sigma>0$ for $d = 3$ and $\sigma = 0$ for $d = 2$. Then there exists $0<\eta<\frac\sigma{10}$ and $C>1$ such that if ${\mathsf A}>0$ and $I\subset\mathbb R$ is a left-closed interval, $t_0\in I$, and $\psi$ is an $L^{2, \sigma}$-strong solution on $I$ satisfying
$$
\|\psi(t_0)\|_{L^{2, \sigma}} \le {\mathsf A}
$$
and
$$
\|\psi\|_{\mathbf D^{-\frac12, \sigma}(I)} \le (C(1+{\mathsf A})^{3-\eta})^{-\frac1\eta},
$$
then $\psi\in F^{0,\sigma}(I)$ and we have the bound
$$
\|\psi\|_{F^{0,\sigma}(I)} \le CA.
$$
\end{thm}
\begin{proof}[Proof of Theorem \ref{data-dis-thm}]
We first consider $I=[t_0,t_1)$ with $t_1\le\infty$. Let $\psi$ be an $L^{2, \sigma}$-solution on $I$ and we define
$$
\delta := \|\psi\|_{\mathbf D^{-\frac12, \sigma}}
$$
and
$$
\mathbf T := \left\{ t_0<T\le t_1 : \sup_{t_0<T'\le T}\|\psi\|_{F^{0,\sigma}([t_0,T'))} \le 2C^*{\mathsf A} \right\}.
$$
The local well-posedness result in Proposition \ref{lwp} implies that $T\in\mathbf T$ provided that $T-t_0$ is sufficiently small, i.e., $\mathbf T \neq\varnothing$. We let
$$
T_{\sup} := \sup\mathbf T.
$$
Our goal is to show that $T_{\sup} = t_1$. We assume that $T_{\sup}<t_1$ and let $T_n\in\mathbf T$ be a sequence converging to $T_{\sup}$. The continuity of the solution $\psi$ at $T_{\sup}$, together with \eqref{energy-str-est} and the definition of $\mathbf T$ implies that
\begin{align*}
\|\psi(T_{\sup})\|_{L^{2, \sigma}} & \le C^* \sup_{t_0<T<T_{\sup}}\|\psi\|_{F^{0,\sigma}([t_0,T))} \\
& \le (2C^*)^2{\mathsf A}
\end{align*}
Applying Proposition \ref{lwp} again, there exists $n$ and $\epsilon_0>0$ such that for all $0<\epsilon<\epsilon_0$ we have on the interval $[T_n,T_{\sup}+\epsilon)$ the bound
\begin{align*}
\|\psi\|_{F^{0,\sigma}([T_n,T_{\sup}+\epsilon))} & \le 2C^*\|\psi(T_{\sup})\|_{L^{2, \sigma}} \\
& \le (2C^*)^3{\mathsf A}.
\end{align*}
We now exploit the smallness assumption on the $L^4_{t,x}$ norm. An application of \eqref{energy-str-est}, triangular inequality with respect to the time intervals, together with the trilinear estimates in Proposition \ref{main-tri-est}, and the fact that $\psi$ is an $L^{2, \sigma}$-strong solution on $[t_0,t_1)$ implies that
\begin{align*}
\|\psi\|_{F^{0,\sigma}([t_0,T_{\sup}+\epsilon))} & \le C^*\|\psi(t_0)\|_{L^{2, \sigma}} \\
& + C^*\|\psi\|_{\mathbf D^{-\frac12, \sigma}([t_0,T_{\sup}+\epsilon))}^\eta (\|\psi\|_{F^{0,\sigma}([t_0,T_n))}+\|\psi\|_{F^{0,\sigma}([T_n,T_{\sup}+\epsilon))})^{1-\eta} \\
& \qquad\qquad\times ( \|\psi\|_{F^{0,\sigma}([t_0,T_n))}+\|\psi\|_{F^{0,\sigma}([T_n,T_{\sup}+\epsilon))} )^2 \\
& \le C^*{\mathsf A} + \delta^\eta (2C^*)^6{\mathsf A}^2{\mathsf A}^{1-\eta}.
\end{align*}
If we take
$$
\delta \le [ (2C^*)^5 {\mathsf A}^{3-\eta} ]^{-\frac1\eta},
$$
then
$$
\|\psi\|_{F^{0,\sigma}([t_0,T_{\sup}+\epsilon))} \le 2C^*{\mathsf A}.
$$
Consequently, we have $T_{\sup}+\epsilon\in\mathbf T$, which contradicts the assumption $T_{\sup}<t_1$. Therefore, we must have $T_{\sup}=t_1$.
\end{proof}
\subsection{Proof of global well-posedness and scattering}
Now we give the proof of Theorem \ref{large-gwp}. We only consider the forward-in-time problem for the system \eqref{main-eq}, since our main system is time reversible. We let $\psi:[t_0,t^*)\times\mathbb R^d\rightarrow\mathbb C^{\tilde d}$ be a forward maximal $L^{2, \sigma}$-solution to \eqref{main-eq} such that
$$
\sup_{t_0\le t<t^*}\|\psi(t)\|_{L^{2, \sigma}} \le {\mathsf A},\quad \|\psi\|_{\mathbf D^{-\frac12, \sigma}([t_0,t^*))} <\infty.
$$
Since the dispersive norm $\|\cdot\|_{\mathbf D^{-\frac12, \sigma}}$ is finite, by the dominated convergence theorem, we see that for every $\delta>0$, there exists an interval $I=[t_1,t^*)$ with $t_1<t^*$ such that
$$
\|\psi\|_{\mathbf D^{-\frac12, \sigma}(I)} \le\delta.
$$
In particular, choosing $\delta$ sufficiently small, depending only on ${\mathsf A}$, an application of Theorem \ref{data-dis-thm} implies that $\psi\in F^{0,\sigma}(I)$. Therefore, by the existence of left limits in $V^2$, there exists $\psi^\infty \in L^{2, \sigma}$ such that
$$
\lim_{t\rightarrow t^*}\|\psi(t) - \psi_\ell^\infty \|_{L^{2, \sigma}} = 0,
$$
where $\psi_\ell^\infty = \sum_{\theta \in \{+, -\}}e^{-\theta i(t-t_0)\Lambda}\Pi_\theta\psi^\infty$.
Then the local theory, together with the definition of maximal $L^{2, \sigma}$-solution implies $t^*=\infty$. In consequence, the solution $\psi$ exists globally in time and scatters as $t\rightarrow\infty$.
\section{Preliminaries for Proposition \ref{main-tri-est}}\label{pre}
This section is devoted to introducing preliminaries for Proposition \ref{main-tri-est}, which play a crucial role in the proof of our main result.
\subsection{Multipliers}\label{multi}
We define $\mathcal Q_\mu$ to be a finitely overlapping collection of cubes of diameter $\frac{\mu}{1000}$ covering $\mathbb R^d$, and let $\{ \rho_{\mathsf q}\}_{\mathsf q\in\mathcal Q_\mu}$ be a corresponding subordinate partition of unity.
For $\mathsf q\in\mathcal Q_\mu$, $h\in 2^{\mathbb Z}$ let
$$
P_{\mathsf q} = \rho_{\mathsf q}(-i\nabla),\quad C^{\theta}_h = \rho\left(\frac{|-i\partial_t + \theta\Lambda|}{h}\right).
$$
We define $C^\theta_{\le h}=\sum_{\delta\le h}C^\theta_\delta$ and $C^\theta_{\ge h}$ is defined in the similar way. For simplicity we also write $C^+_h = C_h$.
Given $0 < \mathtt r \lesssim1$, we define $\mathcal C_{\mathtt r}$ to be a collection of finitely overlapping caps of radius ${\mathtt r}$ on the sphere $\mathbb S^2$. If $\kappa\in\mathcal C_{\mathtt r}$, we let $\omega_\kappa$ be the centre of the cap $\kappa$. Then we define $\{\rho_\kappa\}_{\kappa\in\mathcal C_{\mathtt r}}$ to be a smooth partition of unity subordinate to the conic sectors $\{ \xi\neq0 , \frac{\xi}{|\xi|}\in\kappa \}$ and denote the angular localisation Fourier multipliers by
$
R_\kappa = \rho_\kappa(-i\nabla).
$
\subsection{Analysis on the sphere}\label{an-sph}
We introduce some basic facts from harmonic analysis on the unit sphere. The most of ingredients can be found in \cite{candyherr, ster}. We also refer the readers to \cite{steinweiss} for more systematic introduction to the spherical harmonics.
Since $-\Delta_{\mathbb S^2}y_{\ell, n} = \ell(\ell+1)y_{\ell, n}$, by orthogonality one can readily get
$$\|\Lambda_{\mathbb S^2}^\sigma f\|_{L^2_\omega({\mathbb S^2})} \approx \left\|\sum_{N\in2^{\mathbb N}\cup\{0\}}N^\sigma H_Nf\right\|_{L^2_\omega({\mathbb S^2})}.$$
\begin{lem}[Lemma 7.1 of \cite{candyherr}]\label{sph-ortho}
Let $N\ge1$. Then $H_N$ is uniformly bounded on $L^p(\mathbb R^3)$ in $N$, and $H_N$ commutes with all radial Fourier multipliers. Moreover, if $N'\ge1$, then either $N\approx N'$ or
$$
H_N\Pi_\theta H_{N'}=0,
$$
where
$
\Pi_{\theta} :=\frac{1}{2}\left(\mathbb I + \theta \Lambda^{-1}\Big[\alpha^x \cdot (-i\nabla) + \beta\Big]\right),
$
with $\theta\in\{+,-\}$.
\end{lem}
\subsection{Adapted function spaces}\label{ftn-sp}
Let $\mathcal Z =\left\{ \{t_k\}_{k=0}^K : t_k\in\mathbb R, t_k<t_{k+1} \right\}$ be the set of increasing sequences of real numbers.
We define the $2$-variation of $v$ to be
$$
|v|_{V^2} = \sup_{ \{t_k\}_{k=0}^K\in\mathcal I } \left( \sum_{k=0}^K\|v(t_k)-v(t_{k-1})\|_{L^2_x}^2 \right)^\frac12
$$
Then the Banach space $V^2$ can be defined to be all right continuous functions $v:\mathbb R\rightarrow L^2_x$ such that the quantity
$$
\|v\|_{V^2} = \|v\|_{L^\infty_tL^2_x} + |v|_{V^2}
$$
is finite. Set $\|u\|_{V^2_\theta}=\|e^{-\theta it\Lambda}u\|_{V^2}$. We recall basic properties of $V^2_\theta$ space from \cite{candyherr, candyherr1, haheko}. In particular, we use the following lemma to prove the scattering result.
\begin{lem}[Lemma 7.4 of \cite{candyherr}]\label{v-scatter}
Let $u\in V^2_\theta$. Then there exists $f\in L^2_x$ such that $\|u(t)-e^{-\theta it\Lambda}f\|_{L^2_x}\rightarrow0$ as $t\rightarrow\pm\infty$.
\end{lem}
The following lemma is on a simple bound in the high-modulation region.
\begin{lem}[Corollary 2.18 of \cite{haheko}]
Let $2\le q\le\infty$. For $h\in2^{\mathbb Z}$ and $\theta \in \{+, -\}$, we have
\begin{align}\label{bdd-high-mod}
\begin{aligned}
\|C^{\theta}_h u\|_{L^q_tL^2_x} \lesssim h^{-\frac1q}\|u\|_{V^2_\theta},\\
\|C^{\theta}_{\ge h}u\|_{L^q_tL^2_x} \lesssim h^{-\frac1q}\|u\|_{V^2_\theta}.
\end{aligned}
\end{align}
\end{lem}
We recall the uniform disposability of the modulation cutoff multipliers, which reads for $1\le q, p \le \infty$,
\begin{align}\label{uni-dis1}
\|C^\theta_{\le h}P_\lambda R_\kappa u\|_{L^q_tL^p_x} + \|C^\theta_h P_{\lambda}R_\kappa u\|_{L^q_tL^p_x} \lesssim \|P_\lambda R_\kappa u\|_{L^q_tL^p_x},
\end{align}
if $\kappa\in\mathcal C_{\mathtt r},\ h \gtrsim {\mathtt r}^2\lambda,$ and ${\mathtt r}\gtrsim\lambda^{-1}$.
Since convolution with $L^1_t(\mathbb R)$ functions is bounded on the $V^2$ space, we also have for every $h\in2^{\mathbb Z}$,
\begin{align}\label{uni-dis}
\|C^\theta_{\le h}u\|_{V^2_\theta} \lesssim \|u\|_{V^2_\theta}.
\end{align}
\begin{lem}[Lemma 7.3 of \cite{candyherr}]\label{lem-v-dual}
Let $F\in L^\infty_tL^2_x$, and suppose that
$$
\sup_{\|P_\lambda H_Nv\|_{V^2_\theta}\lesssim1}\left|\int_{\mathbb R} \langle P_\lambda H_Nv(t),F(t)\rangle_{L^2_x}\,dt \right| <\infty.
$$
If $u\in C(\mathbb R,L^2_x)$ satisfies $-(i\partial_t+\theta\Lambda) u=F$, then $P_\lambda H_Nu\in V^2_\theta$ and we have the bound
\begin{align}\label{v-dual}
\|P_\lambda H_Nu\|_{V^2_\theta} \lesssim \|P_\lambda H_Nu(0)\|_{L^2_x} + \sup_{\|P_\lambda H_Nv\|_{V^2_\theta}\lesssim1}\left|\int_{\mathbb R} \langle P_\lambda H_Nv(t),F(t)\rangle_{L^2_x}\,dt \right|.
\end{align}
\end{lem}
\subsection{Auxiliary estimates}
We present several estimates which will play a key role in the proof of our main result.
\begin{prop}[Lemma 3.5 of \cite{choozlee}]\label{stri-2d}
Let $(q,r)$ satisfy that $\frac1q+\frac1p=\frac12$. Then
$$
\|e^{\theta it\Lambda}P_\lambda f\|_{L^q_tL^p_x(\mathbb R^{1+2})} \lesssim \lambda^{\frac2q}\|P_\lambda f\|_{L^2_x(\mathbb R^2)}
$$
for all $\lambda\ge1$.
\end{prop}
\begin{prop}[Proposition 3.1, 3.2 of \cite{cholee}]
Let $P_{\lambda_j} \Pi_{\theta_j}\psi =: \psi_{\lambda_j}\in V^2_{\theta_j}$ for $\theta_j \in \{+, -\}$ and $j = 1,2$. Then we have
\begin{align}
\|P_{\lambda_0}(\psi_{\lambda_1}^\dagger\beta\psi_{\lambda_2})\|_{L^2_tL^2_x(\mathbb R^{1+2})} \lesssim \lambda_1^\eta \lambda_2^{1-\eta}\|\psi_{\lambda_1}\|_{V^2_{\theta_1}}\|\psi_{\lambda_2}\|_{V^2_{\theta_2}}, \label{bi-2d-lh}
\end{align}
for any $0<\eta<1$.
In particular, for High$\times$High interaction, i.e., $\lambda_0\lesssim\lambda_1\approx\lambda_2$, we have the following estimates:
\begin{align}
\|P_{\lambda_0}(\psi_{\lambda_1}^\dagger\beta\psi_{\lambda_2})\|_{L^2_tL^2_x(\mathbb R^{1+2})} & \lesssim \lambda_0^\frac12\left(\frac{\lambda_0}{\lambda_1}\right)^\frac12 \|\psi_{\lambda_1}\|_{V^2_{\theta_1}}\|\psi_{\lambda_2}\|_{V^2_{\theta_2}},\quad \theta_1=\theta_2,\label{bi-2d-hh} \\
\|P_{\lambda_0}(\psi_{\lambda_1}^\dagger\beta\psi_{\lambda_2})\|_{L^2_tL^2_x(\mathbb R^{1+2})} & \lesssim \lambda_0^\frac12 \|\psi_{\lambda_1}\|_{V^2_{\theta_1}}\|\psi_{\lambda_2}\|_{V^2_{\theta_2}},\quad \theta_1\neq\theta_2.
\end{align}
\end{prop}
\begin{prop}[Lemma 3.1 of \cite{behe}]
Let $2<q\le\infty$. If $0<\mu\le\lambda$, and $\frac1q+\frac1p=\frac12$, then for every $\mathsf q\in\mathcal Q_\mu$ we have
$$
\|e^{-\theta it\Lambda}P_{\mathsf q}P_\lambda f\|_{L^q_tL^p_x(\mathbb R^{1+3})} \lesssim (\mu\lambda)^\frac1q \|P_{\mathsf q}P_\lambda f\|_{L^2_x(\mathbb R^3)}.
$$
\end{prop}
\begin{lem}[Lemma 8.5 of \cite{candyherr}]\label{ang-con}
Let $2\le p<\infty$, and $0\le s<\frac2p$. If $\lambda,N\ge1$, ${\mathtt r}\gtrsim\lambda^{-1}$, and $\kappa\in\mathcal C_{\mathtt r}$, then we have
$$
\|R_\kappa P_\lambda H_N f\|_{L^p_x(\mathbb R^3)} \lesssim ({\mathtt r} N)^s \|P_\lambda H_N f\|_{L^p_x(\mathbb R^3)}.
$$
\end{lem}
\section{Multilinear estimates I: Three spatial dimensions}\label{sec-multi-3d}
In this section we shall prove Proposition \ref{main-tri-est} in $\mathbb R^{1+3}$.
\subsection{Trilinear estimates for $s=0,\,\sigma>0$}\label{multi-3d}
In this section we prove \eqref{0-tri}. By trivial extension we may assume that $I = \mathbb R$. Let $\Theta = (\theta, \lambda, N)$ and $$\psi_{\Theta} = P_{\lambda}H_{N} \psi_{\theta}.$$
Let $\Theta_j = (\theta_j, \lambda_j, N_j)$ for $j \in \{1, 2, 3, 4\}$.
By definition of $F^{0, \sigma}$ and $V_\theta^2$ and Lemma \ref{lem-v-dual} we write
\begin{align*}
&\|\mathcal I_{\theta}[V_b*(\varphi^\dagger\beta\phi)\beta\psi](t_0; \cdot)\|_{F^{0, \sigma}}^2\\
&\qquad = \sum_{\lambda_4,N_4\in2^\mathbb N}N_4^{2\sigma}\sum_{\theta' \in \{+, -\}}\|\Pi_{\theta'} P_{\lambda_4} H_{N_4} \mathcal I_{\theta}[V_b*(\varphi^\dagger\beta\phi)\beta\psi](t_0; \cdot)\|_{V^2_{\theta'}}^2 \\
&\qquad = \sum_{\lambda_4,N_4\in2^\mathbb N}N_4^{2\sigma}\|\Pi_{\theta} P_{\lambda_4} H_{N_4} \mathcal I_\theta[V_b*(\varphi^\dagger\beta\phi)\beta\psi](t_0; \cdot)\|_{V^2_\theta}^2 \\
&\qquad = \sum_{\lambda_4,N_4\ge1}N_4^{2\sigma}\sup_{\|\uppsi_{\Theta_4}\|_{V^2_\theta\le1}}\left|\int_{\mathbb R^{1+3}} [V_b*(\varphi^\dagger\beta\phi)] (\uppsi_{\Theta_4}^\dagger\beta\psi )\,dtdx \right|^2 \\
&\qquad \le \sum_{\lambda_4,N_4\ge1}N_4^{2\sigma}\left(\sum_{\Theta_j, j=1,2,3}\sup_{\|\uppsi_{\Theta_4}\|_{V^2_{\theta_4}}\lesssim1}\left|\int_{\mathbb R^{1+3}} \Lambda^{-2}(\varphi^\dagger_{\Theta_1}\beta\phi_{\Theta_2})(\uppsi_{\Theta_4}^\dagger \beta\psi_{\Theta_3})\,dtdx \right| \right)^2 \\
&\qquad=: \mathfrak I_{\lambda,N}.
\end{align*}
We shall decompose $\mathfrak I_{\lambda,N}$ into the modulation $h$, which is the distance to the light cone in the space-time Fourier variables. Then we consider the high frequency and the low frequency cases, compared to the size of the modulation. That is, we are concerned with the following frequency interactions for $\{j,k\}=\{1,2\}$ or $\{3,4\}$,
$$
\min\{\lambda_0,\lambda_j,\lambda_k\} \lesssim \textrm{med}\{\lambda_0,\lambda_j,\lambda_k\} \approx\max\{\lambda_0,\lambda_j,\lambda_k\},
$$
together with
$$
h \gtrsim \lambda_{\max},\quad h\ll \lambda_{\max},
$$
where $\lambda_{\max}=\max\{\lambda_1,\lambda_2,\lambda_3,\lambda_4\}$. We shall write
\begin{align*}
\mathfrak I_{\lambda,N}& = \textrm{(Low frequency)}+\textrm{(High frequency)} \\
&=: \mathfrak I_{h\gtrsim\lambda}+\mathfrak I_{h\ll\lambda}.
\end{align*}
To estimate the low frequency part $\mathfrak I_{h\gtrsim\lambda}$,
we write
\begin{align*}
\mathcal I_{\lambda,N} &:= \left|\int_{\mathbb R^{1+3}} \Lambda^{-2}\Big( \left[ C_{\ge h}^{\theta_1}\varphi_{\Theta_1}\right]^\dagger\beta\phi_{\Theta_2}\Big)(\uppsi_{\Theta_4}^\dagger\beta\psi_{\Theta_3})\,dtdx\right|.
\end{align*}
Then for small $0<\eta\ll1$, we have
\begin{align*}
&\sum_{h \gtrsim\lambda_{\max}}\mathcal I_{\lambda,N} \\
&= \sum_{h \gtrsim\lambda_{\max}}(\mathcal I_{\lambda,N})^\eta (\mathcal I_{\lambda,N})^{1-\eta} \\
& \lesssim \left(\frac{\lambda_0}{\lambda_{\max}}\right)^{\frac12-3\eta}\left(\lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_{t,x}}\|\phi_{\Theta_2}\|_{L^4_{t,x}}\|\psi_{\Theta_3}\|_{L^4_{t,x}}\right)^\eta \\
&\qquad\qquad \times\|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\|\psi_{\Theta_3}\|_{V^2_{\theta_3}}\|\uppsi_{\Theta_4}\|_{V^2_{\theta_4}}.
\end{align*}
Here the estimate
$$
(\mathcal I_{\lambda,N})^\eta\lesssim \left(\frac{\lambda_0}{\lambda_{\max}} \right)^{-2\eta}\left(\lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12}\lambda_4^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_{t,x}}\|\phi_{\Theta_2}\|_{L^4_{t,x}}\|\psi_{\Theta_3}\|_{L^4_{t,x}}\|\uppsi_{\lambda_4,N_4}\|_{L^4_{t,x}}\right)^\eta
$$
follows from H\"older's inequality and $L^4$-Strichartz estimates and the estimate of $(\mathcal I_{\lambda,N})^{1-\eta}$ follows from \cite{chohlee}. This completes the estimate of $\mathfrak I_{h\gtrsim\lambda}$.
On the other hand, the estimate of $\mathfrak I_{h\ll\lambda}$ requires more work. In this case the required estimates \eqref{0-tri} follows from the following $L^2$-bilinear estimates for $\frac1{10}<\delta<\frac12$ and $\eta<\frac\sigma{10}$,
\begin{align}\label{bi1}
\begin{aligned}
\|P_{\lambda_0}H_{N_0}(\varphi^\dagger_{\Theta_1}\beta\phi_{\Theta_2})\|_{L^2_tL^2_x} & \lesssim (\lambda_1^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x}\lambda_2^{-\frac12}\|\phi_{\Theta_2}\|_{L^4_tL^4_x} )^{\eta} \\
& \quad\qquad\times \lambda_0\left(\frac{\lambda_{\min}}{\lambda_{\max}}\right)^\delta (N_{\min})^\sigma \left(\|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\right)^{1-\eta}.
\end{aligned}
\end{align}
Indeed, by \eqref{bi1} we have
\begin{align*}
&\|\mathcal I_{\theta}[V_b*(\varphi^\dagger\beta\phi)\beta\psi](t_0; \cdot)\|_{F^{0, \sigma}}^2\\
&\qquad \lesssim \sum_{\lambda_4,N_4}N_4^{2\sigma}\bigg( \sum_{\Theta_j, j=1,2,3}(\lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12} \|\varphi_{\Theta_1}\|_{L^4_{t,x}}\|\phi_{\Theta_2}\|_{L^4_{t,x}}\|\psi_{\Theta_3}\|_{L^4_{t,x}})^\eta \\
& \qquad\qquad \times \left(\frac{\lambda_{\min}\lambda_{\min}'}{\lambda_{\max}^{12}\lambda_{\max}^{34}}\right)^\delta (N_{\min}N_{\min}')^\sigma (\|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\|\psi_{\Theta_3}\|_{V^2_{\theta_3}})^{1-\eta} \bigg)^2 \\
&\qquad =: \mathfrak S(\varphi,\phi,\psi).
\end{align*}
Even if the square summation seems quite complicated, it would be straightforward with a simple observation: By H\"older inequality in $\ell^p$-space, we get
\begin{align*}
&\sum_{\lambda,N} \left[(\lambda^{-\frac12}N^\sigma\|\varphi_{\Theta}\|_{L^4_{t,x}})^\eta (N^\sigma \|\varphi_{\Theta}\|_{V^2_{\theta}})^{1-\eta}\right]^2\\
&\qquad\lesssim \left\|(\lambda^{-\frac12}N^\sigma\|\varphi_{\Theta}\|_{L^4_{t,x}})\right\|_{\ell^{2}_{\lambda, N}}^\eta \left\|(N^\sigma \|\varphi_{\Theta}\|_{V^2_{\theta}})\right\|_{\ell^{2}_{\lambda, N}}^{1-\eta}.
\end{align*}
Therefore, in order to prove our first main theorem, it suffices to prove the estimates \eqref{bi1}. At the very beginning, we further decompose the frequency-localised bilinear form $\varphi^\dagger_{\lambda_1,N_1}\beta\phi_{\lambda_2,N_2}$ into the modulation localised form:
$$
P_{\lambda_0}H_{N_0}(\varphi^\dagger_{\Theta_1}\beta\phi_{\Theta_2} ) = \sum_{h\in2^\mathbb Z}(\mathcal A_0(h) + \mathcal A_1(h) + \mathcal A_2(h)),
$$
where
\begin{align*}
\mathcal A_0(h) & = C_hP_{\lambda_0}H_{N_0}(C^{\theta_1}_{\le h}\varphi_{\Theta_1})^\dagger\beta C^{\theta_2}_{\le h}\phi_{\Theta_2},\\
\mathcal A_1(h) & = C_{\le h}P_{\lambda_0}H_{N_0}(C^{\theta_1}_{ h}\varphi_{\Theta_1})^\dagger\beta C^{\theta_2}_{\le h}\phi_{\Theta_2},\\
\mathcal A_2(h) & = C_{\le h}P_{\lambda_0}H_{N_0}(C^{\theta_1}_{\le h}\varphi_{\Theta_1})^\dagger\beta C^{\theta_2}_{ h}\phi_{\Theta_2} .
\end{align*}
We first consider the frequency relation in a low modulation regime, i.e.,
$$
\lambda_0\lesssim\lambda_1\approx\lambda_2, \ h \lesssim \lambda_0,\ \textrm{and}\ \lambda_1\lesssim\lambda_0\approx\lambda_2,\ h\lesssim\lambda_1.
$$
We observe that the square-summation over caps and cubes gives that
\begin{align}\label{phi-1}
\begin{aligned}
&\left( \sum_{\mathsf q\in\mathcal Q_{\lambda_0}}\sum_{\kappa\in\mathcal C_{\mathtt r}}\|P_{\mathsf q}R_\kappa\varphi_{\Theta_1}\|_{L^4_tL^4_x}^2 \right)^\frac12 \\
&\qquad\qquad\qquad \lesssim {\mathtt r}^{-2\eta}\left(\frac{\lambda_0}{\lambda_1}\right)^{-2\eta}(\lambda_0\lambda_1)^\frac14\left(\lambda_1^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x}\right)^\eta \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}^{1-\eta}.
\end{aligned}
\end{align}
In particular, we have
\begin{align}\label{phi-2}
\begin{aligned}
\left( \sum_{\mathsf q\in\mathcal Q_{\lambda_0}}\sum_{\kappa\in\mathcal C_{\mathtt r}}\|P_{\mathsf q}R_\kappa\varphi_{\Theta_1}\|_{L^4_tL^4_x}^2 \right)^\frac12 \lesssim {\mathtt r}^{-\frac\epsilon2}\left(\frac{\lambda_0}{\lambda_1}\right)^{-\frac\epsilon2}(\lambda_0\lambda_1)^\frac14 \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}.
\end{aligned}
\end{align}
The estimates of bilinear form $\varphi_{\Theta_1}^\dagger\beta\,\phi_{\Theta_2}$ in the low modulation regime is very similar to \cite{chohlee}. The only difference from our previous paper is that we apply \eqref{phi-1} instead of \eqref{phi-2}. To avoid unnecessary duplication we briefly present the proof of the bilinear form in this case. (See also Theorem 5.1 of \cite{chohlee}.) First we put ${\mathtt r}=(\frac{h\lambda_0}{\lambda_1\lambda_2})^\frac12$ and ${\mathtt r}^*=(\frac{h}{\lambda_0})^\frac12$.
For $\lambda_0\lesssim\lambda_1\approx\lambda_2$ with $h \lesssim \lambda_0$, we have
\begin{align*}
\|\mathcal A_0(h)\|_{L^2_tL^2_x} & \lesssim {\mathtt r}^{\sigma-4\eta}(h\lambda_0)^\frac12\left(\frac{\lambda_0}{\lambda_1}\right)^{\frac12-4\eta}(N_{\min}^{12})^\sigma\left(\lambda_1^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x}\lambda_2^{-\frac12}\|\phi_{\Theta_2}\|_{L^4_tL^4_x}\right)^\eta \\
& \qquad\qquad\times \left(\|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\right)^{1-\eta}.
\end{align*}
We also have
\begin{align*}
\|\mathcal A_0(h)\|_{L^2_tL^2_x} & = \sup_{\|\psi\|_{L^2_tL^2_x}\le 1}\left|\int C_h\psi_{\lambda_0,N_0} (C^{\theta_1}_{\le h}\varphi_{\Theta1})^\dagger\beta(C^{\theta_2}_{\le h}\phi_{\Theta_2})\,dtdx\right| \\
& \lesssim \sup_{\|\psi\|_{L^2_tL^2_x}\le 1}\sum_{\substack{\kappa,\,\kappa'\in\mathcal C_{\mathtt r} \\ |\theta_1\kappa-\theta_2\kappa'|\lesssim{\mathtt r}}}\sum_{\substack{ \kappa''\in\mathcal C_{{\mathtt r}^*} \\ |\theta_1\kappa+\kappa''|\lesssim{\mathtt r}^* }}\sum_{\substack{\mathsf q,\, \mathsf q'\in\mathcal Q_{\lambda_0} \\ |\theta_1\mathsf q-
\theta_2\mathsf q'|\lesssim\lambda_0}} \\
& \qquad\qquad\left|\int_{\mathbb R^{1+3}} R_{\kappa''}C_h \psi_{\lambda_0,N_0} (R_\kappa P_{\mathsf q}C^{\theta_1}_{\le h}\varphi_{\Theta_1})^\dagger\beta(R_{\kappa'}P_{\mathsf q'}C^{\theta_2}_{\le h}\phi_{\Theta_2})\,dtdx\right| \\
& \lesssim {\mathtt r}^{1-2\eta}({\mathtt r}^* N_0)^\sigma (\lambda_0\lambda_1)^\frac12\left(\frac{\lambda_0}{\lambda_1}\right)^{-4\eta} \left(\lambda_1^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x}\lambda_2^{-\frac12}\|\phi_{\Theta_2}\|_{L^4_tL^4_x}\right)^\eta \\
& \qquad\qquad\times \left(\|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\right)^{1-\eta} \\
& \lesssim (\mathtt r^*)^{\sigma-4\eta}(h\lambda_0)^\frac12 \left(\frac{\lambda_0}{\lambda_1}\right)^{\frac12-6\eta}(N_0)^\sigma\left(\lambda_1^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x}\lambda_2^{-\frac12}\|\phi_{\Theta_2}\|_{L^4_tL^4_x}\right)^\eta \\
& \qquad\qquad\times \left(\|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\right)^{1-\eta}.
\end{align*}
In this section $\kappa, \kappa', q, q'$ in $|\theta \kappa + \theta'\kappa'|$ and $|\theta q + \theta' q'|$ denote the center points of the corresponding caps $\kappa, \kappa'$ and cubes $\mathsf q, \mathsf q'$. The summation on $h\lesssim\lambda_0$ gives the required bound.
We can estimate $\mathcal A_1(h), \mathcal A_2(h)$ in a similar way. We omit the details.
For $\lambda_1\lesssim\lambda_0\approx\lambda_2$, with $h\lesssim\lambda_1$, we have
\begin{align*}
\|\mathcal A_0(h)\|_{L^2_tL^2_x} & = \sup_{\|\psi\|_{L^2_tL^2_x}\lesssim1}\left|\int C_h\psi_{\lambda_0,N_0} (C^{\theta_1}_{\le h}\varphi_{\Theta_1})^\dagger\beta(C^{\theta_2}_{\le h}\phi_{\Theta_2})\,dtdx\right| \\
& \lesssim \sup_{\|\psi\|_{L^2_tL^2_x\lesssim1}}\sum_{\substack{\kappa,\,\kappa',\,\kappa''\in\mathcal C_{{\mathtt r}_{*}} \\ |\theta_1\kappa+\theta_2\kappa'|,|\kappa''+\theta_2\kappa'|\lesssim{\mathtt r}_{*}}}\sum_{\substack{\mathsf q',\, \mathsf q''\in\mathcal Q_{\lambda_1} \\ |\mathsf q' + \mathsf q''|\lesssim\lambda_1}}\\
& \qquad\qquad\left|\int R_{\kappa''}P_{\mathsf q''}C_h\psi_{\lambda_0,N_0} (R_\kappa C^{\theta_1}_{\le h}\varphi_{\Theta_1})^\dagger\beta(R_{\kappa'}P_{\mathsf q'}C^{\theta_2}_{\le h}\phi_{\Theta_2})\,dtdx\right| \\
& \lesssim \sup_{\|\psi\|_{L^2_tL^2_x\lesssim1}}\sum_{\substack{\kappa,\,\kappa',\,\kappa''\in\mathcal C_{{\mathtt r}_*} \\ |\theta_1\kappa+\theta_2\kappa'|,|\kappa''+\theta_2\kappa'|\lesssim {\mathtt r}_*}}\sum_{\substack{\mathsf q', \,\mathsf q''\in\mathcal Q_{\lambda_1} \\ |\mathsf q' + \mathsf q''|\lesssim\lambda_1}} \\
&\qquad\qquad {\mathtt r}_*\|R_{\kappa''}P_{\mathsf q''}C_h\psi_{\lambda_0,N_0}\|_{L^2_tL^2_x}\|R_\kappa C^{\theta_1}_{\le h}\varphi_{\Theta_1}\|_{L^4_tL^4_x}\|R_{\kappa'}P_{\mathsf q'}C_{\le h}^{\theta_2}\phi_{\Theta_2}\|_{L^4_tL^4_x} \\
& \lesssim ({\mathtt r}_*)^{1-2\eta}\left(\frac{\lambda_1}{\lambda_2} \right)^{-2\eta}\lambda_1^\frac12(\lambda_1\lambda_2)^\frac14({\mathtt r}_* N_{\min}^{012})^\sigma\left( \lambda_1^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x}\lambda_2^{-\frac12}\|\phi_{\Theta_2}\|_{L^4_tL^4_x} \right)^\eta \\
&\qquad\qquad\qquad\qquad\times \left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}} \right)^{1-\eta} \\
& \lesssim (\mathtt r_*)^{\sigma-2\eta}h_1^\frac12\lambda_1^\frac14\lambda_2^\frac14 \left(\frac{\lambda_1}{\lambda_2}\right)^{-2\eta}(N_{\min}^{012})^\sigma\left( \lambda_1^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x}\lambda_2^{-\frac12}\|\phi_{\Theta_2}\|_{L^4_tL^4_x} \right)^\eta \\
&\qquad\qquad\qquad\qquad\times \left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}} \right)^{1-\eta} ,
\end{align*}
where we chose ${\mathtt r}_*=(\frac{h}{\lambda_1})^\frac12$. The summation with respect to the modulation $h\lesssim\lambda_1$ yields the required estimates. As the case $\lambda_0\lesssim\lambda_1\approx\lambda_2$, we can treat the terms $\mathcal A_1(h)$ and $\mathcal A_2(h)$ in a similar manner. We omit the details.
\subsection{High modulation regime}
In this subsection we shall estimate the bilinear form $\varphi_{\Theta_1}^\dagger\beta\phi_{\Theta_2}$ in the regime: $\lambda_{\min}\ll h\ll\lambda_{\max}$.
We first consider $\lambda_0\lesssim\lambda_1\approx\lambda_2$ with $\lambda_0\ll h \ll\lambda_1$.
We observe that the angle between the diameter of support of $\widehat{\varphi}$ and $\widehat \phi$ is less than $\left(\frac{\lambda_0}{\lambda_1}\right)^\frac12$. We then decompose $\mathcal A_0(h)$ into the following:
\begin{align}
\begin{aligned}
\|\mathcal A_0(h)\|_{L^2_{t,x}} & \lesssim \|C_h P_{\lambda_0}H_{N_0}(C^{\theta_1}_{\approx h}\varphi_{\Theta_1})^\dagger\beta(C^{\theta_2}_{\ll h}\phi_{\Theta_2})\|_{L^2_{t,x}} \\
&\qquad\qquad + \|C_h P_{\lambda_0}H_{N_0}(C^{\theta_1}_{\ll h}\varphi_{\Theta_1})^\dagger\beta(C^{\theta_2}_{\approx h}\phi_{\Theta_2})\|_{L^2_{t,x}} \\
&=: A_{0,1}+A_{0,2}.
\end{aligned}
\end{align}
It is enough to estimate the $A_{0,1}$ term because of symmetry. By almost orthogonal decomposition of small cubes of size $\lambda_0$ we see that
\begin{align*}
A_{0,1}
& \lesssim \left(\frac{\lambda_0}{\lambda_1}\right)^\frac12\sum_{\substack{\mathsf q, \mathsf q'\in\mathcal Q_{\lambda_0} \\ |\mathsf q - \mathsf q'|\lesssim\lambda_0}}\|P_{\mathsf q}C^{\theta_1}_{\approx h}\varphi_{\Theta_1}\|_{L^4_{t,x}}\|P_{\mathsf q'}C^{\theta_2}_{\ll h}\phi_{\Theta_2}\|_{L^4_{t,x}} \\
& \lesssim \left(\frac{\lambda_0}{\lambda_1}\right)^\frac12 \bigg(\sum_{\mathsf q'}\|P_{\mathsf q'}C^{\theta_2}_{\ll h}\phi_{\Theta_2}\|_{L^4_{t,x}}^2 \bigg)^\frac12 \bigg( \sum_{\mathsf q'} \bigg(\sum_{\mathsf q:|\mathsf q - \mathsf q'|\lesssim\lambda_0}\|P_{\mathsf q}C^{\theta_1}_{\approx h}\varphi_{\Theta_1}\|_{L^4_{t,x}}\bigg)^2 \bigg)^\frac12 \\
& \lesssim \left(\frac{\lambda_0}{\lambda_1}\right)^\frac12 \bigg(\sum_{\mathsf q'}\|P_{\mathsf q'}C^{\theta_2}_{\ll h}\phi_{\Theta_2}\|_{L^4_{t,x}}^2 \bigg)^\frac12 \bigg(\sum_{\mathsf q}\|P_{\mathsf q}C^{\theta_1}_{\approx h}\varphi_{\Theta_1}\|_{L^4_{t,x}}^2 \bigg)^\frac12.
\end{align*}
For the square-summation of $\phi_{\Theta_2}$ in the third inequality we apply similar form of \eqref{phi-1} to obtain
\begin{align}\label{square-sum-q}
\bigg(\sum_{\mathsf q'}\|P_{{\mathsf q}'}C^{\theta_2}_{\ll h}\phi_{\Theta_2}\|_{L^4_{t,x}}^2 \bigg)^\frac12 & \lesssim \left(\frac{\lambda_0}{\lambda_2}\right)^{-2\eta}(\lambda_0\lambda_2)^\frac14 \left(\lambda_2^{-\frac12} \|\phi_{\Theta_2}\|_{L^4_{t,x}}\right)^\eta \|\phi_{\Theta_2}\|_{V^2_{\theta_2}}^{1-\eta}
\end{align}
and for the summation of $\varphi_{\Theta_1}$ we write
\begin{align}
\bigg(\sum_{{\mathsf q}}&\|P_{{\mathsf q}}C^{\theta_1}_{\approx h}\varphi_{\Theta_1}\|_{L^4_{t,x}}^2 \bigg)^\frac12\nonumber\\
& \lesssim \left( \sum_{{\mathsf q}}\left(\|P_{{\mathsf q}}C^{\theta_1}_{\approx h}\varphi_{\Theta_1}\|_{L^4_{t,x}}^\eta \|P_{{\mathsf q}}C^{\theta_1}_{\approx h}\varphi_{\Theta_1}\|_{L^4_{t,x}}^{1-\eta} \right)^2 \right)^\frac12 \nonumber\\
& \lesssim \left(\sum_{\mathsf q}\|P_{{\mathsf q}}C^{\theta_1}_{\approx h}\varphi_{\Theta_1}\|_{L^4_{t,x}}^2\right)^\frac\eta2 \left(\sum_{\mathsf q} \|P_{{\mathsf q}}C^{\theta_1}_{\approx h}\varphi_{\Theta_1}\|_{L^4_{t,x}}^2 \right)^{\frac{1-\eta}{2}}\nonumber \\
& \lesssim \left(\sum_{\mathsf q}\|P_{{\mathsf q}}C^{\theta_1}_{\approx h}\varphi_{\Theta_1}\|_{L^4_{t,x}}^2\right)^\frac\eta2 \lambda_0^{\frac34(1-\eta)} \left(\sum_{\mathsf q} \|P_{{\mathsf q}}C^{\theta_1}_{\approx h}\varphi_{\Theta_1}\|_{L^4_{t}L^2_x}^2 \right)^{\frac{1-\eta}{2}} \nonumber\\
& \lesssim \left(\sum_{\mathsf q}\|P_{{\mathsf q}}C^{\theta_1}_{\approx h}\varphi_{\Theta_1}\|_{L^4_{t,x}}^2\right)^\frac\eta2 \lambda_0^{\frac34(1-\eta)}h^{-\frac14(1-\eta)}\|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}^{1-\eta} \nonumber\\
& \lesssim \lambda_1^\frac\eta2\left(\frac{\lambda_0}{\lambda_1}\right)^{-2\eta}\left(\frac{\lambda_0}{h}\right)^{\frac14(1-\eta)}\lambda_0^{\frac12(1-\eta)}\left(\lambda_1^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x}\right)^\eta \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}^{1-\eta}.\label{high-mod-eta}
\end{align}
Here we used the Bernstein inequality, and the bound of high modulation regime \eqref{bdd-high-mod} and then \eqref{square-sum-q}.
Thus we get
\begin{align*}
\sum_{\lambda_0\ll h\ll\lambda_1}A_{0,1} & \lesssim \lambda_0\left( \frac{\lambda_0}{\lambda_1}\right)^{\frac14-5\eta} \left( \lambda_1^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x}\lambda_2^{-\frac12}\|\phi_{\Theta_2}\|_{L^4_tL^4_x} \right)^\eta \\
&\qquad\qquad\qquad\qquad\qquad\times \left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}} \right)^{1-\eta}.
\end{align*}
For $\mathcal A_1(h)$, we have the similar decomposition as follows
\begin{align*}
\|\mathcal A_1(h)\|_{L^2_tL^2_x} & \lesssim \|C_{\approx h}P_{\lambda_0}H_{N_0}(C^{\theta_1}_{ h}\varphi_{\Theta_1})^\dagger\beta C^{\theta_2}_{\ll h}\phi_{\Theta_2}\|_{L^2_tL^2_x} \\
& \qquad\qquad + \|C_{\ll h}P_{\lambda_0}H_{N_0}(C^{\theta_1}_{ h}\varphi_{\Theta_1})^\dagger\beta C^{\theta_2}_{\approx h}\phi_{\Theta_2}\|_{L^2_tL^2_x} \\
&=: A_{1,1}+A_{1,2}.
\end{align*}
The term $A_{1,1}$ can be treated in the identical manner as $A_{0,1}$. For $A_{1,2}$, we write
\begin{align*}
A_{1,2} & \lesssim \left(\frac{\lambda_0}{\lambda_1}\right)^\frac12\sum_{\substack{{\mathsf q},\,{\mathsf q}'\in\mathcal Q_{\lambda_0} \\ |{\mathsf q}-{\mathsf q}'|\lesssim\lambda_0}} \|C_{\ll h}P_{\lambda_0}H_{N_0}(P_qC^{\theta_1}_{ h}\varphi_{\Theta_1})^\dagger\beta P_{{\mathsf q}'}C^{\theta_2}_{\approx h}\phi_{\Theta_2}\|_{L^2_tL^2_x} \\
& \lesssim \left(\frac{\lambda_0}{\lambda_1}\right)^\frac12 \left(\sum_{\mathsf q}\|P_qC^{\theta_1}_h\varphi_{\Theta_1}\|_{L^4_tL^4_x}^2 \right)^\frac12 \left( \sum_{{\mathsf q}'}\|P_{{\mathsf q}'}C^{\theta_2}_{\approx h}\phi_{\Theta_2}\|_{L^4_tL^4_x}^2 \right)^\frac12.
\end{align*}
We use \eqref{high-mod-eta} to obtain
\begin{align*}
A_{1,2} & \lesssim \left( \frac{\lambda_0}{\lambda_1} \right)^\frac12 \lambda_1^\eta \left(\frac{\lambda_0}{\lambda_1}\right)^{-4\eta}\left( \frac{\lambda_0}{h}\right)^{\frac12(1-\eta)}\lambda_0^{1-\eta} \left( \lambda_1^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x}\lambda_2^{-\frac12}\|\phi_{\Theta_2}\|_{L^4_tL^4_x} \right)^\eta \\
&\qquad\qquad\qquad\qquad\times \left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}} \right)^{1-\eta} \\
& \lesssim \lambda_0\left( \frac{\lambda_0}{h}\right)^{\frac12(1-\eta)}\left(\frac{\lambda_0}{\lambda_1}\right)^{\frac12-5\eta}\left( \lambda_1^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x}\lambda_2^{-\frac12}\|\phi_{\Theta_2}\|_{L^4_tL^4_x} \right)^\eta \\
&\qquad\qquad\qquad\qquad\times \left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}} \right)^{1-\eta},
\end{align*}
and hence the summation with respect to the modulation $\lambda_0\ll h\ll\lambda_1$ gives the desired estimate.
Now we consider the case $\lambda_1\lesssim\lambda_0\approx\lambda_2$ with $\lambda_1\ll h\ll \lambda_0$.
\begin{align*}
A_{0,1} & \lesssim \sup_{\|\psi\|_{L^2_tL^2_x}\lesssim1} \sum_{\substack{{\mathsf q}',\,{\mathsf q}''\in\mathcal Q_{\lambda_1} \\ |{\mathsf q}'-{\mathsf q}''|\lesssim\lambda_1}}\|P_{{\mathsf q}''}C_h\psi_{\lambda_0,N_0}\|_{L^2_tL^2_x}\|C^{\theta_1}_{\approx h}\varphi_{\Theta_1}\|_{L^4_tL^4_x}\|P_{{\mathsf q}'}C^{\theta_2}_{\ll h}\phi_{\Theta_2}\|_{L^4_tL^4_x} \\
& \lesssim \|C^{\theta_1}_{\approx h}\varphi_{\Theta_1}\|_{L^4_tL^4_x}\left( \sum_{{\mathsf q}'}\|P_{{\mathsf q}'}C^{\theta_2}_{\ll h}\phi_{\Theta_2}\|_{L^4_tL^4_x}^2 \right)^\frac12.
\end{align*}
Here, for the $\varphi_{\Theta_1}$, we simply use the Bernstein's inequality, boundedness \eqref{bdd-high-mod}, and the uniform disposability \eqref{uni-dis1} to get
\begin{align}\label{stri-mod-eta}
\begin{aligned}
\|C^{\theta_1}_{\approx h}\varphi_{\Theta_1}\|_{L^4_tL^4_x} & = \|C^{\theta_1}_{\approx h}\varphi_{\Theta_1}\|_{L^4_tL^4_x}^\eta\|C^{\theta_1}_{\approx h}\varphi_{\Theta_1}\|_{L^4_tL^4_x}^{1-\eta} \\
& \lesssim \|C^{\theta_1}_{\approx h}\varphi_{\Theta_1}\|_{L^4_tL^4_x}^\eta \lambda_1^{\frac34(1-\eta)} \|C^{\theta_1}_{\approx h}\varphi_{\Theta_1}\|_{L^4_tL^2_x}^{1-\eta} \\
& \lesssim \lambda_1^{\frac\eta2}\left(\lambda_1^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x}\right)^\eta \lambda_1^{\frac34(1-\eta)}h^{-\frac14(1-\eta)}\|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}^{1-\eta} ,
\end{aligned}
\end{align}
We recall \eqref{square-sum-q} and get
\begin{align*}
A_{0,1} & \lesssim \left(\frac{\lambda_1}{h}\right)^{\frac14(1-\eta)}\lambda_1^\frac34\lambda_0^\frac14\left(\frac{\lambda_1}{\lambda_0}\right)^{-2\eta}\left( \lambda_1^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x}\lambda_2^{-\frac12}\|\phi_{\Theta_2}\|_{L^4_tL^4_x} \right)^\eta \\
&\qquad\qquad\qquad\qquad\times \left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}} \right)^{1-\eta} \\
& \lesssim \lambda_0 \left( \frac{\lambda_1}{h} \right)^{\frac14(1-\eta)}\left(\frac{\lambda_1}{\lambda_0}\right)^{\frac34-2\eta}\left( \lambda_1^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x}\lambda_2^{-\frac12}\|\phi_{\Theta_2}\|_{L^4_tL^4_x} \right)^\eta \\
&\qquad\qquad\qquad\qquad\times \left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}} \right)^{1-\eta}.
\end{align*}
Similarly, for $A_{0,2}$ we write
\begin{align*}
A_{0,2} & \lesssim \|C^{\theta_1}_{\ll h}\varphi_{\Theta_1}\|_{L^4_tL^4_x}\left(\sum_{{\mathsf q}'\in\mathcal Q_{\lambda_1}}\|P_{{\mathsf q}'}C^{\theta_2}_{\approx h}\phi_{\Theta_2}\|_{L^4_tL^4_x}^2\right)^\frac12.
\end{align*}
Then we follow a similar argument to the estimate of \eqref{high-mod-eta} to get
\begin{align*}
\left(\sum_{{\mathsf q}'\in\mathcal Q_{\lambda_1}}\|P_{{\mathsf q}'}C^{\theta_2}_{\approx h}\phi_{\Theta_2}\|_{L^4_tL^4_x}^2\right)^\frac12 & \lesssim \lambda_2^\frac\eta2\left(\frac{\lambda_1}{\lambda_0}\right)^{-2\eta}\left(\frac{\lambda_1}{h}\right)^{\frac14(1-\eta)}\lambda_1^{\frac12(1-\eta)}\left( \lambda_2^{-\frac12}\|\phi_{\Theta_2}\|_{L^4_tL^4_x} \right)^\eta \|\phi_{\Theta_2}\|_{V^2_{\theta_2}}^{1-\eta}.
\end{align*}
On the other hand, for $\varphi_{\Theta_1}$, we use the uniform disposability \eqref{uni-dis1} and $L^4$-Strichartz estimates to obtain
\begin{align*}
\|C^{\theta_1}_{\ll h}\varphi_{\Theta_1}\|_{L^4_tL^4_x} & \lesssim \|\varphi_{\Theta_1}\|_{L^4_tL^4_x}^{\eta}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x}^{1-\eta} \\
& \lesssim \lambda_1^\frac12 (\lambda_1^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x})^\eta \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}^{1-\eta}.
\end{align*}
Combining these, we get
\begin{align*}
A_{0,2} & \lesssim \lambda_0\left(\frac{\lambda_1}{\lambda_0}\right)^{1-2\eta}\left(\frac{\lambda_1}{h}\right)^{\frac14(1-\eta)}\left( \lambda_1^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x}\lambda_2^{-\frac12}\|\phi_{\Theta_2}\|_{L^4_tL^4_x} \right)^\eta \\
&\qquad\qquad\qquad\qquad\times \left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}} \right)^{1-\eta}.
\end{align*}
Thus the summation of $A_{0,1}$ and $A_{0,2}$ with respect to $\lambda_1\ll h\ll\lambda_0$ gives the required bound.
We now consider $\mathcal A_1(h)$. We decompose it into
\begin{align*}
\|\mathcal A_1(h)\|_{L^2_tL^2_x} & \lesssim \|C_{\approx h}P_{\lambda_0}H_{N_0}(C^{\theta_1}_{ h}\varphi_{\Theta_1})^\dagger\beta C^{\theta_2}_{\ll h}\phi_{\Theta_2}\|_{L^2_tL^2_x} \\
& \qquad\qquad + \|C_{\ll h}P_{\lambda_0}H_{N_0}(C^{\theta_1}_{ h}\varphi_{\Theta_1})^\dagger\beta C^{\theta_2}_{\approx h}\phi_{\Theta_2}\|_{L^2_tL^2_x} \\
&:= A_{1,1}+A_{1,2}.
\end{align*}
Since $A_{1,1}$ can be treated in the identical manner as $A_{0,1}$, we only treat the $A_{1,2}$ term. By orthogonal decompositions by smaller cubes of size $\lambda_1$ and following the argument in \eqref{stri-mod-eta} and estimate \eqref{high-mod-eta} we have
\begin{align*}
A_{1,2} & \lesssim \|C^{\theta_1}_h \varphi_{\Theta_1}\|_{L^4_tL^4_x}\left(\sum_{{\mathsf q}'\in\mathcal Q_{\lambda_1}}\|P_{{\mathsf q}'}C^{\theta_2}_{\approx h}\phi_{\Theta_2}\|_{L^4_tL^4_x}^2 \right)^\frac12 \\
& \lesssim \lambda_1^{\frac\eta2}(\lambda_1^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x})^\eta \lambda_1^{\frac34(1-\eta)}h^{-\frac14(1-\eta)}\|\varphi_{\Theta_1}\|_{V^2_{\theta_2}}^{1-\eta} \\
&\times \lambda_2^\frac\eta2\left(\frac{\lambda_1}{\lambda_0}\right)^{-\epsilon}\left(\frac{\lambda_1}{h}\right)^{\frac14(1-\eta)}\lambda_1^{\frac12(1-\eta)}\left( \lambda_2^{-\frac12}\|\phi_{\Theta_2}\|_{L^4_tL^4_x} \right)^\eta \|\phi_{\Theta_2}\|_{V^2_{\theta_2}}^{1-\eta} \\
& \lesssim \lambda_1 \left(\frac{\lambda_1}{h} \right)^{\frac12(1-\eta)}\left( \frac{\lambda_1}{\lambda_2} \right)^{-\eta}\left(\lambda_1^{-\frac12}\lambda_2^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x}\|\phi_{\Theta_2}\|_{L^4_tL^4_x} \right)^\eta \\
& \qquad\qquad \times \left(\|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}} \right)^{1-\eta}.
\end{align*}
Thus we have
$$
\sum_{\lambda_1\ll h\ll\lambda_0}A_{1,2} \lesssim \lambda_1\left(\frac{\lambda_1}{\lambda_0}\right)^{\frac12(1-3\eta)}\left(\lambda_1^{-\frac12}\lambda_2^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_tL^4_x}\|\phi_{\Theta_2}\|_{L^4_tL^4_x} \right)^\eta \left(\|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}} \right)^{1-\eta}.
$$
The estimate of $\mathcal A_2(h)$ is then given straightforwardly. Indeed, we decompose $\mathcal A_2(h)$ into
\begin{align*}
\|\mathcal A_2(h)\|_{L^2_tL^2_x} & \lesssim \|C_{\approx h}P_{\lambda_0}H_{N_0}(C^{\theta_1}_{\ll h}\varphi_{\Theta_1})^\dagger\beta C^{\theta_2}_{ h}\phi_{\Theta_2}\|_{L^2_tL^2_x} \\
& \qquad\qquad + \|C_{\ll h}P_{\lambda_0}H_{N_0}(C^{\theta_1}_{\approx h}\varphi_{\Theta_1})^\dagger\beta C^{\theta_2}_{ h}\phi_{\Theta_2}\|_{L^2_tL^2_x} \\
&:= A_{2,1}+A_{2,2}.
\end{align*}
We see that $A_{2,1}$ is essentially the same as $A_{0,2}$ whereas the $A_{2,2}$ is same as $A_{1,2}$. Hence we complete the proof of \eqref{0-tri} when $d=3$.
\section{Multilinear estimates I\!I: Two spatial dimensions}\label{multi-2d}
Since the angular regularity is not involved in our analysis, we use the notation $\Theta = (\theta, \lambda)$ and $\psi_\Theta = P_{\lambda}\psi_\theta$ for $\lambda \in 2^{\mathbb N}, \theta \in \{+, -\}$.
We write
\begin{align*}
\mathfrak I_\lambda&:=\sum_{\lambda_4\ge1}\left( \sup_{\|\uppsi\,\,\|_{V^2_{\theta_4}}\le 1}\sum_{\lambda_0, \Theta_j}\bigg|\int_{\mathbb R^{1+2}}\Lambda^{-2}P_{\lambda_0}(\varphi_{\Theta_1}^\dagger\beta\phi_{\Theta_2})\widetilde P_{\lambda_0}(\uppsi_{\Theta_4}^\dagger\beta\psi_{\Theta_3})\,dtdx \bigg| \right)^2 \\
& \le (\textrm{Low modulation})+(\textrm{High modulation}) \\
& =: \mathfrak I_{\lambda\gg h}+\mathfrak I_{\lambda\lesssim h}.
\end{align*}
The estimate of $\mathfrak I_{\lambda\lesssim h}$ is essentially the same as the three dimensional case. We omit the details.
In advance we note that in the low-modulation-regime, i.e., $\lambda\gg d$ with High$\times$High modulation such as $\lambda_0\lesssim\lambda_1\approx\lambda_2$ and $\lambda_0\lesssim\lambda_3\approx\lambda_4$, we must have $\theta_1=\theta_2$ and $\theta_3=\theta_4$. We also observe that the estimate of $\mathfrak I_{\lambda\gg h}$ is very straightforward in view of the previous result \cite{cholee}. It only needs a small modification in multilinear estimates. Indeed, we write
\begin{align*}
\mathfrak I_{\lambda\gg h} & \lesssim \sum_{\lambda_4\ge1}\left(\sup_{\|\uppsi\,\,\|_{V^2_{\theta_4}}=1}\left(\mathbf I_1+\mathbf I_2+\mathbf I_3\right)\right)^2,
\end{align*}
where
\begin{align*}
\mathbf I_1 = \sum_{\substack{\lambda_1,\lambda_2\ge1 \\ \lambda_0\ll\lambda_3\approx\lambda_4}}|\cdots|,\quad \mathbf I_2=\sum_{\substack{\lambda_1,\lambda_2\ge1 \\ \lambda_3\ll\lambda_0\approx\lambda_4}}|\cdots|,\quad \mathbf I_3=\sum_{\substack{\lambda_1,\lambda_2\ge1 \\ \lambda_4\ll\lambda_0\approx\lambda_3}}|\cdots|.
\end{align*}
We further divide $\mathbf I_j$, $j=1,2,3$ as follows:
\begin{align*}
\mathbf I_1 & \lesssim \sum_{\substack{\lambda_0\ll\lambda_1\approx\lambda_2 \\ \lambda_0\ll\lambda_3\approx\lambda_4}}|\cdots|+\sum_{\substack{\lambda_1\ll\lambda_0\approx\lambda_2 \\ \lambda_0\ll\lambda_3\approx\lambda_4}}|\cdots| + \sum_{\substack{\lambda_2\ll\lambda_0\approx\lambda_1\\ \lambda_0\ll\lambda_3\approx\lambda_4}}|\cdots| \\
& =: \mathbf I_{11}+\mathbf I_{12}+\mathbf I_{13},\\
\mathbf I_2 &\lesssim \sum_{\substack{\lambda_0\ll\lambda_1\approx\lambda_2 \\ \lambda_3\ll\lambda_0\approx\lambda_4}}|\cdots|+\sum_{\substack{\lambda_1\ll\lambda_0\approx\lambda_2 \\ \lambda_3\ll\lambda_0\approx\lambda_4}}|\cdots|+\sum_{\substack{\lambda_2\ll\lambda_0\approx\lambda_1 \\ \lambda_3\ll\lambda_0\approx\lambda_4}}|\cdots| \\
&=: \mathbf I_{21}+\mathbf I_{22}+\mathbf I_{23}, \\
\mathbf I_3 &\lesssim \sum_{\substack{\lambda_0\ll\lambda_1\approx\lambda_2 \\ \lambda_4\ll\lambda_0\approx\lambda_3}}|\cdots|+\sum_{\substack{\lambda_1\ll\lambda_0\approx\lambda_2 \\ \lambda_4\ll\lambda_0\approx\lambda_3}}|\cdots|+\sum_{\substack{\lambda_2\ll\lambda_0\approx\lambda_1 \\ \lambda_4\ll\lambda_0\approx\lambda_3}}|\cdots| \\
&=: \mathbf I_{31}+\mathbf I_{32}+\mathbf I_{33}.
\end{align*}
\subsection{Estimates of $\mathbf I_1$}
If $\lambda_0\ll \lambda_1\approx\lambda_2$, we use in order the H\"older inequality and bilinear estimates \eqref{bi-2d-hh} for the bilinear forms involved in the $(1-\eta)$ exponent and then simply use H\"older inequality for the bilinear forms involved in the $\eta$ exponent. We also use $L^4_{t,x}$-Strichartz estimates Proposition \ref{stri-2d} for $\uppsi_{\Theta_4}$ to obtain the $V^2$ norm. The remaining task is very straightforward as follows:
\begin{align*}
\mathbf I_{11} & \lesssim \sum_{\substack{\lambda_0\ll\lambda_1\approx\lambda_2\\\lambda_0\ll\lambda_3\approx\lambda_4\\\lambda_0 \ge 1}}\lambda_0^{-2}\|P_{\lambda_0}(\varphi_{\Theta_1}^\dagger\beta\phi_{\Theta_2})\|_{L^2_{t,x}}\|P_{\lambda_0}(\uppsi_{\Theta_4}^\dagger\beta\psi_{\Theta_3})\|_{L^2_{t,x}} \\
& \lesssim \sum_{\substack{\lambda_0\ll\lambda_1\approx\lambda_2\\\lambda_0\ll\lambda_3\approx\lambda_4}}\lambda_0^{-2}\left( \|\varphi_{\Theta_1}\|_{L^4_{t,x}}\|\phi_{\Theta_2}\|_{L^4_{t,x}}\|\psi_{\Theta_3}\|_{L^4_{t,x}}\|\uppsi_{\Theta_4}\|_{L^4_{t,x}} \right)^\eta \\
&\qquad\qquad\qquad\times \left(\|P_{\lambda_0}(\varphi_{\Theta_1}^\dagger\beta\phi_{\Theta_2})\|_{L^2_{t,x}}\|P_{\lambda_0}(\uppsi_{\Theta_4}^\dagger\beta\psi_{\Theta_3})\|_{L^2_{t,x}}\right)^{1-\eta} \\
& \lesssim \sum_{\substack{\lambda_0\ll\lambda_1\approx\lambda_2\\\lambda_0\ll\lambda_3\approx\lambda_4}}\lambda_0^{-2}(\lambda_1\lambda_3)^\eta \left( \lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12}\lambda_4^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_{t,x}}\|\phi_{\Theta_2}\|_{L^4_{t,x}}\|\psi_{\Theta_3}\|_{L^4_{t,x}}\|\uppsi_{\Theta_4}\|_{L^4_{t,x}} \right)^\eta \\
& \times \lambda_0^{1-\eta}\left(\frac{\lambda_0}{\lambda_1}\right)^{\frac12(1-\eta)}\left(\frac{\lambda_0}{\lambda_3}\right)^{\frac12(1-\eta)}\left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\|\psi_{\Theta_3}\|_{V^2_{\theta_3}}\|\uppsi_{\Theta_4}\|_{V^2_{\theta_4}} \right)^{1-\eta} \\
& \lesssim \sum_{\substack{\lambda_0\ll\lambda_1\approx\lambda_2\\\lambda_0\ll\lambda_3\approx\lambda_4}}\lambda_0^{-1+\eta}\left(\frac{\lambda_0}{\lambda_1}\right)^{\frac12(1-3\eta)}\left(\frac{\lambda_0}{\lambda_3}\right)^{\frac12(1-3\eta)}\left( \lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_{t,x}}\|\phi_{\Theta_2}\|_{L^4_{t,x}}\|\psi_{\Theta_3}\|_{L^4_{t,x}} \right)^\eta \\
&\qquad\qquad\qquad \times \left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\|\psi_{\Theta_3}\|_{V^2_{\theta_3}} \right)^{1-\eta} \|\uppsi_{\Theta_4}\|_{V^2_{\theta_4}} \\
& \lesssim \sum_{\lambda_1\approx\lambda_2,\lambda_3\approx\lambda_4}\left(\frac{\lambda_{\rm med}}{\lambda_{\max}}\right)^{\frac12(1-3\eta)}\left( \lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_{t,x}}\|\phi_{\Theta_2}\|_{L^4_{t,x}}\|\psi_{\Theta_3}\|_{L^4_{t,x}} \right)^\eta \\
&\qquad\qquad\qquad \times \left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\|\psi_{\Theta_3}\|_{V^2_{\theta_3}} \right)^{1-\eta} \|\uppsi_{\Theta_4}\|_{V^2_{\theta_4}}.
\end{align*}
The estimate of $\mathbf I_{12}$ is very similar to $\mathbf I_{11}$. We apply H\"older inequality and then use bilinear estimates \eqref{bi-2d-lh} for the bilinear form $\varphi_{\Theta_1}^\dagger\beta\phi_{\Theta_2}$ and \eqref{bi-2d-hh} for the form $\uppsi_{\Theta_4}^\dagger\beta\psi_{\Theta_3}$ involved in the $(1-\eta)$ exponent, respectively. We again use H\"older inequality for the remaining term involved in $\eta$ exponent.
\begin{align*}
\mathbf I_{12} & \lesssim \sum_{\substack{\lambda_1\ll\lambda_0\approx\lambda_2 \\ \lambda_0\ll\lambda_3\approx\lambda_4\\\lambda_0 \ge 1}}\lambda_0^{-2}(\lambda_1\lambda_2)^\frac\eta2\lambda_3^\eta \left( \lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_{t,x}}\|\phi_{\Theta_2}\|_{L^4_{t,x}}\|\psi_{\Theta_3}\|_{L^4_{t,x}} \right)^\eta \\
&\times (\lambda_1\lambda_2)^{\frac12(1-\eta)}\lambda_0^{\frac12(1-\eta)}\left(\frac{\lambda_0}{\lambda_3}\right)^{\frac12(1-\eta)}\left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\|\psi_{\Theta_3}\|_{V^2_{\theta_3}} \right)^{1-\eta} \|\uppsi_{\Theta_4}\|_{V^2_{\theta_4}} \\
& \lesssim \sum_{\substack{\lambda_1\ll\lambda_0\approx\lambda_2 \\ \lambda_0\ll\lambda_3\approx\lambda_4}}\lambda_0^{-\frac12(1-\eta)}\left(\frac{\lambda_1}{\lambda_0}\right)^\frac12\left(\frac{\lambda_0}{\lambda_3}\right)^{\frac12(1-3\eta)}\left( \lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_{t,x}}\|\phi_{\Theta_2}\|_{L^4_{t,x}}\|\psi_{\Theta_3}\|_{L^4_{t,x}} \right)^\eta \\
&\qquad\qquad\qquad\times\left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\|\psi_{\Theta_3}\|_{V^2_{\theta_3}} \right)^{1-\eta} \|\uppsi_{\Theta_4}\|_{V^2_{\theta_4}} \\
& \lesssim \sum_{\lambda_1\ll\lambda_2\ll\lambda_3\approx\lambda_4}\left(\frac{\lambda_2}{\lambda_3}\right)^{\frac12(1-3\eta)}\left( \lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_{t,x}}\|\phi_{\Theta_2}\|_{L^4_{t,x}}\|\psi_{\Theta_3}\|_{L^4_{t,x}} \right)^\eta \\
&\qquad\qquad\qquad\times\left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\|\psi_{\Theta_3}\|_{V^2_{\theta_3}} \right)^{1-\eta} \|\uppsi_{\Theta_4}\|_{V^2_{\theta_4}}.
\end{align*}
The estimate of $\mathbf I_{13}$ is identical to the estimate of $\mathbf I_{12}$. The only task is to interchange the input frequencies $\lambda_1$ and $\lambda_2$. In this manner we observe that $\mathbf I_{23}$ and $\mathbf I_{33}$ are identical to $\mathbf I_{22}$ and $\mathbf I_{32}$, respectively. We omit the details.
\subsection{Estimates of $\mathbf I_2$}
The estimate of $\mathbf I_{21}$ is also similar as $\mathbf I_{12}$. We only exchange the role of bilinear forms $\varphi_{\Theta_1}^\dagger\beta\psi_{\Theta_2}$ and $\uppsi_{\Theta_4}^\dagger\beta\psi_{\Theta_3}$.
\begin{align*}
\mathbf I_{21} & \lesssim \sum_{\substack{\lambda_0\ll\lambda_1\approx\lambda_2\\ \lambda_3\ll\lambda_0\approx\lambda_4}}\lambda_0^{-2}\lambda_1^\eta (\lambda_3\lambda_4)^\frac\eta2\left( \lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_{t,x}}\|\phi_{\Theta_2}\|_{L^4_{t,x}}\|\psi_{\Theta_3}\|_{L^4_{t,x}} \right)^\eta\\
&\qquad\times \lambda_0^{\frac12(1-\eta)}\left(\frac{\lambda_0}{\lambda_1}\right)^{\frac12(1-\eta)}(\lambda_3\lambda_4)^{\frac12(1-\eta)}\left( \|\varphi_{\lambda_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\|\psi_{\Theta_3}\|_{V^2_{\theta_3}} \right)^{1-\eta} \|\uppsi_{\Theta_4}\|_{V^2_{\theta_4}} \\
& \lesssim \sum_{\substack{\lambda_0\ll\lambda_1\approx\lambda_2\\ \lambda_3\ll\lambda_0\approx\lambda_4}}\lambda_0^{-1-\eta}\lambda_1^{-\frac12(1-3\eta)}(\lambda_3\lambda_4)^\frac12\left( \lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_{t,x}}\|\phi_{\Theta_2}\|_{L^4_{t,x}}\|\psi_{\Theta_3}\|_{L^4_{t,x}} \right)^\eta \\
&\qquad\qquad\qquad \times \left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\|\psi_{\Theta_3}\|_{V^2_{\theta_3}} \right)^{1-\eta} \|\uppsi_{\Theta_4}\|_{V^2_{\theta_4}} \\
& \lesssim \sum_{\lambda_3\ll\lambda_4\ll\lambda_1\approx\lambda_2}\lambda_1^{-\frac14}\left(\frac{\lambda_4}{\lambda_1}\right)^\eta\left(\frac{\lambda_3}{\lambda_4}\right)^\frac12\left( \lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_{t,x}}\|\phi_{\Theta_2}\|_{L^4_{t,x}}\|\psi_{\Theta_3}\|_{L^4_{t,x}} \right)^\eta \\
&\qquad\qquad\qquad \times \left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\|\psi_{\Theta_3}\|_{V^2_{\theta_3}} \right)^{1-\eta} \|\uppsi_{\Theta_4}\|_{V^2_{\theta_4}}.
\end{align*}
Now we use bilinear estimates \eqref{bi-2d-lh} and get
\begin{align*}
\mathbf I_{22} & \lesssim \sum_{\substack{\lambda_1\ll\lambda_0\approx\lambda_2\\ \lambda_3\ll\lambda_0\approx\lambda_4}}\lambda_0^{-2}(\lambda_1\lambda_2\lambda_3\lambda_4)^\frac12 \left( \lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_{t,x}}\|\phi_{\Theta_2}\|_{L^4_{t,x}}\|\psi_{\Theta_3}\|_{L^4_{t,x}} \right)^\eta \\
&\qquad\qquad\qquad \times \left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\|\psi_{\Theta_3}\|_{V^2_{\theta_3}} \right)^{1-\eta} \|\uppsi_{\Theta_4}\|_{V^2_{\theta_4}} \\
& \lesssim \sum_{\lambda_1,\lambda_3\ll\lambda_2\approx\lambda_4}\left(\frac{\lambda_1\lambda_3}{\lambda_2\lambda_4}\right)^\frac12\left( \lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_{t,x}}\|\phi_{\Theta_2}\|_{L^4_{t,x}}\|\psi_{\Theta_3}\|_{L^4_{t,x}} \right)^\eta \\
&\qquad\qquad\qquad \times \left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\|\psi_{\Theta_3}\|_{V^2_{\theta_3}} \right)^{1-\eta} \|\uppsi_{\Theta_4}\|_{V^2_{\theta_4}}.
\end{align*}
\subsection{Estimates of $\mathbf I_3$}
The estimate of $\mathbf I_{31}$ is similar to $\mathbf I_{21}$. The only difference is to exchange the role of $\psi_{\Theta_3}$ and $\uppsi_{\Theta_4}$. Indeed, we have
\begin{align*}
\mathbf I_{31} & \lesssim \sum_{\substack{\lambda_0\ll\lambda_1\approx\lambda_2 \\ \lambda_4\ll \lambda_0\approx\lambda_3}}\lambda_0^{-2}\lambda_1^\eta \lambda_0^{\frac12(1-\eta)}\left(\frac{\lambda_0}{\lambda_1}\right)^{\frac12(1-\eta)} (\lambda_3\lambda_4)^\frac12 \left( \lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_{t,x}}\|\phi_{\Theta_2}\|_{L^4_{t,x}}\|\psi_{\Theta_3}\|_{L^4_{t,x}} \right)^\eta \\
&\qquad\qquad\qquad \times \left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\|\psi_{\Theta_3}\|_{V^2_{\theta_3}} \right)^{1-\eta} \|\uppsi_{\Theta_4}\|_{V^2_{\theta_4}}.
\end{align*}
The simple use of bilinear estimates \eqref{bi-2d-lh} yields
\begin{align*}
\mathbf I_{32} & \lesssim \sum_{\substack{\lambda_1\ll\lambda_0\approx\lambda_2 \\ \lambda_4\ll \lambda_0\approx\lambda_3}} \lambda_0^{-2}(\lambda_1\lambda_2\lambda_3\lambda_4)^\frac12 \left( \lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_{t,x}}\|\phi_{\Theta_2}\|_{L^4_{t,x}}\|\psi_{\Theta_3}\|_{L^4_{t,x}} \right)^\eta \\
&\qquad\qquad\qquad \times \left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\|\psi_{\Theta_3}\|_{V^2_{\theta_3}} \right)^{1-\eta} \|\uppsi_{\Theta_4}\|_{V^2_{\theta_4}} \\
& \lesssim \sum_{\lambda_1,\lambda_4\ll\lambda_2\approx\lambda_3}\left(\frac{\lambda_1\lambda_4}{\lambda_2\lambda_3}\right)^\frac12\left( \lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_{t,x}}\|\phi_{\Theta_2}\|_{L^4_{t,x}}\|\psi_{\Theta_3}\|_{L^4_{t,x}} \right)^\eta \\
&\qquad\qquad\qquad \times \left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\|\psi_{\Theta_3}\|_{V^2_{\theta_3}} \right)^{1-\eta} \|\uppsi_{\Theta_4}\|_{V^2_{\theta_4}}.
\end{align*}
Finally, by the estimates of $\mathbf I_j$, $j=1,2,3$ we conclude that
\begin{align}
\begin{aligned}
\mathfrak I_{\lambda\gg h}&\lesssim \sum_{\lambda_4\ge1}\bigg( \sum_{\lambda_1,\lambda_2,\lambda_3\ge1}\left(\frac{\lambda_{\rm med}}{\lambda_{\max}}\right)^\delta \left( \lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12}\|\varphi_{\Theta_1}\|_{L^4_{t,x}}\|\phi_{\Theta_2}\|_{L^4_{t,x}}\|\psi_{\Theta_3}\|_{L^4_{t,x}} \right)^\eta \\
&\qquad\qquad\qquad\qquad\qquad\qquad\left( \|\varphi_{\Theta_1}\|_{V^2_{\theta_1}}\|\phi_{\Theta_2}\|_{V^2_{\theta_2}}\|\psi_{\Theta_3}\|_{V^2_{\theta_3}} \right)^{1-\eta} \bigg)^2.
\end{aligned}
\end{align}
Thus we have
\begin{align*}
\|\mathcal I_{\theta}[V_b*(\varphi^\dagger\beta\phi)\beta\psi](t_0; \cdot)\|_{F^{0,0}(I)} & \le C (\|\varphi\|_{\mathbf D^{-\frac12, 0}(I)}\|\phi\|_{\mathbf D^{-\frac12, 0}(I)}\|\psi\|_{\mathbf D^{-\frac12, 0}(I)})^\eta \\
&\qquad\qquad \times (\|\varphi\|_{F^{0,0}(I)}\|\phi\|_{F^{0,0}(I)}\|\psi\|_{F^{0,0}(I)})^{1-\eta},
\end{align*}
which is the desired estimate.
\section{Majorana condition: Proof of Theorem \ref{majorana-gwp}}\label{sec-maj}
We define the \textit{charge conjugation operator} $C^z_\theta$ by
$$
C^z_\theta\psi = \theta z\gamma^2\psi^*,
$$
where $z = e^{i\omega}$ for some $\omega \in \mathbb R$ and $\theta \in \{+, -\}$.
We also define the projection operator $P_\theta^z$ by
$$
P^z_\theta\psi = \frac12\left(\psi+C^z_\theta\psi\right).
$$
Then we get
\[
P_+^z + P_-^z = I.
\]
\begin{prop}
Let $z=e^{i\omega}$ for any $\omega\in\mathbb R$ and $\theta\in\{+,-\}$. The operator $P^z_\theta$ satisfies the following:
\begin{align}\label{eq:proj-z}
(P^z_\theta)^2 = P^z_\theta, \quad P^z_\theta P^z_{-\theta} = 0.
\end{align}
\end{prop}
\begin{proof}
An simple computation gives $P^z_+ + P^z_{-}$ is the identity operator. One can observe that
\begin{align*}
P^z_\theta C^z_\theta\psi &= P^z_\theta(\theta z\gamma^2\psi^*) = \frac12(\theta z\gamma^2\psi^*+\theta z\gamma^2\theta z^*(\gamma^2)^*\psi) = \frac12(\theta z\gamma^2\psi^*+\psi) = P^z_\theta \psi.
\end{align*}
Here we used $(\gamma^2)^\dagger = -\gamma^2$ and $\gamma^2\times(-\gamma^2) = \mathbb I_{\tilde d}$. Then we see that
\begin{align*}
P^z_\theta P^z_\theta\psi & = \frac12 P^z_\theta(\psi+C^z_\theta\psi) = \frac12(P^z_\theta\psi+P^z_\theta C^z_\theta\psi) = \frac12(P^z_\theta\psi + P^z_\theta\psi) = P^z_\theta\psi,
\end{align*}
and hence we conclude that $(P^z_\theta)^2\psi=P^z_\theta\psi$. This implies that $P^z_\theta P^z_{-\theta} = 0$, since
\begin{align*}
P^z_\theta P^z_{-\theta}\psi &= P^z_\theta(I-P^z_\theta)\psi = P^z_\theta\psi - (P^z_\theta)^2\psi = P^z_\theta\psi-P^z_\theta\psi = 0.
\end{align*}
This finishes the proof of \eqref{eq:proj-z}.
\end{proof}
Given a spinor field $\psi:\mathbb R^{1+d}\rightarrow\mathbb C^{\tilde d}$, we have the decomposition
$$
\psi = \sum_{\theta \in \{+, -\}}P^z_\theta \psi.
$$
The following proposition is readily obtained. However it presents the significant observation on the study of large data well-posedness via the Majorana condition.
\begin{prop}
For any spinor field $\psi:\mathbb R^{1+d}\rightarrow \mathbb C^{\tilde d}$, recall that we define $\overline\psi = \psi^\dagger\gamma^0$. Then we have the following identity: $$
\overline{P^z_\theta\psi}P^z_\theta\psi=0
$$
\end{prop}
\begin{proof}
To see this, we first note that $\overline{P^z_\theta\psi}P^z_\theta\psi$ must be real-valued. We write
\begin{align*}
\overline{P^z_\theta\psi}P^z_\theta\psi&= \frac14(\psi^\dagger+\theta z^*\psi^T(-\gamma^2))\gamma^0(\psi+\theta z\gamma^2\psi^*),
\end{align*}
where $\psi^T$ is the transpose of spinor field $\psi$. Then we obtain
\begin{align*}
\overline{P^z_\theta\psi}P^z_\theta\psi&= \frac14\left(\overline\psi\psi+\theta z\psi^\dagger\gamma^0\gamma^2\psi^*-\theta z^*\psi^T\gamma^2\gamma^0\psi+\psi^T(-\gamma^2)\gamma^0\gamma^2\psi^*\right).
\end{align*}
Here, we observe that
\begin{align*}
\psi^T(-\gamma^2)\gamma^0\gamma^2\psi^* &= \psi^T\gamma^0(\gamma^2)^2\psi^* = -\psi^T\gamma^0\psi^* = -(\psi^\dagger\gamma^0\psi)^* = -(\overline\psi\psi)^* = -\overline\psi\psi.
\end{align*}
Similarly, we see that
\begin{align*}
-\theta z^*\psi^T\gamma^2\gamma^0\psi &= \theta z^*\psi^T\gamma^0\gamma^2\psi = -(\theta z\psi^\dagger\gamma^0\gamma^2\psi^*)^*.
\end{align*}
Combining these identities, we deduce that
\begin{align*}
\overline{P^z_\theta\psi}P^z_\theta\psi &= \frac i2\textrm{Im}(\theta z\overline\psi\gamma^2\psi),
\end{align*}
which should be zero, since the LHS is purely real, whereas the RHS is purely imaginary.
\end{proof}
Now we consider the initial value problems for the Dirac equations
\begin{align}\label{eq:c-dirac}
\left\{
\begin{array}{l}
-i\gamma^\mu\partial_\mu\psi+\psi = [V_b*(\psi^\dagger \beta\psi)]\psi, \\
\psi|_{t=0}=\psi_0.
\end{array}
\right.
\end{align}
Recall that we have put the mass parameter $M=1$.
We shall study the time evolution property of solutions to the equations for large data by exploiting the Majorana condition. The first step is to consider the system of cubic Dirac equations instead of the above equation which presents
\begin{align}\label{psi-decomp}
\begin{aligned}
-i\gamma^\mu\partial_\mu\varphi+\varphi = V_b*(\overline{P^z_\theta\varphi}P^z_{-\theta}\phi+\overline{P^z_{-\theta}\phi}P^z_\theta\varphi)\varphi, \\
-i\gamma^\mu\partial_\mu\phi+\phi = V_b*(\overline{P^z_\theta\varphi}P^z_{-\theta}\phi+\overline{P^z_{-\theta}\phi}P^z_\theta\varphi)\phi.
\end{aligned}
\end{align}
for sufficiently smooth $\varphi, \phi$ with initial data
\begin{align}\label{decomp-data}
\varphi|_{t=0} = P^z_\theta\psi_0,\quad \phi|_{t=0} = P^z_{-\theta}\psi_0.
\end{align}
\noindent The aim of this section is to prove the following.
\begin{thm}\label{majorana-gwp1}
Let $z = e^{i\omega}, \omega \in \mathbb R$. Let $\sigma > 0$ for $d = 3$ and $\sigma = 0$ for $d = 2$. Then there exists $0 < \epsilon < 1$ such that for any ${\mathsf A} \ge 1$ and any $0 < \mathtt a \le \epsilon {\mathsf A}^{-1}$, if the initial data satisfy
$$
\|P^z_+\psi_0\|_{L^{2,\sigma}(\mathbb R^d)} \le \mathtt a,\ \|P^z_{-}\psi_0\|_{L^{2,\sigma}(\mathbb R^d)}\le {\mathsf A},
$$
then the equation \eqref{dirac} is globally well-posed. To be precise, the Cauchy problem of \eqref{psi-decomp} is globally well-posed in the sense that
\begin{align}
P^z_+\psi,P^z_{-}\psi \in C(\mathbb R;L^{2,\sigma}(\mathbb R^d)) .
\end{align}
Furthermore, there exist $\varphi_0^{\pm},\phi_{0}^{\pm}\in L^{2,\sigma}$ such that
\begin{align*}
\lim_{t\rightarrow\pm\infty}\|P^z_+\psi(t) - e^{\theta it\Lambda}\varphi_{0}^{\pm}\|_{L^{2,\sigma}(\mathbb R^d)} = 0, \\
\lim_{t\rightarrow\pm\infty}\|P^z_{-}\psi(t) - e^{\theta it\Lambda}\phi_{0}^{\pm}\|_{L^{2,\sigma}(\mathbb R^d)} = 0.
\end{align*}
\end{thm}
\begin{proof}[Proof of Theorem \ref{majorana-gwp1}]
We recall the Banach space $F^{0,\sigma}\subset C(\mathbb R;L^{2,\sigma}(\mathbb R^{d}))$. If $\psi\in F^{0,\sigma}$ is a solution to the cubic Dirac equations \eqref{eq:c-dirac},
then by the multilinear estimates given by putting $\eta=0$ in the multilinear estimates Theorem \ref{main-tri-est}, we have
\begin{align}\label{psi-bdd-maj}
\|\psi\|_{F^{0,\sigma}} &\le \|\psi_0\|_{L^{2,\sigma}}+C\|\psi_1\|_{F^{0,\sigma}}\|\psi_2\|_{F^{0,\sigma}}\|\psi_3\|_{F^{0,\sigma}}.
\end{align}
We consider the set
$$
X = \{ (\varphi,\phi)\in F^{0,\sigma}\times F^{0,\sigma} : \|\varphi\|_{F^{0,\sigma}}\le 2\|\varphi(0)\|_{L^{2,\sigma}},\ \|\phi\|_{F^{0,\sigma}}\le2\|\phi(0)\|_{L^{2,\sigma}} \}
$$
and for ${\mathsf A}, \mathtt a > 0$, we define the norm
$$
\|(\varphi,\phi)\|_{X} = \mathtt a^{-1}\|\varphi\|_{F^{0,\sigma}} + {\mathsf A}^{-1}\|\phi\|_{F^{0,\sigma}}.
$$
Then $X$ is a complete metric space with the metric corresponding to the norm. Now we let $\Phi=(\Phi_1,\Phi_2)$ be the inhomogeneous solution map for \eqref{psi-decomp} given by the Duhamel principle. Then the bound \eqref{psi-bdd-maj} and the definition of the set $X$ give
\begin{align}
\begin{aligned}
\|\Phi_1(\varphi,\phi)\|_{F^{0,\sigma}} & \le \|\varphi(0)\|_{L^{2,\sigma}}+C \|\varphi\|_{F^{0,\sigma}}^2\|\phi\|_{F^{0,\sigma}} \\
& \le \|\varphi(0)\|_{L^{2,\sigma}} +16C \|\varphi(0)\|_{L^{2,\sigma}}^2\|\phi(0)\|_{L^{2,\sigma}}, \\
& \le (1+16C{\mathsf A}\mathtt a)\|\varphi(0)\|_{L^{2,\sigma}},
\end{aligned}
\end{align}
and also
\begin{align}
\begin{aligned}
\|\Phi_2(\varphi,\phi)\|_{F^{0,\sigma}} & \le \|\phi(0)\|_{L^{2,\sigma}}+C \|\phi\|_{F^{0,\sigma}}^2\|\varphi\|_{F^{0,\sigma}} \\
& \le \|\phi(0)\|_{L^{2,\sigma}} +16C \|\phi(0)\|_{L^{2,\sigma}}^2\|\varphi(0)\|_{L^{2,\sigma}}, \\
& \le (1+16C{\mathsf A}\mathtt a)\|\phi(0)\|_{L^{2,\sigma}}.
\end{aligned}
\end{align}
Then we put $\mathtt a \le \dfrac{1}{16C{\mathsf A}}$ and deduce that the map $\Phi$ is the flow map from $X$ into $X$. By multilinear estimates lead us that the map $\Phi$ is a contraction on the set $X$. Indeed, suppose that we have $(\varphi_1,\phi_1),\,(\varphi_2,\phi_2)\in X$. Then we estimate
$$
\|\Phi_1(\varphi_1,\phi_1)-\Phi_1(\varphi_2,\phi_2)\|_{F^{0, \sigma}} \le 16C{\mathsf A}\mathtt a \|\varphi_1-\varphi_2\|_{F^{0,\sigma}} + 8C\mathtt a^2\|\phi_1-\phi_2\|_{F^{0,\sigma}}
$$
and
$$
\|\Phi_2(\varphi_1,\phi_1)-\Phi_2(\varphi_2,\phi_2)\|_{F^{0,\sigma}} \le 16C{\mathsf A}\mathtt a \|\phi_1-\phi_2\|_{F^{0,\sigma}} + 8C{\mathsf A}^{2}\|\varphi_1-\varphi_2\|_{F^{0,\sigma}}.
$$
In consequence we obtain
\begin{align*}
\|\Phi(\varphi_1,\phi_1)-\Phi(\varphi_2,\phi_2)\|_X &\le 24C{\mathsf A}\|\varphi_1 -\varphi_2\|_{F^{0,\sigma}} + 24C\mathtt a\|\phi_1-\phi_2\|_{F^{0,\sigma}}\\
& = 24C{\mathsf A}\mathtt a\|(\varphi_1,\phi_1)-(\varphi_2,\phi_2)\|_X.
\end{align*}
Thus by choosing $\epsilon = \dfrac{1}{48C}$, the solution map $\Phi$ is a contraction on $X$ for any $\mathtt a \le \epsilon{\mathsf A}^{-1}$.
\end{proof}
\section*{Acknowledgements}
This work was supported in part by NRF-2021R1I1A3A04035040(Republic of Korea).
|
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{"url":"https:\/\/tex.stackexchange.com\/questions\/436259\/timeline-in-latex\/436304","text":"# Timeline in LaTeX\n\nI am trying to create a timeline similar to the one displayed below, it does not have to be exactly the same. I already tried looking at other threats about timelines, which made me to believe I am probably gonna have to use the Tikz package. But other than that, they didn't really help me much further.\n\nThis is what I tried:\n\n\\documentclass{article}\n\\usepackage{tikz}\n\\begin{document}\n\\begin{tikzpicture}\n% draw horizontal line\n\\draw[ultra thick, ->] (0,0) -- (\\ImageWidth,0);\n\n% draw vertical lines\n\\foreach \\x in {2,4,6,8,10,12}\n\\draw (\\x cm,3pt) -- (\\x cm,-3pt);\n\n% draw node\n\\draw[ultra thick] (4,0) node[below=3pt,thick] {t-2} node[above=3pt] {};\n\\draw[ultra thick] (6,0) node[below=3pt,thick] {t-1} node[above=3pt] {};\n\\draw[ultra thick] (8,0) node[below=3pt, thick] {t} node[above=3pt] {};\n\\draw[ultra thick] (10,0) node[below=3pt] {t+1} node[above=3pt] {};\n\n\\draw [black, ultra thick ,decorate,decoration={brace,amplitude=5pt},\nxshift=5pt,yshift=-4pt] (4,0.5) -- (8,0.5)\nnode [black,midway,above=4pt,xshift=-2pt] {\\footnotesize Training period};\n\n\\draw [ black, ultra thick,decorate,decoration={brace,amplitude=5pt},\nxshift=8pt,yshift=-11pt] (10,-0.5) -- (8,-0.5)\nnode [black,midway,below=4pt,xshift=8pt] {\\footnotesize Testing period};\n\\end{tikzpicture}\n\n\\end{document}\n\n\nIt ended up looking as follows:\n\nWhich is rather far from my desired result.\n\n\u2022 What have you tried so far? Doing this with the tikz package should be relatively easy (besides the curly braces, but solutions have already been provided for some other questions about them). \u2013\u00a0epR8GaYuh Jun 13 '18 at 8:25\n\u2022 I don't have anything to show as of yet, still trying to understand the tikz package. I'm relatively new to latex. \u2013\u00a0Keith Jun 13 '18 at 8:38\n\u2022 Related question on SO main: stackoverflow.com\/questions\/217834\/\u2026 \u2013\u00a0Marijn Jun 13 '18 at 9:00\n\u2022 I added the code I have until now, but it doesn't really look very neat. \u2013\u00a0Keith Jun 13 '18 at 12:27\n\nYou can also use line width to draw the colored rectangle. And define the color shading within the \\foreach.\n\n\\documentclass{article}\n\\newcommand{\\ImageWidth}{11cm}\n\\usepackage{tikz}\n\\usetikzlibrary{decorations.pathreplacing,positioning, arrows.meta}\n\n\\begin{document}\n\\begin{tikzpicture}\n% draw horizontal line\n\\draw[thick, -Triangle] (0,0) -- (\\ImageWidth,0) node[font=\\scriptsize,below left=3pt and -8pt]{years};\n\n% draw vertical lines\n\\foreach \\x in {0,1,...,10}\n\\draw (\\x cm,3pt) -- (\\x cm,-3pt);\n\n\\foreach \\x\/\\descr in {4\/t-2,5\/t-1,6\/t,7\/t+1}\n\\node[font=\\scriptsize, text height=1.75ex,\ntext depth=.5ex] at (\\x,-.3) {$\\descr$};\n\n% colored bar up\n\\foreach \\x\/\\perccol in\n{1\/100,2\/75,3\/25,4\/0}\n\\draw[lightgray!\\perccol!red, line width=4pt]\n(\\x,.5) -- +(1,0);\n\\draw[-Triangle, dashed, red] (5,.5) -- +(1,0);\n\n% colored bar down\n\\foreach \\x\/\\perccol in\n{3\/100,4\/75,5\/0}\n\\draw[lightgray!\\perccol!green, line width=4pt]\n(\\x,-.7) -- +(1,0);\n\\draw[-Triangle, dashed, green] (6,-.7) -- +(1,0);\n\n% braces\n\\draw [thick ,decorate,decoration={brace,amplitude=5pt}] (4,0.7) -- +(2,0)\nnode [black,midway,above=4pt, font=\\scriptsize] {Training period};\n\\draw [thick,decorate,decoration={brace,amplitude=5pt}] (6,-.9) -- +(-1,0)\nnode [black,midway,font=\\scriptsize, below=4pt] {Testing period};\n\n\\end{tikzpicture}\n\n\\end{document}\n\n\nThere are some problems with your code:\n\n\u2022 You are using lots of ultra thick lines which are not really needed.\n\u2022 You have some xshift values for the braces which move them out of their right position.\n\u2022 Your node syntax to label the axis could be much shorter.\n\u2022 You are missing the decorations library which makes your code not compilable.\n\nI did some cleanup of your code and ended up with this (which probably still is not perfect):\n\n\\documentclass[border=2mm]{standalone}\n\n\\usepackage{tikz}\n\\usetikzlibrary{decorations.pathreplacing}\n\n\\definecolor{myLightGray}{RGB}{191,191,191}\n\\definecolor{myGray}{RGB}{160,160,160}\n\\definecolor{myDarkGray}{RGB}{144,144,144}\n\\definecolor{myDarkRed}{RGB}{167,114,115}\n\\definecolor{myRed}{RGB}{255,58,70}\n\\definecolor{myGreen}{RGB}{0,255,71}\n\n\\begin{document}\n\n\\begin{tikzpicture}[%\nevery node\/.style={\nfont=\\scriptsize,\n% Better alignment, see https:\/\/tex.stackexchange.com\/questions\/315075\ntext height=1ex,\ntext depth=.25ex,\n},\n]\n% draw horizontal line\n\\draw[->] (0,0) -- (8.5,0);\n\n% draw vertical lines\n\\foreach \\x in {0,1,...,8}{\n\\draw (\\x cm,3pt) -- (\\x cm,0pt);\n}\n\n% place axis labels\n\\node[anchor=north] at (3,0) {$t-2$};\n\\node[anchor=north] at (4,0) {$t-1$};\n\\node[anchor=north] at (5,0) {$t$};\n\\node[anchor=north] at (6,0) {$t+1$};\n\\node[anchor=north] at (8.5,0) {years};\n\n% draw scale above\n\\fill[myLightGray] (1,0.25) rectangle (2,0.4);\n\\fill[myDarkGray] (2,0.25) rectangle (3,0.4);\n\\fill[myDarkRed] (3,0.25) rectangle (4,0.4);\n\\fill[myRed] (4,0.25) rectangle (5,0.4);\n\\draw[myRed,dashed,thick,-latex] (5.05,0.325) -- (6.05,0.325);\n\n% draw scale below\n\\fill[myLightGray] (3,-0.4) rectangle (4,-0.55);\n\\fill[myGray] (4,-0.4) rectangle (5,-0.55);\n\\fill[myGreen] (5,-0.4) rectangle (6,-0.55);\n\\draw[myGreen,dashed,thick,-latex] (6.05,-0.475) -- (7.05,-0.475);\n\n% draw curly braces and add their labels\n\\draw[decorate,decoration={brace,amplitude=5pt}] (3,0.45) -- (5,0.45)\nnode[anchor=south,midway,above=4pt] {Training period};\n\\draw[decorate,decoration={brace,amplitude=5pt}] (6,-0.6) -- (5,-0.6)\nnode[anchor=north,midway,below=4pt] {Testing period};\n\\end{tikzpicture}\n\n\\end{document}","date":"2019-04-18 18:29:06","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9085515737533569, \"perplexity\": 8258.256042440704}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-18\/segments\/1555578526228.27\/warc\/CC-MAIN-20190418181435-20190418202419-00047.warc.gz\"}"}
| null | null |
{"url":"http:\/\/erc.eq-j.cn\/html\/2020\/4\/20200405.htm","text":"Earthquake Research in China\u00a0\u00a02020, Vol. 34 Issue (4): 510-522\u00a0\u00a0\u00a0\u00a0\u00a0DOI: 10.19743\/j.cnki.0891-4176.202004009\nEarly Identification of the Jiangdingya Landslide of Zhouqu Based on SBAS-InSAR Technology\nYU Haihua1), CAI Guolin1), GAN Quan2), SHEN Dong1)\n1). Faculty of Geosciences and Environmental Engineering, Southwest Jiaotong University, Chengdu 611756, China;\n2). Sichuan Bureau of Surveying, Mapping and Geoinformation, Chengdu 610000, China\nAbstract: SBAS-InSAR technology is characterized by the advantages of reducing the influence of terrain-simulation error, time-space decorrelation, atmospheric error, thereby improving the reliability of surface-deformation monitoring. This paper studies the early landslide identification method based on SBAS-InSAR technology. Selecting the Jiangdingya landslide area in Zhouqu County, Gansu Province as the research area, 84 ascending-orbit Sentinel-1A SAR images from 2015 to 2019 are collected. In addition, using SBAS-InSAR technology, the rate and time-series results of surface deformation of the landslide area in Jiangdingya during this period are extracted, and potential landslides are identified. The results show that the early landslide identification method based on SBAS-InSAR technology is highly feasible and is a better tool for identifying potential landslides in large areas.\nKey words: SBAS-InSAR\u00a0\u00a0\u00a0\u00a0 Jiangdingya landslide\u00a0\u00a0\u00a0\u00a0 Early identification\u00a0\u00a0\u00a0\u00a0 Deformation rate\u00a0\u00a0\u00a0\u00a0 Sentinel-1A\n\nINTRODUCTION\n\nGeological disasters occur frequently in the mountainous areas in western China, accounting for 60% of national geological disasters (Huang Runqiu, 2009). Among them, landslides account for a large proportion. Due to the lack of large-scale monitoring and identification of landslides, early warning and accurate forecasting cannot be carried out, and the situation of \"there is no landslide in the surveyed area, and there is no mapping in the landslide area\" often appears (Liu Chuanzheng, 2015). In particular, the ability of early identification, monitoring, and early warning systems is far from enough, resulting in serious potential security hazards to people's lives and property, as well as the national infrastructures (Yin Yueping, 2013). Therefore, effective landslide identification technology is urgently needed to identify potential landslide hazard areas in advance.\n\nFor such early identification, it is vitally important to obtain the surface deformation of potential landslides. At present, traditional field surveys, geomorphologic mapping, and other investigative methods cannot be implemented in mountain areas with complex terrains. Furthermore, they suffer from inferior timeliness and small monitoring ranges, which cannot meet the requirements of early identification (Zhang Shijia et al., 2018; Wang Zhihua, 2007). The landslide identification method based on remote sensing has made up for the deficiency of traditional methods and has become one important means of early landslide identification, monitoring, and early warning. However, optical remote sensing is vulnerable to adverse weather, such as clouds and fog, and cannot be deployed at night. In contrast, interferometric synthetic aperture radar (InSAR), as an active remote sensing tool, has the capability of all-weather and all-day operation and higher monitoring accuracy, and it has been widely used in early landslide identification.\n\nThe application of InSAR technology in landslide information extraction mainly involves the use of differential interferometric synthetic aperture radar (DInSAR) and differential interferometry using time series. Due to the influence of terrain simulation, time-space decorrelation, and atmospheric error, the application scope of DInSAR technology is limited by plentiful factors. The method of differential interferometry using time series eliminates the influence of the these three factors. Meanwhile, it can monitor the cumulative micro-deformation of long time series and is thus widely used in volcanic research (Babu A. et al., 2019; Niu Yufen et al., 2019), seismic deformation field inversion (ent\u00fcrk S. et al., 2019), surface subsidence monitoring (Yin Hongjie et al., 2011), landslide identification, and early warning (Colesanti C. et al., 2003; Dong Jie et al., 2018). Among the various methods of differential interferometry using time series, the most representative ones include permanent scatter InSAR (PS-InSAR) (Ferretti A. et al., 2001) and small-baseline subset InSAR (SBAS-InSAR) (Yin Hongjie et al., 2011; Liu Peng et al., 2013). PS-InSAR technology can accurately reverse surface deformation using the displacement of PS points that remain stable for a long time and are less affected by time-space decorrelation. However, PS-InSAR technology requires 15-20 images or even more (Ge Daqing, 2013), and it is difficult to extract permanent scatterers in areas with complex terrain, high vegetation coverage, and limited SAR images, making it difficult to monitor surface deformation (Zhang Yi, 2018). Compared with PS-InSAR, SBAS-InSAR requires fewer images and does not need to extract PS points, solves the problem of decorrelation caused by excessively long time-space baseline, and improves the temporal resolution of monitoring, which is conducive to the analysis of surface deformation time series in non-urban areas (Yin Hongjie et al., 2011; Liu Peng et al., 2013).\n\nIn view of the numerous advantages of SBAS-InSAR technology, many scholars have conducted landslide identification research based on SBAS-InSAR technology. Zhang Shijia et al. (2018) obtained the surface deformation of the Minjiang River Basin and verified the applicability of SBAS technology in landslide identification. Zhang Yi (2018) extracted the deformation information of a landslide and unstable slope in Zhouqu County by combining different multi-period and multi-band SAR images, providing a reference for disaster prevention and mitigation work in Zhouqu County.\n\nAs the Sentinel-1A satellite has wide coverage and a short revisit period, its images have become an important data source for disaster monitoring, giving them considerable application potential in the field of landslide early warning. Mondini A.C. et al., (2019) demonstrated that Sentinel-1A images are a reliable data source by extracting rapid landslides when researching 32 cases of rapid landslide worldwide. Dai Keren et al. (2016) and Zhang Yi (2018) used Sentinel-1A images as a data source to monitor landslides, obtained surface deformation in landslide areas, and further verified the feasibility of applying Sentinel-1A images to landslide early warning.\n\nBased on this research, and using the Jiangdingya landslide area in Zhouqu County, Gansu Province as the research area, 84 ascending-orbit Sentinel-1A SAR images of this area from 2015 to 2019 are selected in this paper. The effectiveness of identification of potential landslides based on SBAS-InSAR technology is examined, the annual mean rate of surface deformation of the Jiangdingya landslide is extracted, and the deformation rate along the radar sight direction is transformed into the deformation rate along the slope direction. The latter is combined with the relationship between radar sight direction and slope direction. As a result, potential landslides in the target area are recognized.\n\n1 EARLY IDENTIFICATION OF LANDSLIDE BASED ON SBAS-INSAR TECHNOLOGY 1.1 Principle of SBAS-InSAR Technology\n\nSBAS-InSAR uses images that meet the time baseline threshold and the space baseline threshold to form several sets for differential interference. Then, high-quality interference pairs are screened based on the coherence of interference patterns, and phase unwrapping is used to solve the surface deformation sequences of each small set using the least-squares method. Finally, singular value decomposition is used to jointly solve multiple small baseline sets to obtain the surface deformation time series (Zhang Yi, 2018; Guo Leping et al., 2017).\n\nAssuming that there are N+1 SAR images covering the experimental area, the acquisition time is t={t0, t1, \u2026, tN}, and all images are freely combined and paired. There are then M image pairs that meet space-time baseline requirements, and the relationship between M and N is as follows:\n\n $\\frac{{N + 1{\\rm{ }}}}{2}{\\rm{ }} \\le M \\le N\\left({{\\rm{ }}\\frac{{N + 1{\\rm{ }}}}{2}{\\rm{ }}} \\right)$ (1)\n\nWithout considering the influence of decorrelation or atmospheric delay, the differential interference phase at a given time ti(i=1, 2, 3, \u2026, N) relative to the reference time t0 is recorded as \u03b4\u03c6(tj) (i=1, 2, 3, \u2026, M). The value of the pixel (m, n) in the I(I=1, 2, 3, \u2026, M) interferogram is then expressed as (Liu Guoxiang et al., 2019):\n\n $\\delta {\\varphi _I}\\left({m, n} \\right) = \\varphi ({t_i}, m, n) - \\varphi ({t_j}, m, n) \\approx {\\rm{ }}\\frac{{4{\\rm{ \\mathsf{ \u03c0} }}}}{\\lambda }\\left[ {{\\rm{d}}({t_i}, m, n) - {\\rm{d}}({t_j}, m, n)} \\right]$ (2)\n\nwhere \u03bb is the radar wavelength, and d(ti, m, n) and d(tj, m, n) are the deformation along the radar line of sight (LOS) at ti and tj, respectively.\n\nIf d(tj, m, n)=0, Equa.(2) can be simplified as:\n\n $\\varphi \\left({t, m, n} \\right) = {\\rm{ }}\\frac{{4{\\rm{ \\mathsf{ \u03c0} }}}}{\\lambda }{\\rm{d}}({t_i}, m, n), i = 1, 2, \\ldots, N$ (3)\n\nSince differential interference is performed on a pixel-by-pixel basis, the vector composed of phases with high coherence points in SAR images of all time nodes is denoted as \u03c6T, and the vector formed by the phase after the interferogram is unwrapped is recorded as \u03b4\u03c6T; as shown in the following formulas:\n\n $\\begin{array}{*{20}{c}} {{\\varphi ^{\\rm{T}}} = \\left[ {\\varphi \\left({{t_1}} \\right), \\varphi \\left({{t_2}} \\right), \\ldots, \\varphi \\left({{t_N}} \\right)} \\right]}\\\\ {\\delta {\\varphi ^{\\rm{T}}} = [\\delta \\varphi ({t_1}), \\delta \\varphi ({t_2}), \\ldots, \\delta \\varphi ({t_M})]} \\end{array}$ (4)\n\nThe time series of the main and slave images are:\n\n ${\\rm{IM = }}\\left[ {{\\rm{I}}{{\\rm{M}}_{\\rm{1}}}{\\rm{, I}}{{\\rm{M}}_{\\rm{2}}}{\\rm{, \\ldots, I}}{{\\rm{M}}_M}} \\right]{\\rm{, IS = }}\\left[ {{\\rm{I}}{{\\rm{S}}_{\\rm{1}}}{\\rm{, I}}{{\\rm{S}}_{\\rm{2}}}{\\rm{, \\ldots, I}}{{\\rm{S}}_M}} \\right]{\\rm{}}$ (5)\n\nSuppose the master and slave images are sorted by a time series, i.e. IMj>ISj, j=1, 2, \u2026, M, then the differential interference phase is expressed as follows:\n\n $\\delta {\\varphi _j} = \\varphi ({t_{{\\rm{IM}}{{\\rm{}}_i}}}) - \\varphi ({tI_{{\\rm{}}{{\\rm{S}}_i}}})$ (6)\n\nFor all interference pairs, Equa. (6) can be simplified as:\n\n $\\delta \\varphi = A\\varphi$ (7)\n\nwhere each row in matrix A[M\u00d7N] corresponds to an interference pair, and each column corresponds to a scene SAR image (Li Shanshan et al., 2013). When MN, i.e., the rank of matrix A is N, Equa.(7) can be solved by the least-squares method:\n\n $\\varphi = {[{A^{\\rm{T}}}A]^{ - 1}}{A^{\\rm{T}}}\\delta \\varphi$ (8)\n\nIn the actual calculation process, due to the poor independence of M equations and the different combinations between the baselines, a deficient rank of matrix A occurs, i.e., the rank of matrix A is less than N. Because of this, the singular value decomposition (SVD) method should be used to solve Equa.(7) (Liu Guoxiang et al., 2019). However, if the phase is solved directly using the SVD method, the phase value will be discontinuous in time, which does not conform to the motion law of the landslide. Therefore, it is necessary to solve the phase change rate as an unknown quantity (Li Shanshan et al., 2013). Given this, the phase change rate vector can be expressed as:\n\n ${v^{\\rm{T}}} = {\\rm{ }}\\left[ {{v_1} = {\\rm{ }}\\frac{{{\\varphi _1}}}{{{t_1} - {t_0}}}, \\ldots, vN = {\\rm{ }}\\frac{{{\\varphi _N} - {\\varphi _{N - 1}}}}{{{t_N} - {t_{N - 1}}}}} \\right]$ (9)\n\nEqua.(7) can be converted to:\n\n $\\sum\\limits_{i = {\\rm{I}}{{\\rm{S}}_j} + 1}^{{\\rm{I}}{{\\rm{M}}_j}} {({t_i} - {t_{i + 1}}){v_i}} {\\rm{ }} = \\delta {\\varphi _j}, j = 1, \\ldots, M$ (10)\n\nEqua.(10) can be simplified as:\n\n $\\begin{array}{*{20}{c}} {Bv = \\delta \\varphi }\\\\ {B\\left[ {i, j} \\right] = {\\rm{ }}\\left\\{ {\\begin{array}{*{20}{c}} {{t_j} + 1 - {t_j}, ({\\rm{I}}{{\\rm{S}}_{j + 1}} \\le j \\le {\\rm{I}}{{\\rm{M}}_j}, {\\forall _i} = 1, \\ldots, M)}\\\\ 0 \\end{array}} \\right.} \\end{array}$ (11)\n\nwhere B is the M\u00d7N order matrix. The phase change rate v is obtained by decomposing B using the SVD method, and then v is integrated to find the phase change value in each period. Finally, the deformation sequence along the radar line of sight is obtained by multiplying the coefficient (\u03bb\/4\u03c0).\n\n1.2 Transformation of Line of Sight Deformation Rate and Slope Deformation Rate\n\nBecause the deformation information monitored by SBAS-InSAR technology lies along the radar line of sight, and the sliding direction of a landslide is downward along the slope face, it is necessary to transform the deformation information along the line of sight into deformation information along the slope direction to express the real deformation situation in the landslide area (Zhang Shijia et al., 2018). Fig. 1 shows that according to the slope and aspect of the landslide area, combined with the line of sight incident angle, the angle between the satellite flight direction and the positive north direction, the transformation of the deformation information along the line of sight direction to the deformation information along the slope direction can be completed. The transformation formula is as follows:\n\n $\\left\\{ {\\begin{array}{*{20}{c}} {{V_{{\\rm{SLOPE}}}} = {V_{{\\rm{LOS}}}}\/{\\rm{cos}}\\beta }\\\\ {{\\rm{cos}}\\beta = (- {\\rm{sin}}\\alpha {\\rm{cos}}\\varphi)(- {\\rm{sin}}\\theta {\\rm{cos}}\\eta) + (- {\\rm{cos}}\\alpha {\\rm{cos}}\\varphi)({\\rm{sin}}\\theta {\\rm{sin}}\\eta) + {\\rm{sin}}\\varphi {\\rm{cos}}\\theta } \\end{array}} \\right.$ (12)\n Fig. 1 Geometric schematics of radar line of sight and slope direction (Cascini L. et al., 2010)\n\nwhere VLOS represents the deformation rate along the radar line of sight direction, VSLOPE stands for the deformation rate along the slope direction, \u03b2 is the angle between the -line of sight direction and the surface of the slope, i.e., \u03c6 is the slope of the slope; \u03b1 the aspect of the slope, \u03b8 represents the incidence angle, and \u03b7 represents the angle between the satellite orbit direction and the due north direction.\n\n2 ANALYSIS AND VERIFICATION OF EXPERIMENTAL RESULTS 2.1 Overview of the Research Area and Experimental Data\n\nIn order to study the applicability of early landslide identification using SBAS-InSAR technology, the Jiangdingya landslide in Zhouqu County, Gansu Province, is selected as the research area, as shown in Fig. 2. The Jiangdingya landslide is located on the left bank of the Bailong River, 9 km away from Zhouqu County, with a height difference of 240 m and a mountain slope of nearly 50\u00b0, providing favorable terrain conditions for the occurrence of landslide disaster (Guo Changbao et al., 2019). The Jiangdingya landslide has experienced substantial slips, especially large-scale slips in 1991 (Mu Peng, 2011) and 2018 (Guo Changbao et al., 2019), resulting in numerous secondary disasters and huge losses.\n\n Fig. 2 Location of research area\n\nThe SAR data used in the experiment are 84 Sentinel-1A ascending-orbit images provided by ESA. For its parameters, the imaging mode is interferometrically wide, the breadth is 250 km, the range resolution is 20 m, and the azimuth resolution is 5 m. The detailed parameters are shown in Table 1. The referenced digital elevation model (DEM) is SRTM DEM provided by NASA with a spatial resolution of 30 m. In addition, all the optical images used in this article are provided by Google Earth. The optical images in Fig. 2(a) are Google Earth \"Grade 13\" images, while the optical images in Fig. 2 (b) and Figs. 5-8 are Google Earth \"Grade 17\" images.\n\nTable 1 Parameters of Sentinel-1A ascending-orbit images\n\n Fig. 3 Time and position of all image pairs\n\n Fig. 4 Time and space baselines of all image pairs\n\n Fig. 5 Surface deformation rate in line of sight from 2015 to 2019\n\n Fig. 6 Surface deformation rate in slope direction from 2015 to 2019\n\n Fig. 7 Deformation time series rate in line of sight\n\n Fig. 8 Surface deformation monitoring results of SBAS-InSAR based on ascending-orbit data\n2.2 Analysis of Experimental Results\n\n84 Sentinel-1A ascending-orbit images with a time baseline span of nearly 4 years are matched in sequence, resulting in 170 interference pairs. Among them, the average absolute space baseline is 58.419 m and the average absolute time baseline is 72.141 days. The image on December 9th, 2016 is selected as the super main image. The time and image positions of all pairs are shown in Fig. 3, and the time baseline and spatial baseline of all pairs are shown in Fig. 4.\n\nThe surface deformation rate along the radar line of sight in the Jiangdingya landslide area from 2015 to 2019 is obtained based on SBAS-InSAR technology treatment, as shown in Fig. 5, and the average surface deformation rate reaches -5.711\u20149.994 mm\/a. The absolute value of the deformation rate indicates its magnitude information, and the positive and negative deformation rate indicate the direction of deformation. In detail, a negative value indicates that the deformation point moves along the satellite line of sight near the SAR direction, while a positive value indicates that the deformation point moves along the satellite line of sight away from the SAR direction.\n\nFig. 5 shows that the deformation points are concentrated at the bottom, middle, and upper parts of the slope, where the terrain is rugged, the mountains are high, and the valleys are deep. Such regions are potential high-risk area for a landslide. The slope and aspect of the research area are extracted according to the reference DEM, and the transformation of deformation information along the line of sight to deformation information along the slope direction is completed by combining the incidence angle of the line of sight and the angle between the satellite flight direction and the positive north direction, as shown in Fig. 6. In order to avoid exaggerating, the following information is observed: when -0.3 < cos\u03b2 < 0, cos\u03b2 =-0.3, when 0 \u2264 cos\u03b2 < 0.3, cos\u03b2 =0.3 (Herrera G. et al., 2013). According to the results, the maximum average deformation rate of the landslide area along the slope direction in 2015-2019 is -31.750 mm\/year. The negative value indicates the deformation point sliding down the slope.\n\n2.3 Verification of Experimental Results\n\nThe latest Jiangdingya landslide occurred on July 12th, 2018. A large number of landslide bodies poured into the river channel with a total volume of approximately 10 000 m3, leading to the water level increase and the subsequent formation of a barrier lake (Guo Changbao et al., 2019). Fig. 7 shows the deformation time series of the Jiangdingya area from 2015 to 2018 before the landslide. All deformations are cumulative deformation which refer to the date of January 1st, 2015. Fig. 7 shows that the surface deformation of the Jiangdingya landslide area gradually accelerates, and the seriously deformed area gradually increases. As of July 8th, 2018, the maximum surface deformation in the main landslide area is 32.931 mm, and the surface deformation process in this area is consistent with the law of landslide body movement.\n\nTo further analyze the relationship between surface deformation and landslide movement in the Jiangdingya area, the annual average surface deformation rate of the region from January 2015 to May 2019 is obtained. In Fig. 8, the red frame regions are those with serious deformation in the surface deformation monitoring results obtained using SBAS-InSAR technology. There are a large number of deformation points in this area, and the annual average deformation variable at the place with the largest deformation has a maximum value of 9.994 mm, indicating that this area is exactly where the Jiangdingya landslide is located.\n\nAt present, the four landslide revival areas in the Jiangdingya landslide area (shown in Fig. 9) are still in a creeping state and are likely to resurrect again under the action of heavy rainfall and river erosion (Guo Changbao et al., 2019). The early identification results of the Jiangdingya landslide based on SBAS-InSAR technology are consistent with existing results, suggesting that SBAS-InSAR technology has high applicability in the early identification of landslides.\n\n Fig. 9 Development characteristics of steep wall at the rear edge of the Jiangdingya landslide (Guo Changbao et al., 2019)\n3 CONCLUSION\n\nLandslides in the mountainous areas in western China are scattered over numerous places, most of which are located at high altitude. These regions are characterized by a slowly deformed surface after geological disasters Relying on traditional investigative methods, the investigation period is long and the risk coefficient is high; moreover, it is challenging to locate and provide early warning of potential landslide hazards in advance. Using the Jiangdingya landslide area as the study area, an examination on the effectiveness of identification of potential landslides based on SBAS-InSAR technology is conducted. In addition, with the support of a large number of long-term sequence ascending-orbit Sentinel-1A images from 2015 to 2019, the surface deformation information of this area is extracted, and the time series results of surface deformation are obtained, successfully identifying the Jiangdingya landslide. Compared with existing research results, the early identification method of landslide hazard based on SBAS-InSAR technology has high applicability and utilization value, which will considerably aid the early identification of landslides in large areas.\n\nREFERENCES\n Babu A., Kumar S.Babu A., Kumar S. SBAS interferometric analysis for volcanic eruption of Hawaii island[J]. Journal of Volcanology and Geothermal Research, 2019, 370: 31-50. Cascini L., Fornaro G., Peduto D.Cascini L., Fornaro G., Peduto D. Advanced low- and full-resolution DInSAR map generation for slow-moving landslide analysis at different scales[J]. Engineering Geology, 2010, 112(1\/2\/3\/4): 29-42. Colesanti C., Ferretti A., Prati C., Rocca F.Colesanti C., Ferretti A., Prati C., Rocca F. Monitoring landslides and tectonic motions with the Permanent Scatterers Technique[J]. Engineering Geology, 2003, 68(1\/2): 3-14. Dai Keren, Li Zhenhong, Tom\u00e1s R., Liu Guoxiang, Yu Bing, Wang Xiaowen, Cheng Haiqin, Chen Jiajun, Stockamp J.Dai Keren, Li Zhenhong, Tom\u00e1s R., Liu Guoxiang, Yu Bing, Wang Xiaowen, Cheng Haiqin, Chen Jiajun, Stockamp J. Monitoring activity at the Daguangbao mega-landslide (China) using Sentinel-1 TOPS time series interferometry[J]. Remote Sensing of Environment, 2016, 186: 501-513. Dong Jie, Zhang Lu, Tang Minggao, Liao Mingsheng, Xu Qiang, Gong Jianya, Ao MengDong Jie, Zhang Lu, Tang Minggao, Liao Mingsheng, Xu Qiang, Gong Jianya, Ao Meng. Mapping landslide surface displacements with time series SAR interferometry by combining persistent and distributed scatterers:A case study of Jiaju landslide in Danba, China[J]. Remote Sensing of Environment, 2018, 205: 180-198. Ferretti A., Prati C., Rocca F.Ferretti A., Prati C., Rocca F. Permanent scatterers in SAR interferometry[J]. IEEE Transactions on Geoscience and Remote Sensing, 2001, 39(1): 8-20. Ge Daqing. Research on the key techniques of SAR interferometry for regional land subsidence monitoring[D]. Beijing: China University of Geosciences (Beijing), 2013 (in Chinese with English abstract). Guo Changbao, Ren Sanshao, Li Xue, Zhang Yongshuang, Yang Zhihua, Wu Ruian, Jin JijunGuo Changbao, Ren Sanshao, Li Xue, Zhang Yongshuang, Yang Zhihua, Wu Ruian, Jin Jijun. Development characteristics and reactivation mechanism of the Jiangdingya ancient landslide in the Nanyu Town, Zhouqu County, Gansu Province[J]. Geoscience, 2019, 33(1): 206-217 (in Chinese with English abstract). Guo Leping, Yue Jianping, Yue ShunGuo Leping, Yue Jianping, Yue Shun. Application of SBAS technique in surface subsidence monitoring of Nanjing Hexi Area[J]. Bulletin of Surveying and Mapping, 2017(3): 26-28, 41 (in Chinese with English abstract). Herrera G., Guti\u00e9rrez F., Garc\u00eda-Davalillo J.C., Guerrero J., Notti D., Galve J.P., Fern\u00e1ndez-Merodo J.A., Cooksley G.Herrera G., Guti\u00e9rrez F., Garc\u00eda-Davalillo J.C., Guerrero J., Notti D., Galve J.P., Fern\u00e1ndez-Merodo J.A., Cooksley G. Multi-sensor advanced DInSAR monitoring of very slow landslides:The Tena Valley case study (Central Spanish Pyrenees)[J]. Remote Sensing of Environment, 2013, 128: 31-43. Huang RunqiuHuang Runqiu. Some catastrophic landslides since the twentieth century in the southwest of China[J]. Landslides, 2009, 6(1): 69-81. Li Shanshan, Li Zhiwei, Hu Jun, Sun Qian, Yu XiaoyingLi Shanshan, Li Zhiwei, Hu Jun, Sun Qian, Yu Xiaoying. Investigation of the seasonal oscillation of the permafrost over Qinghai-Tibet Plateau with SBAS-InSAR algorithm[J]. Chinese Journal of Geophysics, 2013, 56(5): 1476-1486 (in Chinese with English abstract). Liu ChuanzhengLiu Chuanzheng. Awareness of two issues in geological disaster prevention and control[J]. The Chinese Journal of Geological Hazard and Control, 2015, 26(3): 1-2 (in Chinese). Liu Guoxiang, Chen Qiang, Luo Xiaojun, Cai Guolin. Principle and Application of InSAR[M]. Beijing: Science Press, 2019 (in Chinese). Liu Peng, Li Zhenhong, Hoey T., Kincal C., Zhang Jingfa, Zeng Qiming, Muller J.P.Liu Peng, Li Zhenhong, Hoey T., Kincal C., Zhang Jingfa, Zeng Qiming, Muller J.P. Using advanced InSAR time series techniques to monitor landslide movements in Badong of the Three Gorges region, China[J]. International Journal of Applied Earth Observation and Geoinformation, 2013, 21: 253-264. Mondini A.C., Santangelo M., Rocchetti M., Rossetto E., Manconi A., Monserrat O.Mondini A.C., Santangelo M., Rocchetti M., Rossetto E., Manconi A., Monserrat O. Sentinel-1 SAR amplitude imagery for rapid landslide detection[J]. Remote Sensing, 2019, 11(7): 760. Mu PengMu Peng. Analysis on causes and stability of landslide at Jiangdingya in Zhouqu County of Gansu Province[J]. China Water Resources, 2011(4): 50-52 (in Chinese with English abstract). Niu Yufen, Dzurisin D., Lu ZhongNiu Yufen, Dzurisin D., Lu Zhong. Interferometric synthetic aperture radar study of recent eruptive activity at Shrub mud volcano, Alaska[J]. Journal of Volcanology and Geothermal Research, 2019, 387: 106671. ent\u00fcrk S., \u00c7ak\u0131r Z., Ergintav S., Karabulut H.ent\u00fcrk S., \u00c7ak\u0131r Z., Ergintav S., Karabulut H. Reactivation of the Ad\u0131yaman Fault (Turkey) through the Mw 5.7 2007 Sivrice earthquake:An oblique listric normal faulting within the Arabian-Anatolian plate boundary observed by InSAR[J]. Journal of Geodynamics, 2019, 131: 101654. Wang ZhihuaWang Zhihua. Remote sensing for landslide survey, monitoring and evaluation[J]. Remote Sensing for Land & Resources, 2007(1): 10-15, 23 (in Chinese with English abstract). Yin Hongjie, Zhu Jianjun, Li Zhiwei, Ding Xiaoli, Wang ChangchengYin Hongjie, Zhu Jianjun, Li Zhiwei, Ding Xiaoli, Wang Changcheng. Ground subsidence monitoring in mining area using DInSAR SBAS algorithm[J]. Acta Geodaetica et Cartographica Sinica, 2011, 40(1): 52-58 (in Chinese with English abstract). Yin YuepingYin Yueping. Strengthen the thinking of geological disaster prevention and control in the process of urbanization[J]. The Chinese Journal of Geological Hazard and Control, 2013, 24(4): 5-8 (in Chinese). Zhang Shijia, Jiang Jianjun, Miao Yamin, Bai ShibiaoZhang Shijia, Jiang Jianjun, Miao Yamin, Bai Shibiao. Application of the SBAS technique in potential landslide identification in the Minjiang Watershed[J]. Mountain Research, 2018, 36(1): 91-97 (in Chinese with English abstract). Zhang Yi. Detecting ground deformation and investigating landslides using InSAR technique-taking middle reach of Bailong River Basin as an example[D]. Lanzhou: Lanzhou University, 2018 (in Chinese with English abstract).","date":"2021-03-09 06:58:34","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5578179359436035, \"perplexity\": 6340.2466056443345}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-10\/segments\/1614178389472.95\/warc\/CC-MAIN-20210309061538-20210309091538-00586.warc.gz\"}"}
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from django.contrib.auth.forms import AuthenticationForm
from django import forms
class LoginForm(AuthenticationForm):
"""LoginForm
Form inherited from AuthenticationForm used for the user login
Main goal is to use widget to get a good UI
"""
def __init__(self, *args, **kwargs):
super(LoginForm, self).__init__(*args, **kwargs)
self.fields['username'].required = True
self.fields['username'].widget = forms.TextInput(
attrs={
'class': 'form-control',
'placeholder': 'E-mail',
'name': 'email',
'type': 'email',
'autofocus': ''
}
)
self.fields['password'].required = True
self.fields['password'].widget = forms.PasswordInput(
attrs={
'class': 'form-control',
'placeholder': 'Password',
'name': 'password',
'type': 'password',
}
)
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 8,960
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Interview: Patriot Ledger
The Patriot Ledger (MA), Written By Bernard Weinraub
Magnolia is Cinematic Wake-Up Call
It's a day and night of heavy rains in the San Fernando Valley of southern California. Over the course of three hours, in the forthcoming film Magnolia, at least a dozen quite different people intersect in sometimes odd ways. There are a television game-show host and his angry, estranged daughter; there are a boy genius who appears on the game show and his ambitious father; there are a dying old man, his young sexy wife and his lost son; there's a policeman in love.
"It's funny," said the film's writer-director, Paul Thomas Anderson, 29. "I fell in love with movies as an adolescent growing up in the Valley, and I thought I could never be a great filmmaker because I had never lived on the mean streets of New York, or I had never been in a war.
"Once you get past that, and once you think where you're from and what you've seen makes for good stories, too, you realize you can do it too."
No one would dispute that.
Anderson's acclaimed 1997 film, Boogie Nights, about the world of pornography in the Valley in the late 1970s and early '80s, put him on the map as one of Hollywood's most innovative and talented young filmmakers. His newest film, which will open in New York and Los Angeles around Christmas -- and probably after the first of the year elsewhere in the country -- has dazzled early audiences. Executives at New Line Cinema, its distributor, predict that the alternately dark and funny movie will emerge as one of the most discussed and controversial films of the new year.
Magnolia breaks the standard studio mold of the usual prestige melodrama," said Mike DeLuca, New Line's president for production. "It doesn't pander, it doesn't manipulate, and it defies convention with its structure and its imagery. It's a cinematic wake-up call illustrating what ails us at the end of the century."
The film is set on or near Magnolia Boulevard, a main thoroughfare in the San Fernando Valley. Its common theme seems to be the loneliness of so many of its characters and the family relationships and bonds that break and mend over a day and night.
The actors, many of them from Boogie Nights, include Julianne Moore, Philip Seymour Hoffman, William H. Macy, John C. Reilly, Jeremy Blackman, Melinda Dillon and Philip Baker Hall.
Also in the supporting cast are Jason Robards as the dying old man and, of all people, Tom Cruise, as his son. In an unusual secondary, but flashy, role for the movie star, Cruise plays a charismatic sex guru who makes television infomercials for men's empowerment.
Cruise has never before played such an on-the-edge role. (Associates of Cruise said he might avoid promoting the film because he did not want to give the impression that Magnolia was a traditional Cruise movie.)
Anderson said that Cruise had called him after seeing Boogie Nights in London, while making Stanley Kubrick's Eyes Wide Shut. Coincidentally Anderson was in London, and Cruise invited him to the movie set to meet Kubrick.
"It was like meeting J.D. Salinger," Anderson said. "I was thrilled."
Cruise also told Anderson to keep him in mind for his next film. Anderson, who had already begun writing Magnolia, said he had him in mind for a role and would call Cruise in six or seven months.
"I finally sent him the script, and the next day Tom called me and said, `Please come to my house to talk about it,' " recalled Anderson. "And away we went. What I said to him then was, `If you hadn't called me, I never would have thought of you.' A movie star like Tom Cruise was, I thought, out of my reach."
What may be out of reach for some viewers may be the film's length: three hours. (Boogie Nights ran 2 hours 35 minutes.)
"The challenge for us, quite simply, is not only having a three- hour film but also having a film that's not easily described in a single line," said Robert Friedman, co-chairman for marketing at New Line. "The challenge is to get people in to see the movie. Once they do, it's a pretty darn amazing experience."
Anderson said the people at New Line were "cool" to the length and knew that the movie would be this long from the moment they received his script.
"Making a movie at this length does set you up for criticism, and in this day and age it's slightly dangerous to do." He said, "It's slightly arrogant and a little bold to require three hours of someone's time to tell a story. It means you really have to deliver."
He laughed. "Like, if I hear a movie I'm going to see is three hours, I get a little uneasy."
Anderson, whose late father, Ernie Anderson, earned a living making television voice-overs, is now writing a script for Jonathan Demme. But he said he promised Demme he wouldn't discuss it.
"It's not set in the Valley," he said. "I'm getting out of the Valley."
Posted by modage at 10/11/1999 04:15:00 PM
Article Mentions: interview, magnolia, paul thomas anderson
Fiona Apple - Fast As You Can (1999)
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"redpajama_set_name": "RedPajamaCommonCrawl"
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| 2,517
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Victoria Ying The Beauty of Snow in Spring From Disney Beauty And The Beast Original Watercolor on Paper
Fine Artists ›
Fine Art ›
Judy Larson Gallery
Ebenezer and the War Horse Limited Edition Print
Judy Larson Ebenezer and the War Horse Limited Edition Print is eligible for layaway in 3 equal payments of $75.00 over 60 days.
1/15/2021 $75.00 1st payment
2/14/2021 $75.00 2nd payment
3/16/2021 $75.00 3rd & final payment
As an option you may also pay for Judy Larson Ebenezer and the War Horse Limited Edition - Print using Paypal or with your Amazon Account(*select items). Please note that all orders must be delivered to a physical address verified by Paypal or Amazon. These options are not applicable for orders to be delivered to Military or International destinations.
Ebenezer and the War Horse
LIMITED EDITION PRINT - JL00052 Print
NOTES: "Out of the millions of horses that have shared our country's history, only a few have distinguished themselves enough for their names to be remembered. Among these select few is Ebenezer, Chief Joseph's renowned Appaloosa racehorse. Red roan in color, with large blood-red spots on his white rump, he was not the prettiest horse in the herd, but he could run like the wind. So famous was he that newspapers in Walla Walla and Lewiston, Washington recorded each time Chief Joseph rode him into town. Everyone with a good, fast horse of his own, whether frontiersman or Native American, dreamed of the day when his horse would beat Chief Joseph's Ebenezer. But, alas! Ebenezer won all his races well out in front of the best horses in the region.
The black horse portrayed in Ebenezer and the War Horse honors another of Chief Joseph's horses. Although his name has not survived through time, he was special, as well, for he was the horse that Chief Joseph chose to ride to surrender."
Ebenezer and the War Horse Limited Edition Print by Judy Larson is signed by the artist and comes with a certificate of authenticity.
Judy Larson bio
Judy Larson always knew she was going to be an artist. She was surrounded by them as a child, and was particularly inspired by her father, a professional illustrator. Judy received a Bachelor of Science degree in Commercial Art from Pacific Union College in Northern California, then spent the next 17 years as a commercial artist, illustrator and art director. In 1988, influenced by her love of nature and animals, Judy devoted her time to wildlife art. Her primary focus in each of her paintings is the animal, with the horse as a recurring subject. Her unique approach to her work is through the use of scratch board--a technique that can render magnificent detail but one requiring infinite patience. Scratch board, an old, but little used medium, consists of a smooth, thin surface of hardened China clay applied to a board. The subject is then painted solidly with black India ink to create a silhouette. Now the exacting work begins, engraving the image into the surface of the artwork. While many artists use steel nibs or engraving tools, Judy prefers to work with X-acto blades, changing them ever few minutes to produce as fine a line as possible. Once the subject has been totally scratched, it is a finished black and white illustration, ready for the artist to add color. The methods of adding color are diverse. Judy prefers a combination of airbrush, gouache or acrylics for finishing, with frequent rescratching for detail. Scratch board is a demanding medium, one that Judy has used masterfully in developing her unique approach to wildlife art.
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\section{Introduction\label{intro}}
High order harmonic generation (HHG) occurs when atoms or
molecules exposed to an intense femtosecond laser pulse are ionized by
tunneling. The freed electron is then accelerated in the external electric
field. Because of the periodic oscillation of the laser field, the electron is
brought back to the parent ion where it may recombine emitting an XUV photon
\cite{corkum1993}. This XUV radiation has been shown to contain information on
the electronic structure of the emitting molecule and on its internal dynamics.
Attosecond nuclear \cite{baker2006} and electronic dynamics \cite{smirnova2009,
haessler2010} have been extracted from HHG in simple molecules and spectral
features in the harmonic emission have been related to the molecular electronic
structure and have been used for imaging the highest occupied molecular orbital
(HOMO).
The idea of exploiting HHG for the tomographic reconstruction of
molecular orbitals was first introduced by Itatani et al. in 2004 for the
nitrogen molecule \cite{itatani2004}.
Since then, numerous experiments have been realized, addressing the role of the
HOMO in the harmonic spectral intensity \cite{kanai2005,vozzi2005},
in the molecular-frame photo-ionization \cite{le2009a} and in the subsequent
attosecond XUV emission \cite{boutu2008}, as well as in the polarization state
of the emitted radiation \cite{levesque2007}.
The dependence of the HHG process on the HOMO structure has also been exploited
for the characterization in the time domain of the rotational
\cite{miyazaki2005} and vibrational \cite{li2008} molecular excitations.
All these studies rely on two major assumptions: (i) the
molecular HHG is dominated by the HOMO structure; (ii) the relationship between
molecular structure and emitted XUV spectrum is simple and completely captured
by the Strong Field Approximation (SFA), i.e. the electron quiver motion is not
perturbed by the Coulomb potential of the ion.
Both these assumptions have been recently put into question.
Recent experiments enlightened the role of multiple orbital contributions to HHG
emission \cite{smirnova2009,haessler2010}. Furthermore, the influence of the
Coulomb field of the parent ion in the generation of high order harmonics from
molecules has been considered as a serious hindrance to a clear HOMO
reconstruction \cite{walters2008}. These assumptions should then be overtaken to
perform molecular tomography to more complex species.
Besides these two more fundamental obstacles, there are also
additional, more technical difficulties. In order to retrieve the HOMO
structure, one has to record the XUV harmonic spectra for different molecular
orientations with respect to the laser field. Hence, it is necessary to fix the
molecular orientation in space and change the polarization direction of the
HHG-driving field \cite{itatani2004}.
Laser-assisted molecular alignment is a widespread technique able to accomplish
this task \cite{stapelfeldt2003}, but the molecular alignment achieved in this
way is not ideal. Hence the experimental results and the corresponding HOMO
tomography are affected by angular averaging effects. Moreover, in the case of
non-linear molecules, the tomographic procedure requires to fix two or three
angular coordinates of the molecule under investigation. For instance, the study
of linear polar molecules requires to fix the head-tail direction in space. The
feasibility of laser assisted molecular orientation has been recently
demonstrated \cite{le2009b} and exploited in HHG spectroscopy
\cite{frumker2012a,frumker2012b,spanner2012}, but no direct application to
molecular imaging has been yet realized.
The amount of information that can be extracted from the
harmonic emission depends on the spectral extension of the XUV radiation, that
is known to scale with the so-called cut-off law: $E_{max} = I_p + 3.17U_p$,
where $I_p$ is the ionization potential of the molecule and $U_p$ is the
ponderomotive energy of the electron in the laser field. This poses another
important problem when HHG molecular imaging is extended to species with low
ionization potential (i.e. all organic molecules, and in particular those having
important biological functions) as the extension emission spectrum is reduced.
Since $\mathrm{U_p} \propto \lambda^2 \mathrm{I}$, where I is the peak intensity
and $\lambda$ the wavelength of the driving laser pulse, the emission cut-off
may be extended by both increasing the field intensity or the laser wavelength.
In this respect, standard Ti:Sapphire laser sources generally used in HHG are
not ideal candidates for tomography in fragile molecules, since the intense
optical fields needed completely ionize the molecule before a well-developed
XUV spectrum is generated.
To overcome the limitations posed by ionization saturation,
the exploitation of mid-infrared driving sources has been demonstrated to be a
powerful tool to extend harmonic emission far in the XUV
\cite{takahashi2008,vozzi2010,vozzi2010b,vozzi2010c,popmintchev2012}.
With a mid-IR source\cite{vozzi2007} we recently demonstrated
that it is possible to extend the spectral investigation in carbon dioxide
beyond 100 eV in the absence of multielectron effects, thus avoiding any
ambiguity in the reconstructed wavefunction. In addition, by exploiting an
all-optical non-interferometric technique, it was possible to trace both the
spectral intensity and phase of high order harmonics generated by single
molecules as a function of emitted photon energy and molecular angular
orientation, without averaging effects. Furthermore, the tomographic procedure
was generalized in order to take into account the Coulomb potential seen by the
re{-}colliding electron wavepacket\cite{vozzi2011}.
In this work, we extend that approach to more complex molecules, such as N$_2$O
and C$_2$H$_2$ pointing out some strengths and weaknesses of this investigation
technique.
\section{Experimental Setup\label{setup}}
We exploited an optical parametric amplifier (OPA) pumped by an amplified
Ti:sapphire laser system (60 fs, 20 mJ, 800 nm). The OPA is based on difference
frequency generation and provides driving pulses with 1450 nm central
wavelength, pulse duration of 20 fs and pulse energy of 1.2 mJ\cite{vozzi2007}.
High harmonics were generated by focusing the mid-IR pulse in a
supersonic gas jet under vacuum, due to the strong absorption exhibited by air
in the XUV spectral region.
The molecules in the jet were impulsively aligned with a portion of the
fundamental 800-nm beam which was spectrally broadened by optical filamentation
in an argon-filled gas cell and temporally stretched up to 100 fs by propagation
through a glass plate. Such duration is required for achieving a good alignment
of the molecular sample.
In our experimental setup, driving and aligning pulse were collinear and their
polarizations were parallel. The delay between the two pulses was adjusted by
means of a fine-resolution translation stage. The XUV radiation was acquired
by means of a flat-field spectrometer and a multi-channel plate detector coupled
to a CCD camera \cite{poletto2001}.
\section{Results\label{results}}
\begin{figure}[!ht]
\centering
\includegraphics[width=0.8\textwidth]{FD171-hhg-N2O}
\caption{(a) Sequence of harmonic spectra measured in N$_{2}$O as a function of
emitted photon energy and delay between the aligning and the driving pulse (log
scale). (b) Calculated alignment factor for N$_{2}$O in the experimental
conditions (rotational temperature 75 K, aligning pulse duration 100 fs,
aligning pulse intensity $3.32\times 10^{13}$ W/cm$^2$). \label{hhg_N2O}}
\end{figure}
\begin{figure}[!ht]
\centering
\includegraphics[width=0.8\textwidth]{FD171-hhg-C2H2}
\caption{(a) Sequence of harmonic spectra measured in C$_{2}$H$_{2}$ as a
function of emitted photon energy and
delay between the aligning and the driving pulse (log scale). (b) Calculated
alignment factor for C$_{2}$H$_{2}$ in the experimental conditions (rotational
temperature 75 K, aligning pulse duration 100 fs, aligning pulse intensity
$2.16\times 10^{13}$ W/cm$^2$). \label{hhg_C2H2}}
\end{figure}
Harmonic spectra were acquired in N$_{2}$O and C$_{2}$H$_{2}$ as a function of
the delay $\tau$ between the aligning and driving pulse around the first
rotational half revival ($\tau_{N_2O}=19.95$ ps and $\tau_{C_2H_2}=7.08$ ps).
The results are shown in figure \ref{hhg_N2O}(a) and \ref{hhg_C2H2}(a) for
N$_{2}$O and C$_{2}$H$_{2}$ respectively. Figures \ref{hhg_N2O}(b) and
\ref{hhg_C2H2}(b) show the corresponding calculated alignment factor for the
experimental conditions.
In both molecules, the sequence of harmonic spectra shows a strong modulation
with the delay $\tau$ that can be ascribed to the dependence of harmonic yield
on the molecular orbital structure. In particular, a reduction of the harmonic
emission can be observed for the delay corresponding to the maximum of the
alignment factor and an enhancement of the harmonic yield appears for the
minimum of the alignment factor. A major difference between the two cases is
the presence of a region of harmonic enhancement at high photon energy, that
appears in N$_2$O at maximum alignment.
These effects can be naively interpreted in terms of two-center interference
occurring in the re-collision step \cite{lein2002, vozzi2005}. If one consider a
diatomic homo-nuclear molecule with a symmetric electronic state with respect
to the nuclei exchange and assumes the re-colliding electron as a plane wave,
the condition for constructive interference reads $R\cos(\theta) = n \lambda_B$,
where $R$ is the internuclear separation, $\theta$ is the angle between the
molecular axis and the electron wave-vector, $n$ is an integer number and
$\lambda_B$ is the de Broglie wavelength associated to the re-colliding electron
wave-packet. Similarly the condition for destructive interference is
$R\cos(\theta) = (n+1/2) \lambda_B$ and the first destructive interference
occurs for $n=0$. The conditions become reversed for molecules with
antisymmetric electronic structure.
This concept can be extended to the molecules subject of our investigation. The
acetylene molecule has a symmetric $\pi$ HOMO in which the separation between
the carbon atoms is $R_{\mathrm{C \equiv C}}=1.2\ \textrm{\AA}$. This is the
distance that
should be considered for the evaluation of the interference condition. The
N$_2$O HOMO does not have a clear symmetry, however in our experimental
condition the harmonic spectra are acquired in aligned molecules and correspond
to the average between the two possible orientation. The resulting signal can
be
interpreted in terms of emission from an effective molecular orbital similar to
the anti-symmetric $\pi$ orbital of CO$_2$. In this view the overall length of
this ``effective'' orbital is $R_{\mathrm{N_2O}}=2.3\ \textrm{\AA}$. Since
$R_{\mathrm{N_2O}}
\approx 2 R_{\mathrm{C \equiv C}}$, a destructive interference occurs in the
same spectral
region for both molecules, corresponding to $n=1$ for N$_2$O and $n=0$ for
C$_2$H$_2$.
Figures \ref{hhg_N2O}(a) and \ref{hhg_C2H2}(a) show two peculiar advantages
related to the exploitation of mid-IR driving pulses for HHG. Indeed the
harmonic cutoff extension related to the increase in the ponderomotive energy
with respect to standard Ti:sapphire sources allows the observation of spectral
features as the harmonic enhancement for high photon energy visible in N$_2$O
in correspondence of the revival peak. In the framework of the above mentioned
two-center model, this feature can be attributed to the appearance of
constructive interference in that spectral region. Moreover, for the same
emitted photon energy, mid-IR driving wavelengths require a lower pulse peak
intensity thus reducing the ionization saturation in species with relatively
low ionization potential, such as $C_2H_2$ ($\mathrm{I_P}=11.4$ eV).
\section{Reconstruction of Single Molecule XUV Emission \label{SMXEM}}
From the experimental data reported in figures \ref{hhg_N2O}(a) and
\ref{hhg_C2H2}(a) it is possible to retrieve structural information on the
target molecule following the approach introduced by Vozzi et al.
\cite{vozzi2011}. Figures
\ref{macro_N2O}(a) and \ref{macro_C2H2}(a) show the same experimental results
presented in figures \ref{hhg_N2O}(a) and \ref{hhg_C2H2}(a), in which the
harmonic structure due to the periodic re-collision of the electron wave-packet
has been filtered out. These results have been exploited for the reconstruction
of the XUV field emitted from a single molecule and projected on the
polarization direction
of the aligning field as a function of the angle between the molecular
axis and the driving polarization direction. The reconstruction is based on a
combination of a phase-retrieval algorithm and a Kaczmarz algorithm
\cite{popa2004}. The main idea behind this approach is that the macroscopic XUV
emission is the coherent superposition of the XUV field emitted by all
molecules
weighted with their angular distribution. This distribution changes along the
revival in a predictable way,
hence the sequence of harmonic emission contains enough
information for the reconstruction of the harmonic electric field in amplitude
and phase.
\begin{figure}[!ht]
\centering
\includegraphics[width=0.8\textwidth]{FD171-macro-N2O}
\caption{(a) Sequence of XUV spectra measured in N$_{2}$O as a function of
emitted photon energy and delay between the aligning and the driving pulse; the
harmonic structure has been filtered out. Retrieved macroscopic harmonic
emission amplitude (b) and phase (c) corresponding to the data reported in
(a).\label{macro_N2O}}
\end{figure}
\begin{figure}[!ht]
\centering
\includegraphics[width=0.8\textwidth]{FD171-macro-C2H2}
\caption{(a) Sequence of XUV spectra measured in C$_{2}$H$_{2}$ as a function
of
emitted photon energy and delay between the aligning and the driving pulse; the
harmonic structure has been filtered out. Retrieved macroscopic harmonic
emission amplitude (b) and phase (c) corresponding to the data reported in
(a).\label{macro_C2H2}}
\end{figure}
The result of this reconstruction is shown in figure \ref{SMXEM_N2O} for
N$_{2}$O and in figure \ref{SMXEM_C2H2} for C$_{2}$H$_{2}$. In both figures,
panel (a) reports the amplitude of the XUV field and panel (b) shows the
corresponding phase. In N$_{2}$O there is a clear phase jump of about 2 rad,
that changes its position with photon energy and molecular alignment. This
phase jump corresponds to a minimum in the XUV amplitude and its position is
quite in good agreement with the prediction of the naive two-center model
introduced above, which is shown as a dashed line in the figure.
It is worth noting that the reconstruction technique is based on the
interference of XUV emission from different molecular orientations, thus the
phase can be retrieved as a function of $\theta$ at fixed XUV photon energy. In
order to retrieve the phase relationship between contributions at neighboring
energies, it is necessary to introduce an a priori condition that can be
derived
from theoretical considerations or experimental measurements. In the case of
N$_{2}$O we imposed a flat spectral phase of the macroscopic harmonic emission
for the delay corresponding to the molecular anti-alignment. This condition was
chosen in analogy with the CO$_{2}$ case\cite{vozzi2011}, due to the similarity
between the two HOMOs as discussed in the previous section.
The results of this assumption can be observed in figure \ref{macro_N2O}, where
the reconstructed amplitude (b) and phase (c) of the macroscopic XUV emission
from N$_{2}$O are reported. The retrieved
amplitude is in good agreement with the experimental data (figure
\ref{macro_N2O}(a)). The phase of the macroscopic emission shows a steep change
of about 2 rad around 50 eV at the delay $\tau$ corresponding to the maximum
alignment.
\begin{figure}[!ht]
\centering
\includegraphics[width=0.8\textwidth]{FD171-SMXEM-N2O}
\caption{Retrieved single molecule XUV emission map in N$_{2}$O as a function
of
emitted photon energy and the angle between the molecular axis and the aligning
beam polarization direction in amplitude (a) and phase(b). Dashed lines show
the
position of the destructive interference predicted by the two-center model.
\label{SMXEM_N2O}}
\end{figure}
\begin{figure}[!ht]
\centering
\includegraphics[width=0.8\textwidth]{FD171-SMXEM-C2H2}
\caption{Retrieved single molecule XUV emission map in C$_{2}$H$_{2}$ as a
function of emitted photon energy and the angle between the molecular axis and
the aligning beam polarization direction in amplitude (a) and phase(b). Dashed
lines show the position of the destructive interference predicted by the
two-center model. \label{SMXEM_C2H2}}
\end{figure}
In the case of C$_{2}$H$_{2}$ we followed the same approach in the retrieval
procedure.
We imposed in this case a flat spectral phase for the macroscopic harmonic
emission
at the delay $\tau$ corresponding to the molecular alignment.
This assumption was necessary in order to complete the retrieval procedure, but
it is arbitrary
and not supported by theoretical models; it could be however improved by
changing the retrieving
condition according to an experimental spectral phase measurement.
This kind of experiment can be performed for example by RABBIT technique at a
given alignment delay \cite{boutu2008}.
The retrieved single molecule XUV emission in C$_{2}$H$_{2}$, shown in figure
\ref{SMXEM_C2H2}, is very different from the one reported for N$_{2}$O. In
particular a strong contribution comes from molecules with perpendicular
orientation with respect to the driving field polarization direction. In the
retrieved phase (figure \ref{SMXEM_C2H2}(c)) two phase jumps are clearly
observed. The first one appears for small alignment angles and roughly follows
the prediction of the two-center model. The second jump appears at large
alignment angles and may be attributed to the shape of the HOMO seen by the
re-colliding electron. However, since the reconstruction is based on the
arbitrary assumption of flat macroscopic spectral phase at the alignment delay,
the retrieved outcomes should be considered preliminary.
In spite of this, the retrieved macroscopic XUV amplitude (figure
\ref{macro_C2H2}(b))
is in fair agreement with the experimental results.
\section{Molecular Orbital Tomography \label{tomo}}
The results reported in the previous section can be used for the
two-dimensional
reconstruction of molecular orbitals, following the tomographic procedure
proposed by Itatani et al. \cite{itatani2004} and extended by Vozzi et al.
\cite{vozzi2011}. However to proceed with this tomographic reconstruction, it
is necessary to rule out the occurrence of multi-electron effects in HHG.
A simple experimental procedure to check whether spectral modulations in
harmonic emission are due to multi-electron effects is to change the driving
field intensity. As shown by Smirnova et al. \cite{smirnova2009}, one expects
all the features
due to multi-electron effects to shift with the driving field intensity. Figure
\ref{intscan} shows harmonic spectra acquired in aligned N$_{2}$O for a delay
$\tau$ corresponding to the maximum of the alignment for different values of
the driving intensity. The spectral minimum associated to the phase change
retrieved in figure \ref{SMXEM_N2O}(b) appears always around 55 eV and does not
shift with the intensity. This behavior guarantees that the main spectral features in the
harmonic emission are mainly dictated by the HOMO structure. This consistency
check allowed us to exploit the retrieved single molecule harmonic emission for
the reconstruction of N$_{2}$O orbital. The result is shown in figure
\ref{homo_N2O}(a). Figure \ref{homo_N2O}(b) shows the N$_{2}$O orbital
calculated with a quantum chemistry program \cite{dalton}. Even if the overall
dimension of the
molecular orbital is well reproduced, the asymmetry of this orbital is very
clear and cannot be addressed by the tomographic reconstruction, since in the
experiment the molecules were aligned but not oriented. Another departure of
the retrieved orbital with respect to the calculated one is the presence of side
lobes, that can be attributed to the limited working range of the XUV
spectrometer used in these experiments. Since there is a correspondence between
the energy range of harmonic emission and the spatial frequency domain, the
limited spectral range collectible in the experiment corresponds to a spatial
filtering in the Fourier domain, which gives raise to such lobes. These
observations are further confirmed by figure \ref{homo_N2O}(c), which shows the
calculated HOMO corresponding to the average between the two possible
orientations of N$_{2}$O molecular axis and takes into account the limited
spectral bandwidth available in
the experiment. The features of this fictitious orbital are in very good
agreement with the reconstruction of figure \ref{homo_N2O}(a).
It is worth nothing that such limitations can be overcome by extending the
acquired spectral range over all the XUV emission and by exploiting all-optical
impulsive techniques for orientation of polar molecules, such the one
demonstrated by Frumker et al.\cite{frumker2012a,frumker2012b}.
\begin{figure}[!ht]
\centering
\includegraphics[width=0.5\textwidth]{FD171-intscan}
\caption{Harmonic spectra generated in N$_{2}$O at the delay $\tau$
corresponding to the maximum molecular alignment for several driving peak
intensities I between 1 and $1.7\times 10^{14}$ W/cm$^2$. \label{intscan}}
\end{figure}
\begin{figure}[!ht]
\centering
\includegraphics[width=0.8\textwidth]{FD171-homo-N2O}
\caption{(a) Highest occupied molecular orbital of N$_{2}$O as retrieved from
the single molecule XUV emission map. (b) Highest occupied molecular orbital of
N$_{2}$O calculated with a quantum chemistry program \cite{dalton}. (c)
N$_{2}$O
HOMO calculated averaging over the two possible orientations of the molecular
axis and considering the filtering in spectral domain corresponding to the
experimental conditions.\label{homo_N2O}}
\end{figure}
Differently from the case of N$_{2}$O, in C$_{2}$H$_{2}$ it is not possible to
easily rule out the multi-electron contributions.
Because of the smaller cutoff energy, the experimental approach applied in the
case of N$_{2}$O for the exclusion of multi-electron contribution is not feasible.
Nevertheless the application of tomographic approach to the single molecule
emission maps shown in figure \ref{SMXEM_C2H2} provides interesting results. We
show in figure \ref{homo_C2H2}(a) the retrieved C$_{2}$H$_{2}$ HOMO. Also in
this case, a comparison with the result calculated with a quantum chemistry
program (see figure \ref{homo_C2H2}(b)) shows a good agreement in the overall
shape of the orbital. Again the additional lobes are related to the limited
harmonic range detected in the experimental acquisition, as can be seen in
figure \ref{homo_C2H2}(c) where the orbital is calculated taking into account
the spectral filtering.
\begin{figure}[!ht]
\centering
\includegraphics[width=0.8\textwidth]{FD171-homo-C2H2}
\caption{a) Highest occupied molecular orbital of C$_{2}$H$_{2}$ as retrieved
from the single molecule XUV emission map. (b) Highest occupied molecular
orbital of C$_{2}$H$_{2}$ calculated with a quantum chemistry program
\cite{dalton}. (c) C$_{2}$H$_{2}$ HOMO calculated considering the filtering in
spectral domain corresponding to the experimental conditions. \label{homo_C2H2}}
\end{figure}
\section{Conclusions\label{conclusions}}
Since the pioneering work of Itatani et al. on molecular orbital imaging, the
impressive advances in laser technologies
gave the access to new mid-IR sources for driving HHG and pushing the harmonic
emission far towards the soft-X ray range.
These sources allowed the application of HHG spectroscopy to fragile molecules
as hydrocarbons, which play as prototypes
for the study of ubiquitous phenomena in chemistry and material science.
In this work we showed the application of molecular orbital reconstruction based
on HHG to non-trivial samples, such as N$_{2}$O and
C$_{2}$H$_{2}$. These results, though requiring further improvements,
demonstrate the capability of molecular orbital tomography and
represent the first step towards the imaging of dynamical processes in complex
molecules.
\section*{Acknowledgements}
The research leading to these results has received funding from LASERLAB-EUROPE
(grant agreement n° 284464, EC Seventh Framework Programme),
from ERC Starting Research Grant UDYNI (grant agreement n° 307964, EC Seventh
Framework Programme)
and from the Italian Ministry of Research and Education (ELI project - ESFRI
Roadmap).
\noindent\small
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{"url":"https:\/\/fall2022.byucs110.org\/labs\/lab9-scripts\/","text":"Computer Science\n\n# Lab 9 - Scripts\n\n## Pytest\n\nFrom now on, we will provide you with the pytests we use to grade your labs and projects. These tests will provide insight on what errors are occurring in your code and how to fix them.\n\nTo install pytest, go to your terminal and run:\n\nconda activate cs110\n\nThen run:\n\npip install byu-pytest-utils\n\n## What does this print?\n\nDraw it out!\n\ndef divides_by_three(number):\nreturn number % 3 == 0\n\ndef main(number):\nif divides_by_three(number):\nprint('woot!')\nelse:\nprint('boo')\n\nmain(6)\nmain(7)\nmain(8)\nmain(9)\ndef under_ten(number):\nreturn number % 10\n\ndef main(number):\nnumber = under_ten(number)\nnumber = number + (number \/\/ 2)\nprint(number)\n\nmain(7)\nmain(17)\nmain(32)\nn = 0\nwhile n < 4:\nprint(n)\nn = n + 1\ndef increment_and_mod(n, mod):\nreturn (n + 1) % mod\n\niteration = 0\nn = 0\nwhile iteration < 10:\niteration = iteration + 1\nprint(n)\nn = increment_and_mod(n, 3)\ndef add(a, b):\nreturn a + b\n\ntotal = 0\nn = 0\nwhile n < 5:\nprint(total)\ndef print_to_zero(n):\nwhile n >= 0:\nprint(n)\nn = n - 1\n\nprint_to_zero(3)\nprint_to_zero(4)\ndef is_even(number):\nreturn number % 2 == 0\n\ndef print_smaller_evens(n):\nwhile n >= 0:\nn = n - 1\nif is_even(n):\nprint(n)\n\nprint_smaller_evens(10)\nstart = 5\nend = 10\n\nn = start\nwhile n < end:\nprint(n)\nn = n + 1\ndef add_all(a, b, c, d):\nresult = a + b + c + d\n\nprint(add_all(1, 2, 3, 4))\n\n## Activities\n\n### odd_numbers.py\n\nYou are provided with the script odd_numbers.py; however, it isn\u2019t working as desired.\n\nThe function print_lower_odds(number) should print out all odd numbers less than number, from greatest to least, each number being prefixed by >> .\n\nFor example:\n\nprint_lower_odds(10)\n\nShould print:\n\n>> 9\n>> 7\n>> 5\n>> 3\n>> 1\n\nUse the debugger to help you identify the problems and then fix them.\n\n### oddities.py\n\nWrite a function print_oddities_up_to(number) that it prints all \u201coddities\u201d less than number from least to greatest.\n\nIn this case, a number is an \u201coddity\u201d if it is not divisible by 2 or 3.\n\nWrite at least one additional function that determines whether a number is an oddity or not.\n\nFor example:\n\nprint_oddities_up_to(8)\n\nWould print:\n\n1\n5\n7\n\n### math_practice.py\n\nImplement the functions found in math_practice.py following the instructions provided in the docstrings.\n\nodd_numbers.py6 points\noddities.py6 points\nmath_practice.py3 points for each function","date":"2023-03-30 09:11:06","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.3709118664264679, \"perplexity\": 6585.910197651351}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2023-14\/segments\/1679296949107.48\/warc\/CC-MAIN-20230330070451-20230330100451-00081.warc.gz\"}"}
| null | null |
\section{Introduction}
\bigskip Chiral perturbation theory ($\chi $PT) has now been applied to the
problem of nucleon-nucleon (NN) interactions for almost two decades. Since
Refs.~\cite{We90,We91}, chiral potentials have been deduced from $\chi$PT
Lagrangians~\cite{Or96, Ka97, Ep99, EM02,EM03, Ep05} through the computation
of all NN-irreducible diagrams that occur up to a given order in $\chi$PT.
The resulting potential includes a variety of contact interactions, which
parameterize our ignorance of physics coming from energies higher than the
chiral-symmetry-breaking scale, $\Lambda _{\chi}$. Some of these contact
interactions occur in the $\pi$N Lagrangian, and the low-energy constants
(LECs) that multiply these operators can be determined independently from
the pion-nucleon scattering data~\cite{Fe98, Bu00}. More of a
challenge---both conceptual and practical---is posed by the string of
operators that represent the short-distance physics in the NN sector. These
operators renormalize the divergent loops found when calculating the $\chi$%
PT potential, and so encode the contribution to observables from high-energy
NN states. Each such contact interaction is associated with an unknown LEC. $%
\chi$PT power counting applied to the NN potential mandates that at a fixed
order in the chiral expansion only a finite number of LECs enter the
potential. For instance, at $O(P^3)$ in the NN S-waves (which will be where
the bulk of our attention is in this work) there are five contact operators
that must be considered: two act in the ${}^1$S$_0$ channel, two in the ${}^3
$S$_1$ channel, and one causes coupling between the ${}^3$S$_1$ and ${}^3$D$%
_1$ partial waves.
For low partial waves the $\chi$PT potential is non-perturbative, and the
standard approach involves use of the Lippmann-Schwinger equation (LSE) to
reconstruct the full NN amplitude from the NN-irreducible part~\cite%
{Le97,Ge99,EM06,Dj07}. (C.f. Refs.~\cite%
{KSW98A,KSW98B,FMS99A,FMS99B,Be02,Ol03,BM04,BB03,LvK08,Be08} where parts of
the potential are treated in perturbation theory.) Since the NN interactions
obtained in $\chi$PT do not fall off as the nucleon momenta $\mathbf{p}$ go
to infinity, a cutoff, denoted here by $\Lambda$, must be imposed on the
intermediate states in the LSE. The LECs are then fitted to data for a
variety of cutoffs, and should absorb any strong (power-law with power $>0$
or logarithmic) dependence of observables on $\Lambda$. A corollary is that
the predictions of the effective theory should not depend on the particular
quantity chosen to fix the LECs, as long as the kinematic point is within
the domain of validity of the chiral effective theory. A potential
which does not have these desirable properties (approximate cutoff and
renormalization-point independence) is not \textquotedblleft properly
renormalized\textquotedblright. If the set of contact operators employed
cannot achieve this we conclude that this ``chiral effective
theory'' ($\chi$ET) is unable to give reliable results, since its
predictions are too sensitive to the treatment of the unknown short-distance
physics at scale $\Lambda$. In Refs.~\cite{Or96,Ep99,EM02,EM03,Ep05} the
potential $V$ was computed to a fixed order, and then the NN LECs that
appear in $V$ were fitted to NN data for a range of cutoffs between 500 and
800~MeV. The resulting predictions---especially the ones obtained with the $%
O(P^4)$ potential derived in Refs.~\cite{EM03,Ep05}---are approximately $\Lambda-$independent and describe NN data with
considerable accuracy.
However, the range of $\Lambda$ considered in these papers is quite narrow.
Does the fact that these analyses only consider $\Lambda \leq 800$~MeV
represent an intrinsic limitation on the $\chi$ET approach?
In the case of leading-order potential (which is $O(P^0)$ and whose
long-range part consists only of one-pion exchange) the answer has been
shown to be ``No" in the $^1$S$_0$ and and $^{3}$S$_{1}-^{3}$D$_{1}$
channels. The problem can be properly renormalized for arbitrarily large $%
\Lambda$'s with only constant contact interactions present in $V$ in each
channel---in accordance with the $\chi$PT power counting~\cite{SDW94, Fr99,
ES01, Be02, PVRA04B, PVRA05, NTvK05, Bi06, Ya08}~\footnote{%
In the ${}^1$S$_0$ channel the constant must be $m_q$-dependent~\cite{KSW96}
in order to absorb a logarithmic divergence. This violates the power
counting, but is of no practical consequence as regards the description of
laboratory NN data.}. However, once $\Lambda > 600$ MeV the $\chi$PT
counting for the NN potential fails in waves with $L > 0$ where the tensor
force from one-pion exchange is attractive~\cite%
{ES01,NTvK05,Bi06,PVRA06B,Ya09}. That counting predicts no contact
interaction in these channels, but a contact interaction is necessary to
stabilize the LSE predictions at these cutoffs. Furthermore, in these
partial waves it has been found that even when higher-order $\chi$PT
potentials are considered and the phase shifts obtained from the LSE are
approximately cutoff independent, the results can still show significant
renormalization-point dependence~\cite{Ya09}.
At next-to-leading order (NLO, $O(P^2)$) and next-to-next-to-leading order
(NNLO, $O(P^3)$), the $\chi$PT potential includes diagrams with
two-pion-exchange (TPE). Since the loop integrals there diverge as the
square or cube of the momentum transfer $|\mathbf{q}| \equiv |\mathbf{p}%
^{\prime}-\mathbf{p}|$, the TPE must be associated with contact terms up to
order $P^2$---just as one would have predicted using naive-dimensional
analysis in powers of NN momentum.
In Refs.~\cite{Ep99,epsfr,Ti05,En08} the role of various different $O(P^2)$
contact terms in the LSE was examined and the renormalization of the NN
S-waves was discussed. Refs.~\cite{Ep99,epsfr,En08} found that
momentum-dependent contact terms had difficulty absorbing the cutoff
dependence at larger cutoffs, and also that the contribution from the
(supposedly higher-order) TPE was larger than one-pion exchange. (See Ref.~%
\cite{Sh08} for a similar conclusion in the ${}^1$S$_0$ channel using a
co-ordinate-space analysis.) Ref.~\cite{En08} also considered an
energy-dependent $O(P^2)$ contact interaction in the ${}^1$S$_0$ channel and
showed that the use of a contact interaction linear in energy (which
formally should be equivalent to the momentum-dependent interaction at the
orders in the chiral expansion being considered here) leads to the
appearance of resonances there for $\Lambda \rightarrow \infty$. In either
case, the results of these studies suggest that for cutoffs larger than
about 1 GeV the contribution from contact terms dominates over that from the
long-range part of the potential, raising the question of whether $\chi$PT
is acting as a phenomenological fit form, rather than an effective theory.
As argued in Refs.~\cite{EM06,Ya09,Sh08}, the chiral-symmetry-breaking scale
could be responsible for this situation.
This suggests that there may be a critical cutoff $%
\sim \Lambda _{\chi }$ above which it does not make sense to iterate the TPE
inside the LSE\footnote{See \cite{EG09} for a recent analytic approach which yields a similar conclusion.}.
It is the purpose of this
paper to critically examine the behavior of the NN phase shifts predicted by
$\chi$ET at NLO and NNLO in the ${}^1$S$_0$ and ${}^3$S$_1$--${}^3$D$_1$
channels over a range of cutoffs from $\Lambda=500$ MeV to $\Lambda > 2$
GeV. In this way we hope to identify this critical cutoff, if it exists, and
discuss the signatures of, and mechanisms responsible for, its appearance.
In order to address these issues we extend our previously developed
subtractive-renormalization method~\cite{Ya08,Ya09} to evaluate the $^{1}$S$%
_{0}$ and $^{3}$S$_{1}-^{3}$D$_{1}$ phase shifts with the higher-order $\chi$%
ET potentials. If energy-dependent contact interactions are employed in
these potentials we show how to relate the $\chi$ET phase shifts to on-shell
quantities. For momentum-dependent contact interactions we formulate the
scattering equations in terms of two (one) on-shell quantities in the
triplet (singlet) channel and then fit the single remaining unknown
parameter to experimental data. This allows us to easily go to high cutoffs
in the LSE and provides clean information about the effect of the
renormalization point. Our method works for any long-range potential. Here
we consider long-range parts evaluated using both dimensional regularization
(DR) and spectral-function-regularization (SFR)\cite{epsfr} at $O(P^2)$ and $%
O(P^3)$. The forms of these potentials were given in Ref.~\cite{Ya09}. Since
issues like differing extractions of the $\pi$N LECs $c_1$, $c_3$ and $c_4$
could produce sizeable changes in the phase shifts we are not overly focused
on the quality of our fit. Instead, our purpose is to see whether these
potentials behave self-consistently with respect to cutoffs in the LSE.
Thus, our strategy is to evaluate in detail the cutoff-dependence of the
S-wave phase shifts, and examine the renormalization-point dependence. These
features depend much less on the precise choice made for $c_1$, $c_3$ and $%
c_4$.
The structure of our work is as follows. First, in Sec.~\ref{sec-singlet},
we introduce our subtractive-renormalization method and explain how to deal
with both momentum- and energy-dependent contact terms in the singlet
channel. We take as input to the method the ${}^1$S$_0$ scattering length, $%
a_s$, and the phase shift at a particular energy, $E^*$. Then, in Sec.~\ref%
{sec-triplet}, we extend our subtractive-renormalization method to the
coupled-channels case and deal with constant, energy-dependent and
momentum-dependent contact terms. While the constant contact term is, as in
Ref.~\cite{Ya08}, solved by one subtraction with $a_t$ as input, for the
energy-dependent contact term we pin down the three unknown constants with
information with $a_t$, the generalized scattering length $\alpha_{20}$ that
governs the low-energy behavior of $\epsilon_1$, and the phase shift at a
particular energy as inputs. In Sec.~\ref{sec-result1}, we discuss our
results in the $^1$S$_0$ channel and their implications. In Sec.~\ref%
{sec-result2}, we discuss our results in the $^3$S$_1$-${}^3$D$_1$ channel
and their implications. We summarize our findings in Sec.~\ref{sec-con}.
\section{Subtractive renormalization in the singlet channel}
\label{sec-singlet} The short-range part of the potential, i.e., the contact
terms, is a parameterization of unknown high-energy physics. Before
introducing our subtractive renormalization scheme, we need to decide what
types of contact term we should consider. Throughout this work we will
consider chiral two-pion-exchange~\cite{Or96,Ka97,Re99,Ep99} as the
long-range part of the NN potential. This is associated with contact terms
up to $O(P^2)$. However, since for non-relativistic particles we have $%
\mathbf{p}^2/M \sim E$ there is the possibility to consider
energy-dependent, instead of momentum-dependent contact interactions. In
particular, this may have certain advantages in terms of evading theorems
that limit the impact of short-distance potentials on phase shifts~\cite%
{Wi55,PC96}. Thus, in the ${}^1$S$_0$ channel we will consider the following
three types of contact terms:
\begin{itemize}
\item[(A)] $v_{SR,0}=\lambda $
\item[(B)] $v_{SR,0}(E)=\lambda +\gamma E$
\item[(C)] $v_{SR,0}(p^{\prime},p)=\lambda +C_2 \ (p^{2}+p^{\prime 2})$,
\end{itemize}
where $\lambda ,$ $\gamma ,$ $C_2$ are unknown constants. (Of course, the
numerical value of $\lambda $ is different for each case.) The overall
potential is then given by
\begin{equation}
v_0(p^{\prime},p;E)=v_{SR,0}(p^{\prime},p;E) + v_{LR,0}(p^{\prime},p;E),
\end{equation}
with the subscripts $_{SR}$ and $_{LR}$ standing for the short- and
long-range parts of the potential. Specifically, in the ${}^1$S$_0$ channel
we have
\begin{equation}
v_{LR,0}=\langle 000|V_C + W_C - 3 [V_S + W_S + \mathbf{q}^2 (V_T +
W_T)]|000 \rangle,
\end{equation}
with expressions for the S-wave projected central, spin, and tensor
isoscalar and isovector potentials taken from Ref.~\cite{Ka97,epsfr} and
collected in Ref.~\cite{Ya09}.
\subsection{The constant contact interaction}
The constant contact term can be evaluated by one subtraction with the ${}^1$%
S$_0$ scattering length $a_s$ as input~\cite{Ya08}. This is possible because
knowledge of the on-shell amplitude plus the long-range potential yields the
fully off-shell amplitude, as long as only a constant contact interaction is
present~\cite{HM01,AP04,Ya08}. The step from the on-shell amplitude at an
arbitrary energy $E^*$, $t(p_0^*,p_0^*;E^*)$ (with $p_0^* \equiv ME^*$), to
the half-off-shell amplitude at the same energy is explained in Appendix~\ref%
{appendixa}. The symmetry property of the amplitude $%
t(p,p_0^*;E^*)=t(p_0^*,p;E^*)$ then means that the same sequence of
manipulations can be used to construct $t(p^{\prime},p;E^*)$. With the
fully-off-shell amplitude at a single energy in hand, it is well known how
to construct the $t$-matrix at any arbitrary energy~\cite{Fr99}. In operator
form we have
\begin{equation}
t(E)=t(E^*) + t(E^*)[g_0(E) - g_0(E^*)] t(E), \label{eq:EEstar}
\end{equation}
where $g_0$ is the free resolvent of the Lippmann-Schwinger equation,
\begin{equation}
g_0(E;p)=\frac{1}{E^+ - p^2/M}.
\end{equation}
(Here $M$ is the nucleon mass and $E^+=E+i \varepsilon$, with $\varepsilon$
a positive infinitesimal.)
\subsection{Energy-dependent contact interaction}
\label{sec-1S0endep}
In this case there are two unknown constants. The strategy is to use one
subtraction (as in Ref.~\cite{Ya08}) to first eliminate the constant part of
the contact term. For the energy-dependent contact term we then perform a
second subtraction to eliminate the remaining unknown.
We start from the partial-wave LS equation, which is written explicitly in
this channel as
\begin{eqnarray}
t_0(p^{\prime },p;E)=v_{0}(p^{\prime },p;E)+ \frac{2}{\pi }%
M\int_{0}^{\Lambda }\frac{dp^{\prime \prime }\;p^{\prime \prime
}{}^{2}\;v_0(p^{\prime },p^{\prime \prime };E)\;t_{0}(p^{\prime \prime },p;E)%
}{p_{0}^{2}+i\varepsilon -p^{\prime \prime }{}^{2}}. \label{eq:1a}
\end{eqnarray}
The energy-dependent ${}^1$S$_0$ potential is given by
\begin{equation}
v_{0}(p^{\prime },p;E)=v_{LR }(p^{\prime},p)+\lambda + \gamma E.
\label{eq:4.3}
\end{equation}
Here $v_{LR}$ represents the long range part of the $\chi$PT potential,
which is computed up to NLO or NNLO. However, the derivation presented below
holds for any energy-independent long-range potential which is a function of
$p$ and $p^{\prime }$ and satisfies $v_{LR}(p^{\prime
},p)=v_{LR}(p,p^{\prime })$. From now on we drop the partial-wave index,
since this section only describes the singlet channel $^1$S$_0$.
To further simplify the presentation, we adopt the following operator
notation of the LS equation
\begin{eqnarray}
t(E)=\lambda +\gamma E+v_{LR } + \left[\lambda +\gamma E+v_{LR } \right] \
g_0(E) \ t(E). \label{eq:4.4}
\end{eqnarray}
Setting $E=0$ in Eq.~(\ref{eq:4.4}) leads to
\begin{eqnarray}
t(0)=\lambda +v_{LR } + \left[ \lambda +v_{LR} \right] \ g_0(0) \ t(0).
\label{eq:4.5}
\end{eqnarray}
Equation (\ref{eq:4.5}) contains only one unknown, $\lambda $. Therefore,
the matrix element $t(p^{\prime },p;0)$ can be obtained from one
experimental point, i.e., the NN scattering length, $a_{s}$, using the
procedure explained in Subsection A, and given in detail in Ref.~\cite{Ya08}%
. Using properties of the LSE, Eq.~(\ref{eq:4.5}) can be rewritten as
\begin{eqnarray}
t(0)&=&\lambda +v_{LR }+t(0)g_0(0) \left(\lambda +v_{LR }\right) \cr &=&%
\left[1+t(0)g_0(0)\right] \ \left[\lambda +v_{LR }\right]. \label{eq:4.65}
\end{eqnarray}
This allows to reconstruct $\lambda$ if we wish to do so:
\begin{eqnarray}
\lambda +v_{LR }= \left[1+t(0)g_0(0)\right]^{-1} \ t(0).
\label{eq:4.7}
\end{eqnarray}
\noindent Leaving $E$ finite, Eq.~(\ref{eq:4.4}) can be written as
\begin{eqnarray}
t(E)=\left[ \lambda +\gamma E+v_{LR } \right] \ \left[1+g_0(E)t(E)\right],
\label{eq:4.55}
\end{eqnarray}
leading to a formal expression for the potential term
\begin{eqnarray}
\lambda +\gamma E+v_{LR }=t(E) \ \left[1+g_0(E)t(E) \right]^{-1}.
\label{eq:4.6}
\end{eqnarray}
Subtracting Eq.(\ref{eq:4.7}) from Eq.(\ref{eq:4.6}) isolates the term $%
\gamma E$,
\begin{eqnarray}
\gamma E=t(E) \ \left[1+g_0(E)t(E)\right]^{-1}-\left[1+t(0)g_0(0)\right]%
^{-1} \ t(0). \label{eq:4.75}
\end{eqnarray}
Proceeding in a similar fashion as in Ref.~\cite{Ya08}, we can obtain $%
t(E^{\ast })$ at a fixed energy $E^{\ast }$ from the value of the NN phase
shift at $E^\ast $ (for details see Appendix \ref{appendixa}). We find~%
\footnote{%
Note that some care is required if the standard subtraction method is used
to deal with the singularity when the integration momentum becomes equal to $%
p_0$. In that case it is useful to introduce an additional term in the
integral equation that involves the potential evaluated at the point $%
p^{\prime}=p_0$, $p=p_0$. However, if other techniques are used (e.g.
contour rotation) this additional term is not necessary and so we do not
include it in our derivation.}
\begin{eqnarray}
\lambda +\gamma E^{\ast }+v_{LR }= t(E^{\ast }) \ \left[1+g_0(E^{\ast
})t(E^{\ast })\right]^{-1}=[1+t(E^*)g_0(E^*)]^{-1} t(E^*). \label{eq:4.8}
\end{eqnarray}
Taking the difference of Eqs.~(\ref{eq:4.8}) and (\ref{eq:4.6}) leads to
\begin{eqnarray}
\gamma (E^{\ast }-E) = \left[1+t(E^{\ast }) g_0(E^{\ast })\right]^{-1}
t(E^\ast ) - t(E) \ \left[ 1+g_0(E)t(E) \right]^{-1}. \label{eq:4.9}
\end{eqnarray}
To eliminate $\gamma $, we insert Eq.~(\ref{eq:4.75}) into Eq.~(\ref{eq:4.9}%
),
\begin{eqnarray}
\Big(t(E)[1+g_0(E)t(E)]^{-1} &-& [1+t(0)g_0(0)]^{-1}t(0) \Big) \left( \frac{%
E^\ast}{E}-1 \right) \cr & =& \left[1+t(E^{\ast }) g_0(E^{\ast })\right]%
^{-1} t(E^\ast ) - t(E) \ \left[ 1+g_0(E)t(E) \right]^{-1}. \label{eq:4.95}
\end{eqnarray}
Rearranging Eq.~(\ref{eq:4.95}) and multiplying both sides with $1+g_0(E)t(E)
$ from the right and $1+t(0)g_0(0)$ from the left, we arrive at
\begin{eqnarray}
t(E) + t(0) [g_0(0) -g_0(E)] t(E) +\frac{E}{E^*} \Big\{ t(0) - \left[%
1+t(0)g_0(0)\right] \alpha t(E^{\ast })\Big\} \ g_0(E) t(E) \cr = \left(1-%
\frac{E}{E^*} \right) t(0) + \frac{E}{E^*} \Big[1+t(0)g_0(0)\Big]\alpha
t(E^{\ast }), \label{eq:4.96}
\end{eqnarray}
where $\alpha \equiv [1 + t(E^*) g_0(E^*)]^{-1}$, can be calculated
numerically. Since we already obtained the fully off-shell $t(0)$ and $t(E^*)
$, Eq.~(\ref{eq:4.96}) is an integral equation for $t(E)$, which we can
solve by standard methods.
To summarize, we perform two subtractions to the LS equation to eliminate
the two unknown constants $\lambda $ and $\gamma$. The resulting equation
requires as input the scattering length $a_{s}$ and the phase shift at one
specific energy $E^\ast$. The only restriction on $E^*$ is that it must be
within the domain of validity of our theory. Hence one can test the
consistency of the theory by examining the extent to which results depend
upon the choice of $E^*$.
\subsection{Momentum-dependent contact interaction}
The coefficient of the momentum-dependent contact interaction, denoted here by $C_2$, is not
straightforwardly related to any S-matrix element. Thus we cannot apply our
subtraction procedure in the case of a momentum-dependent contact
interaction. Hence we instead adopt a ``mixed" procedure, which involves a
single subtraction plus fitting of $C_2$.~\footnote{%
Note that this is different from the subtractive procedure of Ref.~\cite%
{Ti05} for dealing with the same contact interaction. There the Born
approximation is assumed to be valid for large negative energies.}
This ``mixed" procedure is carried out as follows: First we guess a value
for the constant $C_2$, and then insert it into the half-off-shell and
on-shell LSE:
\begin{eqnarray} \label{eq:1}
t(p,0;0) &=&v_{LR}(p,0)+\lambda +C_2 p^{2} - \frac{2}{\pi }M
\int_{0}^{\Lambda }dp^{\prime }\;\left[ v_{LR}(p,p^{\prime })+\lambda +C_2
(p^{2}+p^{\prime 2}) \right] \;t(p^{\prime },0;0), \notag \\
\\
t(0,0;0) &=&v_{LR}(0,0)+\lambda -\frac{2}{\pi }M\int_{0}^{\Lambda
}dp^{\prime }\; \left( v_{LR}(0,p^{\prime })+\lambda +C_2 p^{\prime 2}
\right) \;t(p^{\prime },0;0). \label{eq:2}
\end{eqnarray}
Taking the difference of the two equations cancels the constant $\lambda$
and leads to
\begin{eqnarray}
t(p,0;0)&=&v_{LR}(p,0)-v_{LR}(0,0)+\frac{a_{s}}{M}+C_2 p^{2} \notag \\
&-&\frac{2}{\pi } M\;\int_{0}^{\Lambda }dp^{\prime }\; \;\left(
v_{LR}(p,p^{\prime })-v_{LR}(0,p^{\prime })+C_2 p^{2} \right) \;t(p^{\prime
},0;0). \label{eq:3}
\end{eqnarray}
Using the already-chosen value of $C_2$, together with the experimental
value of $a_s$, we can solve for $t(p^{\prime},0;0)$ from Eq.~(\ref{eq:3}).
Then there is a consistency condition that determines the value of $\lambda$%
. This equation can be easily derived from Eq.~(\ref{eq:2}):%
\begin{eqnarray}
\lambda =\frac{\frac{a_{s}}{M}-v_{LR}(0,0) + \frac{2}{\pi }M\;
\int_{0}^{\Lambda }dp^{\prime }\;\left( v_{LR}(0,p^{\prime })+C_2 p^{\prime
2} \right) \;t(p^{\prime },0;0)}{1-\frac{2}{\pi } M\;\int_{0}^{\Lambda
}dp^{\prime }\;t(p^{\prime },0;0)}. \label{eq:4}
\end{eqnarray}
The above equation gives the value of $\lambda$ that is consistent with the
experimental value of the scattering length and our choice of $C_2$. It thus
defines a relationship $\lambda=\lambda(C_2;a_s)$. (Note that---in spite of
the form of Eq.~(\ref{eq:4})---the relationship is not linear, since $%
t(p^{\prime},0;0)$ is also affected by the choice of $C_2$.) Therefore, when
trying to determine $\lambda$ and $C_2$ we only need to guess $C_2$ and can
then obtain $\lambda$ from Eqs.~(\ref{eq:3}) and (\ref{eq:4}). These two
constants are then entered into the on-shell LS equation which is solved for
the phase shifts. Finally $C_2$ is adjusted to fit the desired observable.
In Sec.~\ref{sec-result1} we will examine the results obtained when $C_2$ is
adjusted to reproduce the ${}^1$S$_0$ effective range, $r_0$, and those
found when we enforce the requirement that the theory correctly predict the
phase shift at a particular energy $E^\ast$.
\section{Subtractive renormalization in the J=1 triplet channel}
\label{sec-triplet}
Since the NN interaction is non-central, the $S=1$ waves constitute a
coupled-channel problem. Here we consider the ${}^3$S$_1$-${}^3$D$_1$
coupled partial waves, and we again consider three different contact terms:
\begin{itemize}
\item[(A)] $\left(
\begin{array}{cc}
\lambda & 0 \\
0 & 0%
\end{array}
\right) $
\item[(B)] $\left(
\begin{array}{cc}
\lambda +C_2 (p^{2}+p^{\prime 2}) & \lambda_{t} \ p^{\prime 2} \\
\lambda _{t} \ p^{2} & 0%
\end{array}
\right) $
\item[(C)] $\left(
\begin{array}{cc}
\lambda +\gamma E & \lambda_t \ p^{\prime 2} \\
\lambda_t \ p^{2} & 0%
\end{array}
\right). $
\end{itemize}
Here we write the contact terms explicitly in their matrix form, where the
diagonal represents the direct channels $^3$S$_1$-$^3$S$_1$ and $^3$D$_1$-$^3
$D$_1$ and the off-diagonal the channels $^3$S$_1$-$^3$D$_1$ and $^3$D$_1$-$%
^3$S$_1$. The unknown constants are $\lambda $, $\gamma$ (or $C_2$) and $%
\lambda _{t}$, and their value is different in each case. Case A is the
leading-order contact interaction discussed in Refs.~\cite{Be02,Ya08}. Cases
B and C include the structures that appear at NLO [$O(P^2)$] in the standard
chiral counting for short-distance operators.
At this order in the chiral expansion the J=1 coupled-channel problem
acquires two additional contact interactions. One is of the form $%
\boldsymbol{\sigma}_1 \cdot \mathbf{q} \ \boldsymbol{\sigma}_2 \cdot \mathbf{%
q}$~\cite{Ep99}, and hence gives a non-zero matrix element for the $^3$S$_1$-$^3$%
D$_1$ transition. The other can be written as either an energy-dependent or
momentum-dependent piece of the $^3$S$_1$-$^3$S$_1$ potential, although
until now most works on $\chi$ET have considered only the latter~\cite%
{Or96,Ep99,EM03,Ep05}, but see Ref.~\cite{En08}. For the energy-dependent
contact interaction we will show below that three subtractions can be
performed to eliminate the unknown LECs. This leaves us with an integral
equation for the $t$-matrix at an arbitrary energy that takes as input three
experimental quantities. For case B, due to the momentum-dependence in the
contact term, we use a double-subtraction-plus-one-fit scheme to solve the
LSE. The procedures for solving case B and case C are quite similar, as they both
employ subtractions to eliminate $\lambda $ and $\lambda _{t}$.
\subsection{Constant contact interaction}
This case can be solved with a single subtraction, as described in Ref.~\cite%
{Ya08}. Once again, the key fact is that once $a_t$, the triplet scattering
length, is known, the value of $t(0,0;0)$ is fixed, and that, together with
knowledge of the long-range potential, is sufficient to obtain the two
non-zero elements of $t(p,0;0)$. The procedure by which we obtain these two
elements, $t_{00}$ and $t_{20}$, can be reconstructed from Appendix~\ref%
{appendixb} by setting $C_2=0$ in the first part of the derivation presented
there. Hermiticity of $v$ then implies that $t_{00}(p,0;0)=t_{00}(0,p;0)$
and $t_{20}(p,0;0)=t_{02}(0,p;0)$, and the same set of manipulations can
then be used to obtain $t_{l^{\prime}l}(p^{\prime},p;0)$. The operator
equation (\ref{eq:EEstar}), with $E^*=0$, then allows us to reconstruct $t(E)
$, albeit this time as a two-by-two matrix of operators, rather than the
single operator $t$ that is needed to describe the ${}^1$S$_0$ channel.
\subsection{Momentum-dependent central part}
In this case we will eliminate two constants from the integral equation: $%
\lambda$, and the coefficient of the tensor short-distance interaction $%
\lambda_t$. We will then adjust the coefficient $C_2$ to reproduce one piece
of `experimental' information.
As usual, we begin by writing the on-shell and half-off-shell partial-wave
LS equations:
\begin{eqnarray} \label{eq:5}
\left(
\begin{array}{cc}
t_{00}(p_{0},p_{0};E) & t_{02}(p_{0},p_{0};E) \\
t_{20}(p_{0},p_{0};E) & t_{22}(p_{0},p_{0};E)%
\end{array}
\right) & =& \left(
\begin{array}{cc}
\lambda +C_2 (p_{0}^{2}+p_{0}^{2})+v_{LR }^{00}(p_{0},p_{0}) & \lambda
_{t}p_{0}^{2}+v_{LR }^{02}(p_{0},p_{0}) \\
\lambda _{t}p_{0}^{2}+v_{LR }^{20}(p_{0},p_{0}) & v_{LR }^{22}(p_{0},p_{0})%
\end{array}
\right) \notag \\
+ \frac{2}{\pi }M \int_{0}^\Lambda &dp^{\prime } & \frac{p{}^{\prime 2}}{%
p_{0}^{2}-p^{\prime }{}^{2}+i\varepsilon } \left(
\begin{array}{cc}
\lambda +C_2 (p_{0}^{2}+p^{\prime 2})+v_{LR }^{00}(p_{0},p^{\prime }) &
\lambda _{t}p^{\prime 2}+v_{LR }^{02}(p_{0},p^{\prime }) \\
\lambda _{t}p_{0}^{2}+v_{LR }^{20}(p_{0},p^{\prime }) & v_{LR
}^{22}(p_{0},p^{\prime })%
\end{array}
\right) \notag \\
& \times & \left(
\begin{array}{cc}
t_{00}(p^{\prime },p_{0};E) & t_{02}(p^{\prime },p_{0};E) \\
t_{20}(p^{\prime },p_{0};E) & t_{22}(p^{\prime },p_{0};E)%
\end{array}
\right)
\end{eqnarray}
\begin{eqnarray} \label{eq:6}
\left(
\begin{array}{cc}
t_{00}(p,p_{0};E) & t_{02}(p,p_{0};E) \\
t_{20}(p,p_{0};E) & t_{22}(p,p_{0};E)%
\end{array}
\right) &=&\left(
\begin{array}{cc}
\lambda +C_2 (p^{2}+p_{0}^{2})+v_{LR }^{00}(p,p_{0}) & \lambda
_{t}p_{0}^{2}+v_{LR }^{02}(p,p_{0}) \\
\lambda _{t}p^{2}+v_{LR }^{20}(p,p_{0}) & v_{LR }^{22}(p,p_{0})%
\end{array}
\right) \notag \\
+\frac{2}{\pi }M\int_{0}^{\Lambda } & dp^{\prime } & \frac{p{}^{\prime 2}}{
p_{0}^{2}-p^{\prime }{}^{2}+i\varepsilon } \left(
\begin{array}{cc}
\lambda +C_2 (p^{2}+p^{\prime 2})+v_{LR }^{00}(p,p^{\prime }) & \lambda
_{t}p^{\prime 2}+v_{LR }^{02}(p,p^{\prime }) \\
\lambda _{t}p^{2}+v_{LR }^{20}(p,p^{\prime }) & v_{LR }^{22}(p,p^{\prime })%
\end{array}
\right) \notag \\
&\times & \left(
\begin{array}{cc}
t_{00}(p^{\prime },p_{0};E) & t_{02}(p^{\prime },p_{0};E) \\
t_{20}(p^{\prime },p_{0};E) & t_{22}(p^{\prime },p_{0};E)%
\end{array}
\right),
\end{eqnarray}
where $p_{0}^{2}/M=E$. The subscript (superscript) in $t_{l^{\prime}l}
(v^{ll^{\prime}}_{LR})$ indicates the angular-momentum quantum number of the
channels. To simplify notation, we write the $2\times2$ matrix as t.
Subtracting Eq.~(\ref{eq:6}) from Eq.~(\ref{eq:5}) cancels $\lambda $,
\begin{eqnarray}
\mathbf{t}(p,p_{0};E)&-&\mathbf{t}(p_{0},p_{0};E) =\left(
\begin{array}{cc}
C_2 (p^{2}-p_{0}^{2})+v_{LR }^{00}(p,p_{0})-v_{LR }^{00}(p_{0},p_{0}) &
v_{LR }^{02}(p,p_{0})-v_{LR }^{02}(p_{0},p_{0}) \nonumber \\
\lambda _{t}(p^{2}-p_{0}^{2})+v_{LR }^{20}(p,p_{0})-v_{LR }^{20}(p_{0},p_{0})
& v_{LR }^{22}(p,p_{0})-v_{LR }^{22}(p_{0},p_{0})%
\end{array}
\right) \\
&+&\frac{2}{\pi }M\int_{0}^{\Lambda } dp^{\prime } \frac{p{}^{\prime 2}} {%
p_{0}^{2}-p^{\prime }{}^{2}+i\varepsilon } \left(
\begin{array}{cc}
C_2 (p^{2}-p_{0}^{2})+v_{LR }^{00}(p,p^{\prime })-v_{LR
}^{00}(p_{0},p^{\prime }) & v_{LR }^{02}(p,p^{\prime })-v_{LR
}^{02}(p_{0},p^{\prime }) \\
\lambda _{t}(p^{2}-p_{0}^{2})+v_{LR }^{20}(p,p^{\prime })-v_{LR
}^{20}(p_{0},p^{\prime }) & v_{LR }^{22}(p,p^{\prime })-v_{LR
}^{22}(p_{0},p^{\prime })%
\end{array}
\right) \mathbf{t}(p^{\prime },p_{0};E). \notag \\
&& \label{eq:7}
\end{eqnarray}
\noindent Letting $E\rightarrow 0$, i.e. $p_{0}\rightarrow 0$, in Eq.~(\ref%
{eq:7}) leads to
\begin{eqnarray} \label{eq:7.5}
\mathbf{t}(p,0;0)-\mathbf{t}(0,0;0)&=&\left(
\begin{array}{cc}
C_2 p^{2}+v_{LR }^{00}(p,0)-v_{LR }^{00}(0,0) & 0 \\
\lambda _{t}p^{2}+v_{LR }^{20}(p,0) & 0%
\end{array}
\right) \notag \\
-\frac{2}{\pi }M\int_{0}^{\Lambda }&dp^{\prime }& \left(
\begin{array}{cc}
C_2 p^{2}+v_{LR }^{00}(p,p^{\prime })-v_{LR }^{00}(0,p^{\prime }) & v_{LR
}^{02}(p,p^{\prime })-v_{LR }^{02}(0,p^{\prime }) \\
\lambda _{t}p^{2}+v_{LR }^{20}(p,p^{\prime }) & v_{LR }^{22}(p,p^{\prime })%
\end{array}
\right) t(p^{\prime },0;0).
\end{eqnarray}
\noindent Here we used the threshold behavior of the partial-wave potential,
$v_{LR}^{l^{\prime }l}(p,k)$ $\sim p^{l^{\prime }}k^{l}$, to infer that,
e.g. $v_{LR }^{20}(0,p)=0$.
Equation (\ref{eq:7.5}) shows that this feature of $v_{LR}$ has as
consequence that $t_{02}(p,0;0)=t_{20}(0,p;0)=0$ and $%
t_{22}(p,0;0)=t_{22}(0,p;0)=0$. Using Eq.~(\ref{eq:7.5}) we can thus obtain
the first part of the t-matrix equation that needs to be solved, namely
\begin{eqnarray} \label{eq:7.6}
t_{00}(p,0;0)-\frac{a_{t}}{M} &=& C_2 p^{2}+v_{LR }^{00}(p,0)-v_{LR
}^{00}(0,0) \notag \\
-\frac{2}{\pi }M\int_{0}^{\Lambda }&dp^{\prime }& \Big[ \left( C_2
p^{2}+v_{LR }^{00}(p,p^{\prime })-v_{LR }^{00}(0,p^{\prime }) \right)
t_{00}(p^{\prime },0;0) + \left( v_{LR }^{02}(p,p^{\prime}) - v_{LR
}^{02}(0,p^{\prime}) \right) t_{20}(p^{\prime},0;0)\Big],
\end{eqnarray}
where we have used that $t_{00}(0,0;0)=\frac{a_{t}}{M}$. We observe that $%
\lambda_t$ has been eliminated from this equation, but that we do not have a
closed set of equations, since $t_{20}(p^{\prime},0;0)$ appears on the
right-hand side.
To eliminate $\lambda_t$ we set $p=0$ in Eq.~(\ref{eq:7})
\begin{eqnarray}
&&\mathbf{t}(0,p_{0};E)-\mathbf{t}(p_{0},p_{0};E)= \notag \\
&& \qquad \left(
\begin{array}{cc}
-C_2 p_{0}^{2}+v_{LR }^{00}(0,p_{0})-v_{LR }^{00}(p_{0},p_{0}) & v_{LR
}^{02}(0,p_{0})-v_{LR }^{02}(p_{0},p_{0}) \\
-\lambda _{t} p_{0}^{2}-v_{LR }^{20}(p_{0},p_{0}) & -v_{LR
}^{22}(p_{0},p_{0})%
\end{array}
\right) \notag \\
&& \qquad \qquad +\frac{2}{\pi }M\int_{0}^{\Lambda }dp^{\prime } \frac{%
p{}^{\prime 2}}{p_{0}^{2}-p^{\prime }{}^{2}+i\varepsilon } \left(
\begin{array}{cc}
-C_2 p_{0}^{2}+v_{LR }^{00}(0,p^{\prime })-v_{LR }^{00}(p_{0},p^{\prime }) &
v_{LR }^{02}(0,p^{\prime })-v_{LR }^{02}(p_{0},p^{\prime }) \\
-\lambda _{t}p_{0}^{2}-v_{LR }^{20}(p_{0},p^{\prime }) & -v_{LR
}^{22}(p_{0},p^{\prime })%
\end{array}%
\right) \mathbf{t}(p^{\prime },p_{0};E) \notag \\
. \label{eq:8}
\end{eqnarray}
In order to use the threshold behavior of $t_{20}$ we need to divide Eq.~(%
\ref{eq:7}) by $p^{2}-p_{0}^{2}$ and Eq.~(\ref{eq:8}) by $p_{0}^{2}$ and
then add both equations. A quick examination shows that $\gamma$ and $%
\lambda_t$ both drop out of the final result---but the final result is
quite lengthy. Hence, to simplify the notation arising from these
manipulations we define the following map
\begin{equation}
\mathcal{F} \big[O(p,p^{\prime})\big] \equiv \frac{1}{p^2-p_0^2}\Big( %
O(p,p^{\prime}) - O(p_0,p^{\prime}) \Big), \label{Fmap}
\end{equation}
where the operator $O$ can represent either the $t$-matrix or a potential
function. With this notation we obtain
\begin{eqnarray}
&& \mathcal{F}\big[\mathbf{t}(p,p_0;E)\big] - \mathcal{F}\big[\mathbf{t}%
(0,p_0;E)\big]= \notag \\
&& \qquad \left(
\begin{array}{cc}
\mathcal{F}\big[v_{LR }^{00}(p,p_{0})\big] - \mathcal{F}\big[v_{LR
}^{00}(0,p_{0})\big] & \mathcal{F}\big[ v_{LR }^{02}(p,p_{0})\big] -
\mathcal{F}\big[ v_{LR }^{02}(0,p_{0})\big] \\
\mathcal{F}\big[v_{LR }^{20}(p,p_{0})\big] - \mathcal{F}\big[ v_{LR
}^{20}(0,p_{0})\big] & \mathcal{F}\big[ v_{LR }^{22} (p,p_{0})\big] -
\mathcal{F}\big[ v_{LR }^{22} (0,p_{0})\big]%
\end{array}
\right) \notag \\
&& \qquad \qquad + \frac{2}{\pi }M\int_{0}^{\Lambda } dp^{\prime } \frac{%
p{}^{\prime 2}}{p_{0}^{2}-p^{\prime }{}^{2}+i\varepsilon } \left(
\begin{array}{cc}
\mathcal{F}\big[v_{LR }^{00}(p,p^{\prime})\big] - \mathcal{F}\big[v_{LR
}^{00}(0,p^{\prime})\big] & \mathcal{F}\big[ v_{LR }^{02}(p,p^{\prime})\big] %
- \mathcal{F}\big[ v_{LR }^{02}(0,p^{\prime})\big] \\
\mathcal{F}\big[v_{LR }^{20}(p,p^{\prime})\big] - \mathcal{F}\big[ v_{LR
}^{20}(0,p^{\prime})\big] & \mathcal{F}\big[ v_{LR }^{22} (p,p^{\prime})%
\big] - \mathcal{F}\big[ v_{LR }^{22} (0,p^{\prime})\big]%
\end{array}
\right) \mathbf{t}(p^{\prime },p_{0};E) \notag \\
&& \label{eq:9}
\end{eqnarray}
\noindent The second row of the kernel of this equation simplifies
considerably since
\begin{equation}
\mathcal{F}[v_{LR}^{20}(0,p_0)]=\frac{v_{LR}^{20}(p_0,p_0)}{p_0^2},
\end{equation}
due to the threshold behavior of $v_{LR}$. A similar result applies to $%
v_{LR}^{22}$. Thus we also have
\begin{equation}
\mathcal{F}[t_{20}(0,p_0;E)]=\frac{t_{20}(p_0,p_0;E)}{p_0^2}.
\end{equation}
Now, letting $E\rightarrow 0$ (i.e. $p_{0}\rightarrow 0$) in Eq.~(\ref{eq:9}%
), and extracting the $S \rightarrow D$ element of the resulting equation,
we can also employ (again because of threshold behavior)
\begin{equation}
\lim_{p_{0}\rightarrow 0}\mathcal{F}[v^{20}_{LR}(p,p^{\prime})]=\frac{v^{20}_{LR}(p,p^{\prime})}{%
p^2},
\end{equation}
together with the analogous result for $v^{22}_{LR}$, in order to obtain
\begin{eqnarray}
\frac{t_{20}(p,0;0)}{p^{2}} &= & \lim_{p_{0}\rightarrow 0} \left[\frac{
t_{20}(p_{0},p_{0};E)}{p_{0}^{2}}\right]+\frac{v_{LR }^{20}(p,0)}{p^{2}}
-\lim_{p_{0}\rightarrow 0}\left[ \frac{v_{LR }^{20}(p_{0},p_{0})}{p_{0}^{2}}%
\right] \\
-\frac{2}{\pi }M\int_{0}^{\Lambda } & dp^{\prime }& \Bigg[ \left( \frac{%
v_{LR }^{20}(p,p^{\prime })}{p^{2}}-\lim_{p_{0}\rightarrow 0} \left[\frac{
v_{LR }^{20}(p_{0},p^{\prime })}{p_{0}^{2}}\right] \right) t_{00}(p^{\prime
},0;0) \notag \\
&+& \left( \frac{v_{LR }^{22}(p,p^{\prime })}{p^{2}}-\lim_{p_{0}\rightarrow
0} \left[\frac{v_{LR }^{22}(p_{0},p^{\prime })}{p_{0}^{2}}\right] \right)
t_{20}(p^{\prime },0;0) \Bigg]. \label{eq:10}
\end{eqnarray}
This is the key result of our derivation, since it provides a second
integral equation that, together with Eq.~(\ref{eq:7.6}), is a closed set
for the half-off-shell t-matrix at zero energy. These equations require as
input the long-range potential $v_{LR}$, a choice of $C_2$, and the
experimental observables $a_t$ and the value of
\begin{equation}
\lim_{p_{0}\rightarrow 0}\left[\frac{t_{20}(p_{0},p_{0};E)}{p_{0}^{2}}\right]%
. \label{eq:genscat}
\end{equation}
This limit, and all the limits of $v_{LR}$ in Eq.~(\ref{eq:10}), are well
defined due to the threshold behavior of the potential and the t-matrix. The
limit (\ref{eq:genscat}) is a specific case of a ``generalized scattering
length", which is defined as
\begin{eqnarray*} \label{eq:105}
\frac{\alpha _{l^{\prime }l}}{M}=\lim_{p_{0}\rightarrow 0} \left[\frac{%
t_{l^{\prime }l}(p_{0},p_{0};E)}{p_{0}^{l^{\prime }+l}}\right].
\end{eqnarray*}
Since $\alpha_{l^{\prime}l}$ is related to an on-shell element of the
t-matrix it can be obtained from an experimental phase-shift analysis (see,
e.g. Ref.~\cite{PVRA05B}).
Equations~(\ref{eq:7.6}) and (\ref{eq:10}), together with an initial guess
for the value of $C_2$, now yield $t_{00}(p,0;0)$ and $t_{20}(p,0;0)$ by
standard methods for the solution of integral equations. The other elements
of the zero-energy $\mathbf{t}$, $t_{02}(p,0;0)$ and $t_{22}(p,0;0)$, are
equal to zero, as mentioned above.
It is formally possible to eliminate $C_2$ from Eq.~(\ref{eq:9}) by a
similar set of manipulations to the ones that led to Eq.~(\ref{eq:10}). This
yields a pair of coupled integral equations for $t_{00}(p,0;0)$ and $%
t_{20}(p,0;0)$ that is NN LEC free. However, these equations contain the
quantity $\lim_{p_{0}\rightarrow 0}\mathcal{F}\big[t_{00}(0,p_{0};E)\big]$
as part of the driving term. This is a limit of a half-off-shell matrix
element, and so cannot be obtained in a model-independent way from
experimental data.
Therefore we now proceed similarly to the case of the momentum-dependent
interaction in the singlet channel. We guess a value for $C_2$, and solve
Eqs.~(\ref{eq:7.6}) and (\ref{eq:10}) for the half-off-shell $\mathbf{t}$ at
zero energy. We then insert these quantities into the consistency condition
for the LEC $\lambda_t$:
\begin{equation}
\lambda _{t}=\frac{t_{20}(p,0;0)-v_{LR }^{20}(p,0)+\frac{2}{\pi }
M\;\int_{0}^{\Lambda }dp^{\prime }\;\left( v_{LR }^{20}(p,p^{\prime
})t_{00}(p^{\prime },0;0)+v_{LR }^{22}(p,p^{\prime })t_{20}(p^{\prime
},0;0)\right) }{p^{2}(1-\frac{2}{\pi }M\;\int_{0}^{\Lambda }dp^{\prime
}t_{00}(p^{\prime },0;0))}. \label{eq:lambdat}
\end{equation}
Once we have $\lambda_t$, the consistency condition for $\lambda$ reads
\begin{equation}
\lambda =\frac{a_{t}/M-v_{LR }^{00}(0,0)+\frac{2}{\pi }M\;\int_{0}^{\Lambda
}dp^{\prime }\;[\left( C_2 p^{\prime 2}+v_{LR }^{00}(0,p^{\prime
}))t_{00}(p^{\prime },0;0)+(\lambda _{t}p^{\prime 2}+v_{LR
}^{02}(0,p^{\prime }))t_{20}(p^{\prime },0;0)\right) ]}{1-\frac{2}{\pi }%
M\;\int_{0}^{\Lambda }dp^{\prime }t_{00}(p^{\prime },0;0)}.
\label{eq:lambda}
\end{equation}
The derivation of these two constraints for the LECs is given in Appendix~%
\ref{appendixb}. Eqs.~(\ref{eq:lambdat}) and (\ref{eq:lambda}) define the
full potential for the coupled-channels problem for a given trial value of $%
C_2$. This potential is then inserted into the LSE and phase shifts
computed. $C_2$ is adjusted until the desired experimental datum is
reproduced with sufficient accuracy.
This method has the advantage that Eqs.~(\ref{eq:lambda}) and (\ref%
{eq:lambdat}) define functional relationships: $\lambda=\lambda(C_2;a_t,%
\alpha_{20})$, and $\lambda_t=\lambda_t(C_2;a_t,\alpha_{20})$. (The
generalized scattering length $\alpha_{20}$ enters implicitly through its
effect on the half-off-shell t-matrices appearing in the formulae for $%
\lambda$ and $\lambda_t$.) Thus, we need to perform only one-parameter fits
in order to obtain all three of the LECs that are pertinent in the S-wave
J=1 channel.
\subsection{Energy-dependent central part}
Here we could follow the method of Section~\ref{sec-1S0endep} and eliminate
all three constants from the integral equation: $\lambda$, the coefficient
of the tensor short-distance interaction, $\lambda_t$, and the coefficient
of the energy-dependent part, $\gamma$. However, it is simpler to mimic the
steps of the previous Subsection, and reconstruct the underlying contact
interaction from knowledge of the phase shifts. The manner in which this is
done is described in detail in Appendix~\ref{appendixb} but can be
summarized in the following steps:
\begin{enumerate}
\item Use knowledge of $a_t$, the triplet scattering length, to eliminate $%
\lambda$ from the integral equation. Hence obtain $t_{00}(p,0;0)$.
\item Substitute the result into Eq.~(\ref{eq:7.5}), with $\gamma=0$ (since
the energy-dependent piece of the contact interaction has no impact on $t$
at $E=0$) and rearrange the equation so as to obtain $\lambda_t$ from the
on-shell quantity $\lim_{p_0 \rightarrow 0} t_{20}(p_0,p_0;E)/p_0^2 \equiv
\alpha_{20}/M$.
\item The value of $\lambda$ can then be obtained from the requirement that $%
t_{00}(0,0;0)=a_t/M$.
\item Use the ${}^3$S$_1$ and ${}^3$D$_1$ phase shifts and the mixing
parameter $\epsilon_1$ at a specific value of the energy, say $E^*$, to
reconstruct $t(p_0^*,p_0^*;E^*)$. Then, from a suitably modified (\ref{eq:7}%
), solve for $t(p,p_0^*;E)$.
\item With this half-off-shell $t$ at a finite energy in hand, $\gamma$ can
be obtained from the version of Eq.~(\ref{eq:6}) that is suitable for the
energy-dependent contact interaction. The final formula for $\gamma$ is
given in Eq.~(\ref{eq:22}).
\end{enumerate}
It may appear that in this procedure one needs five pieces of input data
to fix the constants ($a_t$, the generalized scattering length $\alpha_{20}$%
, and the three phase shifts at energy $E^*$). However, only one of the four
matrix elements of $t$ is used to extract $\gamma$. Hence, in reality only
three experimental inputs: $t_{00}(0,0;0)$, $t_{00}(p_0^*,p_0^*;E^*)$, and
the derivative of $t_{20}(p_0,p_0;E)$ with respect to $E$ at a specific energy, are needed to
solve this problem.
We then obtain phase shifts for arbitrary $E$ by by substituting the
obtained values of $\lambda $, $\lambda _{t}$ and $\gamma $ into the
potential that appears in the LSE and solving that equation for the on-shell
$t$-matrix.
\section{Results: the Singlet S-wave}
\label{sec-result1}
In this section we present the results obtained in the ${}^1$S$_0$ channel
when we employ the three different types of contact terms introduced in
Sect.~II. First, we only adopt the leading-order contact term, i.e. a
constant. Then, we consider the energy-dependent contact term, i.e., $%
v_{SR,0}=\lambda +\gamma E$. Finally, we use the more standard
momentum-dependent contact term $v_{SR,0}=\lambda +C_2 (p^{2}+p^{\prime 2})$%
. For each of these cases, we examine the results found with the following
three different forms of the long-range potential $v_{LR}$:
1. TPE computed using dimensional regularization up to $O(P^2)$ (denoted as
DR NLO).
2. TPE computed using dimensional regularization up to $O(P^3)$ (denoted as
DR NNLO).
3. TPE computed using spectral-function regularization up to $O(P^3)$
(denoted as SFR NNLO).
For the case of spectral-function regularization, we adopt $\bar{%
\Lambda}=800$~MeV as intrinsic cutoff. The effect that varying the intrinsic cutoff from $600-800$ MeV has on this potential is discussed in reference \cite{PVRA09}. When combined with the
momentum-dependent contact interaction, the $O(P^2)$ TPE yields the usual
NLO calculation in $\chi$ET, while the two $O(P^3)$ forms of $v_{LR}$ both
give calculations of NNLO accuracy.
\subsection{The lowest-order contact term}
A proper renormalization of these TPE potentials entails the presence of
contact terms up to second order in $|\mathbf{q}|$. However, it is still of
interest to investigate what happens if only the lowest-order contact term
is included in the potential (see also Refs.~\cite{PVRA06A,En08}). This
enables us to see how the phase shift converges as the LSE cutoff $\Lambda$
is increased. Then, when higher-derivative contact interactions are added to
the potential, we can see how the cutoff dependence changes.
The renormalization for the leading-order contact term is performed with one
subtraction, with the scattering length $a_{s}=-23.7$ fm~\cite{nnonline}
used as renormalization condition. This is the same as procedure as
described for the ${}^1$S$_0$ channel in Ref.~\cite{Ya08}, except that here
we are using a long-range potential that is higher-order in the chiral power
counting.
The phase shifts $^1$S$_0$ calculated with the DR NLO, DR NNLO and SFR NNLO
potential are shown for different cutoffs $\Lambda $ in Fig.~\ref{fig-fig1}.
The results show that a single, constant, contact interaction is sufficient
to stabilize the $^1$S$_0$ phase shift with respect to the cutoff once $%
\Lambda >800$~MeV for the SFR NNLO TPE and $\Lambda >1000$ MeV for the DR
NLO TPE. The DR NNLO potential requires $\Lambda >1200$ MeV to become stable.
The DR NLO potential exhibits a resonance-like behavior when $\Lambda \leq
700$ MeV, but the phase shift quickly converges after $\Lambda $ exceeds
800~MeV. This indicates that---at least for this potential---a cutoff
smaller than $700$~MeV removes essential physics, and brings the phase shift
rather close to that obtained with $v_{LR}=v_{OPE}$ in Ref.~\cite{Ya08}. On
the other hand, it is amusing to note that after $\delta(^1$S$_0)$ converges
with respect to the cutoff, the DR NLO result is actually closer to the
PWA93 phase shifts~\cite{St93,nnonline} than is the result from the other
two potentials.
This indicates that the pure (i.e. un-compensated by contact interaction)
NNLO TPE produces too large a correction, although the general trend of
providing more repulsion is correct. We note that the phase shift calculated
with the SFR NNLO potential stabilizes as a lower value of $\Lambda$, and
that this $\Lambda \rightarrow \infty$ result is closer to the PWA93 result,
compared to that obtained with the DR NNLO potential.
In the upper panel of Fig.~\ref{fig-t1} we plot the values of the effective range $r_{0}$
obtained from the above potentials. Since $a_s$ is fixed to the same value
in all calculations, this figure shows the influence of $\Lambda$ on the
low-energy phase shifts. Comparing our results for the three different
potentials to the ``experimental'' value $r_{0}=2.7$~fm~\cite{nnonline}, we
see that the SFR~NNLO potential gives a value of $r_{0}$ that is closest to
the experimental value. The results with the DR NLO and SFR NNLO TPE are
stable with respect to $\Lambda$ once $\Lambda \geq 700$~MeV. The DR NNLO
TPE value eventually also stabilizes within small variations once $\Lambda >
1200$~MeV (not shown in Fig.~\ref{fig-t1}). This cutoff is rather
large, reflecting the too-strong energy dependence that was already
discussed in the previous paragraph.
\subsection{Second order: energy-dependent contact term}
\label{sec-results1S0endep}
Next we use a second-order contact term with the simplest energy-dependent
form, namely $v_{SR,0}=\lambda +\gamma E$. This form arises because the
contact operator $N^\dagger i \partial_t N N^\dagger N$ can absorb the
divergences in the two-pion exchange potential, with the difference absorbed
into terms that are proportional to the free-nucleon equation of motion.
(See, e.g. Ref.~\cite{Ka97}, where the amplitude is calculated in Born
approximation, and so the contact terms required are only a function of the
nucleon on-shell momentum $p_0$.) These additional terms can be transformed
into higher-order multi-body operators. Thus, up to $O(P^3)$ in $\chi$PT,
the energy-dependent contact term is just as valid a representation of the
short-distance physics as the more commonly chosen momentum-dependent one.
Of course, the use of energy-dependent potentials engenders additional
complications---especially when one wishes to embed the two-nucleon Hilbert
space in a larger Fock space. Nevertheless, we can choose this form for the
short-distance operators to describe NN scattering, and it has the advantage
that the two unknown constants can be solved by our double-subtraction
method discussed in Sect.~II.
As input this method requires the value of $a_{s}$ and the phase shift at a
specific energy $E^{\ast }$. In principle, this energy can be chosen
arbitrarily. In our calculation we choose $T_{lab}=2E^{\ast }=2.8$ MeV, i.e.
at the energy where the $^{1}S_{0}$ phase shift has its turning point. We
tested other choices $E^{\ast }<10$~MeV and found that they have only a minor
effect on the results. We also tested a relatively high energy, $T_{lab}>100$%
~MeV. Such a choice generates results which fit the phase shift at low
energies (due to $a_{s}$ as input) and at the chosen higher energy (due to
the use of $\delta(E^{\ast })$ as input). However, we did not adopt this
procedure, since in it the ultimate dependence of results on the choice of
renormalization point (i.e. $E^*$) reflects the the cutoff-dependence in the
phase shift in a manner that we shall discuss below.
In Fig.~\ref{fig-fig44} the $^1$S$_0$ phase shift calculated with the TPE
potentials DR~NLO, DR~NNLO and SFR~NNLO with the energy-dependent contact
term is displayed for cutoffs $\Lambda =600-2000$~MeV. A general feature is
that the phase shift oscillates as a function of $\Lambda$. There are also
strongly diverging phase shifts at $\Lambda =800$ and $2000$ MeV for the
DR~NLO potential because a resonance state is created by the contact term.
(A similar behavior for the DR~NLO potential was observed in Ref.~\cite{En08}
for a momentum-dependent contact term with $\Lambda =2000$~MeV.) For the DR
NNLO (SFR NNLO) TPE, at $\Lambda=1000 (2000)$~MeV the phase shift diverges
violently from the previous converged value.
In order to further understand the effect of the energy-dependent contact
term, we calculate the DR NNLO TPE phase shift at $T_{lab}=10$ and $100$ MeV for $%
\Lambda $ up to $19$ GeV and display the results in Fig.~\ref{fig-fig7}. The
phase shifts have the same oscillatory behavior, with the period becoming longer as $%
\Lambda $ becomes larger. Comparison with the results obtained by adopting
only a constant contact term (second panel of Fig.~\ref{fig-fig1}, where
this phase shift converges with respect to $\Lambda$ for $\Lambda >1200$
MeV), suggests that the quasi-periodic behavior of $\delta(T_{lab}=10$ \& $100~%
\mathrm{MeV})$ is caused by the energy-dependent contact term.
The oscillatory behavior of the phase shift at this energy is associated
with resonances at specific $\Lambda$. An argand plot confirms that there is
a resonance for the cutoff at which this phase shift diverges. Tracking the
energy of the resonance as a function of $\Lambda$ shows that it moves in a
regular way, with the real and imaginary parts of $E_R$ decreasing in size
as $\Lambda$ increases, until eventually the resonance becomes a bound state
and the pole in t moves towards an energy of $-\infty$. In order to
understand this better, we perform a two-potential analysis of the t-matrix
resulting from this particular $v_{SR}$. The details are given in Appendix~%
\ref{appendixc}. We find that the appearance of poles is a consequence of
the specific form of the contact interaction. When these poles are in the
domain of validity of the theory the resulting phase shifts are in poor
agreement with the Nijmegen PSA. Only at those cutoffs where the artificial
poles correspond to deeply bound states or high-energy resonances do the
phase shifts match the Nijmegen analysis well.
The effective range $r_{0}$ obtained from all three different potentials is
about $r_{0}=2.68$fm and varies by $< 2\%$ over the range of cutoffs
considered, with the variation being only $\sim 0.5\%$ if the long-range
part is computed to $O(P^3)$. This is a consequence of our choice of a
low-energy phase shift as second input. Comparing with the upper panel of Fig.~\ref{fig-t1},
one can clearly see that $r_{0}$ receives significant correction from the $%
O(P^{2})$ contact term, especially for the DR~NLO potential.
Thus, the energy-dependent contact term improves the overall agreement of
the ${}^1$S$_0$ phase shift, and in particular the effective range $r_{0}$,
with respect to the average value $r_{0}=2.7$~fm extracted from the
different Nijmegen potentials. Once one tries to increase the cutoff $%
\Lambda $ beyond $1000$ MeV, this form of short-distance physics leads to
problems if the DR TPE is used as long-range potential. However, adopting
TPE computed using spectral-function regularization means a wider range of
cutoffs can be employed before the same issues occur as for the DR
TPE.
It is worthwhile mentioning that there exists a particular regularization of
two-pion exchange which avoids resonant behavior in the phase shifts. This
involves using dimensional regularization for the NLO part of TPE, and
spectral-function regularization for the NNLO part\footnote{see \cite{Ya09} for the detail of this potential and the analysis in p-waves.}. Then, even if an
energy-dependent contact term is chosen for the short-range potential, only
minor oscillations of the phase shifts with respect to $\Lambda$ occur for $%
\Lambda > 1$ GeV. However, strictly speaking, a potential which is obtained
using two different regularization schemes is not self-consistent. Thus, in
the later discussion we will only present the results obtained from the
three TPE potentials mentioned at the beginning of this section.
\subsection{Momentum-dependent contact term}
A momentum-dependent contact term associated with the TPE has the form $%
\lambda +C_2 (p^{2}+p^{\prime 2})$, and can be solved by one subtraction
plus the use of a one-parameter fit, as explained in Sect.~II. The input
chosen for the subtraction is again $a_s$. In what follows we will try to
perform two different fits to fix the value of $C_2$. One uses only
low-energy information, namely the effective range, $r_{0}=2.7$~fm. The
other fit fixes $C_2$ by attempting to reproduce the phase shift at $%
T_{lab}=200$~MeV.
The resulting $^1$S$_0$ phase shift is shown in Fig.~\ref{fig-fig8} for the
DR NLO TPE. In this case we cannot reproduce $r_0$, or the phase shift at $%
T_{lab}=200$ MeV, once $\Lambda > 600$ MeV. This pathology of the fit at NLO
was discussed in Ref.~\cite{PVRA06A}, and is related to the Wigner bound
that limits the impact of short-distance physics on the effective range---or
more generally on the energy dependence of phase shifts~\cite{PC96,Sc96}.
Therefore, for the DR NLO TPE we can only perform an overall best fit to the
phase shift. Fig.~\ref{fig-fig8} shows that, within this limitation, the
phase shift quickly becomes cutoff independent once $\Lambda >900$ MeV.
The results for the DR NNLO TPE are shown in Fig.~\ref{fig-fig9}, where we
see that the two different fit procedures generate different results for the
same $\Lambda $. This is especially visible at $\Lambda$~=~500 and 1000~MeV,
where a resonance-like behavior is present in the latter case when $C_2$ is
fitted to $r_{0}$. For values of $\Lambda $ not close to these problematic
cutoffs the phase shift is almost independent of the renormalization point.
This independence of renormalization point is even more apparent when the
SFR NNLO TPE is adopted. As can be seen in Fig.~\ref{fig-fig100}, for $%
\Lambda$ between $700-1800$~MeV, the two different fitting procedures lead
to almost the same $^1$S$_0$ phase shift. The same figure also suggests that
$\Lambda < 600$ MeV is too small and cuts off too much of the potential to
allow a good fit to the Nijmegen analysis. Finally, the agreement between
the two fits breaks down at $\Lambda=2000$ MeV. Thus, by adopting the SFR NNLO
TPE, we achieve renormalization-point independence for a wider range of
cutoffs, but this renormalization-point independence breaks down for higher
$\Lambda$.
Finally, we show in the lower panel of Fig.~\ref{fig-t1} the extracted effective range $r_{0}$
as function of $\Lambda $ for the $^1$S$_0$ phase shift fitted to the
Nijmegen value at $T_{lab}=200$ MeV. This allows to assess the sensitivity
of the results to renormalization point ($E^* \approx 0$ or $E^* \approx 200$
MeV lab. energy). The DR NLO TPE has the least variation of $r_{0}$ with $%
\Lambda$ over the range of $\Lambda$ considered here, but it needs to be
remembered that the experimental value cannot be reproduced with a real
value of $C_2$~\cite{PVRA06A,PVRA08A}. At order NNLO, the SFR TPE gives
values of $r_{0}$ which vary less with $\Lambda$ than do the ones from the
DR TPE once $\Lambda>800$ MeV, and are stable until $\Lambda=2000$ MeV. This
can already be inferred from Fig.~\ref{fig-fig100}.
In summary, the results obtained from the NNLO DR TPE with a
momentum-dependent contact term have a peculiar behavior at $\Lambda =1000$%
~MeV. Adopting the SFR potential increases the validity range of $\Lambda$
as in the energy-dependent case.
\section{Results in the J=1 triplet channel}
\label{sec-result2}
In the J=1 triplet channel we again adopt the three different potentials,
i.e., the DR TPE up to NLO, the DR TPE up to NNLO and the SFR TPE up to
NNLO. To renormalize the potential in this channel we employ three different
contact terms (cases A--C in Sec.~\ref{sec-triplet}). For case B
(energy-dependent) and C (momentum-dependent), the value of the generalized
scattering length
\begin{equation*}
\lim_{p_{0}\rightarrow 0}\frac{t_{20}(p_{0},p_{0};E)}{p_{0}^{2}}\equiv \frac{%
\alpha _{20}}{M}
\end{equation*}%
is required. We first adopt the value given in \cite{PVRA05B} for $\alpha
_{20}$. Then we adjust $\alpha _{20}$ to find the best fit. Since $\alpha
_{20}$ is not given from experiment, we list the values extracted from
several so called \textquotedblleft high-precision" potentials in Table~\ref%
{table-5}.
\subsection{The lowest-order contact term}
We first perform the renormalization with the constant contact term (A) in
the $^{3}$S$_{1}-^{3}$S$_{1}$ channel. The results obtained with this
short-distance potential, and long-distance potentials DR NLO, DR NNLO and
SFR NNLO are shown in Figs.~\ref{fig-fig11}-\ref{fig-fulltc}, respectively.
In the first two cases the $^{3}$S$_{1}$ phase shifts reach cutoff
independence as $\Lambda $ approaches $1000$ MeV, and then diverge once more
at higher $\Lambda $. The $^{3}$D$_{1}$ phase shifts exhibit a similar
feature, although the pattern of convergence is not as obvious as in $^{3}$S$%
_{1}$. For the mixing angle $\epsilon _{1},$ there is no obvious convergence
with respect to $\Lambda$ for the DR NLO and DR NNLO TPE. This can be
explained by the fact that we only adopt a contact term in the $^{3}$S$%
_{1}-^{3}$S$_{1}$ channel and fit to the scattering length $a_{t}=5.428$ fm.
This was sufficient to stabilize the J=1 triplet channel in the case of the
one-pion-exchange potential\cite{Be02,PVRA04B,Ya08}. However, the NNLO TPE is singular and
attractive in both eigen-channels and thus needs two contact interactions
(or, equivalently, two subtractions) in order to stabilize its predictions
with respect to $\Lambda$~\cite{PVRA06A}. In contrast, the NLO TPE has a
repulsive behavior as $r \rightarrow 0$ in both J=1 eigen-channel. This also
leads to predictions that are unstable after one subtraction, albeit for the
opposite reason than at NNLO.
The SFR NNLO potentials do not have the same singular attractive behavior at
short distances, and so it is no coincidence that its phase shifts show the
best overall convergence in $^{3}S_{1}$, $^{3}D_{1}$, and $\epsilon_1$. The
SFR strongly reduces the short-range attraction in the NNLO pieces of the
TPE, and a lower-order contact term is then able to remove the $\Lambda$
dependence of the phase shifts. Fig.~\ref{fig-3sc} shows that the phase
shifts at $T_{lab}=10$ and $50$ MeV obtained with the SFR NNLO TPE and a constant
contact term become stable for $\Lambda>2500$ MeV.
\subsection{Energy-dependent contact term}
The results of using an energy-dependent contact term to renormalize the
three different long-range potentials are shown in Figs.~\ref{fig-fig14}-\ref%
{fig-fulle}. In Fig.~\ref{fig-fig14} we adopt $a_{t}=5.428$ fm, $\alpha
_{20}=2.28\times 10^{-10}$ MeV$^{-3}$~\cite{PVRA05B}, and the Nijmegen PWA
values of $\delta (^{3}S_{1})$, $\delta (^{3}D_{1})$ and $\epsilon _{1}$ at $%
T_{lab}=10$ MeV as our input to renormalize the DR\ NNLO TPE. This choice
does not produce an optimal description of $\epsilon _{1}$ with respect to
the Nijmegen analysis, although it is pretty close.
Actually, $\epsilon_1$ is a notoriously delicate observable, and the choice
made for $\alpha_{20}$ affects it appreciably. Therefore, in the case of DR
NNLO we further adjust $\alpha _{20}$ to $2.25\times 10^{-10}$ MeV$^{-3},$
which yields the best fit of $\epsilon _{1}$ with respect to the Nijmegen
values for $\Lambda =700-1200$ MeV. This result is shown in Fig.~\ref%
{fig-fig15}. Comparison with Fig.~\ref{fig-fig14} shows that this 1.5\%
shift in $\alpha_{20}$ leads to noticeable improvement in $\epsilon_1$ over
this cutoff range, with few other significant changes. The same value, $%
\alpha _{20}=2.25\times 10^{-10}$ MeV$^{-3}$, also gives the best fit for DR
NLO, and this $\alpha_{20}$ produces results for SFR NNLO which coincide
quite well with the Nijmegen data. In addition, $\alpha_{20}=2.25\times
10^{-10}$ MeV$^{-3}$ is still within a few per cent of the value obtained
for this quantity from the potentials constructed in Ref.~\cite{St94}. Such
a difference is certainly within the range of expected N$^3$LO corrections
to our results.
The DR NLO results obtained with $\alpha_{20}=2.25\times 10^{-10}$ MeV$^{-3}$
are shown in Fig.~\ref{fig-fig16}. We see there that the phase shifts
oscillate for $\Lambda \leq 800$ MeV, and then converge for $\Lambda
=900-1800$ MeV. For $\Lambda=2000$ MeV, the phase shifts diverge badly again.
The DR NNLO results present several interesting features. First, there
exists a range of cutoffs $\Lambda =800-900$ MeV where all three phase
shifts agree with the Nijmegen analysis remarkably well\footnote{It is interesting to point out that for $\Lambda=800-900$ MeV, the energy-dependent contact term actually produces phase shifts which fit the Nijmegen analysis better than the momentum-dependent contact term, in both $^1S_0$ and $^3S_1-^3D_1$ channels. This may indicate that this on-shell-energy-dependent contact term represents the short-distance physics quite well in this narrow range of cutoffs.}. However, with
higher cutoffs the phase shifts gradually move away from the Nijmegen
analysis. Second, for $\Lambda =2000(1400) $ MeV, $\delta (^{3}S_{1})(\delta
(^{3}D_{1}))$ diverge and show a resonance-like behavior.
In order to analyze this feature, we calculate the J=1 triplet phase shifts
for a fixed energy of $T_{lab}=10$ and $50$ MeV and a wide range of cutoffs $%
\Lambda =500-5500$ MeV, and plot them in Figs.~\ref{fig-fig18} and \ref%
{fig-fig188}. Both figures show oscillatory behavior of the phase shifts
with respect to $\Lambda $. The S and D-waves have different points of
divergence in this cycle, and (unsurprisingly) affect one another when
either gets large. The oscillation pattern of the mixing angle $\epsilon _{1}
$ is a combination of that appearing in the two phase shifts $\delta
(^{3}S_{1})$ and $\delta (^{3}D_{1})$.
In Fig.~\ref{fig-fulle} we show the results for the SFR NNLO potential with $%
\alpha_{20}=2.25\times 10^{-10}$ MeV$^{-3}$. The phase shifts only show
minor cutoff dependence for $\Lambda <1000$ MeV and converge nicely for
higher $\Lambda$. As far as the fit to the Nijmegen PWA is concerned, a
decent agreement is observed for all $\Lambda $ for $\delta (^{3}S_{1})$ and
$\delta (^{3}D_{1})$. For $\epsilon _{1}$, the converged curves do not fit
as well as those for $\Lambda =600-800$ MeV. This is not a surprise, since
we adopted a value of $\alpha _{20}$ which fits $\epsilon _{1}$\ in this
region of cutoff. We found that at higher cutoffs, i.e., $\Lambda\sim 2300$
MeV, the SFR NNLO TPE presents a divergent pattern in the phase shifts.
Similar to the singlet channel, SFR NNLO can defer, but cannot avoid, the
resonance caused by the energy-dependent contact term.
The effective range $r_{0}$ obtained at NLO is $r_0=1.75 \pm 0.01$ fm for $%
900~\mathrm{MeV} \leq \Lambda \leq 1800$ MeV, with the error representing
the extent of the cutoff dependence in that range. For $\Lambda=2000$ MeV, $%
r_0=1.70$fm, which is due to the resonance-like behavior in the phase shifts
created by the energy-dependent contact term. For DR NNLO, $r_0=1.75 \pm 0.01$
fm except for $\Lambda=1400$ and $2000$ MeV, which are just the cutoffs
where the phase shifts also diverge, as shown in Figs.~\ref{fig-fig18} and %
\ref{fig-fig188}. The same behavior in $r_0$ is present in the SFR NNLO,
but for higher cutoffs.
In summary, in the triplet channel the energy-dependent contact term creates
a similar pattern as in the singlet channel. For the DR TPE (SFR NNLO),
there is a highest cutoff $\Lambda =1200 (2300)$ MeV with which the
potential can be iterated in the LSE before resonances are generated at
energies within the domain of validity of $\chi$ET.
\subsection{Momentum-dependent contact term}
The results of using a momentum-dependent contact term to renormalize the
DR\ NNLO TPE are shown in Fig.~\ref{fig-fig19}. Here we adopt $a_{t}=5.428$
fm and $\alpha _{20}=2.25\times 10^{-10}$ MeV$^{-3}$ as input and then
perform a fit to the phase shift $\delta (^{3}S_{1})$ at $T_{lab}=200$ MeV.
This pins down the three unknown LECs in the potential at this order.
For phase shifts obtained by our two-subtractions-plus-one-fitting method,
once the trial value of $C_2$ is imposed, the other two constants $\lambda $
and $\lambda _{t}$ are fixed by the two pieces of experimental information
that are our input: $\alpha_{20}$ and $a_t$. Thus, the results shown in Fig.~%
\ref{fig-fig19} satisfy the constraints at $T_{lab}=0$ associated with this
information and reproduce the ${}^3$S$_1$ phase shift at $T_{lab}=200$ MeV.
However, they do not provide the best possible fit to the phase shifts. In
contradistinction to the previous section, the fit is not appreciably
improved by variations of $\alpha_{20}$. Compared to the results obtained
using the same contact terms in Refs.~\cite{Ep99,EM03,Ep05}, the overall fit
to the Nijmegen phase shifts is noticeably worse. Furthermore, the results
obtained by fixing $\delta (^{3}S_{1})$ at $T_{lab}=200$ MeV do not show
stability for $\delta (^{3}D_{1})$ and $\epsilon _{1}$\ with respect to $%
\Lambda $. This suggests that the TPE\ associated with a momentum-dependent
contact term depends strongly on the choice of renormalization point once $%
\Lambda > 1$ GeV.
We wish to see up to what value of $\Lambda$ it is possible to ameliorate
these difficulties by a different choice of input. Thus we remove the two
constraints imposed by $a_{t}=5.428$ fm and $\alpha _{20}=2.25\times 10^{-10}
$ MeV$^{-3}$ and allow all three NN LECs to vary freely, so that we can
obtain the best overall fit to the phase shifts. The results for DR NLO, DR
NNLO and SFR NNLO are listed in Figs.~\ref{fig-fig21a} and \ref{fig-fig200},
respectively. The fits are a noticeable improvement over Fig.~\ref{fig-fig19}%
. For all three potentials, starting from $\Lambda =700$ MeV, the $^{3}S_{1}$
and $^{3}D_{1}$ phase shifts fit the Nijmegen analysis quite well and have
only minor variation with respect to $\Lambda $ up to $1000$ MeV. A stronger
cutoff dependence shows up only in the mixing parameter $\epsilon_{1}$.
There we observe a clear decrease in the $\Lambda$ sensitivity of the result
as the long-range potential is improved from DR NLO to DR NNLO and then
again to SFR NNLO. We plot $r_{0}$ extracted from those best fits in Fig.~%
\ref{fig-t4}. Again, we see better agreement with the Nijmegen value $%
r_0=1.833$ fm~\cite{PVRA05B} for DR NNLO than for DR NLO. For $700 \le \Lambda \le 1000$ MeV, $r_0$ varies by less than $3\%$ for all three potentials.
\section{Conclusion}
\label{sec-con}
In this paper we developed subtractive renormalization schemes which can be
applied to NN scattering in $\chi$ET at NNLO and maximize the use of
information from experiment. We showed how to manipulate LSEs containing the
contact interactions of $\chi$ET so as to eliminate constant, tensor, and
energy-dependent contact operators from the equation in favor of pieces of
experimental data. A by-product of our analysis is the result that the
momentum-dependent contact operator cannot be eliminated in this way, and
thus is not straightforwardly related to any NN S-matrix element. We can
still, however, analyze the $\chi$ET potential containing that operator, by
using a mixture of subtractive renormalization and more standard fitting
methods.
The subtractive-renormalization method shows particular advantages in cases
where the LECs undergo limit-cycle or limit-cycle-like behavior, and thus
adjusting several of them to reproduce data represents a difficult fitting
problem. It is also very useful when carrying out calculations at cutoffs $%
\Lambda > 1$ GeV, since fine-tuning of the unknown NN LECs occurs in that
regime. Finally, the direct input of experimental information allows a
straightforward determination of the impact of the renormalization
point---i.e. the choice of \textit{which} experimental information we
input---on the resulting phase shifts.
Looking first at the case of a constant contact interaction, in the ${}^1$S$%
_0$ channel our results confirm the finding of Ref.~\cite{En08}: when a
constant contact term is adopted, the resulting $^{1}$S$_{0}$ phase shift
converges and becomes independent of $\Lambda$ in the limit $\Lambda
\rightarrow \infty$. This occurs once $\Lambda >1000 (1200)$ for the DR NLO
and SFR NNLO (DR NNLO) potentials. However, we found that in this
calculation the contribution from the NNLO($Q^{3}$) part of the TPE is
larger than the NLO part.
In contrast, in the coupled ${}^3$S$_1-{}^3$D$_1$ channels the dimensionally
regularized interactions cannot be renormalized with a single, constant,
contact interaction. However, when the two-pion exchange is computed using
spectral-function regularization $\Lambda$-independent results are again
obtained at large $\Lambda$: presumably because the short-distance behavior
of this potential is the same as that of one-pion exchange, which is
well-known to be renormalized by a single constant contact term in the ${}^3$%
S$_1-{}^3$S$_1$ part of the potential~\cite{Be02,PVRA04B,Ya08}.
We also considered the contact terms that can occur up to O($Q^2$) in $\chi$%
ET, examining both energy-dependent and momentum-dependent contact terms.
For both the ${}^1$S$_0$ and ${}^3$S$_1-{}^3$D$_1$ channels, the phase
shifts obtained from dimensionally regularized two-pion exchange show
oscillatory behavior as a function of $\Lambda$ when energy-dependent
contact terms are used. This is associated with the appearance, and
subsequent movement with energy, of poles in the NN amplitude. The first
cutoff at which such a pole appears in the domain of validity of $\chi$ET is
around $\Lambda=1000$ MeV.
We also found that cutoffs less than about $600$ MeV sometimes produced
resonant features in the phase shifts. We attribute this failure to the
propensity of these cutoffs to eliminate too much of the central attraction
in the TPE that is responsible for reproducing the energy dependence of the
S-wave phase shifts extracted from experiment---especially in the ${}^1$S$_0$%
.
In between these two limits, i.e. for 600 MeV $< \Lambda <$ 1 GeV, we found
that the 1993 Nijmegen PWA phase shifts, in both the ${}^1$S$_0$ and ${}^3$S$%
_1-{}^3$D$_1$ channels, could be quite well reproduced using a
short-distance potential that includes an energy-dependent contact term,
together with a long-distance potential calculated to $O(Q^3)$ in $\chi$PT.
With an energy-dependent contact term good fits to the Nijmegen phase shifts
can be obtained at cutoffs above 1 GeV, but only in fairly narrow windows of
$\Lambda$, where the energy of the spurious bound state produced by the
energy-dependent contact term is large compared to the energy scales of
interest.
Finally, if the spectral-function regularized TPE is adopted, then the range
of the cutoff one can use before the resonance-like behavior occurs is
greatly increased.
Turning to the case of the momentum-dependent contact interaction: we found
that in the case of this type of contact interaction there is a significant
improvement with respect to the Nijmegen phase shift when one improves the
long-range part of the potential from NLO to NNLO. Indeed, if the $\chi$ET
NLO potential is employed there are certain observables (e.g. the effective
range) that \textit{cannot} be reproduced if we restrict ourself to real
values of the LECs in the NN potential. If we wish our NLO calculation to
reproduce the effective range in the ${}^1$S$_0$ we are restricted to $%
\Lambda < 600$~MeV. This problem is absent at NNLO.
We examined the renormalization-point dependence of our results, and found
that in the ${}^3$S$_1-{}^3$D$_1$ channel at $\Lambda=800$ MeV the results
from both the DR and SFR TPE up to NNLO are almost independent of the
renormalization point. However, once $\Lambda =1000$ MeV is reached, the
results with the DR NNLO potential show appreciable dependence on the choice
of experimental input. Once again, the SFR TPE is better behaved in this
regard: there is only minor renormalization-point dependence for the SFR
NNLO potential from $\Lambda=600$ MeV until $\Lambda=1800$ MeV, and the
renormalization-point independence breaks down at $\Lambda=2000$ MeV.
The use of a momentum-dependent potential and the expressions for the NNLO
TPE should allow us to confirm previous results for the triplet channel
(e.g. those of Ref.~\cite{Ep99}). However, we found that if we impose strict
relations between the three unknown LECs in order to reproduce particular
pieces of experimental data (e.g. $a_{t}$ and $\alpha _{20}$), the resulting
fit to the Nijmegen phase shifts is in general worse than that obtained by
direct fitting of data over a finite range of energies~\cite{Ep99}. Only
when all the three constants are allowed to vary independently do we get
best fits, which are comparable in quality to the results in~\cite{Ep99,
PVRA06A}. However, we want be emphasize that this fit can only be obtained
for cutoffs $\Lambda =700$--$900$ MeV for the DR NNLO TPE. (For the SFR NNLO
TPE, it is possible to obtain a reasonable fit up to $\Lambda=2000$ MeV.)
There is thus an appreciable difference between using two subtractions that
fix the zero-energy behavior of the amplitude and performing a best fit
involving all three $J=1$, $S=1$ NN LECs. That difference is an indication
that the triplet S-wave phase shifts obtained from chiral TPE depend on the
renormalization point rather strongly, despite the fact that they are quite
cutoff independent in this window of $\Lambda$.
In summary, we have probed the renormalization-point and cutoff dependence
of chiral potentials which include TPE when these potentials are iterated
using the LSE. We find that energy-dependent contact terms---which might be
considered useful as they allow one to evade certain theorems regarding
short-distance potentials~\cite{Wi55,PC96}---tend to introduce unphysical
resonances in the results once $\Lambda >1$ GeV. These resonances are
generically absent when a momentum-dependent contact term is employed. Since
the $\chi $ET power counting suggests that energy- and momentum dependent
contact interactions are equivalent up to the order to which we work this
implies that the power counting is only a useful guide for $\Lambda <1$ GeV.
Although the momentum-dependent contact interaction does not produce the
dramatic failures to agree with experiment that occur in the
energy-dependent case, it does have some difficulties. Chief among these is
the failure of certain aspects of the NLO calculation. Serious doubt must
exist regarding the convergence of an expansion where (at least for the ${}^1$S$_0$
channel) LO is in poor agreement with experiment, NLO cannot reproduce the
effective range, and it is not until NNLO that anything like a reasonable
picture of the phase shifts emerges.
The less singular interactions obtained with spectral-function
regularization avoid many of the difficulties inherent in the use of their
dimensionally regularized counterparts. The cutoff dependence is almost
always weaker, with the cutoff dependence of phase shifts found with the SFR
NNLO potential and a constant contact term disappearing altogether for $%
\Lambda > 1$ GeV. However, $\epsilon_1$ cannot be reproduced in this range
of cutoffs, and the issue with the effective range at NLO persists.
We conclude that it makes little sense to iterate the two-pion-exchange
potentials obtained from chiral perturbation theory using the
Lippmann-Schwinger equation unless one restricts the momentum in the LSE
integral to lie below $\sim 1$ GeV~\cite{Le97,EM06,Dj07}. In this domain,
and for the case of a momentum-dependent contact term, our
subtractive-renormalization method produces results which agree with
previous analyses~\cite{PVRA06A, PVRA06B,Ep99, epsfr, En08}.
\vfill
\begin{acknowledgments}
This work was performed in part under the auspices of the U.~S. Department
of Energy, Office of Nuclear Physics, under contract No. DE-FG02-93ER40756
with Ohio University. We thank the Ohio Supercomputer Center (OSC) for the
use of their facilities under grant PHS206. D.R.P. is grateful for financial
support form the Mercator programme of the Deutsche Forschungsgemeinschaft,
for the hospitality of the Theoretical Physics group at the University of
Manchester during the initial stages of this work, and for useful
conversations with Evgeny Epelbaum.
\end{acknowledgments}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 9,665
|
AmRev360
AmRev360: Taste the Revolution with Chef Walter Staib
When visiting the Revolutionary sites of historic Philadelphia, how could you pass up a chance to dine where George Washington, Alexander Hamilton, and many more Founders once dined? For decades, a trip to Old City was not complete without a meal at City Tavern Restaurant. And the man behind it for the last 26 years was Chef Walter Staib, Emmy Award-winning TV host of A Taste of History, James Beard-nominated chef, and culinary historian.
Sadly, on Nov. 2, Chef Staib announced the closure of City Tavern. Chef Staib joined the latest episode of the Museum's AmRev360 series, hosted by President & CEO Dr. R. Scott Stephenson, to discuss how a chef from Germany fell in love with colonial American cuisine, what's next for himself and City Tavern, and his Emmy Award-winning TV show, A Taste of History.
Music by Philip Travaline. Images courtesy of City Tavern, Headhouse Media, and Concepts by Staib Ltd.
A Sweet Taste of History
Read an excerpt from Chef Walter Staib's cookbook, A Sweet Taste of History: More than 100 Elegant Dessert Recipes from America's Earliest Days, recently featured in the Museum's Read the Revolution series. See Content
Chef Staib's cookbooks, full seasons of A Taste of History, and pewter and ceramic tableware are available for purchase online. Learn more about the recipes and destinations featured in A Taste of History.
AmRev360 features lively conversations on the American Revolution from all angles between Stephenson and a broad slate of dynamic guests, including authors, actors, community leaders, and more.
Watch lively conversations on the American Revolution from all angles, hosted by Museum President & CEO Dr. R. Scott Stephenson and featuring a broad slate of dynamic guests.
Read an excerpt from Chef Walter Staib's cookbook, A Sweet Taste of History: More than 100 Elegant Dessert Recipes from America's Earliest Days.
A Recipe for a Revolutionary Birthday
Download the recipes for a "whisky cake" favored by George Washington as well as a rum shrub cocktail for your next celebration.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 4,407
|
<?php
namespace OC\App\CodeChecker;
abstract class AbstractCheck implements ICheck {
/** @var ICheck */
protected $check;
/**
* @param ICheck $check
*/
public function __construct(ICheck $check) {
$this->check = $check;
}
/**
* @param int $errorCode
* @param string $errorObject
* @return string
*/
public function getDescription($errorCode, $errorObject) {
switch ($errorCode) {
case CodeChecker::STATIC_CALL_NOT_ALLOWED:
$functions = $this->getLocalFunctions();
$functions = \array_change_key_case($functions, CASE_LOWER);
if (isset($functions[$errorObject])) {
return $this->getLocalDescription();
}
// no break;
case CodeChecker::CLASS_EXTENDS_NOT_ALLOWED:
case CodeChecker::CLASS_IMPLEMENTS_NOT_ALLOWED:
case CodeChecker::CLASS_NEW_NOT_ALLOWED:
case CodeChecker::CLASS_USE_NOT_ALLOWED:
$classes = $this->getLocalClasses();
$classes = \array_change_key_case($classes, CASE_LOWER);
if (isset($classes[$errorObject])) {
return $this->getLocalDescription();
}
break;
case CodeChecker::CLASS_CONST_FETCH_NOT_ALLOWED:
$constants = $this->getLocalConstants();
$constants = \array_change_key_case($constants, CASE_LOWER);
if (isset($constants[$errorObject])) {
return $this->getLocalDescription();
}
break;
case CodeChecker::CLASS_METHOD_CALL_NOT_ALLOWED:
$methods = $this->getLocalMethods();
$methods = \array_change_key_case($methods, CASE_LOWER);
if (isset($methods[$errorObject])) {
return $this->getLocalDescription();
}
break;
}
return $this->check->getDescription($errorCode, $errorObject);
}
/**
* @return string
*/
abstract protected function getLocalDescription();
/**
* @return array
*/
abstract protected function getLocalClasses();
/**
* @return array
*/
abstract protected function getLocalConstants();
/**
* @return array
*/
abstract protected function getLocalFunctions();
/**
* @return array
*/
abstract protected function getLocalMethods();
/**
* @return array E.g.: `'ClassName' => 'oc version',`
*/
public function getClasses() {
return \array_merge($this->getLocalClasses(), $this->check->getClasses());
}
/**
* @return array E.g.: `'ClassName::CONSTANT_NAME' => 'oc version',`
*/
public function getConstants() {
return \array_merge($this->getLocalConstants(), $this->check->getConstants());
}
/**
* @return array E.g.: `'functionName' => 'oc version',`
*/
public function getFunctions() {
return \array_merge($this->getLocalFunctions(), $this->check->getFunctions());
}
/**
* @return array E.g.: `'ClassName::methodName' => 'oc version',`
*/
public function getMethods() {
return \array_merge($this->getLocalMethods(), $this->check->getMethods());
}
/**
* @return bool
*/
public function checkStrongComparisons() {
return $this->check->checkStrongComparisons();
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 7,881
|
Administrative Sciences Association of Canada
Bureau des divisions
Description des postes divisionnaires
Description des divisions de l'ASAC
Conseil d'administration de l'ASAC
Rôles des membres du conseil d'administration
Dirigeants divisionnaires
Nominations et élections
Congrès de l'ASAC passés
Devenir membre de l'ASAC
Prix et subventions
Fellows de l'ASAC
À propos de l'ASAC
Mission et stratégies
Code d'éthique
Assistant Professor Position, Department of Accounting, Edwards School of Business – University of Saskatchewan
octobre 29, 2021 by asaccanada
The Edwards School of Business at the University of Saskatchewan is seeking to appoint a faculty member at the rank of Assistant Professor in the Department of Accounting effective July 1, 2022.
The successful candidate will be expected to contribute to excellence in research and scholarship through high quality peer reviewed publications in their areas of expertise. Preferred candidates will have their primary research interest in the area of financial accounting and reporting, auditing or taxation, along with a desire to teach these subjects at the undergraduate and graduate levels.
Candidates must hold a Ph.D. in Accounting or equivalent. Candidates near the completion of their Ph.D. are eligible to apply providing the requirements for the Ph.D. degree will be completed imminently and with the recognition that commencement of appointment will be contingent on the completion of the Ph.D. Possessing a professional Canadian accounting designation or equivalent will be an asset.
Although the Department is seeking an Assistant Professor, exceptionally qualified and experienced, accomplished candidates may also apply. Ranks and salary will be based on academic and professional qualifications, experience and accomplishments of the successful candidate. Starting salary bands for the 2021-2022 academic year are as follows: Assistant Professor: $98,178 to $117,978; Associate Professor: $117,978 to $137,778; and Professor $137,778 to $160,878, with a higher starting salary in rare and exceptional circumstances pursuant to Article 18.2.6.12 of the 2017-2022 USFA Collective Agreement (https://usaskfaculty.ca/collective-agreement/).
This position includes a comprehensive benefits package which includes a dental, health and extended vision care plan; pension plan, life insurance (compulsory and voluntary), academic long-term disability, sick leave, travel insurance, death benefits, employee assistance program, a professional expense allowance and a flexible health/wellness spending program.
• A letter of application, including a statement of citizenship/immigration status;
• A detailed and current curriculum vitae;
• Statements of research and teaching interests;
• A current working paper from your research;
• A teaching dossier or evidence of teaching effectiveness that will include sample course syllabi/outlines, teaching evaluations and a statement of teaching philosophy and interests;
• Three current letters of reference forwarded by the referees directly to the Department Head at the email listed below.
For more information on any of these opportunities please contact:
Accounting Position
Dr. Ganesh Vaidyanathan, Associate Professor
Department Head, Accounting
vaidyanathan@edwards.usask.ca
Review of applications will commence on December 1 and will continue until position in filled.
The Edwards School of Business has an undergraduate business program with approximately 1,800 students. As an AACSB accredited business school, we also have thriving MBA, M.Sc. Finance, M.Sc. Marketing, and Masters of Professional Accounting programs. For more information on the Edwards School of Business, please visit our website at http://www.edwards.usask.ca. The University of Saskatchewan serves approximately 26,000 students and is located in Saskatoon, Saskatchewan, Canada. Saskatoon is a vibrant university city with a population approaching 275,000. The city is located on a river in the heartland of Saskatchewan. The city is well known for its summer festivals and riverbank events such as Shakespeare on the Saskatchewan, the Jazz Festival, the Fringe Festival and hosts many outdoor activities year-round. For more information please visit http://tourismsaskatoon.com.
The University of Saskatchewan is strongly committed to a diverse and inclusive workplace that empowers all employees to reach their full potential. All members of the university community share a responsibility for developing and maintaining an environment in which differences are valued and inclusiveness is practiced. The university welcomes applications from those who will contribute to the diversity of our community. All qualified candidates are encouraged to apply; however, Canadian citizens and permanent residents will be given priority.
Filed Under: Accounting Tagged With: assistant professor, University of Saskatchewan
Our membership is designed to ensure you find the right fit to experience a truly rewarding and long-term relationship with the ASAC. Find out today how being a part of ASAC can benefit you. More Information
Join ASAC
Check out job postings from Canadian universities.
Copyright © 2022 ASAC
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 3,011
|
using System;
using System.IO;
namespace Thinktecture.Tools.Web.Services.Wscf.Environment
{
public class FileManipulationHelper
{
public static string GetRelativePath(string relativePathOf, string relativeTo)
{
// Remove the leading and trailing back slashs.
relativePathOf = relativePathOf.Trim('\\');
relativeTo = relativeTo.Trim('\\');
// Break the directory paths to arrays.
string[] ofDirectoryStack = relativePathOf.Split('\\');
string[] toDirectoryStack = relativeTo.Split('\\');
int lastMatch = -1;
for(int index = 0; index < ofDirectoryStack.Length; index++)
{
if(index > toDirectoryStack.Length - 1)
{
break;
}
if(ofDirectoryStack[index] == toDirectoryStack[index])
{
lastMatch++;
}
else
{
break;
}
}
string relativePath = "";
// Add the reverse lookup.
int reverse = toDirectoryStack.Length - (lastMatch + 1);
for(int index = 0; index < reverse; index++)
{
relativePath = "../" + relativePath;
}
int nextItem = lastMatch + 1;
if(lastMatch < ofDirectoryStack.Length - 1)
{
relativePath = relativePath + string.Join("/", ofDirectoryStack, nextItem,
(ofDirectoryStack.Length - nextItem));
}
return relativePath;
}
public static string GetAbsolutePath(string relativePath, string currentFolder, string rootFolder)
{
// Trim the leading and trailing / charactors.
relativePath = relativePath.Trim('/');
// Break the path in to an array.
string[] directories = relativePath.Split('/');
string parents = "";
int index = 0;
string cf = currentFolder;
DirectoryInfo pd = null;
for(index = 0; index < directories.Length; index++)
{
if(directories[index] == "..")
{
pd = Directory.GetParent(cf);
if(pd != null)
{
if(pd.FullName.ToLower() == rootFolder.ToLower())
{
parents = "";
index++;
break;
}
else
{
parents = pd.Name;
cf = pd.FullName;
}
}
else
{
throw new DirectoryNotFoundException("Directory not found.");
}
}
else
{
break;
}
}
// Fill the blanks.
if(pd != null && pd.FullName.ToLower() != rootFolder.ToLower())
{
pd = Directory.GetParent(pd.FullName);
while(pd.FullName.ToLower() != rootFolder.ToLower())
{
parents = pd.Name + "\\" + parents;
pd = Directory.GetParent(pd.FullName);
}
}
string children = string.Join("\\", directories, index, directories.Length - index);
string absPath = "";
if(parents != "")
{
absPath = rootFolder + "\\" + parents;
}
else if(pd != null)
{
absPath = rootFolder;
}
else
{
absPath = currentFolder;
}
if(children != "")
{
absPath = absPath + "\\" + children;
}
return absPath;
}
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 7,104
|
package conntrack
import (
"io/ioutil"
"os"
"path"
"strconv"
"strings"
"testing"
"github.com/influxdata/telegraf/testutil"
"github.com/stretchr/testify/assert"
)
func restoreDflts(savedFiles, savedDirs []string) {
dfltFiles = savedFiles
dfltDirs = savedDirs
}
func TestNoFilesFound(t *testing.T) {
defer restoreDflts(dfltFiles, dfltDirs)
dfltFiles = []string{"baz.txt"}
dfltDirs = []string{"./foo/bar"}
c := &Conntrack{}
acc := &testutil.Accumulator{}
err := c.Gather(acc)
assert.EqualError(t, err, "Conntrack input failed to collect metrics. "+
"Is the conntrack kernel module loaded?")
}
func TestDefaultsUsed(t *testing.T) {
defer restoreDflts(dfltFiles, dfltDirs)
tmpdir, err := ioutil.TempDir("", "tmp1")
assert.NoError(t, err)
defer os.Remove(tmpdir)
tmpFile, err := ioutil.TempFile(tmpdir, "ip_conntrack_count")
assert.NoError(t, err)
defer os.Remove(tmpFile.Name())
dfltDirs = []string{tmpdir}
fname := path.Base(tmpFile.Name())
dfltFiles = []string{fname}
count := 1234321
ioutil.WriteFile(tmpFile.Name(), []byte(strconv.Itoa(count)), 0660)
c := &Conntrack{}
acc := &testutil.Accumulator{}
c.Gather(acc)
acc.AssertContainsFields(t, inputName, map[string]interface{}{
fname: float64(count)})
}
func TestConfigsUsed(t *testing.T) {
defer restoreDflts(dfltFiles, dfltDirs)
tmpdir, err := ioutil.TempDir("", "tmp1")
assert.NoError(t, err)
defer os.Remove(tmpdir)
cntFile, err := ioutil.TempFile(tmpdir, "nf_conntrack_count")
maxFile, err := ioutil.TempFile(tmpdir, "nf_conntrack_max")
assert.NoError(t, err)
defer os.Remove(cntFile.Name())
defer os.Remove(maxFile.Name())
dfltDirs = []string{tmpdir}
cntFname := path.Base(cntFile.Name())
maxFname := path.Base(maxFile.Name())
dfltFiles = []string{cntFname, maxFname}
count := 1234321
max := 9999999
ioutil.WriteFile(cntFile.Name(), []byte(strconv.Itoa(count)), 0660)
ioutil.WriteFile(maxFile.Name(), []byte(strconv.Itoa(max)), 0660)
c := &Conntrack{}
acc := &testutil.Accumulator{}
c.Gather(acc)
fix := func(s string) string {
return strings.Replace(s, "nf_", "ip_", 1)
}
acc.AssertContainsFields(t, inputName,
map[string]interface{}{
fix(cntFname): float64(count),
fix(maxFname): float64(max),
})
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 1,310
|
The news site Portnews.ru yesterday reported that the Russian company considers to move its Arctic oil reloading business from Kirkenes, Norway, to neighboring Murmansk.
"We know nothing about this, and hope it is not correct", Stolt-Nielsen says to the Independent Barents Observer. He indicates however that the Russian company over the last three years has been under pressure to move the reloading business back to domestic waters.
"Lukoil is pleased with our services in Kirkenes", Jacob Stolt-Nielsen underlines. "We are well established [in Kirkenes] and are getting good feedback from the customer", the Norwegian company owner adds.
Over the last years, Norterminal, a company fully owned by Stolt-Nielsen, has engaged in extensive reloading of oil from Lukoil's Arctic fields. In 2015, the volumes amounted to about 7 million tons.
The oil comes from Varandey, Lukoil's out-shipment terminal on the Pechora Sea coast. It is reloaded ship-to-ship in the fjord outside Kirkenes, the Norwegian border town, and thereafter shipped to Rotterdam and other west European ports.
Stolt-Nielsen suspects that the speculations over Lukoil's business in Norway could somehow be related with Gazprom-Neft's new reloading facility in the Kola Bay. Gazprom recently announced that the tanker "Umba" will serve as terminal tanker for oil from the Prirazlomnoye and Novy Port fields.
As previously reported, Norterminal has major plans for its cooperation with Lukoil. A projected new land-based terminal near Kirkenes will have an initial oil storage capacity of 700.000 cbm and an annual capacity of 20.000.000 tons. Until the new terminal is built, Norterminal intends to continue with ship-to-ship reloading of the Russian oil.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 9,198
|
Крукля́нки (, ) — село в Польщі, у гміні Круклянкі Гіжицького повіту Вармінсько-Мазурського воєводства. Село розташоване на річці Сапіна.
Населення — (2011).
Історія
Село було засноване 1545 року Яном Бембельником. Поселенці — в основному, поляки (для 28 сімей, 6 було литовською походження) — отримали шість років звільнення від державного збору.
Село було багате і добре йому велося, тому вже в 1574 році збудовано костел, і майже в той же час побудовано школу. Хоча Круклянки горіли кілька разів, вдалося врятувати костел, який став цінною пам'яткою архітектури Східно-Пруської з XVI століття. Під час чуми в 1710 році померло тут 132 осіб, у тому числі остання людина із дворянської сім'ї фон Ґансенів, могила якого була виявлена 1875 року, під підлогою костела на лівій стороні вівтаря.
До початку дев'ятнадцятого століття, більшість жителів волості становили поляки, але до кінця цього століття вони становили лише третину населення.
У даний час деякі родини німецького походження, більшість з них це переїздні з центральної Польщі, переміщених з-за Бугу, і значна частина українського населення, переміщених в рамках операції «Вісла» в роках 1947-1952.
У період сильного економічного зростання в кінці дев'ятнадцятого та початку двадцятого століття було тут дві залізничні лінії. Круклянки опинилися на шляху, що з'єднував два міста Ґіжицко з Венгожевом а Можджани та Юрково Венгожевське на лінії Круклянки-Олєцко. Таким чином, залишки старих залізничних будівель: будівлі залізничного вокзалу в Круклянках та Юркові. Венгожевські насипи, ущелини, шляхопроводів та мостів, у тому числі зруйнований залізничний міст над річкою Сапіна — кілька сотень метрів на південь від Круклянок.
Залізничний міст над річкою Сапіна жителі села називають «зваленим мостом». Він був побудований в 1908 році і був однією з найбільших залізничних будівель на Мазурах.
У серпні 1914, під час Першої світової війни, німецькі сапери підірвали одне крило (східне), щоб не віддати залізничної лінії російським військам. Знищення ці були відремонтовані в ході війни, тоді введено додаткову підтримку арки. Під час війни Круклянки були в помітній мірі знищені.
Під час Другої світової війни у Круклянках була побудована система укріплень, побудована із залізобетонних бункерів і окопів, котрий належав до Ґіжицького укріпленого району (ҐРУ).
Як в цілому ҐРУ, так і на території Круклянок використано природні перешкоди — ліси, озера, затоки, пагорби. Це сильно укріплений район не відігравав велику роль у Другій світовій війні. Використані маневри Червоної Армії призвели до швидкого відступу німецьких військ.
Залізничний міст на річці Сапіна був знова збомбардований в 1945 році зі страху очікуваного приходу радянських військ, пізніше 8 вересня 1945 року він був підірваним місцевими жителями, які сподівалися, що це буде перешкодою грабежів.
Після зайняття цих земель росіянами, більшість з пристроїв захисту вивезено, а решту підірвано. Залишки ҐРУ на сьогоднішній день це мальовничі руїни, на які можна наткнутись під час гуляння лісом та полями.
У 1945 село включено до Польщі. Більшість населення було замінено переміщеними людьми.
Парафіяльний костел Успіння Пресвятої Діви Марії, побудований з польового каменю. Будівля була побудована в 1753, вежа раніше (в кінці шістнадцятого століття), а надбудова на два поверхи близько 1600 р, верхи вежі, виготовлено перед 1648. Костел був реконструйований в 1875, і після пошкодження Першої світової війни. Вівтар був споруджений в 1753, з використанням набору елементів кінця сімнадцятого століття.
У 1975—1998 роках село належало до Сувальського воєводства.
Розташування
Круклянки розташовані у східній частині Великих Мазурських Озер над озером Ґолдапіво і вздовж річки Сапіни.
Пейзаж муніципалітету Круклянкі утворився під час останнього бальтійського заледеніння. Льодовик, який кілька разів насувався, і зникав з цих районів, залишив багато цікавих форм у вигляді моренних підвищень і ерозійних долин.
Туристична пропозиція
Круклянки мають розвинену туристичну інфраструктуру. Є тут традиційні ресторани, громадське харчування на курортах, кафе і пункти туристичної інформації, тенісний корт і спортивний стадіон. Є багато туристичних ферм.
Бібліографія
Andrzej Wakar i Bohdan Wilamowski, Węgorzewo. Z dziejów miasta i powiatu, Pojezierze, Olsztyn, 1968. (str. 128—130)
Rzempołuch A., Przewodnik po zabytkach sztuki dawnych Prus Wschodnich. Agencja Wyd. «Remix», Olsztyn 1993.
Демографія
Демографічна структура станом на 31 березня 2011 року:
Примітки
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Села Гіжицького повіту
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Home / China / Society
Water bottles put missing kids' photos in public's hands
By XIE CHUANJIAO | China Daily | 2016-11-15 07:20
Photos of missing children are being printed on bottled water by a company in Qingdao. [Photo by Zhang Xiaopeng/CHINA DAILY]
A bottled water company in eastern China has begun printing pictures of missing children on its packaging to help parents and authorities find them.
Qingdao Kingtex International said it has sold 500,000 bottles with information on six children missing since July. The labels also include the child's birth date and a hotline number.
"We just wanted to play our part in helping those parents who might spend their lifetime searching for their missing children," said Shi Yiwei, a marketing manager at the company.
"We're mainly selling the water bottles in Qingdao," she said. "Next, we'd like to expand the program to Beijing and Shanghai, to spread the information nationwide, so as to raise more awareness of missing children."
The project was launched in cooperation with Baobei Huijia (Baby Back Home), a volunteer group that supports parents looking for missing children nationwide.
"It's always a good thing to add another method to help search for lost children. At the very least, these efforts will help to draw people's attention to the problem of child trafficking," said Zhang Baoyan, who founded the group with her husband, Qin Yanyou.
Kingtex's water bottles have received a lot of attention on social media. While some applaud the move, others have questioned the company's motives, not least because its bottled water sells for 5.5 yuan (80 US cents), higher than many domestic brands.
"The water is so expensive. Can't the company lower the price so as to improve sales and spread the information to more people?" wrote one netizen on Sina Weibo, the popular Twitter-like platform.
Shi said Kingtex's prices have remained the same since 2014. "We also distribute a number of bottles for free in order to reach more people," she added.
NGOs are playing an important role in assisting the search for missing children and combating human trafficking in China. Baobei Huijia said it has helped to reunite more than 1,700 families since it was established in 2007. "We're still working on about 31,000 unsolved cases," Zhang said.
A traditional preference for male heirs, particularly in rural areas, has led to a black market, with families willing to spend large sums for a baby boy.
Accurate statistics about China's missing children are difficult to obtain, yet authorities have been clamping down on human trafficking.
Data from the Supreme People's Court show judges at all levels heard more than 7,000 cases involving the abduction of women or children between 2010 and 2014. The trials led to nearly 13,000 convictions. In April 2009, the Ministry of Public Security established a national DNA database to help match recovered children with their biological parents.
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\section{Introduction}\label{s:intro}
In short, the purpose of the present work is to investigate long time effects of an algorithm for counterbalancing steps in a random walk. As we shall first explain, our motivation stems from a nearest neighbor process on the integer lattice, know as the elephant random walk.
The elephant random walk is a stochastic process with memory on $\ZZ$, which records the trajectory of an elephant that makes steps with unit length left or right at each positive integer time. It has been introduced by Sch\"utz and Trimper \cite{SchTr} and triggered a growing interest in the recent years; see, for instance, \cite{BaurBer, Bercu, ColGavSch1, ColGavSch2, KuMa, Kur}, and also \cite{Baur, BerLau2, Marco, Buart, GutStadt, Mitak} for related models. The dynamics depend on a parameter $q\in[0,1]$ and can be described as follows. Let us assume that the first step of the elephant is a Rademacher variable, that is equals $+1$ or $-1$ with probability $1/2$. For each time $n\geq 2$, the elephant remembers a step picked uniformly at random among those it made previously, and decides either to repeat it with probability $q$, or to make the opposite step with complementary probability $1-q$. Obviously, each step of the elephant has then the Rademacher law, although the sequence of steps is clearly not stationary.
Roughly speaking, it seems natural to generalize these dynamics and allow steps to have an arbitrary distribution on $\RR$, say $\mu$. In this direction, K\"ursten \cite{Kur} pointed at an equivalent way of describing the dynamics of the elephant random walk which makes such generalization non trivial\footnote{ Note that merely replacing the Rademacher distribution for the first step of the elephant by $\mu$ would not be interesting, as one would then just get the evolution of the elephant random walk multiplied by some random factor with law $\mu$.}. Let $p\in[0,1]$, and imagine a walker who makes at each time a step which is either, with probability $p$, a new independent random variable with law $\mu$, or, with probability $1-p$, a repetition of one of his preceding steps picked uniformly at random. It is immediately checked that when $\mu$ is the Rademacher distribution,
then the walker follows the same dynamics as the elephant random walk with parameter $q=1-p/2$. When $\mu$ is an isotropic stable law, this is the model referred to as
the shark random swim by Businger \cite{Buart}, and more generally, when $\mu$ is arbitrary, this is the step reinforced random walk that has been studied lately in e.g. \cite{Marco2, NRLP, Bescal, UNRBM}.
The model of K\"ursten yields an elephant random walk only with parameter $q\in[1/2,1]$; nonetheless the remaining range can be obtained by a simple modification. Indeed, let again $p\in[0,1]$ and imagine now an averse walker who makes at each time a step which is either, with probability $p$, a new independent random variable with law $\mu$, or, with probability $1-p$, the \textit{opposite} of one of his previous steps picked uniformly at random. When $\mu$ is the Rademacher distribution, we simply get
the dynamics of the elephant random walk with parameter $q=p/2\in [0,1/2]$.
More formally, we consider a sequence $(X_n)$ of i.i.d. real random variables with some given law $\mu$ and a sequence $(\varepsilon_n)_{n\geq 2}$ of i.i.d. Bernoulli variables with parameter $p\in[0,1]$, which we assume furthermore independent of $(X_n)$.
We construct a counterbalanced sequence $(\check X_n)$ by interpreting each $\{\varepsilon_n=0\}$ as a counterbalancing event and each $\{\varepsilon_n=1\}$ as an innovation event.
Specifically, we agree that $\varepsilon_1=1$ for definiteness and denote the number of innovations after $n$ steps by
$${\mathrm i}(n) \coloneqq \sum_{j=1}^n\varepsilon_j\qquad\text{for }n\geq 1.$$
We introduce a sequence $(v(n))_{n\geq 2}$ of independent variables, where each $v(n)$ has the uniform distribution on $\{1, \ldots , n-1\}$, and which is also independent of $(X_n)$ and $(\varepsilon_n)$. We then define recursively
\begin{equation}\label{E:negreinf}
\check X_n \coloneqq\left\{
\begin{matrix}
- \check X_{v(n)}&\text{ if }\varepsilon_n=0, \\
X_{{\mathrm i}(n)}&\text{ if }\varepsilon_n=1.\\
\end{matrix}
\right.
\end{equation}
Note that the same step can be counterbalanced several times, and also that certain steps counterbalance previous steps which in turn already counterbalanced earlier ones. The process
$$\check S(n)\coloneqq \check X_1+ \cdots + \check X_n, \qquad n\geq 0$$
which records the positions of the averse walker as a function of time, is called here a random walk with counterbalanced steps.
Note that for $p=1$, i.e. when no counterbalancing events occur, $\check S$ is just a usual random walk with i.i.d. steps.
In short, we are interested in understanding how counterbalancing steps affects the asymptotic behavior of random walks.
We first introduce some notation.
Recall that $\mu$ denotes the distribution of the first step $X_1=\check X_1$ and f
we write
$$m_k\coloneqq\int_\RR x^k \mu(\mathrm d x)$$ for the moment of order $k\geq 1$ of $\mu$ whenever $\int_\RR |x|^k \mu(\mathrm d x)<\infty$.
To start with, we point out that if the first moment is finite, then the algorithm \eqref{E:negreinf} yields
the recursive equation
$$\EE(\check S(n+1)) =pm_1 + (1-(1-p)/n) \EE(\check S(n)), \qquad n\geq 1,$$
with the initial condition $\EE(\check S(1))=m_1$. It follows easily that
\begin{align*}\EE(\check S(n))
&\sim \frac{p}{2-p}\, m_1 n\qquad \text{as }n\to \infty;
\end{align*}
see e.g. Lemma 4.1.2 in \cite{Durrett}. Our first result about the ballistic behavior should therefore not come as a surprise.
\begin{proposition}\label{P0} Let $p\in[0,1]$. If $\int_\RR |x| \mu(\mathrm d x)<\infty$, then there is the convergence in probability
$$\lim_{n\to \infty} \frac{\check S(n)}{n} = \frac{p}{2-p} m_1.$$
\end{proposition}
We see in particular that counterbalancing steps reduces the asymptotic velocity of a random walk by a factor $p/(2-p)<1$. The velocity is smaller when the innovation rate $p$ is smaller
(i.e. when counterbalancing events have a higher frequency),
and vanishes as $p$ approaches $0+$.
The main purpose of this work is to establish the asymptotic normality when $\mu$ has a finite second moment.
\begin{theorem}\label{T2} Let $p\in(0,1]$. If $\int_\RR |x|^2 \mu(\mathrm d x)<\infty$, then there is the convergence in distribution
$$\lim_{n\to \infty} \frac{\check S(n)-\frac{p}{2-p}m_1n }{\sqrt n} = \mathcal N\left(0, \frac{m_2 -\left(\frac{p}{2-p}m_1\right)^2 }{3-2p} \right) ,$$
where the right-hand side denotes a centered Gaussian variable parametrized by mean and variance.
\end{theorem}
It is interesting to observe that the variance of the Gaussian limit depends linearly on the square $m_1^2$ of the first moment and the second moment $m_2$ of $\mu$ only, although not just on the variance $m_2-m_1^2$ (except, of course, for $p=1$).
Furthermore, it is not always a monotonous function\footnote{ For instance, in the simplest case when $\mu$ is a Dirac point mass, i.e. $m_2=m_1^2$, then
the variance is given by $\frac{4(1-p)m_2}{(3-2p)(2-p)^2}$ and reaches its maximum for
$p=(9-\sqrt{17})/8\approx 0.6$. At the opposite, when $\mu$ is centered, i.e. $m_1=0$, the variance is given by $m_2/(3-2p)$ and hence increases with $p$.} of the innovation rate $p$, and does not vanish when $p$ tends to $0$ either.
Actually, our proofs of Proposition \ref{P0} and Theorem \ref{T2} provide a much finer analysis than what is encapsulated by those general statements.
Indeed, we shall identify the main actors for the evolution of $\check S$ and their respective contributions to its asymptotic behavior.
In short, we shall see that the ballistic behavior stems from those of the variables $X_j$ that have been used just once by the algorithm \eqref{E:negreinf} (in particular, they have not yet been counterbalanced), whereas the impact of variables that occurred twice or more regardless of their signs $\pm$, is asymptotically negligible as far as only velocity is concerned. Asymptotic normality is more delicate to analyze. We shall show that, roughly speaking, it results from the combination of, on the one hand, the central limit theorem for certain centered random walks, and on the other hand, Gaussian fluctuations for the asymptotic frequencies of some pattern induced by \eqref{E:negreinf}.
Our analysis relies on a natural coupling of the counterbalancing algorithm \eqref{E:negreinf} with
a basic linear reinforcement
algorithm which has been introduced a long time ago by H.A. Simon \cite{Simon} to explain the occurrence of certain heavy tailed distributions in a variety of empirical data. Specifically, Simon defined recursively a sequence denoted here by $(\hat X_n)$ (beware of the difference of notation between $\hat X$ and $\check X$) via
\begin{equation}\label{E:reinf}
\hat X_n \coloneqq \left\{
\begin{matrix}
\hat X_{v(n)}&\text{ if }\varepsilon_n=0, \\
X_{{\mathrm i}(n)}&\text{ if }\varepsilon_n=1.\\
\end{matrix}
\right.
\end{equation}
We stress that the same Bernoulli variables $\varepsilon_n$ and the same uniform variables $v(n)$ are used to run both Simon's algorithm \eqref{E:reinf} and \eqref{E:negreinf}; in particular either $\check X_n= \hat X_n$ or $\check X_n= -\hat X_n$.
It might then seem natural to refer to \eqref{E:negreinf} and \eqref{E:reinf} respectively as negative and positive reinforcement algorithms. However, in the literature, negative reinforcement usually refers to a somehow different notion (see e.g. \cite{Pem}), and we shall thus avoid using this terminology.
A key observation is that \eqref{E:negreinf} can be recovered from \eqref{E:reinf} as follows. Simon's algorithm naturally encodes a genealogical forest with set of vertices $\NN=\{1, \ldots,\}$ and edges $(v(j),j)$ for all $j\geq 2$ with $\varepsilon_j=0$; see the forthcoming Figure 1.
Then $\check X_n=\hat X_n$ if the vertex $n$ belongs to an even generation of its tree component, and $\check X_n=-\hat X_n$ if $n$ belongs to an odd generation.
On the other hand, the statistics of Simon's genealogical forest can be described in terms of independent random recursive trees (see e.g. Chapter 6 in \cite{Drmota} for background) conditionally on their sizes.
This leads us to investigate the difference $\Delta(\mathbb T_k)$ between the number of vertices at even generations and the number of vertices at odd generations in a random recursive tree $\mathbb T_k$ of size $k\geq 1$. The law of $\Delta(\mathbb T_k)$ can be expressed in terms of Eulerian numbers, and properties of the latter enable us either to compute explicitly or estimate certain quantities which are crucial for the proofs of Proposition \ref{P0} and Theorem \ref{T2}.
It is interesting to compare asymptotic behaviors for counterbalanced steps with those for reinforced steps. If we write $\hat S(n)=\hat X_1+ \cdots + \hat X_n$ for the random walk with reinforced steps, then it is known that the law of large numbers holds for
$\hat S$, namely $\hat S(n)/n\to m_1$ a.s. when $\int_{\RR}|x|\mu(\mathrm{d}x)<\infty$, independently of the innovation parameter $p$. Further, regarding fluctuations when $\int_{\RR}|x|^2\mu(\mathrm{d}x)<\infty$, a phase transition occurs for the critical parameter $p_c=1/2$, in the sense that
$\hat S$ is diffusive for $p>1/2$ and superdiffusive for $p<1/2$; see \cite{Bescal, UNRBM}. Despite of the natural coupling between \eqref{E:negreinf} and \eqref{E:reinf},
there are thus major differences\footnote{ This should not come as a surprise. In the simplest case when $\mu=\delta_1$ is the Dirac mass at $1$, one has $\hat S(n)\equiv n$ whereas $\check S$ is a truly stochastic process, even for $p=0$ when there is no innovation.} between the asymptotic behaviors of $\check S$ and of $\hat S$:
Proposition \ref{P0} shows that the asymptotic speed of $\check S$ depends on $p$, and Theorem
\ref{T2} that there is no such phase transition for counterbalanced steps and $\check S$ is always diffusive.
The phase transition for step reinforcement when $\mu$ has a finite second moment can be explained informally as follows; for the sake of simplicity, suppose also that $\mu$ is centered, i.e. $m_1=0$. There are $\mathrm i(n)\sim pn$ trees in Simon's genealogical forest, which are overwhelmingly microscopic (i.e. of size $O(1)$), whereas only a few trees reach the size $O(n^{1-p})$. Because $\mu$ is centered, the contribution of microscopic trees to $\hat S(n)$ is of order $\sqrt n$, and that of the few largest trees of order $n^{1-p}$. This is the reason why $\hat S(n)$ grows like $\sqrt n \gg n^{1-p}$ when $p>1/2$, and rather like $n^{1-p}\gg \sqrt n$ when $p<1/2$. For counterbalanced steps, we will see that, due to the counterbalancing mechanism, the contribution of a large tree with size $\ell\gg 1$ is now only of order $\sqrt \ell$. As a consequence, the contribution to $\check S(n)$ of the largest trees of Simon's genealogical forest is only of order $O(n^{(1-p)/2})$. This is always much smaller than the contribution of microscopic trees which remain of order $\sqrt n$.
We further stress that, even though only the sizes of the trees in Simon's genealogical forest are relevant for the analysis of the random walk $\hat S$ with reinforced steps, the study of the random walk $\check S$ with counterbalanced steps is more complex and requires informations on the fine structure of those trees, not merely their sizes.
The rest of this text is organized as follows. Section 2 focusses on the purely counterbalanced case $p=0$.
In this situation, for each fixed $n\geq 1$, the distribution of $\check S(n)$ can be expressed explicitly in terms of Eulerian numbers.
Section 3 is devoted to the coupling between the counterbalancing algorithm \eqref{E:negreinf} and H.A. Simon's algorithm \eqref{E:reinf}, and to the interpretation of the former in terms of a forest of random recursive trees induced by the latter. Proposition \ref{P0} and Theorem \ref{T2} are proved in Section 4, where we analyze more finely the respective contributions of some natural sub-families. Last, in Section 5, we present a stable version of Theorem \ref{T2} when $\mu$ belongs to the domain of attraction (without centering) of an $\alpha$-stable distribution for some $\alpha\in(0,2)$.
\section{Warm-up: the purely counterbalanced case}
This section is devoted to the simpler situation\footnote{ Observe that this case without innovation has been excluded in Theorem \ref{T2}.} when $p=0$. So $\varepsilon_n\equiv 0$ for all $n\geq 2$, meaning that every step, except of course the first one, counterbalances some preceding step. The law $\mu$ then only plays a superficial role as it is merely relevant for the first step. For the sake of simplicity, we further focus on the case when $\mu=\delta_1$ is the Dirac mass at $1$.
The dynamics are entirely encoded by the sequence $(v(n))_{n\geq 2}$ of independent uniform variables on $\{1,\ldots, n-1\}$; more precisely the purely counterbalanced sequence of bits is given by
\begin{equation}\label{E:p=0}
\check X_1= 1 \quad \text{and} \quad \check X_n=-\check X_{v(n)}\quad \text{for all }n\geq 2.
\end{equation}
The random algorithm \eqref{E:p=0} points at a convenient representation in terms of random recursive trees. Specifically, the sequence $(v(n))_{n\geq 2}$ encodes a random tree ${\mathbb T}_{\infty}$ with set of vertices $\NN$
and set of edges $\{(v(n),n): n\geq 2\}$. Roughly speaking, ${\mathbb T}_{\infty}$ is constructed recursively by incorporating vertices one after the other
and creating an edge between each new vertex $n$ and its parent $v(n)$ which is picked uniformly at random in $\{1, \ldots, n-1\}$ and independently of the other vertices.
If we view $1$ as the root of ${\mathbb T}_{\infty}$ and call a vertex $j$ \textit{odd} (respectively, \textit{even}) when its generation (i.e. its distance to the root in ${\mathbb T}_{\infty}$)
is an odd (respectively, even) number, then
$$\check X_n= \left\{ \begin{matrix} 1 & \text{ if $n$ is an even vertex in $\mathbb T_{\infty}$,}\\
-1 & \text{ if $n$ is an odd vertex in $\mathbb T_{\infty}$.}
\end{matrix} \right.
$$
Let us now introduce some notation with that respect. For every $n\geq 1$, we write ${\mathbb T}_n$ for the restriction of ${\mathbb T}_{\infty}$
to the set of vertices $\{1,\ldots, n\}$ and refer to ${\mathbb T}_n$ as a random recursive tree with size $n$. We also write
$\mathrm{Odd}({\mathbb T}_n)$ (respectively, $\mathrm{Even}({\mathbb T}_n)$) for the number of odd (respectively, even) vertices in ${\mathbb T}_n$
and set
$$\Delta({\mathbb T}_n) \coloneqq \mathrm{Even}({\mathbb T}_n) - \mathrm{Odd}({\mathbb T}_n) = n-2 \mathrm{Odd}({\mathbb T}_n).$$
Of course, we can also express
$$\Delta({\mathbb T}_n)= \check X_1+ \cdots + \check X_n,$$
which is the trajectory of a completely averse elephant random walk (i.e. for the parameter $q=0$).
The main observation of this section is that law of the number of odd vertices is readily expressed in terms of the Eulerian numbers. Recall that
$\langle{}^n_k\rangle$ denotes the number of permutations $\varsigma$ of $\{1, \ldots,n\}$ with $k$ descents, i.e. such that $\#\{1\leq j<n: \varsigma(j) > \varsigma(j+1)\}=k$. Obviously $\langle{}^n_k\rangle\geq 1$ if and only if $0\leq k <n$,
and
one has
$$\sum_{k=0}^{n-1} \euler{n}{k}=n!.$$
The linear recurrence equation
\begin{equation}\label{E:recurEuler}
\euler{n}{k}=(n-k)\euler{{n-1}}{{k-1}}+(k+1)\euler{{n-1}}{k}
\end{equation}
is easily derived from a recursive construction of permutations (see Theorem 1.3 in \cite{Peter}); we also mention the explicit formula (see Corollary 1.3 in \cite{Peter})
$$\euler{n}{k}=\sum _{j=0}^{k}(-1)^{j}{\binom {n+1}{j}}(k+1-j)^{n}.$$
\begin{lemma}\label{L1} For every $n\geq 1$, we have
$$\PP(\mathrm{Odd}({\mathbb T}_n)=\ell)= \frac{1}{(n-1)! } \euler{{n-1}}{{\ell-1}},$$
with the convention\footnote{ Note that this convention is in agreement with the linear recurrence equation \eqref{E:recurEuler}.} that $\langle{}^{\ 0}_{-1}\rangle=1$ in the right-hand side for $n=1$ and $\ell=0$.
\end{lemma}
\begin{proof} Consider $n\geq 1$ and note from the very construction of random recursive trees that there is the identity
$$ \PP(\mathrm{Odd}({\mathbb T}_{n+1})=\ell) = \frac{\ell}{n}\, \PP(\mathrm{Odd}({\mathbb T}_n)=\ell) + \frac{n+1-\ell}{n}\, \PP(\mathrm{Odd}({\mathbb T}_n)=\ell-1).$$
Indeed, the first term of the sum in the right-hand side accounts for the event that the parent $v(n+1)$ of the new vertex $n+1$ is an odd vertex (then $n+1$ is an even vertex), and the second term for the event that $v(n+1)$ is an even vertex (then $n+1$ is an odd vertex).
In terms of $A(n,k) \coloneqq n! \PP(\mathrm{Odd}({\mathbb T}_{n+1})=k+1)$, this yields
$$ A(n,k) = (k+1) A(n-1,k) + (n-k) A(n-1,k-1),$$
which is the linear recurrence equation \eqref{E:recurEuler} satisfied by the Eulerian numbers. Since plainly $A(1,0)=\PP(\mathrm{Odd}(2)=1)=1=\langle{}^1_0\rangle$,
we conclude by iteration that $A(n,k)=\langle{}^n_k\rangle$ for all $n\geq 1$ and $0\leq k<n$. Last, the formula in the statement also holds for $n=1$ since $\mathrm{Odd}(1)=0$.
\end{proof}
\begin{remark} Lemma \ref{L1} is implicit in Mahmoud \cite{Mahmoud}\footnote{ Beware however that the definition of Eulerian numbers in \cite{Mahmoud} slightly differs from ours, namely
$\langle{}^n_k\rangle$ there corresponds to $\langle{}^{\ n}_{k-1}\rangle$ here.}. Indeed $\mathrm{Odd}({\mathbb T}_n)$ can be viewed as the number blue balls in an analytic Friedman's urn model
started with one white ball and replacement scheme $\left({}^{0\ 1}_{1\ 0}\right)$; see Section 7.2.2 in \cite{Mahmoud}. In this setting, Lemma \ref{L1} is equivalent to
the formula for the number of white balls at the bottom of page 127 in \cite{Mahmoud}. Mahmoud relied on the analysis of the differential system associated to the replacement scheme via a Riccati differential equation and inversion of generating functions. The present approach based on the linear recurrence equation \eqref{E:recurEuler} is more direct.
Lemma \ref{L1} is also a closed relative to a result due to Najock and Heyde \cite{NaHeyde} (see also Theorem 8.6 in Mahmoud \cite{Mahmoud}) which states that the number of leaves in a random recursive tree with size $n$ has the same distribution as that appearing in Lemma \ref{L1}.
\end{remark}
We next point at a useful identity related to Lemma \ref{L1} which goes back to Laplace (see Exercise 51 of Chapter I in Stanley \cite{Stanley}) and is often attributed to Tanny \cite{Tanny}. For every $n\geq 0$, there is the identity in distribution
\begin{equation}\label{E:tanny}
\mathrm{Odd}({\mathbb T}_{n+1})\,{\overset{\mathrm{(d)}}{=}}\, \lceil U_1+\cdots + U_n\rceil,
\end{equation}
where in the right-hand side, $U_1, U_2, \ldots$ is a sequence of i.i.d. uniform variables on $[0,1]$ and $\lceil \cdot \rceil$ denotes the ceiling function.
We now record for future use the following consequences.
\begin{corollary}\label{C0}
\begin{itemize}
\item[(i)] For every $n\geq 2$, the variable $\Delta({\mathbb T}_n)$ is symmetric, i.e. $\Delta({\mathbb T}_n)\,{\overset{\mathrm{(d)}}{=}} \, -\Delta({\mathbb T}_n)$,
and in particular $\EE(\Delta({\mathbb T}_n))= 0$.
\item[(ii)] For all $n\geq 3$, one has $ \EE(\Delta({\mathbb T}_n)^2)= n/3$.
\item[(iii)] For all $n\geq 1$, one has $\EE(|\Delta({\mathbb T}_n)|^4) \leq 6n^2$.
\end{itemize}
\end{corollary}
\begin{proof} (i) Equivalently, the assertion claims that in a random recursive tree of size at least $2$, the number of odd vertices and the number of even vertices have the same distribution. This is immediate from \eqref{E:tanny} and can also be checked directly from the construction.
(ii) This has been already observed by Sch\"utz and Trimper \cite{SchTr} in the setting of the elephant random walk; for the sake of completeness we present a short argument.
The vertex $n+1$ is odd (respectively, even) in ${\mathbb T}_{n+1}$ if and only if its parent is an even (respectively, odd) vertex in ${\mathbb T}_{n}$. Hence one has
$$\EE(\Delta({\mathbb T}_{n+1})-\Delta({\mathbb T}_{n})\mid {\mathbb T}_{n})= -\frac{1}{n}\Delta({\mathbb T}_{n}),$$
and since $\Delta({\mathbb T}_{n+1})-\Delta({\mathbb T}_{n})= \pm 1$, this yields the recursive equation
$$\EE(\Delta({\mathbb T}_{n+1})^2)= (1-2/n)\EE(\Delta({\mathbb T}_{n})^2) +1.$$
By iteration, we conclude that $ \EE(\Delta({\mathbb T}_n)^2)= n/3$ for all $n\geq 3$.
(iii)
Recall that the process of the fractional parts $\{U_1+\cdots + U_n\}$ is a Markov chain on $[0,1)$ whose distribution at any fixed time $n\geq 1$
is uniform on $[0,1)$. Writing $V_n= 1-2U_n$ and $W_n=2\{U_1+\cdots + U_n\}-1$, we see that $V_1, V_2, \ldots$ is a sequence of i.i.d. uniform variables on $[-1,1]$
and that $W_n$ has the uniform distribution on $[-1,1]$ too.
The characteristic function of the uniform variable $V_j$ is
$$\EE(\exp(\iu \theta V_j))=\theta^{-1}\sin(\theta) = 1-\frac{\theta^2}{6} + \frac{\theta^4}{120} + O(\theta^6)\qquad\text{as }\theta \to 0,$$
and therefore for every $n\geq 1$,
\begin{align*}\EE(\exp(\iu \theta (V_1+\cdots+V_n)))&=\left(1-\frac{\theta^2}{6} + \frac{\theta^4}{120} + O(\theta^6)\right)^n\\
&=1-\frac{n}{6}\theta^2 + \left(\frac{n}{120}+ \frac{n(n-1)}{72}\right) \theta^4+ O(\theta^6).
\end{align*}
It follows that
$$\EE( (V_1+\cdots+V_n)^4)= 24 \left(\frac{n}{120}+ \frac{n(n-1)}{72}\right) \leq n^2/3.$$
We can rephrase \eqref{E:tanny} as the identity in distribution
$$\Delta({\mathbb T}_{n+1})\,{\overset{\mathrm{(d)}}{=}}\, V_1+\cdots+V_n + W_n.$$
Since $\EE(W_n^4)=1/3$, the proof is completed with the elementary bound $(a+b)^4\leq 16(a^4+b^4)$.
\end{proof}
We now conclude this section with an application of \eqref{E:tanny} to the asymptotic normality of $\Delta({\mathbb T}_n)$. Since $\EE(U)=1/2$ and $\mathrm{Var}(U)=1/12$, the classical central limit theorem immediately yields the following.
\begin{corollary}\label{C1} Assume $p=0$ and $\mu=\delta_1$. One has
$$\lim_{n\to \infty} \frac{ \Delta({\mathbb T}_n)}{\sqrt n} = {\mathcal N}(0,1/3)\qquad \text{in distribution.}$$
\end{corollary}
Corollary \ref{C1} goes back to \cite{NaHeyde} in the setting of the number of leaves in random recursive trees; see also \cite{BaurBer, Bercu, ColGavSch1, ColGavSch2} for alternative proofs in the framework of the elephant random walk.
\section{Genealogical trees in Simon's algorithm}
From now on, $\mu$ is an arbitrary probability law on $\RR$ and we also suppose that the innovation rate is strictly positive, $p\in(0,1)$.
Recall the construction of the sequence $(\hat X_n)$ from Simon's reinforcement algorithm \eqref{E:reinf}.
Simon was interested in the asymptotic frequencies of variables having a given number of occurrences. Specifically, for every $n,j\in \NN$, we write
$${N}_j(n)\coloneqq \#\{\ell \leq n: \hat X_{\ell}=X_j\}$$
for the number of occurrences of the variable $X_j$ until the $n$-th step of the algorithm \eqref{E:reinf}, and
\begin{equation}\label{E:nudef}
\nu_k(n)\coloneqq \#\{1\leq j \leq \mathrm{i}(n): {N}_j(n)=k\}, \qquad k\in\NN
\end{equation}
for the number of such variables that have occurred exactly $k$ times. Observe also that the number of innovations satisfies the law of large numbers ${\mathrm i}(n)\sim p n$ a.s.
\begin{lemma}\label{L2} For every
$k\geq 1$, we have
$$\lim_{n\to \infty} \frac{\nu_k(n)}{pn}=
\frac{1}{1-p}
{\mathrm B}(k,1+1/(1-p)) \qquad\text{in probability,}$$
where ${\mathrm B}$ denotes the beta function.
\end{lemma}
Lemma \ref{L2} is essentially due to H.A. Simon \cite{Simon}, who actually only established the convergence of the mean value.
The strengthening to convergence in probability can be obtained as in \cite{BRST} from a concentration argument based on the Azuma-Hoeffding's inequality;
see Section 3.1 in \cite{PPS}. The right-hand side in the formula is a probability mass on $\NN$ known as the Yule-Simon distribution with parameter $1/(1-p)$.
We record for future use a couple of identities which are easily checked from the integral definition of the beta function:
\begin{equation} \label{E:sumbeta}
\frac{1}{1-p} \sum_{k=1}^{\infty} {\mathrm B}(k,1+1/(1-p)) =1
\end{equation}
and
\begin{equation} \label{E:sumkbeta}
\frac{1}{1-p} \sum_{k=1}^{\infty} k {\mathrm B}(k,1+1/(1-p)) = \frac{1}{p}.
\end{equation}
For $k=1$, Lemma \ref{L2} reads
\begin{equation}\label{E:Simon1}
\lim_{n\to \infty} n^{-1}\nu_1(n)= \frac{p}{2-p}\qquad\text{in probability.}
\end{equation}
We shall also need to estimate the fluctuations, which can be derived by specializing a Gaussian limit theorem for extended P\'olya urns due to Bai \textit{et al.} \cite{BHZ}.
\begin{lemma}\label{L5} There is the convergence in distribution
$$\lim_{n\to \infty} \frac{\nu_1(n)- np/(2-p)}{\sqrt n} = \mathcal N\left(0,\frac{2p^3-8p^2+6p}{(3-2p)(2-p)^2} \right).$$
\end{lemma}
\begin{proof} The proof relies on the observation that Simon's algorithm can be coupled with a two-color urn governed by the same sequences of random bits $(\varepsilon_n)$ and of uniform variables $(v(n))$ as follows. Imagine that we observe the outcome of Simon's algorithm at the $n$-step
and that for each $1\leq j \leq n$, we associate a white ball if the variable $\hat X_j$ appears exactly once, and a red ball otherwise.
At the initial time $n=1$, the urn contains just one white ball and no red balls.
At each step $n\geq 2$, a ball picked uniformly at random in the urn (in terms of Simon's algorithm, this is given by the uniform variable $v(n)$).
If $\varepsilon_n=1$, then the ball picked is returned to the urn and one adds a white ball (in terms of Simon's algorithm, this corresponds to an innovation
and $\nu_1(n)=\nu_1(n-1)+1$).
If $\varepsilon_n=0$, then the ball picked is removed from the urn and one adds two red balls
(in terms of Simon's algorithm, this corresponds to a repetition
and either $\nu_1(n)=\nu_1(n-1)-1$ if the ball picked is white, or $\nu_1(n)=\nu_1(n-1)$ if the ball picked is red).
By construction, the number $W_n$ of white balls in the urn coincides with the number $\nu_1(n)$ of variables that have appeared exactly once
in Simon's algorithm \eqref{E:reinf}.
We shall now check our claim in the setting of \cite{BHZ} by specifying the quantities which appear there. The evolution of number of white balls in the urn is governed by Equation (2.1) in \cite{BHZ}, viz.
$$W_n=W_{n-1}+ I_nA_n+(1-I_n)C_n,$$ where $I_n=1$ if a white ball is picked and $I_n=0$ otherwise. In our framework, we further have
$A_n=2\varepsilon_n-1$ and $C_n=\varepsilon_n$. If we write ${\mathcal F}_n$ for the natural filtration generated by the variables $(A_k,C_k, I_k)_{k\leq n}$,
then $A_n$ and $C_n$ are independent of ${\mathcal F}_{n-1}$ with
$$\EE(A_n)= 2p-1, \quad \EE(C_n)= p, \quad \mathrm{Var}(A_n)= 4(p-p^2), \quad \mathrm{Var}(C_n)= p-p^2.$$
This gives in the notation (2.2) of \cite{BHZ}:
\begin{align*}
\sigma^2_M&= \frac{p}{2-p}4(p-p^2)+ \left(1-\frac{p}{2-p}\right) (p-p^2) + (p-1)^2 \frac{p}{2-p} \left(1-\frac{p}{2-p}\right)\\
&= \frac{2p^3-8p^2+6p}{(2-p)^2},
\end{align*}
and finally
$$\sigma^2=\frac{2p^3-8p^2+6p}{(3-2p)(2-p)^2}.$$
Our claim can now be seen as a special case of Corollary 2.1 in \cite{BHZ}.
\end{proof}
We shall also need a refinement of Lemma \ref{L2} in which one does not only record the number of occurrences of the variable $X_j$, but more generally the genealogical structure of these occurrences. We need to introduce first some notation in that respect.
Fix $n\geq 1$ and let some $1\leq j \leq {\mathrm i}(n)$ (i.e. the variable $X_j$ has already appeared at the $n$-th step of the algorithm).
Write $\ell_1< \ldots < \ell_k \leq n$ for the increasing sequence of steps of the algorithm at which $X_j$ appears, where $k={N}_j(n)\geq 1$.
The genealogy of occurrences of the variable $X_j$ until the $n$-th step is recorded as a tree $T_j(n)$ on $\{1, \ldots, k\}$
such that for every $1\leq a < b \leq k$, $(a,b)$ is an edge of $T_j(n)$ if and only if $v(\ell_b)=\ell_a$, that is, if and only if the identity
$\hat X_{\ell_b}=X_j$ actually results from the fact that the algorithm repeats the variable $\hat X_{\ell_a}$ at its $\ell_b$-th step.
Plainly, $T_j(n)$ is an increasing tree with size $k$, meaning a tree on $\{1, \ldots, k\}$ such that the sequence of vertices along any branch from the
root $1$ to a leaf is increasing. In this direction, we recall that there are $(k-1)!$ increasing trees with size $k$ and that the uniform distribution of the set
increasing trees with size $k$ coincides with the law of the random recursive tree of size $k$, ${\mathbb T}_k$. See for instance Section 1.3.1 in Drmota \cite{Drmota}.
\begin{figure}
\begin{center}
\includegraphics[height=6cm]{ForestSimon.png}
\end{center}
\caption{ Example of a genealogical forest representation of Simon's algorithm \eqref{E:reinf} after 18 steps. The dotted edges account for innovation events, i.e. $\varepsilon_j=1$
and the four genealogical trees are rooted at $1,5,7,14$. In each subtree, vertices at even generations are colored in green and vertices at odd generation in white. For instance the genealogical tree $T_2(18)$ is rooted at $5$, it has $3$ even vertices and $2$ odd vertices.
}\label{fig:summary}
\end{figure}
More generally, the distribution of the entire genealogical forest given the sizes of the genealogical trees can be described as follows.
\begin{lemma}\label{L3} Fix $n\geq 1$, $1\leq k\leq n$, and let $n_1, \ldots , n_k\geq 1$ with sum $n_1+ \cdots + n_k=n$. Then conditionally on
${N}_j(n)=n_j$ for every $j=1, \ldots, k$, the genealogical trees $T_1(n), \ldots , T_k(n)$ are independent random recursive trees with respective sizes $n_1, \ldots , n_k$.
\end{lemma}
\begin{proof} Recall that the set $\{(v(j),j)$ for $1\leq j\leq n\}$ is that of the edges of the random recursive tree with size $n$, ${\mathbb T}_n$.
The well-known splitting property states that removing a given edge, say $(v(j),j)$ for some fixed $j$, from ${\mathbb T}_n$ produces two subtrees which in turn, conditionally on their sizes, are two independent random recursive trees. This has been observed first by Meir and Moon \cite{MeirMoon}; see also \cite{BBsurvey} and references therein for more about this property.
The genealogical trees $T_1(n), \ldots , T_k(n)$ result by removing the edges $(v(j),j)$ in ${\mathbb T}_n$ for which $\varepsilon_j=1$ and enumerating in each subtree component their vertices in the increasing order.
Our statement is now easily seen by applying iteratively this splitting property. \end{proof}
We shall also need for the proofs of Proposition \ref{P0} and Theorem \ref{T2} an argument of uniform integrability that relies in turn on the following lemma.
Recall that if $T$ is a rooted tree, $\Delta(T)$ denotes the difference between the number of vertices at even distance from root and that at odd distance.
\begin{lemma}\label{L6} For every $1<\beta<2\wedge \frac{1}{1-p}$, one has
$$\sup_{n\geq 1} \frac{1}{n} \sum_{j=1}^{n}\EE({N}_j(n)^{\beta})< \infty$$
and
$$\sup_{n\geq 1} \frac{1}{n}\EE\left( \sum_{j=1}^{{\mathrm i}(n)} |\Delta(T_j(n) ) |^{2\beta} \right) <\infty.$$
\end{lemma}
\begin{proof} The first claim is a consequence of Lemma 3.6 of \cite{NRLP} which states that for $\beta \in(1,1/(1-p))$ [beware that the parameter denoted by $p$ in \cite{NRLP} is actually $1-p$ here], there exists numerical constants $c>0$ and $\eta\in(0,1)$ such that
$\EE({N}_j(n)^{\beta})\leq c(n/j)^{\eta}$ for all $1\leq j \leq n$.
Next, combining Jensen's inequality with Corollary \ref{C0}(iii), we get that for $k\geq 2$
$$\EE( |\Delta({\mathbb T}_k)|^{2\beta}) \leq \EE(|\Delta({\mathbb T}_k)|^4)^{\beta/2}\leq 6k^{\beta}.$$
Then recall that conditionally on ${N}_j(n)=k\geq 1$, $T_j(n) $ has the law of the random recursive tree with size $k$, ${\mathbb T}_k$, and hence
\begin{align*}
\EE\left( \sum_{j=1}^{{\mathrm i}(n)} |\Delta(T_j(n) ) |^{2\beta} \right)
&= \sum_{j=1}^{n} \left(\sum_{k=1}^{n} \EE( |\Delta({\mathbb T}_k)|^{2\beta}) \PP({N}_j(n)=k)\right)\\
&\leq 6\sum_{j=1}^{n} \left(\sum_{k=1}^{n} k^{\beta} \PP({N}_j(n)=k)\right).
\end{align*}
We know from the first part that this last quantity is finite, and the proof is complete.
\end{proof}
\section{Proofs of the main results}
As its title indicates, the purpose of this section is to establish Proposition \ref{P0} and Theorem \ref{T2}.
The observation that for every $n\geq 1$ and $1\leq j \leq {{\mathrm i}(n)}$, the variable $X_j$ appears exactly
$\mathrm{Even}(T_j(n))$ times and its opposite $-X_j$ exactly $\mathrm{Odd}(T_j(n))$ times until the $n$-step of the algorithm \eqref{E:negreinf},
yields the identity
\begin{equation}\label{E:repres}
\check S(n)\coloneqq \sum_{i=1}^n\check X_i = \sum_{j=1}^{{\mathrm i}(n)} \Delta(T_j(n)) X_j,
\end{equation}
which lies at the heart of our approach. We stress that in \eqref{E:repres} as well as in related expressions that we shall use in the sequel, the sequence of i.i.d. variables $(X_n)$ and the family of genealogical trees $(T_j(n))$ are independent, because the latter are constructed from the sequences $(\varepsilon_n)$ and $(v(n))$ only.
Actually, our proof analyzes more precisely the effects of the counterbalancing algorithm \eqref{E:negreinf} by estimating specifically the contributions of certain sub-families to the asymptotic behavior of $\check S$. Specifically, we set for every $k\geq 1$,
\begin{equation} \label{E:checkSk}
\check S_{k}(n) \coloneqq \sum_{j=1}^{{\mathrm i}(n)} \Delta(T_j(n)) X_j \Ind_{N_j(n)=k},
\end{equation}
so that
$$\check S(n) = \sum_{k=1}^{n} \check S_k(n).$$
\subsection{Proof of Proposition \ref{P0}}
The case $p=1$ (no counterbalancing events) of Proposition \ref{P0} is just the weak law of large numbers, and the case $p=0$ (no innovations) is a consequence of Corollary \ref{C1}. The case $p\in(0,1)$ derives from the next lemma which shows more precisely that the variables $X_j$ that have appeared
in the algorithm \eqref{E:negreinf} but have not yet counterbalanced determine
the ballistic behavior of $\check S$, whereas those that have appeared twice or more (i.e. such that $N_j(n)\geq 2$) have a negligible impact.
\begin{lemma}\label{L7}
Assume that $\int_\RR |x| \mu(\mathrm d x)<\infty$ and recall that $m_1=\int_{\RR} x \mu(\mathrm d x)$. Then the following limits hold in probability:
\begin{enumerate}
\item[(i)] $\lim_{n\to \infty} n^{-1} \check S_1(n) =m_1p/(2-p)$,
\item[(ii)]
$\lim_{n\to \infty} n^{-1} \sum_{k=2}^n \left| \check S_{k}(n) \right| = 0$.
\end{enumerate}
\end{lemma}
\begin{proof}
(i) Recall the notation \eqref{E:nudef} and that the sequence of i.i.d. variables $(X_j)$ is independent of the events $\{{N}_j(n)=1\}$.
We see that there is the identity in distribution
$$\check S_1(n) \,{\overset{\mathrm{(d)}}{=}}\, S_1(\nu_1(n)),$$
where $S_1(n)=X_1+\cdots + X_n$ is the usual random walk.
The claim follows readily from the law of large numbers and \eqref{E:Simon1}.
(ii)
We first argue that for each fixed $k\geq 2$,
\begin{equation}\label{E:fixedk}
\lim_{n\to \infty} n^{-1} \check S_{k}(n) = 0 \qquad\text{almost surely.}
\end{equation}
Indeed, recall that $\nu_k(n)$ denotes the number of genealogical trees $T_j(n)$ with size $k$.
It follows from Lemma \ref{L3} that conditionally on $\nu_k(n)=\ell$, the sub-family of such $T_j(n)$ enumerated in the increasing order of the index $j$, is given by $\ell$ i.i.d. copies of the random recursive tree ${\mathbb T}_k$. Hence, still conditionally on $\nu_k(n)=\ell$, enumerating the elements of the sub-family $\{X_j \Delta(T_j(n)): {N}_j(n)=k\}$ in the increasing order of $j$ yields $\ell$ independent variables, each being distributed as $X_1 \Delta({\mathbb T}_k)$ with $X_1$ and $ \Delta({\mathbb T}_k)$ independent. We deduce from Corollary \ref{C0}(i) that the variable $X_1 \Delta({\mathbb T}_k)$ symmetric, and since it is also integrable, it is centered. Since $\nu_k(n)\leq n$, this readily entails \eqref{E:fixedk} by an application of the law of large numbers.
The proof can be completed by an argument of uniform integrability. In this direction, fix an arbitrarily large integer $\ell$ and write by the triangular inequality
\begin{align*} \frac{1}{n} \sum_{k=\ell}^n \left| \check S_{k}(n) \right| &\leq \frac{1}{n} \EE\left( \sum_{j=1}^{{\mathrm i}(n)} |X_j | {N}_j(n) \Ind_{{N}_j(n)\geq \ell} \right)\\
&=\frac{ \EE(|X_1|)}{n} \sum_{j=1}^{n} \EE\left( {N}_j(n) \Ind_{{N}_j(n)\geq \ell} \right)\\
&\leq \ell^{1-\beta} \frac{ \EE(|X_1|)}{n}\sum_{j=1}^{n} \EE\left( {N}_j(n)^{\beta} \right),
\end{align*}
where the last inequality holds for any $\beta>1$. We see from Lemma \ref{L6} that the right-hand side converges to $0$ as $\ell \to \infty$ uniformly in $n\geq 1$, and the rest of the proof is straightforward.
\end{proof}
\subsection{Proof of Theorem \ref{T2}}
For $p=1$ (no counterbalancing events), Theorem \ref{T2} just reduces to the classical central limit theorem, so we assume $p\in(0,1)$.
The first step of the proof consists in determining jointly the fluctuations of the components $\check S_k$ defined in \eqref{E:checkSk}.
\begin{lemma}\label{L8} Assume that $\int_\RR |x|^2 \mu(\mathrm d x)<\infty$ and recall that $m_2=\int_{\RR} x^2 \mu(\mathrm d x)$.
Then as $n\to \infty$, the sequences of random variables
$$ \frac{ \check S_1(n) -pm_1/(2-p)}{\sqrt n} \qquad \text{(for $k=1$)}$$
and
$$ \frac{ \check S_{k}(n)}{\sqrt n}\qquad \text{(for $k\geq 2$)}$$
converge jointly in distribution
towards a sequence
$$\left( \mathcal N_{k}(0,\sigma^2_{k})\right)_{k\geq 1}$$
of independent centered Gaussian variables, where
$$ \sigma_1^2\coloneqq \frac{p m_2}{2-p} - \frac{p^2m_1^2}{(3-2p)(2-p)^2},$$ $\sigma^2_2\coloneqq0$, and
$$\sigma^2_{k}\coloneqq \frac{ k p m_2}{3(1-p)} \mathrm B(k,1+1/(1-p)) \qquad \text{for $k\geq 3$}.$$
\end{lemma}
\begin{proof} For each $k\geq 1$, let $(Y_k(n))_{n\geq 1}$ be a sequence of i.i.d. copies of $\Delta(\mathbb T_k) X$, where $X$ has the law $\mu$ and is independent of the random recursive tree $\mathbb T_k$. We further assume that these sequences are independent. Taking partial sums yields a sequence indexed by $k$ of independent random walks
$$S_k(n)= Y_k(1)+ \cdots + Y_k(n), \qquad n\geq 0. $$
For each $n\geq 1$, the family of blocks
$$B_{k}(n)\coloneqq \{j\leq {\mathrm i}(n): N_j(n)=k\} \qquad \text{for }1\leq k\leq {\mathrm i}(n)$$
forms a random partition of $\{1, \ldots, {\mathrm i}(n)\}$
which is independent of the $X_j$'s.
Recall that we are using the notation $\nu_k(n)=\#B_k(n)$, and also from Lemma \ref{L3}, that conditionally on the $N_j(n)$'s, the genealogical trees $T_j(n)$ are independent random recursive trees.
We now see from the very definition \eqref{E:checkSk} that for every fixed $n\geq 1$, there is the identity in distribution
$$\left( \check S_k(n)\right)_{k\geq 1} \,{\overset{\mathrm{(d)}}{=}}\, \left( S_k(\nu_k(n))\right)_{k\geq 1},$$
where in the right-hand side, the random walks $(S_k)_{k\geq 1}$ are independent of the sequence of block sizes $(\nu_k(n))_{k\geq 1}$.
Next we write, first for $k=1$,
$$S_1(\nu_1(n)) - \frac{pn}{2-p} m_1 = S_1\left(\left\lfloor \frac{pn}{2-p}\right \rfloor \right) - \frac{pn}{2-p}m_1 + \sum_{j=\lceil pn/(2-p)\rceil}^{\nu_1(n)} Y_1(j) ,$$
second $S_2\equiv 0$ (since $\Delta(\mathbb T_2)\equiv 0$) for $k=2$, and third, for $k\geq 3$,
$$S_k(\nu_k(n)) = S_k\left(\left\lfloor \frac{pn}{1-p}
{\mathrm B}(k,1+1/(1-p))\right \rfloor \right) + \sum_{j=\lceil \frac{pn}{1-p}
{\mathrm B}(k,1+1/(1-p))\rceil}^{\nu_k(n)} Y_k(j) ,$$
with the usual summation convention that $\sum_{j=a}^b = - \sum_{j=b}^a$ when $b<a$.
Since the i.i.d. variables $Y_1(\cdot)$ have mean $m_1$ and variance $m_2-m_1^2$, the central limit theorem ensures that there is the convergence in distribution
\begin{equation} \label{E:clt}
\lim_{n\to \infty} n^{-1/2}\left( S_1\left(\left\lfloor \frac{pn}{2-p}\right \rfloor \right) - \frac{pn}{2-p}m_1\right) = \mathcal N_1\left( 0,\frac{p (m_2-m_1^2)}{2-p}\right ).
\end{equation}
Similarly, for $k\geq 3$, each $Y_k(n)$ is centered with variance $km_2/3$ (by Corollary \ref{C0}(i-ii)) and hence, using the notation in the statement, there is the convergence in distribution
\begin{align} \label{E:clt3}
\lim_{n\to \infty} n^{-1/2}S_k\left(\left\lfloor \frac{pn}{1-p}
{\mathrm B}(k,1+1/(1-p))\right \rfloor \right) & = \mathcal N_k\left( 0,\sigma_k^2 \right)
\end{align}
Plainly, the weak convergences \eqref{E:clt} and \eqref{E:clt3} hold jointly when we agree that the limits are independent Gaussian variables.
Next, from Lemma \ref{L2} and the fact that for $k\geq 3$, the i.i.d. variables $Y_k(j)$ are centered with finite variance, we easily get
$$\lim_{n\to \infty} n^{-1/2} \left | \sum_{j=\lceil \frac{pn}{1-p}
{\mathrm B}(k,1+1/(1-p))\rceil}^{\nu_k(n)} Y_k(j) \right |=0 \qquad \text{in }L^2(\PP).$$
Finally, for $k=1$, we write
$$ \sum_{j=\lceil pn/(2-p)\rceil}^{\nu_1(n)} Y_1(j)= m_1(\nu_1(n) - \lfloor pn/(2-p)\rfloor) + \sum_{j=\lceil pn/(2-p)\rceil}^{\nu_1(n)} (Y_1(j)-m_1).$$
On the one hand, we have from the same argument as above that
$$\lim_{n\to \infty} n^{-1/2} \left | \sum_{j=\lceil pn/(2-p)\rceil}^{\nu_1(n)} (Y_1(j)-m_1)\right |=0 \qquad \text{in }L^2(\PP).$$
On the other hand, we already know from Lemma \ref{L5} that there is the convergence in distribution
$$\lim_{n\to \infty} m_1\frac{\nu_1(n) - \lfloor pn/(2-p)\rfloor}{\sqrt n} = \mathcal N\left( 0,\frac{2p^3-8p^2+6p}{(3-2p)(2-p)^2} m_1^2\right).$$
Obviously, this convergence in law hold jointly with \eqref{E:clt} and \eqref{E:clt3}, where the limiting Gaussian variables are independent. Putting the pieces together, this completes the proof.
\end{proof}
The final step for the proof of Theorem \ref{T2} is the following lemma.
\begin{lemma}\label{L9} We have
$$\lim_{K\to \infty} \sup_{n\geq 1} n^{-1} \EE\left( \left| \sum_{k\geq K} \check S_{k}(n)\right|^2 \right) =0.$$
\end{lemma}
\begin{proof} We write
$$\sum_{k\geq K} \check S_k(n)= \sum_{j=1}^{n} X_j \Delta(T_j(n)) \Ind_{{N}_j(n)\geq K}.$$
Since the $X_j$ are independent of the $T_j(n)$, we get
\begin{align*} \EE\left( \left | \sum_{k\geq K} \check S_k(n)\right|^2 \right)&= \EE\left( \sum_{j,{j'}=1}^{n} X_j X_{j'} \Delta(T_j(n)) \Ind_{{N}_{j'}(n)\geq K} \Delta(T_{j'}(n)) \Ind_{{N}_{j'}(n)\geq K}\right)\\
&\leq m_2 \sum_{j,{j'}=1}^{n} \EE\left( \Delta(T_j(n)) \Ind_{{N}_{j'}(n)\geq K} \Delta(T_{j'}(n)) \Ind_{{N}_{j'}(n)\geq K}\right).
\end{align*}
We evaluate the expectation in the right-hand side by conditioning first on $N_j(n)=k$ and $N_{j'}(n)=k'$ with $k,k'\geq 3$. Recall from Lemma \ref{L3} that the genealogical trees $T_j(n)$ and $T_{j'}(n)$ are then two random recursive trees with respective sizes $k$ and $k'$, which are further independent when $j\neq j'$. Thanks to Corollary \ref{C0}(i-ii)
we get
\begin{align*} &\EE\left( \Delta(T_j(n)) \Ind_{{N}_{j'}(n)\geq K} \Delta(T_{j'}(n)) \Ind_{{N}_{j'}(n)\geq K}\right)
\\&= \left\{\begin{matrix}
\frac{1}{3} \EE(N_j(n)\Ind_{{N}_j(n)\geq K}) &\text{ if }j=j'\,\\
0 &\text{ if }j\neq j'.
\end{matrix}\right.
\end{align*}
We have thus shown that
$$\EE\left( \left | \sum_{k\geq K} \check S_k(n)\right|^2 \right) \leq \frac{m_2 }{3}\sum_{j=1}^{n} \EE(N_j(n)\Ind_{{N}_j(n)\geq K} ),$$
which yields our claim just as in the proof of Lemma \ref{L7}(ii).
\end{proof}
The proof of Theorem \ref{T2} is now easily completed by combining Lemmas \ref{L8} and \ref{L9}.
Indeed, the identity
$$\frac{pm_2}{2-p}+\sum_{k=2}^{\infty}\sigma^2_{k} = \frac{m_2}{3-2p}$$
is easily checked from \eqref{E:sumkbeta}.
\section{A stable central limit theorem}
The arguments for the proof of Theorem \ref{T2} when the step distribution $\mu$ has a finite second moment can be adapted to the case when $\mu$ belongs to some stable domaine of attraction; for the sake of simplicity we focus on the situation without centering. Specifically, let $(a_n)$ be a sequence of positive real numbers that is regularly varying with exponent $1/\alpha$ for some $\alpha\in(0,2)$, in the sense that $\lim_{n\to \infty} a_{\lfloor rn\rfloor}/a_n= r^{1/\alpha}$ for every $r>0$, and suppose that
\begin{equation} \label{E:stable}
\lim_{n\to \infty} \frac{X_1+\cdots + X_n}{a_n} = Z \qquad \text{in distribution},
\end{equation}
where $Z$ is some $\alpha$-stable random variable. We refer to Theorems 4 and 5 on p. 181-2 in \cite{GnKo} and Section 6 of Chapter 2 in \cite{IbLin}
for necessary and sufficient conditions for \eqref{E:stable} in terms of $\mu$ only.
We write $\varphi_{\alpha}$ for the characteristic exponent of $Z$, viz.
$$\EE\left(\exp(\iu \theta Z)\right) = \exp(-\varphi_{\alpha}(\theta))\qquad \text{for all }\theta \in \RR;$$
recall that $\varphi_{\alpha}$ is homogeneous with exponent $\alpha$, i.e.
$$\varphi_{\alpha}(\theta)= |\theta|^{\alpha} \varphi_{\alpha}(\textrm{sgn}(\theta))\qquad \text{for all }\theta\neq 0.$$
Recall the definition and properties of the Eulerian numbers $\langle {}^{n}_{k}\rangle$ from Section 2, and also the Pochhammer notation
$$(x)^{(k)} \coloneqq \frac{\Gamma(x+k)}{\Gamma(x)} =\prod_{j=0}^{k-1} (x+j), \qquad x>0, k\in\NN,$$
for the rising factorial, where $\Gamma$ stands for the gamma function.
We can now claim:
\begin{theorem}\label{T3} Assume \eqref{E:stable}. For each $p\in(0,1)$, we have
$$\lim_{n\to \infty} \frac{\check S(n)}{a_n} = \check Z \qquad \text{in distribution},$$
where $\check Z$ is an $\alpha$-stable random variable with characteristic exponent $\check \varphi_{\alpha}$ given by
$$\check \varphi_{\alpha}(\theta) = \frac{p }{(1-p)} \sum_{k=1}^{\infty} \sum_{\ell=0}^{k-1}\frac{ \varphi_{\alpha}((k-2\ell)\theta)}{ (1+1/(1-p))^{(k)}} \euler{{k-1}}{{\ell-1}}, \qquad \theta \in \RR.$$
\end{theorem}
The proof of Theorem \ref{T3} relies on a refinement of Simon's result (Lemma \ref{L2}) to the asymptotic frequencies of genealogical trees induced by the reinforcement algorithm \eqref{E:reinf}. We denote by ${{\mathcal T}^{\uparrow}}$ the set of increasing trees (of arbitrary finite size), and for any $\tau\in {\mathcal T}^{\uparrow}$, we write $|\tau|$ for its the size (number of vertices) and $\Delta(\tau)$
for the difference between its numbers of even vertices and of odd vertices. Refining \eqref{E:nudef}, we also define
$$\nu_{\tau}(n) \coloneqq \sum_{j= 1}^{{\mathrm i}(n)} \Ind_{\{T_j(n) =\tau\}}, \qquad \tau\in {\mathcal T}^{\uparrow}.$$
\begin{lemma}\label{L2+} We have
$$\sum_{\tau\in{{\mathcal T}^{\uparrow}}}
\frac{ |\tau|+ |\Delta(\tau)|^2}{(1+1/(1-p))^{(|\tau|)}}= \frac{4p}{3(1-p)},
$$
and there is the convergence in probability
$$\lim_{n\to \infty} \sum_{\tau\in{{\mathcal T}^{\uparrow}}} (|\tau|+ |\Delta(\tau)|^2)\left| \frac{\nu_{\tau}(n)}{n} -
\frac{ p}{(1-p) (1+1/(1-p))^{(|\tau|)}} \right| =0.$$
\end{lemma}
\begin{proof}
We start by claiming that for every $k\geq 1$, and every tree $\tau\in {{\mathcal T}^{\uparrow}}$ with size $|\tau|=k$,
we have
\begin{equation}\label{E:excor}
\lim_{n\to \infty} \frac{\nu_{\tau}(n)}{n} =
\frac{ p}{(1-p) (1+1/(1-p))^{(k)}} \qquad\text{in probability.}
\end{equation}
Indeed, the distribution of the random recursive tree $\mathbb T_k$ of size $k$ is the uniform probability measure on the set
of increasing trees with size $k$, which has $(k-1)!$ elements. We deduce from Lemma \ref{L3} and the law of large numbers that
$$\nu_{\tau}(n) \sim \nu_k(n)/(k-1)!.$$
The claim \eqref{E:excor} now follows from Lemma \ref{L2} and the identity
$$ {\mathrm B}(k,1+/(1-p)) = \frac{ (k-1)! }{ (1+1/(1-p))^{(k)}}.$$
We now have to prove that \eqref{E:excor} holds in $L^1(|\tau|+ |\Delta(\tau)|^2,{\mathcal T}^{\uparrow})$.
On the one hand, one has obviously for every $n\geq 1$
$$\sum_{\tau\in{{\mathcal T}^{\uparrow}}} |\tau|\nu_{\tau}(n)= n.$$
On the other hand, there are $(k-1)!$ increasing trees with size $k$ and hence
$$\frac{p}{1-p} \sum_{\tau\in{{\mathcal T}^{\uparrow}}} \frac{|\tau|}{(1+1/(1-p))^{(|\tau|)}}= \frac{p}{1-p} \sum_{k=1}^{\infty} k{\mathrm B}(k,1+1/(1-p))=1,$$
where the second equality is \eqref{E:sumkbeta}.
We deduce from Scheff\'e's Lemma and \eqref{E:excor} that there is the convergence in probability
$$\lim_{n\to \infty} \sum_{\tau\in{{\mathcal T}^{\uparrow}}}|\tau|\left| \frac{\nu_{\tau}(n)}{n} -
\frac{ p}{(1-p) (1+1/(1-p))^{(|\tau|)}} \right| =0.$$
Similarly, we deduce from Corollary \ref{C0}(ii) and Lemma \ref{L3} that, for every $n\geq0$
\begin{align*}
\EE\left( \sum_{\tau\in{{\mathcal T}^{\uparrow}}} \Delta(\tau)^2 \nu_{\tau}(n)\right) & = \EE\left(\sum_{j=1}^{{\mathrm i}(n)} \Delta(T_j(n))^2\right)
= \frac{1}{3} \EE\left(\sum_{j=1}^{{\mathrm i}(n)} |T_j(n)|\right) = n/3,
\end{align*}
and further, since there are $(k-1)!$ increasing trees with size $k$ and $\mathbb T_k$ has the uniform distribution on the set of such trees,
\begin{align*}\frac{p}{1-p} \sum_{\tau\in{{\mathcal T}^{\uparrow}}} \frac{\Delta(\tau)^2}{(1+1/(1-p))^{(|\tau|)}}&= \frac{p}{1-p} \sum_{k=1}^{\infty} \EE(\Delta(\mathbb T_k)^2){\mathrm B}(k,1+1/(1-p))\\
&= \frac{p}{1-p} \sum_{k=1}^{\infty} \frac{k}{3}{\mathrm B}(k,1+1/(1-p))=\frac{1}{3}.
\end{align*}
We conclude again from Scheff\'e's Lemma that
$$\lim_{n\to \infty} \sum_{\tau\in{{\mathcal T}^{\uparrow}}}\Delta(\tau)^2\left| \frac{\nu_{\tau}(n)}{pn} -
\frac{ p}{(1-p) (1+1/(1-p))^{(|\tau|)}} \right| =0,$$
and the proof is complete.
\end{proof}
We now establish Theorem \ref{T3}.
\begin{proof}[Proof of Theorem \ref{T3}]
We denote the characteristic function of $\mu$ by
$$\Phi(\theta)=\int_{\RR} \e^{\mathrm i \theta x} \mu(\mathrm d x) \qquad \text{for }\theta\in \RR.$$
Fix $r>0$ small enough so that $|1-\Phi(\theta)|<1$ whenever $|\theta|\leq r$, and then define the characteristic exponent $\varphi:[-r,r]\to \CC$
as the continuous determination of the logarithm of $\Phi$ on $[-r,r]$. In words, $\varphi$ is the
unique continuous function on $[-r,r]$ with $\varphi(0)=0$ and such that $\Phi(\theta)= \exp(-\varphi(\theta))$ for all $\theta\in[-r,r]$. For definitiveness, we further set
$\varphi(\theta)=0$ whenever $|\theta|>r$.
Next, observe from the Markov's inequality that for any $1<\beta<2\wedge(1-p)^{-1}$ and any $a>0$
$$\PP\left( \exists j\leq {\mathrm i}(n): |\Delta(T_j(n))|\geq a\sqrt n\right) \leq a^{-2\beta} n^{-\beta} \EE\left(\sum_{j=1}^{{\mathrm i}(n)} |\Delta(T_j(n))|^{2\beta}\right)
,$$
so that, thanks to Lemma \ref{L6},
$$\lim_{n\to \infty} \frac{1}{\sqrt n} \max_{1\leq j \leq {\mathrm i}(n)} |\Delta(T_j(n))|=0\qquad \text{ in probability.}$$
In particular, since the sequence $(a_n)$ is regularly varying with exponent $1/\alpha >1/2$, for every $\theta\in \RR$,
the events
$$\Lambda(n,\theta)\coloneqq \{ |\theta \Delta(T_j(n))/a_n|< r\text{ for all } j=1, \ldots {\mathrm i}(n) \}, \qquad n\geq 1$$
occur with high probability as $n\to \infty$, in the sense that $\lim_{n\to \infty} \PP(\Lambda(n,\theta))=1$.
We then deduce from \eqref{E:repres} and the fact that the variables $X_j$ are i.i.d. with law $\mu$ that for every $\theta\in \RR$,
$$\EE(\exp(\mathrm i \theta \check S(n)/a_n)\Ind_{\Lambda(n,\theta)}) =
\EE\left(\exp \left( - \frac{1}{n} \sum_{j=1}^{{\mathrm i}(n)} n\varphi \left( \theta a_n^{-1} \Delta(T_j(n))\right)\right)\Ind_{\Lambda(n,\theta)} \right).$$
We then write, in the notation of Lemma \ref{L2+},
$$\frac{1}{n} \sum_{j=1}^{{\mathrm i}(n)} n\varphi \left( \theta a_n^{-1} \Delta(T_j(n))\right)= \sum_{\tau\in{\mathcal T}^{\uparrow}}n\varphi \left( \theta a_n^{-1} \Delta(\tau)\right) \frac{\nu_{\tau}(n)}{n}.$$
Recall that we assume \eqref{E:stable}. According to Theorem 2.6.5 in Ibragimov and Linnik \cite{IbLin}, $\varphi$ is regularly varying at $0$ with exponent $\alpha$, and since $\alpha<2$, the Potter bounds (see Theorem 1.5.6 in \cite{BGT}) show that for some constant $C$:
\begin{equation}\label{E:Potter}n\varphi \left( \theta a_n^{-1} \Delta(\tau)\right)\leq C |\theta \Delta(\tau)|^2.
\end{equation}
We deduce from Lemma \ref{L2+} that for every fixed $\theta\in \RR$, there is the convergence in probability
$$\lim_{n\to \infty} \sum_{\tau\in{\mathcal T}^{\uparrow}}n|\varphi \left( \theta a_n^{-1} \Delta(\tau)\right) | \left| \frac{\nu_{\tau}(n)}{n} -
\frac{ p}{(1-p) (1+1/(1-p))^{(|\tau|)}} \right|=0 .$$
Furthermore, still from Theorem 2.6.5 in Ibragimov and Linnik \cite{IbLin}, we have
$$
\lim_{n\to \infty}n\varphi(\theta /a_n) = \varphi_{\alpha}(\theta), \qquad \text{for every }\theta\in\RR,
$$
and we deduce by dominated convergence, using Lemma \ref{L2+} and \eqref{E:Potter}, that
$$\lim_{n\to \infty} \sum_{\tau\in{\mathcal T}^{\uparrow}}n|\varphi \left( \theta a_n^{-1} \Delta(\tau)\right) -\varphi_{\alpha}(\theta \Delta(\tau))|
\frac{ p}{(1-p) (1+1/(1-p))^{(|\tau|)}} =0 .$$
Putting the pieces together, we have shown that
$$\lim_{n\to \infty} \EE(\exp(\mathrm i \theta \check S(n)/a_n)) =
\sum_{\tau\in{{\mathcal T}^{\uparrow}}} \varphi_{\alpha}(\theta \Delta(\tau))
\frac{ p}{(1-p) (1+1/(1-p))^{(|\tau|)}}.
$$
It only remains to check that the right-hand side above agrees with the formula of the statement. This follows from Lemma \ref{L1} and the fact that for every $k\geq 1$, $\mathbb T_k$ has the uniform distribution on $\{\tau\in{{\mathcal T}^{\uparrow}}: |\tau|=k\}$.
\end{proof}
\eject
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 9,259
|
Fortune favours the brave
Business, Interviews, Lifestyle | Mon 29 Feb | Author – Colin White
Virginia Fortune talks to Colin White about her progression from front of house manager at the Point Depot to manager of private members clubs at the 3Arena.
The emergence of the O2 – now known as the 3Arena – out of the ashes of the old Point Depot was a bone of contention amongst Irish gig-goers after the venue reopened in 2008.
The façade of much of the old railway station remained, but poor site lines and the infamous sticky floor had been replaced by something far more contemporary.
Personally, I welcomed the new space. It felt monotone, minimal and spacious. The audio experience had also seen an improvement and, to my ears, seemed more vibrant and superior to the previous set-up, which was prone to inconsistencies, and at times, a muddy sound.
The 3Arena is one of the busiest venues in the world and is now well established as the country's undisputed leading entertainment venue. Virginia Fortune's involvement with the venue dates back to 1989 when she was appointed front of house manager.
In 1991 she left to become involved with the start-up of the Michelin-starred Commons Restaurant on St Stephens Green, where she built up a rapport with a large cross section of business clients, and most of Dublin's high fliers of industry at the time.
Then, in 2003, prior to the launch of The O2 Point Club, she was recruited again, this time to assist with membership recruitment. With a vast client list and her many contacts, memberships grew substantially. From what was initially due to be a two-week temporary sales drive, Fortune is still at the helm as manager of private members club experiences at the 3Arena.
TIME FOR A RETHINK
Fortune fondly remembers her foray into exclusive private memberships in 2003. "When the O2 opened, membership was capped at around 100 members at the time and we were successful in quickly doubling our numbers," she says.
"Then the halcyon boom times arrived and we continued to run at complete capacity. At that moment in time, the limited amount of physical space we had to work with at the venue wasn't an issue for our members; everybody who was anybody wanted to be seen at our club."
When the recession hit, Fortune, and the entertainment industry as a whole, were not immune to the damaging effects of a collapsing world economy.
Virginia Fortune
Fortune states: "Memberships dropped by 30% in a short space of time. We had to try and counterbalance what was taking place, so we decided to take on Mairead McElvaney to look after the marketing side of things.
"With Mairead's assistance, we introduced price reductions and added a number of member benefits. Basically, we were trying anything we could to stop the constant fall in revenue," she reveals.
"Direct mails would have been the norm up until about four years ago. Now, we focus on campaigns across Ticketmaster in which we target specific shows and demographics. We're not only building a database. We're harnessing a relationship with our customers through regular communication."
A NEW SPACE
In 2007, the Point Theatre needed a facelift and a decision was made to renovate. Rumours at the time about new emerging Dublin venues also played a part in forcing the upgrade of facilities. Although fond of the old venue, Fortune knew the time was right for a rebrand.
"The old venue certainly had a magic to it, but site lines were poor and there were frequent issues with excessive queuing. In order to keep up with the times and to accommodate the bigger shows, the redevelopment was necessary to increase capacity to cater to larger events," she says.
Many marquee shows were announced for the reopening of the venue in 2008 and Fortune and her team enjoyed a period of success in obtaining impressive membership numbers. Memberships were flying in on the back of the announcement of some high-profile acts, such as Kings of Leon, Tina Turner, Snow Patrol and The Rolling Stones.
There's nothing more special than walking from one of our clubs into the arena when a big show is starting. Equally, it's a phenomenal feeling when everyone returns to the bar after a successful gig to enjoy the rest of their night
"We had to market the club to individuals and families, rather than to the big corporates and we invested a lot of time into getting our price range just right. Our members wanted to see in increase in value, rather than solely a reduction in price," claims Fortune.
"We got it right. Membership numbers were static for a long time, but our alternative system of marketing the clubs proved fruitful. Rather than sending out collateral and ringing people, we became focused on our digital communication."
INCREASED ENTITLEMENTS
So, what does the 1878 offer members today? Membership guarantees a first-class entertainment experience and is suitable for groups up to eight. Membership to the club also grants access to the best seats in the house for all shows at the 3Arena (even when sold out), as well as access to free car parking.
The transferability of membership has been a big selling point for Fortune and her team. This allows the card to be used throughout an organisation or shared privately amongst family and friends.
One of the many added benefits for the 1878's members is a dual membership with another prominent Dublin establishment. Live Nation, the entertainment company that owns the 3Arena, also runs the Bord Gáis Energy Theatre. Customers are also entitled to membership of The Circle Club at the theatre, which guarantees up to four seats for every show at the Bord Gáis Energy Theatre.
The Premium Club offers priority purchase on two premium seats for every show, pre and post-show access to the club on every show night via a private entrance, full transferability of membership and the option to dine on show nights from carefully considered menus.
Fortune is a very much a proactive, hands-on boss. There are ten office staff currently employed across sales and marketing at the 1878, as well as up to 50 hospitality staff to manage each event.
Fortune believes she is working alongside the best member-orientated team in Ireland and when asked about her day-to-day role as general manager, she is keen to highlight the hard work of her support team and the loyalty customers have shown.
"Nothing is too much when it comes to member requests, who are loyal year-upon-year. A large percentage have been with us since year one and see the value of their memberships," says the Greystones native.
"Across the board, our team is finely tuned, particularly with regard to customer service. That's hugely important to us. If you don't treat people well, they simply won't return. We know each member is making a big investment in us," she adds.
REAPING THE BENEFITS
Fortune and her team are really seeing the fruits of their labour as more and more members are returning and renewing their subscriptions. Increased confidence in the economy has obviously also helped in that regard.
Member retention has become an increasingly critical element for Fortune and focusing on current members, instead of purely chasing new business, is paramount to the success of the 1878 and The Premium Club.
"Nowadays, everyone expects bang for their buck, and rightly so," she declares. "It's up to us to ensure each member or guest receives a great service. We've also increased capacity. Everyone now has their own dedicated table and the staff-to-member ratio has improved dramatically."
Virginia Fortune is the driving personality behind the clubs, bringing a passion for people and music to her role as general manager at the 1878 and The Premium Club. She also holds firm in her belief that the pre-show experience on offer is unlike any other in Ireland.
"There's nothing more special than walking from one of our clubs into the arena when a big show is starting. Equally, it's a phenomenal feeling when everyone returns to the bar after a successful gig to enjoy the rest of their night," she enthuses.
"In the past, we had to chase opportunities, but now customers are coming to us to experience the amazing shows and our top-notch hospitality."
"Never give up, never be a bully" – 60 Seconds with Virginia Fortune, Private Members Clubs at 3Arena
Ones to Watch: Tixserve–a track-and-trace ticket delivery system that simultaneously offers a major marketing tool for ticket sellers
TRAVEL: The heart of the Danube
# 2016 FDI 100
Topics: 3Arena, Bord Gáis Energy Theatre, brave, Colin White, Commons Restaurant, Greystones, Kings of Leon, Live Nation, loyalty, Mairead McElvaney, Memberships, O2, O2 Point Club, Point Depot, Point Theatre, private members' clubs, Snow Patrol, the 1878, The Circle Club, The Premium Club, The Rolling Stones, Ticketmaster, Tina Turner, Virginia Fortune
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 7,844
|
A newly refurbished serviced 35 office building, located on Widnes Business Park, an established commercial area in Widnes. Our office space has not only been designed to offer great office suites, but to provide stunning and vibrant communal areas and facilities to enhance the whole working environment experience. Modern meeting rooms offer all the latest technology including interactive touch screen whiteboards, Skype and conferencing facilities along with complimentary tea and coffee for the more formal meetings and for the less formal meetings you could use a more social setting and take advantage of our fantastic break out area.
Superbly located for easy access from Liverpool Airport, Widnes centre, the new Expressway and nationwide motorway network.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 3,914
|
Q: dask to_dataframe repeated index I am loading a list of text files into a dask dataframe. Each text file has multiple rows of dicts (separated by newline). For each row of the text file, I do some small processing defined in "remove_escapes" and returns a list. I call flatten to ensure that I have a list (instead of list of lists).
input_file_list = self.get_file_list()
posts_db = db.from_sequence(input_file_list)
posts_db = posts_db.map(self.remove_escapes).flatten()
posts_df = posts_db.to_dataframe()
posts_df = posts_df.compute()
def remove_escapes(self, chunk_file):
json_list = []
with open(chunk_file, 'r') as fp:
for line in fp:
line = line.strip()
if line:
line = line.replace("\\\\", "\\")
json_data = json.loads(line)
json_list.append(json_data)
return json_list
I get this:
Body Comments Id Title
0 <p>It depends on the context:</p>

<ol... side note: Hash#fetch is not exactly Hash#[]. ... 13935 None
1 <p>It depends on the context:</p>

<ol... @tokland `:c` not found 13935 None
2 <p>It depends on the context:</p>

<ol... "There is also a convention that it is used as... 13935 None
3 <p>I'd like to have a python program alert me ... `import os; os.system('say "Beer time."'); pri... 13941 Python Sound ("Bell")
4 <p>I'd like to have a python program alert me ... the question is answered but... you do need qu... 13941 Python Sound ("Bell")
5 <p>I'd like to have a python program alert me ... Does not seem to be working for me on Mojave 13941 Python Sound ("Bell")
6 <p>Have you tried :</p>

<pre><code>im... I'm on ubuntu, it doesn't work for me. Any idea? 13949 None
7 <p>Have you tried :</p>

<pre><code>im... @kecske it's common [to disable the audible-be... 13949 None
8 <p>Have you tried :</p>

<pre><code>im... Works on Windows XP as well (in a console app). 13949 None
9 <p>I had to turn off the "Silence terminal bel... Seems to work with python 2 only.... 13959 None
0 <p>I want to use a track-bar to change a form'... Also, Decimal can't represent as wide a value ... 4 Convert Decimal to Double?
1 <p>Given a <code>DateTime</code> representing ... what all of the answers so far have missed is ... 9 How do I calculate someone's age in C#?
2 <p>Given a <code>DateTime</code> representing ... No one has considered leap years? or checking ... 9 How do I calculate someone's age in C#?
3 <p>Given a <code>DateTime</code> representing ... Note that for someone less than one year old, ... 9 How do I calculate someone's age in C#?
4 <p>Given a <code>DateTime</code> representing ... why nobody is using TimeSpan? 9 How do I calculate someone's age in C#?
5 <p>Given a specific <code>DateTime</code> valu... What if you want to calculate a relative time ... 11 Calculate relative time in C#
6 <p>Given a specific <code>DateTime</code> valu... moment.js is a very nice date parsing library.... 11 Calculate relative time in C#
7 <p>Given a specific <code>DateTime</code> valu... There is the .net package https://github.com/N... 11 Calculate relative time in C#
8 <p>Here's how I do it</p>

<pre class=... "< 48*60*60s" is a rather unconventional defin... 12 None
9 <p>Here's how I do it</p>

<pre class=... Since all those If..else are just timeslabs, y... 12 None
0 <p>Best solution is to let IIS do it.</p>
... Jeff Atwood List some problems he's run into... 17068 None
1 <p>use <code>System.Xml.Linq.XElement</code> a... I'm working with NET 2.0 17093 None
2 <p>We are developing an application that invol... I fail to see answers for this questions which... 17106 How to generate sample XML documents from thei...
3 <p><a href="http://netbeans.org" rel="nofollow... That era is now over... 17110 None
4 <p><a href="http://www.altova.com/xmlspy.html"... XMLSpy looked good but generated xml that then... 17114 None
5 <p>How do you run an external program and pass... I think you need to rewrite your question - op... 17140 How do you spawn another process in C?
6 <pre><code>#include <stdlib.h>

... Never use system. It is far from multithreadin... 17148 None
7 <p>I know that IList is the interface and List... If anyone is still wondering, I find the best ... 17170 When to use IList and when to use List
8 <p>I don't think there are hard and fast rules... why not make it a just a List in the first pla... 17177 None
9 <p>Here's how I do it</p>

<pre class=... But currently SO only show the "Time ago" form... 12 None
.. ... ... ... ...
0 <p>I'm going to continue my habit of going aga... No, I'm not talking about apps that are that s... 10448 None
1 <p>I'm going to continue my habit of going aga... I don't see how moving business logic into sto... 10448 None
2 <p>If you were on Windows, I'd tell you to use... +1 I've used this named pipe methodology seve... 10450 None
3 <p>The 'click sound' in question is actually a... I had a problem with this line: isEnabled = v... 10456 HowTo Disable WebBrowser 'Click Sound' in your...
4 <p>Ideally, I'm looking for a templated logica... @d03boy: Well it has HashSet<T> now, but after... 10458 Is there a "Set" data structure in .Net?
5 <p>Ideally, I'm looking for a templated logica... See [this question](https://stackoverflow.com/... 10458 Is there a "Set" data structure in .Net?
6 <p>Ideally, I'm looking for a templated logica... Possible duplicate of [C# Set collection?](htt... 10458 Is there a "Set" data structure in .Net?
7 <p><a href="http://msdn.microsoft.com/en-us/li... Matt, +1. That sounds like exactly what he ask... 10459 None
8 <p>I've noticed that if you use WebBrowser.Doc... your suggested solution prevents the control f... 10463 None
As you see above, the indices are repeated. Is there a way to ensure the indices are nicely ordered and increasing?
A: This is inherit to the partitioning in dask.
For additional differences between pandas and dask dataframes see dask dataframe examples
A: From here https://github.com/dask/dask/issues/3788, I learnt that this behavior is as intended.
|
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"redpajama_set_name": "RedPajamaStackExchange"
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| 2,673
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{"url":"https:\/\/developer.mozilla.org\/en-US\/docs\/Mozilla\/Projects\/SpiderMonkey\/JSAPI_reference\/JS_GetPositiveInfinityValue","text":"# JS_GetPositiveInfinityValue\n\nRetrieve a floating-point infinity as a value of type JS::Value.\n\n## Syntax\n\nJS::Value\nJS_GetPositiveInfinityValue(JSContext *cx);\n\nJS::Value\nJS_GetNegativeInfinityValue(JSContext *cx);\n\n\/\/ Obsolete since SpiderMonkey 42\njsval\nJS_GetPositiveInfinityValue(JSContext *cx);\n\njsval\nJS_GetNegativeInfinityValue(JSContext *cx);\nName Type Description\ncx JSContext * A context.\n\n## Description\n\nJS_GetPositiveInfinityValue returns a JS::Value that represents an IEEE floating-point positive infinity. JS_GetNegativeInfinityValue returns the corresponding negative infinity.\n\nInfinities are typically used to represent numbers that are greater in magnitude than the greatest representable finite values. As a value in mathematical calculations, an infinite value behaves like infinity. For example, any nonzero value multiplied by infinity is infinity with the expected sign, and any finite value divided by infinity is zero (again with the expected sign).\n\nTo get a floating-point NaN, use JS_GetNaNValue.","date":"2017-02-23 22:38:06","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9488283395767212, \"perplexity\": 3785.502701846208}, \"config\": {\"markdown_headings\": true, \"markdown_code\": false, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2017-09\/segments\/1487501171232.43\/warc\/CC-MAIN-20170219104611-00479-ip-10-171-10-108.ec2.internal.warc.gz\"}"}
| null | null |
<?xml version="1.0" encoding="UTF-8"?>
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Licensed to the Apache Software Foundation (ASF) under one or more
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date-of-creation="2005-6-30"
timeout="1"
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The test is against initialValue1010 method.
For testing purposes ThreadLocals to test are organized
into two arrays with two object each.
ThreadLocals in each array are created by one thread.
The test performs the following checks on values returned by the method.
1) Being overridden method returns the expected value.
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"redpajama_set_name": "RedPajamaGithub"
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| 2,358
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\section{Introduction} \label{S:intro}
The correlation of redshifted emission line spectra has been proposed as a potentially powerful probe of large-scale structure (LSS) at high redshifts \citep{1999ApJ...512..547S,2008A&A...489..489R,2010JCAP...11..016V}. While traditional LSS probes such as galaxies and quasars for the foreseeable future will be limited to $z<2$ and $z<4$, respectively, mapping emission lines can probe out to $z>6$, potentially mapping the Epoch of Reionization (EoR). In these very high-redshift regions, emission lines can also be cross-correlated with the 21-cm line from the intergalactic medium (IGM), constituting a very complementary probe to 21-cm alone or extragalactic emission lines alone, whose auto-spectra are plagued with foregrounds \citep{2011ApJ...728L..46G,2011ApJ...741...70L}. Three extragalactic emission lines of interest in these types of studies include CO \citep{2008A&A...489..489R,2011ApJ...730L..30C,2011ApJ...728L..46G,2011ApJ...741...70L}, CII \citep{2012ApJ...745...49G}, and HI \citep{2011ApJ...728L..46G}. In this paper, we will focus on the CO lines.
CO is readily produced from carbon and oxygen in star-forming regions. The CO molecule exhibits several rotational transitional states $J\to J-1$ with line frequencies $\nu_J=J\nu_{\rm CO}$ and $\nu_{\rm CO}=115$ GHz. CO lines have been studied both as foreground contaminants to Cosmic Microwave Background (CMB) observations \citep{2008A&A...489..489R} as well as LSS tomography probes \citep{2011ApJ...728L..46G,2011ApJ...741...70L}. In this paper we will follow the formalism presented in \citet{2011ApJ...741...70L} (hereafter L11) and extend it to lower redshifts, producing two models that make different assumptions. The two CO rotational lines we seek to constrain are CO(1-0) and CO(2-1) because they are typically the brightest and both their emitting and redshifted frequencies conveniently fall in the frequency range of CMB observations, where vast amounts of data and expertise have been cultivated over the years. The multiple emission lines of CO due to multiple rotational states allow interlopers to be identified more easily, giving CO an advantage over other emission lines as cosmological probes. These observational prospects have created considerable interest in building future dedicated experiments, even if the exact strength of the signal at high redshift and on large scales is still rather uncertain. As such, we investigate what can be probed with current experiments. For the LSS study of interest, we work in the regime of ``intensity mapping'', measuring aggregated CO emissions associated with the large-scale structure rather than with individual galaxies, as an efficient way of extracting the faint signals (e.g., \citet{2010Natur.466..463C}). This approach is complementary to high-resolution CO observations enabled by the Atacama Large Millimeter/submillimeter Array (ALMA\footnote{\texttt{https://almascience.nrao.edu/}}) and the Jansky Very Large Array (JVLA\footnote{\texttt{https://safe.nrao.edu/evla/nova/index.shtml}}), for example, where individual galaxies have been mapped to high redshifts; their small (sub-arcminute) field-of-view is not well suited for large scale surveys.
One probe that could possibly constrain CO line emission is in fact the Wilkinson Microwave Anisotropy probe (WMAP) \citep{2011ApJS..192...14J}. With band frequencies ranging from 23-94 GHz, WMAP should contain emission up to $z\sim 4$ from CO(1-0) and up to $z\sim 7$ from CO(2-1). Constructing an auto-correlation spectrum in each band is very difficult due to contamination from other astrophysical processes that motivated WMAP in the first place such as the Cosmic Microwave Background (CMB) or galactic emission. However, a cross-correlation with a different LSS tracer would not have the same foregrounds and thus be immune to these contaminants. A suitable choice would be to cross-correlate the WMAP bands with photometric quasars in its corresponding CO redshift range. Specifically, we choose the photometric quasar sample constructed from the Sloan Digital Sky Survey (SDSS) \citep{2000AJ....120.1579Y} Data Release 6 (DR6) \citep{2008ApJS..175..297A} in \citet{2009ApJS..180...67R}. This sample contains quasars with redshifts as high as $z=6.1$, which covers most of WMAP's CO redshift range, allowing us to perform this cross-correlation for both CO(1-0) and CO(2-1) lines. Note that this redshift range could allow us to search for higher CO lines. CO(3-2) would be seen in V and W bands, while CO(4-3) and CO(5-4) would be seen only in the W band. We leave searches for these lines to future work as they are expected to be dimmer. At redshifts $z<1$, we can cross-correlate the WMAP W band with the SDSS Data Release 7 (DR7) \citep{2009ApJS..182..543A} spectroscopic Luminous Red Galaxies (LRGs) sample in \citet{2010ApJ...710.1444K}.
In this paper, we propose cross-correlating brightness temperature maps with quasars (QSOs) and Luminous Red Galaxies (LRGs) to detect fluctuations in CO at high redshifts. We first see what can be done with current data by cross-correlating WMAP temperature maps in each band with photo-quasars from SDSS DR6 and LRGs from SDSS DR7 in the band's CO redshift range. We attempt to remove CMB fluctuations from the maps to isolate the CO signal (see Sec.~\ref{S:method} for details). We report no detection of a cross-correlation between quasars or LRGs and CO temperature due to extragalactic CO line emission, although we place an upper limit that rules out any brightness level much greater than our model. It appears that the weakness of these constraints is mainly due to instrumental noise in the WMAP temperature maps, although the density of quasars plays a lesser role. We do not suspect quasar photo-z errors to play a role since the errors are $\Delta z\sim O(0.1)$ while the redshift bins we use are of order unity. To detect these lines in the future, we look towards a hypothetical future CO experiment to measure CO fluctuations for the line CO(1-0) at $z=3$ over the range $2.9<z<3.1$ with 20 redshift bins. We forecast that this experiment cross-correlated with the full spectroscopic quasar survey from the Baryon Oscillation Spectroscopic Survey (BOSS) \citep{2011AJ....142...72E,2012ApJS..199....3R} will measure the cross-correlation amplitudes
with signal-to-noise ratios (SNR) as high as 4 for each redshift bin and 20 for the entire redshift range, depending on the model. The CO experiment cross-correlated with the {\it Hobby-Eberly Telescope Dark Energy Experiment} (HETDEX) \citep{2008ASPC..399..115H} could do significantly better with an SNR of 13 for each redshift bin and 58 for the entire redshift range. We also set constraints for current and future ground-based CMB polarimeters like SPTPol or ACTPol.
The plan of our paper is as follows: in Sec.~\ref{S:data}, we describe the data products we use, including the WMAP 7-year observations (WMAP7) \citep{2011ApJS..192...14J} temperature maps, the photo-quasar maps from SDSS DR6, and the LRG maps from SDSS DR7. In Sec.~\ref{S:eqns}, we derive the form of the CO-LSS cross-power spectra along with statistical errors for both models. In Sec.~\ref{S:method}, we describe the estimator for the cross-correlation, presenting measurements in Sec.~\ref{S:results}. Finally, we present forecasts for a possible future CO experiment as well as ground-based experiments in Sec.~\ref{S:forecasts} and conclude in Sec.~\ref{S:conclude}. Wherever not explicitly mentioned, we assume a flat $\Lambda$CDM cosmology with parameters compatible with WMAP7.
\section{Data} \label{S:data}
\subsection{WMAP Temperature} \label{S:wmaptemp}
We attempt to extract redshifted CO emission associated with LSS embedded in the WMAP7 \citep{2011ApJS..192...14J}\footnote{http://lambda.gsfc.nasa.gov/product/map/current} temperature maps by means of cross-correlation. Primordial CMB and Galactic foregrounds which contribute to the WMAP temperature maps will not correlate with LSS; however, CMB secondary anisotropy signals, in particular the thermal Sunyaev-Zeldovich effect (tSZ) \citep{1980ARA&A..18..537S}, are expected to correlate with LSS. TSZ has been detected using similar data sets in \citet{2010ApJ...720..299C} and we will discuss it more in Sec.~\ref{S:results:SZ}. Radio and dust emission also should correlate with LSS, and we discuss its contribution in Sec.~\ref{S:results:SZ} as well. WMAP has 5 temperature bands: K (23 GHz), Ka (33 GHz), Q (41 GHz), V (61 GHz), and W (94 GHz) with bandwidths of 5.5, 7.0, 8.3, 14.0, and 20.5 GHz, respectively. We do not use the K map since it is dominated by Galactic emission; thus we are left with 4 bands. We use the HEALPix \citep{2005ApJ...622..759G} $N_{res}=9$ maps with the KQ85 mask to remove the brightest resolved point sources and the bright Galactic emission regions near the galactic plane. These cuts leave each map containing 2,462,208 $N_{res}=9$ pixels, covering 32,289 deg$^2$ (78.3\% of the sky).
One subtlety that must be taken into account is that the temperature values in the WMAP data products are perturbations of {\it physical} temperature, assuming a mean temperature equal to the CMB temperature. The CO temperature, however, is a brightness temperature. These two temperature measurements diverge at the higher frequency bands, mainly the V and W bands. Therefore, before performing our analysis we must convert the CMB physical temperature perturbations to brightness temperature perturbations according to the formula in \citet{1997ApJ...480L...1G}
\begin{eqnarray}\label{E:brcon}
\delta T_b = \frac{x^2e^x}{(e^x-1)^2}\delta T
\end{eqnarray}
where $\delta T_b$ is the brightness temperature, $\delta T$ is the temperature map from the WMAP data product, and $x=h\nu/kT_{CMB}$ with $T_{CMB}=2.725K$ and $\nu$ being the band frequency. The prefactor for $\delta T$ has a different value for each WMAP band, ranging from 0.97 for the Ka band to 0.80 for the W band. Note that this expression neglects the band width.
\subsection{SDSS Data}
We use photometric quasars from the SDSS DR6 \citep{2008ApJS..175..297A} and LRGs from the SDSS DR7 \citep{2009ApJS..182..543A} to trace the matter density and construct its angular power spectrum. The SDSS consists of a 2.5-m telescope \citep{2006AJ....131.2332G} with a 5-filter (\textit{ugriz}) imaging camera \citep{1998AJ....116.3040G} and a spectrograph. Automated pipelines are responsible for the astrometric solution \citep{2003AJ....125.1559P} and photometric calibration \citep{1996AJ....111.1748F, 2001AJ....122.2129H, 2006AN....327..821T, 2008ApJ...674.1217P}. Bright galaxies, LRGs, and quasars are selected for follow-up spectroscopy \citep{2002AJ....124.1810S, 2001AJ....122.2267E, 2002AJ....123.2945R, 2003AJ....125.2276B}. The data used here from DR6 and DR7 were acquired between August 1998 and July 2008.
\subsubsection{SDSS quasars}
We use the photometric quasar catalog composed by \citet{2009ApJS..180...67R} (hereafter RQCat). The entire catalog consists of 1,172,157 objects from 8417 deg$^2$ on the sky selected as quasars from the SDSS DR6 photometric imaging data. We limit our dataset in this analysis to UV-excess quasars (catalog column \textbf{uvxts} = 1) because they have a higher selection efficiency. We also require the catalog column \textbf{good} $> 0$ to reject objects that are likely stars. For the survey geometry we use the union of the survey runs retrieved from the SDSS DR6 CAS server. We omitted runs 2189 and 2190 because many objects in these runs were cut from RQCat. This mask was pixelized using the MANGLE software \citep{2004MNRAS.349..115H,2008MNRAS.387.1391S}. We pixelize the quasars as a number overdensity, $\delta_q=(n_q-\overline{n})/\overline{n}$, onto a HEALPix pixelization of the sphere with $N_{res}=9$, where $n$ is the pixel's number of quasars divided by the pixel's survey coverage $w$ and $\overline{n}=(\sum_in_iw_i)/(\sum_iw_i)$. We then reject pixels with extinction $E(B-V)\geq 0.05$, full widths at half-maximum of its point-spread function (PSF) FWHM $\geq 2$ arcsec, and stellar densities (smoothed with a $2^\circ$ FWHM Gaussian) $n_{stars}\geq 562$ stars/deg$^2$ (twice the average stellar density), similar to \citet{2008PhRvD..78d3519H}. The extinction cut is very important because a high extinction affects the $u$ band, which is crucial to identifying quasars. Also, since stars tend to be misidentified as quasars, it seems prudent to cut regions with high stellar density. We implement these cuts using dust maps from \citet{1998ApJ...500..525S} and stars ($18.0<r<18.5$) from the SDSS DR6 \citep{2008ApJS..175..297A}. We also reject pixels for which the survey region covers less than 80\% of the pixel area. In addition, RQCat contains regions that seem undersampled. We excise angular rectangles around these regions to remove them from the data. The angular rectangles in celestial coordinates that were removed are $(\alpha,\delta)=(122^\circ\mbox{--}139^\circ,-1.5^\circ\mbox{--}(-0.5)^\circ)$, $(121^\circ\mbox{--}126^\circ,0^\circ\mbox{--}4^\circ)$, $(119^\circ\mbox{--}128^\circ,4^\circ\mbox{--}6^\circ)$, $(111^\circ\mbox{--}119^\circ,6^\circ\mbox{--}25^\circ)$, $(111.5^\circ\mbox{--}117.5^\circ,25^\circ\mbox{--}30^\circ)$, $(110^\circ\mbox{--}116^\circ,32^\circ\mbox{--}35^\circ)$, $(246^\circ\mbox{--}251^\circ,8.5^\circ\mbox{--}13.5^\circ)$, $(255^\circ\mbox{--}270^\circ,20^\circ\mbox{--}40^\circ)$,\\ $(268^\circ\mbox{--}271^\circ,46^\circ\mbox{--}49^\circ)$, and $(232^\circ\mbox{--}240^\circ,26^\circ\mbox{--}30^\circ)$. Finally, we cut pixels that appeared to have severe photometric calibration errors ($\gtrsim5$ mag). After these cuts, the survey region comprises 534,564 $N_{res}=9$ pixels covering $7010$ deg$^2$.
\subsubsection{SDSS LRGs}
We use the LRG catalog composed by \citet{2010ApJ...710.1444K}. LRGs are the most luminous galaxies in the universe, making them important for probing large volumes. They are also old stellar systems with uniform spectral energy distributions and a strong discontinuity at 4000 \AA, which enable precise photometric redshifts. The LRG catalog consists of 105,623 spectroscopic LRGs from redshifts $0.16<z<0.47$. We do not make any alterations to this catalog. Our survey region comprises approximately 638,583 $N_{res}=9$ pixels covering $8374$ deg$^2$.
\subsection{WMAP-QSO Cross-Data Set}
We intersect the pixel sets from the temperature and quasar maps to produce two sets of maps that can be cross-correlated. This operation leaves each map with 441,228 $N_{res}=9$ pixels covering 5786 deg$^2$ with $\sim$ 7 arcmin.~pixel resolution. Since each WMAP band probes a separate redshift range for each of the CO emission lines, we also divide the quasar map into 8 maps, two for each of the WMAP bands. Note that some of the bands for CO(1-0) will intersect in redshift with other bands for CO(2-1). We list the properties of the 4 WMAP maps and 8 quasar maps in Table \ref{T:zbin} as well as the probed spatial scale determined by our pixel size. Note that the Ka(2-1) redshift range exceeds the redshifts of the quasars; therefore, we will not determine any limits on the CO(2-1) line with the Ka band.
\subsection{WMAP-LRG Cross-Data Set}
We also intersect the areas of the temperature and LRG maps for cross-correlation, with 619,708 $N_{res}=9$ pixels covering 8126 deg$^2$. Note that the WMAP W band is the only band that overlaps with the LRG sample, and this is true only for the CO(1-0) line. Thus, we will only get one constraint from this cross-correlation. However, the number of LRGs in this redshift range is much more than the number of quasars, so we expect a more significant constraint than that from the quasars.
\begin{table*}[!t]
\begin{center}
\caption{\label{T:zbin} WMAP redshift bins for CO emission lines $J=1\to0$ and $J=2\to1$ with the quasar (QSO) counts from the QSO map (T x QSO intersected map). For reference, we write in parentheses the transverse scale corresponding to the pixel scale in Mpc/$h$ for each $z$ band probed. As stated above, the total number of QSOs in DR6 is 1,172,157.}
\vspace{0.5cm}
\begin{tabular}{ccccc}
\hline
Band&$z(1\to0)$\quad ([Mpc/$h$])&$N_{\rm QSO}(1\to0)$&$z(2\to1)$\quad ([Mpc/$h$]) &$N_{\rm QSO}(2\to1)$ \\
\hline
Ka&2.151--2.898 (9) &90,780 (74,395)&5.302--6.796 (13) &80 (63)\\
Q&1.547--2.121 (7) &146,111 (121,297)&4.094--5.242 (11) &573 (466)\\
V&0.691--1.130 (5) &62,172 (51,818)&2.382--3.260 (9) &27,032 (22,180)\\
W&0.103--0.373 (1) &42,184 (34,400)&1.206--1.746 (6) &140,819 (116,791)\\
\hline
\end{tabular}\end{center}
\end{table*}
\section{Cross-Correlation Power Spectrum} \label{S:eqns}
We wish to correlate fluctuations in CO line emission with quasar and LRG maps. Our CO temperature modeling will follow L11. L11 derived the CO brightness temperature by calculating the specific intensity of CO emission as a line-of-sight integral of the volume emissivity. In L11, the star formation rate density (SFRD) used, given by their Eq.~6, is comparable with the value needed to reionize the universe at high $z$. However, since we are interested in lower redshifts than in L11, the model must be modified, as we explain below.
\subsection{Model A: CO Luminosity -- Halo Mass} \label{S:modela}
The basic strategy is to construct a model that matches three
key observational inputs~\citep{2011ApJ...730L..30C}: the observed correlation between CO
luminosity and far-infrared luminosity, the far-infrared luminosity-star formation rate (SFR) correlation, and the observed SFRD of the Universe. In order to predict the spatial fluctuations in the
CO brightness temperature, we need to
further connect the star formation rate and host halo mass.
In comparison to the high redshift $z \sim 7$ case discussed in
L11, this estimate should be on more solid
ground in several respects. The CO luminosity far-infrared
correlation and the correlation between the far-infrared luminosity and
star formation rate are measured at $z \sim 0-3$; their applicability
at higher redshift is questionable. In particular, the CO luminosity-SFR
correlation may drop at high redshift, since the low mass galaxies of
interest may have low metallicity, as well as an insufficient dust abundance to shield
CO from dissociating radiation. In addition, the increased CMB
temperature at high redshift may significantly reduce the contrast
between CO and the CMB. Furthermore, the overall star formation
rate density is better determined at $z \sim 2$ than
at $z \sim 7$. On the other hand, the simplistic model adopted
in L11 to connect star formation rate to halo mass
is likely less applicable at low redshift, where various feedback effects
such as photoionization heating, supernovae, and AGN feedback have had more time
to operate.
The starting point for the L11 model is the observed empirical
correlation between CO luminosity, $L_{\rm CO(1-0)}$, and far-infrared
luminosity, $L_{\rm FIR}$. Specifically, following L11 we use the
correlation reported in \citet{2010ApJ...714..699W}. Recent work, summarized in
\citet{Carilli:2013qm}, suggests however that rapidly star-forming
`starburst' galaxies may exhibit a somewhat different correlation
between CO and far-infrared luminosity than `main sequence' galaxies
with more gradual, yet longer lived star formation. In this work, for
simplicity, we ignore any trend in this correlation with galaxy
properties, and assume that the Wang et al. (2010) relation applies
globally. Future measurements, further quantifying trends in this
relation with redshift, average galactic metallicity, and other galaxy
properties will help to refine our modeling. As in L11, we assume the
\citet{Kennicutt:1997ng} relation to connect far infrared luminosity and star
formation rate.
Lets turn to some quantitative estimates. The result of combining
the CO luminosity far-infrared correlation with the far-infrared SFR
relation is, in L11,
\begin{eqnarray}
L_{\rm CO(1-0)} = 3.2 \times 10^4 L_\odot \left[\frac{\rm{SFR}}{M_\odot \rm{yr}^{-1}}\right]^{3/5}.
\label{eq:lco}
\end{eqnarray}
In L11 the authors further assumed a very simple model to convert SFR
to host halo mass:
\begin{eqnarray}
\rm{SFR} &=& f_\star \frac{\Omega_b}{\Omega_m} \frac{M}{t_s}
= 0.17 M_\odot \rm{yr}^{-1}\times \nonumber\\ && \left[\frac{f_\star}{0.1}\right] \left[\frac{\Omega_b/\Omega_m}{0.17}\right] \left[\frac{10^8 \rm{yrs}}{t_s}\right]
\left[\frac{M}{10^9 M_\odot}\right].
\label{eq:sfr_mhost}
\end{eqnarray}
They took the fraction of baryons converted into stars to
be $f_\star = 0.1$, and a (constant) star formation time scale of $t_s = 10^8$ yrs.
Star formation was assumed to take place with equal efficiency
in all halos above some some minimum host mass, $M_{\rm sfr, min}$.
It was found that the
implied star formation rate density is comparable to the critical
star formation rate density required to keep the Universe ionized,
suggesting that this simple prescription is a reliable estimate.
Furthermore, \citet{2007ApJ...668..627S} found that a similar model roughly
matches the luminosity function of Lyman Break Galaxies (LGBs) at $z=6$.
Unfortunately, this simple prescription will not provide a good
estimate of the SFR at lower redshifts, as we discuss.
We now explain why the estimate for $\VEV{T_{\rm CO}}$ in L11 cannot be extrapolated to low redshifts.
Taken together, Equations \ref{eq:lco} and \ref{eq:sfr_mhost} imply
that $L_{\rm CO(1-0)} \propto M^{3/5}$. For simplicity, we assumed that
$L_{\rm CO(1-0)}$ instead {\em scaled linearly with halo mass} after matching the implied
CO luminosity at $M=10^8 M_\odot$. Since massive halos are rare at $z \sim 7$,
the CO emissivity in this model is dominated by galaxies in low mass halos close
to the minimum CO luminous halo mass, $M_{\rm co, min}$,
and so adopting a linear scaling rather than a $M^{3/5}$
power law, has relatively little impact on the mean CO brightness temperature.
At $z \sim 7$ the linear scaling leads to only a slight overestimate compared
to the $M^{3/5}$ power law.
Together, these assumptions implied $L_{\rm CO(1-0)}(M) = 2.8 \times 10^3 L_\odot
M/(10^8 M_\odot)$. Note that \citet{2011ApJ...741...70L} focused
on CO(2-1) assuming that the $J=2-1$ and $J=1-0$ lines have the
same luminosity, which
is very conservative. In the optically thick, high temperature limit,
the brightness temperature
would be a factor of $8$ larger. At $z \sim 7$ this more than makes up
for the possible overestimate from the linear scaling.
Now it is easy to see that applying this blindly at $z \sim 2$ may
lead to problems because significantly more massive halos are no longer
rare.
As a result, if we assume a linear scaling when
the true scaling is sub-linear, we will significantly overestimate the
CO emissivity and brightness temperature.
For example, the above $L_{\rm CO(1-0)} - M$ relation gives
$L_{\rm CO}(M = 10^{11} M_\odot) = 2.8 \times 10^6 L_\odot$. However if
we had used the sublinear scaling, we would have obtained
$L_{\rm CO}(M = 10^{11} M_\odot) = 1.8 \times 10^5 L_\odot$, which is lower by a factor of $\sim 15$.
Although the CO luminosity halo mass relation used in L11
seems like a plausible estimate at high redshifts, where low mass galaxies
dominate the SFRD and CO emissivity, blindly extrapolating it to
higher halo masses
is problematic.
We derive a more accurate expression by first adjusting the SFR-M prescription to be
more suitable at low redshifts, where feedback processes will further
suppress star formation in low mass halos.
We follow \citet{2003ApJ...586..693W} in
adopting a halo-mass dependent star formation efficiency, as suggested
by the $z \sim 0$ observations of \citet{2003MNRAS.341...54K}. In particular,
we assume that the fraction of gas turned into stars above $M_{\rm sfr, min}$
scales
as $M^{2/3}$ below some mass $M_0$, and that the star formation efficiency
is independent of mass above the characteristic mass $M_0$. More specifically we assume
\begin{eqnarray}
SFR = && f_\star \frac{\Omega_b}{\Omega_m} \frac{M_o}{t_s}
\left[\frac{M}{M_0}\right]^{5/3}; M \leq M_0 \nonumber \\
SFR = && f_\star \frac{\Omega_b}{\Omega_m} \frac{M}{t_s}; M \geq M_0.
\label{eq:sfrnew}
\end{eqnarray}
In reality $f_\star$, $M_o$ and $t_s$ likely have some redshift dependence.
Here we fix the characteristic mass scale $M_0$ at $M_0 = 5 \times 10^{11} M_\odot$ (close to the mass scale at $z \sim 2$ in the \citet{2003ApJ...586..693W} model),
and normalize the proportionality constant $f_\star \Omega_b/\Omega_m$
to match the observed SFRD at $z \sim 2$, $\dot{\rho_\star} = 0.1 M_\odot$ yr$^{-1}$ Mpc$^{-3}$ \citep{2006ApJ...651..142H}. The mass dependent efficiency reduces the efficiency of star
formation in low mass halos, reflecting the impact of feedback processes.
It is convenient that the SFR scales as $M^{5/3}$ in this model, because
this yields a linear scaling of $L_{\rm CO} \propto M$, {\em although with
significantly lower normalization than the previous relation}. (This is strictly
true only below $M_0$ but these halos dominate the CO emissivity even
at low $z$ and so it is safe to adopt this scaling for all halo
masses here.) This means that the only
thing that changes in this model is the brightness temperature normalization $\VEV{T_{\rm CO}}$,
and not the bias and Poisson terms.
Combining the revised SFR model with Equation \ref{eq:lco}
we find:
\begin{eqnarray}
L_{\rm CO(1-0)} = 1 \times 10^6 L_\odot \left[\frac{M}{5 \times 10^{11} M_\odot}\right].
\label{eq:lco_fix}
\end{eqnarray}
This normalization is a factor of $\sim 14$ lower than in L11
but is likely a better match at $z \sim 2$ where star formation occurs
mostly in substantially more massive halos than at $z \sim 7$.
In order to turn this into an expression for the brightness temperature
we can follow the formulas from L11, but with
small modifications. First, we will assume the optically thick, high-temperature limit. This should be a good approximation for the
$J=1-0$ and $2-1$ lines at the redshifts of interest. We assume that
the duty cycle of CO luminous activity matches the star formation duty
cycle, i.e., we assume $f_{\rm duty} = t_s/t_{\rm age}(z)$, where we
fix $t_s \sim 10^8$ yrs, and $t_{\rm age}(z)$ is the age of the Universe.
The result of this calculation with the normalization of
Eq.~\ref{eq:lco_fix} is
\begin{eqnarray}
\VEV{T_{\rm CO}}(z_J) = 0.65 \mu K \left[\frac{f_{\rm coll}(M_{\rm co, min}; z_J)}{0.3}\right] \left[\frac{3.4 \times 10^9 \rm{yr}}{t_{\rm age}(z)}\right] \times&&\nonumber\\
\left[\frac{H(z_J=2)}{H(z_J)}\right]
\left[\frac{1+z_J}{3}\right]^2,\,\,\,\,\,\,\,&&
\label{eq:tco_lowz}
\end{eqnarray}
where we have scaled to characteristic values at $z \sim 2$.
Note that the redshift scaling is $\propto (1+z_J)^{1/2}$ only
in the high redshift limit where $H(z_J) \propto (1+z_J)^{3/2}$.
The results are plotted in Figure \ref{F:tco}. In this model
the star formation efficiency declines below the characteristic mass scale
$M_0 \sim 5 \times 10^{11} M_\odot$,
which may underestimate the higher redshift signal if low mass galaxies
form stars efficiently at high redshift and are CO luminous.
However, our present
data sets provide little constraint at very high redshift, and so we
are less concerned with the predictions there. Note that $\VEV{T_{\rm CO}}$ in this model is $0.1-1\mu$K for all $z$ (for $M_{\rm co, min}=10^9M_\odot$) and thus will be difficult to see in a CMB experiment like WMAP.
\begin{figure}[!t]
\begin{center}
\includegraphics[width=0.5\textwidth]{co_mean_v_z.eps}
\caption{\label{F:tco} The mean CO brightness temperature in Model A as a function of redshift. The solid, dotted, and dashed lines denote $\VEV{T_{\rm CO}}$ for $M_{\rm co,min}=$10$^9$, 10$^{10}$, and 10$^{11}M_\odot$, respectively. Note the same mean $\VEV{T_{\rm CO}}$ is assumed for both the CO(1-0) and CO(2-1) lines.}
\end{center}
\end{figure}
\subsection{Model B: CO Luminosity -- Star Formation Rate}
We recognize that a weak part of the above estimate is the simplistic model
connecting SFR and host halo mass. In this section we take
a more empirical approach to estimating the spatially-averaged
CO brightness temperature and thereby circumvent the need to connect
SFR and host halo mass. In particular, here we use
recent determinations of the star formation rate (SFR) function from \citet{2012ApJ...756...14S}.
These authors use measurements of the UV luminosity function along with
the Kennicutt \citep{1998ARA&A..36..189K} relation, connecting SFR and UV luminosity to determine
the SFR function, i.e., the number density of galaxies with a star formation rate between SFR and SFR + dSFR.
The observed UV luminosity is corrected for dust attenuation based on the slope of the UV continuum
spectra, which gives a luminosity and redshift-dependent extinction correction.
They fit Schechter functions to the resulting SFR functions using new UV luminosity function and extinction
measurements at $z \sim 4-7$, and tabulate results from the literature at lower redshift (their Table 3).
The Schechter function \citep{1976ApJ...203..297S} is:
\begin{eqnarray}
\Phi(SFR)dSFR = \phi_\star \left(\frac{SFR}{SFR_\star}\right)^\alpha \rm{exp}\left[-\frac{SFR}{SFR_\star}\right] \frac{dSFR}{SFR_\star},
\label{eq:phi_sfr}
\end{eqnarray}
and is parameterized by a characteristic $SFR$, $SFR_\star$, a characteristic number density, $\phi_\star$, and
a faint-end slope, $\alpha$. The star formation rate density, $\dot{\rho_\star}$, can be derived by integrating
the SFR over the Schechter function. Assuming that the Schechter function form is maintained to arbitrarily low
SFRs then gives:
\begin{eqnarray}
\dot{\rho_\star} = \phi_\star SFR_\star \Gamma[2 + \alpha].
\label{eq:sfrd}
\end{eqnarray}
Here $\Gamma[2+\alpha]$ is a Gamma function.
We can further combine the SFR Schechter function with the observed $L_{\rm CO (1-0)} - SFR$ correlation (Equation \ref{eq:lco}), to
estimate the co-moving (frequency-integrated) CO emissivity. As discussed below, we will assume this prescription only for
sources above some minimum SFR, $SFR_{\rm min}$, and that sources below this critical SFR do not contribute
appreciably to the CO emissivity. Combining Equation \ref{eq:lco} and Equation \ref{eq:phi_sfr}, the resulting
co-moving emissivity is:
\begin{eqnarray}
\epsilon_{\rm CO (1-0)} = \phi_\star L_0 \left(\frac{SFR_\star}{1 M_\odot/yr}\right)^{3/5} \Gamma\left[\alpha + 1.6, \frac{SFR_{\rm min}}{SFR_\star}\right].
\label{eq:eps_co}
\end{eqnarray}
In this equation, $L_0$ denotes the luminosity in the CO(1-0) line for a SFR of $SFR= 1 M_\odot/$yr. We fix this
at $L_0 = 3.2 \times 10^4 L_\odot$ as in Equation \ref{eq:lco}, and as suggested by $z \sim 0-3$ CO observations. The factor
$\Gamma[\alpha+1.6, SFR/SFR_\star]$ is an Incomplete Gamma Function. This reveals the importance
of the sub-linear scaling of Equation \ref{eq:lco} for the CO emissivity: combining the sub-linear scaling
and the Schechter form for the SFR function yields a formally divergent CO emissivity as $SFR \rightarrow 0$ for
$\alpha \leq -1.6$. This may be an artifact of extrapolating the Schechter form and/or the sub-linear
$L_{\rm CO (1-0)}$ scaling with SFR to arbitrarily low SFR; presumably one or both of these relations drop-off at low SFR. For instance, low luminosity
dwarf galaxies may have small metallicites and fall below the extrapolation
of the $L_{\rm CO}-SFR$ relation. These relations generalize the approach
of \citet{2011ApJ...730L..30C}, who estimated the CO emissivity from the SFRD assuming the two scale linearly with SFR.
Our more accurate relation requires, however, specifying $SFR_{\rm min}$ to relate the co-moving CO emissivity and the SFRD.
Using the Equations in L11 we can relate the co-moving CO emissivity as calculated above to the spatially-averaged
CO brightness temperature. It is useful to note that this gives:
\begin{eqnarray}
\VEV{T_{\rm CO}}(z_J) = 0.65 \mu K \left[\frac{\epsilon_{\rm CO (1-0)}}{6.3 \times 10^2 L_\odot \rm{Mpc}^{-3}}\right]\times&&\nonumber\\
\left[\frac{H(z_J=2)}{H(z_J)}\right]
\left[\frac{1+z_J}{3}\right]^2.\,\,\,\,\,\,\,&&
\label{eq:tco_eps}
\end{eqnarray}
\begin{figure}
\begin{center}
\includegraphics[width=0.5\textwidth]{co_obs_v_z.eps}
\caption{The average CO brightness temperature in Model B. These results are estimated from
Equations \ref{eq:eps_co} and \ref{eq:tco_eps} using the SFR function fits from \citet{2012ApJ...756...14S}. The
solid pentagon at $z=0$ is the average CO brightness temperature inferred from the CO luminosity function
measured at $z=0$ by \citet{2003ApJ...582..659K}.}
\label{fig:tco_emp}
\end{center}
\end{figure}
We calculate the results of Equations \ref{eq:eps_co} and \ref{eq:tco_eps} from $z=0.2-6.8$ using the best-fit results
in Table 3 of \citet{2012ApJ...756...14S}. We focus on the UV- and MIR-derived SFR functions from that paper, and do not
include the H$\alpha$-based estimates, although these appear to be comparable. We ignore the point at $z=0$
in their Table 3 from \citet{2011MNRAS.415.1815B}, since there appears to be a typo in the fit parameters from this paper,
as remarked in the footnote to Table 3.
The results of these calculations are shown in Fig.~\ref{fig:tco_emp}. The overall redshift evolution shown here is
noisy because $\VEV{T_{\rm CO}}$ is estimated from several discrete observational measurements, each with significant
observational error bars.
As anticipated above,
the brightness temperature estimates are sensitive to $SFR_{\rm min}$, which is somewhat uncertain. The observed
$L_{\rm CO (1-0)}-SFR$ correlation includes galaxies with SFRs down to $\sim 0.5 M_\odot/$yr~\citep{2010ApJ...714..699W}, but
the data are a bit sparse at low SFR. The Schechter function fits probe galaxies down to typical SFRs of $\sim0.1-1 M_\odot/$yr, with some redshift dependence at the faint-end limit (\textit{e.g.}~Fig.~2 of \citet{2012ApJ...756...14S}).
The results are
particularly sensitive to $SFR_{\rm min}$ at $z = 6.8$, where the slope of the faint-end luminosity
function is especially steep ($\alpha = -1.96$).
The more empirical model shown here agrees with the models of the previous section of Figure \ref{F:tco} at the order-of-magnitude level,
although the results in the $SFR_{\rm min} = 0.01 M_\odot/$yr case are larger by a factor of $\sim 4$ at $z \sim 2$
than the earlier $M_{\rm CO min} = 10^9 M_\odot/$yr model. The high-redshift results at $z \gtrsim 5$ are also broadly
consistent with the estimates in L11.
There are still significant uncertainties in the
Schechter function fits to the SFRs, and in the extrapolations to the faint end, but this latter estimate
is probably more secure than the estimates of the previous section which rely on an oversimplified model to connect SFR and halo mass. Note that we will estimate power spectra for Model B by multiplying the Model A spectra by $\VEV{T_{\rm CO}}^2({\rm Model\,B})/\VEV{T_{\rm CO}}^2({\rm Model\,A})$, although strictly speaking this is not accurate since the clustering bias depends on the SFR-M relation, which is not determined for Model B.
As one final sanity check, we can use the measured CO(1-0) luminosity function from \citet{2003ApJ...582..659K} at $z=0$ to estimate the CO emissivity
and brightness temperature at $z=0$. They provide a Schechter function fit with $\rho(L) = d \Phi/d\log_{\rm 10} L
= \rho_\star \ln(10) (L/L_\star)^{\alpha+1} \exp(-L/L_\star)$, and $\rho_\star = 7.2 \times 10^{-4}$ Mpc$^{-3}$ mag$^{-1}$,
$\alpha = -1.30$. \citet{2003ApJ...582..659K} quotes results in terms of velocity-integrated CO luminosities, with a characteristic $L_\star$
value of
$L_{\rm CO, V; \star} = 1.0 \times 10^7$ Jy km s$^{-1}$ Mpc$^2$. We convert this into solar units using
the relations in the Appendix of \citet{2009ApJ...702.1321O}, finding $L_\star = 9.6 \times 10^4 L_\odot$. Using the best-fit Schechter function parameters from \citet{2003ApJ...582..659K}, and integrating to $L_{\rm CO(1-0)} = 0$ gives $\VEV{T_{\rm CO}} = 0.064 \mu K$, broadly consistent with our estimates, extrapolated to $z=0$ (see the solid pentagon in Fig. \ref{fig:tco_emp}).
\subsection{CO Clustering}
In order to model the clustering of CO emitters, following L11, we will assume the standard relation between CO luminosity and host halo mass, linear bias, that the scales of interest are much larger than the virial radius of the relevant halos so that we can neglect the non-linear term, and that the halo shot noise obeys Poisson statistics. With this assumption, the 3D power spectra for CO temperature fluctuations goes as
\begin{eqnarray}
\label{Eq:Pco}
P_{\rm CO}(k,z) &=& \VEV{T_{\rm CO}}^2(z)\left[b^2(z)P_{\rm lin}(k,z)+\frac{1}{f_{\rm duty}(z)}\frac{\VEV{M^2}}{\VEV{M}^2}\right]\nonumber\\
\end{eqnarray}
where $b(z)$ is the effective $z$-dependent halo bias given in Eq.~15 of L11
\begin{eqnarray}
b(z)=\frac{\int_{M_{\rm co,min}}^\infty dM\,M\frac{dn}{dM}b(M,z)}{\int_{M_{\rm co,min}}^\infty dM\,M\frac{dn}{dM}}\, ,
\end{eqnarray}
where $dn/dM$ is the halo mass function from \citet{2008ApJ...688..709T} and $M$ is the associated halo mass, and $P_{\rm lin}(k,z)$ is the linear matter power spectrum. The second term in brackets for $P_{\rm CO}$ is the shot noise, with
\begin{eqnarray}
\VEV{M^2}&=&\int_{M_{\rm co,min}}^\infty dM\,M^2\frac{dn}{dM}\, ,\nonumber\\
\VEV{M}&=&\int_{M_{\rm co,min}}^\infty dM\,M\frac{dn}{dM}\, .
\end{eqnarray}
We implicitly assume that every dark matter halo will host at least one CO emitter with a duty cycle $f_{\rm duty}(z)$. We plot the CO three-dimensional power spectrum in Fig.~\ref{F:pk}.
\begin{figure}
\begin{center}
\includegraphics[width=0.45\textwidth]{pkzA.eps}
\includegraphics[width=0.45\textwidth]{pkzB.eps}
\caption{The 3D power spectrum for CO fluctuations in the redshift range $0<z<4$. The top panel is for Model A with $M_{\rm co,min}=10^9M_\odot$, while the bottom panel is for Model B with $SFR_{\rm min}=0.01M_\odot/$yr. The black dashed (dotted) line shows the clustering (shot noise) term for $z=2$.}
\label{F:pk}
\end{center}
\end{figure}
We also model the clustering of quasars and LRGs. We assume the quasars, LRGs, and CO emitters trace the same LSS such that cross-correlating the temperature maps with either the quasars or LRGs will uncover clustering of CO emitters. Note, however, that we do not require nor assume the quasars or LRGs produce CO emission. The objective of this paper is not to find all CO gas; it is in fact to use halos with CO gas to trace LSS. Since the amount of CO gas in halos should not statistically vary across the sky, CO emission should be a good tracer of LSS, or the density distribution of massive halos, across the sky. Thus, we do not need to claim that quasars trace the CO gas. We only need the quasars to trace LSS and the CO emitters to trace LSS.
We also assume linear biasing so that the 3D quasar and LRG spectra are
\begin{eqnarray}
P_{Q}(k,z) &=&b_Q^2(z)P_{\rm lin}(k,z)\nonumber\\
P_{LRG}(k,z) &=&b_{LRG}^2P_{\rm lin}(k,z)\, ,
\end{eqnarray}
where $b_Q(z)$ is the quasar clustering bias that we compute using the estimate of Eq.~15 in \citet{2005MNRAS.356..415C} and $b_{LRG}$ is the LRG clustering bias. In general, the LRG bias would be redshift-dependent, but we will assume a constant bias for the LRGs. Similarly, the 3D cross power-spectra can be written as
\begin{eqnarray}
P_{CO-Q}(k,z) &=&r_Q\VEV{T_{\rm CO}}(z)b(z)b_Q(z)P_{\rm lin}(k,z)\nonumber\\
P_{CO-LRG}(k,z) &=&r_{LRG}\VEV{T_{\rm CO}}(z)b(z)b_{LRG}P_{\rm lin}(k,z)\, ,
\end{eqnarray}
where we include cross-correlation coefficients $r_Q$ and $r_{LRG}$ between CO emitters and quasars and LRGs, respectively, which we assume to be scale- and redshift-independent throughout this analysis. Note it is possible that quasars and LRGs may live in the same halos as CO emitters, contributing a shot noise term to their cross-correlation. We will neglect this shot noise in the analysis.
Assuming the Limber approximation in the small scale limit (typically $\ell>10$), we can then derive the angular auto/cross power spectra. The cross-power spectra have the form
\begin{eqnarray}
\label{Eq:ClCOQ}
C_\ell^{CO-Q}&=&\int dz \frac{H(z)}{c}\frac{f_{\rm CO}(z)f_Q(z)}{\chi^2(z)}\times\nonumber\\&&P_{CO-Q}[k=\ell/\chi(z),z]\nonumber\\
C_\ell^{CO-LRG}&=&\int dz \frac{H(z)}{c}\frac{f_{\rm CO}(z)f_{LRG}(z)}{\chi^2(z)}\times\nonumber\\
&&P_{CO-LRG}[k=\ell/\chi(z),z]\, ,
\end{eqnarray}
where $H(z)$ is the Hubble parameter and $f_{\rm CO}(z)$, $f_Q(z)$, and $f_{LRG}(z)$ are selection functions of the CO temperature fluctuations, the quasar distribution, and the LRG distribution, respectively. In our analysis, we assume flat, normalized selection functions for CO, quasars, and LRGs as a rough estimate. The three relevant angular auto-power spectra are given by
\begin{eqnarray}\label{E:clauto}
C_\ell^{\rm CO}&=&\int dz \frac{H(z)}{c}\frac{f_{\rm CO}^2(z)}{\chi^2(z)}P_{\rm CO}[k=\ell/\chi(z),z]\nonumber\\
C_\ell^{Q}&=&\int dz \frac{H(z)}{c}\frac{f_{Q}^2(z)}{\chi^2(z)}P_{Q}[k=\ell/\chi(z),z]\nonumber\\
C_\ell^{LRG}&=&\int dz \frac{H(z)}{c}\frac{f_{LRG}^2(z)}{\chi^2(z)}P_{LRG}[k=\ell/\chi(z),z]\, .
\end{eqnarray}
We plot the predicted angular power cross-spectra for Model A with quasars in Fig.~\ref{F:cln} and the cross-spectra with LRGs for $M_{\rm co,min}=10^9M_\odot$ in Fig.~\ref{F:clnlrg}. On the plot we superimposed an estimate for Model B by just rescaling $\VEV{T_{\rm CO}}$. For the LRGs we assume $b_{LRG}=2.48$, which we measured from the data (see Sec.~\ref{S:lrgresult}). Given the finite redshift widths defined by the WMAP bands, the evolution of the peaks of the angular spectra with the band frequency reflects the evolution in angular diameter distance with redshift.
A higher band frequency means a lower redshift, which means a larger angular scale (lower $\ell$) for the peak. For each band, the CO(1-0) spectrum can be higher or lower than the CO(2-1) spectrum depending on the redshift ranges probed, which determine the halo, quasar, and LRG biases and the CO brightness temperature.
Given Model A, we also find that $C_{\ell=200}^{CO(1-0)}$ ($C_{\ell=200}^{CO(2-1)}$)= 3.90$\times10^{-7}$ (2.64$\times10^{-7}$), 3.48$\times10^{-7}$ (3.88$\times10^{-7}$), 1.93$\times10^{-7}$ (3.73$\times10^{-7}$), and 7.80$\times10^{-8}$ (2.83$\times10^{-7}$) $\mu$K$^2$ respectively in the Ka, Q, V and W bands if we neglect the shot noise term, which becomes dominant only at $\ell$ = 579 (N/A), 292 (N/A), all $\ell$ (744), and all $\ell$ (170) for the same bands. Note that these numbers change if $M_{\rm co,min}\neq10^9M_\odot$. For reference, the CMB angular power spectra is $C_\ell^{CMB}\simeq$ 1 $\mu$K$^2$ at $\ell$=200 so it will act as an important source of noise, if not subtracted.
High quasar shot noise causes CO and quasars to not be perfectly correlated even on the largest scales. We note that the shot noise is more important at lower $z$ (lower $\nu$) as the number of massive halos increase more rapidly than the number of small mass halos. As will be discussed below, the instrumental noise at $\ell=200$ is 0.0253, 0.0230, 0.0347, and 0.0501 $\mu$K$^2$. In Figs.~\ref{F:cln} and \ref{F:clnlrg}, we also present the various sources of noise in comparison to the cross-correlation signal. Note that we do not include foreground noise, although it is included in the final results. Other than foreground noise, we see that the spectra are dominated by the WMAP instrumental noise.
Given these numbers, the prospect of measuring a cross-correlation signal are not very high given the current state of our model. However, given the theoretical uncertainties, we will still attempt to measure directly this correlation with currently available data.
\begin{figure}
\begin{center}
\includegraphics[width=0.45\textwidth]{clnoise1.eps}
\includegraphics[width=0.45\textwidth]{clnoise2.eps}
\caption{\label{F:cln} The predicted angular cross-power spectra and noise curves with each WMAP band cross-correlated with its corresponding quasar map. The solid, red line is the cross-correlation signal for CO line emission for Model A with $M_{\rm co,min}=10^9M_\odot$, and the purple, short-dashed line is for Model B with $SFR_{\rm min}=0.01M_\odot/$yr. The blue lines are the noise curves assuming Model A for the following cases: cosmic variance and CO shot noise limited (dotted), including WMAP instrumental noise (dot-dashed) only, including quasar shot noise only (dot-dot-dot-dashed), and including both noise sources (dashed). Note that the noise curves are slightly modified for Model B.}
\end{center}
\end{figure}
\begin{figure}
\begin{center}
\includegraphics[width=0.45\textwidth]{clnlrg.eps}
\caption{\label{F:clnlrg} The predicted angular cross-power spectrum and noise curves for Model A with the WMAP W band cross-correlated with the LRG map for $M_{\rm co,min}=10^9M_\odot$. The solid, red line is the cross-correlation signal for CO line emission, and the purple, short-dashed line is for Model B with $SFR_{\rm min}=0.01M_\odot/$yr. The blue lines are the noise curves for the following cases: cosmic variance and CO shot noise limited (dotted), including WMAP instrumental noise (dot-dashed) only, including quasar shot noise only (dot-dot-dot-dashed), and including all noise (dashed). Note that the noise curves would be slightly modified for Model B.}
\end{center}
\end{figure}
\section{Measurement Methodology}\label{S:method}
We estimate the CO--LSS tracer angular power spectrum, where the LSS tracer can be either quasars or LRGs, using a minimum-variance estimator of the form
\begin{eqnarray}\label{E:clest}
\hat{C}_\ell^{CO-Tr}=\sum_{m=-\ell}^\ell \frac{\bar{a}_{\ell m}^{\rm CO}\bar{a}_{\ell m}^{Tr*}}{(2\ell+1)f_{\rm sky,CO-Tr}}\, ,
\end{eqnarray}
where $f_{\rm sky,CO-Q}=0.140$ and $f_{\rm sky,CO-LRG}=0.197$ are the sky coverage fraction of the WMAPxSDSS map intersections for quasars and LRGs, respectively, and $\bar{a}_{\ell m}^X$ are the inverse-noise-filtered spherical harmonic coefficients for observable $X$, given by
\begin{eqnarray}\label{E:invfilt}
\bar{a}_{\ell m}^X=(C_\ell^X+C_\ell^{n,X})[(\mathbf{S}_X+\mathbf{N}_X)^{-1}\mathbf{d}_X]_{\ell m}\, ,
\end{eqnarray}
where $C_\ell^X$ and $C_\ell^{n,X}$ are the fiducial signal and noise angular auto-power spectra, respectively. We will assume Model A in Sec.~\ref{S:modela} with $M_{\rm co,min}=10^9M_\odot$ for the fiducial signal auto-power spectrum, noting that the estimator should not depend significantly on the fiducial spectrum. The CO thermal noise and quasar shot noise angular power spectra are
\begin{equation}\label{E:clnco}
C_\ell^{n,CO_i}=\Delta\Omega\sigma_{T_{ij}}^2/W_\ell^{T_i}\, ,
\end{equation}
and
\begin{equation}\label{E:shot}
C_\ell^{n,Tr_i}=(\bar{n}_{i_{tr}})^{-1}\, ,
\end{equation}
for the $i$ band, where $j$ is the band being used as the CMB map (see subsequent paragraph). $\Delta \Omega$ is the pixel size, $\sigma_{T_{ij}}^2$ is the average of $\sigma_{T_i}^2+\sigma_{T_j}^2$ over all pixels, $\bar{n}_{i_{tr}}$ is the average number of objects per steradian for the tracer, and $W_\ell^{T_i}$ includes the pixel and WMAP beam window functions. $\mathbf{d}_X$ is the data vector for observable X with the entries in $\mathbf{d}_{\rm CO}$ and $\mathbf{d}_{Tr}$ being the CO fluctuation (in $\mu$K units) and the object number overdensity for the tracer, respectively. $\mathbf{S}_X$ is the fiducial signal covariance matrix and $\mathbf{N}_X$ is the noise covariance matrix with $\VEV{\mathbf{d}_X\mathbf{d}_X^T}=\mathbf{S}_X+\mathbf{N}_X$. $(\mathbf{S}_X+\mathbf{N}_X)^{-1}$ works as an operator that weights each data point by its covariance, effectively filtering out very noisy modes. It is evaluated using a multigrid preconditioner (see Appendix A of \citet{2007PhRvD..76d3510S} for details). The variance for the estimator in Eq.~\ref{E:clest} is given by
\begin{eqnarray} \label{E:clestvar}
{\rm Var}[\hat{C}_\ell^{CO-Q}] = \frac{1}{(2\ell+1)f_{\rm sky}}\left[(\hat{C}_\ell^{CO-Q})^2+\hat{D}_\ell^{\rm CO}\hat{D}_\ell^Q\right]\, ,
\end{eqnarray}
where
\begin{eqnarray}
\hat{D}_\ell^{\rm CO}&=&\sum_{m=-\ell}^\ell \frac{|\bar{a}_{\ell m}^{\rm CO}|^2}{(2\ell+1)f_{\rm sky,CO}}\nonumber\\
\hat{D}_\ell^{Tr}&=&\sum_{m=-\ell}^\ell \frac{|\bar{a}_{\ell m}^{Tr}|^2}{(2\ell+1)f_{\rm sky,Tr}}\, ,
\end{eqnarray}
and $f_{\rm sky,CO}=0.783$, $f_{\rm sky,Q}=0.17$, and $f_{\rm sky,LRG}=0.203$ are the sky coverage fractions of our WMAP and SDSS data samples, respectively. For completeness, we also give
\begin{eqnarray}\label{E:varcotr}
{\rm Var}[\hat{D}_\ell^{\rm CO}] = \frac{2}{(2\ell+1)f_{\rm sky,CO}}\left(\hat{D}_\ell^{\rm CO}\right)^2\nonumber\\
{\rm Var}[\hat{D}_\ell^{Tr}] = \frac{2}{(2\ell+1)f_{\rm sky,Tr}}\left(\hat{D}_\ell^{Tr}\right)^2\, .
\end{eqnarray}
By estimating the errors this way, we include all sources of noise including galactic foregrounds.
As discussed above, the WMAP temperature maps are dominated by the CMB and galactic foregrounds, with CO emission constituting a small contribution. CMB fluctuations would drastically increase statistical errors in our CO search, so we choose to try and subtract out the CMB. Because the WMAP bands are so large, we cannot model the CMB fluctuations to subtract them properly. However, since the V and W maps are dominated by the CMB outside the galactic mask\footnote{Specifically, the foreground contribution to the V and W maps are approximately 20\%.}, we can use these maps to subtract the CMB from all the maps. Specifically, we subtract the W map from the maps Ka and V, and we subtract the V map from the Q and W maps. Note that all the maps have zero mean. Since the WMAP maps are already given in physical temperatures, we can just subtract them directly before converting them to brightness temperatures (see Sec.~\ref{S:wmaptemp}). The noise fluctuations in the V and W maps make the subtractions imperfect, even if we were to assume the foregrounds in these maps were negligible. This causes noise fluctuations to increase due to the noise introduced by the subtracted map according to Eq.~\ref{E:clnco}, but this should not introduce a bias. However, subtracting different bands, even with the extra noise, is worthwhile because keeping the CMB perturbations would contribute much more noise. We give an example of a CMB-subtracted map in Fig.~\ref{F:cocmb}. From the two maps, it is evident that our method removes the hot and cold spots of the CMB. Note that this subtraction does not affect the CO signal we aim at cross-correlating with since the bands do not overlap in CO redshift sensitivity as is clear in Tab.~\ref{T:zbin}.
\begin{figure}
\begin{center}
{\scalebox{.3}{\includegraphics{wmap_samplefQ.eps}}}
{\scalebox{.3}{\includegraphics{map_ap.eps}}}
\end{center}
\caption{\label{F:cocmb} The brightness temperature map in the WMAP Q band (left) and the same map with the brightness temperature map from the W band (estimating the CMB fluctuations) subtracted (right) with galactic emission and point sources masked out. The color legend is given in units of mK.}
\label{fig:wmap_map}
\end{figure}
\section{Results} \label{S:results}
\subsection{Quasar Results}
We present measurements of the CO-Q cross-correlation in Fig.~\ref{F:cross}. Remember that the Ka band measurement for the CO(2-1) line was not performed. Even though the CMB was subtracted from the data, the error bars on the cross-correlation are still much larger than the predicted spectra from Model A. The WMAP temperature noise and resolution, as well as the number density of quasars and any remaining fluctuations due to foregrounds, affect the statistical errors, but from Fig.~\ref{F:cln} we can infer the noise from the brightness temperature residuals is by far the greatest contributor compared with its theoretical noise-free fluctuations. It appears that our measured angular power spectra are consistent with a null signal and Model A in all the redshift bins. Model B is fairly high in some of the bands, but not entirely ruled out. We also show 10x the Model B estimate, which we see is ruled out in several of the bands on the larger scales. This becomes much clearer in the $\VEV{T_{\rm CO}}$ constraints below. Thus we can infer that the CO brightness is not significantly brighter than Model B ({\it e.g.}~there is no abundant population of faint CO emitters that is not included in our model, unless the cross-correlation coefficient $r_Q$ is low). However, a future experiment with an increased temperature sensitivity and quasar density is needed to detect the CO-Q cross-correlation (see Sec.~\ref{S:forecasts}).
\begin{figure}
\begin{center}
\includegraphics[width=0.45\textwidth]{clall1.eps}
\includegraphics[width=0.45\textwidth]{clall2.eps}
\caption{\label{F:cross} The measured, binned CO-Q cross-correlation angular power spectrum measurements with 1-sigma error bars. The solid line is the nonzero theoretical CO-Q correlation from Model A for $M_{\rm co,min}=10^9M_\odot$, the dotted line is the rough estimate for Model B for $SFR_{\rm min}=0.01M_\odot/$yr, and the dashed line is 10x the estimate for Model B.}
\end{center}
\end{figure}
Using the extra simplifying assumption that $\VEV{T_{\rm CO}}$ is constant over each redshift window we can translate our measurement into a constraint on the effective $\VEV{bT_{\rm CO}}$ temperature, $\overline{\VEV{bT_{\rm CO}}}$. We define the $\VEV{bT_{\rm CO}}$ estimator by rewriting Eq.~\ref{Eq:ClCOQ} as
\begin{eqnarray}
\label{Eq:ClCOQ2}
C_\ell^{CO-Q}&=& r\overline{\VEV{bT_{\rm CO}}}\int dz \frac{H(z)}{c}\frac{f_{\rm CO}(z)f_Q(z)}{\chi^2(z)}b_Q(z)\times\nonumber\\
&&P_{\rm lin}(k=\ell/\chi(z),z)\nonumber\\
&=&r\overline{\VEV{bT_{\rm CO}}}F_\ell\, .
\end{eqnarray}
Assuming $r=1$, we then can construct a minimum-variance estimator for $\overline{\VEV{bT_{\rm CO}}}$ according to the expression
\begin{eqnarray} \label{E:btest}
\overline{\VEV{bT_{\rm CO}}}=\frac{\sum_\ell \hat{C}_\ell^{CO-Q}F_\ell/{\rm Var}[\hat{C}_\ell^{CO-Q}]}{\sum_\ell F_\ell^2/{\rm Var}[\hat{C}_\ell^{CO-Q}]}\, ,
\end{eqnarray}
with uncertainty
\begin{eqnarray} \label{E:btunc}
\frac{1}{\sigma_{bT_{\rm CO}}^2}=\sum_\ell\frac{F_\ell^2}{{\rm Var}[\hat{C}_\ell^{CO-Q}]}\, .
\end{eqnarray}
In practice, we limit ourselves to $80<\ell<1000$ to avoid the Integrated Sachs-Wolfe (ISW) ``contamination'' on larger scales \citep{1967ApJ...147...73S}. Using the same formalism as above, our constraints on $\overline{\VEV{bT_{\rm CO}}}$ (assuming $r=1$) are given in Table \ref{T:btco}. If we further assume knowledge of the bias, which in practice means that we know that host halo mass of CO emitting objects, these constraints yield $\overline{T}_{\rm CO}$ constraints, where we rewrite Eq.~\ref{Eq:ClCOQ} as
\begin{eqnarray}
\label{Eq:ClCOQ3}
C_\ell^{CO-Q}= r\overline{T}_{\rm CO}\int dz \frac{H(z)}{c}\frac{f_{\rm CO}(z)f_Q(z)}{\chi^2(z)}b(z)b_Q(z)\times&&\nonumber\\P_{\rm lin}(k=\ell/\chi(z),z)\, .\,\,\,\,\,\,\,\,&&
\end{eqnarray}
Still assuming $r=1$, we plot constraints of $\VEV{T_{\rm CO}}$ for each line over four redshift bins, comparing the result to Model A for $M_{\rm co,min}=10^9M_\odot$, Model B for $SFR_{\rm min}=0.01M_\odot/$yr, in Fig.~\ref{F:tcolim}, and 10x Model B. It is evident from these plots that the $\VEV{T_{\rm CO}}$ signal from Model A and Model B are not detectable from our analysis, but we know that the brightness cannot be an order of magnitude greater than Model B. In fact, a model more than 3 times greater than Model B would be ruled out by the combined constraints of the two higher redshift points for the CO(1-0) line. Another way of saying this is that the signal-to-noise ratio (SNR) for the two points, which is the SNRs of the two individual points added in quadrature, where the signal is the difference between the data point and the model, is more than 3 for a model more than 3 times Model B.
\begin{table}[!t]
\begin{center}
\caption{\label{T:btco} Measured $\overline{bT}_{\rm CO}$ with 1-sigma error bars of CO emission lines $J=1\to0$ and $J=2\to1$ from cross-correlations of WMAP temperature bands with SDSS DR6 quasar (QSO) counts.}
\begin{tabular}{ccc}
\hline
WMAP band&$\overline{bT}_{CO(1-0)}\,(\mu$K)&$\overline{bT}_{CO(2-1)}\,(\mu$K)\\
\hline
Ka&$3.89\pm19.39$&N/A\\
Q&$-11.5\pm9.8$&$-138.\pm207.$\\
V&$1.41\pm8.68$&$18.7\pm24.1$\\
W&$2.98\pm3.94$&$-0.74\pm6.36$\\
\hline
\end{tabular}\end{center}
\end{table}
\begin{figure}[!t]
\includegraphics[width=0.45\textwidth]{tco10.eps}
\includegraphics[width=0.45\textwidth]{tco21.eps}
\caption{\label{F:tcolim} Limits on $\VEV{T_{\rm CO}}$ of CO(1-0) and CO(2-1) over four redshift bins. The solid line is the fiducial temperature from Model A for $M_{\rm co,min}=10^9M_\odot$, the dotted line is the fiducial temperature from Model B for $SFR_{\rm min}=0.01M_\odot/$yr, and the dashed line is 10x the estimate for Model B.}
\end{figure}
\subsection{LRG Results}\label{S:lrgresult}
Before measuring CO parameters using the LRG correlation, we proceed to determine the LRG clustering bias $b_{LRG}$ for the full sample, ignoring the bias redshift evolution. We start by rewriting the LRG angular auto-spectrum in Eq.~\ref{E:clauto} as
\begin{eqnarray}
C_\ell^{LRG}&=&b_{LRG}^2\int dz \frac{H(z)}{c}\frac{f_{LRG}^2(z)}{\chi^2(z)}P_{\rm lin}[k=\ell/\chi(z),z]\nonumber\\
&=&b_{LRG}^2G_\ell^{LRG}\, ,
\end{eqnarray}
where we assume a constant LRG bias. We use a model for the measured LRG auto-spectrum given by $\hat{D}_\ell^{LRG}=b_{LRG}^2G_\ell^{LRG}+C_\ell^{LRG,{\rm shot}}$, where $C_\ell^{LRG,{\rm shot}}$ is the shot noise component for LRGs given in Eq.~\ref{E:shot}. This model allows us to construct the estimator for the LRG bias
\begin{eqnarray}
[\widehat{b^2}]_{LRG}=\frac{\sum_\ell (\hat{D}_\ell^{LRG}-C_\ell^{LRG,{\rm shot}})G_\ell^{LRG}/{\rm Var}[\hat{D}_\ell^{LRG}]}{\sum_\ell (G_\ell^{LRG})^2/{\rm Var}[\hat{D}_\ell^{LRG}]}\, ,
\end{eqnarray}
with uncertainty
\begin{eqnarray}
\frac{1}{[\sigma_{\widehat{b^2}}]_{LRG}^2}=\sum_\ell\frac{(G_\ell^{LRG})^2}{{\rm Var}[\hat{D}_\ell^{LRG}]}\, ,
\end{eqnarray}
where we limit the sum to $10<\ell<100$. Using this method, we find $b_{LRG}=2.482\pm0.055$, which is a little high compared to other analyses but still consistent given that our redshift range is wider. In \citet{2008PhRvD..78d3519H}, the measured LRG bias for the relevant redshift range was $b_{LRG}=2.03\pm$0.07. We list results assuming both values.
With an LRG bias, we can perform the CO-LRG cross-correlation measurement, the results of which we show in Fig.~\ref{F:cldlrg}. These measurements, like the CO-quasar measurements, are consistent with both models, as well as 10x Model B, making it not really useful. We estimate CO parameters for $0.16<z<0.47$ in a similar manner as the previous section. The values we measured for $\overline{bT}_{CO(1-0)}$ and $\overline{T}_{CO(1-0)}$ assuming our LRG bias are $0.76\pm1.06$ $\mu$K and $0.56\pm0.78$ $\mu$K, respectively. Assuming the LRG bias from \citet{2008PhRvD..78d3519H}, the values change to $0.93\pm1.30$ $\mu$K and $0.69\pm0.96$ $\mu$K, respectively. Note that the cross-correlation measurement was performed without including the measured LRG bias to the fiducial signal; however, we repeated the measurement with the bias and confirmed that the final result was unchanged. These constraints are indeed tighter than those from the quasars, due to a much higher areal density of objects. However, at $z\simeq 0.25$, the relevant redshift, the predicted $\VEV{T_{\rm CO}}$ for both models are in the 0.1--0.2 range, making these constraints still not useful.
\begin{figure}
\begin{center}
\includegraphics[width=0.45\textwidth]{cldlrg.eps}
\caption{\label{F:cldlrg} The measured, binned CO-LRG cross-correlation angular power spectrum measurements with error bars. The solid line is the nonzero theoretical CO-Q correlation from Model A for $M_{\rm co,min}=10^9M_\odot$, the dotted line is the rough estimate for Model B for $SFR_{\rm min}=0.01M_\odot/$yr, and the dashed line is 10x the estimate for Model B.}
\end{center}
\end{figure}
\subsection{Possible Contaminants} \label{S:results:SZ}
Our measurements could be contaminated by any extragalactic foregrounds correlated with quasars or LRGs. We already attempted to remove ISW by neglecting large-scale modes, but other foregrounds may contaminate our signal. For example, quasars could live in SZ-contributing clusters or be closely correlated with dust-rich galaxies, which would lead to a bias in our measurements. Since we subtract maps, it is possible for a negative bias to occur, canceling a true signal. Although this is unlikely, if nothing else it is useful to describe these possible contaminants since they will be important in future analyses. We will briefly discuss these possibilities below.
As mentioned in Sec.~\ref{S:wmaptemp}, the thermal Sunyaev-Zel'dovich effect (tSZ) \citep{1980ARA&A..18..537S} is expected to correlate CMB temperature with large scale structure tracers like quasars or LRGs. In fact, there are hints of a SZ-quasar cross-correlation in WMAP/SDSS data \citep{2010ApJ...720..299C}. Thus, it is important to check the level of tSZ contamination in our measurement. Since tSZ has a well-defined frequency dependence, we can rule out tSZ if our cross-correlation measurements do not follow this frequency dependence. Furthermore, we note that while performing the CMB subtraction using the V or W band, we also alter the tSZ signal frequency dependence so that at every frequency, $X$, we will write the tSZ amplitude, as $\Delta
\tilde{T}_{\rm SZ} =\Delta T_{\rm SZ}^X-\Delta T_{\rm SZ}^{V\ or\ W}$ . Even though we do not expect tSZ in our CO analysis to be significant, in the following analysis we confirm whether or not this effect is indeed a problem.
The tSZ effect causes the CMB temperature to receive a secondary perturbation $\Delta T_{\rm SZ}$ every time it scatters with an object, with the perturbation given by
\begin{eqnarray}
\Delta T_{\rm SZ}(x) = yT_{\rm CMB}f(x)\, ,
\end{eqnarray}
where $y$ is the object-dependent Compton $y$ parameter, $x=h\nu/kT$, and $f(x)=x\coth x-4$. This implies that once the CMB reaches us, its tSZ perturbation in a pixel due to a set of objects (of one type) in a redshift bin will be $\delta T=\Delta T_{\rm SZ}(N-\bar{N})$, where $N$ is the number of objects in the pixel and $\bar{N}$ is the average number of objects per pixel. This model allows us to relate the angular cross-correlation between tSZ and a LSS tracer, $Tr$, to the tracer's auto-spectrum according to the form
\begin{eqnarray}
C_\ell^{SZ-Tr}=\Delta \tilde{T}_{\rm SZ}\bar{N}(C_\ell^{Tr}+C_\ell^{Tr,{\rm shot}})\, .
\end{eqnarray}
Using this model, we can construct an estimator for $\Delta \tilde{T}_{\rm SZ}$ using quasars as the LSS tracer, given by
\begin{eqnarray}
\hat{\Delta \tilde{T}_{\rm SZ}}=\frac{1}{\bar{N}}\frac{\sum_\ell \hat{C}_\ell^{SZ-Q}(C_\ell^{Q}+C_\ell^{Q,{\rm shot}})/{\rm Var}[\hat{C}_\ell^{SZ-Q}]}{\sum_\ell (C_\ell^{Q}+C_\ell^{Q,{\rm shot}})^2/{\rm Var}[\hat{C}_\ell^{SZ-Q}]}\, .
\end{eqnarray}
The procedure for measuring the tSZ-Q cross-spectrum is equivalent to measuring the CO-Q cross-spectrum, except that tSZ-Q uses the CMB physical temperature while CO-Q uses the brightness temperature. This allows us to set $\hat{C}_\ell^{SZ-Q}=\hat{C}_\ell^{CO-Q}/f_{\rm br}$, where $f_{\rm br}$ is the prefactor in Eq.~\ref{E:brcon} for converting physical temperatures to brightness temperatures.
We show the measured $\Delta \tilde T_{\rm SZ}$ at each redshift for the quasar samples in Fig.~\ref{F:tsz}. For the LRG sample, we find $\Delta \tilde T_{\rm SZ}=\{0.54\pm1.05,0.61\pm1.12\}$ for $b_{LRG}=\{2.48,2.03\}$, respectively. None of these measurements are significant, but since we did not detect CO emission, we can't rule out the presence of tSZ either. We emphasize that it is necessary to search for tSZ if CO line emission is detected in any future experiment. We note that a spectrograph like the one we propose in Sec.~\ref{S:forecasts} will have a high enough resolution to easily remove tSZ (and the CMB) directly, so this should not be an issue.
\begin{figure}[!t]
\includegraphics[width=0.45\textwidth]{tsz10.eps}
\includegraphics[width=0.45\textwidth]{tsz21.eps}
\caption{\label{F:tsz} Limits on $\Delta \tilde{T}_{\rm SZ}$ for the CO(1-0) and CO(2-1) quasar samples. For the CO(1-0) samples, the points from left to right to right are for the WMAP maps W-V, V-W, Q-W, and Ka-W. For the CO(2-1) samples, the points from left to right to right are for the WMAP maps W-V, V-W, and Q-W. The $z$ axis labels the redshift of the quasar map.}
\end{figure}
We also investigated contamination by dusty galaxies. Dust has a positive spectral index, causing its contribution with the W and V bands to be higher than the other bands. Thus, while correlating the difference of two WMAP bands, e.g. Q-V, with quasars, we may get a negative correlation which could possibly cancel the CO-quasar correlation. We test this possibility by correlating Ka-W, Q-V, and V-W with quasars that would correlate with the CO(1-0) line in the Q, Ka, and Q bands, respectively. Since this correlation cannot come from CO, we know that if dusty galaxies are canceling the CO-quasar correlation, they would create a negative signal. Note that W-V correlated with quasars or LRGs should
not have this problem because the signal would be positive. We find the correlations are indeed consistent with zero, as seen in Fig.~\ref{F:cldust}, so we can conclude that dusty galaxies are not canceling a CO-quasar cross-correlation detection. Similarly, this null result also rules out other possible contaminants such as radio sources.
\begin{figure}
\begin{center}
\includegraphics[width=0.45\textwidth]{clalld.eps}
\caption{\label{F:cldust} The measured, binned CO-Q cross-correlation angular power spectrum measurements for out-of-band quasars with 1-sigma error bars. This measurement shows that the temperature correlation with dust is not significant enough to bias the CO-emission measurements. The dashed line represents a null correlation.}
\end{center}
\end{figure}
\subsection{Another use of WMAP data}
Another method of detecting CO temperature fluctuations is by cross-correlating the CO(1-0) and CO(2-1) lines coming from the same emission redshift. For each WMAP band, the CO(1-0) redshift range and the CO(2-1) redshift range are different; this can cause the CO(1-0) redshift range for one band to overlap with the CO(2-1) redshift range of another band, e.g. Ka for CO(1-0) and V for CO(2-1). Since these two rotational lines would presumably be emitted by the same sources, we can cross-correlate pairs of WMAP bands that overlap in this way to search for CO emission. This could be a way to measure the line ratio as a function of redshift. Specifically, we could search for an angular cross-correlation of the form
\begin{eqnarray}
C_\ell^{CO-CO'}=\int dz \frac{H(z)}{c}\frac{f_{\rm CO}(z)f_{CO'}(z)}{\chi^2(z)}\times&&\nonumber\\sP_{\rm CO}[k=\ell/\chi(z),z]\, ,\,\,\,\,\,\,\,\,&&
\end{eqnarray}
integrated over the intersection of the two redshift ranges where unprimed CO is CO(1-0) and primed CO is CO(2-1). We include a parameter $s$, the ratio between the two CO lines. We can attempt to measure a unitless amplitude $B$ such that $\hat{C}_\ell^{CO-CO'}=BC_\ell^{CO-CO'}$. In this case, $B$ has an estimator similar to Eqs.~\ref{E:btest} and \ref{E:btunc}. In this procedure, there would be other sources of correlation in the WMAP bands coming from various foregrounds, with the CMB as the dominant foreground. To remove this complication, we can either remove these foregrounds by hand or cross-correlate pairs of WMAP bands that do not correspond to the associated CO redshift window as a cross-check to see if the difference in cross-correlation between the two cases is statistically significant. A third option would be to use a three point function, something like Ka-V-QSO, as this would lead to an extra handle on the foreground correlation.
Based on the WMAP band CO redshift ranges in Table \ref{T:zbin}, we find that the Ka band for CO(1-0) should overlap with the V band for CO(2-1), while the Q band for CO(1-0) should overlap with the W band for CO(2-1). The other combinations (Ka-Q, Ka-W, Q-V, and V-W) should not exhibit any cross-correlations that are due to common CO emission. The predicted spectra for Ka-V and Q-W are plotted in Fig.~\ref{F:cl12}. We attempt to subtract the CMB from all four bands, cross-correlate the Ka-V and Q-W pairs, and then look for a detection of an amplitude. However, when we calculate the variance of the cross-correlation $C_\ell^{CO-CO'}$ given by
\begin{eqnarray}\label{E:clttestvar}
{\rm Var}[\hat{C}_\ell^{CO-CO'}] = \frac{1}{(2\ell+1)f_{\rm sky,CO}}\left[(\hat{C}_\ell^{CO-CO'})^2+\right.&&\nonumber\\
\left.\hat{D}_\ell^{\rm CO}\hat{D}_\ell^{CO'}\right]\, ,\,\,\,\,\,\,\,\,&&
\end{eqnarray}
we find extremely large errors for this CO amplitude.
The errors for the amplitude for Model A are $\sigma_B\simeq1400$ for Ka-V and $\sigma_B\simeq 1200$ for Q-W. The numbers are much better for Model B, being $\sigma_B\simeq45$ for Ka-V and $\sigma_B\simeq 68$ for Q-W, although still less powerful than those from the WMAP-quasar cross-correlation.
The limits from the COxQSO analysis greatly outperform the limits from a potential WMAP CO(1-0)xCO(2-1) analysis because the instrumental noise is relatively much higher than the shot noise as shown in Fig.~\ref{F:cln}. In addition, WMAP foregrounds and CO emission from higher $J$ can contaminate the signal, so we do not attempt to measure it in this analysis.
\begin{figure}
\begin{center}
\includegraphics[width=0.45\textwidth]{clmall2.eps}
\caption{\label{F:cl12} The predicted angular cross-power spectra from Model A for $M_{\rm co,min}=10^9M_\odot$ with a WMAP band for CO(1-0) cross-correlated with another WMAP band for CO(2-1). The solid line is the clustering term and the dotted line is the shot noise term.}\end{center}
\end{figure}
\section{CO Experiment Forecasts} \label{S:forecasts}
Now that we have determined WMAP's limited ability to constrain CO temperature, we investigate what can be done in future experiments. We know that the limiting factor in the CO-Q cross-correlation measurement is how well we can reduce temperature noise as well as increase the quasar density. Unlike WMAP, a dedicated experiment should have small frequency bands in order to maximize power from the CO line and to subtract continuum sources, including the CMB and other foregrounds. We consider such a configuration here\footnote{Mike Seiffert, private communication}.
In this section, we seek to determine the signal-to-noise ratio (SNR) of the cross-correlation between the brightness temperature fluctuation of CO(1-0) measured by a hypothetical CO spectrograph and the latest spectroscopic LSS surveys at $z\sim3$. This CO spectrograph measures the CO(1-0) rotational line at observed frequency 28.8 GHz with a 1 GHz bandwidth and 20 50-MHz frequency channels. At a frequency resolution of approximately $R=600$, this experiment should be able to effectively model CMB and foreground emission and remove their contributions. The central frequency of the band would correspond to $z=3$ with channel widths corresponding to $\Delta z = 0.00694$ ($2.9306<z<3.0694$). Table \ref{T:copar} shows sample parameters of this experiment for two cases. For each case, we set a instrumental error $\sigma_{\rm T,fwhm}$, a beam FWHM, and a survey area. Note that $\sigma_{\rm T,fwhm}$ is for a pixel corresponding to the FWHM we choose. We then have a noise angular power spectra of the form
\begin{eqnarray}
C_\ell^{n,CO}=\sigma_{\rm T,fwhm}^2(0.4245\theta_{\rm fwhm})^2/W_\ell^T\, .
\end{eqnarray}
\begin{table}
\begin{center}
\caption{\label{T:copar} CO spectrograph parameters.}
\begin{tabular}{ccc}
\hline
Survey Parameter&Spec 1&Spec 2\\
\hline
$\sigma_{\rm T,fwhm}$ $(\mu$K)&6.7&4.7\\
$\theta_{\rm fwhm}$ (arcmin)&40.5&28.3\\
Survey Area (deg$^2$)&550&270\\
\hline
\end{tabular}\end{center}
\end{table}
We begin with the BOSS final spectroscopic quasar survey. The BOSS full survey will cover 10,200 deg$^2$ by 2014, long before the time when the CO spectrograph would start observing. In the range compatible with our hypothetical spectrograph, based on projections from the BOSS Year One (partial) survey, we expect BOSS to detect 5465 spectro-QSOs per steradian. Based on the redshift distribution in \citet{2012ApJS..199....3R}, we see that the distribution is pretty flat in our range of interest, making us set the density in each redshift bin equal to 5465/20 = 273 str$^{-1}$.
In principle, the spectral resolution is such that the cross-correlation will be performed directly in three dimensions, as is currently performed with 21cm surveys \citep{2010Natur.466..463C,2013ApJ...763L..20M} and L11. For the sake of simplicity, we will however compute here a simple forecast using a bin-by-bin 2D angular power spectra. Using the 2D angular power spectrum to constrain CO temperature neglects modes along the line of sight, lowering the SNR. Thus, we take the SNR forecasts in this section to be lower limits to what could be achieved with a full 3D power spectrum analysis.
The SNR for the experiment is then
\begin{eqnarray}
{\rm SNR}^2 =N_{\rm z-bins}\sum_\ell\frac{(C_\ell^{CO-Q})^2}{{\rm Var}[C_\ell^{CO-Q}]}\, ,
\end{eqnarray}
where $N_{\rm z-bins} = 20$ and we assume each bin contributes approximately equally. Note that we neglect the cross-correlation shot noise in this calculation, as well as for the rest of this Section. We find for Model A with $M_{\rm co,min}=10^9M_\odot$ the forecasts SNR = 1.2 (5.4) for Spectrograph 1 and SNR = 1.8 (8.0) for Spectrograph 2 for each redshift bin (over the full redshift range). Spectrograph 2 may be of interest for the full redshift range, but systematics resulting from not subtracting SZ and dust contamination properly may degrade the signal. An autocorrelation would have an even lower SNR (SNR $\sim7.1$ for Spectrograph 2). Probing Model B should have better prospects; although we can't estimate the cross-spectrum directly, we can estimate that since $\VEV{T_{\rm CO}}$ at $z=3$ for Model B with $SFR_{\rm min}=0.01M_\odot/$yr is about $4/0.85\simeq4.8$ higher than for Model A with $M_{\rm co,min}=10^9M_\odot$, its cross-spectrum will be approximately 4.8 times higher. This changes the SNR values to 4.2 (18.8) for Spectrograph 1 and 4.4 (19.7) for Spectrograph 2. Note that the SNR does not increase by a factor of 4.8 because the noise is model-dependent. The autocorrelation SNR would actually be higher than that for the cross-spectra in this case (SNR $\sim49$ for Spectrograph 2), but foregrounds can degrade this signal. Another option would be to have an instrument that probed CO fluctuations at two frequencies with one twice the frequency of the other. With this, you could search for a cross-correlation signal from CO(1-0)$\times$CO(2-1); the SNR for an ``equivalent Spectrograph 2" for this setup is approximately 59 without the foreground issues of an autocorrelation. However, this would be a much more expensive instrument.
Instead of using BOSS, we could also cross-correlate Spectrographs 1 and 2 with HETDEX \citep{2008ASPC..399..115H}, which will observe 1 million Ly$\alpha$ emitters over 200 square degrees in the redshift range $1.8<z<3.8$. In a $\Delta z=0.007$ redshift bin, HETDEX will have an areal density of about 57,500 str$^{-1}$. This specification cross-correlated with Spectrograph 2 gives us an SNR for each bin (over full redshift range $2.93<z<3.07$) of 5 (22) for Model A and 13 (58) for Model B, which is much higher than we can get with BOSS. Using instead Spectrograph 1 decreases the numbers only slightly, to 3.8 (17) for Model A and 13 (57) for Model B.
Another option is to just cross-correlate the temperature maps from the Planck satellite \citep{2011A&A...536A...1P} with the same photo-quasar map we used in this analysis. However, we find that for Model A none of bands have a SNR greater than one. We find that the highest SNR we get is 0.73 for the CO(2-1) line for Planck's 143 GHz band. We assume subtracting the 100 GHz band from all the other bands and subtracting the 70 GHz band from the 100 GHz band. All the other SNRs for the other Planck bands are much less for the both lines. Model B is more constrainable with SNR=2.0 in the 143 GHz band for the CO(2-1) line, but foregrounds will probably degrade this signal. Using the BOSS spectro-quasar survey would increase $f_{\rm sky}$ to at most 0.25. Since most of the noise errors is due to the temperature maps, Planck x BOSS would not do much better than Planck x SDSS DR6. Moreover, the use of the CII line instead of CO would be more appropriate for Planck high frequencies.
Current ground-based, high-angular-resolution CMB polarimeters offer another promising avenue towards measuring this contribution. Consider an SPTPol-like survey \citep{2012SPIE.8452E..1EA} with 6.5 and 4.5 $\mu$K-arcmin sensitivity at 90 and 150 GHz with 2.0 and 1.2 arcmin FWHM and covering 500 sq.~degrees of a BOSS-like quasar survey. The real SPTPol and BOSS surveys cover opposite hemispheres, but we perform this exercise as an illustration. We assume both bands have a 30\% bandwidth. For $z\sim3$ quasars, the 90 and 150 GHz bands could constrain $\VEV{T_{\rm CO}}$ for $J=$3 and 5, respectively. The BOSS full survey will have quasar areal densities within the relevant redshift bins of about 36,000 and 22,000 str$^{-1}$, respectively. With these parameters, we forecast cross-correlation SNRs for Model A (Model B) of 5 (12) and 8 (13). For the extended SPT-3G, the 220 GHz band is added, which can constrain the $J=8$ line using quasars at $z\sim3$. The survey area is also increased to 2500 sq.~degrees. The sensitivities for the 90, 150 and 220 GHz bands change to 4.2, 2.5, and 4.0 $\mu$K-arcmin and the FWHMs change to 1.7, 1.2, and 1.0 arcmin. In the 220 GHz band, the areal density of quasars in the relevant redshift bin is about 15,000 str$^{-1}$. For SPT-3G, we forecast cross-correlation SNRs for Model A (Model B) of 15 (31), 24 (31) and 54 (59). HETDEX does not do as well as BOSS in this case because it covers such a small area. Of course, SPTPol cannot constrain the same lines as the 28.8 GHz experiment mentioned earlier. Also, the large bands and limiting frequency coverage will make the foreground removal difficult so that these numbers are optimistic. Also, the high optical depth limit is less certain for higher $J$ lines, making the signal more uncertain for this experiment. Finally, the frequency bands for ACPol/SPTPol are wide enough that a full 3D analysis is not feasible.
We summarize the various results for different experiment combinations in Table \ref{T:snr}. Other experiments that could also help with this kind of search are the Primordial Inflation Explorer (PIXIE) \citep{2011JCAP...07..025K}, and the Murchison Widefield Array (MWA) \citep{2009IEEEP..97.1497L}. Although PIXIE is a polarization experiment to detect inflationary gravitational waves, its high frequency resolution over a wide frequency range can constrain CII and CO, particularly for higher $J$ lines, over large redshift ranges, including high redshifts. MWA is currently searching for the 21 cm HI line from the dark ages and reionization. As mentioned in L11, CO x 21 cm can be a powerful probe of the high redshift universe. For example, MWA x SPTPol could constrain CO J=5 or 7 lines and star formation at $z\sim6-7$.
\begin{table*}
\begin{center}
\caption{\label{T:snr} The signal-to-noise ratio (SNR) for measuring the CO brightness temperature with CO$\times$LSS cross-correlations of various experiment combinations. Unless noted otherwise, the first and second value listed for the SNR are for Model A and Model B, respectively.}
\begin{tabular}{cccc}
\hline
&CO line&SNR per $\Delta z=0.007$&SNR over full $z$-range\\
\hline
Spectrograph 1 $\times$ BOSS QSOs&CO(1-0)&1.2, 4.2&5.4, 19\\
Spectrograph 2 $\times$ BOSS QSOs&CO(1-0)&1.8, 4.4&8.0, 20\\
Spectrograph 1 $\times$ HETDEX Ly$\alpha$ emitters&CO(1-0)&3.8, 13&17, 57\\
Spectrograph 2 $\times$ HETDEX Ly$\alpha$ emitters&CO(1-0)&5, 13&17, 58\\
\hline
Planck (143 GHz) $\times$ SDSS DR6 QSOs&CO(2-1)&N/A&2 (Model B)\\
\hline
SPT (90 GHz) $\times$ BOSS QSOs&CO(3-2)&N/A&5,12\\
SPT (150 GHz) $\times$ BOSS QSOs&CO(5-4)&N/A&8,13\\
SPT-3G (90 GHz) $\times$ BOSS QSOs&CO(3-2)&N/A&15,31\\
SPT-3G (150 GHz) $\times$ BOSS QSOs&CO(5-4)&N/A&24,31\\
SPT-3G (220 GHz) $\times$ BOSS QSOs&CO(8-7)&N/A&54,59\\
\hline
\end{tabular}\end{center}
\end{table*}
\section{Conclusions} \label{S:conclude}
We have predicted an angular cross-power spectrum between CO line emission and quasars and LRGs based on $\Lambda$CDM cosmology and the L11 model. We proposed searching for the quasar/LRGs cross-correlation to characterize CO emission in high-redshift galaxies. We have also attempted to detect the cross-correlation in WMAP and SDSS photo-quasars and LRGs up to $z\sim6$. A signal was not detectable, mainly due to the large statistical errors in the WMAP maps. We were able to set upper limits to the brightness temperature of the CO(1-0) and CO(2-1) lines, which rule out models much greater than our Model B. We also explored the CO(1-0)xCO(2-1) cross correlation, another signature of CO emission.
Although current probes appear to be unable to detect CO emission, the potential for future experiments looks considerably greater. Current or soon-to-happen ground based, high-angular-resolution CMB experiments overlapping with BOSS offer a chance to detect higher $J$ lines. In our forecasts for an optimistic model for a future spectrograph to detect CO(1-0) line emission, we found a SNR of 58 for a CO(1-0)x(HETDEX Ly$\alpha$ emitter) analysis at $z\sim3$ and a SNR of 59 for a more expensive CO(1-0)xCO(2-1) analysis. Although these numbers will likely be decreased due to foreground subtraction, this result is still very promising. A future detection of CO brightness temperature perturbations will allow us to model the CO emission-line galaxies at high redshifts, possibly even out to redshifts in the reionization epoch.
\begin{acknowledgments}
We thank D.~Hanson, S.~Furlanetto, and M.~Seiffert for helpful comments and useful discussions. Part of the research described in this paper was carried out at the Jet Propulsion Laboratory, California Institute of Technology, under a contract with the National Aeronautics and Space Administration. AP was supported by an appointment to the NASA Postdoctoral Program at the Jet Propulsion Laboratory, administered by Oak Ridge Associated Universities through a contract with NASA. This work was supported by the Keck Institute of Space Studies and we thank colleagues at the ``First Billion Years'' for stimulating discussions, in particular J. Bowman and A. Readhead for organizing it.
Funding for the SDSS and SDSS-II has been provided by the Alfred P. Sloan Foundation, the Participating Institutions, the National Science Foundation, the U.S. Department of Energy, the National Aeronautics and Space Administration, the Japanese Monbukagakusho, the Max Planck Society, and the Higher Education Funding Council for England. The SDSS Web Site is {\tt http://www.sdss.org/}. The SDSS is managed by the Astrophysical Research Consortium for the Participating Institutions. The Participating Institutions are the American Museum of Natural History, Astrophysical Institute Potsdam, University of Basel, University of Cambridge, Case Western Reserve University, University of Chicago, Drexel University, Fermilab, the Institute for Advanced Study, the Japan Participation Group, Johns Hopkins University, the Joint Institute for Nuclear Astrophysics, the Kavli Institute for Particle Astrophysics and Cosmology, the Korean Scientist Group, the Chinese Academy of Sciences (LAMOST), Los Alamos National Laboratory, the Max-Planck-Institute for Astronomy (MPIA), the Max-Planck-Institute for Astrophysics (MPA), New Mexico State University, Ohio State University, University of Pittsburgh, University of Portsmouth, Princeton University, the United States Naval Observatory, and the University of Washington.
\end{acknowledgments}
|
{
"redpajama_set_name": "RedPajamaArXiv"
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\section{Introduction}\label{sec:intro}
Vertex deletion problems are central in parameterized algorithms and complexity, and they have contributed hugely to the development of new algorithmic methods. The $\Pi$-{\sc Deletion} problem, with input a graph $G$ and an integer $\ell$, asks whether at most $\ell$ vertices can be deleted from $G$ so that the resulting graph satisfies graph property $\Pi$. Its dual, the $\Pi$-{\sc Subgraph} problem, with input $G$ and $k$, asks whether $G$ contains an induced subgraph on at least $k$ vertices satisfying $\Pi$. The problems were introduced already in 1980 by Yannakakis and Lewis \cite{LewisY80}, who showed their \ensuremath{\operatorClassNP}-completeness for almost all interesting graph properties $\Pi$. During the last couple of decades, these problems have been studied extensively with respect to parameterized complexity and kernelization, which has resulted in numerous new techniques and methods in these fields \cite{CyganFKLMPPS15, DowneyF13}.
In many network problems, the size of the {\it boundary} between the subgraph that we are looking for and the rest of the graph makes a difference. A small boundary limits the exposure of the found subgraph, and notions like isolated cliques have been studied in this respect \cite{HuffnerKMN09, ItoI09, KomusiewiczHMN09}. Several measures for the boundary have been proposed; in this work we use the open neighborhood of the returned induced subgraph. For a set of vertices $U$ of a graph $G$ and a positive integer $t$, we say that $U$ is \emph{$t$-secluded} if $|N_G(U)|\leq t$. Analogously,
an induced subgraph $H$ of $G$ is \emph{$t$-secluded} if the vertex set of $H$ is $t$-secluded. For a given graph property $\Pi$, we get the following formal definition of the problem \ProblemName{Secluded $\Pi$-Subgraph}.
\defproblema{\ProblemName{Secluded $\Pi$-Subgraph}}%
{A graph $G$ and nonnegative integers $k$ and $t$.}%
{Decide whether $G$ contains a $t$-secluded induced subgraph $H$ on at least $k$ vertices, satisfying $\Pi$.}
Lewis and Yannakakis \cite{LewisY80} showed that $\Pi$-{\sc Subgraph} is \ensuremath{\operatorClassNP}-complete for every hereditary nontrivial graph property $\Pi$. This immediately implies that \ProblemName{Secluded $\Pi$-Subgraph} is \ensuremath{\operatorClassNP}-complete for every such $\Pi$. As a consequence, the interest has shifted towards the parameterized complexity of the problem, which has been studied by van Bevern et al.~\cite{BevernFMMSS16} for several classes $\Pi$. Unfortunately, in most cases \ProblemName{Secluded $\Pi$-Subgraph} proves to be \classW{1}-hard, even when parameterized by $k+t$.
In particular, it is \classW{1}-hard to decide whether a graph $G$ has a $t$-secluded independent set of size $k$ when the problem is parameterized by $k+t$ \cite{BevernFMMSS16}.
In this extended abstract, we
show that the situation changes when the secluded subgraph we are looking
for is required to be connected, in which case we are able to obtain positive results that apply to many properties $\Pi$. In fact, connectivity is central in recently studied variants of secluded subgraphs, like {\sc Secluded Path} \cite{ChechikJPP13,JohnsonLR14} and {\sc Secluded Steiner Tree} \cite{FominGKK16}.
However, in these problems the boundary measure is the closed neighborhood of the desired path or the steiner tree, connecting a given set of vertices.
The following formal definition describes the problem that we study in this extended abstract, \ProblemName{Connected Secluded $\Pi$-Subgraph}. For generality, we define a weighted problem.
\defproblema{\ProblemName{Connected Secluded $\Pi$-Subgraph}}%
{A graph $G$, a weight function $\omega\colon V(G)\rightarrow \mathbb{Z}_{>0}$, a nonnegative integer $t$ and a positive integer $w$.}%
{Decide whether $G$ contains a connected $t$-secluded induced subgraph $H$ with $\omega(V(H))\geq w$, satisfying $\Pi$.}
Observe that \ProblemName{Connected Secluded $\Pi$-Subgraph} \/ remains \ensuremath{\operatorClassNP}-complete for all hereditary nontrivial graph properties $\Pi$, following the
results of
Yannakakis \cite{Yannakakis79}.
It can be also seen that \ProblemName{Connected Secluded $\Pi$-Subgraph} parameterized by $w$ is \classW{1}-hard even for unit weights, if it is \classW{1}-hard with parameter $k$ to decide whether $G$ has a connected induced subgraph on at least $k$ vertices, satisfying $\Pi$ (see, e.g., \cite{DowneyF13, PapadimitriouY96}).
It is thus more interesting to consider parameterization by $t$, and we show that \ProblemName{Connected Secluded $\Pi$-Subgraph} is fixed parameter tractable when parameterized by $t$ for many important graph properties $\Pi$.
Our main result is given in Section~\ref{sec:fpt-forb} where we consider \ProblemName{Connected Secluded $\Pi$-Subgraph} for all graph properties $\Pi$ that are characterized by finite sets $\mathcal{F}$ of forbidden induced subgraphs and refer to this variant of the problem as \ProblemName{Connected Secluded $\mathcal{F}$-Free Subgraph}. We show that the problem is fixed parameter tractable when parameterized by $t$ by proving the following theorem.
\begin{theorem}\label{thm:forb}
\ProblemName{Connected Secluded $\mathcal{F}$-Free Subgraph} can be solved in time $2^{2^{2^{\mathcal{O}(t\log t)}}}\cdot n^{\mathcal{O}(1)}$.
\end{theorem}
In Section~\ref{sec:fpt-tree} we show that similar results could be obtained for some infinite families of forbidden subgraphs by giving an \ensuremath{\operatorClassFPT}\xspace algorithm for the case when $\Pi$ is the property of being acyclic. Furthermore, in Section~\ref{sec:better} we show that we can get faster algorithms for \ProblemName{Connected Secluded $\Pi$-Subgraph} when $\Pi$ is the property of being a complete graph, a star, a path, and a $d$-regular graph. Finally, in Section~\ref{sec:concl} we briefly discuss kernelization for \ProblemName{Connected Secluded $\Pi$-Subgraph}.
\section{Preliminaries}\label{sec:defs}
We consider only finite undirected simple graphs. We use $n$ to denote the number of vertices and $m$ the number of edges of the considered graphs unless it creates confusion. A graph $G$ is identified by its vertex set $V(G)$ and edge set $E(G)$. For $U\subseteq V(G)$, we write $G[U]$ to denote the subgraph of $G$ induced by $U$.
We write $G-U$ to denote the graph $G[V(G)\setminus U]$; for a single-element $U=\{u\}$, we write $G-u$.
A set of vertices $U$ is \emph{connected} if $G[U]$ is a connected graph.
For a vertex $v$, we denote by $N_G(v)$ the \emph{(open) neighborhood} of $v$ in $G$, i.e., the set of vertices that are adjacent to $v$ in $G$. For a set $U\subseteq V(G)$, $N_G(U)=(\cup_{v\in U}N_G(v))\setminus U$.
We denote by $N_G[v]=N_G(v)\cup\{v\}$ the \emph{closed neighborhood} of $v$; respectively, $N_G[U]=\cup_{v\in U}N_G[v]$.
The \emph{degree} of a vertex $v$ is $d_G(v)=|N_G(v)|$.
Two vertices $u$ and $v$ of graph $G$ are {\it true twins} if $N_G[u]=N_G[v]$, and {\it false twins} if $N_G(u)=N_G(v)$.
A set of vertices $S\subset V(G)$ of a connected graph $G$ is a \emph{separator} if $G-S$ is disconnected.
A vertex $v$ is a \emph{cut vertex} if $\{v\}$ is a separator.
A graph property is {\it hereditary} if it is preserved under vertex deletion, or equivalently, under taking induced subgraphs. A graph property is {\it trivial} if either the set of graphs satisfying it, or the set of graphs that do not satisfy it, is finite.
Let $F$ be a graph.
We say that a graph $G$ is \emph{$F$-free} if $G$ has no induced subgraph isomorphic to $F$. For a set of graphs $\mathcal{F}$, a graph $G$ is \emph{$\mathcal{F}$-free} if $G$ is $F$-free for every $F\in \mathcal{F}$. Let $\Pi$ be the property of being $\mathcal{F}$-free. Then, depending on whether $\mathcal{F}$ is a finite or an infinite set, we say that $\Pi$ is {\it characterized by a finite / infinite set of forbidden induced subgraphs}.
We use the \emph{recursive understanding} technique introduced by Chitnis et al.~\cite{ChitnisCHPP16} for graph problems to solve \ProblemName{Connected Secluded $\Pi$-Subgraph} when $\Pi$ is defined by forbidden induced subgraphs or $\Pi$ is the property to be a forest. This powerful technique is based on the following idea. Suppose that the input graph has a vertex separator of bounded size that separates the graph into two sufficiently big parts. Then we solve the problem recursively for one of the parts and replace this part by an equivalent graph such that the replacement keeps all essential (partial) solutions of the original part. By such a replacement we obtain a graph of smaller size. Otherwise, if there is no separator of bounded size separating graphs into two big parts, then either the graph has bounded size or it is highly connected, and we exploit these properties. We need the following notions and results from Chitnis et al.~\cite{ChitnisCHPP16}.
Let $G$ be a graph. A pair $(A,B)$, where $A,B\subseteq V(G)$ and $A\cup B=V(G)$, is a \emph{separation of $G$ of order $|A\cap B|$} if $G$ has no edge $uv$ with $u\in A\setminus B$ and $v\in B\setminus A$, i.e., $A\cap B$ is an $(A,B)$-separator. Let $q$ and $k$ be nonnegative integers. A graph $G$ is \emph{(q,k)-unbreakable} if for every separation $(A,B)$ of $G$ of order at most $k$, $|A\setminus B|\leq q$ or $|B\setminus A|\leq q$. Combining Lemmas~19, 20 and 21 of~\cite{ChitnisCHPP16}, we obtain the following.
\begin{lemma}[\cite{ChitnisCHPP16}]\label{lem:unbreakable}
Let $q$ and $k$ be nonnegative integers. There is an algorithm with running time $2^{\mathcal{O}(\min\{q,k\}\log (q+k))}\cdot n^3\log n$ that, for a
graph $G$, either finds a separation $(A,B)$ of order at most $k$ such that $|A\setminus B|>q$ and $|B\setminus A|>q$, or correctly reports that $G$ is $((2q+1)q\cdot 2^k,k)$-unbreakable.
\end{lemma}
We conclude this section by noting that the following variant of \ProblemName{Connected Secluded $\Pi$-Subgraph} is \ensuremath{\operatorClassFPT}\xspace when parameterized by $k+t$.
We will rely on this result in the subsequent sections, however we believe that it is also of interest on its own.
\defproblema{\ProblemName{Connected Secluded Colored $\Pi$-Subgraph of Exact Size}}%
{A graph $G$, coloring $c\colon V(G)\rightarrow \mathbb{N}$, a weight function $\omega\colon V(G)\rightarrow\mathbb{Z}_{\geq 0}$ and nonnegative integers $k$, $t$ and $w$.}%
{Decide whether $G$ contains a connected $t$-secluded induced subgraph $H$ such that $(H,c')$, where $c'(v)=c|_{V(H)}(v)$, satisfies $\Pi$, $|V(H)|=k$ and
$\omega(V(H))\geq w$.}
We say that a mapping $c\colon V(G)\rightarrow \mathbb{N}$ is a \emph{coloring} of $G$; note that we do not demand a coloring to be proper. Analogously, we say that $\Pi$ is a \emph{property of colored graphs} if $\Pi$ is a property on pairs $(G,c)$, where $G$ is a graph and $c$ is a coloring.
Notice that if some vertices of the input graph have labels, then we can assign to each label (or a combination of labels if a vertex can have several labels) a specific color and assign some color to unlabeled vertices. Then we can redefine a considered graph property with the conditions imposed by labels as a property of colored graphs.
Observe that we allow zero weights.
We give two algorithms for \ProblemName{Connected Secluded Colored $\Pi$-Subgraph of Exact Size} with different running times. The first algorithm is based on Lemmas~3.1 and 3.2 of Fomin and Villanger~\cite{FominV12}, which we summarize in Lemma \ref{lem:conn-subgr} below. The second algorithm uses Lemma~\ref{lem:derand} by Chitnis et al.~\cite{ChitnisCHPP16}, and we are going to use it when $k\gg t$.
\begin{lemma}[\cite{FominV12}]\label{lem:conn-subgr} Let $G$ be a graph. For every $v\in
V(G)$, and $k,t\geq 0$, the number of connected vertex subsets
$U\subseteq V(G)$ such that
$v \in U$,
$|U| = k$, and
$|N_G(U)|=t$,
is at most $\binom{k+t}{t}$.
Moreover, all such subsets can be enumerated in time $\mathcal{O} (\binom{k+t}{t}\cdot (n+m) \cdot t\cdot (k+t))$.
\end{lemma}
\begin{lemma}[\cite{ChitnisCHPP16}]\label{lem:derand}
Given a set $U$ of size $n$ and integers $0\leq a,b\leq n$, one can construct in time $2^{\mathcal{O}(\min\{a,b\}\log (a+b))}n\log n$ a family $\mathcal{S}$ of at most $2^{\mathcal{O}(\min\{a,b\}\log (a+b))}\log n$ subsets of $U$ such that the following holds: for any sets $A,B\subseteq U$, $A\cap B=\emptyset$, $|A|\leq a$, $|B|\leq b$, there exists a set $S\in \mathcal{S}$ with $A\subseteq S$ and $B\cap S=\emptyset$.
\end{lemma}
\begin{theorem}\label{thm:CSWCS} If property $\Pi$ can be recognized in time $f(n)$, then
\ProblemName{Connected Secluded Colored $\Pi$-Subgraph of Exact Size} can be solved both in time $2^{k+t}\cdot f(k)\cdot n^{\mathcal{O}(1)}$, and in time $2^{\mathcal{O}(\min\{k,t\}\log (k+t))}\cdot f(k)\cdot n^{\mathcal{O}(1)}$.
\end{theorem}
\begin{proof}
Let $(G,c,\omega,k,t,w)$ be an instance of \ProblemName{Connected Secluded Colored $\Pi$-Subgraph of Exact Size}.
First, we use Lemma~\ref{lem:conn-subgr} and in time $2^{k+t}\cdot n^{\mathcal{O}(1)}$ enumerate all connected $U\subseteq V(G)$ with $|U|=k$ and $|N_G(U)|\leq t$. By Lemma~\ref{lem:conn-subgr}, we have at most $\binom{k+t}{t}tn$ sets. For every such a set $U$, we check in time $f(k)+O(k)$ whether the colored induced subgraph $G[U]$ satisfies $\Pi$ and $\omega(U)\geq w$. It is straightforward to see that $(G,c,\omega,k,t,w)$ is a yes-instance if and only if we find $U$ with these properties.
To construct the second algorithm, assume that $(G,c,\omega,k,t,w)$ is a yes-instance. Then there is $U\subseteq V(G)$ such that $U$ is a connected $k$-vertex set such that $|N_G(U)|\leq t$, $\omega(U)\geq w$ and the colored graph $H=G[U]$ satisfies $\Pi$. Using Lemma~\ref{lem:derand}, we can construct in time $2^{\mathcal{O}(\min\{k,t\}\log (k+t))}\cdot n^{\mathcal{O}(1)}$ a family $\mathcal{S}$ of at most $2^{\mathcal{O}(\min\{k,t\}\log (k+t))}\log n$ subsets of $V(G)$ such that the following holds: for any sets $A,B\subseteq V(G)$, $A\cap B=\emptyset$, $|A|\leq k$, $|B|\leq t$, there exists a set $S\in \mathcal{S}$ with $A\subseteq S$ and $B\cap S=\emptyset$. In particular, we have that there is $S\in \mathcal{S}$ such that $U\subseteq S$ and $N_G(U)\cap S=\emptyset$. It implies that $G[U]$ is a component of $G[S]$.
Therefore, $(G,c,\omega,k,t,w)$ is a yes-instance if and only if there is $S\in\mathcal{S}$ such that a component of $G[S]$ is a solution for the instance. We construct the described set $\mathcal{S}$. Then for every $S\in\mathcal{S}$, we consider the components of $G[S]$, and for every component $H$, we verify in time $f(k)+\mathcal{O}(k)$, whether $H$ gives us a solution.
\end{proof}
Theorem~\ref{thm:CSWCS} immediately gives the following corollary.
\begin{corollary}\label{cor:CSWCS}
If $\Pi$ can be recognized in polynomial time, then \ProblemName{Connected Secluded Colored $\Pi$-Subgraph of Exact Size} can be solved both in time $2^{k+t}\cdot n^{\mathcal{O}(1)}$, and in time $2^{\mathcal{O}(\min\{k,t\}\log (k+t))}\cdot n^{\mathcal{O}(1)}$.
\end{corollary}
\section{\ProblemName{Connected Secluded $\Pi$-Subgraph} for properties characterized by finite sets of forbidden induced subgraphs}\label{sec:fpt-forb}
In this section we show that \ProblemName{Connected Secluded $\Pi$-Subgraph} is \ensuremath{\operatorClassFPT}\xspace parameterized by $t$ when $\Pi$ is characterized by a finite set of forbidden induced subgraphs.
We refer to this restriction of our problem as \ProblemName{Connected Secluded $\mathcal{F}$-Free Subgraph}.
Throughout this section, we assume that we are given a fixed finite set $\mathcal{F}$ of graphs.
Recall that to apply the recursive understanding technique introduced by Chitnis et al.~\cite{ChitnisCHPP16}, we should be able to recurse when the input graph contains a separator of bounded size that separates the graph into two sufficiently big parts. To do this, we have to combine partial solutions in both parts. A danger in our case is that a partial solution in one part might contain a subgraph of a graph in $\mathcal{F}$.
We have to avoid creating subgraphs belonging to $\mathcal{F}$ when we combine partial solutions.
To achieve this goal, we need some definitions and auxiliary combinatorial results.
Let $p$ be a nonnegative integer. A pair $(G,x)$, where $G$ is a graph and $x=(x_1,\ldots,x_p)$ is a $p$-tuple of distinct vertices of $G$, is called a \emph{$p$-boundaried graph}
or simply a \emph{boundaried graph}. Respectively, $x=(x_1,\ldots,x_p)$ is a \emph{boundary}. Note that a boundary is an ordered set. Hence, two $p$-boundaried graphs that differ only by the order of the vertices in theirs boundaries are distinct. Observe also that a boundary could be empty. We say that $(G,x)$ is a \emph{properly $p$-boundaried graph} if each component of $G$ has at least one vertex of the boundary. Slightly abusing notation, we may say that $G$ is a ($p$-) boundaried graph assuming that a boundary is given.
Two $p$-boundaried graphs $(G_1,x^{(1)})$ and $(G_2,x^{(2)})$, where $x^{(h)}=(x_1^{(h)},\ldots,x_p^{(h)})$ for $h=1,2$, are \emph{isomorphic} if there is an isomorphism of $G_1$ to $G_2$ that maps each $x_i^{(1)}$ to $x_i^{(2)}$ for $i\in\{1,\ldots,p\}$.
We say that $(G_1,x^{(1)})$ and $(G_2,x^{(2)})$ are \emph{boundary-compatible} if for any distinct $i,j\in\{1,\ldots,p\}$, $x_i^{(1)}x_j^{(1)}\in E(G_1)$ if and only if $x_i^{(2)}x_j^{(2)}\in E(G_2)$.
Let $(G_1,x^{(1)})$ and $(G_2,x^{(2)})$ be boundary-compatible $p$-boundaried graphs and let $x^{(h)}=(x_1^{(h)},\ldots,x_p^{(h)})$ for $h=1,2$. We define the \emph{boundary sum} $(G_1,x^{(1)})\oplus_b(G_2,x^{(2)})$ (or simply $G_1\oplus_b G_2$) as the (non-boundaried) graph obtained by taking disjoint copies of $G_1$ and $G_2$ and identifying $x_i^{(1)}$ and $x_i^{(2)}$ for each $i\in\{1,\ldots,p\}$.
Let $G$ be a graph and let $y=(y_1,\ldots,y_p)$ be a $p$-tuple of vertices of $G$. For a $s$-boundaried graph $(H,x)$ with the boundary $x=(x_1,\ldots,x_s)$ and pairwise distinct $i_1,\ldots,i_s\in \{1,\ldots,p\}$, we say that $H$ is an \emph{induced boundaried subgraph of $G$ with respect to $(y_{i_1},\ldots,y_{i_s})$} if $G$ contains an induced subgraph $H'$ isomorphic to $H$ such that the corresponding isomorphism of $H$ to $H'$ maps $x_j$ to $y_{i_j}$ for $j\in\{1,\ldots,s\}$ and $V(H')\cap \{y_1,\ldots,y_p\}=\{y_{i_1},\ldots,y_{i_s}\}$.
We construct the set of boundaried graphs $\mathcal{F}_b$ as follows. For each $F\in \mathcal{F}$, each separation $(A,B)$ of $F$ and each $p=|A\cap B|$-tuple $x$ of the vertices of $(A\cap B)$, we include $(F[A],x)$ in $\mathcal{F}_b$ unless it already contains an isomorphic boundaried graph.
We say that two properly $p$-boundaried
graphs $(G_1,x^{(1)})$ and $(G_2,x^{(2)})$, where $x^{(h)}=(x_1^{(h)},\ldots,x_p^{(h)})$, are \emph{equivalent (with respect to $\mathcal{F}_b$)} if
\begin{itemize}
\item[(i)] $(G_1,x^{(1)})$ and $(G_2,x^{(2)})$ are boundary-compatible,
\item[(ii)] for any $i,j\in\{1,\ldots,p\}$, $x_i^{(1)}$ and $x_j^{(1)}$ are in the same component of $G_1$ if and only if $x_i^{(2)}$ and $x_j^{(2)}$ are in the same component of $G_2$,
\item[(iii)] for any pairwise distinct $i_1,\ldots,i_s\in\{1,\ldots,p\}$, $G_1$ contains an $s$-boundaried induced subgraph $H\in \mathcal{F}_b$
with respect to the $s$-tuple $(x_{i_1}^{(1)},\ldots,x_{i_s}^{(1)})$ if and only if $H$ is an induced subgraph of $G_2$ with respect to the $s$-tuple $(x_{i_1}^{(2)},\ldots,x_{i_s}^{(2)})$.
\end{itemize}
It is straightforward to verify that the introduced relation is indeed an equivalence relation on the set of properly $p$-boundaried graphs.
The following property of the equivalence with respect to $\mathcal{F}_b$ is crucial for our algorithm.
\begin{lemma}\label{lem:eq}
Let $(G,x)$, $(H_1,y^{(1)})$ and $(H_2,y^{(2)})$ be boundary-compatible $p$-boundaried graphs, $x=(x_1,\ldots,x_p)$ and $y^{(h)}=(y_1^{(h)},\ldots,y_p^{(h)})$ for $h=1,2$.
If $(H_1,y^{(1)})$ and $(H_2,y^{(2)})$ are equivalent with respect to $\mathcal{F}_b$, then $(G,x)\oplus_b (H_1,y^{(1)})$ is $\mathcal{F}$-free if and only if $(G,x)\oplus_b (H_2,y^{(2)})$ is $\mathcal{F}$-free.
\end{lemma}
\begin{proof}
By symmetry, it is sufficient to show that if $G\oplus_b H_1$ is not $\mathcal{F}$-free, then the same hold for $G\oplus_b H_2$.
Suppose that $F$ is an induced subgraph of $G\oplus_b H_1$ isomorphic to a graph of $\mathcal{F}$.
If $V(F)\subseteq V(G)$, then the claim is trivial. Suppose that this is not the case and $V(F)\cap V(H_1)\neq\emptyset$. Recall that $G\oplus_b H_1$ is obtained by identifying each $x_i$ and $y_i^{(1)}$. Denote the identified vertices by $y_1^{(1)},\ldots, y_p^{(1)}$.
Let $F_1=F[V(F)\cap V(H_1)]$ and $F'=F[V(F)\cap V(G)]$; note that $F'$ could be empty.
Let $\{y_{i_1}^{(1)},\ldots,y_{i_s}^{(1)}\}=V(F)\cap \{y_1^{(1)},\ldots, y_p^{(1)}\}$; note that this set could be empty.
Clearly, $(F_1,(y_{i_1}^{(1)},\ldots,y_{i_s}^{(1)}))$ is an $s$-boundaried subgraph of $H_1$ with respect to $(y_1^{(1)},\ldots,y_p^{1})$.
Observe that $\mathcal{F}_b$ contains an $s$-boundaried graph isomorphic to $(F_1,(y_{i_1}^{(1)},\ldots,y_{i_s}^{(1)}))$. Because $H_1$ and $H_2$ are equivalent with respect to $\mathcal{F}_b$, there is an induced $s$-boundaried subgraph
$(F_2,(y_{i_1}^{(2)},\ldots,y_{i_s}^{(2)}))$ of $H_2$ with respect to $(y_1^{(2)},\ldots,y_p^{(2)})$ isomorphic to $(F_1,(y_{i_1}^{(1)},\ldots,y_{i_s}^{(1)}))$. Then $F'\oplus_b F_2$ is isomorphic to $F$, that is, $G\oplus_b H_2$ contains $F$ as an induced subgraph.
\end{proof}
\begin{lemma}\label{lem:eq-check}
It can be checked in time $(|V(G_1)|+|V(G_2)|)^{\mathcal{O}(1)}$ whether two properly $p$-boundaried graphs $G_1$ and $G_2$ are equivalent, and the constant hidden in the $\mathcal{O}$-notation depends on $\mathcal{F}$ only.
\end{lemma}
\begin{proof}
Let $(G_1,x^{(1)})$ and $(G_2,x^{(2)})$, where $x^{(h)}=(x_1^{(h)},\ldots,x_p^{(h)})$ for $h=1,2$, be two boundaried graphs. Clearly, the conditions (i) and (ii) of the definition of the equivalence with respect to $\mathcal{F}$ can be checked in polynomial time.
To verify (iii), let $a=|\mathcal{F}_b|$, $b$ be the maximum size of the boundary of graphs in $\mathcal{F}_b$ and let $c$ be the maximum number of vertices of a graph in $\mathcal{F}_b$.
Clearly, the values of $a$, $b$ and $c$ depend on $\mathcal{F}$ only.
For each $s$-tuple of indices $(i_1,\ldots,i_s)$ where $s\leq b$, we check whether an $s$-boundaried graph $H\in \mathcal{F}_b$ is an $s$-boundaried induced subgraph of $G_1$ and $G_2$ with respect to $(x_{i_1}^{(1)},\ldots,x_{i_s}^{(1)})$ and $(x_{i_1}^{(2)},\ldots,x_{i_s}^{(2)})$ respectively. Since there are at most $bp^b$ $s$-tuples of indices $(i_1,\ldots,i_s)$, at most $a$ graphs
in $\mathcal{F}_b$ and $G_h$ has at most $c|V(G_h)|^c$ induced subgraphs with at most $c$ vertices for $h=1,2$, we have that (iii) can be checked in polynomial time.
\end{proof}
For each nonegative integer $p$, we consider a set $\mathcal{G}_p$ of properly $p$-boundaried graphs obtained by picking a graph with minimum number of vertices in each equivalence class.
We show that the size of $\mathcal{G}_p$ and the size of each graph in the set is upper bounded by some functions of $p$, and this set can be constructed in time that depends only on $p$ assuming that $\mathcal{F}_b$ is fixed.
We need the following observation made by Fomin et al.~\cite{FominGKK16}.
\begin{lemma}[\cite{FominGKK16}]\label{lem:bigdeg}
Let $G$ be a connected graph and $S\subseteq V(G)$. Let $F$ be an inclusion minimal connected induced subgraph of $G$ such that $S\subseteq V(F)$ and $X=\{v\in V(F)|d_F(v)\geq 3\}\cup S$. Then $|X|\leq 4|S|-6$.
\end{lemma}
\begin{lemma}\label{lem:size-eq}
For every positive integer $p$, $|\mathcal{G}_p|=2^{\mathcal{O}(p^2)}$, and for every $H\in \mathcal{G}_p$, $|V(H)|=p^{\mathcal{O}(1)}$, where the constants hidden in the $\mathcal{O}$-notations depend on $\mathcal{F}$ only.
Moreover, for every $p$-boundaried graph $G$, the number of $p$-boundaried graphs in $\mathcal{G}_p$ that are compatible with $G$ is $2^{\mathcal{O}(p\log p)}$.
\end{lemma}
\begin{proof}
Let $a=|\mathcal{F}_b|$, $b$ be the maximum size of the boundary of graphs in $\mathcal{F}_b$ and let $c$ be the maximum number of vertices of a graph in $\mathcal{F}_b$.
Clearly, the values of $a$, $b$ and $c$ depend on $\mathcal{F}$ only.
Assume that the boundary $x=(x_1,\ldots,x_p)$ is fixed.
There are $2^{\binom{p}{2}}$ possibilities to select a set of edges with both end-vertices in $\{x_1,\ldots,x_p\}$.
The number of possible partitions of the boundary into components is the Bell number $B_p$ and $B_p=2^{\mathcal{O}(p\log p)}$.
The number of $s$-tuples of vertices of $\{x_1,\ldots,x_p\}$ that could be boundaries of the copies of $s$-boundaried induced subgraphs $H\in \mathcal{F}_b$ is at most $bp^b$.
Respectively, the number of distinct equivalence classes is at most $2^{\binom{p}{2}}B_p bp^b2^a$, that is, $|\mathcal{G}_p|\leq 2^{\binom{p}{2}}B_pbp^b2^a=2^{\mathcal{O}(p^2)}$.
Let $G$ be a $p$-boundaried graph in one of the classes with minimum number of vertices. Notice that for each $s$-tuple of vertices of $\{x_1,\ldots,x_p\}$, $G$ can contain several copies of the same $H$ as an induced subgraph with respect to this $s$-tuple. In this case we pick one of them and obtain that $G$ contains at most $bp^b2^a$ distinct boundaried induced subgraphs $H\in \mathcal{F}_b$.
Let $S$ be the set of vertices of $G$ that belong to these subgraphs or to the boundary $x$. We have that $|S|\leq bp^b2^ac+p$. Let $X=\{v\in V(G)\mid d_G(v)\geq 3\}\cup S$. By applying Lemma~\ref{lem:bigdeg} to each component of $G$, we obtain that $|X|\leq 4|S|-6$.
By the minimality of $G$, every vertex of degree one is in $S$. Hence, $Y=V(G)\setminus X$ contains only vertices of degree two and, therefore, $G[Y]$ is a union of disjoint paths.
Observe that by the minimality of $G$, each vertex of $Y$ is a cut vertex of the component of $G$ containing it. It implies that $G[Y]$ contains at most $|X|-1$ paths. Suppose that $G[Y]$ contains a path $P$ with at least $2c+2$ vertices. Let $G'$ be the graph obtained from $G$ by the contraction of one edge of $P$. We claim that $G$ and $G'$ are equivalent with respect to $\mathcal{F}_b$. Since the end-vertices of the contracted edges are not the vertices of the boundary, the conditions (i) and (ii) of the definition of the equivalence are fulfilled.
Therefore, it is sufficient to verify (iii). Let
$i_1,\ldots,i_s\in\{1,\ldots,p\}$. Suppose that $G$ contains an $s$-boundaried induced subgraph $H\in \mathcal{F}_b$ with respect to the $s$-tuple $(x_{i_1},\ldots,x_{i_s})$. Then at least two adjacent vertices of $P$ are not included in the copy of $H$ in $G$. It implies that
$H$ is an induced subgraph of $G'$ with respect to $(x_{i_1},\ldots,x_{i_s})$. Suppose that $G'$ contains an $s$-boundaried induced subgraph $H\in \mathcal{F}_b$ with respect to the $s$-tuple $(x_{i_1},\ldots,x_{i_s})$. Then at least one vertex of $P$ is not included in the copy of $H$ in $G'$. Then
$H$ is an induced subgraph of $G'$ with respect to $(x_{i_1},\ldots,x_{i_s})$. But the equivalence of $G$ and $G'$ contradicts the minimality of $G$. We conclude that each path in $G[Y]$ contains at most $2c+1$ vertices. Then the total number of vertices of $G$ is at most $|X|+(|X|-1)(2c+1)=p^{\mathcal{O}(1)}$.
To see that for any $p$-boundaried graph $G$, the number of graphs in $\mathcal{G}_p$ that are compatible with $G$ is $2^{\mathcal{O}(p\log p)}$, notice that if $(H,(x_1,\ldots,x_p))\in \mathcal{G}_p$ and is compatible with $G$, then the adjacency between the vertices of the boundary is defined by $G$.
Then the number of $s$-tuples of vertices of $\{x_1,\ldots,x_p\}$ that could be boundaries of the copies of $s$-boundaried induced subgraphs from $\mathcal{F}_b$ is at most $bp^b$ and for each $s$-tuple we can have at most $2^a$ $s$-boundaried induced subgraphs from $\mathcal{F}_b$.
Taking into account that there $2^{\mathcal{O}(p\log p)}$ possibilities for the verties of the boundary be partitioned according to their inclusions in the components, we obtain the claim.
\end{proof}
Consider now the class $\mathcal{C}$ of $p$-boundaried graphs, such that a $p$-boundaried graph \linebreak $(G,(x_1,\ldots,x_p\})$ $\in\mathcal{C}$ if and only if it holds that for every component $H$ of $G-\{x_1,\ldots,x_p\}$, $N_G(V(H))=\{x_1,\ldots,x_p\}$. We consider our equivalence relation with respect to $\mathcal{F}_b$ on $\mathcal{C}$ and define $\mathcal{G}_p'$ as follows.
In each equivalence class, we select a graph $(G,(x_1,\ldots,x_p\})\in\mathcal{C}$ such that both the number of components of $G-\{x_1,\ldots,x_p\}$ is minimum and the number of vertices of $G$ is minimum subject to the first condition, and then include it in $\mathcal{G}_p'$.
\begin{lemma}\label{lem:size-eq-prime}
For every positive integer $p$, $|\mathcal{G}_p'|=2^{\mathcal{O}(p^2)}$, and for each $H\in \mathcal{G}_p'$, $|V(H)|=p^{\mathcal{O}(1)}$, and the constants hidden in the $\mathcal{O}$-notations depend on $\mathcal{F}$ only.
Moreover, for any $p$-boundaried graph $G$, the number of $p$-boundaried graphs in $\mathcal{G}_p'$ that are compatible with $G$ is $p^{\mathcal{O}(1)}$.
\end{lemma}
\begin{proof}
Let $a=|\mathcal{F}_b|$, $b$ be the maximum size of the boundary of graphs in $\mathcal{F}_b$ and let $c$ be the maximum number of vertices of a graph in $\mathcal{F}_b$.
Clearly, the values of $a$, $b$ and $c$ depend on $\mathcal{F}$ only.
Assume that the boundary $x=(x_1,\ldots,x_p)$ is fixed.
There are $2^{\binom{p}{2}}$ possibilities to select a set of edges with both end-vertices in $\{x_1,\ldots,x_p\}$.
The number of $s$-tuples of vertices of $\{x_1,\ldots,x_p\}$ that could be boundaries of the copies of $s$-boundaried induced subgraphs $H\in \mathcal{F}_b$ is at most $bp^b$.
Respectively, the number of distinct equivalence classes of $\mathcal{C}$ is at most $2^{\binom{p}{2}} bp^b2^a$, that is, $|\mathcal{G}_p'|\leq 2^{\binom{p}{2}}bp^b2^a=2^{\mathcal{O}(p^2)}$.
Let $(G,x)$ be a $p$-boundaried graph in one of the classes such that the number of components of $G-\{x_1,\ldots,x_p\}$ is minimum and the number of vertices of $G$ is minimum subject to (i). Let $Q_1,\ldots,Q_r$ be the components of $G-\{x_1,\ldots,x_p\}$. Let $Q_i'=G[V(Q_i)\cup\{x_1,\ldots,x_p\}]$.
Let $i\in\{1,\ldots,r\}$. Let also $Q_i''$ be the graph obtained from $Q_i$ by the deletion of the edges with both end-vertices in the boundary.
Observe that for each $s$-tuple of vertices of $\{x_1,\ldots,x_p\}$, $Q_i'$ can contain several copies of the same $H$ as an induced subgraph with respect to this $s$-tuple. In this case we pick one of them and obtain that $Q_i'$ contains at most $bp^b2^a$ distinct boundaried induced subgraphs $H\in \mathcal{F}_b$.
Let $S$ be the set of vertices of $Q_i'$ that belong to these subgraphs or to the boundary $x$. We have that $|S|\leq bp^b2^ac+p$. Let $X=\{v\in V(G)\mid d_{Q_i''}(v)\geq 3\}\cup S$. By applying Lemma~\ref{lem:size-eq} to $Q_i''$, we obtain that $|X|\leq 4|S|-6$. Then by the same arguments as in the proof of Lemma~\ref{lem:size-eq}, we obtain that
$Q_i''$ and, therefore, $Q_i'$ has at most $|X|+(|X|-1)(2c+1)=p^{\mathcal{O}(1)}$ vertices. Since $V(Q_i)\subseteq V(Q_i')$, we have that $Q_i$ has $p^{\mathcal{O}(1)}$ vertices.
Suppose that there are $c+1$ pairwise distinct but equivalent $(Q_{j_0}',x),\ldots,(Q_{j_c}',x)$ for $j_0,\ldots,j_c\in\{1,\ldots,r\}$.
Assume that $G$ contains an $s$-boundaried induced subgraph $H\in \mathcal{F}_b$
with respect to an $s$-tuple $(x_{i_1},\ldots,x_{i_s})$ for some $i_1,\ldots,i_s\in\{1,\ldots,p\}$.
Since $|V(H)|\leq c$, there is $h\in\{0,\ldots,c\}$ such that $V(H)\cap V(Q_{j_h})=\emptyset$. Because $(Q_{j_0}',x),\ldots,(Q_{j_c}',x)$ are equivalent, we obtain that $H$ is an $s$-bounderied induced subgraph of $G-V(Q_{j_0})$ contradicting the condition (i) of the choice of $G$.
Therefore, there are at most $c$ pairwise equivalent boundaried graphs in $\{Q_1',\ldots,Q_r'\}$.
We claim that the number of pairwise nonequivalent graphs in $\{Q_1',\ldots,Q_r'\}$ is $p^{\mathcal{O}(1)}$.
Notice that the adjacency between the boundary vertices is defined by $G$.
Then the number of $s$-tuples of vertices of $\{x_1,\ldots,x_p\}$ that could be boundaries of the copies of $s$-boundaried induced subgraphs from $\mathcal{F}_b$ is at most $bp^b$ and for each $s$-tuple we can have at most $2^a$ $s$-boundaried induced subgraphs from $\mathcal{F}_b$. Then the claim follows.
We conclude that $r=cp^{\mathcal{O}(1)}$. Since $|V(Q_i)|=p^{\mathcal{O}(1)}$ for each $i\in\{1,\ldots,r\}$, $|V(G)|=p^{\mathcal{O}(1)}$.
To see that for any $p$-boundaried graph $G$, the number of graphs in $\mathcal{G}_p'$ that are compatible with $G$ is $p^{\mathcal{O}(1)}$, notice that if $(H,(x_1,\ldots,x_p))\in \mathcal{G}_p$ and is compatible with $G$, then the adjacency between the vertices of the boundary is defined by $G$.
Then the number of $s$-tuples of vertices of $\{x_1,\ldots,x_p\}$ that could be boundaries of the copies of $s$-boundaried induced subgraphs from $\mathcal{F}_b$ is at most $bp^b$ and for each $s$-tuple we can have at most $2^a$ $s$-boundaried induced subgraphs from $\mathcal{F}_b$.
\end{proof}
Lemmas~\ref{lem:eq-check}, \ref{lem:size-eq} and \ref{lem:size-eq-prime} immediately imply that $\mathcal{G}_p$ and $\mathcal{G}_p'$
can be constructed by brute force.
\begin{lemma}\label{lem:eq-constr}
The sets $\mathcal{G}_p$ and $\mathcal{G}_p'$
can be constructed in time $2^{p^{\mathcal{O}(1)}}$.
\end{lemma}
To apply the recursive understanding technique, we also have to solve a special variant of \ProblemName{Connected Secluded $\Pi$-Subgraph} tailored for recursion. First, we define the following auxiliary problem for a positive integer $w$.
\defproblema{\ProblemName{Maximum or $w$-Weighted Connected Secluded $\mathcal{F}$-Free Subgraph}}%
{A graph $G$, sets $I,O,B\subseteq V(G)$ such that $I\cap O=\emptyset$ and $I\cap B=\emptyset$, a weight function $\omega\colon V(G)\rightarrow\mathbb{Z}_{\geq 0}$ and a nonnegative integer $t$.}%
{Find a $t$-secluded $\mathcal{F}$-free induced connected subgraph $H$ of $G$ of maximum weight or weight at least $w$
such that $I\subseteq V(H)$, $O\subseteq V(G)\setminus V(H)$ and $N_G(V(H))\subseteq B$ and output $\emptyset$ if such a subgraph does not exist.}
Notice that \ProblemName{Maximum or $w$-Weighted Connected Secluded $\mathcal{F}$-Free Subgraph} is an optimization problem and a \emph{solution} is either an induced subgraph $H$ of maximum weight or of weight at least $w$, or $\emptyset$. Observe also that we allow zero weights for technical reasons.
We recurse if we can separate graphs by a separator of bounded size into two big parts and we use the vertices of the separator to combine partial solutions in both parts. This leads us to the following problem. Let $(G,I,O,B,\omega,t)$ be an instance of \ProblemName{Maximum or $w$-Weighted Connected Secluded $\mathcal{F}$-Free Subgraph} and let $T\subseteq V(G)$ be a set of \emph{border terminals}. We say that an instance $(G',I',O',B',\omega',t')$ is obtained by a \emph{border complementation} if there is a partition $(X,Y,Z)$ of $T$ (some sets could be empty), where $X=\{x_1,\ldots,x_p\}$, such that $Y=\emptyset$ if $X=\emptyset$, $I\cap T\subseteq X$, $O\cap T\subseteq Y\cup Z$ and $Y\subseteq B$, and there is a $p$-boundaried graph $(H,y)\in \mathcal{G}_p$ such that $(H,y)$ and $(G,(x_1,\ldots,x_p))$ are boundary-compatible, and the following holds:
\begin{itemize}
\item[(i)] $G'$ is obtained from $(G,(x_1,\ldots,x_p))\oplus_b (H,y)$ (we keep the notation $X=\{x_1,\ldots,x_p\}$ for the set of vertices obtained by the identification in the boundary sum) by adding edges joining every vertex of $V(H)$ with every vertex of $Y$,
\item[(ii)] $I'= I\cup V(H)$,
\item[(iii)] $O'=O\cup Y\cup Z$,
\item[(iv)] $B'=B\setminus X$,
\item[(v)] $\omega'(v)=\omega(v)$ for $v\in V(G)$ and $\omega'(v)=0$ for $v\in V(H)\setminus X$,
\item[(vi)] $t'\leq t$.
\end{itemize}
We also say that $(G',I',O',B',w',t')$ is a \emph{border complementation of $(G,I,O,B,w,t)$ with respect to $(X=\{x_1,\ldots,x_p\},Y,Z,H)$}.
We say that $(X=\{x_1,\ldots,x_p\},Y,Z,H)$ is \emph{feasible} if it holds that $Y=\emptyset$ if $X=\emptyset$, $I\cap T\subseteq X$, $O\cap T\subseteq Y\cup Z$ and $Y\subseteq B$, and the $p$-boundaried graph $H\in \mathcal{G}_p$ and $(G,(x_1,\ldots,x_p))$ are boundary-compatible.
\defproblema{\ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph}}%
{A graph $G$, sets $I,O,B\subseteq V(G)$ such that $I\cap O=\emptyset$ and $I\cap B=\emptyset$, a weight function $\omega\colon V(G)\rightarrow\mathbb{Z}_{\geq 0}$, a nonnegative integer $t$, and a set $T\subseteq V(G)$ of border terminals of size at most $2t$.}
{Output a solution for each instance $(G',I',O',B',w',t')$ of \ProblemName{Maximum or $w$-Weighted Connected Secluded $\mathcal{F}$-Free Subgraph} that can be obtained from $(G,I,O,B,w,t)$ by a border complementation distinct from the border complementation with respect to $(\emptyset,\emptyset,T,\emptyset)$, and for the border complementation with respect to $(\emptyset,\emptyset,T,\emptyset)$ output a nonempty solution if it has weight at least $w$ and output $\emptyset$ otherwise.}
Two instances $(G_1,I_1,O_1,B_1,\omega_1,t,T)$ and $(G_2,I_2,O_2,B_2,\omega_2,t,T)$ of \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} (note that $t$ and $T$ are the same) are said to be \emph{equivalent} if
\begin{itemize}
\item[(i)] $T\cap I_1=T\cap I_2$, $T\cap O_1=T\cap O_2$ and $T\cap B_1=T\cap B_2$,
\item[(ii)] for the border complementations $(G_1',I_1',O_1',B_1',\omega_1',t')$ and $(G_2',I_2',O_2',B_2',\omega_2',t')$ of the instances $(G_1,I_1,O_1,B_1,\omega_1,t')$ and $(G_2,I_2,O_2,B_2,\omega_2,t')$ respectively of \ProblemName{Maximum or $w$-Weighted Connected Secluded $\mathcal{F}$-Free Subgraph} with respect to every feasible $(X=\{x_1,\ldots,x_p\},Y,Z,H)$ and $t'\leq t$, it holds that
\begin{itemize}
\item[a)] if $(G_1',I_1',O_1',B_1',\omega_1',t')$ has a nonempty solution $R_1$, then $(G_2',I_2',O_2',B_2',\omega_2',t')$ has a nonempty solution $R_2$ with $w_2'(V(R_2))\geq \min\{\omega_1'(V(R_1)),w\}$ and, vice versa,
\item[b)]if $(G_2',I_2',O_2',B_2',\omega_2',t')$ has a nonempty solution $R_2$, then $(G_1',I_1',O_1',B_1',\omega_1',t')$ has a nonempty solution $R_1$ with $\omega_1'(V(R_1))\geq \min\{\omega_2'(V(R_2)),w\}$.
\end{itemize}
\end{itemize}
Strictly speaking, if $(G_1,I_1,O_1,B_1,\omega_1,t,T)$ and $(G_2,I_2,O_2,B_2,\omega_2,t,T)$ are equivalent, then a solution of the first problem is not necessarily a solution of the second. Nevertheless, \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} is an auxiliary problem and in the end we use it to solve an instance $(G,\omega,t,w)$ of \ProblemName{Connected Secluded $\mathcal{F}$-Free Subgraph} by calling the algorithm for \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} for $(G,\emptyset,\emptyset,V(G),\omega,t,\emptyset)$.
Clearly, $(G,\omega,t,w)$ is a yes-instance if and only if a solution for the corresponding instance of \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} contains a connected subgraph $R$ with $\omega(V(R))\geq w$.
It allows us to not distinguish equivalent instances of \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} and their solutions.
\subsection{High connectivity phase}
In this section we solve \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} for $(q,t)$-unbreakable graphs. For this purpose, we use \emph{important separators} defined by Marx in~\cite{Marx06}. Essentially, we follow the terminology given in~\cite{CyganFKLMPPS15}.
Recall that for $X,Y\subseteq V(G)$, a set $S\subseteq V(G)$ is an $(X,Y)$-separator if $G-S$ has no path joining a vertex of $X\setminus S$ with a vertex of $Y\setminus S$.
An $(X,Y)$-separator $S$ is \emph{minimal} if no proper subset of $S$ is an $(X,Y)$-separator. For $X\subseteq V(G)$ and $v\in V(G)$, it is said that $v$ is \emph{reachable} from $X$ if there is an $(x,v)$-path in $G$ with $x\in X$. A minimal $(X,Y)$-separator $S$ can be characterized by the set of vertices reachable from $X\setminus S$ in $G-S$.
\begin{lemma}[\cite{CyganFKLMPPS15}]\label{lem:min-sep}
If $S$ is a minimal $(X,Y)$-separator in $G$, then $S=N_G(R)$ where $R$ is the set of vertices reachable from $X\setminus S$ in $G-S$.
\end{lemma}
Let $X,Y\subseteq V(G)$ for a graph $G$. Let $S\subseteq V(G)$ be an $(X,Y)$-separator and let $R$ be the set of vertices reachable from $X\setminus S$ in $G-S$. It is said that
$S$ is an \emph{important $(X,Y)$-separator} if $S$ is minimal and there is no $(X,Y)$-separator $S'\subseteq V(G)$ with $|S'|\leq |S|$ such that $R\subset R'$ where $R'$ is the set of vertices reachable from $X\setminus S'$ in $G-S'$.
\begin{lemma}[\cite{CyganFKLMPPS15}]\label{lem:imp-sep}
Let $X,Y\subseteq V(G)$ for a graph $G$, let $t$ be a nonnegative integer and let $\mathcal{S}_t$ be the set of all important $(X,Y)$-separators of size at most $t$. Then $|\mathcal{S}_t|\leq 4^t$
and $\mathcal{S}_t$ can be constructed in time $\mathcal{O}(|\mathcal{S}_t|\cdot t^2\cdot (n+m))$.
\end{lemma}
The following lemma shows that we can separately lists all graphs $R$ in a solution of \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} with $|V(R)\cap V(G)|\leq q$ and all graphs $R$ with $|V(G)\setminus V(R)|\leq q+t$.
\begin{lemma}\label{lem:sol-size}
Let $(G,I,O,B,\omega,t,T)$ be an instance of \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} where $G$ is a $(q,t)$-unbreakable graph for a positive integer $q$. Then for each nonempty graph $R$ in a solution of \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph}, either $|V(R)\cap V(G)|\leq q$ or $|V(G)\setminus V(R)|\leq q+t$.
\end{lemma}
\begin{proof}
Let $R$ be a nonempty graph listed in a solution of \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} for an instance $(G',I',O',B',\omega',t')$ \ProblemName{Maximum or $w$-Weighted Connected Secluded $\mathcal{F}$-Free Subgraph}.
Assume that $G'$ is obtained from $(G,(x_1,\ldots,x_p))\oplus_b (H,y)$ for $H\in \mathcal{G}_p$.
Let $U=N_{G'}[V(R)\cap V(G)]$ and $W=V(G)\setminus V(R)$. Clearly $(U,W)$ is a separation of $G$ of order at most $t$. Since $G$ is $(q,t)$-unbreakable, either $|U\setminus W|\leq q$ or $|W\setminus U|\leq q$.
If $|U\setminus W|\leq q$, then $|V(R)\cap V(G)|\leq |U\setminus W|\leq q$.
If $|W\setminus U|\leq q$, then $|V(G)\setminus V(R)|\leq q+t$.
\end{proof}
Now we can prove the following crucial lemma.
\begin{lemma}\label{lem:unbreak}
\ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} for $(q,t)$-unbreakable graphs can be solved in time $2^{(q+t\log(q+t)))}\cdot n^{\mathcal{O}(1)}$ if the sets $\mathcal{G}_p$ for $p\leq 2t$ are given.
\end{lemma}
\begin{proof}
Assume that the sets $\mathcal{G}_p$ for $p\leq 2t$ are given. We consider all possible instances $(G',I',O',B',\omega',t')$ of \ProblemName{Maximum or $w$-Weighted Connected Secluded $\mathcal{F}$-Free Subgraph} obtained from the input instance $(G,I,O,B,w,t)$ as it is explained in the definition of \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph}. To construct each instance, we consider all at most $3^{2t}$ partitions $(X,Y,Z)$ of $T$, where $X=\{x_1,\ldots,x_p\}$, such that $Y=\emptyset$ if $X=\emptyset$, $I\cap T\subseteq X$, $O\cap T\subseteq Y\cup Z$ and $Y\subseteq B$. Then
we consider all $p$-boundaried graph $(H,y)\in \mathcal{G}_p$ such that $(H,y)$ and $(G,(x_1,\ldots,x_p))$ are boundary-compatible. By Lemma~\ref{lem:size-eq}, there are $2^{\mathcal{O}(t\log t)}$ such sets. Consider now a constructed instance $(G',I',O',B',\omega',t')$ and assume that $G'$ is obtained from $(G,(x_1,\ldots,x_p))\oplus_b (H,y)$ for $(H,y)\in\mathcal{G}_p$.
We find a $t$-secluded $\mathcal{F}$-free induced connected subgraph $R$ of $G'$ of maximum weight
such that $I'\subseteq V(R)$, $O'\subseteq V(G')\setminus V(R)$ and $N_{G'}(V(R))\subseteq B'$ if such a subgraph exists.
By Lemma~\ref{lem:sol-size}, either $|V(R)\cap V(G)|\leq q$ or $|V(G)\setminus V(R)|\leq q+t$.
First, we find
a $t$-secluded $\mathcal{F}$-free induced connected subgraph $R$ of $G'$ with $|V(R)\cap V(G)|\leq q$ of maximum weight
such that $I'\subseteq V(R)$, $O'\subseteq V(G')\setminus V(R)$ and $N_{G'}(V(R))\subseteq B'$.
If $|V(R)\cap V(G)|\leq q$, then $|V(R)|\leq |V(H)|+|V(R)\cap V(G)|$. By Lemma~\ref{lem:size-eq}, $|V(H)|=t^{c}$ for some constant $c$. It implies that $|V(R)|\leq t^c+q$. To find $R$, we consider all $k\leq t^c+q$ and find a $t$-secluded $\mathcal{F}$-free induced connected subgraph $R$ of $G'$ of maximum weight such that $I'\subseteq V(R)$, $O'\subseteq V(G')\setminus V(R)$, $N_{G'}(V(R))\subseteq B'$ and $|V(R)|=k$. By Corollary~\ref{cor:CSWCS},
it can be done in time $2^{\mathcal{O}(\min\{t^c+q,t\}\log (t^c+q+t))}\cdot n^{\mathcal{O}(1)}$.
Now we find a $t$-secluded $\mathcal{F}$-free induced connected subgraph $R$ of $G'$ with $|V(G)\setminus V(R)|\leq q+t$ of maximum weight
such that $I'\subseteq V(R)$, $O'\subseteq V(G')\setminus V(R)$ and $N_{G'}(V(R))\subseteq B'$.
Because $|V(R)\cap V(G)|\leq q+t$,
there is a set $O'\subseteq S\subseteq V(G)\setminus V(R)$ such that $|S|\leq q+t$ and $G'-S$ is $\mathcal{F}$-free.
We list all such sets $S$
using the standard branching algorithm for this problem (see, e.g.,~\cite{CyganFKLMPPS15}). The main idea of the algorithm is that if $G'$ has an induced subgraph $F$ isomorphic to a graph of $\mathcal{F}$, then at least one vertex of $F$ should be in $S$. Initially we set $S=O$ and set a branching parameter $h=q+t-|O|$. If $h<0$, we stop.
We check whether $G'-S$ has an induced subgraph $F$ isomorphic to a graph of $\mathcal{F}$. If we have no such graph, we return $S$. If $V(F)\subseteq I'$, then we stop.
Otherwise, we branch on the vertices of $F$. For each $v\in V(F)\setminus I'$, we set $S=S\cup \{v\}$ and set $h=h-1$ and recurse. It is straightforward to verify the correctness of the algorithm and see that it runs in in time $2^{\mathcal{O}(q+t)}\cdot n^{(1)}$, because $\mathcal{F}$ is fixed and each graph from this set has a constant size. If the algorithm fails to output any set $S$, then we conclude that $(G',I',O',B',\omega',t')$ has no solution $R$ with $|V(R)\setminus V(G)|\leq q$. From now on we assume that this is not the case.
For each $S$, we set $O'=O'\cup S$ and find a solution for the modified instance $(G',I',O',B',\omega',t')$. Then we choose a solution of maximum size (if exist).
If $I'=\emptyset$, we guess a vertex $u\in V(G')\setminus O'$ that is included in a solution. We set $I'=\{u\}$ and $B'=B'\setminus \{u\}$ and solve the modified instance $(G',I',O',B',w',t')$. Then we choose a solution of maximum size for all guesses of $u$. From now on we have $I'\neq \emptyset$.
We apply a series of reduction rules for $(G',I',O',B',\omega',t')$. Let $h=q+t$.
\begin{reduction}\label{Ared:one}
If $G'$ is disconnected and has vertices of $I'$ in distinct components, then return the answer no and stop. Otherwise, let $Q$ be a component of $G'$ containing $I'$ and set $G'=Q$, $B'=B'\cap V(Q)$, $O'=O'\cap V(Q)$ and $h=h-|V(G)\setminus V(Q)|$. If $h<0$, then return no and stop.
\end{reduction}
It is straightforward to see that the rule is safe, because a solution is a connected graph. Notice that from now on we can assume that $G'$ is connected. If $O'=\emptyset$, $G'$ is a solution and we get the next rule.
\begin{reduction}\label{Ared:two}
If $O'=\emptyset$, then return $G'$.
\end{reduction}
From now on we assume that $O'\neq\emptyset$.
Let $Q$ be a component of $G'-B'$. Notice that for any solution $R$, either $V(Q)\subseteq V(R)$ or $V(Q)\cap V(R)=\emptyset$, because $N_{G'}(V(R))\subseteq B'$. Moreover, if
$V(Q)\cap V(R)=\emptyset$, then $N_{G'}[V(Q)]\cap V(R)=\emptyset$.
This leads to the following rule.
\begin{reduction}\label{Ared:three}
For a component $Q$ of $G'-B'$ do the following in the given order:
\begin{itemize}
\item if $V(Q)\cap I'\neq\emptyset$ and $V(Q)\cap O'\neq\emptyset$, then return no and stop,
\item if $V(Q)\cap I'\neq\emptyset$, then set $I'=I'\cup V(Q)$,
\item if $V(Q)\cap O'\neq\emptyset$, then set $O'=O'\cup N_{G'}[V(Q)]$.
\end{itemize}
\end{reduction}
The rule is applied for each component $Q$ exactly once.
Now our aim is find all inclusion maximal induced subgraphs $R$ of $G'$ such that $I'\subseteq V(R)$, $O'\cap V(R)=\emptyset$, $N_{G'}(V(R))\subseteq B'$, $|N_{G'}(V(R))|\leq t'$ and all the vertices of $R$ are reachable from $I'$. Then, by maximality, a solution is such a subgraph $R$ that is connected and, subject to connectivity, has a maximum weight. We doing it using important $(N_{G'}[I'],O')$-separators. The obstacle is the condition that $N_{G'}(V(R))\subseteq B'$. Let $Q$ be a component of $G'-O'$. By Reduction Rule~\ref{Ared:three}, we have that exactly one of the following holds: either (i) $V(Q)\subseteq I'$ or (ii) $V(Q)\subseteq O'$ or (iii) $V(Q)\cap I'=V(Q)\cap O'=\emptyset$. To ensure that $N_{G'}(V(R))\subseteq B'$, we have to insure that if (iii) is fulfilled, then it holds that either $V(Q)\subseteq V(R)$ or $V(Q)\cap V(R)=\emptyset$. To do it, we construct the auxiliary graph $G''$ as follows. For each $v\in V(G')\setminus (I'\cup O'\cup B')$, we replace $v$ by $t+1$ true twin vertices
$v_0,\ldots,v_t$ that are adjacent to the same vertices as $v$ in $G$ or to the corresponding true twins obtained from the neighbors of $v$. For an induced subgraph $R$ of $G'$, we say that the induced subgraph $R'$ of $G''$ is an \emph{image} of $R$ if $R'$ is obtained by the same replacement of the vertices $v\in V(R)\setminus (I'\cup O'\cup B')$ by $t+1$ twins. Respectively, we say that $R$ is a \emph{preimage} of $R'$.
We claim that if $R'$ is an induced subgraph of $G''$ such that $I'\subseteq V(R')$, $O'\cap V(R')=\emptyset$,
$|N_{G''}(V(R'))|\leq t'$ and all the vertices of $R'$ are reachable from $I'$, then $R'$ has a preimage $R$ and $N_{G'}(V(R))=N_{G''}(V(R'))\subseteq B'$.
To prove the claim, consider an inclusion maximal induced subgraph $R'$ of $G''$ such that $I'\subseteq V(R')$, $O'\cap V(R')=\emptyset$,
$|N_{G''}(V(R'))|\leq t'$ and all the vertices of $R'$ are reachable from $I'$. Let $v'\in N_{G''}(R'')$ and assume that $v'\notin B'$. Clearly, $v'\notin I'$. Notice that $v'\notin O'$,
because by Reduction Rule~\ref{Ared:three}, we have that for any $w\in O'\setminus B'$, $N_{G'}[w]\subseteq O'$. Since $v'\notin B'\cup I'\cup O'$, $v'\in\{v_0,\ldots,v_t\}$ for $t+1$ true twins constructed for some vertex $v\in V(G')$. Because $|N_{G''}(V(R'))|\leq t'$, there is $i\in\{0,\ldots,t\}$, such that $v_i\notin N_{G''}(R'')$.
As $v_i$ and $v'$ are twins, $v_i\in V(R'')$. Let $R''=G''[V(R')\cup \{v_0,\ldots,v_t\}]$. We obtain that $I'\subseteq V(R')$, $O'\cap V(R')=\emptyset$,
$|N_{G''}(V(R'))|\leq t'$ and all the vertices of $R'$ are reachable from $I'$, but $V(R')\subset V(R'')$ contradicting maximality. Hence, $N_{G''}(V(R'))\subseteq B'$. Then $R'$ has a preimage $R$ and $N_{G'}(V(R))=N_{G''}(V(R'))\subseteq B'$.
Using the claim, we conclude that to find all inclusion maximal induced subgraphs $R$ of $G'$ such that $I'\subseteq V(R)$, $O'\cap V(R)=\emptyset$, $N_{G'}(V(R))\subseteq B'$, $|N_{G'}(V(R))|\leq t'$ and all the vertices of $R$ are reachable from $I'$, we should list inclusion maximal induced subgraphs $R'$ of $G''$ such that $I'\subseteq V(R')$, $O'\cap V(R')=\emptyset$,
$|N_{G''}(V(R'))|\leq t'$ and all the vertices of $R'$ are reachable from $I'$, and then take preimages of the graphs $R'$.
To find the maximal induced subgraphs $R'$ of $G''$ such that $I'\subseteq V(R')$, $O'\cap V(R')=\emptyset$,
$|N_{G''}(V(R'))|\leq t'$ and all the vertices of $R'$ are reachable from $I'$, we use Lemma~\ref{lem:imp-sep}.
In time $4^t\cdot n^{\mathcal{O}(1)}$ we construct the set $\mathcal{S}_{t'}$ of all important $(N_{G''}[I'],O')$-separators of size at most $t'$ in $G''$. Then for each $S\in \mathcal{S}_{t'}$, we find $R'$ that is the union of the components of $G''-S$ containing the vertices of $I'$.
Since Reduction Rules~\ref{Ared:one}--\ref{Ared:three} can be applied in polynomial time and $G''$ can be constructed in polynomial time, we have that the total running time is
$2^{\mathcal{O}(t+q)}\cdot n^{\mathcal{O}(1)}$.
Now we compare the two subgraphs $R$ that we found for the cases $|V(R)\cap V(G)|\leq q$ and $|V(G)\setminus V(R)|\leq q+t$ and output the subgraph of maximum weight or the empty set if we failed to find these subgraph. Taking into account the time used to construct the instances $(G',I',O',B',\omega',t')$, we obtain that the total running time is $2^{\mathcal{O}(q+t\log(q+t)))}\cdot n^{\mathcal{O}(1)}$
\end{proof}
\subsection{The \ensuremath{\operatorClassFPT}\xspace algorithm for \ProblemName{Connected Secluded $\mathcal{F}$-Free Subgraph}}\label{Asec:forb}
In this section we construct an \ensuremath{\operatorClassFPT}\xspace algorithm for \ProblemName{Connected Secluded $\mathcal{F}$-Free Subgraph} parameterized by $t$. We do this by solving \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} in \ensuremath{\operatorClassFPT}\xspace-time for general case.
\begin{lemma}\label{lem:bordforb}
\ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} can be solved in time $2^{2^{2^{\mathcal{O}(t\log t)}}}\cdot n^{\mathcal{O}(1)}$.
\end{lemma}
\begin{proof}
Given $\mathcal{F}$, we construct the set $\mathcal{F}_b$. Then we use Lemma~\ref{lem:eq-constr} to construct the sets $\mathcal{G}_p$ for $p\in \{0,\ldots,t\}$ in time $2^{t^{\mathcal{O}(1)}}$.
By Lemma~\ref{lem:size-eq}, there is a constant $c$ that depends only on $\mathcal{F}$ such that for every nonnegative $p$ and
for any $p$-boundaried graph $G$, there are at most $2^{c p\log p}$ $p$-boundaried graphs in $\mathcal{G}_p$ that are compatible with $G$ and
there are at most $p^{c}$ $p$-boundaried graphs in $\mathcal{G}_p'$ that are compatible with $G$.
We define
\begin{equation}\label{Aeq:q}
q=2^{((t+1)t3^{2t}2^{c2t\log (2t)}+2t)}\cdot 2((t+1)t3^{2t}2^{c2t\log (2t)}+2t)^c t^c+(t+1)t3^{2t}2^{c2t\log (2t)}+2t.
\end{equation}
The choice of $q$ will become clear later in the proof.
Notice that $q=2^{2^{\mathcal{O}(t\log t)}}$.
Consider an instance $(G,I,O,B,\omega,t,T)$ of \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph}.
We use the algorithm from Lemma~\ref{lem:unbreakable} for $G$. This algorithm in time $2^{2^{\mathcal{O}(t\log t)}}\cdot n^{\mathcal{O}(1)}$ either finds a separation $(U,W)$ of $G$ of order at most $t$ such that $|U\setminus W|>q$ and $|W\setminus U|>q$ or correctly reports that $G$ is $((2q+1)q\cdot 2^t,t)$-unbreakable. In the latter case we solve the problem using Lemma~\ref{lem:unbreak} in time $2^{2^{2^{\mathcal{O}(t\log t)}}}\cdot n^{\mathcal{O}(1)}$.
Assume from now that there is a separation $(U,W)$ of order at most $t$ such that $|U\setminus W|>q$ and $|W\setminus U|>q$.
Recall that $|T|\leq 2t$. Then $|T\cap (U\setminus W)|\leq t$ or $|T\cap (W\setminus U)|\leq t$. Assume without loss of generality that $|T\cap (W\setminus U)|\leq t$. Let $\tilde{G}=G[W]$, $\tilde{I}=I\cap W$, $\tilde{O}=O\cap W$, $\tilde{\omega}$ is the restriction of $\omega$ to $W$, and define $\tilde{T}=(T\cap W)\cup (U\cap W)$. Since $|U\cap W|\leq t$, $|\tilde{T}|\leq 2t$.
If $|W|\leq (2q+1)q\cdot 2^t$, then we solve \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} for the instance $(\tilde{G},\tilde{I},\tilde{O},\tilde{B},\tilde{\omega},t,\tilde{T})$ by brute force in time $2^{2^{2^{\mathcal{O}(t\log t)}}}$ trying all possible subset of $W$ at most $t+1$ values of $0\leq t'\leq t$. Otherwise, we solve $(\tilde{G},\tilde{I},\tilde{O},\tilde{B},\tilde{\omega},t,\tilde{T})$ recursively. Let $\mathcal{R}$ be the set of nonempty induced subgraphs $R$ that are included in the obtained solution for $(\tilde{G},\tilde{I},\tilde{O},\tilde{B},\tilde{\omega},t,\tilde{T})$.
For $R\in \mathcal{R}$, define $S_R$ to be the set of vertices of $W\setminus V(R)$ that are adjacent to the vertices of $R$ in the graph obtained by the border complementation for which $R$ is a solution of the corresponding instance of \ProblemName{Maximum or $w$-Weighted Connected Secluded $\mathcal{F}$-Free Subgraph}. Note that $|S_R|\leq t$.
If $\mathcal{R}\neq\emptyset$, then let $S=\tilde{T}\cup_{R\in\mathcal{R}}S_R$, and $S=\tilde{T}$ if $\mathcal{R}=\emptyset$.
Since \ProblemName{Maximum or $w$-Weighted Connected Secluded $\mathcal{F}$-Free Subgraph} is solved for at most $t+1$ of values of $t'\leq t$, at most $3^{2t}$ three-partitions $(X,Y,Z)$ of $\tilde{T}$ and at most $2^{c2t\log (2t)}$ choices of a $p$-boundaried graph $H\in \mathcal{F}_b$ for $p=|X|$, we have that $|\mathcal{R}|\leq (t+1)3^{2t}2^{c2t\log (2t)}$. Taking into account that $|T'|\leq 2t$,
\begin{equation}\label{Aeq:s}
|S|\leq (t+1)t3^{2t}2^{c2t\log (2t)}+2t.
\end{equation}
Let $\hat{B}=(B\cap U)\cup (B\cap S)$. We claim that the instances $(G,I,O,B,\omega,t,T)$ and $(G,I,O,\hat{B},\omega,t,T)$ of \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} are equivalent.
Recall that we have to show that
\begin{itemize}
\item[(i)] $T\cap B=T\cap \hat{B}$,
\item[(ii)] for the border complementations $(G',I',O',B',\omega',t')$ and $(G',I',O',\hat{B}',\omega',t')$ of the instances $(G,I,O,B,\omega,t')$ and $(G,I,O,\hat{B},\omega,t')$ respectively of \ProblemName{Maximum or $w$-Weighted Connected Secluded $\mathcal{F}$-Free Subgraph} with respect to every feasible $(X=\{x_1,\ldots,x_p\}$, $Y,Z,H)$ and $t'\leq t$, it holds that if $(G',I',O',B',\omega',t')$ has a nonempty solution $R_1$, then $(G',I',O',\hat{B}',\omega',t')$ has a nonempty solution $R_2$ with $\omega'(V(R_2))\geq \min\{\omega'(V(R_1)),w\}$ and, vice versa, if $(G',I',O',\hat{B}',\omega',t')$ has a nonempty solution $R_2$, then $(G',I',O',B',\omega',t')$ has a nonempty solution $R_1$ with $\omega'(V(R_1))\geq \min\{\omega'(V(R_2)),w\}$.
\end{itemize}
The condition (i) holds by the definition of $\hat{B}$. Because $\hat{B}\subseteq B$, we immediately obtain that if $(G',I',O',\hat{B}',\omega',t')$ has a nonempty solution $R_2$, then $(G',I',O',B',\omega',t')$ has a nonempty solution $R_1$ with $\omega'(V(R_1))\geq \min\{\omega'(V(R_2)),w\}$.
It remains to prove that for a border complementation $(G',I',O',B',\omega',t')$ and $(G',I',O',\hat{B}',\omega',t')$ of $(G,I,O,B,\omega,t')$ and $(G,I,O,\hat{B},\omega,t')$ respectively of \ProblemName{Maximum or $w$-Weighted Connected Secluded $\mathcal{F}$-Free Subgraph} with respect to a feasible $(X=\{x_1,\ldots,x_p\},Y,Z,H)$ and $t'\leq t$, it holds that if $(G',I',O',B',\omega',t')$ has a nonempty solution $R_1$, then $(G',I',O',\hat{B}',\omega',t')$ has a nonempty solution $R_2$ with $\omega'(V(R_2))\geq \min\{\omega'(V(R_1)),w\}$.
If $V(R_1)\cap V(G)\subseteq U\setminus W$, then $N_{G'}(V(R_1))\subseteq \hat{B}'$. Therefore, for a solution $R_2$ of $(G',I',O',\hat{B}',\omega',t')$, $\omega'(V(R_2))\geq \min\{\omega'(V(R_1)),w\}$. Assume that $V(R_1)\cap W\neq \emptyset$.
Let $\tilde{X}=\tilde{T}\cap(V(R_1)\cap W)=\{y_1,\ldots,y_r\}$, let $\tilde{Y}$ be the set of vertices of $\tilde{T}\setminus V(R_1)$ that are adjacent to vertices of $R_1$
outside $W\setminus U$ and $\tilde{Z}=\tilde{T}\setminus (\tilde{X}\cup \tilde{Y})$.
Let $(R_1',(y_1,\ldots,y_r))$ be the $r$-bounderied graph obtained from $R_1$ by the deletion of the vertices of $W\setminus U$ (note that the graph could be empty).
We have that $\mathcal{G}_r$ contains an $r$-boundaried graph $\tilde{H}$ that is equivalent to $(R_1',(y_1,\ldots,y_r))$.
Recall that we have a solution of \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} for $(\tilde{G},\tilde{I},\tilde{O},\tilde{B},\tilde{\omega},t,\tilde{T})$. In particular, we have a solution $\tilde{R}\in \mathcal{R}$ for the instance
$(\tilde{G},\tilde{I},\tilde{O},\tilde{B},\tilde{\omega},\tilde{t})$ of \ProblemName{Maximum or $w$-Weighted Connected Secluded $\mathcal{F}$-Free Subgraph} obtained by the border complementation with respect to $(\tilde{X}=(y_1,\ldots,y_r),\tilde{Y},\tilde{Z},\tilde{H})$, where $\tilde{t}$ is the number of neighbors of $R_1$ in $W$.
Recall also that the neighbors of the vertices of $\tilde{R}$ are in $S$.
Denote by $(\tilde{R}',(y_1,\ldots,y_r))$ the $r$-bounderied subgraph obtained from $\tilde{R}$ by the deletion of the vertices that are outside of $W$.
By Lemma~\ref{lem:eq}, $R_2=(R_1',(y_1,\ldots,y_r))\oplus_b (\tilde{R},(y_1,\ldots,y_r))$ is $\mathcal{F}$-free. Observe also that $\omega'(R_2)\geq \min\{\omega'(R_1),w\}$. It implies that
$(G',I',O',\hat{B}',\omega',t')$ has a nonempty solution $R_2$ with $\omega'(V(R_2))\geq \min\{\omega'(V(R_1)),w\}$.
Since, $(G,I,O,B,\omega,t,T)$ and $(G,I,O,\hat{B},\omega,t,T)$ of \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} are equivalent, we can consider $(G,I,O,\hat{B},\omega,t,T)$. Now we apply some reduction rules that produce equivalent instances of \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} or report that we have no solution. The ultimate aim of these rules is to reduce the size of $G$.
Let $Q$ be a component of $G[W]-S$. Notice that for any nonempty graph $R$ in a solution of $(G,I,O,\hat{B},w,t,T)$, either $V(Q)\subseteq V(R)$ or $V(Q)\cap V(R)=\emptyset$, because $N_{G[W]}(V(R))\subseteq S$. Moreover, if
$V(Q)\cap V(R)=\emptyset$, then $N_{G[W]}[V(Q)]\cap V(R)=\emptyset$.
Notice also that if $v\in N_{G[W]}(V(Q))$ is a vertex of $R$, then $V(Q)\subseteq V(R)$.
These observation are crucial for the following reduction rules.
\begin{reduction}\label{Ared:one-rec}
For a component $Q$ of $G[W]-S$ do the following in the given order:
\begin{itemize}
\item if $N_{G[W]}[V(Q)]\cap I\neq\emptyset$ and $V(Q)\cap O\neq\emptyset$, then return $\emptyset$ and stop,
\item if $N_{G[W]}[V(Q)]\cap I\neq\emptyset$, then set $I=I\cup V(Q)$,
\item if $V(Q)\cap O\neq\emptyset$, then set $O=O\cup N_{G[W]}[V(Q)]$.
\end{itemize}
\end{reduction}
The rule is applied to each component $Q$ exactly once. Notice that after application of the rule, for every component $Q$ of $G[W]-S$, we have that either $V(Q)\subseteq I$ or $V(Q)\subseteq O$ or $V(Q)\cap (I\cup O\cup\hat{B})=\emptyset$.
Suppose that $Q_1$ and $Q_2$ are components of $G[W]-S$ such that $N_{G[W]}(V(Q_1))=N_{G[W]}(V(Q_2))$ and $|N_{G[W]}(V(Q_1))|=|N_{G[W]}(V(Q_2))|>t$. Then if $V(Q_1)\subseteq V(R)$ for a
nonempty graph $R$ in a solution of $(G,I,O,\hat{B},\omega,t,T)$, then at least one vertex of $N_{G[W]}(V(Q_1))$ is in $R$ as $R$ have at most $t$ neighbors outside $R$. This gives the next rule.
\begin{reduction}\label{Ared:two-rec}
For components $Q_1$ and $Q_2$ of $G[W]-S$ such that $N_{G[W]}(V(Q_1))=N_{G[W]}(V(Q_2))$ and $|N_{G[W]}(V(Q_1))|=|N_{G[W]}(V(Q_2))|>t$ do the following in the given order:
\begin{itemize}
\item if $(V(Q_1)\cup V(Q_2))\cap I\neq\emptyset$ and $(V(Q_1)\cup V(Q_2))\cap O\neq\emptyset$, then return $\emptyset$ and stop,
\item if $(V(Q_1)\cup V(Q_2))\cap I\neq\emptyset$, then set $I=I\cup (V(Q_1)\cup V(Q_2))$,
\item if $(V(Q_1)\cup V(Q_2))\cap O\neq\emptyset$, then set $O=O\cup N_{G[W]}[V(Q_1)\cup V(Q_2)]$.
\end{itemize}
\end{reduction}
We apply the rule for all pairs of components $Q_1$ and $Q_2$ with $N_{G[W]}(V(Q_1))=N_{G[W]}(V(Q_2))$ and $|N_{G[W]}(V(Q_1))|=|N_{G[W]}(V(Q_2))|>t$, and for each pair the rule is applied once.
If $V(Q)\subseteq O$ for a component $Q$ of $G[W]-S$, then $N_{G[W]}(V(Q))\subseteq O$. It immediately implies that the vertices of $Q$ are irrelevant and can be removed.
\begin{reduction}\label{Ared:del-rec}
If there is a component $Q$ of $G[W]-S$ such that $N_{G[W]}(V(Q))\subseteq O$, then
set $G=G-V(Q)$, $W=W\setminus V(Q)$ and $O=O\setminus V(Q)$.
\end{reduction}
Notice that for each component $Q$, we have now that either $V(Q)\subseteq I$ or $V(Q)\subseteq W\setminus (I\cup O\cup\hat{B})$.
To define the remaining rules, we construct
the sets $\mathcal{G}_p'$ for $p\in \{0,\ldots,|S|\}$ in time $2^{2^{\mathcal{O}(t\log t)}}$ using Lemma~\ref{lem:eq-constr}.
Let $Q$ be a component of $G[W]-S$ and let $N_{G[W]}(V(Q))=\{x_1,\ldots,x_p\}$. Let $G'$ be the graph obtained from $G$ by the deletion of the vertices of $V(Q)$ and let $x=(x_1,\ldots,x_p)$. Let $(H,y)$ be a connected $p$-boundaried graph of the same weight as $G[N_{G[W]}[V(Q)]]$. Then by Lemma~\ref{lem:eq}, we have that the instance of \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} obtained from
$(G,I,O,\hat{B},\omega,t,T)$ by the replacement of $G$ by $(G',x)\oplus_b (H,y)$ is equivalent to $(G,I,O,\hat{B},\omega,t,T)$. We use it in the remaining reduction rules.
Suppose again that $Q_1$ and $Q_2$ are components of $G[W]-S$ such that $N_{G[W]}(V(Q_1))=N_{G[W]}(V(Q_2))$ and $|N_{G[W]}(V(Q_1))|=|N_{G[W]}(V(Q_2))|>t$.
Then, as we already noticed, if $V(Q_1)\cup V(Q_2)\subseteq V(R)$ for a
nonempty graph $R$ in a solution of $(G,I,O,\hat{B},\omega,t,T)$, then at least one vertex of $N_{G[W]}(V(Q_1))$ is in $R$.
It means that if we are constructing a solution $R$, then the restriction of the size of the neighborhood of $R$ ensures the connectivity between $Q_1$ and $Q_2$ if we decide to include these components in $R$.
Together with Lemma~\ref{lem:eq} this shows that the following rule is safe.
\begin{reduction}\label{Ared:three-rec}
Let $L=\{x_1,\ldots,x_p\}\subseteq S$, $p>t$, and let $x=(x_1,\ldots,x_p)$. Let also $Q_1,\ldots,Q_r$,$r\geq 1$, be the components of $G[W]-S$ with $N_{G[W]}(V(Q_i))=L$ for all $i\in\{1,\ldots,r\}$.
Let $Q=G[\cup_{i=1}^rN_{G[W]}[V(Q_i)]]$ and $w'=\sum_{i=1}^r\omega(V(Q_i))$. Find a $p$-boundaried graph $(H,y)\in\mathcal{G}_p'$
that is equivalent to $(Q,x)$ with respect to $\mathcal{F}_b$ and denote by $A$ the set of nonboundary vertices of $H$.
Then do the following.
\begin{itemize}
\item Delete the vertices of $V(Q_1),\ldots,V(Q_r)$ from $G$ and denote the obtained graph $G'$.
\item Set $G=(G',x)\oplus_b (H,y)$ and $W=(W\setminus\cup_{i=1}^rV(Q_i))\cup A$.
\item Select arbitrarily $u\in A$ and modify $\omega$ as follows:
\begin{itemize}
\item keep the weight same for every $v\in V(G')$ including the boundary vertices $x_1,\ldots,x_p$,
\item set $\omega(v)=0$ for $v\in A\setminus\{u\}$,
\item set $\omega(u)=w'$.
\end{itemize}
\item If $V(Q_1)\subseteq I$, then set $I=I\setminus(\cup_{i=1}^rV(Q_i))\cup A$.
\end{itemize}
\end{reduction}
To see the safeness of the rule, observe additionally that the neighborhood of each component of $H[A]$ is $L$, because $(H,y)\in \mathcal{G}_p'$.
The rule is applied exactly once for each inclusion maximal sets of components $\{Q_1,\ldots,Q_r\}$ having the same neighborhood of size at least $t+1$.
We cannot apply this trick if we have several components $Q_1,\ldots,Q_r$ of $G[W]-S$ with the same neighborhood $N_{G[W]}V(Q_i)$ if $|N_{G[W]}V(Q_i)|\leq t$. Now it can happen that there are $i,j\in \{1,\ldots,r\}$ such that $V(Q_i)\subseteq V(R)$ and $N_{G[W]}[V(Q_j)]\cap V(R)=\emptyset$ for $R$ in a solution of $(G,I,O,\hat{B},\omega,t,T)$. But if $N_{G[W]}[V(Q_j)]\cap V(R)=\emptyset$ , then by the connectivity of $R$ and the fact that $G[W]-S$ does not contain border terminals, we have that $R=Q_i$. Notice that $I=\emptyset$ in this case and, in particular, it means that $R$ is a solution for an instance of \ProblemName{Maximum or $w$-Weighted Connected Secluded $\mathcal{F}$-Free Subgraph} obtained by the border complementation with respect to $(\emptyset,\emptyset,T,\emptyset)$. Recall that we output $R$ in this case only if its weight is at least $w$.
Still, we can modify Reduction Rule~\ref{Ared:three-rec} for the case when there are components $Q$ of $G[W]-S$ such that $V(Q)\subseteq I$. Notice that if there are components
$Q_0,\ldots,Q_r$ of $G[W]-S$ with the same neighborhood and $V(Q_0)\subseteq I$, then for any nonempty $R$ in a solution of $(G,I,O,\hat{B},\omega,t,T)$, either $R=Q_0$ or $\cup_{i=0}^rV(Q_i)\subseteq V(R)$. Applying Lemma~\ref{lem:eq} , we obtain that the following rule is safe.
\begin{reduction}\label{Ared:four-rec}
Let $L=\{x_1,\ldots,x_p\}\subseteq S$, $p\leq t$, and let $x=(x_1,\ldots,x_p)$. Let also $Q_0,\ldots,Q_r$,$r\geq 0$, be the components of $G[W]-S$ with $N_{G[W]}(V(Q_i))=L$ for all $i\in\{0,\ldots,r\}$ such that $V(Q_0)\subseteq I$. Let $Q=G[\cup_{i=1}^rN_{G[W]}[V(Q_i)]]$ and $w'=\sum_{i=1}^r\omega(V(Q_i))$.
Find a $p$-boundaried graph $(H_0,y)\in\mathcal{G}_p'$
that is equivalent to $(Q_0,x)$ with respect to $\mathcal{F}_b$ and denote by $A_0$ the set of nonboundary vertices of $H_0$, and
find a $p$-boundaried graph $(H,y)\in\mathcal{G}_p'$
that is equivalent to $(Q,x)$ with respect to $\mathcal{F}_b$ and denote by $A$ the set of nonboundary vertices of $H$.
Then do the following.
\begin{itemize}
\item Delete the vertices of $V(Q_0),\ldots,V(Q_r)$ from $G$ and denote the obtained graph $G'$.
\item Set $G=(((G',x)\oplus_b (H_0,y)),y)\oplus_b (H,y)$ and $W=(W\setminus\cup_{i=0}^rV(Q_i))\cup A_0\cup A$.
\item Select arbitrarily $u\in A_0$ and $v\in A$ and modify $\omega$ as follows:
\begin{itemize}
\item keep the weight same for every $z\in V(G')$ including the boundary vertices $x_1,\ldots,x_p$,
\item set $\omega(z)=0$ for $z\in (A_0\setminus\{u\})\cup(A\setminus\{v\})$,
\item set $\omega(u)=\omega(V(Q_0))$ and $\omega(v)=w'$.
\end{itemize}
\item If $V(Q_i)\subseteq I$ for some $i\in\{1,\ldots,r\}$, then set $I=I\setminus(\cup_{i=1}^rV(Q_i))\cup A$.
\end{itemize}
\end{reduction}
To see the safeness, notice additionally that $H_0[A_0]$ is connected, because $(H_0,y)\in \mathcal{G}_p'$ and $Q_0$ is connected.
The rule is applied exactly once for each inclusion maximal set of components $\{Q_1,\ldots$, $Q_r\}$ having the same neighborhood of size at most $t$ such that $V(Q_i)\subseteq I$ for some $i\in\{1,\ldots,r\}$.
Assume now that we have an inclusion maximal set of components $\{Q_1,\ldots$ ,$Q_r\}$ of $G[W]-S$ with the same neighborhoods $N_{G[W]}=\{x_1,\ldots,x_p\}$ such that $(G[N_{G[W]}[V(Q_i)]],(x_1,\ldots,x_p))$ and $(G[N_{G[W]}[V(Q_j)]],(x_1,\ldots,x_p))$ are equivalent with respect to $\mathcal{F}_b$ for each $i,j\in\{1,\ldots,p\}$. Suppose also that $V(Q_i)\cap I=\emptyset$ for $i\in\{1,\ldots,r\}$.
Let $\omega(V(Q_1))\geq \omega(V(Q_i))$ for every $i\in\{1,\ldots,r\}$. Recall that if $R$ is a nonempty graph in a solution, then either $R=Q_i$ for some $i\in \{1,\ldots,r\}$ or $\cup_{i=1}^r V(Q_i)\subseteq V(R)$. Recall also that $R$ is a solution for the instance of \ProblemName{Maximum or $w$-Weighted Connected Secluded $\mathcal{F}$-Free Subgraph} obtained by a border complementation with respect to $(\emptyset,\emptyset,T,\emptyset)$ and we output it only if $\omega(V(R))\geq w$. Since
all $(G[N_{G[W]}[V(Q_i)]],(x_1,\ldots,x_p))$ are equivalent, we can assume that if $R=Q_i$, then $i=1$, because $Q_1$ has maximum weight. Then by Lemma~\ref{lem:eq}, our final reduction rule is safe.
\begin{reduction}\label{Ared:five-rec}
Let $L=\{x_1,\ldots,x_p\}\subseteq S$, $p\leq t$, and let $x=(x_1,\ldots,x_p)$. Let also $Q_0,\ldots,Q_r$,$r\geq 0$, be the components of $G[W]-S$ with $N_{G[W]}(V(Q_i))=L$ for all $i\in\{0,\ldots,r\}$ such that $\omega(V(Q_0))\geq \omega(V(Q_i))$ for every $i\in\{1,\ldots,r\}$ and
the $p$-boundaried graphs $(G[N_{G[W]}[V(Q_i)]],(x_1,\ldots,x_p))$ are pairwise equivalent with respect to $\mathcal{F}_b$ for $i\in\{0,\ldots,r\}$.
Let $Q=G[\cup_{i=1}^rN_{G[W]}[V(Q_i)]]$ and $w'=\min\{w-1,\sum_{i=1}^r\omega(V(Q_i))\}$.
Find a $p$-boundaried graph $(H_0,y)\in\mathcal{G}_p'$
that is equivalent to $(Q_0,x)$ with respect to $\mathcal{F}_b$ and denote by $A_0$ the set of nonboundary vertices of $H_0$, and
find a $p$-boundaried graph $(H,y)\in\mathcal{G}_p'$
that is equivalent to $(Q,x)$ with respect to $\mathcal{F}_b$ and denote by $A$ the set of nonboundary vertices of $H$.
Then do the following.
\begin{itemize}
\item Delete the vertices of $V(Q_0),\ldots,V(Q_r)$ from $G$ and denote the obtained graph $G'$.
\item Set $G=(((G',x)\oplus_b (H_0,y)),y)\oplus_b (H,y)$ and $W=(W\setminus\cup_{i=0}^rV(Q_i))\cup A_0\cup A$.
\item Select arbitrarily $u\in A_0$ and $v\in A$ and modify $\omega$ as follows:
\begin{itemize}
\item keep the weight same for every $z\in V(G')$ including the boundary vertices $x_1,\ldots,x_p$,
\item set $\omega(z)=0$ for $z\in (A_0\setminus\{u\})\cup(A\setminus\{v\})$,
\item set $\omega(u)=\omega(V(Q_0))$ and $\omega(v)=w'$.
\end{itemize}
\end{itemize}
\end{reduction}
Notice that we upper bound the weight of $A$ by $w-1$ to prevent selecting $R=G[A]$ as a graph in a solution. To see that it is safe, observe that if $\omega(V(Q_0))>0$, then the total weight of $A_0$ and $A$ is at least $w$ and recall that by the definition of \ProblemName{Maximum or $w$-Weighted Connected Secluded $\mathcal{F}$-Free Subgraph}, we output nonempty graphs of maximum weight or weight at least $w$. Therefore, if we output $R$ that includes $A_0\cup A$, then we output a graph of weight at least $w$. If $\omega(Q_0)=0$, then $w'=\sum_{i=1}^r\omega(V(Q_i))=0$.
The Reduction Rule~\ref{Ared:five-rec} is applied for each inclusion maximal sets of components $\{Q_1,\ldots$, $Q_r\}$ satisfying the conditions of the rule such that Reduction Rule~\ref{Ared:four-rec} was not applied to these components before.
Denote by $(G^*,I^*,O^*,B^*,\omega^*,t,T)$ the instance of \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} obtained from $(G,I,O,\hat{B},\omega,t,T)$ by Reduction Rules~\ref{Ared:one-rec}-\ref{Ared:five-rec}. Notice that all modifications were made for $G[W]$.
Denote by $W^*$ the set of vertices of the graph obtained from the initial $G[W]$ by the rules. Observe that there are at most $2^{|S|}$ subsets $L$ of $S$ such that there is a component $Q$ of $G[W]-S$ with $N_{G[W]}(V(Q))=L$. If $|L|>t$, then all $Q$ with $N_{G[W]}(V(Q))=L$ are replaced by one graph by Reduction Rule~\ref{Ared:three-rec} and the number of vertices of this graph is at most $|L|^c$ by Lemma~\ref{lem:size-eq} and the definition of $c$. If $|L|\leq t$, then we either apply Reduction Rule~\ref{Ared:four-rec} for all $Q$ with $N_{G[W]}(V(Q))=L$ and replace these components by two graph with at most $|L|^c$ vertices or we apply Reduction Rule~\ref{Ared:five-rec}. For the latter case, observe that there are at most $t^c$ partitions of the components $Q$ with $N_{G[W]}(V(Q))=L$ into equivalence classes with respect to $\mathcal{F}_b$
by Lemma~\ref{lem:size-eq}. Then we replace each class by two graphs with at most $|L|^c$ vertices.
Taking into account the vertices of $S$, we obtain the following upper bound for the size of $W^*$:
\begin{equation}\label{Aeq:w*}
|W^*|\leq 2^{|S|}2|S|^c t^c+|S|.
\end{equation}
By (\ref{Aeq:q}) and (\ref{Aeq:s}), $|W^*|\leq q$.
Recall that $|W\setminus U|>q$. Therefore, $|V(G^*)|<|V(G)|$. We use it and solve \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} for $(G^*,I^*,O^*,B^*,\omega^*,t,T)$ recursively.
To evaluate the running time, denote by $\tau(G,I,O,B,\omega,t,T)$ the time needed to solve \linebreak \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} for \linebreak $(G,I,O,B,\omega,t,T)$.
Lemmas~\ref{lem:eq-check} and \ref{lem:size-eq} imply that the reduction rules are polynomial. The algorithm from Lemma~\ref{lem:unbreakable} runs in time $2^{2^{\mathcal{O}(t\log t)}}\cdot n^{\mathcal{O}(1)}$. Notice that the sets $\mathcal{G}_p$ and $\mathcal{G}_p'$ can be constructed separately from the algorithm for \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph}. Then we obtain the following recurrence for the running time:
\begin{equation}\label{Aeq:run}
\tau(G,I,O,B,\omega,t,T)\leq \tau(G^*,I^*,O^*,B^*,\omega^*,t,T)+\tau(\tilde{G},\tilde{I},\tilde{O},\tilde{B},\tilde{\omega},t,\tilde{T})+2^{2^{\mathcal{O}(t\log t)}}\cdot n^{\mathcal{O}(1)}.
\end{equation}
Because $|W^*|\leq q$,
\begin{equation}\label{Aeq:vert}
|V(G^*)|\leq |V(G)|-|V(\tilde{G})|+q.
\end{equation}
Recall that if the algorithm of Lemma~\ref{lem:unbreakable} reports that $G$ is $((2q+1)q\cdot 2^t,t)$-unbreakable or we have that $|V(G)|\leq ((2q+1)q\cdot 2^t$, we do not recurse but solve the problem directly in time $2^{2^{2^{\mathcal{O}(t\log t)}}}\cdot n^{\mathcal{O}(1)}$. Following the general scheme from~\cite{ChitnisCHPP16}, we obtain that these condition together with (\ref{Aeq:run}) and (\ref{Aeq:vert}) imply that the total running time is $2^{2^{2^{\mathcal{O}(t\log t)}}}\cdot n^{\mathcal{O}(1)}$.
\end{proof}
Now have now all the details in place to be able to prove Theorem \ref{thm:forb}, stating that \ProblemName{Connected Secluded $\mathcal{F}$-Free Subgraph} is \ensuremath{\operatorClassFPT}\xspace when parameterized by $t$.
\medskip
\noindent
{\bf Theorem~\ref{thm:forb}.}{\it
\ProblemName{Connected Secluded $\mathcal{F}$-Free Subgraph} can be solved in time $2^{2^{2^{\mathcal{O}(t\log t)}}}\cdot n^{\mathcal{O}(1)}$.
}
\begin{proof}
Let $(G,\omega,t,w)$ be an instance of \ProblemName{Connected Secluded $\mathcal{F}$-Free Subgraph}. We define $I=\emptyset$, $O=\emptyset$, $B=V(G)$ and $T=\emptyset$. Then we solve \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} for $(G,I,O,B,w,t,T)$ using Lemma~\ref{lem:bordforb}
in time $2^{2^{2^{\mathcal{O}(t\log t)}}}\cdot n^{\mathcal{O}(1)}$. It remains to notice that $(G,\omega,t,w)$ is a yes-instance of \ProblemName{Connected Secluded $\mathcal{F}$-Free Subgraph} if and only if $(G,I,O,B,\omega,t,T)$ has a nonempty graph in a solution.
\end{proof}
\section{Large Secluded Trees}\label{sec:fpt-tree}
In this section we show that \ProblemName{Connected Secluded $\Pi$-Subgraph} is \ensuremath{\operatorClassFPT}\xspace when parameterized by $t$ when $\Pi$ is defined by a infinite set of forbidden induced subgraphs, namely, by the set of cycles.
In other words, a graph $G$ has the property $\Pi$ considered in this section if $G$ is a forest.
We refer to this problem as \ProblemName{Large Secluded Tree}. We again apply the recursive understanding technique introduced by Chitnis et al.~\cite{ChitnisCHPP16} and follow the scheme of the previous section.
In particular, we solve a special variant of \ProblemName{Large Secluded Tree} tailored for recursion.
We define the following auxiliary problem for a positive integer $w$.
\defproblema{\ProblemName{Maximum or $w$-Weighted Secluded Tree}}%
{A graph $G$, sets $I,O,B\subseteq V(G)$ such that $I\cap O=\emptyset$ and $I\cap B=\emptyset$, a weight function $\omega\colon V(G)\rightarrow\mathbb{Z}_{\geq 0}$ and a nonnegative integer $t$.}%
{Find a $t$-secluded induced connected subtree $H$ of $G$ of maximum weight or weight at least $w$
such that $I\subseteq V(H)$, $O\subseteq V(G)\setminus V(H)$ and $N_G(V(H))\subseteq B$ and output $\emptyset$ if such a subgraph does not exist.}
Let $(G,I,O,B,\omega,t)$ be an instance of \ProblemName{Maximum or $w$-Weighted Secluded Tree} and let $T\subseteq V(G)$ be a set of \emph{border terminals}.
We say that a 4-tuple $(X,Y,Z,\mathcal{P})$, where $(X,Y,Z)$ is a partition of $T$ (some sets could be empty) and $\mathcal{P}=(P_1,\ldots,P_s)$ is a partition of $X$ into nonempty sets if $X\neq\emptyset$ and $\mathcal{P}=\emptyset$ otherwise, is \emph{feasible} if $G[X]$ is a forest,
$Y=\emptyset$ if $X=\emptyset$, $I\cap T\subseteq X$, $O\cap T\subseteq Y\cup Z$ and $Y\subseteq B$,
and if $X\neq\emptyset$, then the vertices of each component of $G[X]$ are in the same set of the partition $\mathcal{P}$.
We say that an instance $(G',I',O',B',\omega',t')$ is obtained by the \emph{border complementation (with respect to feasible 4-tuple $(X,Y,Z,\mathcal{P})$ )} for $X\neq\emptyset$ and $\mathcal{P}=(P_1,\ldots,P_s)$ if
\begin{itemize}
\item[(i)] $G'$ is obtained from $G$ by adding vertices $u_1,\ldots,u_s$ and making $u_i$ adjacent to an arbitrary vertex of each component of $G[X]$ in $G[P_i]$
and to the vertices of $Y$ for $i\in \{1,\ldots,s\}$,
\item[(ii)] $I'= I\cup\{u\}$,
\item[(iii)] $O'=O\cup Y\cup Z$,
\item[(iv)] $B'=B\setminus X$,
\item[(v)] $\omega'(v)=\omega(v)$ for $v\in V(G)$ and $\omega'(u)=0$,
\item[(vi)] $t'\leq t$.
\end{itemize}
If $X=\emptyset$, then $(G',I',O',B',w',t')$ is obtained by the \emph{border complementation (with respect to $(X,Y,Z,\mathcal{P})$ )} if
$G'=G$, $I'=I$, $O'=O\cup T$, $B'=B$, $\omega'(v)=\omega(v)$ for $v\in V(G)$ and $t'\leq t$.
\defproblema{\ProblemName{Bordered Maximum or $w$-Weghted Secluded Tree}}%
{A graph $G$, sets $I,O,B\subseteq V(G)$ such that $I\cap O=\emptyset$ and $I\cap B=\emptyset$, a weight function $\omega\colon V(G)\rightarrow\mathbb{Z}_{\geq0}$, a nonnegative integer $t$, and a set $T\subseteq V(G)$ of border terminals of size at most $2(t+1)$.}{Output a solution for each instance $(G',I',O',B',\omega',t')$ of \ProblemName{Maximum or $w$-Weighted Secluded Tree} that can be obtained from $(G,I,O,B,w,t)$ by a border complementation distinct from the border complementation with respect to $(\emptyset,\emptyset,T,\emptyset)$, and for the border complementation with respect to $(\emptyset,\emptyset,T,\emptyset)$ output a nonempty solution if it has weight at least $w$ and output $\emptyset$ otherwise.}
Two instances $(G_1,I_1,O_1,B_1,\omega_1,t,T)$ and $(G_2,I_2,O_2,B_2,\omega_2,t,T)$ of \ProblemName{Bordered Maximum or $w$-Weghted Secluded Tree} (note that $t$ and $T$ are the same) are said to be \emph{equivalent} if
\begin{itemize}
\item[(i)] $T\cap I_1=T\cap I_2$, $T\cap O_1=T\cap O_2$ and $T\cap B_1=T\cap B_2$,
\item[(ii)] for the border complementations $(G_1',I_1',O_1',B_1',\omega_1',t')$ and $(G_2',I_2',O_2',B_2',\omega_2',t')$ of the instances $(G_1,I_1,O_1,B_1,\omega_1,t')$ and $(G_2,I_2,O_2,B_2,\omega_2,t')$ respectively of \ProblemName{Maximum or $w$-Weighted Connected Secluded $\mathcal{F}$-Free Subgraph} with respect to every feasible $(X,Y,Z,\mathcal{P})$ and $t'\leq t$, it holds that if $(G_1',I_1',O_1',B_1',\omega_1',t')$ has a nonempty solution $R_1$, then $(G_2',I_2',O_2',B_2',\omega_2',t')$ has a nonempty solution $R_2$ with $\omega_2'(V(R_2))\geq \min\{\omega_1'(V(R_1)),w\}$ and, vice versa, if $(G_2',I_2',O_2',B_2',\omega_2',t')$ has a nonempty solution $R_2$, then $(G_1',I_1',O_1',B_1',\omega_1',t')$ has a nonempty solution $R_1$ with $\omega_1'(V(R_1))\geq \min\{\omega_2'(V(R_2))$, $w\}$.
\end{itemize}
As in the previous section we not distinguish equivalent instances of \ProblemName{Bordered Maximum or $w$-Weghted Secluded Tree} and their solutions.
\subsection{High connectivity phase}
In this section we solve \ProblemName{Bordered Maximum or $w$-Weghted Secluded Tree} for $(q,t+1)$-unbreakable graphs. We need the following folklore lemma.
\begin{lemma}\label{lem:balance}
Every tree $T$ has a separation $(A,B)$ of order 1 such that $|A\setminus B|\leq \frac{2}{3}|V(T)|$ and $|B\setminus A|\leq \frac{2}{3}|V(T)|$.
\end{lemma}
\begin{lemma}\label{lem:unbreak-tree}
\ProblemName{Bordered Maximum or $w$-Weghted Secluded Tree} for $(q,t+1)$-unbreakable graphs can be solved in time $2^{\mathcal{O}((t+\min\{q,t\})\log(q+t))}\cdot n^{\mathcal{O}(1)}$.
\end{lemma}
\begin{proof}
Consider an instance $(G',I',O',B',\omega',t')$ of \ProblemName{Maximum or $w$-Weighted Secluded Tree} be obtained from $(G,I,O,B,\omega,t)$ by the border complementation with respect to some feasible $(X,Y,Z,\mathcal{P})$.
Assume that $H$ is a nonempty solution of $(G',I',O',B',\omega',t')$. We claim that $|V(H)|\leq 3q+8$.
To obtain a contradiction, assume that $|V(H)|\geq 3q+9$. By Lemma~\ref{lem:balance}, there is a separation $(U,W)$ of $H$ of order 1 such that $|U\setminus W|\leq \frac{2}{3}|V(H)|$ and $|W\setminus U|\leq \frac{2}{3}|V(H)|$, that is, $|U\setminus W|\geq \frac{1}{3}|V(H)|-1\geq q+2$ and $|W\setminus U|\geq \frac{1}{3}|V(H)|-1\geq q+2$.
Let $U'=U\cap V(G)$ and $W'=W\cap V(G)$. We have that $|U'\setminus W'|\geq q+1$ and $|W'\setminus U'|\geq q+1$.
Let $U''=U'\cup N_{G'}(V(H))\subseteq V(G)$ and $W''=V(G)\setminus (U'\setminus W')\subseteq V(G)$.
We have that $(U'',W'')$ is a separation of $G$.
Since $U''\cap W''=N_{G'}(V(H))\cup (U'\cap W')$, we obtain that the order of the separation is at most $t+1$.
Observe that $U''\setminus W''=U'\setminus W''$ and
$W'\setminus U'\subseteq W''\setminus U''$ and, therefore, $|U''\setminus W''|\geq q+1$ and $|W''\setminus U''|\geq q+1$ contradicting the $(q,t+1)$-unbreakability of $G$.
The claim implies that to solve \ProblemName{Maximum or $w$-Weighted Secluded Tree} for \linebreak $(G',I',O',B',\omega',t')$, it is sufficient to consider $k\leq 3q+8$ and for each $k$, find $t$-secluded induced subtree $H$ in $G'$ of maximum weight
such that
$I'\subseteq V(H')$, $O'\subseteq V(G')\setminus V(H)$, $N_{G'}(V(H))\subseteq B'$ and $|V(H)|=k$. By Corollary~\ref{cor:CSWCS},
it can be done in time $2^{\mathcal{O}(\min\{q,t\}\log (q+t))}\cdot n^{\mathcal{O}(1)}$.
To solve \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} for $(G,I,O,B,\omega,t,T)$, we consider all possible partitions $(X,Y,Z)$ of $T$ and partitions $\mathcal{P}$ of $X$ such that $(X,Y,Z,\mathcal{P})$ is feasible.
Since $|T|\leq 2(t+1)$, and the set of $\mathcal{P}$ together with $X$ and $Y$ form a partition of $T$, we conclude that
there are $2^{\mathcal{O}(t\log t)}$ feasible $(X,Y,Z,\mathcal{P})$. Hence, the total running time is $2^{\mathcal{O}((t+\min\{q,t\})\log (q+t))}\cdot n^{\mathcal{O}(1)}$.
\end{proof}
\subsection{The \ensuremath{\operatorClassFPT}\xspace algorithm for \ProblemName{Large Secluded Tree}}\label{Asec:tree}
In this section we construct an \ensuremath{\operatorClassFPT}\xspace algorithm for \ProblemName{Large Secluded Tree} parameterized by $t$. To do it, we solve \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} in \ensuremath{\operatorClassFPT}\xspace-time for general case.
\begin{lemma}\label{lem:bordtree}
\ProblemName{Bordered Maximum or $w$-Weghted Secluded Tree} can be solved in time $2^{2^{\mathcal{O}(t\log t)}}\cdot n^{\mathcal{O}(1)}$.
\end{lemma}
\begin{proof}
There is a constant $c$ such that the number of partitions of a $k$-element set into subsets such that at most two of them could be empty is at most $2^{ck\log k}$.
We define
\begin{equation}\label{eq:q-t}
q=2\cdot 2^{(t+1)t2^{c2(t+1)\log(2(t+1))}+2(t+1)}+(t+1)t2^{c2(t+1)\log(2(t+1))}+2(t+1).
\end{equation}
Notice that $q=2^{2^{\mathcal{O}(t\log t)}}$.
Consider an instance $(G,I,O,B,\omega,t,T)$ of \ProblemName{Bordered Maximum or $w$-Weghted Secluded Tree}.
We use the algorithm from Lemma~\ref{lem:unbreakable} for $G$. This algorithm in time $2^{2^{\mathcal{O}(t\log t)}}\cdot n^{\mathcal{O}(1)}$ either finds a separation $(U,W)$ of $G$ of order at most $t+1$ such that $|U\setminus W|>q$ and $|W\setminus U|>q$ or correctly reports that $G$ is $((2q+1)q\cdot 2^{t+1},t+1)$-unbreakable. In the latter case we solve the problem using Lemma~\ref{lem:unbreak} in time $2^{2^{\mathcal{O}(t\log t)}}\cdot n^{\mathcal{O}(1)}$.
Assume from now that there is a separation $(U,W)$ of order at most $t+1$ such that $|U\setminus W|>q$ and $|U\setminus W|>q$.
Recall that $|T|\leq 2(t+1)$. Then $|T\cap (U\setminus W)|\leq t+1$ or $|T\cap (W\setminus U)|\leq t+1$. Assume without loss of generality that $|T\cap (W\setminus U)|\leq t+1$. Let $\tilde{G}=G[W]$, $\tilde{I}=I\cap W$, $\tilde{O}=O\cap W$, $\tilde{\omega}$ is the restriction of $w$ to $W$, and define $\tilde{T}=(T\cap W)\cup (U\cap W)$. Since $|U\cap W|\leq t+1$, $|\tilde{T}|\leq 2(t+1)$.
If $|W|\leq (2q+1)q\cdot 2^{t+1}$, then we solve \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} for the instance $(\tilde{G},\tilde{I},\tilde{O},\tilde{B},\tilde{\omega},t,\tilde{T})$ by brute force in time $2^{2^{\mathcal{O}(t\log t)}}$ trying all possible subset of $W$ at most $t+1$ values of $0\leq t'\leq t$. Otherwise, we solve $(\tilde{G},\tilde{I},\tilde{O},\tilde{B},\tilde{\omega},t,\tilde{T})$ recursively. Let $\mathcal{R}$ be the set of nonempty induced subgraphs $R$ that are included in the obtained solution for $(\tilde{G},\tilde{I},\tilde{O},\tilde{B},\tilde{\omega},t,\tilde{T})$.
For $R\in \mathcal{R}$, define $S_R$ to be the set of vertices of $W\setminus V(R)$ that are adjacent to the vertices of $R$ in the graph obtained by the border complementation for which $R$ is a solution of the corresponding instance of \ProblemName{Maximum or $w$-Weighted Secluded Tree}. Note that $|S_R|\leq t$.
If $\mathcal{R}\neq\emptyset$, then let $S=\tilde{T}\cup_{R\in\mathcal{R}}S_R$, and $S=\tilde{T}$ if $\mathcal{R}=\emptyset$.
Since \ProblemName{Maximum or $w$-Weighted Secluded Tree} is solved for at most $t+1$ of values of $t'\leq t$, at most $2^{c(2(t+1))\log (2(t+1)) }$ feasible 4-tuples $(X,Y,Z,\mathcal{P})$, we have that $|\mathcal{R}|\leq (t+1)2^{c2(t+1)\log (2(t+1))}$. Taking into account that $|T'|\leq 2(t+1)$,
\begin{equation}\label{eq:s-t}
|S|\leq (t+1)t2^{c2(t+1)\log (2(t+1))}+2(t+1).
\end{equation}
Let $\hat{B}=(B\cap U)\cup (B\cap C)$. We claim that the instances $(G,I,O,B,\omega,t,T)$ and \linebreak $(G,I,O,\hat{B},\omega,t,T)$ of \ProblemName{Bordered Maximum or $w$-Weghted Secluded Tree} are equivalent.
Recall that we have to show that
\begin{itemize}
\item[(i)] $T\cap B=T\cap \hat{B}$,
\item[(ii)] for the border complementations $(G',I',O',B',\omega',t')$ and $(G',I',O',\hat{B}',\omega',t')$ of the instances $(G,I,O,B,\omega,t')$ and $(G,I,O,\hat{B},\omega,t')$ respectively of \ProblemName{Maximum or $w$-Weighted Connected Secluded $\mathcal{F}$-Free Subgraph} with respect to every feasible $(X,Y,Z,\mathcal{P})$ and $t'\leq t$, it holds that if $(G',I',O',B',\omega',t')$ has a nonempty solution $R_1$, then \linebreak $(G',I',O',\hat{B}',\omega',t')$ has a nonempty solution $R_2$ with $\omega'(V(R_2))\geq \min\{\omega'(V(R_1)),w\}$ and, vice versa, if $(G',I',O',\hat{B}',w',t')$ has a nonempty solution $R_2$, then $(G',I',O',B',w',t')$ has a nonempty solution $R_1$ with $\omega'(V(R_1))\geq \min\{\omega'(V(R_2)),w\}$.
\end{itemize}
The condition (i) holds by the definition of $\hat{B}$. Because $\hat{B}\subseteq B$, we immediately obtain if $(G',I',O',\hat{B}',\omega',t')$ has a nonempty solution $R_2$, then $(G',I',O',B',\omega',t')$ has a nonempty solution $R_1$ with $\omega'(V(R_1))\geq \min\{\omega'(V(R_2)),w\}$.
It remains to prove that for a border complementation $(G',I',O',B',\omega',t')$ and $(G',I',O',\hat{B}',\omega',t')$ of $(G,I,O,B,\omega,t')$ and $(G,I,O,\hat{B},\omega,t')$ respectively of \ProblemName{Maximum or $w$-Weighted Connected Secluded $\mathcal{F}$-Free Subgraph} with respect to a feasible $(X,Y,Z,\mathcal{P})$ and $t'\leq t$, it holds that if $(G',I',O',B',\omega',t')$ has a nonempty solution $R_1$, then $(G',I',O',\hat{B}',\omega',t')$ has a nonempty solution $R_2$ with $\omega'(V(R_2))\geq \min\{\omega'(V(R_1)),w\}$.
If $V(R_1)\cap V(G)\subseteq U\setminus W$, then $N_{G'}V(R_1)\subseteq \hat{B}'$. Therefore, for a solution $R_2$ of \linebreak $(G',I',O',\hat{B}',\omega',t')$, $\omega'(V(R_2))\geq \min\{\omega'(V(R_1),w)\}$.
Assume that $V(R_1)\cap W\neq \emptyset$.
Let $\tilde{X}=\tilde{T}\cap(V(R_1)\cap W)$, let $\tilde{Y}$ be the set of vertices of $\tilde{T}\setminus V(R_1)$ that are adjacent to vertices of $R_1$ laying outside $W\setminus U$ and $\tilde{Z}=\tilde{T}\setminus (\tilde{X}\cup \tilde{Y})$.
Notice that $R_1-(W\setminus U)$ is a forest and denote it by $F$. Then there is a partition $\tilde{\mathcal{P}}=(P_1,\ldots,P_s)$ of $\tilde{X}$ into nonempty sets such that two vertices of $\tilde{X}$ are in the same set $P_i$ if and only if they are in the same component of $F$.
Consider the border complementation of $(\tilde{G},\tilde{I},\tilde{O},\tilde{B},\tilde{\omega},\tilde{t})$ with respect to $(\tilde{X},\tilde{Y},\tilde{Z},\tilde{\mathcal{P}})$, where $\tilde{t}$ is the number of neighbors of $R_1$ in $W$. Recall that in the border complementation we have new vertices $u_1,\ldots,u_s\in \tilde{I}$ such that each $u_i$ is adjacent to one vertex of $P_i$ and the vertices of $\tilde{Y}$ for $i\in\{1,\ldots,s\}$. Consider the subgraph $F'$ of $\tilde{G}$ induced by $(V(R_1)\cap W)\cup\{u_1,\ldots,u_s\}$. It is straightforward to see that $F'$ is a tree that has $\tilde{t}$ neighbors in $\tilde{G}$.
It implies that there is $\tilde{R}\in \mathcal{R}$ for the instance
$(\tilde{G},\tilde{I},\tilde{O},\tilde{B},\tilde{\omega},\tilde{t})$ of \ProblemName{Maximum or $w$-Weighted Secluded Tree} obtained by the border complementation with respect to $(\tilde{X},\tilde{Y},\tilde{Z},\tilde{\mathcal{P}'})$ and $\tilde{\omega}(V(\tilde{R}))\geq\min\{\tilde{\omega}(V(R_1)\cap W)\}$.
Recall also that the neighbors of the vertices of $\tilde{R}$ are in $S$.
Now let $R_2=G'[(V(R_1)\cap U)\cup (V(\tilde{R})\cap W)]$. We have that $R_2$ is a tree of weight at least $\min\{\omega'(R_1),w\}$ that has at most $t'$ neighbors in $G'$.
Since, $(G,I,O,B,\omega,t,T)$ and $(G,I,O,\hat{B},\omega,t,T)$ of \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} are equivalent, we can consider $(G,I,O,\hat{B},\omega,t,T)$. Now we apply some reduction rules that produce equivalent instances of \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} or report that we have no solution. The aim of these rules is to reduce the size of $G$.
Let $Q$ be a component of $G[W]-S$. Notice that for any nonempty graph $R$ in a solution of $(G,I,O,\hat{B},\omega,t,T)$, either $V(Q)\subseteq V(R)$ or $V(Q)\cap V(R)=\emptyset$, because $N_{G[W]}(V(R))\subseteq S$. Moreover, if
$V(Q)\cap V(R)=\emptyset$, then $N_{G[W]}[V(Q)]\cap V(R)=\emptyset$.
Notice also that if $v\in N_{G[W]}(V(Q))$ is a vertex of $R$, then $V(Q)\subseteq V(R)$.
These observation are crucial for the following reduction rules.
\begin{reduction}\label{red:one-rec-t}
For a component $Q$ of $G[W]-S$ do the following in the given order:
\begin{itemize}
\item if $N_{G[W]}[V(Q)]\cap I\neq\emptyset$ and $V(Q)\cap O\neq\emptyset$, then return $\emptyset$ and stop,
\item if $N_{G[W]}[V(Q)]\cap I\neq\emptyset$, then set $I=I\cup V(Q)$,
\item if $V(Q)\cap O\neq\emptyset$, then set $O=O\cup N_{G[W]}[V(Q)]$.
\end{itemize}
\end{reduction}
The rule is applied to each component $Q$ exactly once. Notice that after application of the rule, for every component $Q$ of $G[W]-S$, we have that either $V(Q)\subseteq I$ or $V(Q)\subseteq O$ or $V(Q)\cap (I\cup O\cup\hat{B})=\emptyset$.
Observe that if a component $Q$ of $G[W]$ contains a cycle, then $Q$ cannot be a part of any solution. It leads us to the following rule that is applied to each component.
\begin{reduction}\label{Ared:two-rec-t}
If for a component $Q$ of $G[W]-S$, $Q$ contains a cycle, then
\begin{itemize}
\item if $V(Q)\subseteq I$, then return $\emptyset$ and stop, otherwise,
\item set $O=O\cup N_{G[W]}[V(Q)]$.
\end{itemize}
\end{reduction}
Suppose that there is a component $Q$ containing two distinct verices $u$ and $v$ that are adjacent to the same vertex $x\in N_{G[W]}(V(Q))$. If there is a graph $R$ in a solution of $(G,I,O,\hat{B},\omega,t,T)$ with $u,v\in V(R)$, then $x\notin V(R)$, because $R$ is a tree. If $u,v\notin V(R)$, then $x\notin V(R)$, because otherwise $u,v\notin \hat{B}$ would be adjacent to a vertex of $R$. Hence, the next rule is safe.
\begin{reduction}\label{red:three-rec-t}
If for a component $Q$ of $G[W]-S$, there is $x\in N_{G[W]}(Q)\setminus O$ adjacent to two distinct vertices of $Q$, then
set $O=O\cup\{x\}$.
\end{reduction}
We apply the rule exhaustively while it is possible.
After applying Reduction Rules~\ref{red:one-rec-t}-\ref{red:three-rec-t}, we can safely replace each $Q$ by a single vertex.
\begin{reduction}\label{red:four-rec-t}
If for a component $Q$ of $G[W]-S$, $|V(Q)|>1$, then
\begin{itemize}
\item modify $G$ by deleting the vertices of $V(Q)$ and constructing a new vertex $u$ adjacent to $N_{G[W]}(V(Q))$,
\item set $W=(W\setminus V(Q))\cup \{u\}$,
\item set $\omega(u)=\omega(V(Q))$,
\item if $V(Q)\subseteq I$, then set $I=(I\setminus V(Q))\cup\{u\}$,
\item if $V(Q)\subseteq O$, then set $O=(O\setminus V(Q))\cup\{u\}$.
\end{itemize}
\end{reduction}
The rule is applied for each $Q$ with at least two vertices. Notice that now $W\setminus S$ is an independent set.
Note that if there is $u\in W\setminus S$ such that $u\in O$, then it is safe to delete $u$, because $N_{G[W]}(u)\subseteq O$ by Reduction Rule~\ref{red:one-rec-t}.
\begin{reduction}\label{red:five-rec-t}
If there is $u\in W\setminus S$ such that $u\in S$, then delete $u$ from $G$ and the sets $W$ and $O$.
\end{reduction}
From now we have that $(W\setminus S)\cap O=\emptyset$.
Suppose that $L$ is a set of verices of $W\setminus S$ that have the same neighborhoods, i.e, they are false twins of $G[W]$. If there are distinct $u,v\in L$ such that $u,v\in V(R)$ for
a graph $R$ in a solution of $(G,I,O,\hat{B},\omega,t,T)$, then $L\subseteq V(R)$, because $R$ should contain exactly one vertex in the neighborhoods of $u$ and $v$.
Hence, either $L\cap V(R)=\emptyset$ or exactly one vertex of $L$ is in $R$ or $L\subseteq V(R)$. In particular, if $|L\cap I|\geq 2$, then $L\subseteq V(R)$.
Suppose $L\cap I=\emptyset$ and
$u\in L$ is the unique vertex of $L$ in $R$, then we can safely assume that $u$ is a vertex of maximum weight in $L$.
Notice also that in this case $u$ is the unique vertex of $R$. It means that $R$ is obtained for a border complementation of $(G,I,O,\hat{B},\omega,t)$ with respect to $(\emptyset,\emptyset,T,\emptyset)$ and we output $R$ only if $\omega(u)\geq w$.
These observations give us the two following rules that are applied for all inclusion maximal sets
$L\subseteq W\setminus S$ of size at least 3.
\begin{reduction}\label{red:six-rec-t}
If for an inclusion maximal set of false twin vertices $L\subseteq W\setminus S$ such that $|L|\geq3$, $L\cap I\neq\emptyset$, then let $u\in L\cap I$, $v\in L\setminus \{u\}$, and
\begin{itemize}
\item delete the verices of $L\setminus\{u,v\}$ from $G$ and the sets $W,I$,
\item if $(L\setminus \{u\})\cap I\neq \emptyset$, then set $I=I\cup \{v\}$,
\item set $\omega(v)=\sum_{x\in L\setminus\{u\}}\omega(x)$.
\end{itemize}
\end{reduction}
\begin{reduction}\label{red:seven-rec-t}
If for an inclusion maximal set of false twin vertices $L\subseteq W\setminus S$ such that $|L|\geq3$, $L\cap I=\emptyset$, then let $u$ be a vertex of maximum weight in $L$ and let $v\in L\setminus \{u\}$, and then
\begin{itemize}
\item delete the verices of $L\setminus\{u,v\}$ from $G$ and the set $W$,
\item set $\omega(v)=\min\{\sum_{x\in L\setminus\{u\}}\omega(x),w-1\}$.
\end{itemize}
\end{reduction}
Notice that $\omega(v)\leq w-1$. It implies that $v$ cannot be selected as a unique vertex of a solution.
To see that it is safe, observe that if $\omega(u)>0$, then the total weight of $u$ and $v$ is at least $w$ and recall that by the definition of \ProblemName{Maximum or $w$-Weighted Secluded Tree}, we output nonempty graphs of maximum weight or weight at least $w$. Therefore, if we output $R'$ that includes $u$ and $v$ instead of $R$ containing all the vertices of $L$, then we output a graph of weight at least $\min\{\omega(R),w\}$.
Denote by $(G^*,I^*,O^*,B^*,\omega^*,t,T)$ the instance of \ProblemName{Bordered Maximum or $w$-Weghted Connected Secluded $\mathcal{F}$-Free Subgraph} obtained from $(G,I,O,\hat{B},\omega,t,T)$ by Reduction Rules~\ref{red:one-rec-t}-\ref{red:seven-rec-t}. Notice that all modifications were made for $G[W]$. Denote by $W^*$ the set of vertices of the graph obtained from $G[W]$ by the rules. Observe that there are at most $2^{|S|}$ subsets $L$ of $S$ such that there is a components $Q$ of $G[W]-S$ with $N_{G[W]}(V(Q))=L$. Notice that for every $L$ all such components $Q$ are replaced by at most 2 vertices by the reduction rules.
Taking into account the vertices of $S$, we obtain the following upper bound for the size of $W^*$:
\begin{equation}\label{eq:w*-t}
|W^*|\leq 2\cdot 2^{|S|}+|S|.
\end{equation}
By (\ref{eq:q-t}) and (\ref{eq:s-t}), $|W^*|\leq q$.
Recall that $|W\setminus U|>q$. Therefore, $|V(G*)|<|V(G)|$. We use it and solve \ProblemName{Bordered Maximum or $w$-Weghted Secluded Tree} for $(G^*,I^*,O^*,B^*,\omega^*,t,T)$ recursively.
To evaluate the running time, denote by $\tau(G,I,O,B,\omega,t,T)$ the time needed to solve \ProblemName{Bordered Maximum or $w$-Weghted Secluded Tree} for $(G,I,O,B,\omega,t,T)$.
Clearly, all reduction rules are polynomial. The algorithm from Lemma~\ref{lem:unbreakable} runs in time $2^{2^{\mathcal{O}(t\log t)}}\cdot n^{\mathcal{O}(1)}$. Then we obtain the following recurrence for the running time:
\begin{equation}\label{eq:run-t}
\tau(G,I,O,B,\omega,t,T)\leq \tau(G^*,I^*,O^*,B^*,\omega^*,t,T)+\tau(\tilde{G},\tilde{I},\tilde{O},\tilde{B},\tilde{\omega},t,\tilde{T})+2^{2^{\mathcal{O}(t\log t)}}\cdot n^{\mathcal{O}(1)}.
\end{equation}
Because $|W^*|\leq q$,
\begin{equation}\label{eq:vert-t}
|V(G^*)|\leq |V(G)|-|V(\tilde{G})|+q.
\end{equation}
Recall that if the algorithm of Lemma~\ref{lem:unbreakable} reports that $G$ is $((2q+1)q\cdot 2^{t+1},t+1)$-unbreakable or we have that $|V(G)|\leq ((2q+1)q\cdot 2^{t+1}$, we do not recurse but solve the problem directly in time $2^{2^{\mathcal{O}(t\log t)}}\cdot n^{\mathcal{O}(1)}$. Following the general scheme from~\cite{ChitnisCHPP16}, we obtain that these condition together with (\ref{eq:run-t}) and (\ref{eq:vert-t}) imply that the total running time is $2^{2^{\mathcal{O}(t\log t)}}\cdot n^{\mathcal{O}(1)}$.
\end{proof}
We have now all the details that are necessary to prove the main theorem of this section.
\begin{theorem}\label{thm:tree}
\ProblemName{Large Secluded Tree} can be solved in time $2^{2^{\mathcal{O}(t\log t)}}\cdot n^{\mathcal{O}(1)}$.
\end{theorem}
\begin{proof}
Let $(G,\omega,t,w)$ be an instance of \ProblemName{Large Secluded Tree}. We define $I=\emptyset$, $O=\emptyset$, $B=V(G)$ and $T=\emptyset$. Then we solve \ProblemName{Bordered Maximum or $w$-Weghted Secluded Tree} for $(G,I,O,B,\omega,t,T)$ using Lemma~{lem:bordtree}
in time $2^{2^{\mathcal{O}(t\log t)}}\cdot n^{\mathcal{O}(1)}$. It remains to notice that $(G,\omega,t,w)$ is a yes-instance of \ProblemName{Large Secluded Tree} if and only if $(G,I,O,B,\omega,t,T)$ has a nonempty graph in a solution.
\end{proof}
\section{Better algorithms for \ProblemName{Connected Secluded $\Pi$-Subgraph}}
\label{sec:better}
We applied the recursive understanding technique introduced by Chitnis et al.~\cite{ChitnisCHPP16} for \ProblemName{Connected Secluded $\Pi$-Subgraph} when $\Pi$ is defined by a finite set of forbidden subgraphs and $\Pi$ is the property to be a forest in Sections~\ref{sec:fpt-forb} and \ref{sec:fpt-tree} respectively. We believe that the same approach can be used in some other cases. In particular,
the results of Section~\ref{sec:fpt-tree} could be generalized if $\Pi$ is the property defined by a finite list of \emph{forbidden minors} that include a planar graph.
Recall that a graph $F$ is a \emph{minor} of a graph $G$ if a graph isomorphic to $F$ can be obtained from $G$ by a sequence of vertex and edge deletion and edge contractions.
Respectively, a graph $G$ is $F$-minor free if $F$ is not a minor of $G$, and for a family of graphs $\mathcal{F}$, $G$ is $\mathcal{F}$-minor free if $G$ is $F$-minor free for every $F\in\mathcal{F}$. If $F$ is a planar graph, then by the fundamental results of Robertson and Seymour~\cite{RobertsonS86} (see also~\cite{RobertsonST94}), an $F$-minor free graph $G$ has bounded \emph{treewidth}. This makes it possible to show that if $\Pi$ is the property to be $\mathcal{F}$-minor free for a finite family of graphs $\mathcal{F}$ that includes at lease one planar graph, then \ProblemName{Connected Secluded $\Pi$-Subgraph} is \ensuremath{\operatorClassFPT}\xspace when parameterized by $t+k$.
Nevertheless, the drawback of the applying the recursive understanding technique for these problems is that we get double or even triple-exponential dependence on the parameter in our \ensuremath{\operatorClassFPT}\xspace algorithms. It is natural to ask whether we can do better for some properties $\Pi$. In this section we show that it can be done if $\Pi$ is the property to be a complete graph, a star, to be $d$-regular or to be a path.
\subsection{Secluded Clique}
We begin with problem \ProblemName{Secluded Clique}, defined as follows.
\defproblema{\ProblemName{Secluded Clique}}%
{A graph $G$, a weight function $\omega\colon V(G)\rightarrow \mathbb{Z}_{>0}$, an nonnegative integer $t$ and a positive integer $w$.}%
{Decide whether $G$ contains a $t$-secluded clique $H$ with $\omega(V(H))\geq w$.}
We prove that this problem can be solved in time $2^{\mathcal{O}(t \log t)} \cdot n^{\mathcal{O}(1)}$. The result uses the algorithm of Lemma \ref{lem:derand} and the following simple observations.
\begin{lemma}\label{lem:SCtruetwins}
Let $(G,\omega, t, w)$ be an input of \ProblemName{Secluded Clique} and let $H$ be a solution such that $V(H)$ is maximal by inclusion. Let $L$ be an inclusion maximal set of true-twins of $G$. Then $L\cap V(H) \neq \emptyset$ implies that $L \subseteq V(H)$.
\end{lemma}
\begin{proof}
Let $L$ be an inclusion maximal set of twins, and let $u,v \in L$ be such that $u\in V(H)$ and $v \notin V(H)$. Consider the graph $H' = G[V(H) \cup \{v\}]$. Since $H$ is a $t$-secluded clique, and $u, v$ are true twins, we have that $H'$ is also a $t$-secluded clique, and $\omega(V(H')) = \omega(V(H)) + \omega(v) \geq w$. Therefore $H'$ is also a solution of \ProblemName{Secluded Clique}, contradicting the maximality of $H$.
\end{proof}
Let $\mathcal{L}$ be the family all of maximal sets of true twins in a graph $G$. Note that a vertex can not belong to two different maximal sets of true twins, so $\mathcal{L}$ induces a partition of $G$. Consider $\tilde{G}$ the graph obtained from $G$ contracting each maximal set of true-twins $L$ into a single node $x_L$, and removing multiple edges. In other words, $\tilde{G}$ contains one node for each element of $\mathcal{L}$. Two nodes $x_1$ and $x_2$ in $\tilde{G}$ are adjacent if there is an edge in $G$ with one endpoint in $L_1$ and the other one in $L_2$, where $L_1$ and $L_2$ are elements of $\mathcal{L}$ corresponding to $x_1$ and $x_2$, respectively.
We say that a node $x \in \tilde{G}$ is a \emph{contraction} of $L \in \mathcal{L}$ if $x$ is the node of $\tilde{G}$ corresponding to $L$. We say that a set $U \subseteq V(G)$ is the \emph{expansion} of~$\tilde{U} \subseteq V(\tilde{G})$ if the all the nodes of $\tilde{U}$ represent maximal sets of true-twins contained in $U$. We also say in that case that $\tilde{U}$ is the \emph{contraction} of $U$.
\begin{lemma}\label{SCsize}
Let $G$ be a graph, $t$ be a positive integer and let $U$ be a subset of vertices of $G$ inducing a inclusion maximal $t$-secluded clique in $G$. There exists a set $\tilde{U}$ of vertices of $V(\tilde{G})$ such that:
\begin{itemize}
\item[1)] $U$ is the expansion of $\tilde{U}$,
\item[2)] $\tilde{U}$ induces a $t$-secluded clique on $\tilde{G}$, and
\item[3)] $|\tilde{U}| \leq 2^t$.
\end{itemize}
\end{lemma}
\begin{proof}
Let $\tilde{U}$ be the set of nodes consisting in the contraction of the maximal sets of true-twins in $G$ intersecting $U$. We claim that $\tilde{U}$ satisfies the desired properties.
\begin{itemize}
\item[1)] From Lemma \ref{lem:SCtruetwins}, we know that if a maximal set of true-twins $L$ intersects $U$, then it $L$ is contained in $U$. Therefore $\tilde{U}$ is a contraction of $U$ (so $U$ is an expansion of $\tilde{U}$).
\item[2)] Let $x_1$ and $x_2$ be two nodes in $\tilde{U}$ that are contractions of $L_1$ and $L_2$, respectively. Since $U$ induces a clique in $G$, $L_1$ and $L_2$ must contain adjacent vertices, so $x_{1}$ and $x_{2}$ are adjacent in $\tilde{G}[\tilde{U}]$. On the other hand, $|N_{\tilde{G}}(\tilde{U})|$ equals the number of maximal sets of true-twins intersecting $N_G(U)$, so $|N_{\tilde{G}}(\tilde{U})|\leq |N_G(U)| \leq t$. We conclude that $\tilde{U}$ induces a $t$-secluded clique in $\tilde{G}$.
\item[3)] Let $x_1$ and $x_2$ be two different nodes in $\tilde{U}$ that are contractions of $L_1$ and $L_2$, respectively. From definition of maximal sets of true-twins, $N_G(L_1) \neq N_G(L_2)$, so $N_{\tilde{G}}(x_1) \neq N_{\tilde{G}}(x_2)$. Since $\tilde{U}$ is a clique, necessarily $N_{\tilde{G}}(x_1) \cap \tilde{U} = N_{\tilde{G}}(x_2) \cap \tilde{U}$. Therefore, every vertex on $\tilde{U}$ has a different neighborhood outside $\tilde{U}$. Since $|N_{\tilde{G}}(\tilde{U})| \leq t$, we obtain that $|\tilde{U}| \leq 2^{|N_{\tilde{G}}(\tilde{U})|} \leq 2^t$.
\end{itemize}
\end{proof}
\begin{theorem}
\ProblemName{Secluded Clique} can be solved in time $2^{\mathcal{O}(t \log t)} \cdot n^{\mathcal{O}(1)}$.
\end{theorem}
\begin{proof}
The algorithm for \ProblemName{Secluded Clique} on input $(G,\omega, t, w)$ first computes the family $\mathcal{L}$ of all inclusion maximal set of true-twins of $G$, and then computes $\tilde{G}$ in time $\mathcal{O}(n^2)$. Then, the algorithm uses Lemma \ref{lem:derand} to compute in time $2^{\mathcal{O}(t \log t)}n \log n$ a family $\mathcal{S}$ of at most $2^{\mathcal{O}(t \log t)}\log n$ subsets of $V(\tilde{G})$ such that: for any sets $A, B \subseteq V(\tilde{G})$, $A \cap B = \emptyset$, $|A| \leq 2^t$, $|B| \leq t$, there exists a set $S \in \mathcal{S}$ with $A\subseteq S$ and $B\cap S = \emptyset$.
Let $U$ be a set of vertices of $G$ inducing an inclusion maximal solution of \ProblemName{Secluded Clique} on instance $(G,\omega, t, w)$, and let $\tilde{U}$ be the contraction of $U$. From Lemma \ref{SCsize}, we know that $N_{\tilde{G}}(\tilde{U}) \leq t$ and $|\tilde{U}| \leq 2^t$. Then, there exists $S\in \mathcal{S}$ such that $\tilde{U} \subseteq S$ and $N_{\tilde{G}}(U) \cap S = \emptyset$. In other words $\tilde{U}$ is a component of $\tilde{G}[S]$. Therefore, the algorithm checks for every $S \in \mathcal{S}$ and every component $C$ of $\tilde{G}[S]$ if the expansion of $C$ is a solution of the problem.
\end{proof}
\subsection{Secluded Star}
Another example of a particular problem where we have better running times is \ProblemName{Secluded Star}, defined as follows.
\defproblema{\ProblemName{Secluded Star}}%
{A graph $G$, a weight function $\omega\colon V(G)\rightarrow \mathbb{Z}_{>0}$, an nonnegative integer $t$, and a positive integer $w$.}%
{Decide whether $G$ contains a $t$-secluded induced star $S$ with $\omega(V(S))\geq w$.}
In this case, a faster FPT algorithm can be deduced via a reduction to the problem \textsc{Vertex Cover} parameterized by the size of the solution.
For a graph $G$ and $x$ in $V(G)$, we call $N^2(x)$ the set of vertices at distance $2$ from $x$, i.e., $N^2(x)$ is the set of vertices $u \in V(G) $ such that $ u \notin N[x]$ and there exists $ v \in N(x)$ such that $ u \in N(v)$. We also call $N^2[x]$ the set $N^2(x) \cup N[x]$. Let now $F_x = (N^2(x) \cup N(x), E')$ be the subgraph of $G[N^2(x) \cup N(x)]$ such that $E' = E(G[N^2(x)\cup N(x)]) - E(G[N^2(x)])$, i.e., $F_x$ is the graph induced by the vertices in $N^2(x) \cup N(x)$ after the deletion of all edges between nodes in $N^2(x)$. Note that $x$ is not a vertex of $F_x$.
A vertex $x$ is the center of a star $S$ if $x$ is the vertex of maximum degree in $S$.
The following lemma relates the center $x$ of a $t$-secluded star $S$ of a graph $G$, with a vertex cover of size at most $t$ of $F_x$.
\begin{lemma}\label{lem:SStVC}
Let $S$ be a $t$-secluded star on a graph $G$ and let $x$ be the center of $S$. Then $N_G(V(S))$ is a vertex cover of $F_x$. Moreover, if $S$ is an inclusion maximal $t$-secluded star with the center $x$, then $N_G(V(S))$ is an inclusion minimal vertex cover of $F_x$.
\end{lemma}
\begin{proof}
Let $u,v$ two adjacent vertices of $F_x$. Since $F_x[N^2(x)]$ is edgeless, we assume w.l.o.g. that $u$ belongs to $N_G(x)$. If $u$ is contained in $N_G(x) \setminus V(S)$, then $u$ is in $N(V(S))$ (because $x$ belongs to $S$, so $N_G(x) \setminus V(S) \subseteq N_G(V(S))$). If $u$ is in $S$, then either $v$ is in $N_G^2(x)$ or $v$ is in $N_G(x) \setminus V(S)$ (because $F_x[V(S)]$ is edgeless). In both cases $v$ is in $N_G(V(S))$. We conclude that either $u$ or $v$ is contained in $N_G(V(S))$.
Assume that $S$ is an inclusion maximal $t$-secluded star with the center $x$. Suppose that, contrary to the second claim, $N_G(V(S))$ is not an inclusion minimal vertex cover of $F_x$, that is, there is $u\in N_G(V(S))$ such that $X=N_G(V(S))\setminus\{u\}$ is a vertex cover of $F_x$. Because $X$ is a vertex cover of $F_x$ and $X\subseteq N_G(V(S))$, we have that $u\in N_G(x)$ and $N_G(u)\setminus \{x\}\subseteq X$. It implies that $S\rq{}=G[V(S)\cup \{u\}]$ is a $t$-secluded star contradicting the maximality of $S$.
\end{proof}
A basic result on parameterized complexity states that it could decided if a graph contains a vertex cover of size at most $t$ in time
$\mathcal{O}(c^t(n+m))$ for some constant $c$ (see, e.g., \cite{CyganFKLMPPS15,DowneyF13}) by branching algorithms. These algorithms could be adapted to output the list of all inclusion minimal vertex covers of size at most $t$ within the same running time. In fact, it could be done for $c<2$ but this demands some discussion. Hence, we use the following straightforward observation.
\begin{proposition}[\cite{CyganFKLMPPS15,DowneyF13}]\label{prop:vc-enum}
There is an algorithm computing the list of all the inclusion minimal vertex covers of size at most $t$ of a graph $G$ in time $\mathcal{O}(2^t(n+m))$.
\end{proposition}
\begin{theorem}
\ProblemName{Secluded Star} can be solved in time $\mathcal{O}(2^t \cdot n^{\mathcal{O}(1)})$.
\end{theorem}
\begin{proof}
Let $(G,\omega, t, w)$ be an input of \ProblemName{Secluded Star}. The algorithm starts computing, for every $x \in V(G)$, the list of all inclusion minimal vertex covers of size at most $t$ of $F_x$ using Proposition~\ref{prop:vc-enum}. Then, for every vertex cover $U$ of size at most $t$ of $F_x$, the algorithm checks if $N_G[x]\setminus U$ induces in $G$ a solution of the problem. We know from Lemma \ref{lem:SStVC}, that if $S$ is a solution of \ProblemName{Secluded Star} on input $(G, \omega, t, w)$, then $N_G(V(S))$ is an inclusion minimal vertex cover of size at most $t$ of $F_x$. Note also that $S$ is an induced star with center $x$ in a graph $G$, then $V(S) = N_G[x] \setminus N_G(V(S))$.
\end{proof}
\subsection{Secluded Regular Subgraph}
Another example of a problem with single-exponential FPT algorithm is \ProblemName{Connected Secluded Regular Subgraph}, defined as follows.
\defproblema{\ProblemName{Connected Secluded Regular Subgraph}}%
{A graph $G$, a weight function $\omega\colon V(G)\rightarrow \mathbb{Z}_{>0}$, an nonnegative integer $t$, and positive integers $w$ and $d$.}%
{Decide whether $G$ contains a connected $t$-secluded $d$-regular induced subgraph $H$ with $\omega(V(H))\geq w$.}
Let $(G,\omega,t,w,d)$ be an input of \ProblemName{Connected Secluded Regular Subgraph} and let $U$ be a set of vertices of $G$ such that $G[U]$ is a solution of the problem. Note first that any vertex of degree greater than $t+d$ can not be contained in $U$, otherwise $G[U]$ is not $t$-secluded. Let $W$ be the set of vertices of high degree, i.e., $x\in W$ if $|N(x)| \geq t+d+1$. Suppose that $N(U) \subseteq W$. This implies that $U$ is a component of $G-W$. Therefore, our algorithm will first compute the components of $G-W$ and check if some of them is a solution. In the following we assume that $N(U) \setminus W \neq \emptyset$.
We call $L = N(U)\setminus W$, $U_1 = N(L) \cap U$, $U_2 = N(U_1) \cap U$ and $\tilde{U} = U_1 \cup U_2$. Note that $|L| \leq t$, $|U_1| \leq t\cdot (t+d)$ and $|U_2| \leq d t \cdot (t+d)$. Therefore $|\tilde{U}| \leq t(d+1)(t+d)$.
A set of vertices $C$ is called \emph{good} for $U$ if
\begin{itemize}
\item $C \subseteq U$, and
\item For all $u \in C$, $u\in U_1$ implies $N(u)\cap U \subseteq C$.
\end{itemize}
Note that every node $u$ in a good set $C$ satisfies $|N(u)\cap C| \leq d$. Moreover if $u \in U_1 \cap C$ then $|N(u)\cap C| = d$. Note also that if $C_1$ and $C_2$ are good for $U$ then $C_1 \cup C_2$ is good for $U$.
\begin{lemma}\label{lem:dregulargrow}
Let $S$ be a set of vertices of $G$ satisfying $\tilde{U} \subseteq S$ and $S\cap L = \emptyset$. Let now $C$ be a component of $G[S]$ such that $C \cap U \neq \emptyset$, then:
\begin{itemize}
\item[1)] $C$ is good for $U$, and
\item[2)] if $u \in C$ is such that $|N(u) \cap C| < d$, then $S' = S \cup N(u)$ satisfies $\tilde{U} \subseteq S'$ and $S'\cap L = \emptyset$.
\end{itemize}
\end{lemma}
\begin{proof} \hspace{1cm}
\begin{itemize}
\item[1)] Let $u \in C \cap U$ and $v \in N(u) \cap (C \setminus U)$. Then $v$ is contained in $L$, which contradicts the fact that $S\cap L = \emptyset$. Therefore $C \subseteq U$. On the other hand, if $u \in U_1 \cap C$ then $N(u) \cap U \subseteq C$, because $U_2$ is contained in $S$ and $C$ is a connected component of $G[S]$. We conclude that $C$ is good for $U$.
\item[2)] Let $u \in C$ be such that $|N(u) \cap C| < d$. Since $C$ is good for $U$ we know that $u$ is not contained in $U_1$, so $N(u) \cap L = \emptyset$. Therefore $\tilde{U} \subseteq S \subseteq S'$ and $S' \cap L = \emptyset$.
\end{itemize}
\end{proof}
\begin{theorem}
\ProblemName{Connected Secluded Regular Subgraph} can be solved in time $2^{\mathcal{O}(t\log (td))} \cdot n^{\mathcal{O}(1)}$.
\end{theorem}
\begin{proof}
The algorithm for \ProblemName{Connected Secluded Regular Subgraph} on input $(G,\omega, t, w, d)$ first computes the set $W$ of vertices $v$ satisfying $|N(v)| > t+d+1$. Then computes the connected components of $G-W$ and checks if some of them is a solution. If a solution is not found this way, the algorithm computes $\tilde{G} = G-W$. Then, the algorithm uses Lemma \ref{lem:derand} to compute in time
$2^{\mathcal{O}(t \log (td))}n \log n$
a family $\mathcal{S}$ of at most $2^{\mathcal{O}(t\log (td))}\log n$
subsets of $V(\tilde{G})$ such that: for any sets $A, B \subseteq V(\tilde{G})$, $A \cap B = \emptyset$, $|A| \leq t(d+1)(t+d)$, $|B| \leq t$, there exists a set $S \in \mathcal{S}$ with $A\subseteq S$ and $B\cap S = \emptyset$.
For each set $S \in \mathcal{S}$, the algorithm mark as \emph{candidate} every component $C$ of $\tilde{G}[S]$ that satisfy that for all $u\in C$, $|N(u) \cap C| \leq d$. For each candidate component $C$, the algorithm looks for a node $u$ in $C$ such that $|N(u) \cap C| <d$. If such node is found, the corresponding component is enlarged adding to $C$ all nodes in $N(u)$. If $N(u)$ intersects other components of $G[S]$, we merge them into $C$. We repeat the process on the enlarged component $C$ until it can not grow any more, or some node $u$ in $C$ satisfies $|N(u) \cap C| >d$. In the first case we check if the obtained component is a solution of the problem, otherwise the component is unmarked (is not longer a candidate), and the algorithm continues with another candidate component of $\tilde{G}[S]$ or other set $S'\in \mathcal{S}$.
Let $U$ be a set of vertices such that $G[U]$ is a solution of \ProblemName{Connected Secluded Regular Subgraph} on input $(G,\omega, t, w, d)$ such that $N(U) \setminus W \neq \emptyset$. From construction of $\mathcal{S}$, we know that there exists some $S \in \mathcal{S}$ such that $\tilde{U} \subseteq S$ and $S \cap L = \emptyset$. From Lemma \ref{lem:dregulargrow} (1), we know that any component of $\tilde{G}[S]$ intersecting $U$ is good for $U$, so it will be marked as candidate. Finally, from Lemma \ref{lem:dregulargrow} (2) we know that when a component that is good for $U$ grows, the obtained component is also good for $U$.
Indeed, if $C$ is a good component of $\tilde{G}[S]$ containing a node $u$ such that $|N(u)\cap C| <d$, then the component of $\tilde{G}[S \cup N(u)]$ containing $C$ is also good for $U$. We conclude that we correctly find $U$ testing the enlarging process on each component of $\tilde{G}[S]$, for each $S \in \mathcal{S}$.
\end{proof}
\subsection{Secluded Long Path}
Our last problem is \ProblemName{Secluded Long Path}, defined as follows.
\defproblema{\ProblemName{Secluded Long Path}}%
{A graph $G$, a weight function $\omega\colon V(G)\rightarrow \mathbb{Z}_{>0}$, an nonnegative integer $t$, and a positive integer $w$.}%
{Decide whether $G$ contains a $t$-secluded induced path $P$ with $\omega(V(P))\geq w$.}
In this case the reasoning is very similar to the one for problem \ProblemName{Connected Secluded Regular Subgraph} when $d=2$. Let $(G,\omega,t,w)$ be an input of \ProblemName{Secluded Long Path} and let $P$ be a set of vertices of $G$ such that $G[P]$ is a solution of the problem. Note first that any vertex of degree greater than $t+2$ can not be contained in $P$, otherwise $G[P]$ is not $t$-secluded. Let $W$ be the set of vertices of high degree, i.e., $x\in W$ if $|N(x)| \geq t+3$. Suppose first that $N(P) \subseteq W$. This implies that $P$ is a component of $G-W$. Therefore, our algorithm will first compute the components of $G-W$ and check if some of them is a solution. In the following we assume that $N(P) \setminus W \neq \emptyset$.
We call again $L = N(P)\setminus W$ and define $P_1 = N(L) \cap P$, $P_2 = N(P_1) \cap P$ and $\tilde{P} = P_1 \cup P_2$. Note that $|L| \leq t$, $|P_1| \leq t\cdot (t+3)$ and $|P_2| \leq 2t\cdot (t+3)$. Therefore $|\tilde{P}| \leq 3t\cdot (t+3)$. For this case, it will be enough to define good sets as paths. A path $P'$ is called \emph{good} for $P$ if satisfies the following conditions:
\begin{itemize}
\item $P' \subseteq P$, and
\item For all $u \in P'$, $u\in P_1$ implies $N(u)\cap P \subseteq P'$.
\end{itemize}
Note that every vertex $u$ in a path $P'$ good for $P$ satisfies $|N(u)\cap C| \leq 2$. However, unlike the $d$ regular case, where the nodes in $U_1$ had $d$ neighbors in a good set, in this case we may have a node $u \in P_1 \cap P'$ such that $|N(u)\cap P'| = 1$, when it is an endpoint of~$P'$.
\begin{lemma}\label{lem:pathgrow}
Let $S$ be any set of vertices of $G$ satisfying $\tilde{P} \subseteq S$ and $S\cap L = \emptyset$. Let now $C$ be a component of $G[S]$ such that $P \cap U \neq \emptyset$. Then the following holds
\begin{itemize}
\item[1)] $C$ is good for $U$ (so $C$ is a path).
\item[2)] Suppose that $\omega(C) < w$. Then there exists an endpoint $u$ of $C$ satisfying $N(u)\setminus C = \{v\}$, where $v$ is a vertex of $V(G) \setminus S$ contained in $P$ such that $N(v)\setminus W \subseteq P$.
\end{itemize}
\end{lemma}
\begin{proof} \hspace{1cm}
\begin{itemize}
\item[1)] Let $u \in C \cap P$ and $v \in N(u) \cap (C \setminus P)$. Then $v$ is contained in $L$, which contradicts the fact that $S\cap L = \emptyset$. Therefore $C \subseteq P$. On the other hand, if $u \in P_1 \cap C$ then $N(u) \cap P = N(u) \cap \tilde{P} \subseteq C \cap S$. Then $N(u)\cap P \subseteq C$. We conclude that $C$ is good for $P$.
\item[2)]
Since $\omega(C) < w$ and $C$ is good for $P$, $C$ must be strictly contained in $P$. Hence one of the endpoints of $C$ can be extended, i.e., $P \cap (N(u_1) \setminus C) \neq \emptyset$ or $P \cap (N(u_2) \setminus C) \neq \emptyset$. Suppose w.l.o.g. that $v \in P \cap (N(u_1) \setminus C)$ (so $v \in P \setminus S$) and that $N(u_1)$ is not contained in $P$. If $u'$ is a vertex in $N(u_1) \setminus P$, then necessarily $u'$ belongs to $L$. Therefore $u_1$ is in $P_1$, which implies that $v$ is in $\tilde{P}$. This contradicts the fact that $\tilde{P}$ is contained in $S$. We conclude that $N(u_1) \subseteq P$, so $N(u_1)\setminus C = \{v\}$. Clearly $v$ is not in $P_1$, again because otherwise $\tilde{P}$ would not be contained in $S$. We conclude that $N(v)\cap L = \emptyset$, so $N(v) \setminus W \subseteq P$.
\end{itemize}
\end{proof}
\begin{theorem}
\ProblemName{Secluded Long Path} can be solved in time $2^{\mathcal{O}(t \log t)} \cdot n^{\mathcal{O}(1)}$.
\end{theorem}
\begin{proof}
The algorithm for \ProblemName{Secluded Long Path} on input $(G,\omega, t, w)$ first computes the set $W$ of vertices $v$ of $G$ satisfying $|N(v)| > t+2$. Then computes the connected components of $G-W$ and checks if some of them is a solution. If a solution is not found this way, the algorithm computes $\tilde{G} = G-W$.Then, the algorithm uses Lemma \ref{lem:derand} to compute in time $2^{\mathcal{O}(t \log t)}n \log n$ a family $\mathcal{S}$ of at most $2^{\mathcal{O}(t \log t)}\log n$ subsets of $V(\tilde{G})$ such that: for any sets $A, B \subseteq V(\tilde{G})$, $A \cap B = \emptyset$, $|A| \leq 3t(t+3)$, $|B| \leq t$, there exists a set $S \in \mathcal{S}$ with $A\subseteq S$ and $B\cap S = \emptyset$.
For each set $S \in \mathcal{S}$, the algorithm mark as \emph{candidate} every component $C$ of $G[S]$ that is a path and check if one of them is a solution of the problem. Suppose that none of the marked components is a solution. The then algorithm runs a \emph{growing} process on each candidate component $C$ such that $\omega(C) < w$. For each such component $C$, and each endpoint $u$ of $C$, the algorithm checks if $u$ has only one neighbor $v$ outside $C$. If it does, the algorithm add $v$ to the path if $v$ has at most $2$ neighbors not intersecting $N(C)$. Then the growing process is repeated in the new component $C \cup \{v\}$, eventually through the other endpoint of $C$, and through a neighbor of $v$ outside $C$ (the algorithm test both options). If $N(v)$ intersect another path $C'$, we merge $C$ and $C'$. The process stops when the resulting component $C$ is a path such that $\omega(C) \geq w$ or it can not grow any more. In the first case the algorithm checks if $C$ is a solution of the problem,
and otherwise continues with another endpoint of the original component $C$, with another candidate component of $G[S]$ or other set $S' \in \mathcal{S}$.
Let $P$ be a set of vertices such that $G[P]$ is a solution of \ProblemName{Secluded Long Path} on input $(G,\omega, t, w)$, such that $N(P)\setminus W \neq \emptyset$. From construction of $\mathcal{S}$, we know that there exists some $S \in \mathcal{S}$ such that $\tilde{U} \subseteq S$ and $L \cap S = \emptyset$. From Lemma \ref{lem:pathgrow} (1), we know that any component of $\tilde{G}[S]$ intersecting $P$ is good for $U$, so it will be marked as candidate. We know from Lemma \ref{lem:pathgrow} (2) that each component $C$ of $\tilde{G}[S]$ that is good for $P$, such that $\omega(C) < w$, can be extended to a node $v\in P$ such that $N(v) \cap L = \emptyset$. Hence $S' = S \cup \{v\}$ satisfies $\tilde{P} \subseteq S'$ and $S' \cap L = \emptyset$. Then, from Lemma \ref{lem:pathgrow} (1) the component of $\tilde{G}[S']$ containing $C$ is good for $P$. We conclude that we correctly find $P$ testing the growing process on each endpoint of each component of $\tilde{G}[S]$, for each $S \in \mathcal{S}$.
\end{proof}
\section{Concluding remarks}\label{sec:concl}
Another interesting question concerns kernelization for \ProblemName{Connected Secluded $\Pi$-Subgraph}. We refer to~\cite{CyganFKLMPPS15} for the formal introduction to the kernelization algorithms. Recall that a \emph{kernelization} for a parameterized problem is a polynomial algorithm that maps each instance $(x,k)$ with the input $x$ and the parameter $k$ to an instance $(x',k')$ such that i) $(x,k)$ is a yes-instance if and only if $(x',k')$ is a yes-instance of the problem, and ii) $|x'|+k'$ is bounded by $f(k)$ for a computable function $f$. The output $(x',k')$ is called a \emph{kernel}. The function $f$ is said to be a \emph{size} of a kernel. Respectively, a kernel is \emph{polynomial} if $f$ is polynomial.
For \ProblemName{Connected Secluded $\Pi$-Subgraph}, we hardly can hope to obtain polynomial kernels as it could be easily proved by applying the results of Bodlaender et al.~\cite{BodlaenderDFH09} that,
unless $\ensuremath{\operatorClassNP}\subseteq\ensuremath{\operatorClassCoNP}/\text{\rm poly}$,
\ProblemName{Connected Secluded $\Pi$-Subgraph} has no polynomial kernel when parameterized by $t$ if \ProblemName{Connected Secluded $\Pi$-Subgraph} is \ensuremath{\operatorClassNP}-complete.
Nevertheless, \ProblemName{Connected Secluded $\Pi$-Subgraph} can have a polynomial \emph{Turing kernel}.
Let $f\colon\mathbb{N}\rightarrow \mathbb{N}$. A \emph{Turing kernelization} of size $F$ for a parameterized problem is an algorithm that decides whether a given instance $(x,k)$ of the problem, where $x$ is an input and $k$ is a parameter, is a yes-instance in time polynomial in $|x|+k$, when given the access to an oracle that decides whether an instance $(x',k')$, where $|x'|+k'\leq f(k)$, is a yes-instance in a single step. Respectively, a Turing kernel is polynomial if $f$ is a polynomial.
We show that \ProblemName{Connected Secluded $\Pi$-Subgraph} has a polynomial Turing kernel if $\Pi$ is the property to be a star.
\begin{theorem}
\ProblemName{Secluded Star} problem admits a polynomial Turing kernelizaiton.
\end{theorem}
\begin{proof}
Let $S$ be a solution of \ProblemName{Secluded Star} on input $(G, \omega, t, w)$. Remember that for each $x\in V(G)$, we called $F_x$ the subgraph of $G[N_G(x) \cup N_G^2(x)]$ obtained by the deletion of all edges with both endpoints in $N_G^2(x)$. From Lemma \ref{lem:SStVC} if $S$ is a $t$-secluded star of $G$ with center $x$, then $N_G(S)$ is a vertex cover of size at most $t$ of $F_x$. Our kernelization algorithm will first compute, for each $x \in V(G)$, the graph $F_x$. Then, it performs a Buss's kernelization on each graph $F_x$.
For a vertex $x \in V(G)$, let $W_x$ the set of vertices of degree greater than $t$ in $F_x$. Note that every vertex cover of size at most $t$ of $F_x$ must contain $W_x$. Hence, if $|W_x|$ is greater than $t$, then $x$ can not be the center of a $t$-secluded star.
Suppose now that $|W_x|\leq t$ and let $t' = t - |W_x|$. Note that $F_x$ contains a vertex cover of size $t$ if and only if $F_x - W_x$ contains a vertex cover of size at most $t'$. Since $F_x - W_x$ is a graph of degree at most $t$, a vertex cover of size $t'$ can cover at most $t\cdot t'$ edges. In other words, if $F_x - W_x$ contains more than $ 2t\cdot t'$ non isolated vertices, then $x$ can not be the center of a $t$-secluded star.
Suppose now that $F_x - W_x$ has at most $2t\cdot t'$ non isolated vertices. Let $F^+_x$ be the subgraph of $G[N_G^2[x]]$, obtained from $F_x - W_x$ removing all the isolated nodes, and then adding $x$ with all its incident edges. Note that $F^+_x$ contains at most $2t^2 +1$ vertices. Now let $I_x$ be the set of isolated vertices of $F_x - W_x$ contained in $N_G(x)$. Since $N_G(I_x) \setminus \{x\}$ is contained in $W_x$, and $W_x$ is contained in $N_G(S)$ for every $t$-secluded star $S$ with center $x$, we deduce that $I_x$ may be contained in any such star $S$. In other words, if $S$ is a star with center $x$, then $S$ is a solution of \ProblemName{Secluded Star} on input $(G, \omega, t, w)$ if and only if $S - I_x$ is a solution of \ProblemName{Secluded Star} on input $(F^+_x, \omega^+, t', w')$, where $\omega^+(v) = \omega(v)$ for all $v\in V(F^+_x)$ and $w' = w - \omega(I_x)$.
Now let $F'_x$ be the graph obtained from $F^+_x$, attaching to each node $u$ in $N_G^2(x) \cap V(F^+_x)$ a clique $K_u$ of size $2t$, where all vertices of $K_u$ are adjacent to $u$. Note that $F^+_x$ is a graph of size at most $ 4t^3 +1$. Moreover, every $t'$-secluded star in $F'_x$ has center $x$. Indeed, if the center is some vertex in $N(x)$ or $N_G^2(x)$, then $S$ intersects or has as neighbors more than $t$ vertices in some clique $K_u$. Let $\omega'$ be a function on $V(F'_x)$ such that $\omega'(v) = \omega(v)$ if $v \in V(F'_x) \cap V(G)$ and $\omega'(v) = 1$ if $v$ is in one of the cliques adjacent to a node of $N_G^2(x) \cap V(F'_x)$.
We conclude that $(F'_x, \omega', t', w')$ is a yes-instance of \ProblemName{Secluded Star} for some $x \in V(G)$ if and only if $(G, \omega, t, w)$ is a yes-instance of \ProblemName{Secluded Star}.
The kernelization algorithm computes for each $x \in V(G)$ the graph $F_x$ and the set $W_x$. If $W_x \cap N_G^2[x]$ contains more than $t$ nodes the algorithm rejects $x$ and continue with another vertex of $V(G)$. If $|W_x|\leq t$ it computes $F^+_x$ deleting all the isolated vertices of $F_x - W_x$. If $F^+_x$ contains more than $2t\cdot t_1 +1$ nodes the algorithm rejects $x$ and continue with another vertex of $V(G)$. Finally, the algorithm computes $F_x'$, $t'$, $w'$ and calls the oracle on input $(F'_x, \omega', t', w')$. If the oracle answers that $(F'_x, \omega', t', w')$ is a yes-instance, the algorithm decides that $(G, \omega, t, w)$ is a yes-instance. Otherwise the algorithm continue with another vertex of $V(G)$. The algorithm ends when some oracle accepts, or all nodes are rejected.
\end{proof}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 735
|
Cet article liste les députés (Teachta Dála) de la 31e législature irlandaise (Dáil Éireann) de 2011 à 2016.
Députés par circonscription
Coalition gouvernementale
Fine Gael
James Bannon
Seán Barrett
Tom Barry
Pat Breen
Richard Bruton
Ray Butler
Jerry Buttimer
Catherine Byrne
Ciarán Cannon
Joe Carey
Paudie Coffey
Áine Collins
Seán Conlan
Paul Connaughton, Jnr
Noel Coonan
Marcella Corcoran Kennedy
Simon Coveney
Michael Creed
Lucinda Creighton
Jim Daly
John Deasy
Jimmy Deenihan
Pat Deering
Regina Doherty
Paschal Donohoe
Andrew Doyle
Bernard Durkan
Damien English
Alan Farrell
Frank Feighan
Frances Fitzgerald
Peter Fitzpatrick
Charles Flanagan
Terence Flanagan
Brendan Griffin
Noel Harrington
Simon Harris
Brian Hayes
Tom Hayes
Martin Heydon
Phil Hogan
Heather Humphreys
Derek Keating
Paul Kehoe
Enda Kenny
Seán Kyne
Anthony Lawlor
Peter Mathews
Shane McEntee
Nicky McFadden
Dinny McGinley
Joe McHugh
Tony McLoughlin
Olivia Mitchell
Mary Mitchell O'Connor
Michelle Mulherin
Dara Murphy
Eoghan Murphy
Denis Naughten
Dan Neville
Michael Noonan
Kieran O'Donnell
Patrick O'Donovan
Fergus O'Dowd
John O'Mahony
Joe O'Reilly
John Perry
John Paul Phelan
James Reilly
Michael Ring
Alan Shatter
David Stanton
Billy Timmins
Liam Twomey
Leo Varadkar
Brian Walsh
Parti travailliste
Tommy Broughan
Joan Burton
Eric Byrne
Michael Conaghan
Ciara Conway
Joe CostelloRobert Dowds
Anne Ferris
Eamon Gilmore
Dominic Hannigan
Brendan Howlin
Kevin Humphreys
Colm Keaveney
Alan Kelly
Seán Kenny
Ciarán Lynch
Kathleen Lynch
John Lyons
Eamonn Maloney
Michael McCarthy
Michael McNamara
Gerald Nash
Derek Nolan
Aodhán Ó Ríordáin
Jan O'Sullivan
Willie Penrose
Ann Phelan
Ruairi Quinn
Pat Rabbitte
Brendan Ryan
Seán Sherlock
Róisín Shortall
Arthur Spring
Emmet Stagg
Joanna Tuffy
Jack Wall
Alex White
Opposition
Fianna Fáil
John Browne
Dara Calleary
Niall Collins
Barry Cowen
Timmy Dooley
Seán Fleming
Billy Kelleher
Michael Kitt
Brian Lenihan
Micheál Martin
Charlie McConalogue
Michael McGrath
John McGuinness
Michael Moynihan
Éamon Ó Cuív
Willie O'Dea
Seán Ó Fearghaíl
Brendan Smith
Robert Troy
Sinn Féin
Gerry Adams
Michael Colreavy
Seán Crowe
Pearse Doherty
Dessie Ellis
Martin Ferris
Mary Lou McDonald
Sandra McLellan
Pádraig Mac Lochlainn
Jonathan O'Brien
Caoimhghín Ó Caoláin
Aengus Ó Snodaigh
Brian Stanley
Peadar Tóibín
People Before Profit Alliance
Richard Boyd Barrett
Joan Collins
Parti socialiste
Clare Daly
Joe Higgins
Workers and Unemployed Action Group
Séamus Healy
Indépendants
Stephen Donnelly
Luke 'Ming' Flanagan
Tom Fleming
Noel Grealish
John Halligan
Michael Healy-Rae
Michael Lowry
Finian McGrath
Mattie McGrath
Catherine Murphy
Maureen O'Sullivan
Thomas Pringle
Shane Ross
Mick Wallace
31
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 6,053
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A Powerful Prayer - Northstar Knights win in more ways than one
|By [email protected]
(Frisco, TX) Alexandria's Northstar Knights continue to prove that they are a definite force in 18U hockey on a national scale, but it's what they are doing after the whistle that's drawing even more attention.
By Joe Korkowski via Voice of Alexandria. See original article from Voice of Alexandria: HERE.
On Monday afternoon, amid a chaotic day because of the snow and ice in Texas, the Knights were able to come through with a 4-3 overtime win over the Philadelphia Little Flyers to claim the Dixon Cup. The two teams were competing in the North American Prospects Hockey League (NAPHL) Championship game, not long after the Knights 16U team had lost out in their championship game to the Florida Alliance 5-2.
During the game there were plenty of plays on the ice that are worthy of note. Among them was the 3-goal game performance by the Knight's Dylan Cook (Princeton, MN); including the game-tying goal in the third to help send it into overtime. Cody Croal (North Branch, ND) also had a "good day" on the ice getting two assists and the game-winning goal in overtime off of a Reese Laubach assist. The power, or lack thereof, provided its own effect on the game. Controlled blackouts in the area to conserve electricity amid the excess demands for heat caused delays in the Comerica Center Arena. Each time the blackouts occurred it stopped the game, requiring about a 30 minute wait to return to action.
Following the exciting overtime win, Northstar captain, and Philadelphia area native, Chris Brown asked to take the jersey of Little Flyers injured player, Brian Page and lay it in the center of ice. Both teams gathered around and prayed for Brian. along with the safe travels for all players and family.
Brian Page, 17 absorbed an on-ice hit on Nov. 15, 2020 leaving him paralyzed from the chest down. Brian has had a tremendous career playing offense for Delaware Ducks, Philadelphia Jr. Flyers, Philadelphia Little Flyers and most recently signed a tender agreement to play in the North American Hockey League. Brian is in his senior year of High School and has been a player of interest to top college hockey programs.
A post on the Brian Page #87Strong twitter account says that they watched, via video, Chris (Brown) skate to the Philly Flyer bench and grab Brian's jersey. The post commented on how Chris is going through his own journey filled with challenges and yet he thought of another. "His selfless act of leadership has gone beyond the bounds of hockey and onto a greater calling. This moment wasn't by accident, but by design, as God puts us where He wants us, when he wants us there. We all needed this moment to happen, to allow it (to) move each of us for different reasons."
You can watch the game in its entirety by subscribing to Hockey TV at this link.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 3,044
|
package de.iritgo.aktario.core.io.i18n;
import de.iritgo.aktario.core.Engine;
/**
* I18NStrings hold a key to a string resource instead of the string itself.
* The getter method automatically translates the key to this string resource.
*/
public class I18NString
{
/** The string resource key */
private String key;
public I18NString()
{
}
public I18NString(String key)
{
this.key = key;
}
public String getKey()
{
return key;
}
public void setKey(String key)
{
this.key = key;
}
/**
* Get the translated string. If no string resource was found under the
* specified key, the key is returned itself.
*
* @return The translated key or the key itself
*/
public String get()
{
return Engine.instance().getResourceService().getStringWithoutException(key);
}
/**
* Same as @see de.iritgo.aktario.core.io.i18n.I18NString#get() but
* additional parameters are supplied to replace {n} placeholders in the
* string resource.
*
* @param params The string parameters
* @return The translated key or the key itself
*/
public String get(Object... params)
{
return Engine.instance().getResourceService().getStringWithParams(key, params);
}
/**
* Calls @see de.iritgo.aktario.core.io.i18n.I18NString#get() to return
* the translated string.
*
* @return The translated string resource
*/
@Override
public String toString()
{
return get();
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 3,502
|
Эль-Аббасия () — город на юге Судана, расположенный на территории штата Южный Кордофан. Входит в состав округа Рашад.
Географическое положение
Город находится в северо-восточной части штата, на плато Кордофан, на высоте 651 метра над уровнем моря.
Эль-Аббасия расположена на расстоянии приблизительно 210 километров к северо-востоку от Кадугли, административного центра штата и на расстоянии 383 километров к юго-юго-западу (SSW) от Хартума, столицы страны.
Конфликт в Южном Кордофане
28 января 2012 года, в окрестностях Эль-Аббасии, бойцами Народной армии освобождения Судана были захвачены в плен 29 китайских рабочих, участвовавших в строительстве дороги.
См. также
Города Судана
Примечания
Города Судана
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 2,807
|
/**
*/
package org.afplib.afplib;
import org.afplib.base.Triplet;
/**
* <!-- begin-user-doc -->
* A representation of the model object '<em><b>Color Specification</b></em>'.
* <!-- end-user-doc -->
*
* <p>
* The following features are supported:
* </p>
* <ul>
* <li>{@link org.afplib.afplib.ColorSpecification#getColSpce <em>Col Spce</em>}</li>
* <li>{@link org.afplib.afplib.ColorSpecification#getColSize1 <em>Col Size1</em>}</li>
* <li>{@link org.afplib.afplib.ColorSpecification#getColSize2 <em>Col Size2</em>}</li>
* <li>{@link org.afplib.afplib.ColorSpecification#getColSize3 <em>Col Size3</em>}</li>
* <li>{@link org.afplib.afplib.ColorSpecification#getColSize4 <em>Col Size4</em>}</li>
* <li>{@link org.afplib.afplib.ColorSpecification#getColor <em>Color</em>}</li>
* </ul>
*
* @see org.afplib.afplib.AfplibPackage#getColorSpecification()
* @model
* @generated
*/
public interface ColorSpecification extends Triplet {
/**
* Returns the value of the '<em><b>Col Spce</b></em>' attribute.
* <!-- begin-user-doc -->
* <!-- end-user-doc -->
* <!-- begin-model-doc -->
* <p>mandatory<br>fixed length</p>
* <!-- end-model-doc -->
* @return the value of the '<em>Col Spce</em>' attribute.
* @see #setColSpce(Integer)
* @see org.afplib.afplib.AfplibPackage#getColorSpecification_ColSpce()
* @model required="true"
* @generated
*/
Integer getColSpce();
/**
* Sets the value of the '{@link org.afplib.afplib.ColorSpecification#getColSpce <em>Col Spce</em>}' attribute.
* <!-- begin-user-doc -->
* <!-- end-user-doc -->
* @param value the new value of the '<em>Col Spce</em>' attribute.
* @see #getColSpce()
* @generated
*/
void setColSpce(Integer value);
/**
* Returns the value of the '<em><b>Col Size1</b></em>' attribute.
* <!-- begin-user-doc -->
* <!-- end-user-doc -->
* <!-- begin-model-doc -->
* <p>mandatory<br>fixed length</p>
* <!-- end-model-doc -->
* @return the value of the '<em>Col Size1</em>' attribute.
* @see #setColSize1(Integer)
* @see org.afplib.afplib.AfplibPackage#getColorSpecification_ColSize1()
* @model required="true"
* @generated
*/
Integer getColSize1();
/**
* Sets the value of the '{@link org.afplib.afplib.ColorSpecification#getColSize1 <em>Col Size1</em>}' attribute.
* <!-- begin-user-doc -->
* <!-- end-user-doc -->
* @param value the new value of the '<em>Col Size1</em>' attribute.
* @see #getColSize1()
* @generated
*/
void setColSize1(Integer value);
/**
* Returns the value of the '<em><b>Col Size2</b></em>' attribute.
* <!-- begin-user-doc -->
* <!-- end-user-doc -->
* <!-- begin-model-doc -->
* <p>mandatory<br>fixed length</p>
* <!-- end-model-doc -->
* @return the value of the '<em>Col Size2</em>' attribute.
* @see #setColSize2(Integer)
* @see org.afplib.afplib.AfplibPackage#getColorSpecification_ColSize2()
* @model required="true"
* @generated
*/
Integer getColSize2();
/**
* Sets the value of the '{@link org.afplib.afplib.ColorSpecification#getColSize2 <em>Col Size2</em>}' attribute.
* <!-- begin-user-doc -->
* <!-- end-user-doc -->
* @param value the new value of the '<em>Col Size2</em>' attribute.
* @see #getColSize2()
* @generated
*/
void setColSize2(Integer value);
/**
* Returns the value of the '<em><b>Col Size3</b></em>' attribute.
* <!-- begin-user-doc -->
* <!-- end-user-doc -->
* <!-- begin-model-doc -->
* <p>mandatory<br>fixed length</p>
* <!-- end-model-doc -->
* @return the value of the '<em>Col Size3</em>' attribute.
* @see #setColSize3(Integer)
* @see org.afplib.afplib.AfplibPackage#getColorSpecification_ColSize3()
* @model required="true"
* @generated
*/
Integer getColSize3();
/**
* Sets the value of the '{@link org.afplib.afplib.ColorSpecification#getColSize3 <em>Col Size3</em>}' attribute.
* <!-- begin-user-doc -->
* <!-- end-user-doc -->
* @param value the new value of the '<em>Col Size3</em>' attribute.
* @see #getColSize3()
* @generated
*/
void setColSize3(Integer value);
/**
* Returns the value of the '<em><b>Col Size4</b></em>' attribute.
* <!-- begin-user-doc -->
* <!-- end-user-doc -->
* <!-- begin-model-doc -->
* <p>mandatory<br>fixed length</p>
* <!-- end-model-doc -->
* @return the value of the '<em>Col Size4</em>' attribute.
* @see #setColSize4(Integer)
* @see org.afplib.afplib.AfplibPackage#getColorSpecification_ColSize4()
* @model required="true"
* @generated
*/
Integer getColSize4();
/**
* Sets the value of the '{@link org.afplib.afplib.ColorSpecification#getColSize4 <em>Col Size4</em>}' attribute.
* <!-- begin-user-doc -->
* <!-- end-user-doc -->
* @param value the new value of the '<em>Col Size4</em>' attribute.
* @see #getColSize4()
* @generated
*/
void setColSize4(Integer value);
/**
* Returns the value of the '<em><b>Color</b></em>' attribute.
* <!-- begin-user-doc -->
* <!-- end-user-doc -->
* <!-- begin-model-doc -->
* <p>mandatory<br>variable length</p>
* <!-- end-model-doc -->
* @return the value of the '<em>Color</em>' attribute.
* @see #setColor(byte[])
* @see org.afplib.afplib.AfplibPackage#getColorSpecification_Color()
* @model required="true"
* @generated
*/
byte[] getColor();
/**
* Sets the value of the '{@link org.afplib.afplib.ColorSpecification#getColor <em>Color</em>}' attribute.
* <!-- begin-user-doc -->
* <!-- end-user-doc -->
* @param value the new value of the '<em>Color</em>' attribute.
* @see #getColor()
* @generated
*/
void setColor(byte[] value);
} // ColorSpecification
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 6,187
|
## Victoria Alexander
## The Virgin's Secret
The book is dedicated with affection and thanks
to Mariah Stewart
for the loan of Alistair McGowan and Shandihar;
to Amy Mayberry,
who tries her best to keep me on track;
and to my friend Irene Mercatante,
who is Most Excellent in so many ways.
## Contents
Prologue
We aren't supposed to be up here," Sterling Harrington said...
One
It appears the natives are particularly restless this year." Nathanial...
Two
Only the top half of her face was visible over...
Three
Gabriella's heart thudded in her chest. Most annoying, as she'd...
Four
Excellent," Miss Montini murmured, her gaze still on the papers...
Five
There was indeed a servant waiting to escort her to...
Six
What did you do to Mr. Dennison?" Nate strode into...
Seven
You—You—You—" Gabriella sputtered as if she couldn't quite catch her...
Eight
Damnation, she knew that voice.
Nine
I must say, this is all very interesting." Merrill Beckworth...
Ten
Share the credit for discovery of the seal?" Gabriella paced...
Eleven
They were plying her for information, that's what they were...
Twelve
Gabriella perched nervously on the edge of a red velvet...
Thirteen
If I leave to fetch you some refreshment, will you...
Fourteen
Shall we?" Lord Rathbourne offered his arm.
Fifteen
What in the name of all that's holy did you...
Sixteen
My apologies once again, Mr. McGowan." Gabriella smiled up at...
Seventeen
Nate casually stepped up to the refreshment table beside McGowan.
Eighteen
Lord Rathbourne will be with you shortly." His lordship's stern-faced...
Nineteen
She slammed the door in his face? How dare she?
Twenty
What do you mean this is not Montini's seal?" Disbelief...
Twenty-One
What are you doing here?" Nathanial asked sharply.
Twenty-Two
You asked to see me, sir."
Twenty-Three
Even the vast number of treasures that might well take...
Twenty-Four
Would you be so good as to tell Lord Rathbourne...
Twenty-Five
Well?" Nate demanded the moment Quint stepped into the library.
Twenty-Six
You what?" At once he was wide-awake.
Twenty-Seven
Dr. Crenshaw stepped out of Gabriella's room the next morning,...
Twenty-Eight
Caroline," Mother called with an enthusiastic wave.
Twenty-Nine
This was the final day.
Epilogue
You should have taken the seal back from Montini when...
About the Author
Other Books by Victoria Alexander
Copyright
About the Publisher
## Prologue
London, 1867
We aren't supposed to be up here," Sterling Harrington said in the superior tone of an older brother, a tone Nathanial had heard all too often in his eight years. Even at age eleven, Sterling already sounded like the earl he would one day be, at least to his youngest sibling.
"I daresay it won't be a problem if we don't get caught." Quinton Harrington pushed past Sterling, moved away from the stairs and farther into the attic. As Quinton carried the candle, his brothers weren't far behind.
Quinton was two years younger than Sterling and a year older than Nathanial. While Sterling was always the leader in any of the trio's escapades, Quinton was more often than not the brother who came up with the idea in the first place. Their governess, Miss Thompson, said Quinton had an adventurous soul, and even though it didn't sound like a compliment to Nathanial, Quinton took it as such. Sterling, however, was the responsible brother, as Miss Thompson often said in an approving manner; as befit the future Earl of Wyldewood, she would add.
Sterling was the one who took the blame when things went awry. He said it was his duty, although why anyone would want to take the blame made no sense to Nathanial. It was another one of those things he assumed he'd probably understand when he was older.
When Nathanial had suggested they grow frogs in the bathtub on the third floor, Sterling claimed it was his idea when maids found a tub full of squirming tadpoles. Girls certainly did make a fuss over things like that. When the boys lost a ball down the old well in the back garden, Quinton was the one who proposed lowering Nathanial into the well to retrieve it because he was the smallest. Nathanial never would have told his brothers that it was much darker in the well than he had expected, and more than a bit scary. But it was Sterling who told Father it was his idea, and Sterling who was punished, even if Quinton did then come forward.
Miss Thompson said that at least the scamp had a conscience, whatever that was, although she apparently thought it was a good thing. And when the governess read them a story about a Greek boy who tried to fly with wings made of feathers and wax, it was Sterling who scoffed and said they should have used glue. But it was Quinton who managed to find the glue and feathers and sticks they needed to build their own wings.
It had taken them nearly a week. When they finished, they used an old rose trellis to climb to the roof of the gardener's cottage. Nathanial, of course, was picked to make the flight. Being the youngest and smallest did have its drawbacks. If they hadn't tied a rope around his waist—to make certain he didn't fly away—he might have been hurt. As it was, he ended up dangling from the roof in need of adult rescue. They were all punished for that adventure. The trellis had now been removed, bathtubs were only to be used for bathing, and the well had been filled in. But Nathanial would still follow his older brothers anywhere.
"It's rather too dark to see anything," he said now, as if stating a fact and not at all bothered by the dark.
The rain thrummed against the roof of Harrington House, a sound not nearly as ominous on the lower floors. If he had been alone, Nathanial might have found the dim, cavernous attic a little frightening. On a sunny day, the garden of the family's London home and the parks nearby provided ample opportunity for adventure, but it had been raining off and on for three days now, and the boys were confined to the house. As was Miss Thompson. Perhaps it was that last prank they pulled that had been the final straw. Miss Thompson was the only girl they knew who didn't seem at all bothered by frogs, but finding one in her desk drawer today did seem to upset her, oddly enough. She'd sent the boys off with orders to read, and then retreated to her private sitting room. She did that on occasion. Usually when it rained.
"So." Quinton surveyed the attic, holding the candle high. He already looked like a pirate, and as soon as they found pirate clothes, they all would. "Where should we begin?"
"The trunks," Sterling said. "There will be pirate clothes in the trunks." He led the way toward the far recesses of the attic, a rather dark and scary place, Nathanial thought, if truth were told. But his brothers were with him so he needn't worry.
"Which one?" Sterling studied the various assorted trunks that looked exactly like pirate treasure chests. Only bigger.
"The biggest of course." Quinton flashed a grin at his younger brother. "The biggest always has the best treasure."
"Very well." Sterling lifted the lid on the largest trunk and the boys peered inside.
"There's only clothes in there." Nathanial grimaced. He had rather hoped they would indeed find treasure.
"These aren't just clothes." Quinton handed Nathanial the candle, then reached into the trunk and pulled out a red coat that looked like one on their painted tin soldiers. "These are clothes for pirates and knights."
"And adventurers," Sterling said. "And explorers."
"I want to be an explorer," Nathanial said quickly. "Or an adventurer."
"Look at this." Sterling pulled something else out of the trunk.
Quinton grimaced. "It's a book."
"It's a journal." Sterling moved closer to the candle and flipped through the journal. "It's great-grandmother's."
"It's still just a book," Quinton said.
"I know," Sterling murmured. "But it might be a good book."
Quinton scoffed. "How good can a book be?"
"You like books about pirates," Nathanial offered.
"This one is about smugglers." Sterling paged through the journals.
Quinton brightened. "Great-grandmother knew smugglers?"
"I think," Sterling said slowly, "Great-grandmother might have been a smuggler."
"Read it," Nathanial said.
"Very well." Sterling nodded.
The boys sat cross-legged on the floor. Sterling took the candle from Nathanial and positioned it to cast the best light on the pages. For the next hour or so he read to his brothers of the adventures of their great-grandmother, who apparently was indeed a smuggler, pursued by a government agent—a previous Earl of Wyldewood.
At last it stopped raining and Sterling closed the journal. "I don't think we should tell Mother about this," he said firmly.
"Because then we'd have to tell her we were in the attic?" Nathanial asked.
"No." Quinton scoffed. "Because she might not like having a smuggler in the family."
"Oh." Nathanial thought it was rather interesting to have a smuggler in the family. It might be rather interesting to be a smuggler. "Let's be smugglers instead of pirates."
"We can't today," Sterling said. "Miss Thompson will be wondering what became of us. But we can come up here again and read and play smuggler perhaps."
"Can we have smuggler names as well?" Eagerness rang in Nathanial's voice.
"Smuggler names." Quinton laughed. "What are smuggler names?"
"They're like pirate names only for smugglers," Nathanial said in a lofty manner. "And I shall be Black Jack Harrington."
The two older brothers traded glances. Sterling shook his head. "We don't think that's quite right for you."
"Why not?"
"Because your real name isn't Jack, for one thing. We're not just playing you know," Quinton explained with the superiority of an older brother. "It's quite a serious thing to have new names. Even smuggler names. Your smuggler name has to make sense with your real name."
"Nate," Sterling said abruptly. "Sounds like a smuggler, and you can be Quint."
Quinton frowned. "It's not very exciting." He thought for a moment. "What about Peg Leg Quint or Quint the Wicked?"
"More likely Quint the Scamp." Sterling smirked.
"And who will you be?" Nathanial—now Nate—asked. "What will your smuggler name be?"
"I shall remain Sterling."
Quint snorted. "Not much of a name for a smuggler."
"Oh, I shan't be a smuggler." Sterling grinned. "I shall be the intrepid Earl of Wyldewood, agent of the crown, fearless hunter of smugglers. And I shall be the rescuer of the fair maiden, her hero."
"Girls can't play," Nate said firmly. "They're girls."
"Then I shall be Quint." Quinton planted his fists on his hips and puffed out his chest. "Daring, bold King of the Smugglers."
"Who am I to be?" Nate looked from the intrepid earl to the king of the smugglers. It wasn't at all fair. No matter what the game, he always had the last choice.
"Very well." Sterling heaved a long suffering sigh. "I shall give up fearless. You may be the Fearless Smuggler Nate."
"I'd rather like to keep 'daring,' but I shall give you 'bold.'" Quint grinned. "You are now the Fearless Smuggler, Nate the Bold."
The Fearless Smuggler Nate the Bold. He quite liked it.
"We shall have a grand time playing smuggler and smuggler hunter," Sterling said in a most serious manner, as if it were in fact a most serious matter. "And we shall amass great treasures and have grand adventures and rescue fair maidens."
"And wander the world and discover new places," Quint added.
"And...and..." Nate couldn't think of anything. Once again he was last. But it didn't matter. He too could have grand adventures and wander the world.
"We need a pact, I think," Sterling said thoughtfully. "A smugglers' pact."
Nate frowned. "Do smugglers have pacts?"
"I don't know." Quint shrugged. "You mean like musketeers? One for all and all for one?"
"That's a motto." Sterling scoffed. "Besides, we're brothers. We'll always be one for all and all for one."
Nate studied him. "Forever and ever?"
"As we ever have and ever will be." Sterling nodded in a solemn manner, as if he were making a promise that would indeed last forever. "Brothers one for the other."
"One for the other," Quint murmured.
"One for the other." Nate grinned.
It was a very good pact.
## One
They had the look of men who would have disregarded society entirely if they could. If they did not enjoy its comforts and its pleasures. No, not merely society but civilization itself. They shared a similarity of appearance that marked them as brothers, but it was more in the look in their eye and the set of their chin and the confidence in their walk than the coloring of their hair or the breadth of their shoulders or their taller than ordinary height.
There was a look in the eye of the youngest, of intelligence and amusement. Even the least sensible woman knew, upon meeting his gaze, that here was a man who was more than he might at first appear. And knew as well that he was a man who might steal the heart of even the most resistant woman.
But oh, what a lovely theft.
Reflection of a female observer upon meeting Nathanial Harrington and his brother
London, 1885
It appears the natives are particularly restless this year." Nathanial Harrington gazed over the crowd below from his vantage point on the mezzanine balcony.
"It is spring, after all," his older brother, Quinton, said, an amused note in his voice. "The mating rituals have begun."
"I daresay the cream of London society would not be at all pleased at your referring to the season's festivities as mating rituals," Nate said wryly.
"As accurate as the observation might be."
"Accuracy has never played a significant role in the activities of society." Nate glanced at his brother. "Nor, fortunately for you, has punctuality."
Quint shrugged. "I am merely fashionably late."
"You left Egypt a full fortnight before I did, and yet I've been back in London for five days now." Nate eyed his brother. "What kept you? Where have you been?"
"Here and there. As for what kept me, it's remarkable, the number of—" Quint grinned in the wicked manner that had been the downfall of more than one unsuspecting woman. "—'diversions' a man without the accompaniment of his conscience might encounter."
Nate raised a brow. "When you say 'conscience,' are you referring to me?"
"Absolutely, little brother." Quint chuckled. "You are my conscience, the custodian of my morals, the guardian of my virtue, the—"
Nate laughed. "I don't seem to do a very good job of it."
"And for that I am eternally grateful."
"As am I." As much as he hated to admit it, given that trouble seemed to nip incessantly at Quint's heels, Nate knew his life would have been extraordinarily dull were it not for his brother's penchant for adventure.
When Nate had finished his studies, it was Quint who suggested that he join him on his travels and quests for the lost treasures of the ages. Together they had been to lands and places Nate never dreamed he'd see with his own eyes. The day might find them in Egypt or Persia or Asia Minor, where the Nile or the Tigris or the Euphrates flowed. Wherever men had once lived and built cities and aspired to forever.
If truth were told, he'd rather expected to spend his days in the dusty bowels of museum libraries or the hallowed halls of one university or another. He had anticipated his life would consist of merely searching for the knowledge of the ancients. Instead, he now studied yellowed manuscripts and carved stone fragments for clues to finding the tangibles left behind by history. For Nate, the artifacts and antiquities he and his brother found breathed life into long dead civilizations and made them real. Quint was more concerned with the fine price they would bring from museums or collectors. Yet despite their differences in philosophies, or perhaps because of them, they made an excellent and accomplished team.
"Did you..." Quint paused, the question unasked, but then it didn't need to be said aloud.
Nate cast his brother a resigned look. "The fines were paid, the permits arranged for the appropriate—if fictitious—dates to avoid further fines, all necessary authorities received the usual—and in a few cases, more generous than usual—bribes. And the French counsel is now certain it was not you seen leaving his wife's rooms. Attention was diverted toward one of the Americans." Nate shook his head. "It's a pity really. I rather liked them."
"I daresay their morals in matters of this nature are no better than mine. And certainly no better than the French counsel's wife." Quint flashed him an unrepentant smile. "Your help is most appreciated, you know."
"I do." Nate sighed. "However, you should be prepared for Mother's ire. I can't help you there. She was concerned that you wouldn't make it home at all."
"Come now, I would never miss our little sister's coming out ball." Quint adjusted the cuffs at his wrists. He had the look of a man who'd dressed in a hurry, as he no doubt had. "Reggie would cut my heart out, as would Mother and, probably, Sterling as well."
"It does seem a requirement to have all family members present when launching a sister on the seas of society." Nate gazed over the crowd below them. "When did you finally arrive in London?"
"What time is it now?" Quint grinned. "Obviously, I haven't missed anything of importance, nor does it sound as if I missed anything of interest in Alexandria."
"Not really." Nate paused. "Oh, there was someone asking about you."
Quint's grin widened. "Someone is always asking about me."
"Yes, well, this was not a suspicious husband or outraged father. Do you recall Enrico Montini?"
Quint shrugged. "Vaguely."
"Surely you remember him. He claimed to have discovered a seal, ancient—Akkadian, if I remember—that made reference to the Virgin's Secret, the lost city of Ambropia. He was very cautious and wouldn't show us the seal itself, only the clay impression made by the seal." Nate stared at his brother. Quint had worked with the professor who was the leading authority on Ambropia years ago. "You can't possibly have forgotten. It was a remarkable find."
"Yes, of course."
"Apparently he died rather suddenly a few months ago."
"How unfortunate," Quint murmured.
"Indeed. His brother, odd little fellow, accosted me a few days after you left. He was quite irate and accused us, really you—"
"Me?"
"Your reputation precedes you." Nate grimaced. While he worked hard to keep their activities legitimate, there had been incidents before he joined Quint that had been, at the very least, questionable. "Montini's brother suspects someone substituted a seal of lesser quality and age for his, which he then unknowingly presented to the Antiquities Society Validation Committee. Needless to say, they were not amused."
"Very little does amuse them," Quint said under his breath.
"Montini was discredited. His brother claims the shattering of his reputation somehow led to his death, and he wants to find those responsible."
The Validation and Allocation Committee of the London Antiquities Society was charged with determining the significance of the finds of its members who hunted for artifacts in the far corners of the world as well as evaluating proposals for future work. The society's board used the committee's decisions to determine whether to lend support to an expedition. Support that might be as minimal as the use of the society's influential name or as consequential as financial backing.
"You should know I told his brother you had left Egypt for Turkey. I suspect he intended to follow you."
"Most appreciated."
"One does what one can for one's brother." Nate shook his head. "Pity about Montini, though."
"No doubt he simply made a mistake," Quint said.
"Still, if I recall the impressions he showed us—"
"Such things happen all the time. You and I have on occasion believed a find to be more significant than it was." Quint paused, nodded at the gathering below them and abruptly changed the subject. Not that it really mattered. "Whose idea was it to have this ball out of doors?"
Nate chuckled. "Who do you think?"
"And Mother allowed it?"
"She fretted all week about the possibility of rain and what would we do then? But you know how Reggie is when she sets her mind on something." Nate shrugged. "And this is, after all, her party."
Even at age eighteen, Regina Harrington had a strength of character that would be some poor man's undoing one day. Their sister was the youngest child and only girl, and neither her mother nor her brothers had ever managed to say no to her. Reggie had gotten it into her head that it would be a grand idea to have dancing on the terrace under the stars and reserve the ballroom for tables for dinner and conversation. She had ignored her mother's concerns with the blithe confidence known only to young women in their first season. Besides it wouldn't dare rain on Lady Regina Harrington's coming out ball, and it hadn't. It was a perfect spring night.
Nate leaned on the balustrade and studied the crowd. "When was the last time we were in England in the spring?"
"I'm not sure." Quint thought for a moment. "This time last year we were in Persia, and the year before that Egypt, I think, or perhaps Turkey. I really can't say but it's been a long time."
It had been at least six years by Nate's estimation since he and his brother had resided for more than a handful of months at a time in England, at their family's London home or their country estate. They were more likely to be found searching for a lost city in Turkey or a pharaoh's vanished tomb in Egypt or a forgotten temple in Persia and the treasure that would surely accompany such a find. These days they were more at home sleeping under the stars than dancing under them. Nate tugged at the scratchy, starched collar imprisoning his neck. And they'd be far more comfortable as well. Still, it was good to be home.
"As much as I hate to admit it, I have rather missed the London season," Quint said thoughtfully.
Nate scoffed. "I find that hard to believe. I thought you hated all this."
"Nonsense, brother dear." Quint scanned the crowd below them. "I've never especially liked the unrelenting rules governing it all. The 'You must do this' and 'You absolutely cannot do that.' But the array of English beauty on display during the season is unmatched. It's a grand feast and well worth the effort."
Nate chuckled. "A feast?"
"Absolutely." Quint rested his forearms on the balustrade, clasped his hands together and scanned the gathering. He nodded toward a group of fresh-faced, hopeful young females in white gowns.
Nate followed his brother's gaze but his eye caught on a dark-haired young woman. She wore a dress the deep color of ripe apricots and casually circled the terrace as if she were looking for something or someone.
"There you have the debutantes, those in their first season. They are a first course, light and teasing to the appetite. No more than a suggestion of the offerings to come."
"And the second course?" The woman carried herself with the self-assurance borne of beauty, but Nate had the most absurd notion that she was somehow out of place. It was a silly thought. He didn't know half the guests in attendance and wouldn't have known who belonged here and who didn't. Nor did he care.
"There." Quint indicated another group of pastel-clad young ladies. "This is no doubt their second or third season or more. They are somewhat more substantial to the palate but again nothing more than a prelude. As for the main course..." He narrowed his gaze thoughtfully. "Presentation of a plate, its appeal to the eye, is as important as flavor. One wouldn't be tempted by an offering that did not whet one's appetite." He continued to study the crowd. "Those in more vibrant colors are married or widows many years out of mourning. Here, brother, you must make your selection of which dish to sample carefully. While an unhappily married woman makes an excellent main course, an outraged husband does tend to produce unpleasant aftereffects."
"Indigestion?" Nate said absently, still watching the unknown lady meander around the perimeters of the terrace. He couldn't clearly make out her features but had the oddest sense of familiarity. Had he met her before? Years ago perhaps? Or on one of his rare visits home? Nonsense, from the balcony he couldn't clearly see her face.
"At the very least. But a widow who is content in her widowhood and has no desire to become a wife again can be a most substantial and satisfying—" Quint grinned."—dining experience."
"Very tasty," Nate murmured.
Quint slanted him a suspicious glance. "Are you listening to me?"
"What? Yes, of course," Nate said quickly, and straightened. "I am hanging on every word. I believe you have come to—" He cleared his throat. "—dessert."
"A most important and delightful addition to a meal." Quint shrugged. "Although dessert is entirely dependent upon one's taste. A light and frothy confection of spun sugar and air—"
"Similar to the first course?"
Quint nodded. "Quite. While tasty upon the tongue, such a sweet can lead to a permanent diet, which I personally prefer to avoid. And a heavier offering, say a pudding, can be thoroughly enjoyable as long as one is careful not to develop a taste for it."
"Or one might find oneself eating pudding for the rest of one's life?"
"Exactly. And as much as I might like pudding, I can't imagine having it every day until I breathe my last."
"Nor can I." Although Nate suspected he would be ready for a steady diet of pudding long before his brother was. Not that he was ready for pudding—or rather, marriage—as of yet. Still, the idea was not nearly as repugnant to him as it was to Quint. He himself was confident he would know the right woman when she stepped into his life. Until then, he was more than willing to try whatever desserts were offered.
"It appears Sterling has noted my arrival," Quint said out of the corner of his mouth, directing a smile and a brief wave to their brother, who stood off to one side of the terrace beside their mother. The Earl of Wyldewood's annoyed glare was as unyielding as the legendary beacon from the long vanished Pharos of Alexandria. "Shall we join the others?"
"I don't think we can avoid it." Nate chuckled.
Quint stepped through the door onto the mezzanine that overlooked the ballroom. Nate cast a last glance over the crowd below, then followed his brother. He had lost the woman in the apricot dress but had no doubt he would find her. He smiled to himself, noting the same sense of anticipation he always had at the start of any quest, be it for the lost treasures of an ancient people or an intriguing female. Would this be a find of great importance? Or like that poor wretch Montini, would it be nothing more than a dreadful mistake?
Regardless, he had always been fond of apricots.
It wasn't as if she'd never been to a ball before. Why, when her brother had been in London, they always attended the annual ball of the Antiquities Society and on occasion others hosted by organizations affiliated with a university or museum.
She wandered along the edges of the crowd on the terrace in as casual a manner as she could muster, as if she belonged here, her confidence bolstered by the knowledge that she looked her best. Her gown was the latest French fashion and something of an extravagance, even if she could well afford it. Regardless, her world did not demand an excess of fashionable gowns. Still, it did enhance her appearance, and she had just enough vanity to appreciate that. She was well aware that she was considered pretty, with her dark hair and deep blue eyes, although it had never been of particular concern.
Gabriella Montini smiled and nodded at people she had never met nor ever expected to meet. Certainly, this would be easier if she'd ever before attended a ball given by an earl. And considerably less, well, awkward if she had actually been invited instead of quietly slipping in through the back garden gate.
This was the home of those vile Harrington brothers, and this was where she hoped to find evidence that one or, more likely, both of them had stolen the Ambropia seal from her brother. Not that she had any real proof yet, but they were at the top of Enrico's list of possibilities and an excellent place to start. She stepped through the tall French doors thrown open to the terrace and walked into the ballroom. Should the opportunity ever present itself, she would have to thank whoever had the odd idea to have the dancing out of doors. It made her task much less difficult. And this time she had a plan.
Gabriella accepted a glass of punch from a passing footman and inquired as to the location of the ladies' retiring room. Not that she had any intention of retiring, but it would provide an excellent excuse should she be discovered. All part of her plan. Admittedly, it wasn't an excellent plan, but it was far better than the last, which hadn't involved the least bit of sensible forethought and could have had disastrous consequences. Disaster was inevitable when one acted on emotion and impulse rather than rational thought.
She should have learned that lesson years ago, and thought she had. But she'd never anticipated how sorrow and anger could build inside a person for months, until it banished sanity from even the most sensible head. Still, it was something of an adventure, and ended without serious incident, though it was not especially successful. It had been years since she'd had any kind of adventure whatsoever that could not be found between the pages of aged, dusty manuscripts and the yellowed notebooks of long dead explorers. And she did so long to get away from books. For that alone it was perhaps worth the deception involved.
"Emma, my dear girl!" An older woman swept up to her in a flurry of satin skirts and exuberance. "How are you? It's been simply forever since we've seen one another. I heard you and your mother were in Paris."
Gabriella ignored the panic twisting her stomach. The lady had obviously mistaken her for someone else, and it seemed wise not to correct her. The last thing she needed was for anyone to realize she didn't belong here. She forced her brightest smile. "It has been a long time."
"You are as lovely as always. At least I think you are." The older woman squinted her eyes and peered at Gabriella. "Do forgive me, my dear, I have misplaced my spectacles once again." She heaved a dramatic sigh. "It's one of the banes of growing older. All sorts of things that used to work quite well no longer perform even adequately. I won't bore you with a long list. Suffice it to say eyesight and forgetting where I've put something are among them."
The woman couldn't clearly see her? Relief and a touch of gratitude for this stranger washed through Gabriella. Not enough, however, to tell her that her spectacles dangled from a jeweled broach pinned to her expansive bosom. "Nonetheless, you do appear well."
"Oh, I am. Quite well, thank you. And I have always been dreadful about misplacing my things so I really can't blame that on age." She leaned closer and laid a hand on Gabriella's arm in a confidential manner. "Age is a lovely excuse, you know. One is allowed to be eccentric rather than merely scatterbrained." She straightened and glanced around the room, which was rather pointless Gabriella thought. "Is your charming husband with you this evening?"
"Yes, of course. He's..." She paused. Not having a husband, she wasn't at all sure where one might be found. But she did know where she wished to put her plan in motion. "In the library, I believe. Yes, I think that's what he said. Do you know where it is?"
"Through the main doors into the corridor and then just a few doors down."
Just past the ladies' receiving room. How convenient. "I really should find him."
"Yes, indeed, you should be getting back to him." The older woman shook her head. "I wouldn't let a husband as handsome as your Lord Carpenter wander about freely. I should find my husband as well. Not as handsome as yours and certainly not as young, but age looks better on him then it does on me."
"I can't believe that."
"Neither can I." She laughed. "Do pay a call on me soon, my dear. It has been far too long." She smiled, nodded, and took her leave.
Gabriella did hope someone would tell her the location of her spectacles. Preferably after Gabriella had left the ball. She headed toward the library and hoped she didn't run into anyone's husband, or anyone at all for that matter. Fortunately, there was no one in the corridor. She found the library door, pressed her ear against it, heard nothing, then drew a deep breath and pushed it open as if she had every right to be there. As if she was simply another invited guest.
She stepped into the room and closed the door behind her. Thankfully, the library was indeed empty and well lit. She would hate to have had to stumble around in the dark. Antique swords and pistols were mounted on the walls on either side of the doors. A large desk sat at the far end of the room. Flanked by floor-to-ceiling windows, it was the dominant feature in the room, as she imagined the desk of an earl would be. The remaining walls were covered with tall book-lined shelves interspersed with portraits of long dead ancestors. She sniffed in disdain. Pirates and thieves the lot of them, no doubt. A smaller desk, probably for use of the earl's secretary, was placed off to one side of his lordship's.
She crossed the room to the smaller desk and wondered where to begin. It had been remarkably easy to learn that the earl's secretary also handled whatever paperwork his younger brothers' activities required. A few casual conversations with some of the older members of the Antiquities Society bemoaning how terribly complicated verification of finds and requests for funding had become. And hadn't it been so much easier in those long ago days when they were the ones uncovering the artifacts and treasures of forgotten civilizations? Why, one could scarcely accomplish anything these days without hiring clerical help, which certainly was a financial burden unless one was independently wealthy. Or had a clever sister who could handle such matters, or an earl for a brother who was willing to provide the services of his own secretary.
If the Harrington brothers had Enrico's seal, there could well be correspondence regarding it. It was a fabulous find. One of the few pieces ever discovered that might lend credence to the existence of the legendary city of Ambropia, if properly authenticated, of course. The discoverer of such an artifact would reap great fame, his reputation and his future assured.
The muscles in her jaw tightened. A reputation and a future that should have been her brother's, that would have been had not someone stolen the seal. It was a little more than a year ago that Enrico had returned to London with the seal. She had lived with her half brother since she was ten years old, after he found her residing in Italy with distant relatives of their father, two years after his death. But she couldn't recall ever seeing Enrico more excited about a discovery. Not that he had shown her the piece, only the impression made by rolling the cylindrical seal over wet clay. Her brother was remarkably superstitious about such things. He'd said it would be bad luck to reveal the seal prematurely. After all, Ambropia was clouded in mystery and legend, which included a curse placed by the city's virgin goddess protector on the heads of those who would disturb its sleep. Now she wondered if he hadn't been right.
When Enrico had unwrapped the seal in front of the Antiquities Society's Verification and Allocation Committee, he found a seal of far lesser significance. His claims that someone had stolen his seal and replaced it with a relatively common one did not sway the committee. Especially as Enrico had lost his temper and charged the society itself with trying to ruin him.
Her brother was never the same after that. Recovering the lost seal had consumed him. Competition for an artifact such as this was intense, and Enrico was certain that one of his rivals had stolen the Ambropia seal. He left London to pursue those he'd suspected responsible. His letters to her had detailed his progress as well as listed the names of the men he thought might have taken the seal or hired someone to steal it.
But the letters grew progressively less rational, less lucid, even a touch mad, although Gabriella had refused to see them that way at the time. A mistake she later deeply regretted. If he had taken Xerxes—the manservant who usually accompanied him—or if she had gone with him herself, perhaps...But she hadn't accompanied her brother on an expedition of any kind for years, not since what she thought of as the "incident," and she knew then he would never have allowed her to do so.
Then, six months ago, she received word that he'd died in Egypt, allegedly of a fever. The impersonal notice from a minor British foreign officer was accompanied by a crate containing her brother's possessions. She'd been devastated, of course. Enrico was twenty years older than she and as much father as brother. Aside from relatives of her English mother, whom she'd never met, he was the only family she had. She'd vowed then to find those who were responsible and restore her brother's reputation.
Now, the answers might be within reach. Absently, she chewed on her lower lip and studied the desk. It was probably locked. Damnation, she should have thought of that and come prepared. This plan was not going substantially better than the last one, and was probably not much smarter.
It wasn't until after Enrico's death that Gabriella had discovered they were far better off financially than she'd ever suspected. She was shocked to learn that their father had left the bulk of his significant fortune to her. Indeed, from the statements she had seen, it appeared it was her money that provided not only her support, but funding for Enrico's work as well. Enrico had never mentioned any of this, nor was it necessary for him to do so. As he was away more often than he was in London, his solicitor handled their finances. The solicitor arranged payment of all their expenses, including the fees for her initial schooling, the costs of continuing her studies at Queen's College, the modest London house where she resided, and the salary of Miss Henry. Florence Henry served as companion, chaperone, and friend, and had been by Gabriella's side since she first took up residence in London.
But her discoveries weren't merely financial. She had also found a packet of letters addressed to her mother—the mother who had died giving her birth. She thought they might prove useful someday to find her English relations, should she ever be so inclined. But as they had never sought her out, why should she look for them? Still, one letter in particular might prove useful. Her newfound wealth certainly had.
Though not exorbitantly rich, Gabriella now found herself in command of a sizable fortune. She wasn't at all used to having money. While it was nice to realize she could afford to do whatever she wished, the very idea of frivolous expenditures brought on a queasy feeling in the pit of her stomach. Even so, the now impressive state of her finances made it that much easier when anger and grief prompted her to impulsively travel to Egypt to confront the Harringtons. She was still bothered by twinges of guilt about having deceived Florence on that score. Regardless, it couldn't be helped, and what Florence didn't know wouldn't upset her. Florence believed that she had spent those few months coming to grips with her grief in the peaceful, contemplative setting of a convent in France. And believed as well that Xerxes and his wife, Miriam, had enjoyed a much deserved holiday, waiting for Gabriella in a nearby village rather than on a futile quest to Egypt.
Gabriella wished Xerxes was with her now. Among a number of unique abilities, Xerxes Muldoon could open any lock. She had no idea how he had acquired such a skill, but it was a useful one to have. Still, it was one thing for her to slip into this party alone, and quite another to be accompanied by Xerxes. The product of an Egyptian mother and an Irish father, he was tall and powerfully built, with an exotic look about him. He would not have gone unnoticed. At this very moment Xerxes was waiting with her carriage near the back gate.
No, she would have to do this by herself. She tried the desk drawers; they were indeed locked. It was pointless to look for a key. People who locked drawers would certainly not leave a key lying around in plain sight. There was scarcely anything on the desk at all save an inkstand, complete with inkwell, several pens, and a letter opener with an Egyptian faience scarab affixed to the handle. A gift from the brothers, no doubt, and probably stolen. Gabriella picked up the letter opener and hefted it in her hand. It would prove useful.
She knelt down and studied the center drawer. There was a keyhole in the middle that more than likely released the locks on all the drawers. If she could wedge the letter opener in the thin crack between the drawer and the desk itself, perhaps she could pop the lock and—
"May I be of some assistance?"
## Two
Only the top half of her face was visible over the edge of the desk but her blue eyes widened in surprise.
Good. Nate liked surprising a woman, it gave him the upper hand. He had spotted her leaving the ballroom and had assumed she was headed to the ladies' receiving room. He had planned to wait by the door for her return, but glanced down the corridor to see the library door closing and decided this was the opportunity to make her acquaintance. If, of course, she wasn't meeting someone else in the library.
He stepped toward her. "May I help you?"
"No, but thank you." She straightened. She was taller than he had thought when he first saw her but not overly so. She stood about half a head shorter than he, the perfect height.
"May I ask what you are doing in here?"
"What I'm doing in here?" She shrugged graceful shoulders left bare by the apricot gown. "As you have caught me, I suppose I must confess."
He smiled the slow, slightly wicked smile that had always served him well. While not quite as accomplished as Quint when it came to the fairer sex, Nate had no lack of confidence in his own ability to charm. And this particular smile was his most effective weapon. "Oh, I am fond of confession. Especially when it comes from a beautiful woman."
She stared at him for a moment, then laughed. "I'm afraid you'll be disappointed." She moved around the desk, a letter opener in her hand. "It's not an especially exciting confession."
His gaze slipped over her. He knew little about fashion, but thanks to the ravings of his mother and sister since his return home, he could see her gown was French and in the latest fashion. The silk of the dress molded nicely to curves no doubt enhanced by a corset. Even so, the swell of her breasts revealed by the low cut of her bodice needed no enhancement. Thank God for the French. "I can't imagine anything you say to be less than exciting."
She cast him a seductive smile of her own and his mouth went dry. "What a delightful thing to say."
"Oh, I can say any number of delightful things." He moved toward her. "I can say how the color of that dress is most becoming with the color of your eyes."
"My, that is delightful."
"I can do better. I can say—" His glance fell to the letter opener in her gloved hand. "What are you doing with that?"
She shifted it in her hand, and for the oddest moment he thought she intended to use it as a weapon. "I saw it on the desk and wanted a closer look. Clumsy fool that I am, I dropped it and it fell under the desk." She handed it to him. "Is the scarab real?"
"As real as something purchased in a market in Cairo can be." He turned it over in his hand. "I picked it up last year as a gift for my brother's secretary."
"Then you are thoughtful as well as charming?"
He laughed. "I can be." He tossed the letter opener on to the desk. "But you promised me a confession as to why you are here in the library."
"I have changed my mind." She raised a shoulder in a casual shrug. "It doesn't seem quite fair for me to confess to you without you confessing to me in return." Her eyes narrowed slightly. "Surely you have something to confess? Some misdeed that has weighed heavily on your conscience?"
"Nothing that comes to mind." He grinned. "Although I will confess I hope you are not here for a clandestine liaison with another gentleman."
She paused, then heaved a dramatic sigh. "You have found me out. How very embarrassing, especially as it appears he is not coming."
"But how fortunate for me." He took her hand and drew it to his lips, his gaze meshing with hers.
"Do you think so?"
"I do." He kept her hand in his and studied her. "Forgive me but have we met? You look remarkably familiar."
"You don't remember?" An odd note sounded in her voice. Nate wasn't sure if she was offended or relieved.
"My apologies." He shook his head. "I can't imagine not remembering you but—"
She pulled her hand from his. "I must say this isn't the least bit delightful."
"I am sorry—"
"You don't remember dancing together?"
"No, I'm afraid—"
"A few flirtatious moments during a walk in a garden very much like yours?"
"I can't recall—"
"A kiss stolen in the moonlight?"
He swallowed hard. "I must be an idiot."
"Yes, you must." She flipped open the fan dangling from her wrist and studied him thoughtfully. "Although I suppose you have danced with many women, had many flirtations in gardens, stolen many kisses in the moonlight. It must be difficult to recall every incident, every woman."
"Yes. No!" Indignation washed through him. "I have never once forgotten—"
She raised a brow.
"Until now." He huffed. "You have me at a distinct disadvantage."
"Do I?" she laughed, the sound engaging and infectious. "Now that is delightful."
He smiled reluctantly. "Who are you?"
"Now, now, if I told you it would quite spoil my fun. And as you don't remember my name, I think it's only right that you should have to earn that knowledge." Amusement glittered in her eyes. "Perhaps if we were to dance again..."
"Or walk in the garden." He moved closer and gazed into her eyes. "Or kiss in the moonlight."
"Perhaps," she said softly.
He lowered his lips to hers.
"But I never kiss anyone a second time who cannot recall kissing me a first time." She stepped out of his reach and started toward the door.
"A dance, then," he said quickly. "At least allow me the opportunity to recall a dance."
She glanced over her shoulder and considered him. "Very well. But I would prefer that we leave the library separately. I should hate to be the subject of gossip."
"Then I will meet you on the terrace?"
She cast him a brilliant smile, and his heart shifted in his chest. "You may count on it."
With that she swept from the room, leaving him to stare at the door she'd closed in her wake.
Who was she? She did seem familiar, but for the life of him, he couldn't place her. Surely he would remember a woman that lovely. He had always been fond of pretty women with dark hair and blue eyes, especially if they were intelligent. And there was no doubt she was clever. She certainly wasn't one of this year's new crop of debutantes. Her manner was far too assured. Besides, she looked only a few years younger than he. Perhaps he had met her on his travels. There was the vaguest suggestion of an accent in her voice. It—She was indeed delightful. No, he would remember kissing her in the moonlight.
And with any luck at all, he'd soon have another kiss in the moonlight. And this one he would not forget.
Good Lord, what had come over her?
Gabriella hurried down the corridor, forced herself to adopt a calm air and stepped into the ballroom. She mingled with the crowd, staying toward the perimeter of the room until she reached the open doors to the terrace.
Certainly she had flirted with men before, but never with such abandon. She hadn't planned to flirt with Nathanial Harrington, it simply happened, almost of its own accord. It wasn't as if she liked the man. She despised him and his brother. Still, she couldn't deny he was charming and handsome, with his dark hair streaked by the sun, the devilish glint in his brown eyes, and his broad shoulders. He had the sort of wicked smile that made a woman wonder exactly what wicked things he was thinking. And wonder as well why those improper thoughts were most intriguing.
Nor had she planned to speak to him at all. In truth she hadn't considered what she might say if she were discovered in the library. The letter opener provided the perfect excuse. If he had come in a moment later she would have been trying to pry the lock open. And that she would have been unable to explain.
She slipped through the doors and out onto the terrace, then began making her way to the steps that led to the garden. He thought she looked familiar, which was disturbing. Hopefully all that nonsense about dancing and kisses in the moonlight would divert his attention. It wouldn't do for him to realize she was the one who had accosted him in Egypt. He had thought then that she was a man, her brother's brother, and she preferred he continue to think that.
She reached the stairs to the garden and paused, stepping back to allow a young couple who obviously had improper thoughts of their own to pass. The worst part of the encounter with Mr. Harrington was that she had enjoyed it. There had been an element of danger in their meeting that was intoxicating. And toying with him had been great fun. The uncomfortable look in the man's eyes when she said they had once kissed was most satisfying. And didn't he deserve it? Hadn't he told her his brother had gone to Turkey? She'd had every intention of going after him when Xerxes learned that both brothers were separately headed back to England. Regardless, that venture had proven no more successful than tonight's.
"Was I mistaken?" Nathanial Harrington emerged from the crowd. "Were we to rekindle my memory with a walk in the garden rather than a dance?"
"I think a walk in a garden with a man who can't remember a lady's name would be rather dangerous." Damnation, she should have left when she had the chance. Still, there was no harm in a single dance. A tiny voice in the back of her head suggested that's why she had lingered. Nonsense. Gabriella brushed aside the thought that she might well want to dance with him.
"Yes, of course." He nodded. "A lady would truly be foolish—"
"I was thinking dangerous for the gentleman." Lord help her, this was fun.
He stared at her, then chuckled. "Very well, then." He gestured at the dance floor. "Shall we?"
"I do so love to waltz," she murmured, and took his arm. It was the most honest thing she'd said to him thus far. He led her onto the floor, and a moment later she was lost.
She did indeed love to waltz. Loved how the music wrapped around her soul and swept her away to a place and time and a life that existed only in her dreams. And only for people like Regina Harrington, who had an earl for a brother and a family willing to do whatever necessary to assure her future, her happiness, and a place where she belonged. Not for people like Gabriella Montini, who had been lost among relatives who cared nothing for her until she was found by a brother who dragged her from one expedition to another, one treasure hunt to the next.
Not that she'd minded. She had loved the life she lived with Enrico. Relished dressing like a boy for safety and being treated like one. She had hated it when her brother realized that his life was no life for a young lady. He had abandoned her in England while he went on with his work, although it was as much her fault as his. And abandoned was not entirely accurate. He had provided for her needs, arranged for her schooling and her expenses. If she had no real home and a family that consisted only of two longtime servants and a paid companion, it was her lot in life, and she'd never been especially discontent. Why it bothered her at this moment, she couldn't say, save to blame it on the waltz. On the promises inherent in the melody and the rhythm and the warmth of the man whose arm encircled her and who held her hand in his.
And hadn't she made the best of it? Hadn't she spent these past nine years studying languages and ancient civilizations, all with an eye toward eventually rejoining her brother? And hadn't whoever destroyed his life destroyed her future as well?
"I fear you are very far away." The firm pressure of Harrington's hand against her back increased and her gaze jumped to his. "Although I daresay I deserve it."
"My apologies, Mr. Harrington," she said lightly. "You may blame it on the music."
"Blame it on the Blue Danube?" He grinned down at her. "Excellent. I would much prefer to think your pensive state is attributable to the beauty of the waltz rather than the dullness of my character."
"I can't imagine any lady considering your character dull."
"Ah, but I suspect you are not just any lady, are you?"
"No." She smiled up at him. "I am the one you cannot remember."
His brows drew together and he studied her face. "I assure you, it will come to me." They executed a perfect turn. "We dance well together."
"As if we have danced together before?"
"Exactly." He shook his head. "I cannot believe I have met the woman of my dreams and I cannot remember her name."
"The woman of your dreams?" Her breath caught but she forced a teasing smile. She had no desire to be the woman of his dreams, not that she thought for a moment that his words were sincere. They came far too easily. "No doubt in the course of your travels there have been many dreams and many women."
"They pale in comparison." He gazed into her eyes, the moment between them at once fraught with unspoken meaning and promise. Dear Lord, he was dangerous.
"The dreams or the women?" she said without thinking.
"Both."
She drew a deep breath and ignored the tremulous feeling inside her brought on by his words and the gleam in his eye. "You are an adventurer, sir, a treasure hunter. Such men are not to be trusted."
"I assure you, I can be quite trustworthy," he said in a lofty manner.
"I admit you can be charming and even perhaps thoughtful, but trustworthy? I doubt that." She shook her head. "Besides, it's been my experience that trust needs to be earned."
"I think I would like the opportunity to prove that I can indeed be trusted." He stared down at her, the look in his eyes abruptly serious.
She ignored it. "I suspect the opportunity you seek has nothing to do with trust."
A slow smile spread across his face. "You are a beautiful woman and a mystery, even if perhaps of my own making. Do you blame me for seeking any opportunity whatsoever?"
"Not at all." She cast him a pleasant smile. "Your reputation, and that of your brother, precedes you. You are a scoundrel, Mr. Harrington, and scoundrels are rarely worthy of trust."
His hand tightened around hers. "Was I a scoundrel when I kissed you in the moonlight?"
"Never more so than then." The music ended and he led her off the dance floor. "Well?"
"Well?"
"Has our dance restored your memory?"
"No." Frustration sounded in his voice. "Will you give me another chance? If you will not walk in the garden, then do allow me another dance."
She raised a brow. "You are a persistent sort."
"Indeed I am." He grinned, and she steeled herself against the charm of it. "It's one of my better qualities."
"I'm not sure I would boast about that."
"It's not boasting, it's simply a fact." He leaned close and spoke softly. "I will remember, you know. I promise you that."
"You shouldn't make promises you can't keep."
"I never do."
"We shall see." She stepped away from him. "I find I am rather parched. Would you fetch me a glass of punch?"
"Only on the condition that when I return, you give me at least a hint as to where we have met before."
She shrugged. "I will consider it."
"Very well, then." He took her hand and raised it to his lips. His gaze locked with hers and for the briefest moment she wondered what might happen next if they had indeed met before, if he had once kissed her in the moonlight. If she were part of his world. Would this then be the beginning of something extraordinary and not merely a game she played? He released her hand. "I shall be but a moment."
She smiled but said nothing.
He turned and made his way across the terrace. She watched him for a few seconds, ignored the oddest feeling of regret, then quickly slipped down the stairs and headed toward the back garden gate. Within a minute she was back at her carriage.
"Well?" Xerxes assisted her into the vehicle.
She shook her head. "The desks were locked. I shall have to return when the household is asleep."
"Tonight?"
"A ball like this will go on for hours. Tomorrow, I think, or the day after would be best."
"It's not a good idea, girl." Xerxes fairly growled the words. From the time she had first come to live with her brother, Xerxes had called her "girl." She had long thought it was his effort to remind her that she was one. It was most endearing.
"And yet, it's the only one I have, so it shall have to do." She settled back against the cracked leather seat, and Xerxes closed the door, muttering something she couldn't quite make out. Which was probably for the best. He had agreed to tonight's endeavor reluctantly and only because he had no better plan himself, save beating the truth out of Harrington and his brother. Gabriella preferred to avoid that, at least until they had some sort of proof as to who had taken the seal.
While she was no closer, tonight's excursion hadn't been a complete failure. At least she knew the arrangement of the earl's library. And her flirtation with Nathanial Harrington had been, well, fun and surprisingly exhilarating. Still, it would never happen again, she would not allow it. Why, just in their brief encounter tonight, the man made her want things she could never have, long for what would never happen. He was indeed dangerous. To her purpose and possibly her heart.
The worst thing wasn't that she had found him so charming. That, she should have expected. But rather that she had found herself so very charmed.
Nate rested his hip against the terrace balustrade and surveyed the crowd. He wasn't the least bit surprised that the lady with the blue eyes of an angel wasn't where he had left her. Indeed, he would have been surprised if she had. She did seem familiar, but while he may well have forgotten a dance, he would never have forgotten a kiss. He had no idea what seductive little game she'd been playing, but he was more than willing to play. Tonight and whenever they next met.
He chuckled to himself and sipped her punch. There was no doubt in his mind that her game was indeed seductive, and most effective as well. Seduction had crackled in the air between them. Nor did he doubt that he would see her again. And when next they met, she would not be the only one playing.
Nor would she be the victor.
## Three
Gabriella's heart thudded in her chest. Most annoying, as she'd always considered herself quite courageous. Still, she'd never stolen into someone's house in the middle of the night before, and a certain amount of trepidation was to be expected, even if all was going well.
She flattened her back against the wall and inched her way down the corridor toward the library. A window at the far end glowed faintly with starlight but the corridor remained heavily shadowed and she took each step with care.
She and Xerxes had decided that the best way to proceed was to retrace her steps from the night before last, starting at the back garden gate. Once again that lock had proven no challenge. They had then made their way through the garden, grateful for the numerous trees, shrubs, and other plantings, to provide hiding spots should the need arise. It was far easier now than it had been when she'd worn a gown. Tonight she had donned the men's clothes she wore in Egypt, her hair tucked up under the same worn felt hat. She had grown up wearing men's clothing and, in spite of the circumstances, loved the sense of freedom they gave her.
Thus far it appeared the household was fast asleep. Save for a few lamps lighting windows here and there, the house was dark. Xerxes assured her, thanks to his surveillance in advance of the ball, there were always a few lamps left burning in the house. A dreadful waste of money in her opinion.
They had decided to enter by way of the French doors leading from the terrace to the ballroom. Xerxes was confident the locks there would prove no greater obstacle than the one on the gate. A check of all the doors had found two in such disrepair that they scarcely kept the doors closed. Perhaps if the Harringtons weren't squandering their money on lamps burning all night...
However, Xerxes wasn't at all happy when Gabriella insisted that he remain outside. She pointed out that, if caught, the earl would be much more likely to turn him over to the authorities than he would her, and Xerxes could scarcely argue with that. Especially when Gabriella enlightened his lordship regarding the legitimate complaint she had with his brothers. From the little she'd heard of the earl, he was known to be an honorable man. Pity his brothers weren't more like him.
Aside from all else, Gabriella was half British and far less imposing than Xerxes. And while she hadn't mentioned it to him, she had a weapon he knew nothing about, which she thought might prove useful.
Regardless, at this particular moment she regretted not letting Xerxes take on this task. She reached the library door, paused for a moment to gather her faltering courage, then slowly pushed the door open and stepped into the room. A small gas lamp glowed faintly on the earl's desk. She sniffed in disdain at the extravagance, although it did serve her purposes.
Gabriella closed the door quietly, crossed the room and took the lamp, then moved it to the secretary's desk. She did need to see what she was doing, after all. She pulled a long thin piece of flexible metal hooked on one end, similar to a flattened crochet needle, from where it was tucked into the lining of her coat. Xerxes had spent much of the day teaching her how the insignificant tool could be used to open a lock. It was a handy skill to have.
It took her but a minute or two to trip the lock. She grinned with satisfaction. This had been surprisingly easy. She slipped the metal tool back in its hiding place and pulled open the center drawer. It was filled with neatly arranged pens and stationary and the other accoutrements a man who dealt in the correspondence and business of an earl might need, but there was nothing of significance. She pulled open the larger of two drawers on the right. Here were files, well organized, tidy and clearly labeled. Her confidence surged. Thank God the earl had the intelligence to hire an efficient secretary.
She flipped through the files. They all had to do with the earl's affairs. None of the precisely labeled files indicated anything regarding the work of the younger Harringtons. Perhaps she would have better luck with the drawers on the other side of the desk. She shut the open drawer, reached for the next—
"Sterling?" The door swung open. "Are you still—"
Gabriella jerked her head up and met the startled gaze of Regina Harrington.
"Good Lord!" Lady Regina called to a point behind her. "Come quick! We're being robbed!"
Gabriella's heart lodged in her throat. One thing they hadn't planned was an escape route. But then she hadn't planned on being caught either. She raced for the nearest window.
"Help! He's getting away!" the girl screamed.
Gabriella fumbled with the window sash.
"Oh no you're not!" Lady Regina yanked an ancient broadsword off the wall.
"Oh yes I am." Damnation, why wouldn't it open?
"Don't think you can break into my home, take whatever you wish and waltz off! Not bloody likely!"
"Your language, Lady Regina," Gabriella muttered, pounding on the sash. It was bad enough to be caught, but to be caught by a spoiled brat was an added insult. "Your mother would be appalled."
"My mother would do exactly what I'm doing," Lady Regina said staunchly, struggling to brandish the heavy sword with both hands. "Apprehending a brigand!"
"Oh, for goodness sakes." And stupid as well as spoiled. "Aren't you afraid?" Gabriella tugged at the window sash. It wouldn't budge. "I could be dangerous."
"I doubt that." Lady Regina scoffed, lowering the sword, which seemed too much for her. "You're no more than an inch or so taller than I and you're a rather frail looking sort."
"I'm not the least bit frail," Gabriella said under her breath, and beat her fist against the sash. "But I am desperate."
"Nonetheless, someone will be here at any moment to assist me." The faintest trace of unease sounded in the girl's voice. "A servant or my brothers or someone." She glanced over her shoulder. "Nathanial and Quinton were right behind me."
"Well, they're not behind you now, and I suspect most of your servants are asleep." For the first time, Gabriella noticed the girl was wearing a ball gown. "Are you just now coming in? At this hour?"
Miss Harrington stared. "I was accompanied by my brothers. It's not at all uncommon for a ball to last—that's none of your concern! You're a common thief!"
"I'm not the least bit common." Gabriella sniffed, and tried the window once more.
"There will be no escape that way. That window sticks."
"Regardless, unless you are willing to stand aside and allow me to leave by the door, I am going out the window." Gabriella looked around for something with which to break the glass.
"It's a nasty drop to the ground," Lady Regina warned, still gripping the sword handle, but making no effort to lift it. "You'll likely break your neck."
"I shall take my chances." She spotted a poker by the fireplace and started toward it.
"Stay right where you are!" The girl's voice rose.
"You'll have to run me through to stop me." Gabriella reached the fireplace, grabbed the poker, and turned just in time to see the girl hurl a vase at her. She ducked. The vase skimmed past her head, knocked off her hat, and shattered on the mantel behind her.
Lady Regina gasped. "You're a woman!"
"Yes, I'm a woman!" Gabriella snapped, and started back toward the window.
The girl advanced. "If you don't drop that this very instant, I assure you, I shall indeed run you through!"
"Hah. You can barely lift that sword, let alone wield it." Gabriella gripped the poker in both hands and pointed it at the younger woman. "And I assure you, I have a fair amount of skill with a...a poker."
Lady Regina narrowed her eyes. "Do I know you?"
"No." Lady Regina gestured with the poker. "Now stand back."
The girl studied her closely. "You look very much like Emma—Lady Carpenter, that is."
"Obviously, I'm not." Who was this Emma person?
"No." Miss Harrington shook her head. "Your hair is much darker. The light in here is weak but you do bear a striking—"
"As much as I hate to interrupt you, I must be on my way." Gabriella clenched her teeth. She didn't have much time before the rest of the household finally responded to the young woman's calls. It was something of a miracle no one had come yet. "I am going to break this window, so I suggest you stand back."
"I cannot allow you to do that!"
"You cannot stop me." Gabriella turned away from the girl, aimed the poker, and drew it back.
"But I can." A familiar male voice rang through the room. "Drop the poker! Now!"
Gabriella sucked in a hard breath. This was it, then. Even if she broke the window, she would never be able to climb through it before he grabbed her. She released the poker and let it clatter to the floor, then turned to meet Nathanial Harrington's gaze.
He gasped. "You!"
Gabriella resisted the urge to drop a sarcastic curtsy. "We meet again, Mr. Harrington."
"And again in the library." He had taken the broadsword from his sister, and for a fraction of a second she saw him as a knight of old. Strong and powerful and menacing. And for an even briefer moment, regretted that he was her enemy. He studied her through narrowed eyes. "How very interesting."
"Where have you been?" his sister snapped. "Why didn't you come when I called? I could have been murdered, assaulted, kidnapped!"
"I'm very dangerous, you know." Gabriella stared at him with a bravado she didn't quite feel.
"Oh, I am well aware of that," he said coolly. He addressed his words to his sister, but his gaze remained on Gabriella. "One of the servants thought he saw someone on the grounds. Quint and I went with him to look but he was mistaken. We found no one."
Gabriella fought to keep relief from showing on her face. At least Xerxes was safe. At once her confidence returned. She could take care of herself, she always had. She smiled in a pleasant manner. "I assure you, I am quite alone."
Harrington raised a brow. "Forgive me if I find anything you say to be less than trustworthy under the circumstances."
"I say, what is going on here?" Quinton Harrington stepped into the room, followed closely by a woman Gabriella recognized as his mother, the Countess of Wyldewood, and his older brother, the earl.
"What is all the commotion about?" Lady Wyldewood asked. She and the earl were both dressed in nightclothes and had obviously just awakened. "And who is this?"
"That's the question, isn't it?" The earl stepped forward to stand beside his brothers. Even in a dressing gown and with his hair disheveled, there was an air of command about him. "Who are you and what are you doing in my library?"
Under other circumstances, Gabriella might have found the three Harrington brothers—all sharing a similarity of height and build, and all undeniably handsome and dashing—to be an enticing display of the best of British manhood. If only two of them weren't the scoundrels she knew them to be.
"She's a thief and I caught her." Regina smirked, and nodded at Nathanial. "And he knows her."
"All I know," Nathanial said slowly, his gaze locked with Gabriella's, "is that I have never kissed her in the moonlight."
"Pity." A half smile curved Quinton Harrington's lips, and his gaze traveled over Gabriella in a most improper manner. She'd never especially considered how revealing men's clothes were before, but now had the uneasy feeling he was seeing her without benefit of any clothing at all. "I would have."
"Quinton," his mother said sharply. "This is not the time for your nonsense. And do light some additional lamps so that we may see her properly."
"I saw her rather well," Quinton said under his breath, and proceeded to comply with his mother's request.
"Now, then," the earl began. "Who are you and what are you doing here?"
Gabriella hesitated. Apparently her courage wasn't entirely up to this task, after all. She squared her shoulders and met the earl's gaze firmly. "I am here to find proof that your brothers stole an artifact of no little significance from my brother. That theft ultimately led to his death." She drew a deep breath. "My name is Gabriella Montini."
Nathanial stared. "That's why you look so familiar. You bear a striking resemblance to your brother."
"Don't be absurd." Quinton snorted. "She doesn't look the least bit like Enrico Montini. He was a good twenty years older and considerably darker in coloring than she."
"Montini," Lady Wyldewood murmured, more to herself than the others, the oddest look on her face.
"Not that brother." Nathanial waved off Quinton's comment. "The brother who confronted me in Egypt."
Good Lord, he still didn't realize she was the brother who had confronted him. Gabriella sent a silent prayer of gratitude heavenward. "The brother you lied to and sent on a wild goose chase to Turkey?"
"You what?" The earl glared at his younger brother.
"I was protecting my brother," Nathanial said sharply.
Quinton shrugged. "Not necessary but appreciated nonetheless."
"Regardless, she looks exactly like that brother," Nathanial said. "Are you twins?"
"We are...like one," Gabriella said, ignoring a twinge of conscience. It wasn't exactly a lie but it certainly wasn't the truth.
Regina glanced toward her mother. "I thought she looked remarkably like Lady Carpenter."
"Yes, of course," the older woman said thoughtfully.
"I fear, Miss Montini," the earl said in the cool tone of someone used to being obeyed without question, "that I require a more detailed explanation for your presence here tonight. Unless you would prefer that I send for the authorities."
"You can't have her arrested," Nathanial said without warning.
Surprise widened Gabriella's eyes. "Why not?"
The earl stared at his brother. "Why not indeed?"
"She has a legitimate complaint. Not with us," Nathanial added quickly. "We did not take the seal. But someone did and has caused her irreparable harm. It doesn't seem quite right to have her arrested."
Regina scoffed. "She did break into the house."
"Frankly, my lord, I would prefer to avoid arrest," Gabriella added.
"What seal?" Impatience rang in the earl's voice. "I want to know what this is all about, and then I shall decide what should be done with her." He turned to Gabriella. "Miss Montini, if you will."
She nodded, then paused to gather her thoughts. "My brother, Enrico, spent his life engaged in the same sort of work as your brothers. The study of archeology and the search for the lost treasures of the ancients."
"Is this going to be a long story?" Regina said under her breath.
"Perhaps it would be best if we adjourned to the parlor, where we could all be seated." Lady Wyldewood smiled at Gabriella. "I find my mind works much better at this time of night if I am not shifting from foot to foot."
"Yes, of course," Gabriella murmured.
Within a few minutes they were all seated in a large parlor, somewhat extravagant but tastefully furnished. A hastily dressed servant the earl addressed as Andrews, who Gabriella assumed was a butler, appeared with brandy. Lady Wyldewood and her daughter settled on one sofa, Gabriella sat on another. The earl and Nathanial each took a chair. Quinton remained standing, leaning idly against the mantel piece.
Once everyone was served, the earl glanced at Gabriella. "Miss Montini, if you will continue."
"Very well." She thought for a moment. "My brother found an ancient Akkadian seal, made of greenstone I believe. This sort of seal is cylindrical in shape. It's incised with symbols. They were quite common in the ancient world. When rolled across wet clay, the carving produces an impression. It can be a message or a story or have religious significance."
The earl nodded. "I am familiar with them. My father had a collection. It's in a case here somewhere."
"The seal Enrico found made a reference to the Virgin's Secret, and had a symbol carved in it for the lost city of Ambropia," Gabriella continued. "While it's been mentioned in the writing of the ancient Greeks, there has never been solid evidence of its existence. The very name means 'immortal place,' and even that is a Greek interpretation of a far older name that has since been lost. It has long been thought to be nothing more than a legend or a myth."
"As much as Troy or Atlantis or Shandihar have been or still are believed to be nothing more than stories," Nathanial said.
"But Enrico believed his seal went beyond merely being the oldest discovered written reference to Ambropia." She leaned forward and addressed the earl. "He thought it was one of a set of seals that together would reveal the location of the city itself."
"The Virgin's Secret," Nathanial said softly.
Lady Wyldewood raised a brow.
"The city was said to be under the protection of an ancient virgin goddess," Nathanial explained. "Her name too has been lost." He shrugged. "Until now, it's only been a story."
"That is a find," the earl murmured.
"Is there a great deal of treasure in this city?" Lady Regina asked.
"The treasure, dear sister, is in the knowledge of history to be obtained," Nathanial said firmly.
Gabriella stared. Was it possible that she had misjudged him?
"Although gold and jewels and items that will fetch small fortunes are always nice." Quinton grinned and sipped his brandy.
Obviously she had not misjudged that brother.
"Go on, Miss Montini," Lady Wyldewood said.
"When Enrico went to present the seal to the Antiquities Society," Gabriella said, and took a bracing sip of her brandy, "he discovered it had been stolen and a seal of no great significance substituted." She paused. "It was most distressing."
The earl nodded. "I can well imagine."
"From what I heard of the story," Quinton said, "Montini became somewhat enraged. Made all sort of wild charges and accusations." He shook his head and looked at Gabriella. "The members of the Verification Committee do not take that sort of thing well."
"No, they don't." She blew a long breath. "His behavior, coupled with the fact that the seal in his possession was not as he had claimed...well, his reputation was shattered. He became determined to find whoever had taken the seal and recover it."
She got to her feet and paced the floor, absently wringing her hands. "That was over a year ago. Enrico left London for Egypt, Turkey, Persia, wherever those few he had told about the seal might be found."
"Including my brothers?" the earl asked.
"Yes, among a handful of others. His letters grew..." She hesitated. Was it disloyal to reveal just how odd Enrico's letters had become? Or, at this point, was it necessary? Perhaps it no longer mattered. "They became less and less rational. His search consumed him. Then six months ago I learned he had died."
"Do you suspect foul play?" the earl asked.
"I was told he died of a fever, but yes, I suspect everything," Gabriella said simply. "I have become a most suspicious person." She resumed her seat. "Now, it is up to me to recover the seal and restore my brother's reputation."
Regina scoffed. "But you're a woman."
"Regardless," Lady Wyldewood said, "I suspect Miss Montini is up to the task."
Gabriella's gaze met the older woman's. "It is my responsibility."
"I see," the earl said thoughtfully, then glanced at Nathanial.
"I can assure you, I have stolen nothing." Sincerity rang in Nathanial's voice. Even so, it was her experience that the very best liars were those who sounded sincere.
The earl's gaze, along with that of everyone else in the room, shifted to Quinton.
"Why are you all staring at me? I did not steal Montini's seal. I haven't stolen anything." Quinton sipped his drink, then added in a low voice, "Recently."
"Sterling." Lady Wyldewood turned to her oldest son. "Can't you do something about this? You're on the board of the Antiquities Society."
The earl shook his head. "It's little more than an honorary position, Mother. And only because Father held it before me. Were it not for the significant funding we provide, as well as the possibility of funding in the future, my welcome on the board would be limited."
"Well, we should do something to help her," Lady Wyldewood said firmly.
Gabriella drew her brows together. "Why?"
Nathanial studied her. "Yes, Mother, why?"
"It seems to me that until this situation is resolved, and the reputation of Miss Montini's brother restored, it hangs over all our heads. The longer Miss Montini continues her efforts, the more likely it is that her brother's suspicions as to the possible identity of the thief or thieves will become common knowledge. The Antiquities Society would not look kindly upon that." Lady Wyldewood met Nathanial's gaze. "While you are not dependent upon the society for funding, you do need to remain in its good graces for reasons of credibility. Your own reputations are at stake." She shifted her gaze to Quinton. "And yours has never been entirely spotless."
Quinton shrugged.
"Aside from the especially unwise decision to try to find information in our house in the middle of the night—"
Unexpected heat washed up Gabriella's face.
"—and being discovered—"
"Not part of the plan," Gabriella said quickly.
Lady Wyldewood pinned her with a firm gaze. "And yet it may well work to your advantage." She addressed the others. "As I was saying, Miss Montini strikes me as an intelligent young woman. If I were her, I would use a weapon I don't believe she realizes she has to ensure our assistance."
The earl's brow furrowed. "What weapon, Mother?"
"If it were to become publicly known that two of the members of this family are suspected of thievery, it would bring scandal down upon us all." Lady Wyldewood shook her head. "It is in our best interest to resolve this quickly. Even a hint of something like this would have a devastating effect."
Lady Regina gasped. "It would ruin my prospects for a good marriage! My very life! We would all be disgraced!"
"I see." The earl considered Gabriella for a moment. "Would you accept our help? It would be an uneasy truce, of course. I very much doubt that you trust us, and I cannot say we trust you."
Gabriella shook her head. "Quite honestly, my lord, I don't know what to say. I did not expect such an offer."
"Let me ask you this, my dear," Lady Wyldewood said. "Was Enrico your sole financial support?"
"Yes," Gabriella said without thinking. Again, it was neither an actual lie nor the complete truth.
The older woman studied her. "You have no other family?"
"No." Aside from English relations she'd never so much as heard from, that, at least, was the truth.
"Except for the brother I encountered in Egypt," Nathanial said.
"And I have had no word from him since then," Gabriella quickly said. "He is no doubt still in Turkey. Although he too may be..." She paused. She'd always considered herself an honest sort. Yet the lies that fell from her lips came with surprising ease. "...gone forever."
Suspicion glittered in Nathanial's eyes. He obviously wasn't stupid. It was only a matter of time before he realized the truth about their first meeting. What would he think of her then? She ignored the annoying question. Nathanial Harrington's opinion of her was of no consequence.
"Then it's agreed." Lady Wyldewood nodded. "As they are the ones most affected, Quinton and Nathanial will assist Miss Montini to find the seal and restore her brother's good name."
Quinton scoffed. "I have better things to do than help her."
"I don't," Nathanial said, then shook his head. "I do, but I can't think of anything more important than recovering an artifact of such significance. To prove Ambropia actually existed would be to rewrite history." He blew a long breath. "It's the stuff careers and reputations are made of."
"My brother's reputation," Gabriella pointed out.
Nathanial met her gaze. "Without question. It is his find."
"I suspect such an artifact would have a great monetary value as well," the earl said in an overly casual manner. "It would bring a small fortune from museums or collectors."
"Yes, it would." Gabriella bristled. "But I fully intend to donate it to the collection of the Antiquities Society."
"Very admirable, my dear," Lady Wyldewood said, "and I have no intention of trying to convince you otherwise, but I am well aware of the precarious nature of the finances of men who follow the path your brother did. Unless they have independent wealth or family money, as my sons do, they depend upon grants and funding from museums or organizations like the Antiquities Society." Concern shone in Lady Wyldewood's eyes. "With one brother dead and another missing, your finances must be uncertain at best."
"I will admit..." Gabriella chose her words with care. "...my discovery of the state of our finances after Enrico's death did come as something of a shock."
"I'm not surprised." The older woman nodded. "My husband was not merely a patron of the society, but he had a passion for the study and the artifacts of ancient man as well. Many a dinner here included men like your brother and long discussions late into the night about their work and their adventures. It has been my observation that such men are more concerned with the past than the present and give little thought to financial stability. I doubt that your brother was substantially different.
"Therefore..." Lady Wyldewood cast her children a decided look. "I propose that until this situation is resolved and the seal recovered, Miss Montini stay here as our guest."
"What?" the earl's brow rose.
Lady Regina scoffed. "How absurd."
"Insane but interesting," Quinton said.
Nathanial nodded slowly. "I think it's an excellent idea. We don't trust her, she doesn't trust us. How better to keep an eye on one another than if we resided in the same house?"
"How better indeed," Quinton said under his breath.
Gabriella scarcely heard him. The very idea of living in the same house with Nathanial Harrington—who danced as if he had been her partner always, and brought out a flirtatious demeanor in her she hadn't known she had, and made her wish, if only for a moment, that they were not on opposite sides—struck her as exceptionally dangerous. "I don't know..."
"Mother." The earl's brow furrowed. "I can't believe you're inviting a complete stranger to stay in our home."
"She's not exactly a complete stranger, Sterling." Lady Wyldewood's gaze met Gabriella's. "I knew your mother."
Gabriella lifted her chin. "I know."
"Did you?"
Gabriella nodded. "After my brother died, I found a packet of letters written to my mother. One was from you."
"I was so sorry to learn of her death, and then your father's."
"It was a very long time ago," Gabriella said with a shrug, as if it didn't matter.
The countess's expression remained serene, but there was the tiniest glimmer of amusement in her eyes. "You should have come to me directly about all this, you know, rather than resort to the clandestine methods you employed tonight."
Gabriella had the good grace to blush. "My apologies."
"I look forward to having a long talk with you about your mother. I suspect you have a lot of questions."
A lump formed in Gabriella's throat. She hadn't considered that Lady Wyldewood might want to talk about the mother who had died giving her birth. She rarely thought about her mother at all, and only dimly remembered a portrait her father had that, along with the rest of his furnishings and possessions, vanished among his relations after his death. The only thing his relatives had no use for was a little girl. She swallowed hard. "That would be most appreciated."
"It is decided, then." Lady Wyldewood rose to her feet, and the others followed suit.
Living with the Harringtons hadn't been part of her plan, but it would certainly serve her purposes. What better way to find the secrets of scoundrels than to live among them? Although, she amended the thought, Lady Wyldewood was very likely as genuinely nice as she appeared.
"The hour is late and I for one would like to retire," the countess said. "I will have a room prepared for you. You may send for your things in the morning." She cast an appraising glance over Gabriella. "I assume you have more appropriate clothing?"
Gabriella nodded. "But I should send a note now. The...lady I have been residing with is a very old friend...of my brother's." Which was, at least, partially true. "She will worry if she finds me missing in the morning."
"Most thoughtful of you, my dear." The countess nodded approvingly, then glanced at her daughter. "Are you coming?"
"Yes, Mother." Lady Regina followed her mother out of the room.
"We should retire as well," the earl said. "You will find paper and pens in Mr. Dennison's desk, although apparently you have already discovered that."
Gabriella smiled weakly.
"Quinton?" The earl glanced at his brother.
Quinton downed the rest of his brandy, set the glass on the mantel, and stepped to Gabriella. "Miss Montini." He took her hand and raised it to his lips. "It has been a most enlightening evening." His gaze never left hers, in precisely the same polished manner as his brother's had when he'd kissed her hand. And yet, with this brother it was nothing more than overly practiced. "I look forward to many more." He released her hand and started for the door.
"Andrews will see you to your room when you're ready," the earl said. "Good evening, Miss Montini."
"Lord Wyldewood," she murmured.
He traded glances with Nathanial, and the younger brother followed him out of the room. A moment later Nathanial returned. "You should write your note, Miss Montini," he said coolly.
She moved to the secretary's desk—Mr. Dennison's desk—sat down in the chair and opened the top drawer, knowing full well Nathanial watched her every move, and ignoring a distinct twinge of embarrassment. After all, she wouldn't know where the paper and pens were if she hadn't broken into the desk.
She would write two notes. One to Xerxes and Miriam assuring them that all was well, and another to Florence. She'd put both notes in the same envelope. Xerxes was no doubt watching the house, and he would certainly intercept her notes and make certain Florence saw only what she should.
She would tell Florence the truth: she had been invited to stay with an old friend of her mother's. It struck her that, at this point, her lies might well be at an end. Aside from the true state of her finances, and that nonsense about a second brother, there was little left to lie about.
She scribbled a quick message to Xerxes, then took a fresh sheet of paper and started the note to Florence.
"You're writing rather a lot, aren't you?"
She resisted the urge to look at him. "I have rather a lot to say. I don't want her to worry."
"Then I imagine you're not telling her you broke into my house."
"No," she said sharply. It was remarkably difficult to concentrate on her writing, knowing his unflinching gaze was fixed on her. "Are you going to keep staring at me?"
"I assure you, Miss Montini," he said smoothly, "I don't intend to let you out of my sight."
## Four
Excellent," Miss Montini murmured, her gaze still on the papers before her. "That will save me the effort of keeping you in my sight."
It was obvious, even from where Nate stood, that she was writing more than one note. He could question her about that, again, but she would no doubt evade his question. Again. "I do not appreciate being lied to."
She folded her notes, slid them into an envelope and sealed it. "I would imagine few people do."
Still, her dishonesty made her no less attractive, and somewhat more intriguing. Odd, as he had always thought he valued honestly above all else. Apparently not as much as he valued deep blue eyes and a nicely curved figure.
"We have never met before, have we?" It was a statement more than a question.
She addressed the envelope. "I never said we did. You said I looked familiar and asked if we had met. I asked if you remembered, and you did not."
"I didn't remember because there was nothing to remember." Hah. He had her there.
"Regardless." She finished writing the address with a refined flourish, set the pen down and looked up at him. "You didn't realize that. You thought you had kissed me and couldn't recall it. It was most insulting."
"How could it possibly be insulting?" He stared. "I couldn't remember because it didn't happen."
"If it had and you couldn't remember, I would have been insulted."
"If it had, I would have remembered!"
"No doubt," she said in a tone that indicated she had a great many doubts about his ability to remember the women he'd kissed. She didn't know anything about him but had made assumptions based on nothing more than her own suspicious nature. Although admittedly the fact that her dead brother had named himself and Quint among those who might have stolen the find of a lifetime from him could have adversely influenced her opinion.
She held out the envelope. "Given the lateness of the hour, it would be best if this were delivered to Mr. Muldoon. He and his wife have been in my—in Miss Henry's—employ for years. He is most discreet and trustworthy and will see that she receives it in the morning. I should hate to wake her up at this time of night, and I'm certain Mr. Muldoon will be awake."
Nate glanced at the address. It was in a respectable if not especially fashionable neighborhood. "I suspect the footman I send with this might well arrive before he does."
She smiled in a pleasant manner. "Oh?"
"Come now, Miss Montini, you strike me as an intelligent woman. And an intelligent woman would not roam the streets of London alone late at night." He waved the envelope. "Therefore it is logical to assume the trustworthy, discreet Mr. Muldoon accompanied you."
"I assure you, Mr. Harrington," she said smoothly, "I am quite alone."
He raised a brow. "Are you?"
"I have never been more alone in my life than I am at this very minute." She rose to her feet and continued as if she had just said nothing of significance, rather than made a comment as enigmatic as it was perhaps revealing. There was a great deal more to Gabriella Montini than met the eye. "If you have no objections, I should like to retire now." But what met the eye was most desirable. "It has been an eventful evening."
He raised a brow. "Then I take it you do not attempt to rob houses every night?"
"Not every night," she said in a casual manner. "No."
"Or ever before?"
"Or ever before." She sighed. "There now, are you reassured that robbing houses is not my chosen profession?"
"I never imagined for a moment that you were a skilled burglar. A professional would not have been caught by a mere girl."
"A random act of circumstance." She met his gaze directly. "You may be confident, the next time I attempt to rob a house I shall take additional precautions against discovery."
Was that a slight hint of amusement in her eye or was she mocking him? He bit back a smile of his own. "That is good to know, since the next time we might well be working together."
"Do you foresee our breaking into houses?"
"I suspect making any prediction regarding you and I would be a mistake." He stepped to the door and opened it. "Now, I would be happy to escort you to your rooms."
"I thought the earl said the butler would show me to my rooms?" She swept past him into the corridor with the same aplomb as if she wore a ball gown rather than somewhat shabby men's attire. He had never before considered just how enticing men's clothing might be on the right woman. While not an improper inch of skin was revealed anywhere—although the mere nature of the trousers themselves were improper—there was something about the vague suggestion of what the loosely fitting clothing concealed that was distinctly...exciting.
"We thought it would be best if I accompanied you rather than a servant." He started off, confident she would stay by his side.
"So that you may keep an eye on me?"
He paused for a moment, then nodded. "Exactly."
She smiled in a superior manner, as though somehow this had been her plan all along, but said nothing. He could well imagine what she might be thinking. Miss Montini did not appear to be the kind of woman to take well to anything less than total victory.
He led her up the stairs to the next floor. While his mother had told Andrews to prepare a room in the wing where she and Regina and Sterling had their rooms, Nate and his older brother had thought it better to put her into the unoccupied rooms next to Quint and across the hall from Nate. It was an excellent idea, he thought now, in so many ways.
He stopped at the door to her rooms and opened it. "I hope this will be satisfactory."
She glanced inside. "It's lovely."
"A footman will be here in the morning to escort you to breakfast."
Miss Montini slanted him an annoyed look. "Will a servant be stationed at my door all night as well?"
He raised a brow. "Is that necessary?"
"I wouldn't think so." Her tone was sharp. "Unless I am to be considered a prisoner."
"Not at all, Miss Montini. You are our guest." He narrowed his eyes. "And I fully expect you to behave like one."
"I do know how to behave properly, Mr. Harrington."
"Oh, so these past few days of house breaking and attending parties you have not been invited to were an aberration?"
"I believe I have already said that." Her jaw clenched. "My actions were necessary."
"No, in truth they weren't. You could have come to my mother or Sterling. You could have come to me. I would have listened to you."
"You didn't listen in Egypt."
"Perhaps not. And when I next see your brother, he shall have my apologies." He paused. "I'm sure you will hear from him soon."
"Yes, well, he's never been very good about that sort of thing." She shrugged as if her brother's lack of communication didn't worry her.
"Miss Montini, if we are to work together, it's not too soon to develop a certain level of trust between us. I suggest, in the morning, you tell me all of your older brother's suspicions as well as everything regarding his search for the seal." He thought for moment. "It might also be wise if I took a look at the letters you received from him. In spite of their questionable nature, there might be something you have overlooked."
"Very sensible." She stared at him thoughtfully. "I must confess, I had the same thought. I requested his letters be sent along with my things."
He waved the sealed note in his hand. "I shall have this delivered at once."
"Thank you. Mr. Harrington?"
"Yes?
She pulled the door closed and leaned her back against it, as if concerned that he might be overcome by the mere sight of a bed and ravish her on the spot. With a little encouragement he would be more than willing to try. He brushed the intriguing thought from his mind. "Might I ask you a question?"
"Go on."
"Why are you doing this?"
He shrugged. "It is an exceptionally large house, and as I assume you did not explore the entire building—"
"No, I didn't mean escort me to my rooms." She huffed in disdain. "I understand why your mother might wish to help me. She wishes to avoid even a hint of scandal, and I suspect that her tentative connection to my mother plays a part as well. But you." She shook her head. "Why would you want to help me?"
"A number of reasons." At least one of which had to do with her blue eyes and the curve of her shoulder in an apricot dress. "To begin with, I have worked far too hard to improve my own brother's reputation to have speculation damage it now. Nor do I wish to have my own honesty in question. Secondly, we are talking about a discovery of immense importance. I should like to play a role in that." He rested his shoulder against the door frame and gazed down at her. "And because I wish to know you better."
She studied him suspiciously. "Do you?"
"I do indeed."
"Why?"
"Because I have thought of nothing but you since we first met. You see, Miss Montini, you made a rather serious error the other night."
"Did I?" Her eyes widened.
"You did. You made me think, however briefly, that we had shared a kiss. The thought that I couldn't recall such a kiss was driving me mad. However," he leaned closer, "even before you vanished from the ball, I realized we had never met because I would never have forgotten kissing you."
"Nonsense," she said weakly.
"And I cannot get the idea of kissing you out of my head."
She swallowed hard. "Are you thinking of kissing me now?"
"I can think of little else."
"Why?"
"Surely, Miss Montini, you have looked in a mirror. You are quite lovely. The delicate curve of your cheek is only emphasized by the defiant tilt of your chin. I appreciate defiance almost as much as intelligence in a woman. Your eyes flash with fire, Miss Montini, when you're angry or indignant or trapped. And your lips..." His gaze dropped to her mouth, then back to her eyes. "...beg to be kissed. Often and quite thoroughly. In short, Miss Montini, you are entirely..." He lowered his mouth to hers. "...irresistible." His lips brushed across hers, and for a moment she stilled.
Then she laughed. "Good Lord, is that really effective with women?"
He straightened slowly. "What?"
"All that 'your eyes shimmer like stars and your lips are like cherries' nonsense."
"I don't believe I mentioned stars or cherries." He smiled slowly.
"You would have if you had thought of it."
"Probably."
Her eyes narrowed. "But you didn't answer my question. Do you find that sort of thing is successful with women?"
"Quite often, yes."
She shook her head. "They must be very stupid women."
"I'll have you know I don't like stupid women. I find them..." He searched for the right word. "...stupid. And not the least bit enjoyable."
"Oh?" She raised a brow. "Are you the type of man who likes a bit of a challenge in his carnal pursuits, then?"
"Yes. No." He paused. She had him so confused he had no idea what he meant. "I'm afraid however I answer that now will be wrong. But do tell me, Miss Montini, do you intend to be a challenge?"
"I do not intend to be the object of your pursuit, if that's what you're asking. Furthermore, if by some wild alignment of the stars and the disappearance of all rational behavior in this world I were to become said object, I am not a challenge you can overcome with pretty words." She crossed her arms over her chest, and he was compelled to step back. "Poor Mr. Harrington. Apparently you have never before encountered a woman who is not merely not stupid, but of an intelligence superior to yours."
"Nor have I done so now," he said with a shrug, although he wasn't entirely sure she might not be as intelligent as he.
She scoffed. "We shall see."
"Indeed we shall."
"You do need to understand and acknowledge this, Mr. Harrington—I am not here to be your friend, and I am certainly not here to be your...your lover." Again her eyes flashed at her words. It was most fetching. "We are uneasy allies in the pursuit of justice. We are together for the express purpose of recovering the seal. Nothing more than that."
"Yes, of course." His gaze drifted to her lips once again. They did indeed beg to be kissed. "Yet it seems to me one relationship does not preclude the other."
"Nor does one ensure the other."
"You do owe me a kiss, you know." He bent closer. "Preferably in the moonlight."
"I see no moonlight at the moment."
"Save that in your eyes."
"I thought there was fire in my eyes."
He chuckled. "They are most remarkable eyes."
"Ah, Mr. Harrington." She rested her hand on his cheek and her voice softened. "You are a scoundrel, but a charming scoundrel. I can well imagine your words would make any number of women melt at your feet. Fortunately..." She gazed into his eyes, and his breath caught. "...I am not one of them." She dropped her hand and pushed open the door. "Good evening."
"Miss Montini." He grabbed her hand and pulled it to his lips. "You should know I am not the type of scoundrel to easily accept defeat. Be it the pursuit of artifacts..." He kissed her palm, then met her gaze firmly. "Or the pursuit of something far more exciting." He released her hand and stepped back. "You may consider that fair warning."
"I shall consider it as nothing more than it was. A frivolous statement from an admitted scoundrel. As such, I see no need to take your warning as anything other than the inconsequential comment that it was." She nodded. "Good evening." She stepped into her room and closed the door firmly in his face. He heard the lock click into place.
He knocked sharply on the door. "I did not admit to being a scoundrel."
Her voice on the other side of the door was faint. "You did not deny it."
"You do realize I fully intend to kiss you in moonlight or elsewhere."
"I wouldn't wager on it, Mr. Harrington."
"Regardless, you do owe me a kiss."
Muffled laughter was the only response.
He stared at the closed door for a moment, then smiled slowly. This—She—was no doubt going to be far more than he had bargained for. But there was time enough to consider her, and all that went with her, in the morning.
At the top of his list of items needing consideration was his brother's role in all this. He needed to make certain Quinton had had no part in this theft. Not that he didn't believe his brother's denial, but Quint had a habit of twisting facts to suit his own purposes. He wouldn't put it past Quint to know more about this than he had said thus far.
Regardless of his brother's involvement or lack thereof, he felt that odd sense of anticipation, of impending excitement, that he always had at the beginning of a new adventure.
Still, he couldn't help but wonder what would be the greater adventure. The quest or the woman?
And ultimately, which would be the greater find?
Gabriella pressed her ear against the door, She heard his footsteps retreat, then another door opened. Apparently, Nathanial Harrington's rooms were directly across the hall. Not that it mattered. He could be sleeping in the next bed for all the good it would do him.
She blew a long breath and looked around. The room was far larger than her own and far more luxurious. There was something to be said for immense wealth. She moved to the bed and saw that nightclothes had been laid out for her. They no doubt belonged to the Lady Regina, who probably wasn't at all pleased about the loan of them.
Gabriella changed quickly, extinguished the lamp, and climbed into bed. And stared unseeing at the ceiling.
Well, this hadn't turned out as she had planned. The plan, as ill-conceived as it now appeared, was simply to find some sort of proof of the Harringtons' involvement with the seal. Now, she had agreed to accept their assistance, and was furthermore installed in their home as a guest. Which, admittedly, could prove to her advantage.
She had not anticipated Lady Wyldewood's response to her situation. Not her insistence that the family provide their help, and certainly not the countess's intention to talk to her about her mother. In truth, she had thought the fact of that long ago friendship might well prove a valuable weapon to keep her out of the hands of the authorities. Lady Wyldewood surely wouldn't have the daughter of an old friend arrested. It was the one thing Gabriella hadn't revealed to Xerxes about tonight's plot.
Admittedly, on occasion, she'd wondered about her mother. Who she had been and who her family was. The letters she'd found gave no clue, even though nearly all the notes were from her mother's sisters. They were the kind of letters sisters wrote to one another, bits of gossip and that sort of thing. One mentioned a necklace her mother had apparently left in England. But while the letter from Lady Wyldewood had been on her stationary, in the letters from her mother's sisters there were no surnames used, no addresses included, and no clue as to where these sisters might be found. Not that she cared beyond simple curiosity. Her jaw clenched. After all, they had had no interest in their sister's only child, so why should she have any interest in them?
Gabriella closed her eyes and willed herself to sleep. How different her life might have been if her aunts had taken her in after her father's death. She would have grown up as a proper English lady in a proper English home, rather than the Gypsylike existence she had lived with Enrico. Not that she would have changed any of it, she amended quickly. Her years of traveling with her brother, sharing his work, had been the grandest of adventures. Still, it would be nice to feel one had a permanent place in the world. A home, a family, people who cared if you existed or not. She had lived in London for nearly a decade now yet still felt out of place. She did not belong here, not really, and there was nowhere she did belong.
She couldn't help but wonder if her mother had lived whether she would be like Lady Wyldewood—kind, generous, and very nice. She rather hoped so, not that it made any difference now. It was pointless to consider what might have been. Her thoughts tonight, here in the dark, were probably attributable to seeing the bond shared by the countess and her family, and of no more significance than that. But if her mother hadn't died or if her aunts had taken her in and she'd been raised as a proper English lady...
She would be the perfect match for Nathanial Harrington.
Where on earth had that come from? It was absurd. She huffed, rolled over and punched the pillow, then curled up and tried to push the annoying thought from her head. No doubt it was all his fine words that had made her think such a ridiculous thing in the first place. She had never wanted to be a proper English lady. She had accepted long ago that she would never be any man's wife, let alone the wife of the son of a noble family. No matter what path Nathanial's life took, he would always be the brother of an earl, and she was not part of that world. She simply did not belong and never would.
And if she had wondered, if only for an instant, when he'd held her in his arms to dance, how lovely it would be to dance with him always, it was a ridiculous notion. And if, when he had kissed her hand, she felt the oddest longing deep within her, if just for a second, for him to keep her hand in his forever, it was an absurd thought. And if, when he'd come so very close to kissing her tonight, she'd wanted him, if for no more than a moment, to truly kiss her, often and quite thoroughly, it was insanity and best forgotten.
No, Nathanial would be...well, he would be her partner in this quest, and when it was done, they would go on with their separate lives. Even if, at the moment, she had no idea what hers might entail and no plans beyond recovering the seal. She had more than enough time to decide her future when Enrico's legacy was assured. Right now, that was all that was important and the only thing she wanted.
And in that last moment before sleep claimed her, when all rational thought had faded and dreams beckoned, the oddest thought drifted through her head. Just possibly Nathanial Harrington might well be all she'd ever wanted.
## Five
There was indeed a servant waiting to escort her to breakfast. Gabriella wondered briefly if he had been outside her room all night, then discarded the idea. It made no difference, after all. She had no intention of leaving the house and had been far too tired to resume any kind of search last night.
No more than a minute or two after she climbed out of bed, a maid had appeared at her door bearing a fashionable dress and appropriate undergarments. No doubt these too were Lady Regina's. The maid, a young woman named Edith, helped Gabriella dress. In spite of the look of curiosity in her eye as to who this stranger was who had taken up residence in the middle of the night, Edith was apparently too well trained to speak unless spoken to. And this morning Gabriella had no desire to explain herself again.
Morning sun flooded the elegant breakfast room. It was already half past nine, far later than Gabriella usually slept. Even so she found herself alone save for the butler—Andrews, if she recalled correctly—and a maid who was leaving as she arrived. Andrews filled her plate from an enticing array of dishes laid out on the sideboard and set it before her.
"Will there be anything else, miss?" the butler said, pouring her a cup of tea.
"No, thank you." Gabriella stared at the heaping offering. There were kippers and kidneys, coddled eggs, bacon and toast, and far more than she was used to. "This will do, I think."
"Very well, miss."
She took a bite of the eggs and realized just how hungry she was. She had been too nervous last night to eat much of anything, and this was delicious. She finished half her plate before she realized she was eating entirely too quickly. In spite of concealing the true state of her finances, she wouldn't want anyone to think she was starving.
"Mr. Andrews?" She glanced at the butler.
He had taken up his position beside the breakfast room door. "Just Andrews, miss."
"Andrews, then." She nodded. "I rarely sleep this late. Am I the last to come down for breakfast?"
"No, miss."
"Oh." Apparently he was also too well trained to offer information that had not been requested. "Who else is about this morning?"
"His lordship rides in the park every morning. Today, Master Nathanial joined him."
"And Lady Wyldewood?"
"She has not yet come down."
"And the others?"
"It was an exceptionally late evening, miss." A hint of chastisement so vague she might have been mistaken sounded in his voice. "Neither Lady Regina nor Master Quinton have yet arisen."
"I see." Gabriella slathered some jam on a piece of toast. "Will they return soon? Lord Wyldewood and his brother, that is."
"I cannot say, miss. However, Master Nathanial suggested you might enjoy waiting for his return in the library. The collection includes a great number of books regarding ancient civilizations he thought you might appreciate." The butler paused. "He also requested that I remind you of your position here as a guest."
"Tell him—" She bit back the words. Andrews was nothing more than a messenger, and it really wasn't fair to take any annoyance triggered by Nathanial's comments out on the butler. She forced a pleasant smile. "I shall tell him myself."
"As you wish, miss."
Obviously, Master Nathanial had no intention of allowing her to forget for a moment that while she was to be considered a guest, she was certainly not to be trusted. Nor, if truth were told, could she blame him. If the circumstances were reversed—if he had been caught trying to find evidence of her wrongdoing in her own home—she would be hard-pressed to trust him.
She took a bite of her toast and considered exactly what she'd gotten herself into. She was now committed to searching for the seal with a man she didn't trust who didn't trust her. In that respect it did seem a fitting partnership. And admittedly, it would work to her advantage. While she was familiar with some of the older members of the Antiquities Society, as well as the director and his wife, she didn't know those men who followed the same path as her brother had. Like Enrico, they were more likely to be found perched on the back of a camel or sleeping beneath the stars than on the streets of London. For most of them, their infrequent return to the seat of the civilized world was a chore to be avoided as long as possible. Necessary only to acquire funding or negotiate with museums or consult with scholars. Even though she'd spent a great deal of time in the society's library, she had never so much as seen any of the men, save one, that her brother suspected of having the seal.
Besides, as much as she hated to admit it, as a man, Nathanial had entrée to places she did not, and could move about far more freely than she as well. It was at times like this that she longed for the days of her childhood when she had dressed as a boy and Enrico treated her as one. It could have gone on forever had her waist not narrowed and her chest blossomed, and had a young man not much older than she discovered that Enrico's little brother was in truth his sister.
She brushed aside the faintest touch of regret. There was no point in looking backward. Life unfolded as it would, as it was meant to be. The ancients knew that. Even in the Bible it said there was a season for everything. Which didn't mean, of course, that one should sit back and wait for life to happen. One needed to pursue one's destiny. Even if one was female.
Gabriella finished her breakfast and a footman led her to the library. As irritating as it was to be accompanied everywhere, she had to admit, if grudgingly, that this was an exceptionally large house and she would have been lost if left to find her own way through its many twists and turns.
The footman opened the library door. She stepped inside and pulled up short. "My apologies, I didn't realize anyone was in here."
A gentleman sitting behind the secretary's desk rose to his feet. "Miss Montini, I presume?"
She walked toward him. "And you must be Mr. Dennison."
The earl's secretary was not especially handsome, yet not unattractive either. Rather, he was one of those unassuming men one might pass on the street and never notice. He nodded in a curt manner. "I have been instructed to give you whatever assistance you require."
"How very accommodating of his lordship."
Mr. Dennison pulled open the drawers on one side of his desk, then the other, and indicated them with a wave of his hand. "Perhaps you would like to go through my files? Again."
Heat washed up her face and she ignored it. "A very thoughtful offer, Mr. Dennison, and most appreciated."
"Would you care to look through the drawers of the earl's desk, then?"
"I'm not sure at this point it's necessary," she murmured.
"It most certainly is not. I can assure you, Miss Montini, I am not aware of any correspondence, documentation, or anything else in reference to the Montini seal."
She raised a brow. "The Montini seal?"
"That's what Mr. Harrington called it."
"I see." The Montini seal. "Nathanial Harrington, you mean?"
"Of course."
"Well, isn't that a surprise," she said under her breath. Apparently the man had meant it when he had assured her that credit for finding the seal would go to her brother. Again she wondered if Nathanial might be a far better man than she had expected.
"While I do handle all manner of paperwork for Mr. Harrington and his brother, as I was saying, I had never heard of this seal before this morning." His eyes narrowed. He no longer appeared the least bit unassuming, but rather, looked very much like a man one would not want to cross. "However, you may certainly search through both desks, as well as anywhere else in the library."
"That's not necessary, Mr. Dennison," she said in as contrite a manner as she could muster. It wouldn't do to continue to irritate the earl's secretary. He could prove an ally at some point. Although given the way he glared at her, that did not seem likely. "I wouldn't dream of questioning your word."
Mr. Dennison snorted in disdain. No, she and Mr. Dennison would probably not be friends.
"Now then, Miss Montini—"
A knock sounded at the door, and it opened without pause. Andrews stepped into the room and cast her an almost apologetic look. "Miss Montini, you have a—"
"Stand aside, my good man." Florence prodded him out of her way with the umbrella she routinely carried—because one could never be too prepared—and stormed into the room. "Gabriella Montini, what do you have to say for yourself?"
"Good morning, Florence?" Gabriella said weakly.
"It's not the least bit good. It's confusing and more than a little upsetting." Florence's gaze slid to the earl's secretary. "And who are you?"
"Edward Dennison, miss." Mr. Dennison drew himself up straighter. "Secretary to the Earl of Wyldewood."
"Hmph." Florence cast him a disdainful look, then turned back to her. "Gabriella, I insist you tell—"
"And you are?" Mr. Dennison cut in.
"I am Miss Florence Henry. Miss Montini's com—"
"Friend," Gabriella said quickly. "She is a very old and very dear friend. She has been so good as to allow me to live with her."
Florence glared at her, then huffed. "Yes, I am her friend." She fixed Gabriella with a firm look. "Now then, Gabriella, tell me what on earth is going on here."
"What is going on, Miss Henry," Mr. Dennison said coolly, "is that your friend was caught breaking into the house in the middle of the night like a common criminal."
Gabriella winced.
Florence stepped toward Mr. Dennison like a lioness defending her cub. "I can scarcely believe that. Furthermore, Miss Montini is not now, nor has she ever been, a criminal of any sort, let alone common!"
"Florence..." Gabriella started.
Mr. Dennison crossed his arms over his chest. "Ask her for yourself, then."
Florence glared at the secretary but directed her words to Gabriella. "Is what this man says true?"
"Didn't I mention that in the note?" Gabriella said under her breath.
"You most certainly did not." Florence continued to stare at Mr. Dennison. "Regardless of the inappropriate nature of her actions, I am certain she had good cause for her behavior."
Mr. Dennison snorted. "Her criminal behavior, you mean."
"Her necessary behavior to uncover the criminal activity of others!" Florence aimed her umbrella at him. "Activity you, no doubt, had a hand in!"
Gabriella leaned toward Andrews. "Perhaps you should send for help."
"Before they kill one another, you mean?" Andrews shook his head. "Mr. Dennison is a gentleman, he'd never strike a lady."
"I wasn't worried about Mr. Dennison's behavior but his safety," Gabriella murmured. She should do something to stop this, but the sight of dear, mild-mannered Florence filled with fury directed at the heretofore unassuming Mr. Dennison was so unimaginable that she couldn't seem to do more than stare.
Florence's eyes narrowed. "You—"
Mr. Dennison glared. "You—"
"You are the most sanctimonious man it has ever been my displeasure to meet! You, sir, are—are—" Florence raised her chin. "—no gentleman!"
Gabriella gasped. In Florence's view of the world, not being a gentleman was the ultimate failing.
"And you," Mr. Dennison's eyes narrowed, "are probably the most overbearing and irritating female I have ever encountered."
"I have never seen anyone so—so—" Florence shook her umbrella at him.
"And I have never seen..." Mr. Dennison grabbed the umbrella and, as Florence would never surrender her umbrella, her along with it. He stared down at her. "I have never seen..."
"What, Mr. Dennison?" Florence snapped. "What have you never seen?"
"I—" He huffed. "I don't believe I have ever seen—" He squared his shoulders. "—eyes quite as remarkable as yours, Miss Henry."
"Flattery, Mr. Dennison, will not win my favor," Florence said in a lofty manner. "However, oddly enough, I was thinking the very same thing about your eyes."
They stared at each other for a long moment as if there was no one else in the room. The initial tension between them changed abruptly, to something far more...intimate. It was distinctly uncomfortable.
At last Mr. Dennison drew a deep breath and released her umbrella. "Miss Henry."
"Mr. Dennison," Florence said coolly.
"You may wish to speak with Miss Montini privately, so I shall take my leave." He paused. "Regretfully."
"That would be most appreciated, Mr. Dennison." The tiniest hint of a smile curved the corners of Florence's lips. Gabriella wasn't sure she had ever seen a smile like that from Florence. "And equally regretted."
Good Lord, Florence was flirting!
Mr. Dennison flushed.
"I should need someone to escort me to my carriage when I am finished speaking with Miss Montini. If you would be so good as to return in a few minutes?"
"It would be my honor." Mr. Dennison nodded a bow and left the room. Andrews followed, obviously stifling a smile.
Gabriella stared at Florence. "What on earth was that all about?"
"I'm not sure." A smug smile danced on Florence's lips. "Did you see that? From the moment I stepped through the door, that man was flirting with me."
"He was arguing with you."
"Call it what you will, he was extremely flirtatious."
"You were extremely flirtatious!"
"Yes, I was, wasn't I?" A twinkle sparked in Florence's eye. "I've never been flirtatious before and I certainly didn't intend to be so now. Quite honestly, I didn't know I knew how. But apparently I find a battle of wills to be rather...stimulating."
"Florence!"
"He was quite dashing, don't you think?"
"Mr. Dennison? I think..." In truth, the gentleman who had been ordinary only a few minutes ago now did seem rather dashing. "Yes, I think he was. And I think he was quite taken with you."
"Well, well, fancy that," Florence murmured.
Gabriella stared. Enrico had hired Florence when Gabriella began living in London, to be both chaperone and companion. Florence had shared Enrico's—now her—modest London house for more than nine years. A mere ten years older than her, Florence was very much the sister—the family—she had never had. But in all their years together, Gabriella had never seen sparks between Florence and any man.
"It simply indicates that Mr. Dennison has extremely good taste," Gabriella said firmly.
"My dear Gabriella." Florence turned a knowing eye on the younger woman. "Your flattery will work no better than Mr. Dennison's."
"I thought his worked rather well."
"Which will not help you." Florence settled in one of the wing chairs positioned before the desk. "Your note said you were staying at a friend of your mother's?"
Gabriella nodded. "Lady Wyldewood knew my mother."
"Did you know that when you attempted to rob her house?"
"Yes." Gabriella sank into the other chair.
"I cannot approve of your methods. However," reluctance sounded in Florence's voice, "that was excellent knowledge to have at hand."
Gabriella studied her. "You don't sound very angry."
"Oh, I am furious with you. But I should have expected something of this nature. You have never been the sort of person who would let sleeping dogs lie, as it were."
"No, I can't. This is something I and I alone have to do."
Florence raised a brow. "A quest, then?"
"Exactly. I have to...to right this wrong." It was indeed a quest, as noble as any of those of the knights of old. "Someone ruined my brother, destroyed his life's work."
"And your hopes for the future."
Gabriella started. "You know?"
"You've never admitted it but I have long suspected your plans."
"My hopes were absurd, and more so now of course with Enrico's death. But still, I suppose the important word there is hope." She shrugged. "Hope that until the moment came that Enrico actually said I could not rejoin him and share in his work—"
"His adventures, you mean?"
"Yes, I suppose I do. I am not so foolish as to think convincing Enrico to take me with him would have been easy. In recent years I tried to subtly encourage him to come to the realization on his own that I could be of great help to him. That's why I studied and learned and worked—to become indispensable." For a moment, a sense of loss, for her brother and for herself as well, threatened to overwhelm her. She ignored it, as she always did. "I was so sure that the discovery of the seal would be the start of a hunt for Ambropia itself and that he would need me."
"I am so sorry, Gabriella." Florence reached over and patted the younger woman's hand. "Tell me, my dear." She straightened and fixed her with a steady gaze. "As you went to all this trouble, rightly or wrongly, did you find anything of significance?"
"No." Gabriella blew a resigned breath. "They all swear they have no knowledge of the whereabouts of the seal."
"And do you believe them?"
"Certainly I believe Lady Wyldewood and the earl. Mr. Dennison, as well, has assured me he has no information about the seal and has offered—really encouraged—me to search through his files myself."
Florence nodded. "He seems an honest man to me."
Gabriella resisted the urge to comment on Florence's assessment of the dashing Mr. Dennison. Files could be removed; his offer meant little. "Both Quinton and Nathanial Harrington say they know nothing about the seal's disappearance. I am not certain about the older of the two, but Nathanial—"
"Nathanial?" Florence's brow rose.
"I believe I can trust him, to a point at any rate." She met Florence's gaze. "He is going to help me find the seal."
"Is he now?" Florence considered her curiously. "I'm surprised that you don't find that suspicious."
"It was his mother's idea. She fears that the longer I continue my efforts to find the seal, the more likely it is that others will learn of my search. Which could bring suspicion and then scandal upon the entire family."
"I see." Florence thought for a moment. "Perhaps you should tell me everything that has occurred thus far, as I am, after all, your friend."
"My dearest friend," Gabriella said in a firm manner. "Yes, well, perhaps I should." She quickly related all that had happened: her appearance at Lady Regina's ball, and the details of what had transpired last night, leaving out the nonsense about kisses in the moonlight or dancing with Nathanial or the odd, longing way he made her feel.
"I see," Florence said when she was finished. "That matches what Xerxes told me, although he didn't have all the pieces."
Gabriella's eyes widened. "You already knew all of this?"
"I just said Xerxes didn't have all the pieces. Goodness, Gabriella, after all these years, you don't think I can't tell when something is amiss?" Florence snorted. "I knew the minute I read your note this morning that all was not quite right. And when I read the one you wrote to Xerxes—"
"You read the note I wrote to him?"
"Did you think I wouldn't? Did you think I wouldn't demand to read it after I read mine?"
"What I didn't think was that you would know I'd written a note to Xerxes in the first place."
"Then you should have arranged for him to be present when the note arrived." Florence shook her head. "When a servant arrives in the wee hours of the morning and says he bears a note from Miss Montini—"
Gabriella winced.
"I don't care whose name is on it, I will read it." Florence rose to her feet. "Gabriella, when we began this journey together you were sixteen and I was a governess without a position. Admittedly, that was due in part to my not being overly fond of small children." She shuddered as she always did when the subject of small children reared its head. "Regardless, your brother thought you and I would suit, as you were nearly an adult."
"And I always thought you were the only one willing to accept the position."
"There was that, of course. My point is that with your brother's death, I am now in your employ."
"We don't need to talk about this now."
"We haven't discussed it at all, and we do need to talk about it. Especially as you are about to embark on something—well, on this quest of yours. You are twenty-five years of age and now of independent means. Your brother hired me to be your chaperone, companion, and in many ways guardian, as, Lord knows, he was so infrequently in London. I had hoped that you would be married long before now."
"I have never planned to marry."
Florence ignored her. "Your failure to do so is as much my fault as it is yours. I did think, though, that amidst those libraries and museums you would find someone who would suit. It's not too late, of course—"
"Florence," Gabriella said firmly. "Marriage is not in my future. It never has been."
Again Florence paid her no mind. But then, she'd always dismissed Gabriella's opinion of her prospects for marriage. "However, until that time comes—"
"It won't."
"Or the time comes that you no longer desire my services—"
"Never," Gabriella said staunchly. "You are as much my family as Enrico was. As Xerxes and Miriam are now."
Florence shook her head. "It's a poor excuse for a family, but I suppose it is better than nothing."
"It's not nothing." For a moment, panic flashed through her at the very thought of being without Florence, Xerxes, and Miriam, who in many ways were more of a family to her than her half brother had ever been. Gabriella pushed the disloyal thought away. Enrico had been a wonderful brother.
Florence smiled. "We love you too, dear. But as I was saying," her voice hardened, "as long as I remain in your employ I shall do all in my power to keep you from harm. To continue to attempt to guide you along a path that will keep you from total disaster and, God willing, to avoid scandal." She shook her head. "Since your brother's death, you have not made it easy."
"I daresay you wouldn't like easy." Gabriella grinned.
"I seem to recall liking easy quite a lot."
"That wouldn't be any fun at all for you."
"Fun, my dear girl, is relative." Florence huffed. "I see no particular difficulty with you staying here. As long as Lady Wyldewood is in residence, you will be well chaperoned. However, I do not intend to leave you here alone."
Gabriella drew her brows together. "You're not planning on staying here—"
"Don't be absurd."
"Then what do you mean?"
"You shall see." Florence cast her a pleasant smile, but there was a determined look in her eye. "You are not the only one who can devise clever plans."
"What are you—"
"Furthermore, I expect a note from you daily and a visit every other day. You shall come to the house or I shall come here. To ensure that, I have brought only enough of your things for a few days."
"And Enrico's letters?"
"I brought those as well." Florence nodded. "Do be careful, my dear. And do attempt to be as honest and forthright as possible."
"Of course," Gabriella said in an overly innocent manner.
Florence studied her skeptically. "The ends do not always justify the means, Gabriella. Remember that. But you will follow your heart, I suppose. You always have."
"You have always encouraged me to do so."
"Yes, well, that might have been my mistake." She gave Gabriella a quick hug and started toward the door. "I shall see you soon, my dear." She paused and looked back. "And try not to refer to the younger Mr. Harrington by his given name."
"It's simply a way to distinguish one brother from another," Gabriella said, shrugging in an offhand manner. "It has no particular significance."
"No?" A skeptical smile creased Florence's lips.
"No," Gabriella said firmly. "It means nothing at all. I don't trust the man completely and I certainly don't like him. Admittedly, I will have to spend a great deal of time in his company, but it..." She set her chin. "...it can't be helped."
"A necessary evil, then?"
"Exactly."
"Yes, of course, he would be," Florence murmured. "Always the worst kind."
"What do you mean by that?"
"Nothing of significance. I shall see you soon, and I expect to receive a note from you tomorrow." Florence pulled the door open. "Mr. Dennison, how good of you to..." The door closed behind her.
Well, that was certainly a surprise. All of it. From Florence's reaction to her flirtation with Mr. Dennison to her mention of a plan of her own. But then hadn't the last two days been fraught with surprise?
Gabriella sighed and sank down in the nearest chair. Nothing was quite as she'd expected. Lady Harrington was very nice. The earl was suspicious and rather stodgy but not unkind. As for his youngest brother—Nathanial Harrington was the greatest surprise of all. He wasn't at all what she'd expected.
And she had no idea if that was good or very, very bad.
## Six
What did you do to Mr. Dennison?" Nate strode into the library, his tone far harder than he had intended.
"I didn't do anything to Mr. Dennison," Miss Montini said coolly. She sat at the secretary's desk as if she owned it, which in and of itself might well annoy Mr. Dennison, but certainly wouldn't disconcert him in any way.
"Someone did." Nate drew his brows together. "He looks both preoccupied and puzzled."
"Does he?" Miss Montini's tone was noncommittal and she continued examining what appeared to be letters arrayed before her on the desk.
"It is not like him to be either." Nate narrowed his eyes. "I have never known Mr. Denison to be anything other than competent and assured. And I have never seen him the least bit disconcerted."
"To every thing there is a season," she said under her breath, her gaze still on the papers on the desk. No doubt these were the letters from her brother.
She was ignoring him, that's what she was doing. Oh certainly she was responding, but in nothing more than a cursory manner. And with biblical quotes, no less. In truth, she was paying him no attention whatsoever. It was most annoying.
Admittedly, his mood was already somewhat foul. He wasn't sure how it was her fault—he hadn't even seen her today—but clearly it was. Usually when he joined his older brother on his morning ride, it was an excellent way to start the day, invigorating and refreshing. There was nowhere on earth as green and lush as England in the spring, even here in London. Today, Sterling had been full of questions about the legend of the Virgin's Secret and what scant factual evidence existed about Ambropia.
In and of itself, Nate enjoyed detailing what little was known of the lost city. But he knew that the Earl of Wyldewood was nothing if not thorough. It was just a matter of time before he would want to know about Enrico Montini. Which might well lead to Miss Montini learning more about her brother than he suspected she knew, or at least that was his impression, given the passion of her quest. He had the oddest desire to protect her from that knowledge. Absurd, of course, since he barely knew her.
He drew a deep breath and forced a cordial note to his voice. Regardless of the circumstances, she was still their guest. "I trust you slept well?"
"Quite."
He had scarcely slept at all, and even the most rational man could indeed place the blame for that squarely at the feet and well-turned ankles of Miss Montini. He had tossed and turned all night. In those few moments when he had dozed, he dreamed of kisses shared with blue-eyed beauties in the moonlight. Little wonder he awoke in a foul mood.
"Your rooms were acceptable, then?"
"More than acceptable."
If they were to accomplish anything together, he should probably set aside all thoughts of kissing the delectable Miss Montini, as difficult as that might be.
"And breakfast? Was it satisfactory?"
"It was excellent."
Still, even now with her hair pinned neatly into place, wearing a gown that was more than proper—indeed one might even call it virginal—her attention focused firmly on the letters before her, he had the most insane desire to vault over the desk, yank her to her feet, pull her into his arms and press his lips to hers. Lips he had no doubt would be firm and warm and pliant beneath his and would respond to his ardor with immediate enthusiasm as her shapely, seductive body pressed—
"And the weather, Mr. Harrington?"
"What?" His attention jerked back to reality, and in his mind she reluctantly slipped out of his arms.
"The weather, Mr. Harrington." She turned over a page then lifted her gaze to his. "I assume that was the next inconsequential topic."
"Inconsequential?" He stared. What was it about this woman that made him want to at once kiss her and turn her over his knee?
A slight, knowing smile touched the corners of her mouth as if she knew exactly what he had been thinking. Damnable creature. Well, two could play at that game.
"I'd scarcely call a fine spring day such as this inconsequential, Miss Montini."
She shrugged. "It is a spring day like any other."
"Not at all. It could be cloudy or rainy or blustery. But today the sun shines, the birds sing, and flowers are in bloom, their fragrance wafted about on the mere caress of a breeze." He propped his hip on the desk and smiled at her. "Indeed, Miss Montini, 'What is so rare as a day in June?'"
"Poetry, Mr. Harrington?" She scoffed. "I would not have suspected you were a poetic sort of man."
"I daresay, there are many things about me you do not suspect." He wagged his brows at her in a wicked manner.
"And many that I do." She settled back in her chair and studied him. "For example, I suspect you are a man who does not let minor inconsistencies like facts stand in your way."
"And why do you suspect that?"
"For one thing," she smiled in a smug manner, "it's only May."
"Are you not impressed, then?" He forced a mock serious note to his vote. "That I can bend the months of the year to suit my purposes?"
"You did no such thing. What you did was quote a line of poetry in hopes of impressing me because any number of women would fall at the feet of poetry spouting, handsome, exciting men who make their living in the pursuit of treasure and adventure."
He grinned. "You think I'm handsome?"
Her eyes widened in obvious dismay at what she'd said and the most delightful blush swept up her face. She leaned forward and directed her attention back to the letters on the desk. "Goodness, Mr. Harrington," she said under her breath. "A mirror would tell you no less, and I cannot imagine you are surprised."
"It's not the observation." He laughed. "It's the observer that has shocked me."
"Hmph."
"I am flattered that you think so highly of me."
"I don't think highly of you," she muttered, her gaze still on the papers before her. "I don't think of you at all."
"You think I'm handsome."
"It was an observation, Nathanial," she said with a shrug. "Nothing more significant than that."
His grin widened. "You think I'm exciting as well."
"I didn't say that." She glanced up at him, her expression again composed and cool. "I was speaking in general terms about men who make their living as you do."
"Nonsense, Gabriella." He laughed. "You think I'm handsome and exciting."
"I most certainly—"
"As we are making confessions." He leaned toward her. "I find you exciting as well as quite lovely."
"I am not the least bit exciting."
He grinned. "But you will not protest my observation as to your beauty?"
"It seems rather pointless; I am well aware of my appearance. Not that it matters."
"It matters to most women."
"I am not most women."
"No, you are not." He chuckled. "Most women would not treat a compliment as though it were an insult."
"You're right." She heaved an exasperated sigh. "It was rude of me of me. Thank you for the compliment, Mr. Harrington, it was very nice of you. Why, I am flattered beyond words."
He snorted back a laugh.
She pushed away from the desk and rose to her feet. "You have no idea how wonderful it is to know that a gentleman"—she cast him a skeptical look, as if questioning whether he was worthy of the title—"thinks you are lovely."
"Quite lovely." He nodded in a somber manner and slipped off the desk to his feet.
She crossed her arms over her chest. "Why, it has quite made my life worth living."
"Well." He shrugged modestly. "One does what one can."
Her eyes narrowed. "I daresay I should weep into my pillow each and every night if you did not think I was lovely."
He grinned. "There's no need for sarcasm."
"I cannot imagine a worse fate than not being lovely in your eyes."
He laughed, and she ignored him.
"Now then, I suggest we dispense with discussion of the fire in my eyes or the tilt of my chin, as I recall they were thoroughly discussed last night." She waved at the papers on the desk. "These are my brother's letters. I have read them countless times but you should go through them. You might see something I've missed. The letters contain the names of four men, including you and your brother—what are you staring at?"
"Your lips, Gabriella." His gaze flicked to her eminently kissable lips and back to her eyes, which did indeed flash with at least annoyance if not fire. "We have not discussed your lips."
"The lips that beg—" She bit her bottom lip as if to hold back the words.
He bit back a smile of his own. "That beg to be kissed? Yes, those lips."
She stared at him, then rolled her gaze toward the heavens. "Very well, then." She stepped toward him, closed her eyes and raised her chin. "Go on."
He grinned down at her. "Go on what?"
Her eyes remained closed but her shoulders heaved with a resigned sigh. "Kiss me. It's what you want. Go on, then."
He bit back a laugh. "Now?"
"Yes, of course now." Her eyes snapped open. "It seems to me we will never get anything at all accomplished if all you can think about is kissing me."
"That's not entirely all I'm thinking about," he said under his breath.
She cast him a glance designed to wither the confidence of even the most arrogant man. "That, Mr. Harrington, is not my problem."
"You called me Nathanial a moment ago."
She paused. "Did I?"
"Indeed you did, and I liked it."
"It was a slip of the tongue." She shrugged. "Not the least bit important. I certainly didn't intend for you—"
"There's nothing like hearing your given name from the lips of a beautiful woman. Lips I might add that are—"
"Yes, yes, begging to be kissed." She waved off his words with an impatient gesture.
"Regardless, I believe it would be most expedient for our purposes if we dispense with formalities altogether. You may call me Nathanial, I shall call you Gabriella."
"Mr. Harrington," she said firmly.
He raised a brow.
"Very well, then, I suppose it does make a certain amount of sense. And I have already been thinking of you as Nathanial. But only to differentiate you from your brother," she added quickly.
"Exactly as I thought."
"And it shall be no more importance that that of a...a...sister—yes that's it—a sister calling a brother by his given name. Not the least bit significant. Now, then." She again closed her eyes and raised her chin. "If you would be so good as to kiss me, we can put this nonsense behind us."
"Behind us?"
"This too is a matter of expediency. Nothing more."
"Expediency." He nodded. "And efficiency too, I would imagine?"
"Yes, yes." Impatience sounded in her voice. Her shoulders stiffened. "Go on with it."
"It's tempting," he said in a low voice, and stared down at her. This was indeed an opportunity. But one only a fool would take. "I daresay I cannot remember when last I encountered anything this...irresistible."
"I am flattered," she said in a cool voice that nonetheless sounded just a touch breathless. Her chin rose another notch.
"But I think not."
Her eyes snapped open. "What do you mean—you think not? How could you possibly think not?"
"It might have been that business about you calling me by my given name in the manner of a sister." He shook his head. "For future reference, Gabriella, when asking a man to kiss you, you should not put him in mind of his sister. It does tend to spoil the mood."
"I did not ask you to kiss me!"
"No." He shrugged. "You told me. That too tends to destroy the ambience of the moment. A man likes to believe—even if it's not true—that he is in command of such things."
She stared in disbelief. "Then you are not going to kiss me?"
"Oh, I am most certainly going to kiss you, but not at this particular moment."
"Don't be absurd. This is your opportunity, and I warn you, Nathanial, there shall not be another. Now." She huffed, stepped toward him and once more closed her eyes and lifted her chin. "Let's get this over and done with."
He bit back a laugh. "My dear Gabriella, a kiss is not something one gets over and done with. It is not a foul tasting medicine one is forced to take."
Her eyes opened. "I do know—"
"Surely you have been kissed?"
"Of course I have been kissed," she said sharply. "Any number of times."
He raised a brow. "Have you?"
She blushed yet again, and he noted how there was something quite compelling about an intelligent, confident woman who blushed so easily. "I am not a child."
Still, he'd wager his next big find that she had not been kissed often and probably not well. "And were those previous kisses such that you simply wished for them to be over and done with?"
"Well, ye—no!" She forced an awkward laugh. "Each and every kiss was quite enjoyable. Really, very nice."
"Very nice?" He shook his head in a somber manner. "A kiss, Gabriella, should never be merely nice."
She opened her mouth to protest.
"Even very nice is not good enough," he said before she could utter a word. It did seem best. "First of all, a kiss is...an overture, if you will, to the grander symphony to come. A prologue to the rest of the story." He clasped his hands behind his back and slowly circled her. Her wary gaze followed him. "A taste of the banquet yet to be savored."
"Mr. Harrington—Nathanial." She jerked her gaze back to a point directly in front of her and squared her shoulders. "There will be no symphony, no story, and certainly no banquet."
Nate smiled. "You are taking my words in a manner in which they were not intended. I am explaining the nature of a kiss in general terms, not the nature of our kiss." He paused. "Unless, of course, you see our kiss as the first step toward you joining me in my bed."
She shot him a look of disdain over her shoulder. "I most certainly do not! And would you stop circling me. I feel like a chicken being marked by a fox."
"Regardless." He casually moved to stand in front of her. "A kiss is still a beginning. As well as a turning point. A kiss should make you feel as if it were the first moment of something wonderful."
She snorted.
"You don't agree?"
"No." Her foot twitched as if she were resisting the urge to stamp it. "A kiss is..."
"Yes?" His brow rose.
"It's..." She raised a shoulder in an offhand shrug. "It's a momentary loss of control of one's senses. Yes, that's it. It's nothing more than an instant of surrender to one's baser instincts."
"Oh dear, Gabriella." He shook his head in a mournful manner. "You may have been kissed but you have obviously not been well or properly kissed. And you have never been kissed by me."
"Come now, Nathanial."
"Do you doubt me?"
"I do not doubt your arrogance."
"A kiss is not something one closes one's eyes and braces oneself for as if one were England preparing for a Viking invasion." He cast her a slow, wicked smile. "A women who has been well kissed does not think of a kiss as merely a kiss."
She stared for a moment then accepted his challenge. "A woman who has kissed you, you mean?"
He shrugged in a modest manner. "I have yet to hear a complaint."
"Very well, then." She smiled pleasantly. "Prove yourself."
He hadn't quite foreseen that. Caution edged his voice. "What do you mean?"
"I mean, Nathanial, I have offered you a kiss for the one you feel I owe you. And even I can understand how you might have a legitimate claim. Therefore, as I can see you will be like a dog with a bone and not let this go, I shall give you another chance." She crossed her arms over her chest, a gleam of triumph in her eye. "Kiss me."
"I don't know that I should." He shook his head slowly. "A kiss—especially a first kiss—is to be savored and enjoyed. And remembered always."
She raised a brow. "Not up to the challenge, then?"
"Oh I am certainly up to the challenge," he murmured, and studied her for a moment. "I'm simply not certain if I wish to be commanded to kiss you."
She shrugged. "It remains your choice."
"Indeed it does." He paused. "And were I to kiss you, I should begin by stepping very close to you." He moved closer and stared down at her, close enough to see the satisfaction in her eyes fade to uncertainty. "So that I may take you in my arms."
"No doubt. Go on."
He wrapped his arms around her and gently drew her closer. "I would then gaze into your eyes, your endless blue eyes that could hold a man, even a man of strength, spellbound. Lost, if you will."
"Nonsense," she said weakly. "They're simply blue."
"There's nothing simple about them. They are the color of a mountain lake, the calm waters before the storm. Eyes that carry within them secrets, and promises of something wonderful for a moment or forever."
"Utter rubbish." Nonetheless her arms slipped around his neck and he bit back a smile
"Then my gaze would slip to your lips." He glanced at her mouth. She bit her bottom lip in the nervous manner he had already noticed, and his stomach tightened. "Just for a moment, just long enough to anticipate the soft, ripe warmth of them against my own. To wonder at the taste of you. Will you taste of bold, erotic spices or will you taste as sweet and delicious as new picked berries? Or as intoxicating as champagne? Anticipation, Gabriella." His gazed shifted and locked with hers. "Anticipation in a first kiss is most important."
She swallowed hard. "How absurd."
"And then I would lean closer, until my senses are awash with the scent of you." He angled his head toward hers until his lips were no more than a breath from hers. "Fresh and vaguely like lavender, with the merest touch of something more. Something exotic, unknown as yet but exciting and completely irresistible."
"Oh..." The word was no more than a sigh, the merest breath of air against his lips. Her eyes drifted closed.
"It would be very nearly perfection itself."
"Yes..." Her body pressed closer to his with a movement so slight he doubted she was aware of it. But he was. "Perfection..."
He had her now. She wanted to kiss him as badly as he wanted to kiss her. And he couldn't remember ever wanting to kiss a woman more. But as much as he wanted this, he knew one kiss with Gabriella Montini would never be enough.
"Very nearly." Nate drew a deep breath and summoned every bit of self-control he possessed. "But without moonlight, it is not the kiss you promised." He straightened and released her, ignoring the stunned look on her face. He moved around the desk and settled in the chair. "Now then, we should get on with these."
She sucked in a sharp breath and glared. There wasn't a doubt in his mind, he would pay for this.
He grinned to himself. He couldn't wait.
## Seven
You—You—You—" Gabriella sputtered as if she couldn't quite catch her breath. As if she'd been hit in the face with a pail of cold water. Not that she was going to let him know how shocked and, well, possibly disappointed she was. As if she had wanted him to kiss her, which she certainly hadn't and never would. Regardless, his behavior was nothing short of dastardly. "Nathanial Harrington, you are an arrogant ass!"
"My, my, Gabriella, your language." He shuffled through the letters on the desk, his gaze firmly on the papers in front of him as if she hadn't said a word. As if she wasn't there!
There was nothing to be done about it. She was going to have to kill him. Slowly.
"Your brothers have obviously been a bad influence on you."
My brother and men exactly like you. She bit back the words and drew a deep calming breath. It hadn't been easy to discard the manners, or rather, lack of manners, she acquired in the years spent with her brother. Propriety, especially when it came to language, had always been something she'd had to work at, much like any of her other studies.
"One would have thought one was back among his comrades in the deserts of Egypt," Nathanial said mildly.
She clenched her fists by her side. "My apologies, Nathanial."
"Accepted."
"I am eternally sorry—"
He smiled in a benevolent manner. "Not at all."
"—that you are such an arrogant ass."
He glanced up at her, his eyes wide with feigned innocence. "I don't know why you're glaring at me like that, it's not as if you wanted to kiss me."
"I'm not glaring," she said in a clipped tone.
"My dear Gabriella, if looks could kill, I would be lying on the floor dead by now, shot through the heart by your gaze alone."
"That would be a very great pity." She sniffed.
"I'm glad you think so."
"It would be entirely too fast." Gabriella braced her hands on the desk and leaned toward him. "No, you deserve something much, much slower. Tied to stakes and stretched out over a hill of African ants perhaps."
He rose to his feet. "Tied, did you say?"
"Under the hot, blistering tropical sun."
He planted his hands on the desk, mirroring her stance, a distinctly wicked smile on his lips. "Naked, no doubt?"
Naked? Why on earth did he have to use the word naked? At once the image of a naked Nathanial Harrington staked over an ant hill popped into her mind. Not that she knew exactly what an adult male would look like in that position, but between her limited experience and the paintings and sculptures she'd seen, well, she did have a fairly vivid imagination. She pushed the thought firmly aside.
"Or perhaps torn from limb to limb by savages in the jungles of South America."
"Savages who have first ripped my clothes to shreds, do you think?" The gleam in his eye matched his smile.
Again a naked Nathanial Harrington filled her head, savages pulling on every limb, tattered remnants of his clothes dripping off him like icing from a cake. She winced and shook her head. "Or...or...eaten by cannibals. Yes, that's exactly what you deserve."
"Boiled alive probably." He nodded in a solemn manner belying the look in his eye. "Naked, of course."
"Would you stop that!" A naked Nathanial Harrington sat in a large iron pot over a fire surrounded by cannibals. She straightened with a jerk. "Stop that this minute!"
He raised a brow. "Stop what?"
"Stop using that word!"
"What word?"
"You know what word!"
He shook his head and grinned. "I have no idea what you're talking about."
She huffed. "Naked, Nathanial!" Good Lord, had she just used naked and Nathanial in the same sentence? Aloud and in front of him? "The word is naked! Naked, naked, naked!" And she couldn't seem to stop. "As you well know."
His grin widened. "The examples were yours."
"Not the way you embellished them with...with..." She closed her eyes and sent a quick prayer heavenward to beg for calm and to give thanks that she wasn't armed. "It was highly improper, most suggestive, entirely too...too intimate and...and..." Erotic, exciting, seductive. She drew a deep breath and met his gaze. "Uncomfortable."
"Come now, Gabriella. You can't—" Realization dawned on Nathanial's face and his smile vanished. "You are embarrassed, aren't you? Why, you're blushing again."
"Yes, well..." The ease with which she blushed was the bane of her existence, and there didn't seem to be anything she could do about it. But of course she was embarrassed. Not as much by his words as by the explicit images her own mind had created. Even so, it was his fault.
"I am sorry." He winced. "I didn't intend—I had no idea that you—that is to say—"
"No idea? And why not?" The words came without thinking. "Because women who break into houses and pretend to be someone they're not and are every bit as clever as you, who want to restore the good name of their family and have a sense of honor, would, of course, not be the sort to be embarrassed by crass, improper comments? That such women do not deserve the common courtesies you would give to a lady on the street? Because my family, my background, my circumstances are not such that they warrant respect?"
It was his turn to look as if he'd been dashed by cold water. "I assure you, Gabriella, my intent was only to tease—flirt, if you will. I never meant—"
"Enough, please." She pushed out her hand to stop him. Where had her outburst come from? In truth she was far more annoyed than embarrassed. Just like her tendency to blush, not keeping her mouth shut when she was angry had always been another unfortunate character flaw. "Now, I must apologize." Certainly, the differences between her family and his, her life and his, had been brought home to her last night. And yes, she might have felt a twinge of what could possibly be called resentment or even jealousy. But it was absurd. Life was what one made of it regardless of the hand one had been dealt. "Your family has been nothing but kind and generous to me, far more so than I deserve. My remarks were uncalled for."
"No, I am to blame. I baited you, and for that I must beg your forgiveness. I am most sorry. I lost my head." He took her hand. "In my defense, Gabriella, you should know..." He raised her hand to his lips, his gaze never leaving hers. "...you were not the only one who was disappointed."
"I wasn't—" She paused, then drew a deep breath. "Your apology is accepted. I would prefer that we never bring this incident up again."
"Oh I agree," Nathanial said somberly, but the wicked twinkle had returned to his eye.
She stared for a moment. "I can't trust you at all, can I?"
"Of course you can. In most matters, I can be most trustworthy. Now, then." He gestured at the letters. "Where do you suggest I begin?"
"Here." She stepped to the desk, of necessity standing far too close to him than was proper. Still, they were going to work together, and her standards of what was and was not acceptable would have to change, or at least bend. She reached in front of him, her arm brushing against his. Without warning the feel of being in his arms washed through her. She firmly set it aside. Now was not the time, nor, she amended at once, would there ever be a time. She arranged the letters in chronological order. "There are only seven. The first few came rather quickly, and as you will see, are the most lucid of the lot. The last..." She shrugged. "I suggest you read them in order."
"Very sensible." He sat down and picked up a letter. "This is the first?"
She nodded. He started reading, then glanced up. "Do you plan to watch me read every word?"
"Not every word."
"It makes me most uncomfortable." He grimaced. "This is a library, Gabriella. I would think you could find something to read. There are a great many reference works here that you might enjoy. Or better yet, a novel."
She scoffed. "I never read novels."
"That explains a great deal," he said under his breath.
"What do you mean?" She drew her brows together. "What exactly does that explain?"
"Your manner. Your attitude toward life, as it were."
"My attitude is just fine."
"You, Gabriella Montini, take the world entirely too seriously."
"You don't know that. You don't know me."
"Nonetheless." He shrugged. "This was not difficult to ascertain."
"Just because a woman doesn't leap into your arms, doesn't long for your embrace, doesn't ache to feel your lips upon hers—"
He arched a brow.
She ignored him. "Does not mean she takes the world too seriously."
"If you say so."
She huffed. "The world is a serious place, Nathanial Harrington."
"Indeed it is."
"And my life is a serious matter. My brother is dead, his reputation is shattered. I have no real family save a handful of serv—friends. And the one thing I truly wanted in my life is now—" She blew a resigned breath. "—out of the question."
He settled back in the chair and studied her. "What is the one thing you truly wanted?"
"It scarcely matters." She waved off his question and wandered to a bookshelf. "As you think a novel will somehow make my manner more frivolous—"
He laughed. "I never used the word frivolous."
She cast him a haughty glance. "It was implied."
"I should have said...lighthearted. Yes, that's it."
"My heart is anything but light at the moment, nor has it ever been."
"What a shame," he said softly.
"Not at all, Nathanial. It's simply how life is." She turned back to the shelves. "Do you have a recommendation? As to a novel, that is?"
"Come now, Gabriella, there must be some author's works you like? You cannot tell me you've never read a novel? Not even in a youthful misspent moment perhaps?"
"My youth was not especially misspent." Unless one considered being dressed as a boy and accompanying your brother from one exotic location to another on a quest for antiquities misspent.
"Still, you must have a favorite?"
"I don't think so," she said under her breath. Now that she thought about it, she couldn't remember ever having read an account of fiction, although surely she must have. Her brother had taught her to read, but in that employed the use of the manuals and historical references he routinely carried with him, and the Bible, of course. When she began her schooling in England, she had been made to memorize a great deal of poetry, and recalled studying the plays of Mr. Shakespeare, but not a single work of narrative fiction came to mind.
"Not Mr. Dickens? Or Mr. Trollope or Miss Austen?"
"Apparently my education has lacked in that respect." She perused the titles on the shelf. "Besides, I've never had the time."
"How do you spend your time?"
"I study, Nathanial. I study ancient civilizations, history, archeology, myths, legends, and anything else that might prove useful for my brother's work. I have earned certificates at Queen's College, have already been awarded one university degree, and I continue my studies. I have as well committed to memory most of the books and papers in the Antiquities Society library and archives." She glanced at him over her shoulder. "I have an excellent memory."
"I'm not surprised."
She arched a brow. "A compliment, Nathanial? One that has nothing to do with the tempting nature of my lips or the hypnotic quality of my eyes?"
"I don't know what came over me." He grinned. "I shall try not to let it happen again."
In spite of herself, she returned his smile. The man was quite engaging. "In addition, I am fluent in nine languages, including Coptic, Persian, Turkish, and Arabic."
He stared. "Nobody speaks Coptic. It's extinct."
"Not entirely. It's still used in the Church of Alexandria."
"Even so, why learn something so obscure?"
"Because it's the closest thing we have to any knowledge of an ancient Egyptian spoken language."
"I suppose it makes sense in a strictly scholarly sense. But why would you learn Turkish, Arabic, Persian? Most women of my acquaintance—even those few engaged in scholarly pursuits—learn French, a smattering of Italian, perhaps German. Even if one wished to travel extensively, that would certainly be sufficient."
"I thought we had already agreed I am not like most women of your acquaintance."
"Still, it seems rather unusual."
"Perhaps it is." She studied him for a long moment. Telling him her plans hardly mattered now. Nothing would come of them. Surely she should trust him enough to tell him this, and trust as well that he wouldn't laugh at her ambitions. She drew a deep breath. "I had hoped to become knowledgeable enough to join my brother in his work. To be indispensable to him."
"I see." He nodded thoughtfully. "I would say it is a farfetched aspiration for a woman, but we have established you are not like most women." He paused and considered her. "This, then, is what you wanted most in your life, isn't it?"
"It's of no consequence now." She shrugged and turned back to the shelf. Somehow, telling him, saying it aloud now twice today when she'd never admitted it to anyone before, made her loss all the more real. "Besides, I'm not sure I ever could have convinced Enrico to let me join him. I had hoped if I learned enough, if I made myself...well, essential, important to his work, he would allow me to come with him."
"Those obscure, remote areas of the world where your brother and the rest of us search for the treasures of the past are not places for western women," he said slowly, as if treading lightly.
"I know that."
"And yet that did not deter you?"
"It sounds rather silly, I suppose. I know the proper place of a woman in this day and age. Still, women do travel the world and go all sorts of places not substantially more civilized than those regions you frequent. Besides, I would prefer to be considered an expert in the field of archeology rather than a mere woman."
He chuckled. "There is nothing 'mere' about you."
"Nonetheless, as a practical matter, I am well aware I can do nothing as a woman alone. It doesn't seem especially fair, but it is the way of the world." She didn't have to turn to know he had risen to his feet and crossed the room to stand behind her. "So, my studies, my training, has all been for nothing."
"I am sorry, Gabriella." Genuine regret sounded in his voice. "I can only imagine what it must be like to lose something you had worked for. Something you had wanted."
"And I did," she said softly, "want it very much." For a moment, misery swept through her. She had some time ago laid to rest the grief she'd felt for her brother. This was for her, her dreams, her hopes. She drew a deep breath. Ridiculous, of course. Her dreams never had any chance of coming true. "It was quite foolish to think it was ever a legitimate possibility." She turned to face him. He was less than the width of her hand away. Her heart sped up. She ignored it. "There you have it, Nathanial. My frivolous hope. As fictional as anything one might read in a novel, I suspect. My—" She sighed. "—secret, as it were." She cast him a deprecating smile.
He smiled back as if he did indeed understand. At once it struck her that no matter what else he might be, he was a nice man. A very nice man. The kind of man one might be able to depend on. The kind of man one might possibly trust.
She stared into his brown eyes and abruptly the moment between them changed. Without warning, an odd tension snapped in the air between them, charged with an intensity and awareness as unexpected as it was irresistible.
The kind of man one could love.
He stared at her. "And do you have many secrets?"
Where on earth did that come from? She had no business loving any man, let alone this one. She pushed the thought away and forced a cool note to her voice. "Yes, of course. We all have secrets."
He moved imperceptibly closer, bracing one hand on the shelves to the left of her head. "Any you wish to share?"
"They wouldn't be secrets then, would they?" Her gaze slipped from his eyes to his mouth. Hers weren't the only lips that begged to be kissed, not that she intended to do anything of the sort. "I should hate for you to know everything about me. Where would be the mystery? The excitement? The challenge?"
"I suspect that will not be a problem," he said under his breath.
She could feel the books on the shelves behind her pressing into her back. Still, why not kiss him? Just once. What harm could it do? "You're nicer than I expected you to be."
"Excellent." He smiled in a wicked manner that should have seemed silly or overly dramatic or far too arrogant instead of making her breath catch and her knees week.
"Are you going to kiss me?" She swallowed hard.
"I think I might, yes."
"There is no moonlight now, Nathanial."
"I may be willing to forgo that condition."
"Would you?" She raised her lips toward his.
"I can't seem to help myself." He leaned closer.
"You said a first kiss should be savored and remembered always."
"I shall remember it forever." His lips were within a breath of her own.
"Nathanial?" She fairly sighed his name.
He paused. "Yes?"
She tossed caution aside and brushed her lips lightly across his. "As will I."
"Mmm." He pressed his lips firmly to hers. Her stomach clenched with newfound desire.
"Ahem." Someone cleared his throat at the doorway. "Beg pardon, my lord."
## Eight
Damnation, she knew that voice.
Nathanial straightened reluctantly, cast her a quick smile and turned toward the newcomer. "Yes?"
Xerxes stood in the door, garbed in the same apparel as every other servant she'd seen thus far in the household, holding a silver slaver bearing a letter. "This just arrived for Miss Montini, sir. I was told to deliver it at once."
"Very well." Nathanial took the letter, glanced at it in passing, and handed it to Gabriella. "Are you new here?" he asked Xerxes. "I was under the impression that John Farrel was the footman on duty in the morning."
"I am serving in John's stead, my lord," Xerxes said smoothly.
Gabriella clenched her teeth.
"I hope nothing is wrong," Nathaniel said.
"He was called to the country on a family matter of some urgency, sir."
A family matter—hah! Gabriella glared. "Was he?"
Xerxes met her gaze firmly. "Yes, miss."
She narrowed her eyes. "How urgent?"
"Gabriella." Nathanial shook his head. "I really don't think—"
"It's his younger sister, miss. My cousin fears she could be getting herself into some difficulty and may need his assistance." Xerxes's gaze locked with hers. "Even perhaps rescue."
Gabriella crossed her arms over her chest. "Is she a child?"
"In spite of her behavior on occasion, no, miss, she is an adult."
"Then I'm sure she's more than competent to handle the situation on her own," Gabriella said.
Nathanial's confused gaze slid between her and the older man. "Gabriella?"
"I've no doubt she thinks she is, miss. However, she has been mistaken about her competence in the past." His eyes narrowed slightly. "The entire family is most concerned." He turned his attention to Nathanial. "Will there be anything else, sir?"
"No, thank you." Nathanial cast an amused glance at Gabriella. "Unless you had something more?"
"Not at the moment," she muttered.
"Then you may go." Nathanial nodded.
Xerxes headed for the door.
"Oh, I don't know your name," Nathanial said.
"John Farrel, sir."
"Like your cousin?"
Gabriella choked back a snort of disgust.
Nathanial glanced at her.
"It's a family name, sir."
"I see." Nathanial nodded, and Xerxes took his leave. "Do you know him?"
"No," she said shortly. "He reminded me of someone—something in his manner, I think." She cast him an apologetic smile. "I seem to be rather on edge today."
"Not surprising, really." He nodded at the letter in her hand. "Aren't you going to read that?"
"Yes, of course." Gabriella opened the letter and scanned it. "It's from my friend, Miss Henry."
"Wasn't she just here?"
"Apparently there were some things she failed to mention." Among them the fact that she, or Xerxes, had paid John the footman to go on a bit of a holiday so Xerxes could take the man's place. Apparently that was the plan Florence had referred to.
"Now that we're alone..."
To keep an eye on her. Gabriella's jaw clenched. No one in her household seemed to understand that she was no longer a child.
"Yes?" she said absently. Still, it wasn't a bad idea to have Xerxes within reach should she need his assistance.
Nathanial cleared his throat. "Now that we're alone..."
"You said that," she murmured. Indeed she should have thought of it herself. She refolded the note and glanced at Nathanial. "Now that we're alone, did you still want to kiss me?"
"And get it over with, you mean?"
"I didn't mean that at all." Even so, the moment wasn't quite as electric as it had been. The desire to press her lips to his not as urgent, although she was certain it wouldn't take much more than a heated glance for her to again want what she'd never imagined she'd want. Regardless, the opportunity had passed.
"I should..." He nodded toward the desk. "...finish the letters."
"By all means." She suspected as well that it would not take much to reignite his desire either.
He sat back down and picked up a letter. The first, she noted. He certainly hadn't progressed very far. She crossed her arms over her chest, bookshelf at her back, and watched him read.
"I thought we had established that I find your observation somewhat unsettling," he said without looking up.
She bit back a smile. "Then I suggest you read quickly."
"Hmph."
She probably shouldn't stand here and watch him, but she couldn't seem to help herself. The man was an enigma and not at all as she had expected. She hadn't expected his nature. He was nice and funny and wicked all at the same time. Beyond that, he did seem, well, honest. A man who possibly could be trusted.
She'd never trusted more than a handful of people in her life. And hadn't Enrico told her over and over that men who coveted ancient treasures were, on the whole, an unscrupulous lot and not—no, never—to be trusted? Still, there was something about Nathanial Harrington that made her want to trust him. Want to believe that he would never betray her.
What had this man done to her? She'd always considered herself a completely honest person. But from very nearly the first moment they met, Nathanial had her saying things and doing things she never would have considered doing. As when he caught her in the library and she'd come up with that ridiculous story about his once having kissed her. Which resulted in his insisting she owed him a kiss, preferably in the moonlight, although that no longer seemed a consideration. She ignored the voice in the back of her head that pointed out her actions before then had not been especially legitimate.
And now, God help her, she wanted him to kiss her. Wanted to feel the warmth of his arms around her, the pressing of his body against hers. Wanted the heat of his desire to burn into her very soul. Wanted the—
"Interesting," he said under his breath.
Gabriella blinked in surprise. "Well, yes, that's not quite what I would..." She uttered an odd, uncomfortable sort of laugh, heat rushing up her face. Again. "I'm not sure interesting, while somewhat accurate, in that, yes it is indeed interesting if completely unexpected and not at all distasteful, but rather...quite..."
He grinned.
She stared, then winced. "You're talking about the letters, aren't you?"
"I am indeed." His grin widened as if he knew the answer before he asked the question. "What are you talking about?"
"The letters, of course." She adopted a brisk tone and moved to the chair positioned in front of the desk, a safe distance from him. He certainly couldn't reach across the desk and pull her unresistingly into his arms. Not that she would be the least bit unresisting. Dear Lord, she groaned to herself, what had he done to her? She drew a calming breath. "Well?"
He picked up a pencil and scribbled on a piece of paper. "As you have said, your brother considered only four possible suspects. These were all men to whom he had shown the clay impression made from the missing seal. The list includes an American, Alistair McGowan, and one Javier Gutierrez, a Spaniard." He shuffled through the letters. "Although, he regards Gutierrez only as an agent for Viscount Rathbourne."
She nodded. "Lord Rathbourne is a member of the Antiquities Society and a well-known collector. I have heard of him and I have seen him on occasion, although I have never met him myself."
"His reputation is such that I don't doubt he would sanction whatever means possible to get what he wanted. If Gutierrez took the seal, he has no doubt turned it over to Lord Rathbourne by now. The last two names, of course, are mine and my brother's." He glanced at her. "Have we been taken off the list?"
She hesitated.
His eyes narrowed. "You still suspect me?"
She met his gaze. "Yes."
"I see." He paused. "I thought we agreed to trust one another?"
"I don't remember agreeing to that at all. I recall you saying we should begin to develop trust between us, which was in reference to my brother's letters." She shrugged. "Trust, Nathanial, has to be earned."
He studied her for a long moment. "Indeed it does. On both sides. My name remains, then, and I assume my brother's as well?"
She nodded. "I have not eliminated him, no."
"I daresay I cannot blame you, given Quentin's reputation. However, I am confident he has had nothing to do with this."
She chose her words with care. "And would you know if he had?"
"Perhaps not. But I do know that I will do all that is necessary to recover the seal, regardless of who has it." His voice was hard and she had no doubt he was as good as his word. "Very well, then." He again wrote on the paper, and she could see now it was a list of the men her brother had suspected and included the names of Nathanial and Quinton Harrington. Her stomach twisted.
"You don't need to add your name," she said without thinking.
"Why?"
"If we are to work together, trust is indeed essential. I should give you the benefit of the doubt. Besides, thus far, aside from my brother's suspicions, you have less reason to trust me than I do to trust you." She drew a deep breath. "When it comes to the theft of the seal, I am willing to attempt precisely that."
"Why?" he said again.
"I have no choice, do I?"
"I should think—"
"I can either trust that you are being forthright and honest, that you are truly trying to help me, or I can be suspicious of your every word." She shook her head. "I am by nature suspicious, and no more so than this past year, but I am also practical. If I spend all of my time doubting, we will not accomplish anything. Therefore, in this endeavor, you have my trust, Nathanial Harrington, for good or ill."
"For good or ill." He shook his head. "Not a ringing endorsement."
"Now and again one must take a leap of faith." She met his gaze. "I'm not sure I ever have before."
"Then it is doubly appreciated." He smiled, then with a flourish crossed his name off the list. "Now, only three names remain, and as I am confident one is innocent, I suggest we concentrate on the other two."
She nodded. "First, the American." She glanced at him. "Do you know him?"
Nathanial shrugged. "Not well but I have made his acquaintance. He seems a decent enough sort."
"But is he the type of man to steal another man's find?"
"It's hard to say. A find like this would tempt even the most honest of men. If McGowan has it..." He thought for a moment. "It might well have come into his hands indirectly. He doesn't strike me as a thief."
Gabriella got to her feet and paced the room. "He hasn't arrived in London yet but is expected to arrive any day now."
Nathanial's brows drew together. "Why would McGowan be in London?"
"For the same reason you're in London at this time of year."
"His sister is coming out?"
She rolled her gaze toward the ceiling. "I tend to forget that your family's wealth means you don't have the same concerns as others in your field."
He stared in confusion, then smacked his palm against his forehead. "Of course. The Verification Committee begins its meeting this week. Anyone with any significant finds or proposals for funding will be in London to present their case." He stared at her. "It was a year ago that your brother—"
"Yes," she said simply, and continued, "As McGowan is not yet in London, I propose we start with Lord Rathbourne. I would imagine you know him?"
"To say I know Lord Rathbourne would be an overstatement." Nathanial chose his words with care. "I am aware of his status as a collector. Not merely of artifacts, but of art and other valuables. Beyond that..."
"Yes?"
"He married the woman my brother loved."
Gabriella widened her eyes. "The earl?"
He nodded.
"I thought he was a widower?"
"He is, but..." Nathanial drummed his fingers on the desk as if deciding how much to tell her. "It's public knowledge for the most part, I suppose, and it's all firmly in the past. It's been, oh, ten years now. Sterling loved Olivia—Lady Rathbourne. It was assumed, at least in this family, that they would marry. Then one day they were no longer seeing one another, the next day she had married Lord Rathbourne, and a few days after that Sterling was engaged to Alice. Which pleased both her family and ours."
"But she died within a year of their marriage."
"Yes." He cast her a suspicious frown. "How did you know that?"
"I probably heard it at the Antiquities Society." She shrugged. "As you said, it is common knowledge."
"I don't think Sterling ever got over it."
She nodded. "His wife's death."
"Yes, of course," he said quickly. "That's what I meant."
Gabriella thought for a minute. "Then you know Lady Rathbourne?"
"I suppose I do, although I haven't spoken to her in years."
"Don't you think it's time, then, to pay a call on her and resume your acquaintance?"
"And what do you propose I say?" He stared at her. "'Good afternoon, Lady Rathbourne. I trust you're well today. Oh, did you know you broke my brother's heart, he's never quite recovered, and by the way, we are curious as to whether or not your husband—the man you left my brother for—is a thief.'"
"Now you're being absurd, Nathanial," she scoffed. "We wouldn't want to put it quite like that."
"Oh." He raised a brow. "How then would you put it?
"I think we should pay a call on Lady Rathbourne and ask her if her husband's collections include a recently acquired ancient Akkadian cylinder seal." She smiled.
"Are you insane?"
"I don't think so," she said coolly. "You think this is insane?"
"Under what pretext would we say we were calling?" He clenched his teeth. "Aside from the absurd idea of resuming our acquaintance."
"I don't know." She resumed pacing and tried to think. "Surely between the two of us we can come up with something plausible. We are fairly intelligent, after all."
"You don't have a plan, do you?" He rose to his feet, his brow furrowed. "You have no idea how to go about locating this seal at all, do you?"
She winced to herself. "Well, I suppose one might say, if one was particularly concerned with minor details..."
"One might say what?"
"One might say," she said slowly, "the answer to that is..."
"Yes?"
"No."
"No!" He shook his head as if he couldn't quite believe his ears. "No?"
"I believe I said that," she said under her breath.
"Say it again!" he snapped.
"No, I don't have a plan. There, now are you happy?"
"Ecstatic!" He drew a deep breath. "So you are saying that you have no plan, no idea where to start, nothing beyond a list of possible suspects?"
"I did once have a plan," she muttered.
"Oh?" He crossed his arms over his chest. "Was that the one that involved breaking into my house and searching my brother's library?"
Along with an ill-fated jaunt to Egypt. She shrugged. "Apparently I'm not very good at plans."
"You've just now realized that?"
"One doesn't learn such things about oneself until one attempts them."
"Perhaps we should simply break into Lord Rathbourne's house and see for ourselves if the seal is there. After all, you have experience in such things now."
"There's no need for sarcasm, Nathanial." She paused. "Still, I suppose—"
"Absolutely not!" He circled the desk. "I forbid it!"
"You what?"
"I forbid it." He stepped toward her. "I will not allow you to pull the kind of stunt you pulled here ever again. You could be in jail by now, or worse, shot. People do tend to shoot people they find breaking into their houses in the middle of the night, you know."
He was right, she hadn't thought of that. Regardless, she straightened her shoulders. "I wouldn't have been caught if people were in their beds in the middle of the night instead of finally returning home at a most indecent hour!"
"If you were half as clever as you think you are, you would have been well aware that a majority of the members of this household were still out for the evening!" He stepped to within inches of her and glared.
"Well, then." She planted her hands on her hips. "If we can't break into his house and we can't call on his wife, how do you propose we find out if Rathbourne has the seal? Do you have a plan?"
"I didn't say we couldn't call on Lady Rathbourne. I asked what reason we would have for such a..." He paused.
She studied him. "You have an idea, don't you?"
He nodded slowly. "Perhaps."
"Possibly a plan?"
"Possibly."
She grinned. "I knew you would come up with something."
His brow rose. "Did you?"
"Well, not until a moment ago."
A reluctant smile curved the corners of his mouth. "And how did you know that?"
"Faith, Nathanial." Her grin widened. "I leapt."
## Nine
I must say, this is all very interesting." Merrill Beckworth narrowed his eyes behind his gold-rimmed glasses and considered them curiously.
Nate slanted a quick glance at Gabriella, seated in the chair beside him. She wore her own clothes today, a dress not quite shabby but certainly well worn, a sensible hat, serviceable gloves, and appeared as poised, serene, and collected as if she sat in the office of the director of the Antiquities Society every day. And why shouldn't she be calm? For her, sitting in front of the massive desk in a room filled with dark woodwork and shadows, surrounded by shelves crammed with books, the odd artifact here and there and the occasional travel souvenir, was not the least bit reminiscent of sitting in front of one's father's desk, waiting for one's duly deserved punishment to be meted out. Nate resisted the urge to squirm in his chair.
"More tea, Miss Montini?" Mrs. Beckworth asked.
"Yes, please." Gabriella held out her cup and the director's wife efficiently refilled it. Still, some of Gabriella's composure might well be attributed to keeping a tight rein on her annoyance with him. She hadn't been at all pleased to discover their destination when they had arrived. He'd hurried her out of the house this morning without telling her, arranging for a maid to accompany them for propriety's sake, but without telling her where they were going. It had taken him nearly two full days simply to arrange the meeting, and he was not about to let any reservations she might have interfere. It wasn't a great idea, but it was better than nothing.
In spite of Gabriella's impatience, the past two days hadn't been a complete waste. She had found any number of things among the library bookshelves of interest to her. She seemed somewhat fond of memoirs. And he had found any number of reasons to stay in the library by her side. He wasn't sure he'd ever met a woman like her. She was lovely, of course, as well as brilliant, not to mention stubborn and headstrong and annoyingly independent. But prying anything of a personal nature out of her was bloody well impossible. He'd never known a female to be so reticent about her life. It was as intriguing as it was frustrating. As was her resistance to all his attempts to kiss her again. Which only made him want her more.
"Mr. Harrington?" Mrs. Beckworth offered the pot.
Nate shook his head. "Thank you but no."
Mrs. Beckworth smiled, refilled her own cup, then settled in a chair a bit behind and off to one side of the director. She was a good twenty years younger than her husband, somewhere in her mid-to late thirties probably. In spite of the severity of her tightly pinned hair and the simplicity of her nondescript gown, Nate suspected that she was the kind of woman who had once been considered a beauty, and even now was still quite lovely. Although there was no accounting for attraction, he couldn't help but wonder what had drawn her to the older, somewhat portly scholar.
"Gabriella, why didn't you come to me before now?" Merrill Beckworth pinned her with a firm look. "Indeed, it seems to me we have seen little of you this past year."
"My studies have kept me busy, sir." She paused then met his gaze directly. "And, frankly, given the nature of my brother's last appearance here..."
"My dear girl." Mrs. Beckworth leaned forward in her chair. "No one would ever think ill of you for your brother's behavior."
"My absence is not due to the society's opinion of me under these circumstances," Gabriella said firmly, setting her teacup on the table between them and folding her hands in her lap. "But rather my opinion of the society."
"Oh." Mrs. Beckworth's eyes widened and she sat back in her chair.
Nate winced.
The director chuckled. He directed his words to Nate but his gaze remained on Gabriella. "Were you aware of Miss Montini's outspoken nature?"
"I have noticed it, sir," Nate said wryly.
Gabriella smiled in a polite manner. "I prefer the term 'forthright' to 'outspoken,' sir."
"I have known Miss Montini for some years now. Since she began her studies at Queen's College, I believe." The director cast her an affectionate smile. "Despite her gender, she has one of the finest scholarly minds I have ever run across. She's quite remarkable. Did you know, Harrington, this young woman remembers everything she has ever read?"
Nate glanced at Gabriella. In spite of the ambiguous nature of the compliment, and the blush that colored her cheeks, she remained completely composed. "No sir, but it does not surprise me."
"Then you are more intelligent than you look," Beckworth said in a dismissive manner, and turned his attention back to Gabriella. "I have long thought the manner in which your brother and the society parted company last year was a great shame."
"Such a pity," Mrs. Beckworth said under her breath.
"Indeed, if he had not been quite so...irrational—"
Gabriella didn't so much as flinch at the word. There was a great deal about her Nate didn't know, but Beckworth was right, she was remarkable.
"—the situation might well have been salvageable. Still, while Mr. Montini's behavior was not excusable, it was somewhat understandable given the circumstances." Beckworth paused. "You should know, my dear, that once the committee members' ruffled feathers were smoothed—no easy task, I assure you—"
"They can be quite stubborn." Mrs. Beckworth sighed in a long suffering manner.
"There was a great deal of interest in your brother's claim. A find of this magnitude, the possibility that Ambropia might have actually existed, well, you can imagine the excitement." He glanced at Nate. "Once the committee calmed down, of course. Still, there was no proof."
"As it had been stolen," Gabriella said pointedly.
"Now you tell me your brother had several men he suspected of either taking or engineering the theft of this seal?"
She nodded.
The director's eyes narrowed. "You realize you cannot make charges of wrongdoing without proof of some sort. And, as the only proof that exists is your brother's unsubstantiated claim, and as he was—"
"Not at all, sir," Nate cut in quickly. "I saw the clay impression made by the Montini seal myself."
The older man raised a brow. "The Montini seal?"
Nate nodded. "It appeared quite genuine."
"Still, such things can be fraudulent." The director shrugged. "And, as I imagine you do not have that impression—"
"I do," Gabriella said. "Enrico left it with me."
"There you have it, sir," Nate said, stifling any show of surprise at her announcement, though it was the first time he'd heard of it. She could have mentioned she had the impression. She'd had the opportunity when they'd read and reread her brother's letters, picking them apart word by word to determine if there was anything they had missed. She'd had the chance during dinners with his family when the conversation turned to their search. Most of the talk either centered on the Antiquities Society's annual general meeting and the events associated with that, including the convening of the Verification Committee and the ball, or minor family matters. She'd seemed fascinated by what he considered not at all unusual.
And after dinner each night, when they gathered in the parlor, and later when he escorted her to her rooms and resisted what was fast becoming more of a need than a mere urge to kiss her, again she had the opportunity to tell him about the impression. And again she kept silent. Still, he supposed it scarcely mattered at the moment.
"These ancient cylinder seals were carved by hand," he said to Beckworth now, "and no matter how perfectly crafted, there are always subtle differences between them. If we can find the seal that matches her impression, we have the Montini seal."
"And the thief," Gabriella added.
"My, that is clever," Mrs. Beckworth said under her breath. "I wouldn't have thought of that."
"Regardless." The director shook his head. "Even should you recover the seal, proving whoever has it in his possession is the same man who actually stole it might well be impossible." He glanced at Nate. "You know how difficult culpability would be to prove in a matter like this. And how quickly artifacts might change hands. Furthermore, when you consider that Lord Rathbourne, with his influence and resources, is among—"
"Mr. Beckworth." Gabriella's hands, folded in her lap, tightened, the knuckles white. Abruptly, Nate realized her calm had a price. His heart twisted for her. "I came to the realization some time ago that whoever physically stole the seal—an act which I believe led to my brother's death—will never be punished for that particular crime. As you say, it would be impossible to prove. Then too is the question of jurisdiction, of where it was stolen. England? Egypt? Somewhere in between?"
"Then I don't see—"
"All I want is to recover the seal and prove it is the same one my brother had." She drew a deep breath. "Thus restoring his reputation. I want him to be credited with the find. Nothing more than that."
Beckworth studied her carefully. "But the seal alone is worth a great deal of money. So is the possibility that it is part of a puzzle that will solve the Virgin's Secret, the location of the lost city itself."
"I don't care," Gabriella said simply.
"Others might."
She shrugged. "Let them."
Beckworth's gaze shifted back to Nate. "This search of yours could prove to be dangerous."
While neither he nor Gabriella had mentioned the idea of danger, Nate was well aware of the possibility. It was among the reasons why he thought coming to the society made sense.
"I fully intend to donate the seal to the society," Gabriella said quickly. "As for the location of the city..." She paused, and Nate wondered if she was again letting go of her own dreams. "...that is not my concern. And while there may well be some risk involved, that shall not dissuade me. Risk, sir, is always a possibility when one attempts to do what is right."
"What a courageous girl you are," Mrs. Beckworth said softly.
"Not at all." Gabriella sat a little straighter, if possible. "I am simply..." She blew a long breath. "...angry, I think. I want my brother's legacy restored."
"Understandable, of course." Beckworth considered her thoughtfully. "But I don't see how I can be of help."
"Sir." Nate leaned forward. "We are here today because it strikes me that even something as simple as Miss Montini and I making inquires about the seal is somewhat awkward. We have no authority in the matter other than a personal interest. I thought that perhaps if the society, if you—"
"You want me to make this an official inquiry?" Beckworth's bushy brow rose.
Gabriella cast Nate a sharp look.
"Exactly." Nate nodded. "As Miss Montini fully intends to give the seal to the society should we prove successful, it seems to me the society has a vested interest in its recovery."
"Well, I suppose..." the older man said slowly.
"If she and I could act as representatives of the society—agents, if you will—it would ease our way to speak to those who might be involved."
"You plan to talk to those you suspect?" Beckworth scoffed.
"Well, we simply can't break into their lodgings," Gabriella said under her breath.
"Confronting the culprit might well force him to show his hand," Nate went on. "We might then be able to come to an agreement about credit for the find, perhaps even sharing it." Gabriella slanted him a disbelieving glance. "It seems to me, sir, that whoever has the seal will want to present it to the Verification Committee just as Miss Montini's brother planned to do last year. It's worthless otherwise."
"Unless they were planning to use it to find the lost city," Mrs. Beckworth said thoughtfully.
Gabriella shook her head. "My brother believed the seal was one of a set and that there are probably at least two more. He thought together they would somehow reveal the location of the city. One alone would not be sufficient."
"How very interesting," Mrs. Beckworth murmured.
Beckworth ignored his wife. "Why not wait until it is presented to the committee to make your claim? Why go to all this trouble before then?"
"That would be too late," Gabriella replied. "Someone else would have claimed the credit for finding the seal. Regardless," she raised her chin, "I do not intend to let this matter drop. Should the seal be presented and its legitimacy verified, I should be forced to make the circumstances surrounding it public. Such a revelation would certainly create a scandal, at least in scholarly circles, as well as among the society's benefactors. As the society has a reputation of integrity to maintain, and as it is financially dependent upon donations, it would seem to me such a revelation would be most unfortunate."
The director narrowed his eyes. "That sounds suspiciously like blackmail, my dear."
"Not at all, sir. It doesn't 'sound' like blackmail." Gabriella met the older man's gaze directly. "I believe it is blackmail."
Nate stared at her. One never would have imagined the pretty, dark-haired woman with the eyes of an angel and the straight-backed bearing of a proper English lady had it in her. Oh, yes, there was much more to the delectable Gabriella Montini than met the eye. She was definitely a woman with secrets. It was very nearly irresistible.
"I wouldn't have thought it of you, Gabriella." The director settled back in his chair.
The faintest hint of a smile quirked her lips. "I wouldn't have thought it of myself, sir, but one does what one must."
"Very well, then." The director tapped his pen on the desk in a thoughtful manner. "You may consider yourselves agents of the society in respect to this matter, and you may use the society's name accordingly. However," his gaze shifted pointedly to Nate, "should there be any activity of a...shall we say, less than lawful nature—"
"Sir!" Nate sat up straighter. "We would never—" At once, Quint's lingering reputation sprang to mind, as well as Gabriella's apparent tendency toward larceny. "You have my assurances—"
"Yes, yes, I'm certain I do." Beckworth waved his comment away. "Nonetheless, should such an incident occur, I would be forced to disavow any knowledge of your actions. Furthermore." his gaze bored into Nate's, "I will hold you personally responsible. Should this quest of Miss Montini's bring any dishonor whatsoever upon this august institution, I will see to it that you and that renegade brother of yours are never allowed to so much as cross the threshold here again."
"Which brother sir?" Nate drew his brows together, although he knew full well which brother. "Surely you're not speaking of the earl? The Earl of Wyldewood?"
"You know exactly which brother I am talking about. I am well aware of the earl's position on the board here, as well as your father's before him. And I am equally aware of your family's financial support. Regardless," Beckworth's eyes narrowed, "this is not an idle threat. If you bring so much as a modicum of disgrace or scandal, not only will I ban you from these premises but I will use every bit of influence at my disposal to make certain no reputable university, museum, or private collector will so much as accept your calling card."
"That's rather harsh, dear," Mrs. Beckworth said.
"Sir." Gabriella's brow furrowed. "Mr. Harrington and his family are assisting me, but this is entirely my endeavor. Therefore, if there are any unforeseen consequences, they should fall entirely on my head. Not Mr. Nathanial Harrington, nor Mr. Quinton Harrington, nor any member of their family."
"Gabriella." Beckworth's expression softened. "In spite of your rather surprising willingness to resort to actions that are beneath you, I fully understand the emotional nature of the situation. As brilliant as I have always thought you to be, you are a member of the fairer sex, after all, and therefore such things—while not condoned—may be overlooked."
Gabriella choked. "I am not—"
"Sir," Nate said quickly, to forestall her saying something they would both regret, or at least he would. "You have my word I shall not do anything to cast this institution in a disreputable light. Nor will I allow Miss Montini to...well...allow her feminine emotions, as it were, to overrule her head."
Gabriella's jaw clenched.
"And I will do all in my power as well to ensure her safety."
"See that you do." Beckworth gestured at his wife. "Mrs. Beckworth will check our files. We should have information as to where Mr. McGowan will be staying and when he is expected to arrive in London that will prove helpful to you." Mrs. Beckworth nodded and hurried out of the office. "I assume you know where to find Lord Rathbourne?"
"Yes, sir," Nate said.
"He is..." The director thought for a moment. "...'ruthless,' I think is the word, when it comes to his collections, and extremely possessive. It's my understanding he does not display them but rather keeps them locked away. There have been any number of rumors through the years regarding the manner of his acquisitions." The older man's gaze met Nate's. "I suspect there are no lengths he would not go to protect what he considers his."
Nate acknowledged the warning in Beckworth's eyes with a nod. Beckworth rose to his feet, Nate and Gabriella following suit.
His wife returned and handed Nate a piece of paper. Her gaze met his. Her eyes were the cool, pale blue color of ice, at odds with the warmth in her voice. "If there is anything else you need, do not hesitate to call on us. We have always been fond of Gabriella."
"If there is nothing else," Beckworth said, "I have a great deal of work to do, what with the Verification Committee as well as the meeting of the general membership bearing down upon us. Mr. Harrington, Gabriella." He cast her an affectionate smile. "Do be careful, my dear."
"And we will see you both in a few days, I assume." Mrs. Beckworth's gaze shifted between Nate and Gabriella in a speculative manner. "At the ball?"
The Antiquities Society Ball marked the beginning of the ten day meeting of the Verification Committee. The committee would adjourn at noon on its final day, the annual general membership meeting following an hour or so later. Reggie and his mother had been talking about the ball ever since he returned home. This would be the first year his sister was old enough to attend. His mother and Sterling, of course, made an appearance every year. Nate couldn't remember the last time he and Quinton had attended.
Gabriella shook her head. "I really don't think—"
"Of course we will." Nate favored the older woman with his most charming smile. "We wouldn't think of missing it." He took her hand and raised it to his lips. Again the look in her eyes struck him as cold. Nonsense, he told himself. It was simply the pale color that made them appear so. "And I do hope you will save a dance for me."
She smiled. "I shall indeed, Mr. Harrington."
Nate grinned. "Good day, then." He nodded at the director. "Sir."
Gabriella murmured a polite farewell. Nate took her elbow and steered her firmly out of the office. The moment they were out of the Beckworths' presence, her jaw tensed, her posture stiffened, and her eyes narrowed. As much as he didn't know about Gabriella Montini, only a man long dead in his grave would fail to recognize the timeless signs of a woman who was not happy, especially with the man at her side. He braced himself.
When they stepped out of the building, she shook off his hand and turned to him, her blue eyes glittering with anger in the sunlight. "What on earth were you thinking?" She glared. "Coming here? To them?"
"You didn't have a plan," he said calmly, and looked down the street for his carriage. "You had no idea where we should begin."
"You could have mentioned this was where we were going! You didn't tell me this was your plan." She fairly spat the word.
"You wouldn't have come."
"Of course I wouldn't have come. This is the last place—"
"Gabriella." He met her gaze directly. "We needed a certain amount of authority, credibility. Legitimacy, if you will."
Her eyes narrowed. "My claim is perfectly legitimate."
"I understand that—"
"And as for credibility, even if I am a female without a brain in her head—"
"No one said anything of the sort. In fact, your intelligence was highly praised by Beckworth."
She snorted with disdain. "For a woman."
"For anyone. You needn't be so indignant about Beckworth's appraisal."
"Perhaps I am allowing my feminine emotions to overrule my head!"
"Perhaps you are." He gritted his teeth. "You know full well the limitations on women in this world. You acknowledged them yourself when we talked about your desire to follow your brother's path in life."
"You suggested sharing credit! I have no intention of doing so."
"It was a suggestion, nothing more." Where was his blasted carriage?
"He charged you with my protection! I don't need—"
"You most certainly do." His patience snapped. "You are irrational on this subject. Thus far your actions have been anything but sensible and well thought out."
She gasped.
"Can you deny it?" He grabbed her elbow and stared into her eyes. "You could be in jail right now. First you attempt to search my brother's library at my sister's ball. Then you break into my house—" A thought struck him and he paused. "Have you done anything else I should know about?"
She hesitated for no more than a fraction of a second, but it was enough. He could see there was something she still kept from him. She squared her shoulders. "No, of course not."
He didn't believe her for a second, and vowed to himself to find out what else she might have done in this quest of hers. "Why didn't you tell me that you had the impression?"
"Oh." Her eyes widened. "That."
"Yes, that!"
She shrugged. "You didn't ask."
"So much for trust and leaps of faith," he said sharply.
"It's not that I don't trust you," she said quickly. "It simply slipped my mind, that's all."
"I don't believe you." He released her and waved at his carriage that had just turned onto the street.
Her forehead furrowed. "Why not?"
"As you said, Gabriella." The carriage pulled to a stop in front of them and he jerked open the door. The waiting maid stared at him with widened eyes and slid back into the farthest corner of the carriage. "Trust has to be earned."
He helped her in and snapped the door shut.
"Furthermore I have no intention of attending that ball." She huffed.
"Oh yes you will. It's in your best interests to make an appearance, and you will do so."
She glared. "You cannot order me as if I were—"
"As if you were the woman I could have arrested?"
She sucked in a sharp breath. "Nathanial Harrington, I can't believe you—"
"Would resort to blackmail?" He narrowed his eyes. "One does what one must. I shall see you at home."
She leaned out of the window. "Aren't you coming?"
"No," he said firmly. "I have an errand to attend to."
"What kind of errand?" Suspicion rang in her voice.
"Trust, Gabriella. Try to have a little faith in me. I will not fail you." Nate signaled to the driver and the carriage rolled off. He heaved a frustrated sigh. "You have my word, Gabriella Montini."
He turned and started off down the street. If he had any hope of keeping his word, he needed to know her secrets. His family's solicitor had long employed an excellent and reputable investigation agency. According to Sterling, its operatives were fast and efficient and had proved most useful in the past. Nate had never needed them before, but if ever there was a time, it was now.
With every moment spent in her company, he discovered there was much he didn't know about Gabriella Montini. And much he needed to discover.
## Ten
Share the credit for discovery of the seal?" Gabriella paced the parlor of her house, noting in the back of her mind how very small it was. Obviously residing with the Harringtons had changed her perceptions. That too was annoying. "Can you imagine such a thing?"
Florence glanced up from the work in her hand. "It seems rather sensible to me."
Gabriella stopped in mid-step and glared. "Sensible?"
"Gabriella." Florence sighed and dropped the pillow cover she'd been embroidering into her lap. "You know as well as I—as well as Mr. Harrington and Mr. Beckworth, apparently—that whoever is in possession of the seal is not necessarily the same person who stole it. Whoever has it now might well have come by it in a relatively legitimate manner. If so, he would be hard-pressed to give up recognition of the find at all, let alone share it."
"I am well aware of that. I simply prefer not to think about it." Gabriella blew a long breath. "Still, to have Nathanial suggest it, well, it smacked of betrayal."
Florence raised a brow. "I thought you said you trusted him?"
"I did. I do. Somewhat." She sighed. "I can't completely. I am trying." She resumed pacing. "It's not that I don't want to trust him. I want to trust him more than anything." The very idea of trusting Nathanial was almost irresistible. Of not having to watch every word she said. Of trusting him with her confidence, her secrets. Maybe even her heart. Although that was absurd.
"I should think it would be a great relief for you to trust someone completely."
Gabriella widened her eyes. "I trust you completely."
"Do you?" Florence said, picking up her embroidery. "Always?"
"Yes, of course." Gabriella ignored the thought of her trip to Egypt. "I trust you implicitly."
"Implicitly?"
"Yes." Gabriella nodded. "Without question."
"Yet you did not trust me enough to tell me anything about your intended misdeeds at Mr. Harrington's home."
"You would have stopped me."
Florence cast her a chastising look. "That is my job."
"And you do it well. Which is why I didn't tell you."
"Hmph." Florence paused, no doubt to compile more examples. "You didn't tell me you have the clay impression of the seal."
"Yes, well..."
Florence glanced at her sharply. "Gabriella?"
"That is a bit of a problem," Gabriella murmured.
"A problem?"
"In definition." Gabriella shrugged. "Nothing more significant than that."
Florence narrowed her eyes. "An explanation, if you please."
"The impression is not actually in my possession at the moment." Gabriella held her breath.
"I see." Florence thought for a moment. "Do you know where it might be?"
"I am certain it is in London and probably under our very noses."
"London is a very big place."
"I'm sure it is here in the house," Gabriella said with far more conviction than she felt.
"But you don't know."
"No, but I am fairly confident." She sank down on the sofa beside Florence. "It doesn't make sense for it to be anywhere else. Enrico told me he was leaving it in the one place where he knew it would be safe. Where, he said, he kept everything he valued. It has to be the house, there is nowhere else." She smiled ruefully. "My brother, if you recall, was even less trusting than I."
"I could scarcely forget." Florence paused. "But is there a chance he would have left it in a box at the bank?"
"He had no box, as far as I can determine. I contacted the bank to confirm that."
"After his death." Florence nodded. "Very sensible."
"Actually, before." Heat washed up Gabriella's face. "When his letters began to ramble, it seemed like a good idea to find the impression. I should have known better. Enrico barely trusted the bank with his money." My money. "Did you have any idea how much money we had?"
"Not at all." Florence sniffed. "I certainly would have asked for an increase in my wages if I had so much as suspected I wasn't employed by an archeologist who could barely pay his mortgage, but by a treasure hunter with an impressive fortune. He never said a word," she added under her breath. "Your brother was a man of many secrets."
"Yes, I know."
Florence paused for a long moment as if considering her words. "On those rare occasions when he was in London, he and I would frequently have long talks. Sometimes we would talk about you or occasionally about the politics inherent in dealing with museums or the Antiquities Society, but usually we talked about his life, his work. About things he had done or seen in his quest for artifacts. I think I was the only one he could talk to about such things. He had few friends, you know, and few he could confide in. Indeed, there were times when I thought of myself as his father confessor." She drew a deep breath. "There was even a moment once, long ago, when I fancied myself in love with him."
Gabriella stared. "Did you?"
Florence smiled. "As I said, it was a moment and not much more than that. I was wise enough not to lose my heart to a man like your brother."
Florence didn't say it, but then it wasn't necessary to say it aloud. Enrico had a passion for women of all sorts. Even when she was a child there was often a women in his room or his tent. They seemed as necessary to his existence as food and drink. It wasn't until years later that she had understood his behavior in regards to women was not that of an honorable man.
"If anything, your brother and I were friends of a sort. As he had charged me with your care, I believe he felt I was worthy of his trust, although he trusted people even less than you do. I believe it was only his excitement about the seal that led him to ignore his usual guarded nature in such matters and show the impression to the men you now suspect of involvement in its theft."
"I do trust you," Gabriella said firmly. "And I need your help."
"Oh?"
"A prolonged absence from the Harrington household on any given day would surely arouse suspicion. I am only here now because Nathanial, in his arrogance, put me in his carriage, ordered the driver to return me to his house, and assumed I would do so."
"Foolish man," Florence murmured.
Gabriella ignored her. "I need you to search the house. Every nook, every cranny." She got to her feet and resumed pacing. "The impression has to be here. This is the only place it could possibly be."
"It's not an especially big house, Gabriella, but I imagine there are any number of hiding places. If it is here, it might well be impossible to find. However, I shall enlist Miriam's help and we shall do our best." She studied Gabriella thoughtfully. "But why on earth did you say you had it when you didn't?"
"I don't know." Gabriella sighed and brushed an errant stand of hair away from her face. "I needed a way to prove the seal, once found, was Enrico's. The words just seemed to come out of my mouth of their own accord."
"That's the problem with deceit, dear. The first lie is awkward, difficult, and often carries a great deal of guilt. The second is a bit easier, the third easier yet. And eventually..." Florence's knowing gaze met Gabriella's. "...deceit becomes far easier than truth."
Gabriella crossed her arms over her chest in an effort to disguise her unease. Indeed, the lie about possessing the impression had been remarkably easy, without thought or guilt. "I shall not let that happen."
Florence's cast her a skeptical look.
"I won't," Gabriella said firmly, resolving to at least try. "I have always been an honest sort, it's just that now...well, honesty is somewhat awkward."
"It always is, Gabriella." Florence shook her head. "Do remember the ends do not always justify the means."
"You needn't keep saying that."
"Oh but I do. At least until you understand its meaning as more than just a saying embroidered on a pillow." Florence heaved a long suffering sigh. "You come by it naturally, I'm afraid, your brother never understood it."
Gabriella narrowed her eyes in confusion. "What do you mean?"
"Simply what I said. For your brother it was the acquisition that counted, not the method by which it was acquired."
"I don't understand."
"Nor do you need to," Florence said in a firm manner, then deftly changed the subject. "Now, will you wear your new gown to the ball?"
The same gown she had worn to Lady Regina's ball. "I don't particularly wish to go."
"Nonetheless, Mr. Harrington is right. You have nothing to be ashamed of and nothing to hide. My dear, you have gone to that ball for a good six years that I can recall, and as you will be in the company of the Harrington family, there is no reason for you not to go this year."
"But last year..." Last year the ball had been glorious. Enrico was excited about presenting his seal to the committee; she had been confident that she could at last convince him to take her with him, and she'd had no end of eager partners. This year...
This year there would be Nathanial.
"And should you need another friendly face—"
"Which reminds me," Gabriella interrupted. "I was not at all pleased to see Xerxes—or John, as he is now known at Harrington House. I gather that was the plan you mentioned?"
"Not quite as deceitful as yours, but then I have not had as much practice nor do I seem to have the natural gift for it that you do." Florence fixed her with a firm look. "I am only grateful this tendency of yours did not surface in your younger days."
"It does seem to be recent," Gabriella murmured. She couldn't very well deny it, as much as she might wish to. No, the lies and deceptions did seem to be piling up, although soon there would be no more need for them.
The rules of the Verification Committee were both clear and unyielding. Once an artifact had been presented and ruled on unfavorably, the presenter had only until the end of the next year's meeting to challenge that decision. Last year the committee had decided that Enrico's claim was not legitimate; that the seal he had was not the one for which he presented evidence was considered irrelevant. Only extraordinary circumstances could prompt a reopening of a case after the time limit had passed, and Gabriella knew of few instances where that happened. The committee did not like to reverse itself or admit mistake. No, she had a sure and certain conviction, in her heart, that if the seal wasn't recovered this year for this meeting of the Verification Committee, it never would be. Her chance to restore her brother's good name would be lost forever.
"How do you think Mr. Harrington will respond when he learns of your deceptions?" Florence asked.
"He will understand the necessity of my actions," Gabriella said with a confidence she didn't quite feel. What if he didn't understand? What if her actions disgusted him? Her stomach lurched at the thought that she might lose him. Not that she had him or wanted him or that he mattered at all.
Florence considered her in an assessing manner, as if she knew exactly what Gabriella was thinking. It was—it always was—most unnerving.
Still, Gabriella should at least be honest with herself, if with no one else. In spite of Nathanial's arrogance, the way he seemed to have taken over her life, she had to admit he was indeed beginning to matter. Quite a lot. She was not looking forward to telling him all the truths about herself, all her secrets. When he knew everything...She firmly set the thought aside. Now was not the time to dwell on what might—what would—happen then.
"As I was saying, I shall be at the ball as well." Florence glanced at a large bouquet of roses in a vase on a side table that Gabriella had noted but to which she'd paid no attention until now. "Mr. Dennison has invited me to join his sister and her husband's party. They shall accompany me to the ball and I shall see him there."
Gabriella raised a brow. "I gather the invitation came with flowers?"
"No. Mr. Dennison came with flowers." Florence smiled in a decidedly smug manner. "Yesterday evening."
"Oh?"
"We had a lovely chat." A dreamy look drifted across the older woman's face. Gabriella realized that Florence was really quite lovely. She wasn't sure why she'd never noticed before. Florence shook her head as if to clear away thoughts of the dashing Mr. Dennison. "Now then, I will see you at the ball."
"Yes, I suppose." Gabriella sighed and again sank down beside Florence. "Still, the society, those people, treated my brother like—"
"Like a man who made claims he could not substantiate." Florence's voice was surprisingly hard. "Like a man who then behaved like a madman and blamed the very people he hoped to win over for his loss."
"He wasn't mad," Gabriella said quickly
"No dear, just obsessed." Florence studied her closely. "Much as you are with finding the seal."
"I'm not obsessed. It's simply something that is left undone." She drew a calming breath. "It seems to me when one dies, one's loose ends should be tied up."
Florence shook her head. "Life is scarcely as tidy as that. Nor is death."
"Pity it can't be more tidy. More certain, if you will."
"The only thing certain about death is that it is inevitable. As for life..." Florence smiled. "I consider its very uncertainty one of the best things about life. One never knows what might happen."
"For the worst, no doubt," Gabriella said darkly.
Florence laughed. "Or for the best. Usually when one least expects it."
"Are you talking about Mr. Dennison?"
"I don't know, Gabriella." Again Florence's eyes took on a far off look. "I rather hope so." Her gaze met Gabriella's. "But then, my dear girl, what would become of you?"
"Of me?" Gabriella laughed. "You needn't worry about me. I shall always have Xerxes and Miriam."
Regardless of her words, she couldn't help but wonder what indeed would become of her. With Enrico's death, any chance she had for the kind of life she'd wanted to lead had vanished. And once the seal was found, she really had no more purpose to her life.
Would she spend the rest of her days poring over old books in the society's library, storing knowledge that she would never put to practical use? Would she grow old in this house, alone save for those who, while more family than servant, still had each other?
"Perhaps a husband," Florence said under her breath.
Gabriella smiled. "I don't think I'm suited for marriage."
"We shall see. Regardless, you shall always have me," Florence said firmly. "Unless of course your Mr. Harrington—"
"He's not my Mr. Harrington." A firm note sounded in Gabriella's voice. "And he never can be."
She ignored the persistent voice murmuring in the back of her head.
But oh, wouldn't she like him to be?
## Eleven
They were plying her for information, that's what they were doing.
Nate gritted his teeth and resisted the urge to wring the neck of very nearly every member of his family. Oh, they were subtle enough, if one didn't know them. If one did, their intentions were obvious, and given the nature of their casual inquires, one might have thought they had a coordinated plan of attack. They had attempted it before, but tonight, somehow, they seemed more determined. He ignored the inconvenient fact that only this afternoon he had employed a firm to do very much the same thing, to find out more about Gabriella Montini.
"And you have lived in London, then, for nine years now?" Sterling sipped his wine in an offhand manner.
Gabriella nodded. "It was thought London would be best for my studies. Even though he was by birth Italian, Enrico preferred London, which was sensible given the Antiquities Society, the universities, and museums here." She shrugged. "London had become home for him, as much as anyplace could."
"London is the center of all things stolen," Quint said with a wry smile. "We have been spiriting antiquities away from their countries of origin for generations."
Sterling cast him a chastising look. "It has never bothered you before."
"And it doesn't bother me now." Quint lifted his glass to his brother. "In fact, I should drink most happily to the arrogance of those modern seats of civilization. And not just London, but Paris and Berlin and Vienna, as well as to all the museums and institutions and private collectors who believe the ancient treasures of any country are better off in our care than in their place of origin. And as they are all willing to pay nicely to acquire more, my dear sanctimonious brother, it doesn't bother me in the least."
"That is a discussion for another time." His mother's firm gaze slipped from Quinton to Sterling and then to Nate, no doubt simply for good measure, as he had yet to join in. "I do not wish to open that particular kettle of fish tonight."
It was a ongoing debate within the Harrington household, as well as among scholars and, God help them all, politicians, and it was certainly not new. The continuing question as to whether the treasures of antiquity should be saved by foreigners spiriting away artifacts to institutions far from their point of origin or whether such activity constituted theft of a nation's heritage had been a topic in this house for as long as Nate could remember.
Influenced by intellectual scholarly articles or something as simple as a conversation on a train, current members of the family switched sides of the debate nearly as often as it lifted its head. All except Reggie, of course, who thought it was dreadfully boring and couldn't they talk about something else for a change? Mother had often said the ease with which they all changed their minds and the passion with which they then pursued their new positions had nothing to do with the issue itself, but with their love of a good argument.
Better to argue about something they could do little about, Nate thought, rather than Sterling's continuing failure to find a new wife, or Quint's disregard for anything that smacked of proper behavior, or his own...well, whichever flaw of his was uppermost in the others' minds at the moment.
Mother turned to Gabriella. "This particular discussion has been raging in this household for generations."
"A philosophical matter of debate." Regina rolled her gaze toward the ceiling. "That's what they call it."
Mother cast her a chastising look, then continued. "My late husband, Charles, said even in his own childhood the question of Britain's possession of the Elgin marbles had been a subject of heated debate around this very table."
"Said discussion no doubt prompted by our great-grandparents' search for lost gold in Egypt." Nate leaned toward Gabriella. "As we understand it, it was quite an adventure, with kidnappings and murderous suitors and that sort of thing."
"It sounds most exciting," she murmured.
Nate wasn't sure if she was bored by his family's less than perfectly proper demeanor at the dinner table or overwhelmed. It would not have surprised him. Aside from her brothers, she was apparently alone in the world.
"Regardless," Quint continued, returning to the topic at hand, "one would think if countries were truly concerned about the loss of their artifacts, they would make it more difficult to spirit them across borders. Hire civil servants perhaps who did not see bribery as an expected portion of their incomes."
Mother winced. "That is a problem."
"It's simply the way things work in much of the world," Nate said. "A necessary evil, if you will."
Gabriella choked back what sounded like a gasp but was probably just a cough.
"But rest assured, Mother," Quint said. "Nate is keeping me within the confines of legality as well as upright behavior."
"And I am most grateful to him," she replied. "It eases my mind to know that your brother is watching you."
Nate scoffed. "I scarcely watch him, Mother."
"Watch over him then," she continued. "I know Quinton is the older brother and should be the one watching over you—"
Quint cast her his most unrepentant grin.
"—but his nature is not conducive to responsibility of that sort."
Quinton laughed. "Or responsibility of any sort."
She fixed her middle son with a firm look. "I am confident that will change someday."
Reggie snorted in a most unladylike manner.
Mother sighed. "I had once thought my youngest sons would become scholars like their father."
"Like Father?" Sterling smiled. "Father was scarcely more than an amateur scholar, Mother. And there was no one more delighted than he when Quinton first abandoned the path of scholarly pursuit to accompany Professor Ashworth on his journeys. And delighted too when Nathanial joined him."
"It was the adventure, you understand," his mother said to Gabriella. "I suspect my husband always rather longed for adventure. It was different, you know, in the past. Charles grew up on stories of the Earls of Wyldewood and their exploits." She glanced at Sterling in a speculative manner. "Today, the earl has little opportunity to chase smugglers or battle pirates or rescue fair maidens."
"However, I keep myself busy," Sterling said mildly.
His mother considered her two youngest sons. "At best, this is a questionable business you are engaged in. And, I suspect, often dangerous and certainly disreputable on occasion."
"Can't be helped, Mother," Nate said.
"It certainly has its moments." Quint chuckled and turned toward Gabriella. "That's something your brother no doubt well understood, Miss Montini."
Nate would have kicked him under the table if he could have reached. The last thing he wanted was a discussion of Enrico Montini.
"I beg your pardon?" she said.
"He understood that there was a fine line between a discovery and a theft. A lauded archeologist or a thief." Quint shrugged. "Enrico Montini was certainly not above doing whatever was necessary to acquire what he wanted. He understood that deceit, illegalities, ignoring moral standards, and so forth are often necessary to achieve the ultimate goal."
"In that we all understand as much," Nate said quickly, and cast his brother a warning glance, "that's what he meant."
"Yes." Quint took a sip of his wine. "That's what I meant."
"Has there been any word from your brother, Gabriella?" His mother turned to Gabriella, thankfully changing the subject. "The one Nathanial met in Egypt. What was his name?"
"Antonio," Gabriella said.
"Ah." Mother nodded. "Named for your father then."
"Yes and no, I have not heard from him. But he has never been good about that sort of thing."
"Perhaps tomorrow we can talk about your mother." Mother smiled. "And her family."
"Lady Wyldewood, while I would like to know something of my mother, as I understand it, her family had no use for her, nor for me. Besides, I suspect we will be rather busy for the next few days." Gabriella's tone was polite, but Nate had the distinct impression she wished to avoid that particular chat. "What with our plans and the ball."
"I should have thought of that." His mother looked at her youngest sons. "As you are in London this year, I shall expect you both to attend."
"Wouldn't miss it," Quint said under his breath.
"Come now, Quinton, it's quite exciting," Sterling said in a wry manner. "Upward of six hundred people all discussing the newly excavated ruins of somewhere or other. Most enjoyable."
"It's an obligation, Sterling, as you well know. As a board member and as benefactors of the society it is our duty to make an appearance," Mother said firmly and directed her words to Gabriella. "My oldest son is not overly fond of events like this."
Sterling grimaced.
"In deference to him, we rarely stay very long."
"This year we shall stay longer," Reggie announced. Sterling cast her an annoyed glance. "Well, it's a ball. A grand ball, and I am quite looking forward to it."
"I have always enjoyed the Antiquities Society Ball," Gabriella said with a smile. "My brother and I and Miss Henry have attended every year since I have been old enough to do so."
"Have you? And yet I have never noticed you." His mother winced. "Forgive me, that sounded dreadful."
Gabriella laughed. "Not at all, Lady Wyldewood. The ball is a huge crush, and as you don't stay very long, it's not at all surprising that our paths have never crossed."
"Every year, hmm. Imagine that. And right under my very nose." Mother studied Gabriella thoughtfully. "We shall save our talk about your mother for another time, then. A few more days will scarcely matter."
Gabriella smiled. "I shall look forward to it."
The remainder of the meal was uneventful, and the feeling Nate had had earlier—that his family was trying to glean information from Gabriella—did not recur. Dinner concluded without major incidents, disclosure, or arguments. The ladies retired for the evening, leaving Nate to follow his brothers onto the back terrace for cigars.
The moment they stepped through the doors, Nate turned to Quint.
"What on earth were you thinking?"
"I probably wasn't." Quint took a cigar from the humidor Andrews placed on a table on the terrace every evening. Cigar smoke was not allowed in the house when Mother was in residence. "What, precisely, are you talking about?"
"I'm talking about your comments about Enrico Montini."
Quint trimmed his cigar. "Why shouldn't we talk about Montini?"
"Because I don't think Miss Montini is aware of the type of man he was."
Sterling selected a cigar. "What kind of man was he?"
"Montini was..." Nate chose his words with care. "Not well liked."
"He was cold, callous," Quint said, lighting his cigar. "Merciless, as it were, when it came to acquiring what he wanted. My reputation may have once—"
Nate snorted.
"—been 'questionable,' but no one has ever suspected me of resorting to whatever means possible to get what I wanted."
"Whatever means possible?" Sterling said slowly.
Quint nodded, a grim look in his eye. "If this seal was stolen from anyone else, and the alleged owner were dead, Montini would be at the top of my list as a suspect. For theft and murder."
Sterling studied his youngest brother. "Why do you think she isn't aware of her brother's nature?"
"I don't know." Nate plucked a cigar from the humidor. "There's something about the way she talks about him. She adored him—idolized him, I think—and she will do whatever necessary to restore his professional reputation." He shook his head. "I can't imagine she would feel the same if she knew the type of man he was."
"And yet we don't know she isn't exactly like him," Sterling said mildly, lighting his own cigar.
"She isn't the least bit like him," Nate said staunchly.
Quint and Sterling traded glances. Sterling chose his words with care. "Still, we really know nothing about her."
"Mother knew her mother," Nate said quickly, ignoring the fact that he had already come to the same conclusion and was taking steps to learn more about the intriguing stranger in their midst.
Quint lit Nate's cigar. "And yet Mother has said nothing more about that. Don't you find that odd?"
"She is up to something." Sterling's eyes narrowed. "She has been preoccupied since Miss Montini arrived. And she studies her with a look in her eye that is most curious."
"As interesting as that is, it's of no concern at the moment," Nate said firmly. "I would prefer, and request, that there be no more discussion of Miss Montini's brother in her presence."
Quint leaned against the terrace balustrade and blew a perfect smoke ring. "You honestly believe she doesn't know what kind of man her brother was?"
"I do." Nate ignored the niggling thought that he might be wrong. He might be wrong regarding any number of things about the lovely Gabriella. There was a reserve around her that she carried like a shield. Even so, there was something about the woman that called to something deep inside him. From the moment he met her, he had the oddest feeling of inevitability, of anticipation perhaps. The vague sense that something extraordinary and unique and wonderful had stepped into his life. It was an absurd idea with nothing whatsoever to base it on save the ridiculous feeling that washed through him when he so much as thought of her.
There was lust, of course. With the fire in her blue eyes and the fervor to right what she considered a grievous wrong, one couldn't help but wonder what other passions might lie just beneath the surface. He had known lust before, but this was tempered with something as yet unknown. And whereas she would do whatever she had to do to recover the seal, he would do whatever necessary to protect her from harm. Besides, he had given his word.
"It scarcely matters, the man is dead now and we have promised to help her." Nate pinned Quint with a hard look. "The lady has been through a lot this past year. I do not wish to upset her further by discussion of her brother's character."
"Or lack of it," Quint muttered.
"I do have to wonder, though..." Sterling blew a stream of blue smoke then met his youngest brother's gaze. "...why you are so vehement about this. You scarcely know the woman."
"I was wondering the very same thing." Quint studied Nate, then snorted back a laugh. "You want her! I should have known."
Nate's jaw clenched. "That's enough, Quint."
"You devil." Quint grinned "You want her in your bed."
"I—" Quint had done this to him most of his life. Bait him until he inevitably blurted out whatever truth he was trying to conceal. He had long ago learned there was only one way to handle Quint's teasing. He forced a wicked grin to his face. "Wouldn't you?"
"No, not at all." Quint shook his head. "She's pretty enough, with those deep blue eyes of hers and that luscious figure and that seductive hint of an accent—"
Nate narrowed his eyes.
"—but she's too bloody damn smart for me. God save me from an intelligent woman. She'd do for you, though." Quint's eyes widened. "Good God you don't just want her—you like her!"
"She's very...nice." Nate tried and failed to hide the defensive note in his voice. "She's quite easy to like."
"Really?" Sterling murmured. "I haven't found her particularly easy to like."
"I have spent a great deal of time with her," Nate said. "I have come to know her better than anyone else."
"She's stubborn and independent and has a streak of larceny in her," Quint said. "No man in his right mind would 'like' her." He laughed. "Want her, definitely, but not like her."
"And yet I do," Nate said defiantly, and glanced at Sterling. "You don't think it's too fast, do you? To like her, that is?"
"Admittedly, you still know little about her." Sterling puffed on his cigar thoughtfully. "So yes, in a rational sense it may well be too soon. However, I suspect rational thought has little influence here. I would, however, be cautious if I were you until you know more."
"I would say it all depends on what you have in mind." Quint studied his younger brother. "Seduction and a short but passionate affair is one thing. I know you are familiar with that concept."
Nate gestured with his cigar. "Go on."
"It's quite another if you have in mind something that will last the rest of your life."
Sterling scoffed. "Nonsense."
Nate nodded, the oddest sinking sensation settling around his heart. His brothers were right, of course. "And it's entirely too soon for that."
"In my opinion..." Quint paused for a moment. "...it's just the opposite."
Sterling stared. "You can't mean that."
"Oh, but I do." Quint nodded. "I have long suspected that if I ever meet the right woman, a woman I would be content to spend the rest of my days with, I will be struck with the certain knowledge that she is right with the efficiency and speed of a bolt of lightning." He met Sterling's gaze. "You know what I mean."
Sterling paused, then nodded.
Abruptly, Quint grinned. "Although I admit it is a somewhat trite and overly romantic idea."
"And unbelievable as well, given its source," Sterling said.
Quint shrugged.
"Take care, little brother." Sterling's gaze met Nate's. "Miss Montini might not be as you see her."
"But then again she might be." Quint blew another smoke ring. "And if so, yes, I think she'd do nicely for you."
"Well, I'm not looking for anyone to do for me," Nate said quickly. "Not at the moment."
"Of course not," Sterling said without an ounce of conviction, and changed the subject. "Have you noticed, by the way, the number of bouquets that have arrived for Reggie in recent days?"
Quint chuckled. "She has certainly made an impression on the eligible young men of society. Still, I suspect Reggie is in no hurry to select a husband. Although I suppose we—and when I say 'we,' I really mean Sterling—should keep a close eye..."
The conversation between the brothers droned on until late in the night. Nate told them what little he and Gabriella had thus far uncovered. Usually, they would have joined the ladies when they had finished their cigars. Tonight, however, Mother had said she wished to retire early, and both Gabriella and Reggie had taken that as their cue to do the same. Regretfully so. He had hoped to again escort Gabriella to her room.
In spite of the absorbing nature of the discussion, ranging from Reggie's potential suitors to the current state of politics to the latest scandals, his thoughts returned again and again to Quint's comments about knowing the right woman at once when she came along. He couldn't help but wonder if—as odd as it might seem at first glance, and given the unusual circumstances they found themselves in—Gabriella might well be the right woman for him.
Or if she was very, very wrong.
## Twelve
Gabriella perched nervously on the edge of a red velvet sofa. Nathanial stood beside the fireplace looking substantially less apprehensive than she felt. And why not? He at least knew Lady Rathbourne.
The parlor they had been shown into was, if possible, even more elegant than the Harringtons'. Whereas their home had a feeling of warmth to it, this house seemed cold and unwelcoming. The parlor was perfectly appointed, in the height of fashion, but it struck her as rather more like a stage setting than a place where living people resided. It was far and away too, well, perfect. The temperature was warm, but a chill shot through her.
"Nathanial Harrington." A tall woman glided into the room and held her hand out to Nathanial. "You were just a boy when last I saw you."
He chuckled and raised her hand to his lips. "Lady Rathbourne, you are as beautiful as ever."
She was indeed beautiful, startlingly so. Lady Rathbourne might well have been the loveliest creature Gabriella ever seen. She was nearly as tall as Nate, her blond hair meticulously styled, an air of grace and elegance lingering about her. Her gown was in shades of red, the latest in French fashion, and Gabriella had the most absurd feeling that it had been selected to compliment the room.
"Am I?" Lady Rathbourne tilted her head in a manner that might have appeared artificial with anyone else yet was completely natural to her, and studied Nate. "How very kind of you to say so."
Lady Rathbourne was, all in all, perfect. Gabriella didn't like her one bit, and liked even less the way she stared into Nathanial's eyes and held onto his hand. Gabriella rose to her feet and, although she'd never considered herself clumsy, wished she could have done so with a bit more grace.
"Not at all," Nathanial said in a gallant manner. "You have not changed one bit. You are exactly as I remember you."
"Well, you have most definitely changed. You have become quite charming, no doubt dangerously so." Lady Rathbourne studied him for a moment. "You are very nearly a foot taller than when we last we met. Indeed, the boy I remember has become quite a handsome man. You very much resemble..." A shadow crossed her face so quickly Gabriella thought she might have been mistaken, "And how is that rascal of a brother of yours, Quinton?"
Nathanial smiled. "Quinton too does not change."
"And...the rest of the family? Your mother and Regina? Regina must be grown as well."
"And just launched upon the seas of society." Nathanial shook his head. "One must fear for those hapless unmarried men who have not been forewarned."
"I can well imagine." Lady Rathbourne laughed. It struck Gabriella very much like a laugh not well used, but then this room too seemed like a place that had heard little laughter.
She cleared her throat softly.
"And this is?" Lady Rathbourne smiled at her.
"My apologies," Nathanial said quickly. "Lady Rathbourne—"
"You used to call me Olivia," she chided.
"I was an impertinent scamp." Nathanial grinned. "Olivia, allow me to introduce Miss Gabriella Montini."
Gabriella nodded. "A pleasure to meet you, Lady Rathbourne."
"Do call me Olivia." To Gabriella's surprise, genuine warmth colored the lady's green eyes, warmth at odds with the surroundings. At once she decided that she liked Lady Rathbourne after all. "I don't receive many callers, especially not old friends. I have known Nathanial's family for much of my life, although I have not seen any of them for some time." She waved at the sofa. "Please sit down. I have requested tea for us, it should be here any moment."
"Thank you." Gabriella retook her seat on the sofa. In this room, with this woman, her serviceable blue gown and practical hat seemed both out of place and rather shabby. In the back of her mind she resolved to purchase some new clothes.
Olivia joined her on the sofa and waved Nathanial to a nearby chair. "Now then, your note said you had a matter of some importance to discuss."
"Yes, that." He looked a shade uneasy. "It's a long story and I'm not sure you can actually be of help. Still, because I knew you..."
Olivia raised an amused brow. "Yes?"
"Perhaps I should explain," Gabriella said quickly.
Nathanial nodded with obvious relief.
"Lady Rathbourne—Olivia," Gabriella began. "My brother was an archeologist who discovered a rare, ancient cylinder seal. A seal that made reference to the lost city of Ambropia, the first such reference ever found. But the seal was stolen from him, and while searching for it he died."
"Oh dear," Olivia murmured. "My condolences."
"Thank you." Gabriella drew a deep breath. "Now it is up to me to recover the seal and prove it was his discovery. I intend to donate it to the Antiquities Society, and we have been authorized by the society to use its name in our queries."
"The Antiquities Society? How very prestigious."
Gabriella nodded and winced to herself. Authorized was perhaps not entirely accurate.
Olivia drew her brows together. "And how can I can be of help?"
"One of those my brother suspected of taking the seal was a man named Javier Gutierrez." Gabriella paused. How did one accuse a woman's husband of theft and possibly worse? "He apparently often acted as an agent in the gathering of antiquities for Lord Rathbourne."
"I see," Olivia said coolly, the warmth of her demeanor abruptly gone.
A maid entered bearing a tray with tea and biscuits. Olivia waited until the maid had left the room to continue.
"As much as I wish otherwise, I'm afraid I can't help you." She poured a cup for Gabriella, handed it to her, then poured one for Nathanial. "I know little about my husband's activities in regards to his collections."
"Olivia." Nathanial leaned forward eagerly. "We were hoping that you might know if he had come into possession of such a seal. If he might have mentioned something that might help us. Or if you had noticed a new acquisition to his collections or—"
"My dear Nathanial." Olivia filled her own cup, her tone offhand, as if she were discussing nothing of particular importance. "I do not share my husband's passion for such things. He never discusses his acquisitions with me and I would not know a seal such as the one you are looking for if it were to sprout legs and walk into the room. Beyond that..." The warmth in her eyes had now vanished along with her demeanor, replaced by something hard and resolute. "My husband collects for the joy of acquiring. He particularly enjoys besting another collector or, better yet, a museum or institution. It is something of a game for him, and his fortune is such that he can demand nothing less than total victory, regardless of cost.
"He is not a man to be trifled with." The vague hint of a warning sounded in her voice. "I would permit you to view his collections, to determine if indeed what you seek is there, but that's impossible. His acquisitions—be they of art or sculpture or antiquities—are displayed in a locked room for his eyes alone. I do not have access to them nor do I wish to." She sipped her tea. "I have not seen most of what he has accumulated and I have no interest in doing so. It is, as I said, not something we share."
Disappointment washed through Gabriella at her words. And more, the oddest feeling of sympathy for this beautiful woman who would appear to have everything yet seemed as well to have nothing. "I see."
"As for your Mr. Gutierrez..." Olivia shook her head. "The name is not familiar. I would be surprised if it was. I never meet anyone who has a business arrangement of any sort with Lord Rathbourne. However..." She paused. "...I would certainly not be surprised to learn that he and my husband were involved in something, shall we say, less than aboveboard. If this seal is as rare as you say, it is exactly the type of item that would indeed arouse Lord Rathbourne's desire." She shrugged. "His fine hand in this incident would not be unexpected."
"But you don't know," Nathanial said.
"No, I am sorry. And I would not ask him." Olivia hesitated as if choosing her words. "Lord Rathbourne and I essentially live separate lives. Occasionally I appear with him at social events, but for the most part we rarely even share the same dwelling." She smiled at Nathanial. "I say this only because we are old friends and I cannot remember the last time I have had a visit from an old friend. I have always been sentimental about such things and fear I am becoming more so with every passing year."
"I'm sure my mother would enjoy it very much if you were to call on her."
"That's not possible," she said simply. "Nathanial." Her gaze pinned his. "I trust the confidences I have shared with you will go no farther than the three of us?"
"Of course not." Nathanial nodded
"Do I have your word on that?"
"You do."
She glanced at Gabriella, who nodded her agreement as well.
Olivia smiled. "Thank you both. Your discretion is most appreciated."
"Will we see you at the Antiquities Society Ball?" Gabriella said impulsively.
"I'm afraid not. I rarely attend such events in London. I prefer to spend most of my time in the country. When I'm in London, I tend to be something of a recluse." She smiled, but this time it did not reach her eyes. "Someday, in my dotage, I suspect I will be called eccentric. Oh dear." She cast Gabriella a rueful glance. "My future sounds dreadful. I shall be an eccentric as well as a sentimental doddering old creature."
"Never," Gabriella said staunchly.
"You are too kind." Olivia directed her attention back to Nathanial, her tone brisk. "Now, do tell me more of your mother. She had any number of charitable pursuits, if I recall."
They chatted for a few minutes about Lady Wyldewood and Regina, about Nathanial's and Quinton's adventures. Gabriella noted Olivia never asked about Lord Wyldewood nor did Nathanial bring up his oldest brother's name.
A quarter of an hour later they took their leave and settled in the carriage they'd had wait for them, Nathanial sitting on the seat beside her rather than across from her. At her insistence, they'd dispensed with the ever-present maid. It seemed wise, given the nature of their mission. It was most improper and really quite nice.
He blew a long breath. "That was worthless."
"I suppose it was." She shrugged. "Still, we do know the seal is something that would arouse Lord Rathbourne's interest. So we have not eliminated Gutierrez. And we know his lordship's collections are kept in a locked room in the house."
Nathanial raised a brow. "So that if we break in we will know where to look?"
"Don't be absurd." She scoffed. "We don't know where the room is. And as we should have to break in at night, we'd spend most of our time stumbling about in the dark looking for—" She caught Nathanial's grin and sighed. "You're teasing me, aren't you?"
"Yes, I am." His grin widened. "It's the most fun I've had today."
"I'm glad I can provide you with some amusement." She huffed and settled back against the worn leather seat. "I daresay Lady Rathbourne has few amusements."
"You noticed that, did you?"
"It was hard not to."
"Mother says she's rarely seen in public." Nathanial shook his head. "She used to be quite a social creature."
"She said she has known your family for a long time."
He nodded. "Her father's country estate borders ours. Although she and Sterling had known each other most of their lives, it was a casual acquaintance, not at all significant. As I remember, they did not fall in love until she was out in society and they ran into one another at a ball in London. It was as if they had never met before. If you recall, I told you that my family assumed they would marry, but then, abruptly, they were no longer seeing one another, and almost at once she married Lord Rathbourne.
"She is as lovely today as she was then." He paused. "But today, she looked like a woman afraid."
"Not at all." Gabriella shook her head. "She looked like a woman resigned. She seems as much a possession as anything else her husband has collected."
"I noticed that as well," he murmured.
"In spite of her obvious strength, she seemed very sad to me. She is not happy with the choices she has made in her life."
"And yet they were her choices."
"As far as you know."
He paused, then nodded. "As far as I know."
"What if they weren't?" Gabriella said slowly. "Her choices, I mean."
"It scarcely matters now and we will probably never know." Nathanial sighed. "We all thought she was the love of Sterling's life. That they were fated to be together—soul mates, if you will. But she shattered his heart. I have always thought the speed with which he married Alice was a direct result of that."
Gabriella hesitated. "Perhaps he should know."
"Who should know what?"
"Your brother, the earl. Perhaps he should know how unhappy she is."
Nathanial turned to her, disbelief in his eyes. "Are you mad?"
"Possibly," she murmured.
"What good would that do? For good or ill she has made her choice. She is married and he has gone on with his life. There is nothing to be done about it."
"Still." An odd wistful note sounded in her voice. "It does seem a shame."
Nathanial narrowed his eyes suspiciously. "What?"
"That one might find the love of one's life, one's soul mate, and lose him because of circumstances."
"She lost him because she chose to marry the man who had the biggest fortune." Nathanial shrugged. "One could argue if she was willing to make such a choice, she could not have been the love of his life after all."
"One could argue as well that one suitor having more money than another is a circumstance." Gabriella had no idea why she was arguing the point at all, why she cared, and yet she did. "Do you think there are such things as soul mates? As two people who are fated to be together against all odds?"
He looked into her eyes in a firm manner. "Yes."
"As strictly a...a..." She thought for a moment. "...a philosophical matter of debate, mind you...What if they are from completely different backgrounds?"
"It's of no concern."
"What if they don't trust one another?"
"They need to learn to do so. Leaps of faith and that sort of thing."
"What if one has a large, boisterous family, traditions, and heritage, and the other has no family whatsoever?"
He smiled. "A large, boisterous family very nearly always welcomes an additional member."
"What if she isn't what he thinks she is?" She gazed into his eyes. "What if she can't be what he needs? What he deserves? What he wants?"
His voice was low, measured, and fraught with meaning. His words a statement as much as a question. "What if she already is?"
Her breath caught. Sheer panic warred with the most wonderful feeling of surrender within her.
She swallowed hard. "It was a philosophical matter of debate, Nathanial, nothing more."
"I know." He pulled her into his arms. "Everyone in my family loves a good debate."
"This is not a good idea."
"And I thought it was an excellent idea," he murmured, his lips against the side of her neck.
"Well, that is...indeed..." Shivers skated down her spine. "...excellent."
"I thought so." He angled his mouth over hers, and she lost herself in his kiss. Her mouth tentatively opened to his and his tongue met hers. Desire and need exploded within her and she clung to him. Sensation swept through her, and her toes curled in her sensible shoes.
The most scandalous thought occurred to her. Why not enjoy Nathanial's touch? Why not surrender to him? She had no idea if he was her soul mate, she wasn't even sure she believed in such nonsense. But she did know he made her feel things she had never suspected she could feel. Made her want things she could never have. And she knew as well that Nathanial Harrington, the youngest brother of the Earl of Wyldewood, would never marry Gabriella Montini, who had no family or position and was, in fact, with her brother's death, no one and nothing of note. Even if he could overlook all else, there was one thing he could not overlook. No honorable man could. If she had learned nothing else in her years in England she had learned this. There was no future in his arms.
And when he knew all her secrets, he would know that as well. Her heart would be left broken and she would be alone. Better to stop thinking, hoping, wanting, now—before it was too late.
The carriage drew to a stop and she pushed out of his embrace.
"We're here." She scrambled out without waiting for his assistance.
"Yes, I was aware of that." He chuckled, climbed out of the carriage and ambled after her. "Unfortunately."
She turned to him and squared her shoulders. "That cannot happen again. There will be no more kisses shared in carriages or in libraries or at the door to my room."
"I haven't kissed you at the door to your room."
"But you've thought about it."
"Indeed, I have." He grinned in an unrepentant manner. "I find I have a hard time thinking about anything but kissing you."
"It has to stop," she said firmly. "All of it."
His grin widened. "I don't see why."
"No." She steeled her resolve against the need to throw herself back into his arms and forced a collected note to her voice. "I don't suppose you do."
Gabriella turned and walked into the house, leaving him staring after her.
She acknowledged Xerxes standing by the door with a slight nod and ignored the questioning look in his eye. She didn't pause until she reached the sanctity of her room. Closing the door firmly behind her, she rested with her back against it and closed her eyes.
The meeting of the Verification Committee was fast approaching. Either they found the seal and restored Enrico's reputation or they failed and all was lost. Regardless, one way or the other, this would be over then. She would leave this house and this family and Nathanial forever.
The irony of it all hit her. She and Nathanial were searching for a seal that might reveal the location of the lost city. After centuries it could well disclose the Virgin's Secret, which had been sought by so many for so long. The search had brought her to the one man she would be willing to give up everything for. But it was her own secret that would keep them forever apart.
Apart like the earl and Lady Rathbourne. Through circumstances and bad choices and life-changing decisions one had no say in.
And things that could never be undone.
## Thirteen
If I leave to fetch you some refreshment, will you be here when I return?" Nathanial smiled down at her and her heart fluttered.
Gabriella ignored it and favored him with an impersonal smile. "I don't know. I haven't decided yet."
She was in his arms, a waltz was playing, and she had already realized that in spite of her resolve not to encourage his affections—if indeed affection was what he harbored for her, and not simply lust—this was the best Antiquities Society Ball she had ever attended. The music was somehow richer tonight, the gowns of the ladies more exquisite, even the flickering gaslight that illuminated the society's ballroom cast an air of magic over the proceedings. In the apricot-colored gown she'd worn to Reggie's ball, she felt like a princess in a fairy tale. In truth, if she had been a more fanciful sort of woman the entire evening would have seemed enchanted.
He laughed and pulled her a bit closer than was proper. "I expect you to be here. I allowed you to vanish from a ball once before, I shall not allow you to do so again."
She raised a brow. "And how would you intend to stop me?"
He cast her a wicked grin.
Gabriella stared in annoyance. She had scarcely seen him at all since they'd returned from Lady Rathbourne's the day before yesterday. She'd begged off dinner one night, claiming a headache. Last night he and his brothers had gone to the earl's club. She heard them return late, more than a little inebriated, she suspected. Today she'd stayed in her room preparing for tonight's ball. No one seemed to think it odd that such a thing would take an entire day. At least no one mentioned it. Nathanial had not sought her out, which was at once a relief and more than a little maddening.
She had spent some of the last few days rereading her brother's letters. She remembered every word, as she always did, but still hoped to see them with a fresh eye. They were, of course, unchanged. The tone gradually shifting from the first to the last. From anger and rational determination to an almost giddy expectation of impending triumph and something rather less than sane. Rantings, actually, in the final letter.
"I want your promise that you will not vanish in the night," Nathanial said.
"Very well, then, I promise."
"Ah, but is it a promise I can trust?"
"Leaps of faith, Nathanial," she said wryly.
He laughed, and she cast him a reluctant smile. The man was incorrigible, his persistence very nearly irresistible. As was the way he held her and gazed into her eyes and made her feel as if they were alone in the ballroom, in the world, with nothing but the music and the magic.
She and Lady Wyldewood, the earl and his sister, had come tonight in a separate carriage from Nathanial and Quinton. She had not seen Nathanial until his arrival a quarter of an hour later. There was something about putting formal attire on a man used to living in the least formal of circumstances that would make the heart of even the most resistant woman skip a beat. And the look in his eye when he saw her had very nearly taken her breath away. She suspected, in her later years, the memory of that look would be like a flower pressed in a book. Something to take out on occasion and remember and savor. She thrust the thought aside, the fragile dried blossom crumbling to dust.
"You paid no attention whatsoever to what I told you, did you?" she said with a resigned sigh.
"That nonsense about not kissing you?" He shrugged. "Not really."
"Well, then heed my words now," she said firmly. Not being with him every minute had strengthened her resolve, but now that she was in his arms, she found it difficult to maintain her determination. Difficult to think of anything but the warmth of his body next to hers, and the hard feel of his shoulder beneath her hand, and the vague scent of something spicy and completely masculine. Nonetheless, she mentally shook her head to clear it. This was for the best. "I have decided from this moment on we should keep our relationship on a strictly professional level. We should be colleagues, if you will."
"Colleagues?"
"Yes." She nodded. "Colleagues."
He laughed. "I don't have any colleagues who look like you."
She sniffed in distain. "My appearance is irrelevant."
"Not to me."
"Nathanial—"
"Nor do I have any colleagues who smell as good as you."
"The way I smell is scarcely—"
"You smell like I imagine heaven would. Of exotic flowers and summer skies and promises left unsaid."
"Most poetic, Nathanial," she said coolly. "Utter nonsense, of course. Summer skies don't smell. As for promises left unsaid..." She scoffed. "I thought we had already established I am not the sort of woman who would be swayed by such flowery sentiment."
"No, of course not. You wish to be my colleague. Although I would be remiss in my responsibilities as a 'colleague' if I failed to point out," he gazed into her eyes, "that I have never had a colleague who fit so perfectly in my arms."
"Have you danced with many of them, then?"
He laughed.
"You are not taking this seriously." She sighed. "This is a serious matter. I would appreciate it if you gave it serious attention."
"Oh, but I am. I am taking it most seriously. I simply intend to ignore it."
"We will not get anything accomplished if we—you—are continuously distracted."
"Then, my dear Gabriella, we will not get anything accomplished." He smiled down at her. "You are a most distracting colleague and I cannot help myself."
"Don't be absurd. Surely you are a stronger man than that."
"Ah, but you have weakened me. Sapped my strength. I am the frailest of men in your presence." He executed a complicated turn, and she followed his lead perfectly.
"Beyond that, I don't want to be your colleague."
"Well, what do you want?" she said before she could stop herself. Before she could realize what a dangerous question it was.
He stared at her and once again the look in his eyes stole her breath, and possibly her heart. "Everything." His hand tightened around hers. "I want everything."
"I have no idea what you mean," she said in a lofty manner. This was dangerous ground. "Besides, that's absurd."
"How can it be absurd if you don't know what I mean?"
"I mean..." She huffed. "I don't know what I mean. But I do know what I want and what I don't want."
"Oh?" He smiled. "And what do you want?"
"I want to find the seal."
"That goes without saying." He nodded. "What else do you want?"
I want...
"I want you to listen to what I say."
"You wound me deeply, Gabriella." He shook his head. "I listen to every word."
"And then you do precisely as you please."
He grinned in an unrepentant manner. "But I do listen."
"You are a stubborn creature, Mr. Harrington."
"No more so than you. It's one of the things I love about you."
"Now you're being ridiculous. We barely know each other. How can you love anything about me?"
"Oh but I do. I love your passion for justice for your brother. And the way the least thing makes you blush in a most delightful way. I love the independence of your nature and how you try to be terribly proper even when you're considering acts that are somewhat less than legal."
She steeled herself against the desire to melt against him. "Nonetheless, you shouldn't use words like love unless you—"
"Mean them?" He nodded thoughtfully. "You're absolutely right." The music drew to a close and he escorted her off the dance floor. "Punch?"
That was it, then? He was to fetch her punch as though he hadn't mentioned love? Hadn't implied he had, well, feelings for her? Not that she wanted him to. Not that it made any difference whatsoever. No, it would only make everything more difficult.
She smiled politely. "That would be most welcome."
He chuckled as if he had the upper hand, and she watched him walk off. He was indeed a fine figure of a man, tall and handsome and dashing. The kind of man that made women want...She sighed. That simply made women ache with a newfound need for something they couldn't quite put their finger on. Made her want all sorts of things she could never have. That she never imagined she'd want. Most of all...him.
She glanced around the ballroom noting that whatever magic might have been here before was gone. The lights were overly bright, the gowns no more than pretty, the music only passable. It was an Antiquities Society Ball like any other. No more special today than it had been in the past.
Still, she acknowledged reluctantly, she had always loved attending this ball. Perhaps because she attended so very few. Last year she had been filled with hope here. Enrico was in an excellent mood, and she'd been optimistic that she could convince him to allow her to accompany him, to assist him in his work. She had danced very nearly every dance, even though she only knew a handful of people.
It struck her for perhaps the first time that her world was extraordinarily narrow. Limited to acquaintances she had made at the college, the older gentlemen who tended to spend their days in the society's clubroom whose paths she crossed on her way to or from the society library, the director and his wife. She had no friends beyond Florence and Xerxes and Miriam. She'd never noticed before, but her brother had had no real friends either, at least none that she knew of. There had been no letters of condolence, no true expressions of sympathy. It was to be expected, of course. Enrico only had his work and nothing more.
Regardless, he had been a good brother. She bit her bottom lip absently. It would be the height of disloyalty to think otherwise. Still, how many times had she said that to herself? Had she reassured herself that he was a good brother? Not merely since his death, but for years before that. And if he'd never been around, if he had not quite protected her as he should have, if he had not taught her all those things someone should have taught her about proper behavior and temptations and how a single unthinking act could effect the rest of her life, well, it scarcely mattered now.
She raised her chin. Enrico was dead and she was essentially no more alone now than when he was alive. Once she found the seal, she would go on with whatever life she might manage to find. At least, she thought wryly, she was not poor.
"You appear too pensive for so grand an evening," Florence's voice sounded beside her.
"Do I?" Gabriella forced a light note to her voice and turned toward her hired companion and friend. She cast her a genuine smile. "I can't imagine why. It is indeed a grand evening."
"I thought perhaps you were thinking of last year's ball."
"It's inevitable, isn't it?" Gabriella's gaze wandered idly around the room. As always, the crowd was an odd mixture of those who studied or searched for the treasures of the ancients, and those who provided financial support for their efforts. Society benefactors mingled with professors, archeologists chatted with board members, elderly scholars danced with titled matrons. "Last year held so much promise, so much lay before us. Enrico would present the seal, his reputation would be solidified, and who knows? Even without the other seals, he might have been able to unravel the Virgin's Secret, find the location of the lost city, and—"
"And he would never have allowed you to come along," Florence said in a hard tone.
"I have read everything ever written about Ambropia, not that there is very much. I have studied languages and maps and histories and—" She met Florence's gaze. "I would have been indispensable."
Florence stared at her, then drew a deep breath. "You would have been nothing more than you ever were to him."
A debt to a dead father. An obligation that was easily met with little inconvenience. A means to control a fortune.
Thoughts she'd preferred for years not to think at all, and the one she'd tried to ignore since his death, throbbed in her head.
Florence's expression softened. "You have facts to face, my dear Gabriella, but now is not the time to do so. This is an evening to put aside all thought beyond enjoyment of the moment." Her eyes twinkled. "I suspect Mr. Harrington is having an agreeable evening. He's scarcely let you out of his sight."
"He's afraid I'll disappear if he does."
"And will you?"
"No." Still, wouldn't that be the easiest way to end whatever it was happening between them? Once they had resolved the question of the seal, she could simply vanish from his life. She certainly had the money to do so.
"I daresay he wouldn't take that well. Not given the way he looks at you."
"Nonsense."
Florence raised a brow.
"I don't know what to do about him." Gabriella shook her head.
"What do you want to do about him?"
I want...
"I don't know," she snapped, then sighed. "My apologies. I didn't mean—"
"It's quite all right." Florence chuckled. "I can well see why you might be confused. Mr. Harrington certainly is handsome enough. And from what Mr. Dennison has said, he seems a good sort. Honest, honorable—"
Gabriella scoffed. "Then we are obviously well-matched."
"Your less than completely honest behavior of late is an aberration. It is not your nature." Florence raised a shoulder in a casual shrug. "I suspect a man like Mr. Harrington would understand and overlook this temporary flaw in your character."
"Perhaps he could overlook that." Gabriella shook her head. "But as you said, he is an honorable man. I daresay an honorable man couldn't possibly overlook—"
"Not if you don't give him the opportunity to do so."
"I am..." It was hard to even say the words. Ruined. Fallen. Soiled goods.
"Gabriella, you were scarcely more than a child. You were fifteen."
"And I should have known better."
"Yes, you should have, and if you had been raised properly instead of being hauled around from one uncivilized place to another, pretending to be a boy, surrounded by men not substantially better than your brother, you would have known better."
"I know better now."
It had come as a shock to her, when she began attending school in England, how a single incident with a boy scarcely older than she would impact the rest of her life. When she realized that by allowing her own seduction—even though she had not comprehended the significance of the act at the time—she would never be an acceptable match for a decent man. When she understood that what she had done with the son of a German archeologist was the same thing she'd heard bandied about in snippets of conversation from men like her brother in lewd and vulgar terms. Shame had filled her then, and she abandoned any thoughts of love and marriage and ever finding someone to trust with her heart.
"But enough about me." She drew a calming breath. "How is your Mr. Dennison this evening?"
"He's not my Mr. Dennison." Florence grinned. "Yet."
Gabriella raised a brow. "Oh?"
"I rather thought any chance I had to marry had passed me by long ago, and I never thought I'd meet a man like Mr. Dennison. He is good and kind and clever. And he makes me feel...well, special." Florence sighed. "As if I were the most beautiful, most clever, most special woman in the entire world. And I must confess, when he kissed me—"
"You allowed him to kiss you?"
"Allowed him? My dear child, I encouraged him." Florence leaned toward her in a confidential manner, a smug smile on her lips. "And he did so exceptionally well."
Gabriella laughed.
Florence looked across the room. "I sent him for refreshments and I see he is back. I should return to him." She studied Gabriella for a moment. "Are you all right, my dear?"
"Of course." She mustered her brightest smile. "I am having nearly as lovely an evening as you. You needn't worry about me tonight. Go back to your Mr. Dennison and have a grand time. I shall send you a note tomorrow and tell you all about my evening."
"Are you any closer to finding the seal?"
"Probably not." Gabriella shrugged. "But we are no farther away, at least."
"The Verification Committee begins its meeting tomorrow." A warning note sounded in Florence's voice.
"And concludes in ten days." Gabriella shook her head. "I am well aware of that."
"You don't have much time left."
"I know."
Florence paused. "Might I make a suggestion?"
Gabriella smiled. "Could I stop you?"
"If you don't find the seal, if you cannot present it to the committee..." Florence paused and her gaze met Gabriella's directly. "Abandon your quest. Go on with your life. Do not let this haunt you for the rest of your days."
"Florence, I—"
"Enrico is dead. He's gone, and you must lay him to rest. Redeeming his professional reputation will not now change the way he felt about you." Florence laid her hand on her arm. "It will not make him love you."
Gabriella's eyes widened. "What an absurd thing to say. I have no doubt as to my brother's feelings for me. There is no question in my mind whatsoever that he did indeed care for me. Why, he rescued me and provided for my home and my studies and everything."
And did so with my money.
The unspoken charge hung in the air between them. She ruthlessly shoved it aside. It was disloyal and unfair.
"My apologies," Florence murmured. "I don't know what I was thinking to have said such a thing. Of course he cared for you." She leaned close and brushed a kiss across Gabriella's cheek. "Have a lovely time this evening, my dear. I shall see you soon." She cast the younger woman an affectionate smile and took her leave.
Gabriella watched her circle the ballroom until she reached Mr. Dennison. Even from where she stood, she could see the manner in which Florence seemed to light up in the secretary's presence. And the way he lit up in hers.
Gabriella smiled. Florence had apparently found love. Love. Her smile faded. How could Florence say such a thing about Enrico? He did love her, of course he did. He was a good brother.
As for the money, she was a child when he had found her. She certainly could not have managed her fortune. And if, as she grew up, he never mentioned it to her, it was no doubt because it wasn't important. He'd used it to support them both as well as fund his travels and his work. Besides, even if she had known, she wouldn't have protested. Aside from all else, he was her brother, her only male relative, and in the eyes of society he had every right to use her resources as he saw fit.
Still, it would have been nice to know.
Not that there was anything that could be done about it now. No one knew better than she that the past was the past, and aside from a few artifacts, a few crumbling relics and the occasional memory, it was best to put the past firmly in the past.
"May I have the honor of this dance, Miss Montini?" A voice sounded behind her.
"Certainly," she said with a sigh of relief. A dance would be just the thing to set her spirits right. "I should be delighted."
She turned to face her new partner and froze.
## Fourteen
Shall we?" Lord Rathbourne offered his arm.
He was tall and imposing, with dark hair touched by gray at the temples. She'd seen him before, of course, but never this close. He looked younger than she had thought, somewhere in his fifties perhaps. On first glance he would have been considered distinguished and extremely handsome, until one noted the cold look in his eyes.
"Yes, of course," she murmured, and allowed him to escort her onto the dance floor. Words like ruthless and whatever means possible popped to mind and a chill shivered through her.
They took their place on the floor, and for once the music didn't sweep her away.
"I understand you paid a call on my wife," he said coolly.
"Yes, I did." She forced her most pleasant smile. He knew about their visit, she had no idea what else he might know. "She was gracious enough to talk with me for a few minutes."
"I know."
"She is quite lovely."
He smiled. "I know that too."
"You are a lucky man."
He glanced down at her, the smile on his lips never reaching his eyes. "I do not depend on luck, Miss Montini. With enough money and determination and power, one makes one's own luck."
"Oh." She uttered a weak laugh. "How resourceful of you."
"I am extremely resourceful." He paused. "I understand you are looking for the ancient seal your brother once claimed to have in his possession."
"Yes?"
"Come now, my dear, you needn't look so surprised. Surely you understand what an incestuous community this is—this world of treasures and those who hunt for or study them. The only way to keep a secret is not to share it. You have been asking questions. It has not gone unnoticed."
A frisson of fear skated up her spine. It was absurd, of course. She was safe enough here in the middle of a crowded ballroom. "As you are aware of my search, perhaps you would be so good as to answer one of my questions."
"How wonderfully direct of you, Miss Montini," he said smoothly. "I should be happy to answer any question you have."
"Excellent." She didn't quite believe him. Still, it would do no harm. "Do you have the seal?"
"Alas, to my eternal regret I do not. I had, however, arranged to acquire it."
Her heart sped up. "Oh?"
"I shall not bore you with the details. Suffice it to say, the efforts taken on my behalf—"
"You mean the attempt to steal the seal by a man in your employ," she blurted, indignation in her voice.
He raised a brow. "My, you are direct. I would not have put it so bluntly." He chuckled, a mirthless sound. "My methods may be considered—"
"Nefarious?"
He glanced down at her. "Again not the word I would have chosen. Regardless, my methods have always proved most efficient. This time, unfortunately, that was not the case."
She stared. "Then you admit you tried to have my brother's seal stolen."
"Miss Montini, whether I admit anything or not scarcely matters. I could admit any number of misdeeds—"
"Misdeeds?" She could barely choke out the word.
"Crimes if you prefer, although that does seem an arbitrary term. And might I point out, Miss Montini, that as we are dancing, and dancing is supposed to be an enjoyable activity, you should try very hard not to look as if you are either furious with me or terrified."
She jerked her chin up and adopted her most brilliant smile. "Does this meet with your approval?"
"Not entirely but it will do."
"Do continue, my lord. You were about to confess everything to me."
"What I was about to say is that I could admit any number of transgressions to you here in the middle of this ballroom. And while the world would prefer to believe the word of a beautiful young woman over an aging collector, that is not the way these things work." He shrugged. "Furthermore, you have no proof of anything. My efforts were futile and I do not have your seal."
She studied him closely. "Why should I believe you?"
"You have no reason to do so, but then I have no reason to lie to you." He pulled her closer and spoke into her ear. "Still, it might be great fun to lie to you."
She shivered but refused to let her unease show. Indeed, this was her opportunity to get information from him. And if a little banter—even flirtatious banter—was required, why not? Besides, here and now she had nothing to fear.
"And what, my lord, would you lie to me about?"
"The usual things one tends to lie about to a lovely woman, I imagine. The extent of my estates. The quality of my stables. The size of my..." He smiled. "...fortune."
"And would you need to lie about such things?"
"One could always own more land or have a larger fortune." He chuckled. "But no, I have no need to lie about those matters."
"Then the seal remains the only thing to lie about?"
"You are a clever girl. But I have not lied to you about the seal. You have my word, which I rarely give, and never lightly."
"I see." It's not that she trusted him, she had no reason to do that. But he was right when he'd said he could confess nearly anything to her and few would take her word over his. So why go to the bother of lying?
They danced on in silence for a few moments.
"I have a proposition for you, Miss Montini," he said at last.
"Do you, my lord? Dare I ask what such a thing would involve?"
"Nothing nefarious, I assure you. It has come to my attention in recent days that many of my acquisitions have not been catalogued as thoroughly they should be. In the past, those I have charged with the care of my collections have been somewhat lax, which is why they did not maintain their positions for any length of time."
She stared. "You're offering me a position?"
"I am indeed."
She narrowed her eyes. "Why?"
"My, you are a suspicious creature, but then I should consider you rather foolish if you weren't. The why is obvious. I have a task that needs to be done."
She shook her head. "But why me?"
"Any number of reasons. First of all you are extremely qualified for a position of this nature. One might think you had been training for it for much of your life."
"Go on."
"One gets to a point in life when one realizes there are more days behind you than before you. In addition, I have recently realized that my friends are far fewer in number than my enemies. Indeed, I am hard pressed to name a true friend. It's nothing new, it has always been that way. I am a difficult person to know and even more difficult to like. It's the price one pays, I suppose, for taking what one wants in life without apology. Still..." A hard light shone in his eyes. "If I had to live my days over, I would do exactly the same things again.
"I make no excuses for who I am, Miss Montini. No doubt it is my advancing years that cause me now to take notice of my failings in life in regards to those things other people take for granted—friends, family, and the like." He paused. "But these are treacherous times, my dear girl, for both of us perhaps. You would beware."
She widened her eyes. "What on earth do you mean by that?"
"Only that there are those who search for the same thing you do. Those who would not hesitate to do whatever they deem necessary to acquire your seal."
"Oh." She swallowed hard. It wasn't a completely new thought, of course, but coming from Lord Rathbourne, it seemed somehow less of a possibility and more of a certainty.
"I want my collections put in order so that when the time comes that I am no longer on this earth, they will make a certain amount of sense. I want my genius, if you will, acknowledged."
"My lord," Gabriella said cautiously. "Are you ill?"
"No more so than any man of my age. My collections are priceless, and I intend for them to remain together always regardless of where they might end up. Beyond that..." He cast her an assessing look, as if she were a horse he intended to purchase. "...I rather like the idea of a beautiful woman among my other possessions."
"Your wife is beautiful."
"But she does not share my interests."
"Nonetheless, I'm not at all sure this is a good idea."
"Allow me to tell you why you cannot pass up this opportunity."
"I can scarcely contain my enthusiasm," she said wryly.
"Sarcasm, Miss Montini, is most unbecoming. I do not permit it among my employees."
"My apologies," she murmured.
"As I was saying, with your brother's death, you cannot help but be at loose ends."
"Not at all," she said staunchly. "I have much to keep myself busy."
"Really? Aside from your search for this seal, may I be so bold as to ask what precisely is filling your days, Miss Montini?"
"No, you may not, my lord." She huffed. He might be a bit frightening but he was not yet her employer, if indeed she decided to accept his offer. She couldn't imagine why she would, although the idea of simply getting to see the art and artifacts he'd collected had an unexpectedly strong appeal. To see all that he had and make certain of what he said he didn't have.
He shrugged. "And I suspect you have considered the fact that, as you will be allowed total access to my collections, you will be able to make certain I have not lied to you. That I do not have the seal you search for."
"I hadn't thought of that," she lied.
"Of course not," he said smoothly.
The music faded to a close. He offered her his arm and escorted her off the floor.
"May I ask you one other thing, my lord?"
"You may ask."
"Other than its intrinsic value as a rare and priceless artifact, why were you interested in the seal?"
"My dear Miss Montini." He cast her a disillusioned look. "I should think you of all people would understand that."
"And if I don't?"
"Now I see. You wish to ascertain exactly what I know and what I don't know." He cast her a chastising look. "I did expect you to be a bit more subtle than this, however."
"I apologize if I have disappointed you."
"Do not let it happen again."
She stared at him. She could never work for such a man. "Yes, my lord."
"As you know, Ambropia was said to have been a city of great riches, great treasure. Whoever finds the city would claim a treasure beyond measure, not merely in a monetary sense but in terms of antiquities as well." His eyes gleamed. "The finder of the city would have his choice of artifacts, hidden from mortal view for eons. Unique in today's world. Irreplaceable. Priceless. For a collector of any stature, items from the lost city would make his collection the finest in all creation. The finest ever known.
"A legacy of that magnitude, Miss Montini, would make a man of my nature go to nearly any length to acquire it."
"I see."
"Yes, I imagine you do." He considered her closely. "Perhaps, before you make your decision as to whether or not to accept my offer, you might wish to see my collections for yourself."
Behind Lord Rathbourne, she noted Nathanial staring at them, a look of concern on his face. It was absurd. She wasn't an idiot, she could certainly take care of herself. She met the older man's dark gaze directly. "I would like that very much."
"Excellent. Let us say the day after tomorrow, then." A knowing smile curved his lips. "You may bring Mr. Harrington along, if you wish. If it would make you less apprehensive."
"I am not the least bit apprehensive, my lord," she said firmly, and held out her hand.
"Perhaps you should be." He raised her hand to his lips but his gaze locked on hers. "I understand your brother thought his seal, along with others, might hold the clue to the Virgin's Secret."
"Two additional seals, possibly." She thought for a moment. "His assumption was based on the pattern on the seal he had, although he suspected even with the other seals, the message would still need to be deciphered."
"How very interesting," he murmured.
"Is it?" She studied him carefully. "Why?"
"Because, my dear Miss Montini, from all that I have been able to gather, not having seen your brother's seal for myself, of course, I have one that appears very much to be its mate."
## Fifteen
What in the name of all that's holy did you think you were doing?" Nate plastered a pleasant smile on his face, took Gabriella's elbow and firmly steered her toward the door.
"And what do you think you're doing?" She smiled at him through gritted teeth.
"I am getting you out of here so that we may talk privately." He escorted her briskly through the door and across the corridor.
"Where are we going?"
"The courtyard," he muttered.
It was not the most private place, since anyone could walk in on them at any time, but he was not familiar enough with the society's building to think of anywhere better. He directed her through the French doors, which had been thrown open to catch the breeze, and down three steps into the courtyard.
There were benches and potted trees, urns overflowing with flowers and a bit of ancient statuary artistically placed as if to remind visitors that this was a place dedicated to such things. It was a charming setting and, as the night was exceptionally mild, would have been the perfect place for a momentary assignation. Indeed, if he hadn't been so angry, that's precisely why he would have brought her here. Thankfully, the courtyard was empty.
She shook off his arm and glared at him. "Explain yourself, Nathanial."
"Explain myself?" He fairly choked on the words. "You want me to explain myself?"
"I most certainly do." She crossed her arms over her chest, a gesture that emphasized the full, ripe nature of her bosoms. It would have been most distracting if he wasn't so annoyed. "Go on."
"Very well." He clenched his jaw. "What were you doing with Lord Rathbourne?"
"I believe we were dancing." She shrugged in an offhand manner. "He is an excellent dancer."
"Oh no." He shook his head. "You weren't merely dancing. You were questioning him."
"Have you ever met Lord Rathbourne?"
"No."
"He is not the type of man one questions."
"Regardless, I have known you long enough to recognize that look on your face."
Her brow arched upward. "And what look is that?"
"You know the look."
"I'm afraid I have no idea what you're talking about," she said in a lofty manner.
"This is not a game, Gabriella." He lowered his voice and leaned closer to her. "Lord Rathbourne is a dangerous man."
"I think his reputation is..." She raised her chin defiantly. "...exaggerated."
"Have you no sense whatsoever?"
"What do you mean by that?"
"Lord Rathbourne is the kind of man who gets what he wants."
"And?"
"And I saw the way he was looking at you. What he wanted on that dance floor was you."
A blush washed up her face. She stared at him for a moment, then laughed. "That's absurd."
"Is it?" He narrowed his eyes. "Do you have any idea how delicious you look in that dress? How the fire of your passion lights up your eyes? How you appear both vulnerable and determined at the same time? Completely irresistible and somehow out of reach?"
"Do you really think—" Her eyes widened. "You're jealous!"
"I most certainly am not." He couldn't possibly be jealous. Jealousy would mean all sorts of things he wasn't prepared to admit. Or perhaps all sorts of things he was afraid to admit. In spite of Quint's comments, it was too soon. Even so, he'd never had his heart broken, but if he knew nothing else, he knew that this was a woman who could do exactly that. "I am concerned for you. For your safety."
"My safety is not your concern."
"It most certainly is." He grabbed her arm and glared down at her. "I have promised to help you find your brother's seal. That promise extends to making certain you come to no harm in the process."
"I shall do exactly what I determine is necessary," she snapped.
"You cannot go off doing precisely as you please without concern for the consequences."
"I danced with him, Nathanial." She shook off his arm. "There is nothing more to it than that." She hesitated. "For the moment."
Apprehension caught at his throat. "What do you mean?"
"Lord Rathbourne has offered to let me see his collections."
"To my knowledge, that is an offer he makes rarely if ever, and never lightly. Don't you find that suspicious?"
"Not at all." She tossed her head back. "It makes perfect sense. He wishes to employ me—"
He drew his brows together. "To do what?"
"To catalogue his collections." A determined light shone in her eye.
"Alone? In his house?"
"I imagine there will be servants about. And Lady Rathbourne."
"Absolutely not. I forbid it."
"You what?"
He glared at her. "I cannot allow you to do something so reckless, so potentially perilous—"
"Nonsense. I know Lord Rathbourne has a certain reputation but I can't imagine I'd be in any real peril. Besides," her eyes narrowed dangerously, "you cannot forbid me to do anything."
"And yet..." He crossed his arms over his chest. He knew she would not take this well but it scarcely mattered. Injuring her sensibilities was well worth it to ensure her safety. "...I am."
"You have no right. Or are you going to threaten to have me arrested again?"
"If necessary to keep you safe..." He nodded. "...I would do exactly that."
"I see. So now you show your true colors."
"My true colors?" Anger raised his voice. "Let us speak of truth for a moment." This was treacherous ground but right now he didn't care. "The truth is that no matter what plans you may have had for your life, no man in his right mind who did what your brother did would let you assist him in his work. Go to the places he went. The truth is that while you are brilliant and knowledgeable, you are still a woman. A beautiful woman, which would only be more of a problem, headstrong and stubborn and independent as well, but a mere woman nonetheless, and it's past time you understood that."
She stared at him. "I thought my independent nature was one of the things you loved about me?"
"I was wrong!"
"I suspected as much." She sniffed. "A man like you has no understanding of the word love. You've probably said it hundreds of times to dozens of different women."
He clenched his jaw. "Hundreds."
"What?"
"Hundreds of different women. Why not say that? You barely know me at all but that's what you think of me."
"I know men of your nature." She shrugged. "I have seen any number of men exactly like you. Men who use women as playthings. You are just like—"
"Your brother?"
Shock washed across her face.
"Understand this, Gabriella, there have been any number of women in my life, but none that didn't want from me exactly what I wanted from them." He grabbed her and pulled her into his arms. "And I have never before used the word 'love' in any manner whatsoever with any of them."
She glared up at him. "Oh?"
"Furthermore, I am not anything like your brother in any number of ways I will refrain from mentioning now. But know this, Gabriella, I would never abandon you."
She gasped. "He didn't—"
"And I would give up my own life before I would allow you to come to any harm." His gaze locked with hers and he watched as her anger faded to acceptance, to belief, and then to something warmer, deeper, more important. His heart thudded in his chest. What had this woman done to him? Damn it all if he didn't indeed love her.
"Gabriella," he moaned, and lowered his mouth to hers.
"He said you could come," she whispered against his lips.
"Now is not the time," he murmured. He had no idea what she meant, nor did he care. All he wanted was to press—
"Lord Rathbourne." She pulled away from him. "He said you could come. To see his collections."
"Excellent..." He drew a steadying breath. The last thing he wanted was to talk, but apparently he had no choice. "As I had no intention of allowing you to go alone. As for this alleged position—"
"He told me..." She paused as if choosing her words. "He told me he had arranged to have my brother's seal stolen."
"He told you that?" The admission caught him unawares. Obviously Rathbourne would never have admitted such a thing if he had the seal.
She nodded.
"By Javier Gutierrez?"
"He didn't mention a name." Her brows drew together. "But something went awry and he did not get the seal."
"Which doesn't mean Gutierrez didn't steal it."
"Do you know where he is?"
"No idea whatsoever."
"But, as everyone else is in London," she began, excitement in her voice, "it stands to reason that Gutierrez—"
"No." He shook his head. "Gutierrez has no legitimate standing with the society or anyone else. He masquerades as an archeologist but he is a thief, nothing more than that. Admittedly, he is knowledgeable about the artifacts he procures for whoever will pay his price, but he will not show his face here. He is far too clever for that."
"Nathanial." Her forehead furrowed. "Why would my brother show the seal to a man like that in the first place?"
Because they were two of a kind. Because he was already touched by madness. Which only begged the question, then, of why he had shown it to Nate and Quint.
"I don't know," he said simply.
She paused. "Have you considered the possibility that perhaps my brother was mistaken about those he suspected? That the seal might have been taken by someone unknown to him? By someone whose name we might never know?"
His gaze searched hers. She wanted reassurance that their efforts would not be for nothing. He couldn't give her that. He drew a deep breath. "Are you prepared to end it, then? To put this behind you and go on with your life? To admit defeat?"
"No." She shook her head. "Not yet."
He smiled. "Well, then."
"I should be getting back." Renewed determination sparked in her eyes and she started toward the courtyard door.
"Where are you going?"
"I am returning to the ballroom. I don't often have the chance to dance, and it is one of the few things I have always done simply for the joy of it. Besides, one never knows what kind of information one might acquire during a dance." She glanced back at him. "And don't think I have forgiven you for your high-handed manner or the vile things you have said."
"The truth is often vile."
She ignored him. "I simply have other matters that concern me at the moment."
"And my manner is in your best interest!"
"Hah." She scoffed and stepped into the corridor.
He started after her and pulled up short. Damnation, he was too late.
"Miss Montini?" A tall, handsome man stood in her path.
"Yes?" she said coolly.
"I was afraid you had decided to leave before we had our dance," he said. Bloody hell. He would have recognized that accent, if not the face, anywhere.
"Our dance?" She shook her head. "My apologies but I fear I don't remember promising you a dance."
"Then my heart will surely break." The American chuckled. "Last year we only danced once, and you promised to save a dance for me this year. Unless, of course..." He paused. "You're not married, are you?"
She laughed. "No, I am most certainly not married."
Not yet!
"Excellent." He offered his arm. "Then shall we?"
"Yes, of course. But I am sorry. This has been a very long year and I'm afraid I don't recall your name."
"Yet another wound to my heart, but in truth, I'm not surprised. I was only one of a multitude of partners you favored last year." He tucked her hand into the crook of his arm, and Nate resisted the urge to leap after them and wrench the intruder from Gabriella's side. "Allow me then to introduce myself once again. I am Alistair McGowan."
"Mr. McGowan." Surprise sounded in her voice, and she cast a smug smile over her shoulder at Nate. "There is no one I would rather dance with."
Nate clenched then relaxed his fists at his sides. He certainly wasn't jealous of Rathbourne, but the viscount's attentions to Gabriella were a cause of great concern. The man had enough money and power to do precisely as he pleased. The very idea of Gabriella going to his house alone, being in his employ, sent tremors of fear through him. There was little he could do to protect her there. Admittedly, there was probably little he could do to keep her away.
But Alistair McGowan was a different matter entirely. From what he knew of the man, and the few times they'd crossed one another's path, he was a decent enough sort. For an American. He no more seriously considered McGowan a suspect in the theft of the seal than he did Quint. If McGowan had the seal, he probably came by it in a relatively honest manner.
He stared after Gabriella and McGowan. The American inclined his head toward her, and a faint ripple of laughter drifted back to him. He was making her laugh? Damned colonist.
Didn't McGowan realize she was taken? Didn't she realize she was taken?
Of course not. He had barely begun to realize it himself.
## Sixteen
My apologies once again, Mr. McGowan." Gabriella smiled up at him, which wasn't at all difficult.
He was an adequate dancer, or at least she was in no fear of having her feet trampled, as so often happened at this particular gathering. McGowan was handsome as well, with blond hair and broad shoulders. He had the greenest eyes she'd ever seen, crinkled at the corners, no doubt from staring across the desert sands. Wickedly attractive, was the phrase that came to mind. Good. When Nathanial watched her dance with this man, perhaps he would indeed be jealous. Not that she cared. "I can't imagine how I might have forgotten you."
"It has been a long year," he said with a smile. "I would have been surprised if you had remembered. It was, after all, only one dance and not as if we had shared a kiss in the moonlight."
She stared at him. "Why did you say that?"
"Because, Miss Montini," he grinned, "you make my thoughts to turn to things like kisses in the moonlight."
"Are all Americans this forward?"
"Yes," he said in a somber manner, though amusement twinkled in his eyes. "As well as charming, each and every one of us. Even the ladies." He thought for a moment. "Although they do, all in all, tend to be prettier than the gentlemen." He leaned toward her ear in a confidential manner. "We much prefer it that way."
She laughed. "I must confess, I didn't expect you to be so dashing."
"No?" He held her a bit firmer and performed a complicated step to avoid another couple who appeared to be careening out of control. She followed him easily. Perhaps he was better than adequate. "You didn't expect a man you think might be a thief to be enjoyable company?"
Caution edged her voice. "How did you know about that?"
He shrugged. "Word does tend to travel, Miss Montini. And while you have my condolences for your brother's death, I assure you I had nothing to do with the disappearance of his seal."
She wasn't sure what to say. It was one thing to accuse Lord Rathbourne of misdeeds. He was, after all, a not especially pleasant person. And quite another to voice her suspicion of Alistair McGowan to his handsome, smiling face. Still, she had no reason to trust him.
"And why should I believe you?"
"I don't know. It's harder to prove one's innocence than one's guilt, I suppose." He heaved a frustrated sigh. "Forgive me, Miss Montini. You should know there is nothing I would like better than to continue this dance, but I fear I am not especially good at dancing and talking at the same time. One takes all my concentration, leaving the other lacking in substance if not style. And I suspect you have a great number of questions for me. Would you mind if we stopped dancing to talk?"
"Not at all." She smiled, and he escorted her off the floor, to chairs arranged by a potted palm. While in plain view, the plant still provided a modicum of privacy. She took a seat and he settled in the chair beside her.
"I ran across your brother more than a year ago now." McGowan began without preamble. "You should be aware, although we had known each other for years, it was in no more than a casual manner. We were nothing more than acquaintances, really. We would cross paths on occasion, share a meal together, trade a story or two, that sort of thing."
"Go on."
"He had recently found the seal, and needless to say, was extremely excited about it."
She leaned forward. "Where exactly did he find it, Mr. McGowan?"
"He never said." He thought for a moment. "At the time, that struck me as somewhat odd, but your brother always did keep things like that to himself. At least that was my experience with him."
She nodded. "He was always reticent to give specific details about his finds."
"Yes, well, many of us are." McGowan shrugged. "It's a very competitive field, Miss Montini. It's not at all unusual to hear of someone who has lost a find because they opened their mouth to the wrong person. Still, it's often hard to keep one's enthusiasm to oneself. The Ambropia seal was the kind of discovery that elicited that type of excitement."
"I fear I am somewhat confused. If you and he were not especially close, why did he show you the seal?"
"Proximity played a part. We happened to be in the same place at the same time. The thrill of discovery is often greater when one can share it with someone who will appreciate its magnitude. We all have a tendency to brag about such things. There's little that warms a competitive heart more than seeing a flicker of envy in the eyes of a colleague." He paused for a moment. "Beyond that, your brother and I shared a similar passion. I too wish to find a city lost to the ages."
She raised a brow. "Ambropia?"
"No, although if a clue to its location fell into my hands, I certainly wouldn't walk away from it." He chuckled. "No, Miss Montini, there are men who search for Ambropia and Hattusha and Knossos today much as they once searched for Babylon and Troy and Ephesus. They do so because there is something about a city lost in time, abandoned, forgotten, relegated to myth and legend, that grabs one's imagination. And buries itself in one's soul." He glanced at her. "Are you familiar with Shandihar?"
She nodded. "It was on the silk route in southern Turkey, Asia Minor, the crossroads of the world at one time. Reputed to be a city of great wealth and glory, it was described in writings from the sixth century. It is believed the people of Shandihar worshiped only one god, or rather, goddess—Ereshkigal, the queen of the night."
He stared at her. "How do you know all that?"
"I remember everything I've ever read." She smiled. "It's a useful skill."
"I can well imagine," he murmured, studying her with a mixture of admiration and possibly envy.
"About Shandihar?" she prompted.
"Ah yes. The discovery of Shandihar, Miss Montini, is the quest that has captured my heart. And I will find her one day." Absolute confidence shone in his eyes. "It is my destiny, I have no doubt of that."
"At least you know Shandihar did indeed exist." Gabriella blew a long breath. "The writings about Ambropia are so obscure, the very name of the goddess who protected it is still as yet unknown. She is only known as the Virgin Goddess."
"And the location of the city is the Virgin's Secret." He nodded. "Which is why your brother's find was so important a discovery. Never before has there been reference either to Ambropia or the Virgin's Secret on so ancient an artifact."
"No, Ambropia was only mentioned by the Greeks, and those writings are vague and minimal."
"That the symbols for both the city and the Virgin's Secret were found on an Akkadian seal would seem to indicate that the city was more than mere legend."
"One would hope, but quite honestly, Mr. McGowan," she met his gaze firmly, "that does not concern me. If my brother were still alive, I am certain he would want to pursue the search for the city itself. I want only to recover the seal and give my brother the credit due him. I don't want him to be remembered as..." She paused to find the right words. "I want to restore his reputation. His good name."
"His good name. Yes, of course," McGowan murmured. His gaze slid past her, then returned to meet hers. "Your quest strikes me as both noble and honorable, but I do hope you understand there are others to whom those words do not apply. Miss Montini." He stared into her eyes. "Ambropia would be a find that would bring untold fame and fortune and glory to its discoverers. Your brother's seal is the first step toward that discovery. There are those who would not hesitate to use whatever means possible to acquire it."
"I am well aware of that, Mr. McGowan."
"Then you are aware as well that your journey could be a dangerous one."
"I am." She nodded. "But I'm not worried."
"Perhaps you should be. I wish I could be of further help." He grinned. "Indeed I can think of nothing I would like better than to help you."
"Why, Mr. McGowan." She widened her eyes in an innocent manner. "Are you trying to sway me with flirtatious banter?"
"I am trying." He smiled, then sobered. "You have no reason to believe me, but I do not have the seal." He paused. "Nor do I know who does."
"And would you tell me if you did?"
"Ah, Miss Montini." He took her hand and raised it to his lips, his gaze never leaving hers. "I daresay I would tell you very nearly anything to see gratitude light up those lovely blue eyes of yours."
She laughed. "Mr. McGowan, you are past trying now. I think you have succeeded."
"Good." He grinned. "I find all this talk has left me parched." He got to his feet. "May I fetch you a cup of punch?"
"That would be lovely." She smiled, and he took his leave. She watched him circle the room, heading toward the alcove where refreshments were arranged.
Blast it all, she did believe him. Not that she was swayed by his flirtatious manner or his handsome face, but there was an air of honesty about him. And his manner struck her as forthright. He seemed the kind of man who would not lie well. She did not consider herself a particularly good judge of character, but there was something about McGowan that elicited trust. She could be wrong but didn't think so.
If she eliminated McGowan as a possibility, as well as Rathbourne—although he'd admitted he had tried to acquire the seal—that left only Gutierrez. Who may or may not have stolen it for Rathbourne, although his lordship claimed he didn't have it. And was willing to let her view his collections to prove it. Which might be rather pointless, all things considered. Still, it would be interesting to see the seal Rathbourne said might match her brother's, as well as his other artifacts.
Which left only Nathanial and Quinton of those her brother suspected. She still wasn't sure she entirely trusted Quinton, and in truth, what woman would? As for Nathanial, she'd had no choice but to trust him. Now, without quite noticing how or when it had happened, she did trust him. Certainly with her quest. Perhaps with her heart as well, although that was not possible.
She shook her head to clear the absurd thought. Even if she had her heart to give, she would think long and hard about giving it to a man who thought she was a mere woman. Women were doing all sorts of things these days that men didn't think they should or could. Why, hadn't Amelia Edwards traveled Egypt for years and then written A Thousand Miles up the Nile? A book archeologists—men—found praiseworthy and most helpful in their own pursuits. Indeed, she'd read several different memoirs, written by women, that were their accounts of traveling in remote parts of the world. There was nothing mere about those women. The fact that she couldn't think of a woman actively engaged in archeological pursuits didn't mean there wasn't one somewhere. Perhaps she could be the first? It was an intriguing idea. A new hope for the future, possibly to replace the one that had vanished with her brother's death.
I would never abandon you.
Enrico had died, he hadn't abandoned her. At least, not with his death. Regardless, Nathanial had no right to imply otherwise.
And I would give up my own life before I would allow you to come to any harm.
How could one stay angry at a man who would say things like that? Still, anger was one way to keep him at arm's length. But with each passing day spent in his presence, she wanted to be in his arms. She wanted to be in his life for the rest of her days, and that simply was not going to happen. Even dwelling on it was absurd.
One way or another, their time together would soon be at an end. Then she would disappear from his life. It was for the best, really. He was obviously beginning to feel some affection for her. And she...
She shoved the thought aside.
It scarcely mattered at the moment. The only important thing now was finding the seal. Before it was too late for her brother's redemption. Before Nathanial's offhand talk of what he loved about her became something more significant. Before she lost her heart to him and it was too late to save herself.
If it wasn't already too late.
## Seventeen
Nate casually stepped up to the refreshment table beside McGowan.
"Harrington," McGowan said coolly.
Nate nodded. "McGowan."
"I'm surprised to see you here. I find myself indulging in this nonsense every year, if only to make certain the society realizes I am still alive and have not abandoned my work and returned permanently to America." McGowan raised a brow. "Yet I cannot recall the last time I saw you in attendance."
"I have no need to let anyone know of my continuing existence on this earth," Nate said in a lofty manner, then grinned. "Besides, I have always found this event to be somewhat deadly in and of itself."
"You're late, by the way." McGowan helped himself to a glass of punch.
"What do you mean, I'm late?"
"I've been here for a good minute now. I expected you, oh, at least thirty seconds ago."
"Don't be absurd." Nate paused. "Why would you expect me?"
"Why? Excellent question." McGowan slanted him an amused glance. "Perhaps because when I encountered Miss Montini, you were right behind her in the courtyard, glaring at me. Or perhaps because you then proceeded to follow us into the ballroom, again glaring at me. Or it might be because when Miss Montini and I were dancing you continued to glare."
Nate scoffed. "I do not glare."
"Harrington, you looked at me as though I was about to steal an artifact of great worth and rare value from right under your very nose."
"I do not glare," Nate muttered.
"And you looked at her as if she was of great worth and rare value."
"We are..." Nate paused, then said the first thing that came to mind. "Colleagues."
McGowan snorted. "That's what I thought." He sipped his punch and studied Nate. "Do you have questions for me as well?"
"Did she?"
McGowan chuckled. "She did indeed. She's very knowledgeable and quite determined."
"You noticed that, did you?"
"It would have been impossible to miss." McGowan paused for a moment. "I had the distinct impression she doesn't realize the kind of man her brother was."
Nate shook his head. "I don't think so.
"I knew Montini for probably, oh, a dozen years or so. Not well, of course, no one knew him well. He was too..." McGowan thought for a moment. "Competitive, I think, to make friends. He didn't really trust anyone. When I first knew him, he had a boy who traveled with him. Perhaps a brother but I'm not certain."
"I've met the brother," Nate said.
McGowan considered Nate closely. "Might I ask you a question?"
Nate nodded.
"Miss Montini asked me why her brother had shared his find with me. I think it was more than likely because I was available. He told me about the seal and showed me the impression in Cairo. I believe it had just come into his possession. That was in January of last year." Curiosity glittered in McGowan's eye. "I was wondering when he shared the impression with you and your brother. And why."
"It was a few weeks later, if I recall correctly." McGowan's question had already occurred to him but it wasn't something he wished to dwell on or examine. It was somewhat disquieting to consider, but he hadn't voiced his unease to anyone. Not to Sterling, definitely not to Gabriella, nor had he talked about it to Quint. Still, it was a minor detail. Scarcely worth mentioning. "I think he showed it to us because of my brother's work years ago with Professor Ashworth."
"Isn't Ashworth an expert on Ambropia?"
Nate nodded. "He was Quint's mentor once. My brother idolized him, and he taught Quint everything he knew. At some point they had a falling out. Quint never said what it was about and they went their separate ways."
"You think Montini just wanted to flaunt his find?"
"To be honest, I don't know." Nate shrugged. "That was Montini's nature. It might also be that, because of Quint's history with Ashworth, Montini thought he would be more likely to recognize the validity of the seal." Nate drew his brows together. "I have tried over and over to recall the exact details of that meeting. What was said and by whom. It seems to me Quint showed little interest in the seal."
"Given his background with Ashworth," McGowan said slowly, "didn't that strike you as odd?"
"Not at the time." In fact, he hadn't paid much attention to Quint's relative lack of enthusiasm for Montini's find. It hadn't seemed important. Now, with all that had happened, it did indeed seem more than odd. Perhaps it was time he had a long chat with his brother.
McGowan sipped his punch and considered Nate thoughtfully. "There's something I didn't tell Miss Montini that might have some bearing on all this. Or it might be nothing at all." He lowered his voice. "It was some time ago—autumn, if I recall correctly. I was in Crete, near the Kephala mound. I didn't see him myself but I heard your brother was in the region as well."
"Go on." Nate and Quinton often went their separate ways for reasons of expediency or personal interests, or simply because on occasion they preferred to travel alone. Quint more so than himself.
"I heard a rumor—and mind you, it was nothing more than that—about a game of chance—cards, I believe. Gossip had it that your brother had won an antiquity of great value."
"How intriguing."
"The loser in that game was a Spaniard." McGowan's brow furrowed. "A man named—"
"Gutierrez?"
"Yes, that's it. I heard Gutierrez was furious. Nasty sort, from what I know of him, and dangerous as well no doubt. He accused your brother of taking advantage of his inebriated state." McGowan paused. "I assume you know Gutierrez often procured antiquities for Lord Rathbourne?"
Nate nodded.
"I heard about Montini's death a few weeks later." He shrugged. "As I said, I don't know if it's significant or of any interest at all in terms of the missing seal."
"Neither do I." Nate shook his head. "I suspect it's not, but I do appreciate it nonetheless."
"There's one other thing."
"Yes?"
"Does Miss Montini know how her brother died?"
"She was told he died of a fever."
"I see. My information is probably inaccurate then." His gaze met Nate's directly. "I heard he was killed. That his throat had been slit." McGowan grimaced. "You might wish to keep that to yourself."
Nate nodded. "She doesn't need to know."
"I'm not sure she realizes how dangerous this game is. But I thought you should know."
"I do appreciate your candor."
"Well, as I have no real way to prove my own innocence, it seemed the thing to do."
"You could simply be trying to steer us in another direction."
"I suppose I could." McGowan grinned. "Would that I were that clever." He filled another glass and offered it to Nate. "I was supposed to be bringing Miss Montini a glass of punch but I assume you'd prefer to do that yourself."
"Yes, thank you," Nate said absently, accepting the glass, his thoughts returning to McGowan's comments.
Was it possible that Quint knew far more about the missing seal than he had let on? Surely not. Quint might have been many things, but he would never steal another man's discovery for himself. No, he knew his brother better than that. But it was indeed past time to have that long chat.
"Does she know?" McGowan asked in an offhand manner.
Nate jerked his attention back to the American. "Does who know what?"
"Does Miss Montini know you're in love with her?"
"I'm not..." Why continue to deny it when even a stranger could see it? Nate blew a long breath. "No, she doesn't."
"You might want to tell her." McGowan chuckled. "Before someone else realizes what a prize she is."
"I am surprised no one has noticed before now."
"Oh, I imagine many have noticed. She is hard not to notice. And now that her brother is dead..." McGowan paused for a long moment as if debating his words. "She did all of his research for him, you know, prepared all of his paperwork, maintained his correspondence, that sort of thing. From what I understand, he was extremely possessive of her. I saw something of that here last year. He was most intimidating toward any man who showed her the least bit of continued attention.
"He used to have a manservant, a big, powerful sort, rather on the exotic side in appearance, who traveled with him for years. Then I heard Montini had left the man here in London as protection for his sister. To be honest, up until then I hadn't known he had a sister. Only the boy, who I assumed had gone to London as well."
"I see." The description sounded familiar. Nate narrowed his eyes. "Was his name John?"
"No, it was Greek, I believe." McGowan frowned. "Or possibly Irish."
"Xerxes Muldoon," Nate said slowly, also known as John, no doubt. How very interesting.
"That's it. At any rate, now that her brother is dead, she is free." McGowan chuckled. "I know if I were looking for a wife—"
"I'm not looking for a wife."
"And that, Harrington, is exactly when you tend to find one." McGowan nodded in Gabriella's direction. "I would go now, if I were you. Before anyone else realizes the delectable and brilliant Miss Montini is now available."
Nate stared at the American. "Excellent advice, McGowan." He grinned. "I shall take it to heart."
McGowan raised his glass. "See that you do." His gaze sobered. "And do not let your guard down."
Nate nodded and started back toward Gabriella.
The American had certainly been full of unexpected revelations. Most significantly, of course, about Quint. But about Montini and Gabriella as well.
He had suspected from the first that she was not being completely honest with him about her life. As for that other brother of hers, it was more than a little odd that she wasn't more concerned about his whereabouts. Indeed, she'd scarcely spoken of him. She'd only said they were close. As one, if he remembered right. And he looked so very much like her...
She stood up when he approached. "Your mother says we should be leaving now."
"I'm surprised Sterling agreed to stay this long," he murmured, and handed her the punch. "McGowan sends his regrets along with refreshment." She was exactly the same height as the brother who had accosted him in Egypt. The brother she rarely mentioned.
"How very kind." She sipped at the punch.
And her eyes, that unique deep blue, could of course run in the family. Her brother had worn a hat pulled low, but Nate had noticed his eyes. They were the same in hue and shape.
"Why are you staring at me?"
And the same fire flashed in them when angered.
"Am I?" he said coolly.
"Yes you are." Her brows drew together.
What he was thinking was absurd.
She huffed. "It's most disconcerting."
If he'd had the same thought about another woman, it would have been ridiculous.
"My apologies."
Absolutely impossible.
Unease shaded her eyes. "Nathanial, what are you thinking?"
But when it came to Gabriella, it wasn't the least bit improbable. Stupid and reckless perhaps but not out of the realm of possibility.
"Nathanial?"
It was exactly the sort of impulsive, dangerous, foolish thing she would have done.
"It's nothing of any importance." He plucked the glass from her hand and set it on a chair, then tucked her hand in the crook of his arm and started toward the door. "I was just thinking about family resemblances and how very much you look like your brother."
"Really." She shrugged. "I never noticed any similarity at all."
Indeed she looked nothing like Enrico Montini, the only brother Nate now suspected she had. And he knew exactly how to confirm it.
## Eighteen
Lord Rathbourne will be with you shortly." His lordship's stern-faced butler nodded a bow and took his leave, closing the library doors behind him.
Gabriella resisted the urge to shift from foot to foot nervously, although she would have happily slit her own wrists before letting Nathanial know she was the least bit apprehensive. He looked somewhat uneasy himself. He hadn't wanted to come; no, in truth he hadn't wanted her to come. And he certainly hadn't hidden his feelings.
They had discussed it yesterday morning, which was a continuation of the discussion they'd had the night before, upon their arrival home from the ball. And they'd been discussing it in the afternoon when the note arrived from Lord Rathbourne inviting them here this morning. She suspected there were any number of people in the household, Xerxes included, who might have called their exchanges something more akin to argument than discussion. But arguing with Nathanial kept him at arm's length, which was exactly where she wanted him. No—not where she wanted him—where she needed him.
Aside from their differing opinions regarding Lord Rathbourne, Nathanial seemed somewhat cross overall. She had caught him studying her when he thought she wasn't looking, as if to determine the answer to a question he hadn't asked. Or to learn her secrets. It was most unnerving. In addition, it seemed that Quinton had vanished, which apparently wasn't at all unusual. Xerxes heard from the other servants that Master Quinton often disappeared for days at a time, which everyone attributed to dissolute living, to drink and gambling and women. Regardless, Nathanial was obviously not pleased by his brother's absence.
"Good day, Miss Montini." Lord Rathbourne stepped into the room and crossed the floor to her, nodding at Nathanial in passing. "Mr. Harrington."
"Good day, sir," Nathanial said in a remote but nonetheless polite manner.
"Miss Montini." Lord Rathbourne took her hand and raised it to his lips. His gaze bored into hers, and a chill ran up her spine. "You have no idea how delighted I am that you could join me today."
"Thank you for the invitation, my lord." Gabriella cast him a reserved smile and pulled her hand from his. "Will Lady Rathbourne be joining us?"
"She has already returned to the county," he said in a dismissive manner. "She has no interest in my collections, but I am hoping you will find my acquisitions so intriguing you will not hesitate to accept my offer of employment."
"I am somewhat busy at the moment."
"Very busy," Nathanial said pointedly.
Lord Rathbourne ignored him. "Ah yes, your search for the missing seal. Come now, Miss Montini, it cannot possibly take up all your time. And I must confess that when I have a pressing problem, I find doing something entirely different clears my mind, thus paving the way for a solution. And one way or another your search will soon be over."
"What makes you think that?" Suspicion sounded in Nathanial's voice.
Lord Rathbourne cast him a long suffering look. "My dear boy, everyone knows a request for validation of any artifact has a life span of one year, from the start of one committee meeting to the end of the meeting on the following year. As the Verification Committee began its meeting yesterday and will conclude eight days from now, the clock is ticking, as it were."
He returned his attention to Gabriella. "Once you have given my proposal due consideration, Miss Montini, I am confident you will find it irresistible." He stepped to a wall of shelves and glanced at Nathanial. "You will find this interesting, Harrington."
"No doubt, sir."
The older man reached inside a bookshelf. "There is a lever here that is released by a combination lock. I dialed the combination a few minutes before you arrived." He flipped the unseen lever and a wall of shelves slid to one side, revealing a large opening. "The lever releases a spring which opens the door. I had the system designed expressly for my needs." He bowed to Gabriella. "After you, my dear."
She took a deep breath and stepped through the opening, his lordship a step behind her, leaving Nathanial to trail after them.
"Welcome to my treasure room."
It seemed at first a small room, lit only by the light from the library behind them. There were no windows and no daylight whatsoever. Gabriella had a disturbing sense of solidity, as if she were in a tomb, the last resting place of a pharaoh. Lord Rathbourne quickly lit gas sconces positioned on either side of the opening on the wall behind her, and she saw that the room was much larger than she had at first thought. In front of her, what she'd assumed was a wall, in fact was a series of narrow panels, each about a foot wide. They extended from the floor to a few inches below the ceiling, an ornate brass knob positioned in exactly the same place on each panel. Lord Rathbourne moved to one, grasped the knob and pulled. The panel was actually one end of a long glass display case that slid outward into the room, revealing narrow shelves filled with Egyptian antiquities.
Gabriella gasped and moved closer. Even Nathanial stepped up to get a better look.
"Here we have the funerary urns of the ancient Egyptians..."
It was indeed a collection to rival any she'd seen in a museum. His lordship pulled out case after case of artifacts. The assortment of Egyptian sepulchral items alone was varied and endless. Here were the figures of deities carved from carnelian and lapis lazuli and jasper, or fashioned out of blue porcelain. She recognized treasures unique to Thebes, Abydos, and Karnac. There were amulets and scarabs cut of stone or semiprecious gems, originally intended to be tucked inside the folds of a mummy's wrappings. And jewelry of gold and silver and bronze from long dead pharaohs and their queens.
There were cases filled with the remnants of other civilizations as well, from the Etruscans, the Lycians, and the Assyrians. There were artifacts of Greek and Roman origin, black and red vessels and sculpted marble remnants, along with coins marked with the likeness of caesars and emperors and kings. At first glance it seemed there was little of the ancient world that was not represented in the varied and extensive collections of Viscount Rathbourne. It was the work of a lifetime, and indeed it might take a lifetime to examine it all. It was extraordinary, and her heart sped up at the sight of it all.
There was as well a case filled with cut and natural gems. Large diamonds, rubies, emeralds. A king's ransom in precious stones.
"These are some of the finest and rarest stones in the world." A note of pride sounded in Lord Rathbourne's voice. "They are as fine as the crown jewels themselves."
"One could support a small country with these," Nathanial murmured.
"A large country, Mr. Harrington," Lord Rathbourne corrected.
"But where—"
"Here and there, Miss Montini. A true collector does not reveal all his secrets." His lordship chuckled dryly. "In addition, there is a room similar to this one on the floor directly above us that is strictly for the storage of my paintings. Renaissance masters, primarily." Rathbourne smiled in a satisfied manner. "I have a fondness for such things and I have the means to indulge myself." He raised a questioning brow. "Would you care to see those as well?"
"Our time is limited today," Nathanial said sharply.
"Nonsense," Gabriella murmured, her gaze darting from one treasure to another. "We have plenty of time."
Lord Rathbourne chuckled. "Perhaps another day would be best." He pulled out one more case. "I think this will be of particular interest to you."
This case held ancient cylinder seals, dozens, perhaps as many as a hundred. Made of stone or clay, they were in no particular order and looked to be Babylonian and Assyrian, Akkadian and Egyptian.
"This one, I think." He pointed at a cylinder carved of greenstone toward the middle of the case.
She stepped closer and peered through the glass. Her breath caught. It did indeed resemble the impression she had seen of her brother's seal. Still...She shook her head. "It's made of the same material and looks similar in size. But without comparing an impression with my brother's, it's impossible to say."
"I can arrange for an impression."
"This is all quite remarkable, my lord, but we should be leaving," Nathanial said.
"I've never seen anything like this in other than a museum." Gabriella met Lord Rathbourne's gaze directly. "It's a pity not to share it with the world."
"I am a selfish man, Miss Montini. I make no apologies for my nature." He shrugged. "When we spoke at the ball, I told you I wished to have my collections put in order so that when I am no longer here, what I have accumulated will be acknowledged. However, since we last spoke I have had further thoughts."
"Yes?" Gabriella said.
"It seems something of a shame to have my life's work relegated to a shelf here and there in a museum full of such things, with only a brass plate as recognition of my achievement. I would much prefer to have all that I have accumulated remain as one. I am considering keeping my collections together in one place, in this very house, displaying them here after I'm gone." His eyes took on a far off expression, as if he were seeing into the future. "The Rathbourne Collection at Rathbourne House. It has a nice sound to it, don't you think?"
Nathanial looked as though he were about to say or do something embarrassing. Gabriella shot him a threatening glance.
"And I will need a curator." Lord Rathbourne's cool gaze met hers. "You, my dear."
She widened her eyes. "Surely you can't be serious."
"I am never less than serious."
"She's not qualified for something like that," Nathanial said quickly.
"She most certainly is," Lord Rathbourne said to him, but his gaze stayed on her. "She has spent years studying antiquities, ancient civilizations, history, languages—"
"How did you know that?" Gabriella drew her brows together.
"I made it my business to know." He shrugged. "It was not difficult information to uncover. You have not lived in secrecy. In addition, you are familiar with current finds and discoveries and research. While you are not well known, you have crossed paths through the years with scholars and collectors, museum directors and archeologists."
"But she's a—" Nathanial started, then obviously thought better of it.
"If you were going to say she's a woman, I commend you for your restraint in not pointing out the obvious. Indeed she is a woman." Lord Rathbourne's tone was deceptively casual. "Surely you are not suggesting her gender should preclude her from consideration for this position?"
Nathanial had the faintest look about him of a rat caught in a trap. Good. Perhaps he would have to chew his leg off to escape. "No sir," he said weakly. "Of course not."
"That's very kind of you, my lord, but—"
"Kindness has nothing to do with it, Miss Montini. I am rarely if ever kind. I find the idea of a beautiful and brilliant woman being the curator of my collection—the public face, as it were—of the Rathbourne Collection, to be a stroke of genius."
"Admittedly," she said choosing her words with care, "it is an intriguing—"
"Allow me to be blunt, if you will," Lord Rathbourne interrupted. "Your life to this point has been filled with your studies and the work you did for your brother. He is now dead. If you were a man, you would have worked by his side. Even a cursory examination of your life up to now would indicate that is something that surely has crossed your mind. An intelligent and imaginative creature such as yourself cannot immerse herself in the legacy of the past without some desire to see where it all began. To have a hand in its rebirth."
"My lord, I—"
"You might well have aspired to follow in your brother's footsteps, even though that aspiration was absurd given the restrictions of your gender. If such desires have indeed occurred to you, it is time to put them to rest and move forward." He leaned closer and his gaze trapped hers. "I am offering you an opportunity that will never come your way again, Miss Montini. The chance to create a prominent, if not preeminent, private museum. Unfettered by the meaningless tyrannies of a board of trustees who can't tell the difference between an exquisite twenty-four-hundred-year-old Greek amphora and a worthless flower vase. You would have unlimited funds at your disposal to acquire new pieces, to complete what I have only begun.
"Think of it, Miss Montini." His voice lowered in a seductive manner, and in the back of her mind she wondered if this indeed was what it felt like to be seduced by a man, not a boy. To feel your resistance fade with every word. To feel an ache, a longing build slowly but inevitably toward surrender. To know, even as you denied it, even as you knew it was a dreadful mistake, that you would succumb. "You will never be among those who search for treasure, but you could be the one who brings those treasures to the world. With your knowledge and my fortune, you and I together could—"
"But won't you be dead?" Nathanial blurted.
"Nathanial," Gabriella snapped.
"He said he wanted this to happen after his death." Nathanial shrugged. "Which would indicate to me there would be no 'together' in any way unless he plans to oversee this from beyond the grave."
"Quite right, Mr. Harrington," Lord Rathbourne said in a cold, clipped tone. "However, while I am increasingly aware of my own mortality, I do not intend to depart this earth in the immediate future." His gaze returned to Gabriella. "As much as I have implicit faith in leaving my acquisitions in Miss Montini's capable hands, I would like to begin planning my legacy while I am still able to do so. And I would like you to begin cataloguing my collections as soon as possible."
She thought for a moment. It was, as he had said, irresistible.
"Gabriella?" Nathanial said.
And what else was she to do with her life? She squared her shoulders. "How soon?"
Nathanial's brow furrowed. "Surely you're not thinking of accepting this offer?"
The older man ignored him. "You may begin as early as tomorrow if you wish, at least to get some idea of the immensity of the job. I do not expect your complete attention until you have resolved the matter of the missing seal."
Nathanial stared. "Gabriella—"
"That is most gracious of you, my lord." She narrowed her eyes thoughtfully. "I imagine the salary would be commensurate with the position?"
"Not at all, Miss Montini. I had planned on paying you an exorbitant amount, far more than any comparable position anywhere." Lord Rathbourne smiled. "I find nothing ensures loyalty better than overpaid employees. We can discuss specific terms now if you wish or later if you prefer."
"Another day would be fine, there is no particular hurry. Indeed, until the matter of my brother's seal has been put to rest, I would be reluctant to accept any monetary compensation. Let us consider anything I do until then to be no more than preparatory." Gabriella extended her hand. "I shall return tomorrow, then."
Lord Rathbourne took her hand in his. "I foresee nothing but success in this venture we are engaging upon, Miss Montini."
She had the distinct impression this was a man who would permit nothing but success. His touch still triggered distaste, but then he didn't want her in the manner in which other men might. He wanted her skill, her knowledge, her mind. Nonetheless, it would be an uneasy alliance. She certainly didn't trust him, and she suspected he didn't trust anyone.
"Good day, Miss Montini." He released her hand. "Mr. Harrington."
"Sir." Nathanial nodded and they took their leave.
They didn't say a word to one another in the carriage ride back to his house, which suited her perfectly. She didn't wish to hear his admonitions, his warnings, his arrogant insistence that he knew better than she how she should live her life. He had no right to do so. Besides, she fully intended never to see him again when the matter of her brother's seal was resolved. It would be better that way. For them both.
The moment they entered the house, he took her elbow and steered her toward the library. The line of his jaw was tight with tension.
She huffed. "Where are we going?"
His voice was low, barely under control, and she realized she might have pushed him just a touch too far. Not that it wasn't his own fault.
"Short of taking you to my rooms, which I would dearly love to do—"
"What? And turn me over your knee?"
"That too," he snapped. "The only place in this house that I can ensure privacy that would not be considered highly improper—"
"And we wouldn't want that."
"No, Gabriella, we wouldn't. I have the sensibilities of my mother and my sister to consider, as well as your reputation. Not that you seem to be giving it any thought whatsoever."
"When did you become so concerned with propriety?"
"When I met you." He kicked open the library door and fairly hauled her into the room.
Mr. Dennison jumped to his feet behind his desk. "Master Nathanial! Is something amiss?"
"You might say that, Dennison." Nathanial jerked his head toward the door. "Now, get out."
Gabriella folded her arms over her chest and glared.
Mr. Dennison's gaze skipped from Nathanial to Gabriella and back. "If there is something I can do to be of assistance—"
"I will call you." Nathanial blew a long breath. "My apologies for my rude behavior, but—"
"None necessary, sir." Mr. Dennison gathered up some papers on his desk, then quickly crossed the room to the door, casting a curious glance at Gabriella as he passed. She didn't doubt there would a note on its way to Florence within the hour. "I will be in the back parlor, sir, if you have need of me."
"If you will simply make sure we are not disturbed." Nathanial mustered a weak smile. "I would be most appreciative."
"Of course, sir." Dennison took his leave, closing the door firmly behind him.
Nathanial narrowed his eyes and stared at her in silence. One minute stretched to a second and then a third. She resisted the urge to stamp her foot on the floor.
"Well, go on. Say it."
"Say what?" He practically growled the words.
She shifted uneasily. "Whatever it is you have to say."
His eyes narrowed even more, if possible. "What makes you think I have anything to say?"
"Come now, Nathanial. You are very nearly about to explode." She sniffed. "Your restraint is not that good."
"My restraint." His voice rose. "My restraint?"
"Yes, your restraint," she said in a lofty manner, and started toward the door. Perhaps this was not a good time to talk about Lord Rathbourne or anything else. Besides, there wasn't anything he could say that hadn't already crossed her mind.
"Oh, no." He stepped directly in her path. "We are going to discuss this and we are going to discuss this now."
"Very well." She turned away from him to take one of the chairs in front of the desk, sitting pointedly with her back to him. And realized it was not a good idea. "If you're going to ask if I'm insane again—"
"Oh, I no longer think there's a question about your sanity."
"I wasn't mad the last time you asked and I daresay I am not mad now."
She heard him behind her, and without warning he grabbed the chair she was in and spun it around to face him. "I won't allow it."
"You have no say in the matter."
"As you are in my home—"
"And I needn't be! I have my own—I have elsewhere I can reside and do so precisely as I think best."
He ignored her. "Nonetheless, I promised to protect you and I cannot do so if you are in that house. Rathbourne is a dangerous man." He leaned closer and braced his hands on the arms of the chair. His eyes blazed with anger. Without thinking, she shrank back. "He wants to add you to his collections the same way he has added his wife."
She scoffed. "Don't be absurd."
"He wants you to be the beautiful and brilliant curator of his collections. As much an acquisition as his art and his artifacts."
"Even if you're right..." She pushed him aside and got to her feet. "...why shouldn't I do this? I am more than qualified. Lord Rathbourne said it and he was right. I have indeed been training for this very position most of my life. Why shouldn't I be the curator of his collections?"
"It's not something that a woman—"
"I am so tired of that argument!" She crossed her arms over her chest. "And what am I supposed to do? I cannot—no—I will not spend the rest of my days with my nose in a book gaining knowledge I will never put to any useful purpose. It's all very well and good for you to stand there and say I cannot do this and I cannot do that because I happened to be born female. You can do whatever you want simply because you're a man. So you tell me, Nathanial, with the benefit of your male wisdom, what am I to do with the rest of my life?"
"You could do what other women do." He stared at her as if she had indeed lost her mind. "Get married and have children."
"No," she said sharply. "I can't."
"That's right because you're not like other women!" He shook his head. "You certainly can't if you work for Rathbourne. Gabriella, your reputation will be ruined."
"I don't have a reputation."
"You will. Do you know what people will say?"
"Bloody hell," she said for the first time in years, the outburst as startling to her as it was to him. Regardless, she was past reason now. She raised her chin and lied. "They'll say what a clever, competent woman she is."
"They'll say you were bought and sold!" His tone was grim. "They'll say that you were as much an acquisition and are as much a possession as everything else in his collection. And such talk would inevitably include speculation about your personal duties with regard to Rathbourne."
She gasped. "There will be no personal duties!"
"No one will believe that!"
She shrugged. "I have never particularly cared what people have thought of me."
"I have always thought it absurd that we care so much about what other people think at all." Nathanial's gaze locked with hers. "And yet, despite what you say—which I don't believe, by the way—you will care."
"All right, then." Her voice rose. "I admit it! Yes, I know exactly what people will say. And yes, I understand it won't be especially pleasant. And yes, it will concern me and I will care!"
"Your reputation will be shattered!"
"No more so than—"
"You will be ruined!"
"I'm already ruined!" The words were out of her mouth before she realized what she'd said. "And that is why I will never marry."
He stared at her. "What do you mean by ruined?"
She widened her eyes in disbelief. "Surely you don't need me to explain? This is difficult enough as it is."
Realization dawned on his face and he paused. "How ruined?"
She choked. "I didn't know there were degrees!"
"Certainly there are degrees." He huffed. "Was it a single indiscretion or were you..."
"What? A whore in a brothel?" How could he possibly ask such a question? "Now who is insane? And furthermore it's none of your concern!"
"Of course it's my concern. I want to know how many men have come before me."
"Before you?" She scoffed. "There has been no you. Nor will there ever be!"
"I wouldn't wager on it!"
"Your confidence, Nathanial..." She strode to the door, yanked it open, and stepped through. "...is exceeded only by your arrogance." She slammed the door behind her.
And almost at once regretted it.
## Nineteen
She slammed the door in his face? How dare she? How could she?
Not that he didn't deserve it. His heart sank. Asking just how ruined she was might not have been the wisest thing to say.
But he'd never been in this situation before. He ran his hand through his hair. What in the name of all that was holy was a man supposed to say when the woman he loved, the woman he fully intended to marry—even if he hadn't quite accepted it yet himself or mentioned it to her—told him she'd shared someone else's bed? It wasn't the kind of thing a man wanted to hear. One wanted—no, expected—to be the only man to ever share the bed of the love of his life.
He should have said he didn't care.
Damnation! If he'd taken a minute to think, perhaps his brain would have come up with just that, or at least something considerably better than his mouth had. He should have said it didn't matter to him if there had been a hundred men or just one. That no matter what had happened in her life before him, it was of no significance. He should have said the only thing that mattered was here and now and forever after.
Blast it all, that's what he should have said. Well, he would say it now if it wasn't too late.
He took a step toward the door and it swung open.
Gabriella stepped into the room, closed the door behind her and pressed her back against it. "I am not used to running away in a cowardly manner. I find I don't like it." Resolve lit her eyes. "When I was fifteen, I met a boy not much older than I who, for want of a better word, seduced me. I was young and foolish. And that, Nathanial, is the degree of my ruin."
Relief washed through him, and guilt. "You didn't have to tell me."
She studied him. "I know."
"Why did you?"
She shrugged, "I suppose I didn't want you to think even worse of me than you do. Now..." She folded her arms over her chest. "...it's your turn."
"My turn for what?"
"I made any number of assumptions about you before we even met. I must confess that in most of those I have been proved wrong. However, I am fairly certain that you too are not a virgin."
He sucked in a sharp breath. "Gabriella!"
"I simply want to know what the degree is when it comes to your own fallen status."
He huffed. "Men do not fall!"
"I know, Nathanial." She heaved a resigned sigh. "And I consider it a great pity. Another example of the inequities in this world." She paused. "I am going to my room now. I have quite a lot to think about, what with the question"—she gestured with her left hand—"of Lord Rathbourne's offer of employment and"—she waved in his direction with her right—"you." With that, she nodded, turned, and swept from the room.
He stared after her. She didn't have to tell him about her past, and if he hadn't been such a fool, she wouldn't have felt compelled to do so. It was obviously something she didn't want to discuss. He'd always considered himself fairly successful with women. Probably not as successful as Gabriella imagined but successful nonetheless. But with her, his foot was lodged firmly and permanently in his mouth and he was a complete and total idiot.
And again he had missed the opportunity to tell her it didn't matter. He resisted the urge to smack his hand against his forehead and started after her.
Without warning the answer struck him and he pulled up short. She didn't have to tell him, but she had. She trusted him! She cared what he thought! He grinned. She couldn't live without him.
Thank God.
He hurried along the corridor to the main staircase. First he would tell her that he didn't care about anything that had happened before the moment they met. He started up the stairs. What she'd done or—he snorted to himself—who she'd pretended to be was of no consequence to him. He turned into the wing that housed their rooms. Then he'd tell her he loved her. He reached her door, was about to knock, instead grabbed the handle, flung it open, and said the first thing that came to mind.
"Did you love him?"
She stood near the window. Her eyes widened with indignation. "What are you doing here? You can't just come in here without my permission. And why didn't you knock?"
He started toward her. "I want to know if you loved him. This...this...boy. Not that it matters," he added quickly. "I simply want to know."
"Very well, then." She rolled her gaze toward the ceiling. "As I said, I was quite young. I knew nothing about love. I cannot say I know anything about love now, to be honest although I do know—" She shook her head and continued. "It was exciting and dangerous. As stupid as it sounds, I didn't even understand it was wrong. But I can say no, I was not in love with him. I've never been in love before. It was..." She thought for a moment. "Oh, I don't know, a first taste perhaps, of desire or passion."
"A first taste?" He moved closer.
She eyed him suspiciously. "What are you doing?"
"I'd like to discuss passion." He stepped toward her. "And desire."
A flicker of panic showed in her eyes. "I've never talked to anyone about this, Nathanial. Never. I don't know why I did so now."
"Because you trust me." He cast her a smug smile.
"Yes I suppose, but..." She shook her head. "Trust is a fragile thing that can be easily shattered." She stepped back. "I do hope you don't think you can now take advantage of me because I am—"
"I don't think that at all. I would never think such a thing." Indignation sounded in his voice, and before she could bolt he caught her and pulled her into his arms. "And I resent you thinking that I would. And I further think there is an excellent possibility that you will take advantage of me."
"Do you indeed?" She raised a brow. "And yet it is your arms that are around me."
"Convenient, isn't it?"
"If I intended to take advantage of you." She pushed against him in a token and ineffectual manner. "And I can't imagine why you would think such a thing is even possible."
"I don't." He gazed into her eyes and smiled slowly. "I simply hope."
She stared at him and heaved a sigh of surrender. "You are so annoying," she muttered, threw her arms around his neck and pressed her lips to his. He gathered her closer, slanted his lips across hers.
Her mouth opened to his, and his tongue met hers. She tasted exactly like she smelled, of spice and heat and all the things he'd ever loved. Desire welled within him and hunger deepened his kiss. And she responded in kind, sharing his hunger, his greed, his need.
Abruptly, she wrenched her lips from his. "This is a dreadful mistake, Nathanial."
"And yet it seems so right," he murmured, his lips trailing along the side of her neck. "We are made one for the other, Gabriella. I cannot imagine why it would be a mistake, but do feel free to tell me."
"Because I have grown to like you, and yes, to trust you and possibly..." She sighed, and he nuzzled that lovely spot where neck met shoulder. "I am not a fool, Nathanial. Anything more between us and I shall surely lose my heart. And you will most certainly break it."
"Nonsense." She could scoff all she wanted but she did indeed smell like a summer day. And he'd always loved summer.
"There isn't a doubt in my mind." She pushed out of his arms and moved away. "I have never experienced heartbreak, except the kind one feels at the death of a brother. I never knew my mother. I was too young at the death of my father to understand the depth of my loss. I have always thought of myself as a person of strength, but this, you..." She waved at him. "I find it terrifying and therefore best avoided. I suspect my heart is a fragile thing."
"I would never hurt you." He reached for her but she stepped out of the way.
"You would never mean to hurt me, but you would. It's inevitable."
"I don't believe that."
"What you believe isn't nearly as important as what I know about...the way the world works, if you will. There is no future for us." She raised her chin. "Now, please leave."
"Do you really want me to go?"
"Yes, I do." She waved at the door. "Go. Now. Please."
"Very well." He studied her for a long moment. "But this isn't over between us, Gabriella."
"Of course not." She brushed an escaped tendril of hair away from her face. "We still need to find the seal."
"In addition," he said firmly, "there is much that remains unsettled. The question of Lord Rathbourne's offer has not been resolved—"
"Oh, I do think—"
"Nor has the matter of your future."
"I daresay, my future has nothing to do with—"
His gaze met hers directly. "I have a great deal I need to say to you, and I'm not sure this is the best time. You're not especially rational at the moment—"
She huffed. "I am unfailingly rational."
"Yes, that is yet another thing I love about you." He grinned and left the room, leaving her staring after him with something that might have been confusion or apprehension or...hope.
He still hadn't told her he didn't care about her past. He'd probably care even less when he knew all there was to know. But right now he knew the only thing that really mattered.
She'd never been in love before. Before. What a glorious word. His grin widened.
Until now.
He strode down the hall with a swagger in his step he would have considered obnoxious in another man. Certainly there were problems to overcome beyond the question of the seal. First and foremost was that nonsense about Rathbourne and his ridiculous position. Still, given her education and intelligence, he could see why she might find it appealing. And then there was the question of money. With the death of her brother, it was obvious to him she didn't have any. Aside from that apricot gown of hers, the rest of her clothes, though well cared for, were decidedly worn. Rathbourne's offer would tempt even the most financially sound.
Nate passed Quint's door and heard sounds of occupation inside. He clenched his jaw. Good. His brother was back. It was past time he and Quint had a long talk, not that he thought Quint knew anything about the missing seal. Still, the queasy feeling that had settled in the pit of his stomach with McGowan's disclosures continued to linger.
Nate rapped sharply on the door. A faint voice sounded from the other side. He pushed opened the door and didn't see his brother in the sitting room that opened on one side of his bed chamber, a mirror image of his own rooms.
"Quint?"
"Here." His brother sauntered out of the adjoining dressing room, half dressed, drying his face with a towel.
Nate raised a brow. "Just now shaving for the day?"
Quint grinned.
Nate studied him. Quint was the only person he'd ever met who could spend two days drinking and whoring and who knew what else, and look more refreshed than tired.
"We have to talk," Nate said firmly.
"Do we?" Quint tossed the towel onto a chair. "I don't like the sound of that." He moved to the wardrobe and perused its contents. "What do you wish to talk about, little brother?"
"Miss Montini—"
"Ah yes, the delectable Miss Montini." He selected a shirt, moved to the cheval mirror, and pulled it on. Both brothers had long ago dispensed with the services of valets, even in London. "Have you kissed her yet?"
"That's neither here nor there and none of your concern."
Quint caught his brother's eye in the mirror and raised a brow.
"Once or twice perhaps," Nate muttered.
"And just a few minutes ago, no doubt."
Nate narrowed his eyes suspiciously. "Why would you say that?"
"I heard you go into her room." Quint chuckled. "It has been my experience that when there is a discussion between a man and a woman that involves screaming at the top of their lungs and then silence, it means either they have killed one another or fallen into each other's arms. You don't appear to be dead."
"No, well..." Nate grinned in a sheepish manner, then paused. "We weren't screaming."
Quint grinned.
"So you couldn't have heard us."
Quint's grin widened.
"Yes, I kissed her," Nate said, his tone harder than he had intended.
"And you're in love with her."
For a moment he considered denying it, but to what purpose? He drew a deep breath. "Yes, I am."
Quint chuckled. "I knew she would do for you."
"I've never met anyone like her."
"There's never been anyone like her."
Nate ignored him. "She is equal parts intelligence and foolishness, honesty and secrets. From the moment I first saw her at Reggie's ball," he shook his head, "I have not been able to get her out of my mind."
Quint raised a brow. "So this is a permanent state?"
"Yes it is." Nate nodded. "Although convincing her of that..."
"How difficult can it be? You say, 'I love you, marry me, and I shall spend the rest of my days doing everything in my power to make you blissfully happy.'"
Nate shook his head. "It's not that easy."
"Have you tried it?"
"Admittedly, I haven't—"
"Then you should."
Nate eyed his older brother. "Should I be taking advice from a man who has never uttered such words himself?"
"The fact that I haven't doesn't mean I don't know the proper way to go about it. Besides, she already knows how you feel."
"How could she—"
"Everyone in the house knows how you feel."
"Still." Nate shook his head. "I don't know that it matters."
"Make it matter." Quint rolled his gaze at the ceiling. "It's obvious she feels the same way about you."
Nate grinned. "I hope so." He blew a long breath. "She has come to trust me to a certain extent but not completely. She has secrets. There is much she has not told me."
Quint shrugged. "We all have secrets."
"Yes, we do." He studied his brother carefully, then drew a deep breath. "Did you steal her brother's seal?"
Quint met his gaze directly. "No."
"Very well," Nate said slowly. "Let me rephrase that. Do you have Montini's seal?"
Quint paused for a long moment. "Not on me."
"Did you win it from Javier Gutierrez in a game of chance in Crete?"
Quint's eyes narrowed. "Lucky guess, brother?"
Nate grimaced. "Unlucky, I would say."
Quint heaved a sigh of surrender. "Yes, I wrested the seal away from Gutierrez." He snorted in disgust. "The man is an idiot."
"The man is dangerous."
"So am I when necessary." He waved at a nearby chair. "You might as well sit down. It's a long story."
Nate sat in the indicated seat and stared at his brother. "Then you should begin."
"Very well." He thought for a moment. "First you should know it's not Montini's seal."
"Oh?" Nate raised a brow.
"Years ago, when I was working with Professor Ashworth, he purchased a crate in Athens of...well, mostly trash. Bits and pieces of pottery, marbles, ancient tools, that sort of thing. There were several cylinder seals in the crate as well. One caught my attention." He met his brother's gaze. "It looked Akkadian and was carved from greenstone."
Nate held his breath. "And?"
"And, from a cursory examination it appeared to have the symbols for Ambropia and the Virgin's Secret. However," he clenched his jaw, "to my eternal regret, I put it back in the crate intending to study it further at a later time. I never saw it again. The crate was stolen."
"What did the professor say?"
"I didn't tell him." Quint shook his head in disgust. "It was what he'd spent much of his life looking for. I wanted to surprise him with it. I was such a fool. I never should have let it out of my sight."
"You think Montini stole it?"
"No, although I wouldn't have put it past him. Besides, if he had stolen it then, it wouldn't have taken him years to announce his find. I have no idea who took it originally, nor how many hands it might have passed through before it came into Montini's possession." He shook his head. "But I knew it was the seal I'd seen—had in my hands—the moment I saw Montini's impression."
"And?" Nate prompted.
"And." His gaze met Nate's without so much as a glimmer of remorse. "And I had every intention of stealing it from him."
Nate drew his brows together. "But you didn't?"
"No." Quint blew a long breath. "I was about to, but Gutierrez did so before I could."
"You saw Gutierrez take the seal?"
Quint chuckled. "He didn't see me but I was practically right behind him. It was no secret how superstitious Montini was about his finds. I knew, I'm assuming Gutierrez knew as well. There was every reason to believe that Montini wouldn't unwrap the cloth around the seal until it came time to present it to the Verification Committee. Which meant he wouldn't discover the theft until then."
"We think Gutierrez might have been in the employ of Lord Rathbourne," Nate said. "His lordship admitted to Gabriella that he had tried to acquire the seal." He thought for a moment. "But why didn't Gutierrez bring the seal to Rathbourne as soon as he had it?"
Quint shook his head. "Who knows why a man like Gutierrez does what he does? Besides, that would have meant a trip to London, and I wouldn't be surprised if Rathbourne isn't the only one employing Gutierrez for less than legitimate purposes. But it worked in my favor. I watched Gutierrez carefully, waiting for the opportunity to take the seal." He glanced at his brother. "Didn't you wonder at the somewhat meandering path our travels have taken this past year?"
"Not really." Nate grimaced. "It didn't seem out of the ordinary at the time."
"At any rate, my opportunity arose in Crete. From my observations of Gutierrez, I knew he was an insatiable gambler and had as well a taste for drink. He is the kind of man who does not know his own limits when it comes to spirits and thinks he is in control of his faculties when he is not." Quint shrugged. "It was remarkably easy to get him inebriated, engage him in cards, and win the seal from him." He chuckled. "I think it took him a few days to realize what he had lost."
"I heard he was furious."
"Yes, I suppose he was." He paused. "I understand Montini had an encounter with him in Crete as well. Shortly after that I heard Montini had been killed, his throat slashed."
Nate stared. "You never mentioned that."
"You didn't want me to talk about the kind of man Montini was. I assumed you wouldn't want me mentioning how I had heard he'd died either."
"You could have told me. Although it scarcely matters now, I suppose. So..." Nate chose his words with care. "Where is the seal?"
Quint hesitated for a long moment, then sighed. "It's in the attic."
Nate got it his feet. "Let's go, then."
"May I at least finish getting dressed?" Quint tucked his shirt into his trousers.
"No." Nate started toward the door.
"Mother will not like it if she sees me without a coat," Quint warned.
"Then we'll take care that she doesn't see you."
Nate led the way up the stairs to the servants' quarters and to the final flight of stairs to the attic, Quint a few steps behind him.
"What did you intend to do with the seal? Try and find Ambropia?"
"No," Quint said in a tone that indicated that was all he would say on the subject.
Nate opened the attic door and turned to his brother. "Well?"
Quint brushed past him. "I put it in the trunk with Great-grandmother's things." He chuckled. "It seemed appropriate."
Nate trailed after him. His heart sped up with excitement. He'd give the seal to Gabriella, she'd give it to the Antiquities Society, and he could then proceed to convince her that her future from this point on was with him. And why not? After all, he'd be her hero.
"I hid it up here the day I got home."
"The day of Reggie's ball?"
Quint nodded. He skirted around the leavings of generations of Harringtons, furniture discarded in favor of something more in style, only to be pushed farther back into the recesses of the massive attic when its replacements were in turn discarded. Paintings were stacked against the walls, trunks and crates and boxes hindered their path. He reached the trunk in question in very nearly the same spot it had occupied since they'd last played up here as children. He flipped open the lid, bent down, and fished around inside it.
"Here it is." Quint pulled out a small, cloth-wrapped bundle tied with a string. He pulled off the string, unwrapped the seal and stared. A moment later his wry laughter rang through the attic.
Nate stared at him. "What do you find so amusing?"
"Irony, dear brother. The world is full of irony. And jokes perpetrated by a god far more whimsical than I. This." Quint thrust the bundle at him and grinned. "This is not Montini's seal."
## Twenty
What do you mean this is not Montini's seal?" Disbelief twisted Nate's stomach.
"Here. See for yourself." Quint handed the seal to him. "It's not greenstone."
Nate held it up to the faint light from the far off attic windows. "It's chalcedony. And it looks..." His heart sank. "Late Assyrian."
"The one we're looking for, the one I took from Gutierrez, was Akkadian, a mere fifteen hundred year difference."
Nate glared at his brother. "How could this happen?"
"How would I know?" Quint snapped.
"The seal was in your possession!" Nate narrowed his eyes. "Wasn't it?"
"It was when I took it from Gutierrez! I examined it thoroughly. Bloody hell." Quint stalked back and forth across the attic. "I've done it again. I had it in my hands! And I lost it! Again! How could I have been so stupid? How could I have—"
"How did you lose it?"
Quint stopped in mid-step and glared at his brother as though he was an idiot. "I didn't lose it."
"You just said—"
"I didn't lose it! Somebody must have stolen it from me and substituted that one. The same way Gutierrez stole it from Montini. And someone else stole it from Ashcroft, and God knows how many other people stole it from who knows how many other people through the years. Through the centuries!" He gritted his teeth. "It's the curse, that's what it is."
Nate shook his head. "The seal isn't cursed."
"No, but the city is, remember? The Curse of the Virgin's Secret? 'He whosoever disturbs the sleep of the Virgin's city,'" he gestured in a wild manner, "and so on and so forth?"
Nate scoffed. "You don't believe in that."
"I'd rather believe in a curse than my own stupidity." Quint sank down on top of a crate, rubbed his forehead and muttered more to himself than to his brother. "But I cannot believe this. I had it in my hands."
Nate stared. His brother's distress was not only foreign to his nature but struck Nate as out of all proportion to the loss. They'd lost far greater treasures before. There was more here than Quint had admitted thus far. Damnation, did everyone around here have secrets but him?
Quint's brow furrowed in thought. "I examined the seal after I won it, then I wrapped it back up and tucked it in my bag." He got to his feet and paced. "Every now and then I would check to make certain it was still there, but I never unwrapped it again."
"Could Gutierrez have taken it?"
"It's a definite possibility." Quint paused and looked pointedly at his brother. "As is Montini." He blew a long breath. "As are any number of other people."
"It doesn't make any sense." Nate shook his head. "If Montini had it, he would have mentioned it in his letters. Probably his last, given you had the seal not long before his death."
"One would think."
"His last letter..." Nate thought for a moment. "While not incoherent, it was little more than ramblings."
"If Montini had it, wouldn't he have wanted to let his sister know?"
"The more I learn of Montini, the more I hesitate to assume anything he might have done. It is a possibility, though. You said he was in Crete when you were."
Quint nodded.
"Then if indeed Montini was the one who switched the seals..." Nate blew a long breath. "Where is the seal now?"
Gabriella stared at the door that had closed behind Nathanial and finally sank down on the bed. Nathanial Harrington was a constant surprise.
She'd never imagined that any man, upon hearing of her ruined state, would act as if it was of no real consequence. Surely he must care somewhat, as evidenced by his stupid question about the degree of her ruin. Regardless, it was probably no more than curiosity on his part. It wasn't as if he planned to marry her. He'd said her future was still unresolved and hadn't corrected her when she told him they had no future together. Of course not. The brothers of earls did not marry the ruined sisters of treasure hunters. It might well happen that way in fairy tales, but life was a far different matter.
She'd fully intended to disappear from his life when their search had ended. To perhaps travel. But now Rathbourne had presented her with another possibility. And disappearing from Nathanial's life didn't mean she had to leave London. Besides, he would no doubt soon be off on his travels with his brother, back to those parts of the world where man had lived eons ago and left remnants of those lives behind. Their own lives would take different directions, and it was entirely likely they would never cross paths again. An awful ache welled up inside her at the thought.
How different her life might have been. As pointless as it was to consider it now, her thoughts couldn't help but stray to her mother and the family she had never known. She got to her feet and paced the room. It was too late. One couldn't go back and start one's life anew. She was who and what she was and nothing could change that.
Despite what she'd said to Rathbourne and Nathanial, she wasn't nearly as confident about the wisdom of accepting his lordship's offer as she knew she sounded. There wasn't anything Nathanial had said that hadn't already crossed her mind. If one wished to speak in terms of degrees of ruin, being in Rathbourne's employ would certainly put her past any hope of true respectability, even though the position would increase her stature in the rarified world of antiquities. She would be what no other woman had dared to be before. And she would pay the price. But it would give her life purpose, and put to some practical use all the knowledge she had worked so hard to acquire. And if, with the passage of the years, her heart became no different than the relics in Rathbourne's collections, brittle and fragile and ancient, who would notice? Who would care?
It was a distressing thought, and she brushed it from her mind. Time enough to deal with the rest of her life later. She and Nathanial still had the seal to find. She'd known from the start that recovering it would be difficult if not impossible, even if she hadn't wished to accept it. But perhaps it was time to face the truth. They'd found no significant information as to who might have taken the seal or where it might be now. Tomorrow she would have complete access to Rathbourne's collections and could verify for herself his claim that he didn't have it. She blew a long breath. Other than that, she was simply out of ideas.
Still, admitting defeat, giving up the search, would mean it was time as well to give up Nathanial. She knew that day was fast approaching. Somehow, she knew as well this longing inside her—this sense of inevitability when she looked into his eyes, the feeling of perfection when he held her in his arms—surely this was love. If not for a youthful indiscretion...
No, it wasn't just her ruin that kept them apart. They were from different worlds and nothing could change that. She moved to the window and gazed unseeing over the street below. Once, she had been seduced by a boy, a cursory, not especially pleasant experience that was over very nearly as soon as it had begun. Today, Rathbourne's words and promises had seduced her as well. Now, she wanted yet another seduction.
If she was to live the rest of her life alone, it seemed a shame not to have at least one memory to sustain her through the long years ahead. One night of passion and desire and lying in the arms of the man she loved. One night of imagining it was not the end but merely the beginning.
One night would never be enough. But if she couldn't have Nathanial forever, one night would have to do.
Dinner had been an odd affair. Everyone seemed preoccupied by their own thoughts. Quint was obviously annoyed at himself for losing the seal. Gabriella had been even more reticent than usual; probably Rathbourne's offer dwelled on her mind. Sterling was always somewhat reserved. Mother continued to study Gabriella in a thoughtful manner. And he didn't know what his next step should be. He was certain if he did not tread warily, he would surely lose whatever hope he might have of winning Gabriella's heart and her hand. If not for Reggie's constant and excited chatter about the season and who she had met and the endless social activities ahead, dinner would have been a dismal and disquieting affair.
Now, Nate paced his room, his dressing gown thrown over his nightclothes, a glass of brandy in his hand. Sterling had insisted that Quint accompany him and Reggie to a musicale that had sounded rather dreadful to everyone but his sister. Mother said she had correspondence to attend to and retired to her rooms. Gabriella refused to meet his gaze throughout dinner and then retreated to her room, claiming it had been an eventful day and she was tired. He had fully intended to retire as well but found himself too restless to even attempt to sleep.
He hadn't had a chance to speak privately with Gabriella, to tell her what he'd learned from Quint. In truth, he wasn't sure exactly what he would say. He now knew who had taken the seal from her brother, and knew as well that it had been in his own brother's possession. But where the seal was now—that was still an unanswered question. A question that might never be answered.
And that wasn't the only question. He took a sip of the brandy. There was still much he didn't know about Gabriella. He was fairly certain now that she was indeed the brother he had met in Egypt. He was a fool not to have realized it sooner. And he was certain as well that "John," the footman, was in truth Xerxes Muldoon. But since the Antiquities Society Ball he'd had no opportunity to question the big man. No, he'd been too busy trying to convince Gabriella of the folly of working for Lord Rathbourne. Or too busy getting the truth from his brother.
Beyond that, he still hadn't told her of his feelings, and in that too he wasn't sure what to say. Did one just blurt out declarations of love? Proposals of marriage? In the back of his mind he had the most awful conviction that if he didn't say something soon, it would be too late.
He blew a long breath. He'd never considered himself a coward, but the fear of not having her by his side for the rest of his days lay like a heavy weight in the pit of his stomach. No, it was better to put off saying anything at all than to run the risk that she didn't share his feelings. Still, Gabriella had said that she'd never been in love before. Which did seem to indicate she was in love now. He clung to the word like a shipwrecked sailor hanging onto a floating spar.
This was absurd, he told himself. All that was unresolved between them was driving him mad. He downed the last of his drink and set the glass beside a decanter. It needed to be settled and it needed to be settled now.
He stalked to his door and flung it open.
To his surprise, Gabriella stood in the corridor, a wrapper worn over her nightclothes, eyes wide, hand poised to knock.
## Twenty-one
What are you doing here?" Nathanial asked sharply.
At once her confidence faltered. She squared her shoulders. "I wish to..." Wish to what? Experience true seduction by a man? By him? "Talk to you."
"You do?" His gaze slid over her. "Dressed like that?"
She ignored him. "You said today there was much that remained unsettled. I am here to...to settle it."
"It?"
"It. Everything." She pulled her brows together. "Are you always this obtuse or is there something about me that encourages you to be annoying?"
"There's something about you." He chuckled. "Did you wish to come in?"
"No," she said in a sharper tone than she'd intended. "I wish to stand here in the corridor."
"It would be most improper, you know, for you to be in my rooms, especially dressed as you are. Most of my family is gone, Mother's rooms are in the other wing, and the servants are all abed. We shall be quite alone."
"I am aware of that."
"Very well, then." He stepped aside and waved her in. "Do come in."
She stepped into the room. He closed the door behind her with a snap and she jumped.
"Are you nervous?"
She scoffed, belying the churning in her stomach. "Not at all."
His quarters consisted of a small sitting room with a writing desk and two comfortable looking wing chairs that flanked a love seat, the chairs positioned before a fireplace. An archway off to one side revealed a bedroom and a bed of massive proportions. Obviously Jacobean, it was dark and heavy and masculine and, she swallowed hard, most appropriate for seduction.
"Would you care to sit down?" He waved at the love seat.
"I would prefer to stand at the moment." She clasped her hands together and drew a deep breath. "First of all, you should know, I have no intention of marrying you."
He grinned. "I don't recall asking you."
"I realize that, but should you decide, out of some misplaced sense of honor, to do so, you should know the answer would be no."
"Thank you for settling that." He studied her curiously. "Why exactly are you here?"
"I have been giving...that is, I have been thinking..."
He shook his head. "Oh, that's never good. Please don't tell me you have a plan."
She glared at him. "You are making this most difficult."
"Making what most difficult?"
"This." Without thinking, she waved at the bed in the next room. "All of this."
"All of what?"
"I want you to seduce me," she blurted, then winced. She hadn't planned to simply announce it, but then her plans never seemed to go well anyway.
His eyes widened. "Do you?"
"Yes, I do." She paused. "You seem surprised."
"I suppose I shouldn't be, given your state of undress, but yes, I am surprised."
She shrugged. "I wasn't sure what to wear to a seduction. I've never been seduced by..."
"A man?"
"Yes, a man before. It seemed to me the fewer clothes I had on, the more...expedient it would be."
"Yes, well, expediency is always a consideration in seduction." He snorted back a laugh.
"Do you think this is amusing?"
"No." He smiled. "I think it's delightful."
"Excellent. You should probably begin by kissing me." She raised her chin and closed her eyes. And waited. And waited. She opened her eyes. "Well?"
He grinned. "Well what?"
Heat rushed up her face and she started for the door. "If you don't wish to—"
He moved to block her way. "There is nothing, Gabriella, absolutely nothing, I wish to do more."
"Well, then." Again she raised her chin and closed her eyes. And again waited.
"Open your eyes." He sighed. "I am not about to seduce a woman who looks like she's bravely going to the gallows."
"My apologies. I don't intend to look like I'm going to the gallows, bravely or otherwise. I am, in truth..." She thought for a moment. "Eager—yes, that's what I am. Perhaps we should..." She took a step toward the bedroom.
"Good Lord, Gabriella, I am not going to throw you on the bed and have my way with you." He cast her a disgusted look, crossed the floor to a decanter, filled a glass and brought it to her. "Here, drink this."
She took the glass and eyed it with suspicion. "What is it?"
"It's an ancient elixir that will cause you to fling yourself at my feet, begging for my very touch."
"It's seems to me I have very nearly done that already," she murmured, and took a long sip, the liquor burning a path down her throat, washing her in an immediate sensation of warmth. "It's brandy."
"Disappointed?"
"No, I like brandy. I don't have it very often but I do like it." She took another sip. "It's quite tasty and very warming."
"It's also very potent." He plucked the glass from her hand and eyed it. "And already half gone."
"Then you should refill it." She did feel more relaxed. And she hadn't lied to him, in spite of some apprehension, which was to be expected after all—she was indeed eager.
He refilled the glass and handed it to her. "Only one more sip. I don't want you inebriated."
"Really?" She took another long sip and gave him back the glass. "Why not?"
"Because while it might be 'expedient,' it won't be nearly as much fun for either of us if one of us isn't aware of what is happening."
"Oh." She drew her brows together. "That makes sense. I do wish to be aware." She met his gaze firmly. "I fully intend for the memories of tonight to sustain me through the rest of my life."
He swirled the brandy in his glass and studied her, his expression unreadable. "I do hope it lives up to expectations."
"I can't imagine otherwise." She smiled. "Seduce me, Nathanial."
"Oh, I have every intention of doing so. But seduction, my dear Gabriella, is an art. And like any art, cannot be rushed."
"Are you an artist, then?"
"Tonight I am," he murmured, and circled around behind her. She held her breath. The lightest kiss fluttered against the back of her neck, and she sucked in a sharp breath. His voice was low and enticing. "There are especially sensitive places on a women that I suspect most women are not even aware of."
"Are there?" she said weakly.
"There are indeed. The back of the neck." Again she felt the barest whisper of a kiss. His arm curled around her and he pulled the tie of her wrapper free, then pushed the garment off her shoulders and it fell to the floor. His hands rested on her shoulders. "The curve where neck meets shoulder." His lips caressed that very spot, and she tilted her head to one side and sighed.
"Oh my..." Her eyes closed. He gently turned her to face him.
"The base of the throat." His lips murmured against her throat, the sensation delightful and thrilling. She felt him fumble with the ties that closed her gown, and a drift of cooler air when it opened. He slid the gown off her shoulders and down her arms until it too joined the wrapper on the floor. And she stood completely naked in front of him.
She opened her eyes. He framed her face with his hands, and his lips met hers in a kiss soft and gentle. A prelude. She opened her mouth to his, and his tongue traced the edges of her lips, slowly and deliberately. She heard a faint moan and realized it had come from her.
She rested her hands on his chest, feeling the heat of his body through the silk of his dressing gown. His hands released her face but his mouth lingered on hers. He skimmed his hands lightly down her arms, then wrapped his arms around her and pulled her closer.
His mouth slanted over hers, demanding now, insistent. Heat gathered in her midsection and lower. She responded to his kiss with a heretofore unknown need, an urgency that seemed to come from somewhere deep inside her. Her tongue dueled with his and she reveled in the taste of him, of brandy and desire and everything she'd ever wanted.
One hand splayed across the small of her back, the other drifted to caress her derriere, and she pressed herself closer to him. The evidence of his arousal was hard against her, and without conscious thought she ground her hips against him.
He groaned. "Good God, Gabriella."
"Nathanial." She fairly sighed his name. "I want..."
"I know, my love." He scooped her into her arms, carried her to the bed, and laid her gently down. As if she were a fragile, delicate treasure.
She watched him quickly strip off his clothes through eyes half closed with desire. Her gaze drifted from broad shoulders to firmly muscled chest and lower, over his taut stomach to his erection, swollen and enormous. Heat flooded her face and she thought her blush odd, since she wasn't the least bit embarrassed at seeing him naked. Or at him seeing her naked. Nothing mattered but the longing for his body to join with hers.
Nathanial lay down beside her, kissed the base of her throat and trailed kisses lower to a point between her breasts. His hand cupped one breast and his fingers grazed her nipple. She gasped with unexpected pleasure. His mouth moved to her other breast and he ran his tongue in a circle around her nipple until her breath came faster. He took her nipple carefully in his mouth and sucked, his hand still toying with her other breast until she writhed with the sheer pleasure of the sensations he produced with his teeth and his tongue and his hands. Moisture gathered between her legs and she shifted her hips upward wanting...more.
"Oh...yes..."
He shifted his mouth to her other breast and slowly ran the tips of his fingers over her stomach in easy circles. His hand drifted lower, the touch of his fingers igniting a blaze deep inside her. He caressed the tops of her thighs, then slipped his hand between her legs, and they fell open as if of their own accord.
He touched her then, in that most intimate spot, and pleasure she'd never suspected shot through her. And she moaned and wanted more. Much more.
"You are so beautiful," he said softly.
He slid his fingers over her, over that point of exquisite pleasure. And toyed with her and teased until her hands clutched at the bedclothes and she squirmed with unbridled need.
Her hips arched upward and she moaned. "Please."
She wondered if she'd lost her mind and thought if this was madness, what a glorious thing it was. She scarcely noticed when his mouth left her breast and followed the path taken by his hand, lower across her stomach and lower still. Until his fingers parted her and she felt the touch of his tongue.
"Dear Lord." She gasped under his touch. "Nathanial." She could barely choke out his name. "You can't. You shouldn't. You—"
"Seduction, my sweet..." His raised his head and looked at her, his eyes dark with passion. "...is a complicated art."
His head nestled back between her legs and his tongue continued what his fingers had begun. And she was lost. Any reluctance to allow what was surely most depraved was dashed aside by never imagined delight. She knew only the pleasure, the sweet sensation of his touch. And it was not enough. A yearning ache built inside her as if she were struggling toward something just out of reach. She needed...she wanted...
Abruptly, Nathanial moved, and she gasped. Surely this wasn't all he intended? She knew full well there was more to seduction. Although, thus far it had been so much more than she'd ever dreamed. Still, surely it couldn't end like this with her needing, wanting, something more.
He positioned himself between her knees and she bit back a sob of relief. She lifted her hips up to meet him, a whimper of need escaping from her lips. His arousal nudged her and she tensed. She wanted this...no, she needed this. Still, she'd only done this that once and it was so long ago. She couldn't recall the boy's erection but certainly it was far smaller and insignificant than this massive implement that was about to impale her.
He nuzzled the side of her neck. "Gabriella, my love."
He eased himself into her. It was an odd sensation, She didn't remember this feeling at all. He entered her slowly and she felt tight around him and stretched and filled. He lay quietly for a long moment then began to move, with an easy, measured stroke. And the strangeness of it faded to a new and unique pleasure. She moved with him and her body throbbed with his.
She clutched at his shoulders and urged him on. His thrusts came deeper, faster, harder, and she wanted more. Her legs wrapped around his. Tension, odd and insistent and unyielding, coiled within her. With every stroke of his, her need built, until she wondered if she might burst with the intensity of the growing pressure. And longed for it. Her hips rose to meet him, welcome him, take him.
Without warning her body exploded around his. She arched upward and dug her fingers into his flesh. Wave after wave of unexpected pleasure coursed through her. Vaguely, through a fog of sensation, she heard someone cry out and realized it was her. He gripped her tighter and pumped faster. With a groan, he thrust deeply into her and shuddered. Warmth filled her and he moaned against her ear and stilled.
For a long moment they lay together, still joined, legs entwined, hearts beating in tandem. It was quite the most extraordinary thing she'd ever experienced. The intimacy of his body joining with hers, the spiraling desire, the joy of release. And now, still together as one, her breathing in rhythm with his, utter peace and complete contentment enveloped her. She could stay like this, with him, forever.
At last he shifted and pulled out of her. A sense of loss gripped her, but he drew her close against him, as if never to let her go. She reveled in a satisfaction she'd never known, never expected to know. She rested her head on his chest, his hand stroking her hair.
"So." A smile sounded in his voice. "Was your seduction acceptable?"
"Yes, Nathanial, it was acceptable." She smiled. "More than acceptable." She sighed against him. "It was delightful."
Indeed, she'd never felt quite so delighted in her life. And cherished. And even, at least for the moment, loved. That, no doubt, was an illusion brought on by intimacy. Still, even an illusion could be enjoyed, as long as one accepted that it was nothing more than fantasy.
"Now you are truly ruined, you know."
"And I wasn't truly ruined before?"
He chuckled. "Degrees, Gabriella, it's all a question of degrees."
"Well, then I suppose I am." She snuggled against him. Against the solid warmth of him. There was something about laying pressed against a naked man who had just made you feel things you'd never suspected possible that was unlike anything she'd ever imagined. There was obviously something to be said for being truly, gloriously ruined. "I certainly feel truly ruined."
"Gabriella..." His voice was deceptively causal. "About this offer of Rathbourne's..."
At once the feeling of contentment vanished. She pulled away from him, sat up and clutched the covers around her. "Yes?"
He sat up, resting his back against the pillows. His gaze met hers directly. "I think it's a mistake."
"I think it's an opportunity," she said slowly. "And aside from other considerations, I now have the chance to make certain Rathbourne does not have the seal."
"I don't care. He's a dangerous man."
"I'm not concerned."
"That in itself is a problem." His eyes narrowed. "You should be concerned. And wary."
She studied him carefully. "Do you now plan to forbid me to accept his offer? Again?"
"If I have to." His voice was grim.
"I thought we had established that you had no right to forbid me to do anything."
"This," he gestured in an angry manner, "gives me the right."
"This," she mimicked his gesture, "doesn't give you any rights whatsoever. I was already ruined, remember?"
"I can scarcely forget," he snapped.
Of course he couldn't forget, what man could? "This discussion is at an end," she said in a cold manner. "I need to return to my room. My clothes—"
"I'll get them," he said through gritted teeth, threw off the covers, slid out of bed, and strode into the sitting room.
"You're naked!" She clapped her hands over her eyes, the annoying feeling of heat again flushing her face.
"I was just as naked a minute ago," he muttered from the sitting room. "It didn't seem to bother you then."
"A minute ago it didn't!"
"You can uncover your eyes." He had slipped on his dressing gown. "Here." He dropped her nightclothes on the bed.
"Turn around."
"Whatever you want." He turned around. "Because everything is apparently about whatever you want."
"That's absurd." She scrambled out of bed and quickly slipped on her gown. "I have no idea what you are talking about."
"You want to find the seal! You want to restore your brother's name! You want to work for Rathbourne! And tonight," his voice hardened, "you wanted me."
"I apologize if it was an inconvenience for you." She pulled on her wrapper and tied it. "I shall not bother you again." She started for the door.
"Damnation, Gabriella, do not put words in my mouth." He grabbed her and pulled her into his arms. "It wasn't an inconvenience and only you would say something like that."
She struggled against him but he held her tight.
"I have wanted you from the first moment I saw you. You should know that if you don't already. But I want more. And one of the things I want is to keep you safe."
"I can't imagine any harm will come to—"
"I can!" he said sharply. "I can imagine all sorts of harm that might come to you." He drew a deep breath. "If you insist on going to Rathbourne's, I would insist—"
She narrowed her eyes. "Insist?"
"Fine, a request, then."
"And if I don't agree, will you threaten to have me arrested again?"
"Perhaps! Or maybe I'll just tie you to a chair and keep you right here until you learn some sense!"
"Oh, you'd like that wouldn't you?"
A hint of wicked amusement glinted in his brown eyes, and the most inappropriate frisson of desire shot through her. She ignored it. "Probably." He shook his head. "I don't want you to go to that house alone."
"You're not—"
"No, I'm not. I would like you to take one of the footman along." He studied her intently. "The new one, I think. John Farrell."
Xerxes? She nodded. "Very well."
He raised a brow. "What? No argument?"
"It seems a reasonable enough request. Now..." She pushed against him. "Are you going to release me?"
"For the moment." He gazed into her eyes and her breath caught. "But only for the moment." He kissed her hard and released her from his embrace. "You should go before someone sees you."
She nodded. Why was it that when he kissed her, she forgot everything else? And she wanted nothing more than to stay in his arms forever? She stepped out of Nathanial's room, crossed the corridor, and opened her door.
Before she could go in, the door to Quinton's room opened. "Miss Montini?"
She winced to herself but adopted a pleasant smile. "Mr. Harrington. I trust you had a pleasant evening."
"It was a thrill beyond measure," he said wryly, then paused. "I have a favor to ask of you."
"Oh?" What on earth would he want from her?
"My brother has never been in love, and I suspect his heart could be easily broken. I watched Sterling's heart break and should prefer not to see the same thing happen to Nate. I would appreciate it if you would take care that such a thing does not happen."
"I have no idea what you're talking about, Mr. Harrington."
"No? My mistake. Apparently you are not as intelligent as I have heard." He nodded. "Good evening, Miss Montini."
"Mr. Harrington," she murmured, and stepped into her room, closing the door behind her.
Surely he was wrong. Oh, it was obvious Nathanial did indeed have some feelings for her. He wished to protect her, of course, but that was part and parcel of the obligation he had taken on to help her find the seal. He couldn't possibly...Her heart fluttered at the thought.
No! She thrust the idea firmly aside. He would have said something by now. He'd had every opportunity. And Nathanial was not the type of man to keep something like that to himself.
She climbed into bed and tried to sleep. But she couldn't get the question out of her head.
What if Quinton was right?
## Twenty-two
You asked to see me, sir."
Nate sat in his brother's chair behind the desk in the library and studied the footman. Xerxes Muldoon was a big, broad-shouldered man, probably some twenty years older than himself, with a slightly exotic appearance about the eyes. He looked nothing like a footman. Nate wondered why he hadn't noticed before.
"Miss Montini insists on doing something I think is exceptionally foolish," he began.
Muldoon waited expectantly, the perfect footman.
"This morning, she is to begin cataloguing the collections of Lord Rathbourne. There have long been rumors about Rathbourne, some of the most vile nature. Whether they are true or not, I don't know." He shook his head. "And frankly I don't care."
"Sir?"
"I think he's dangerous. I don't want her at his house without protection." He met Muldoon's gaze directly. "I want you to accompany her."
"Yes sir." Muldoon shifted uneasily. "Does Miss Montini know?"
"She does. She's not especially happy about it but she's agreed." He blew a long breath. "She is the most independent female I have ever met."
"She does appear so, sir."
"And she thinks nothing of the consequences of her actions."
"No sir."
"She is probably the most brilliant woman I have ever encountered as well."
"Yes sir."
"And stubborn." Nate huffed and shook his head. "Has she always been this stubborn, Muldoon?"
"Yes sir, she—" Realization dawned in Muldoon's eyes. "I beg your pardon sir, I..." Muldoon sighed and crossed his arms over his chest in a most unfootman-like manner. "How did you know?"
"I didn't until a few days ago. Then I put two and two together. I am not as stupid as I might appear."
"You couldn't possibly be," Muldoon said mildly. "I have heard of Rathbourne on occasion, when I traveled with Miss Montini's brother." He raised a brow. "I assume you know that as well."
Nate nodded.
"I think you're right to be concerned. You needn't worry. I won't let her out of my sight. I have always protected her." His eyes narrowed and an implicit threat sounded in his voice. "From everyone."
"I love her," Nate said simply.
"Good." Muldoon's voice softened. "I suspected as much. She needs someone to love her."
"Didn't her brother?"
"In his own manner, I think. He didn't mistreat her in any way." Muldoon paused to choose his words. "No doubt you're aware of Montini's reputation, the kind of man he was?"
Nate nodded.
"He was a very selfish man. He had his work and his own needs. There was nothing left for a little girl." He shook his head. "He never even noticed when she stopped being a little girl, but others did."
"And that's when..."
Muldoon shot him a sharp glance. "She told you about that, did she?"
"Yes."
"How interesting." He studied Nate thoughtfully. "I knew, of course, and my wife and Miss Henry, but to my knowledge she has never told anyone else."
"I see." Nate couldn't quite hide a smile.
"I wouldn't be too confident if I were you." Muldoon shook his head. "I have known Gabriella much of her life. Even so, I find her to be one of the most confusing people I have ever met. She is convinced that because of the incident she will never marry."
"I gathered as much."
"She has, as well, a suspicious nature. Trust does not come easily to her."
Nate nodded. "I noticed that too."
"Nor does love." Muldoon paused. "My wife and I have often discussed this. We have long been concerned about her." He cast Nate a wry smile. "We may be in her employ but we have always thought of her with the fondness one feels for a daughter." He fixed Nate with a hard look. "With the exception of my wife, Miss Henry, and myself, everyone Gabriella has ever truly trusted or loved in her life has left her. It is not a great leap of the imagination to understand that is why she does not trust or love easily."
"I would never hurt her." Nate met the other man's gaze. "Nor would I ever abandon her."
"See that you don't."
"There is another thing that puzzles me."
Muldoon chuckled. "Just one?"
Nate grinned. "No, but I am curious. About her trip to Egypt?"
"You figured that out too."
Nate shook his head. "Again, not until recently."
"It wasn't as difficult as you might think. As pretty as she is, she still makes a passable young man." Muldoon grinned. "And then of course she had me to make sure nothing went awry."
"Even so." Nate shook his head. "It was a dangerous venture."
"As well I know. She threatened to go by herself if I didn't accompany her." He shrugged. "And I wouldn't put it past her to do just that."
"I have a number of other questions."
"No doubt you do, but I've probably said too much already. Anything else you will have to ask her yourself." Muldoon paused. "Am I to assume I will be allowed to stay here, then? In my position as John Farrell?"
"For as long as necessary." Nate nodded toward the door. "Gabriella is probably ready to go. You should join her."
Muldoon nodded.
"Oh, before you do, might I ask what you did with the other John Farrell?"
Muldoon grinned. "Paid him handsomely to take a holiday and visit his family in the country."
"Excellent." Nate laughed, then sobered. "It might be best, as well, if you don't tell Gabriella about this chat of ours."
"I would never lie to her," Muldoon said firmly. "However, if she doesn't ask me directly, I would see no need to say anything. For the moment," he added pointedly.
"Agreed." Nate paused. "Watch out for her, Muldoon."
"I always have." The older man smiled and took his leave.
Nate chuckled to himself. Secrets, everywhere he looked someone had secrets. He suspected he had only scratched the surface of Gabriella's. But his talk with Muldoon explained a great deal. If he was going to keep Gabriella in his life—and after last night there wasn't a doubt in his mind that he wanted to do exactly that, not that he'd had any doubts before—he would obviously need all the help he could get. Muldoon might just turn out to be an unexpected ally. The big man had only her best interests at heart. After all, he had known her much of her life and was currently in her employ—
Her employ?
Surely he meant her brother's employ.
But her brother was dead. What did Muldoon mean?
Nate got to his feet and started after him.
The library door swung open and his mother stepped into the room. "Oh, good, Nathanial, you're here."
"I was just about to leave."
"No, you're not." Sterling appeared behind her, stepped into the library and closed the door behind him in a resolute manner. He had his earl's face on, which usually meant someone had done something to annoy the staid, controlled structure of his existence.
"Do sit down, both of you," Sterling said coolly. He carried a sheaf of papers in his hand and moved to stand behind his desk.
Nate leaned toward his mother. "What have we done?"
"I have no idea, dear. My conscience is clear." She thought for a moment, then nodded. "Yes, completely clear."
"As is mine," Nate murmured. Not that it was ever completely clear, but he couldn't think of anything he had done recently to which his brother might take umbrage. Of course, Sterling might wish to chastise him for having Gabriella in his rooms last night, but then, Nate knew Sterling would never involve Mother in a discussion of that nature.
"This..." Sterling tossed the papers on his desk. "Is a report."
"How nice, dear," Mother said with a pleasant smile. "What does that have to do with us?"
"It's a report about Miss Montini."
Nate and his mother exchanged glances.
Sterling seated himself behind the desk. "Might I suggest in the future, when each of us feels the need to hire an agency of investigation in regards to the same person, we do so together. Thus incurring only one bill rather than three."
"You needn't take that tone, Sterling." His mother fixed her oldest son with a firm look. "I had no idea that you would take it upon yourself to investigate Miss Montini."
"Regardless of whether or not you knew her mother, did you think I would simply allow a woman who was caught breaking into the house to reside with us without wanting to know more about her?" Sterling drummed his fingers on the desk.
Mother sniffed. "When you put it that way, it makes a certain amount of sense."
He picked up the papers and shuffled through the report. "It seems each of us requested information in a slightly different area. Mother wished to know more about family. You, Nate, wanted her past. And I wanted her current state of affairs."
Nate grinned. "We are nothing if not thorough."
Sterling glanced at the report. "This paints an interesting picture of Miss Montini. I would say some of it is not unexpected, although much of it is quite surprising. Don't you think so, Mother?"
"I don't know dear," she said in an offhand manner. "I haven't seen the report yet."
Sterling's eyes narrowed. "But you know more about Miss Montini than you have thus far mentioned."
"She knew Gabriella's mother." Nate's gaze slid from his brother to his mother. "But she told us that."
"Well, perhaps one of us should have thought to ask who her mother was, or why Miss Montini looks so much like Emma Carpenter."
"There was no need for that to be in the report," their mother said under her breath. "I knew all that." She cast her oldest son a stern look. "You should not let them charge you for that."
"Well, I don't know what you're talking about." Impatience sounded in Nate's voice.
"Allow me to explain, dear," Mother said. "As you know, Emma's mother, Lady Danworthy—Caroline—has been a friend of mine since we were girls. She had two sisters, one older, one younger. The younger of the two, Helene—a lovely girl, by the way—was disinherited when she married against their father's wishes. She married a wealthy man, significantly older than she and, God forbid, Italian."
"Montini?" Nate said slowly.
Mother nodded. "The situation was most distressing. Her father declared that as far as he was concerned, he had only two daughters. He relented rather quickly after Helene married and left for Italy, but what with the distances involved and whatnot, it was too late by then. Helene died giving birth to Gabriella. The family tried to maintain contact with Mr. Montini, but from what Caroline has told me through the years, he had little interest in keeping in touch with them. He died when Gabriella was eight years old and somehow she was lost."
Nate drew his brows together. "What do you mean, lost?"
"By the time Caroline's family learned of his death, the little girl had vanished. They tried for years to find her. Ultimately they were told the girl had died." Mother heaved a heartfelt sigh. "Needless to say, they stopped all efforts to find her. It has been one of the great sorrows of Caroline's life that she wasn't able to save her sister's child. And that," she turned a firm gaze to Sterling, "is why I insisted Gabriella stay here. I did not want her lost again."
"Why haven't you simply told Lady Danworthy that you've found her niece?"
"I had my reasons," Mother said in a lofty manner.
"And those are?" Sterling prompted.
"I am not used to being quizzed like a common criminal by my own son."
"My apologies, Mother," Sterling muttered.
"You don't sound the least bit sincere, dear. You need to work on that. First of all, Caroline is in Paris with her sister, and Emma as well. In her most recent letter she said they will be returning in a few days. Secondly, I was not entirely certain of Gabriella's intentions. After all, we did meet her when she was caught breaking into the house. There is also a sizable inheritance involved. Quite frankly, I didn't want to tell Caroline about her niece's existence if Gabriella's motives were less than honorable. It would quite break her heart.
"However..." Mother smiled in a satisfied manner. "Aside from a certain penchant for what might be called impulsive, even illegal, behavior, I am quite impressed with Gabriella's character." She turn a steady eye on her youngest son. "You could do far worse."
Nate grinned. "But I could not do better."
"No, darling, I don't think you could."
"According to this," Sterling waved at the report, "after her father's death, Gabriella was literally passed from one distant relative of his to another." His expression darkened. "From what I've read here, it was not a pleasant experience for her. She was not especially wanted and was treated more like a servant than a relative."
"Oh dear," Mother murmured.
"Approximately two years later, Enrico Montini, her half brother from her father's first marriage, found the girl." Sterling glanced at his mother. "I assume that's when Lady Danworthy's family was told she was dead?"
Mother nodded. "That sounds accurate."
"The trail would appear to end there. However, the report says Montini was thereafter accompanied by a young boy that he claimed was his brother. However..." Sterling's gaze met Nate's. "Gabriella has no other brother."
Nate nodded. "I had come to much the same conclusion, but I hadn't put this particular piece of the puzzle together."
"So you're saying that Enrico Montini hauled his sister around to all those dreadful, dangerous places you and Quinton frequent on your search for antiquities and pretended she was a boy?" Anger flashed in Mother's eyes. "How could he? Had he no thought for the consequences of raising a girl that way?"
Nate shook his head. "Apparently Montini had no thought for anything other than his own concerns."
"And there was a significant fortune involved," Sterling added.
"Yes, of course," Mother said thoughtfully. "The inheritance from her mother's father."
"I doubt that he knew about that." Sterling tapped the report with his finger. "Apparently, Gabriella's father left her a sizable fortune and it was entirely in Montini's hands." He shrugged. "According to this, Montini's father was not pleased with his son's choice of vocation and left very nearly everything to his daughter. As long as Montini had Gabriella, he had the means to support his work." He looked at his brother. "Your Miss Montini is a very wealthy woman, although I suspect she didn't know that until her brother died."
"That explains a lot," Nate said. "Go on."
"Nine years ago, the alleged younger brother vanished and Miss Montini began attending school here in London. She is remarkably well educated for a woman—"
Nate snorted. "Don't let her hear you say that."
Sterling continued. "She owns a small house in a respectable if not particularly fashionable section of the city and employs..." He flipped through the pages. "A Miss Florence Henry as a companion and—"
"Xerxes Muldoon and his wife." Nate nodded. "I recently discovered that."
"Well, this is all very interesting," Mother said.
"Indeed it is." Sterling clasped his hands together on the desk, his gaze shifting between his mother and his brother. "Now that I am confident we know very nearly all there is to know about Gabriella, what do the two of you intend to do?"
"I intend to reunite Gabriella with her family," Mother said firmly.
"I intend to make her a member of this family." Nate grinned.
Sterling raised a brow. "Because she has a fortune and a respectable family?"
"No," Nate said sharply. "Because she is the most remarkable woman I've ever met. Because I cannot imagine living my life without her. And because regardless of who she is or what she has or doesn't have, she holds my heart in her hands."
Mother beamed. "Wonderful, Nathanial."
Sterling cast him an approving smile. "Best wishes, little brother. Now..." He leaned forward and studied Nate. "What is the progress regarding the Montini seal?"
## Twenty-three
Even the vast number of treasures that might well take as long to document as Lord Rathbourne had taken to collect could not diminish the unease that settled more firmly about Gabriella with every minute in his house.
Now, she sat on a wrought-iron bench in the tiny walled garden off the library. When the butler, Franks, had shown them in, he pointed out the French doors, hidden behind velvet drapes, that led to the garden. He had said as well that Lord Rathbourne suggested she might like to make use of the garden to avail herself of fresh air should the treasure room become too stuffy. It was unexpectedly thoughtful. A gravel path led from the door to a tall hedge. On the other side of the hedge, hidden from the house, the wrought-iron bench faced a small tiered fountain.
Sitting here, one might well imagine they were somewhere far from the dark, brooding house looming behind them. Gabriella wondered if Lady Rathbourne ever sat here and imagined just that.
She wasn't sure she could spend day after day in this house, in the windowless room she couldn't stop comparing to a tomb. Lord Rathbourne was not at home today and she was grateful for his absence. When she'd been in the treasure room, Xerxes positioned himself just outside the opening in the viscount's library. Right now he stood behind her in the doorway, vigilant and watchful. Even though the hedge obscured her view, she knew he was there. Any other time, she would have found his protectiveness annoying. Here and now, it was a comfort.
Today, she had planned to start making a preliminary list of the separate and varied collections, if only to begin to determine the enormity of the job. Instead, however, she'd spent most of her time studying the seal that was so similar to the one her brother once had. While she was certain it wasn't the same, she would very much have liked to see an impression and compared it with her brother's. Which reminded her, she needed to stop by her own house to see if Florence had found the impression on her way home.
Home. When did she start thinking of Harrington House as home? It wasn't that it was grander than anything she'd ever known; size and affluence had nothing to do with it. There was a sense there of family and tradition and belonging. An air of continuation. The feeling that no matter what else happened in the world, regardless of where family members might wander, this would always be a place that welcomed them.
Perhaps if she hadn't experienced the easy affection in that house, Rathbourne's might not seem as grim, although she doubted it. The few servants she had met thus far, while not overtly unpleasant, were not particularly friendly either. She couldn't quite put her finger on why, but it struck her that there was no warmth in this house. No feeling of occupancy. No sense that this was something other than a showplace, a display cabinet. She shivered at the thought.
In spite of the opportunity Rathbourne's collections offered for her future, she knew now she had to reconsider her decision to accept the position.
That would please Nathanial. Not that she cared. She blew a long breath. Of course she cared. She was lying to herself if she thought otherwise. At least being at Rathbourne's house took her mind off Nathanial. And off Quinton's comments.
Was it at all possible that Nathanial's heart was at stake? Some of the things he'd said—and so much that he'd almost said—might lead one to believe, if one were silly and foolish—
"What are you thinking, girl," Xerxes said, rounding the path.
"Nothing of significance." She shrugged.
"I thought you might have been thinking about Mr. Harrington."
She started to deny it, then couldn't. Besides, Xerxes always seemed to know when she was less than truthful.
"Perhaps you should—think about Mr. Harrington, that is."
"Thinking about Mr. Harrington in any manner whatsoever is pointless."
"Why?"
"Why?" She stood and turned toward him. "Because he and I..." She shook her head. "We would never suit."
"Oh?" Xerxes raised a brow. "From my observation, there is no one who would suit you better."
"Regardless." She shook her head. "It's really not possible."
"Not if you won't let it be possible." He studied her carefully. "It's up to you, girl. It's all in your hands. I would hate to see you let happiness slip through your fingers."
"Happiness?" She thought for a long moment. She wasn't sure she'd ever been truly happy, and not sure she'd know happiness if it came her way. She'd been content enough, she supposed, although now it struck her that most of her life thus far had been spent in preparation for something that would never come. Had been spent waiting. "Do you really think so?"
"I think," he chose his words with care, "Nathanial Harrington is the best thing that has ever come into your life. And I think you are the best thing that has come into his. I further think if you don't understand that and accept it, then you are not as intelligent as I have always known you to be. However," he shrugged, "what I think isn't nearly as important as what you know."
"What I know?" What did she know? She knew she loved Nathanial. She knew she didn't want to leave him. As for his feelings...Didn't she know those as well? Didn't she know when he called her "my love" or when he tried to protect her or when he held her in his arms or when she gazed into his brown eyes—didn't she know then that he shared her feelings? She met Xerxes's gaze. "What if I'm wrong?"
"What if you're not?" Xerxes smiled. "You should talk to the man. I suspect you haven't done that."
She shook her head. "We talk all the time."
"About how you feel? What you want?"
"Nathanial says everything is about what I want."
"And?"
"And." She drew a deep breath. "And Nathanial Harrington is what I want."
"Then perhaps it's time to do something about that."
"Perhaps...you're right, it is indeed." She smiled. "And I will. Now, it's time to go." She started back into the house, her voice brisk, her tone decided. "We need to stop at my house. I must speak to Florence and you should probably see your wife."
Xerxes chuckled. "Your Mr. Harrington doesn't stand a chance."
"He's not my—" Resolve washed through her and she grinned. "But he will be."
She would talk to Nathanial as soon as possible. Confess her feelings and pray he felt the same.
By the time their carriage rolled to a stop in front of her house, the most intoxicating sense of hope bubbled within her. She had learned to trust him, she loved him, and it was past time to take the greatest leap of faith of all.
"I was just about to send you a note." Florence hooked her arm through Gabriella's and steered her into the parlor. "Miriam and I are going to take the first train north in the morning. We have received word that her mother is ill."
"Oh dear," Gabriella murmured. "How bad is it?"
"We're not certain. The note we received was rather vague." Florence seated herself and indicated Gabriella should sit beside her. "But it did say we should waste no time. Miriam is quite concerned."
"I can well imagine." Gabriella frowned. "But shouldn't Xerxes go with you, then?"
"Absolutely not," Florence huffed in indignation. "He needs to stay here with you." Her eyes narrowed. "Especially as you insist on working for that dreadful viscount."
"I see you have been talking to Mr. Dennison."
Florence's expression remained firm but a smile lit her eyes. "Why yes, I have. Now about this position of yours with Lord Rathbourne—"
"You needn't worry about that." Gabriella blew a resigned breath. "As great as the opportunity is, I have decided not to accept his offer. Being there today, however, did give me the chance to look for the seal."
"And?"
"And he has one that might well be its mate, but I found nothing more than that. Nor did I really expect to. Lord Rathbourne's arrogance is such that if he had my brother's seal, I doubt that he would hesitate to tell me." She shrugged. "Unless I could prove it was Enrico's—have you found the impression yet?"
Florence shook her head.
"I doubt it will matter. I am afraid we might never recover the seal. As for Lord Rathbourne, I intend to tell him tomorrow that I will not accept his offer."
"Excellent." Relief washed across Florence's face. "I cannot tell you how worried I have been. As has your Mr. Harrington."
"Mr. Dennison again?"
"He is a fount of information as well as..."
Gabriella raised an amused brow. "As well as?"
"Gabriella." Florence took the younger woman's hands. "I suspect Mr. Dennison may soon ask me to marry him."
Happiness for her friend washed through her. "How wonderful."
"Yes, it is," Florence said slowly. "However, I am not certain what my answer will be."
"Why ever not?"
"If I marry Mr. Dennison..." Florence paused. "I am concerned as to what will happen to you."
"To me?" Gabriella widened her eyes with surprise. "You should not concern yourself with me."
"Your welfare has concerned me for nearly a decade," Florence said firmly. "I do not intend to abandon you now."
"Don't be absurd." Gabriella scoffed. "You wouldn't be abandoning me at all. Besides, you and Xerxes and Miriam are my family. Even if we no longer reside in the same house, that will never change. And we are not so large a family that we cannot include one more."
"But things will change."
"For the best," Gabriella said firmly. "You must follow your heart, my dear friend. You have always told me to do so."
"And are you following it now?" Florence's gaze searched hers.
"I think I always have really." Gabriella thought for a moment. "I followed my heart when I studied to make myself indispensable to Enrico. In many ways the search for the seal was following my heart. As for anything else...yes, Florence." She cast her friend a brilliant smile. "I am going to follow my heart."
But it seemed the moment she had decided to confess her feelings, there was no opportunity to do so. It was late afternoon by the time she arrived back at Harrington House, and Nathanial was nowhere to be seen. Still, his absence gave her time to decide exactly what she would say. Simply blurting out her feelings didn't seem quite right. She'd never been especially good at being coy, so anything other than a forthright approach might be even more awkward than necessary. She should tell him as well that she wasn't impoverished, and perhaps it was also time to mention her childhood...and oh, yes, the fact that she was the brother he had met in Egypt. She cringed at the thought of all she had kept from him. Still, if he did love her, it might not matter. If he didn't, then it wouldn't make any difference at all.
She was hopeful that after dinner there might be time for a moment alone. But Nathanial and his brothers had again gone to the earl's club for the evening. Lady Wyldewood had invited her to accompany her and her daughter to some event or other, but she'd begged off. Instead she retired to her room with a book. She had selected, of all things, a work of fiction from the library, something about a young woman attempting to find the perfect matches for her friends. Utter nonsense, really, but surprisingly engrossing. She had intended to read until she heard Nathanial in the hall, and left her door open a crack to ensure that she would. Then she'd talk to him.
And if her visiting his room again led to something other than conversation and confession, well, apparently once one was truly ruined, one wished for nothing more than to be truly ruined again. And again.
But before long the words swam before her eyes and the book fell from her hand and she slept, to dream of dark-eyed men with hair streaked by the sun, and kisses in the moonlight, and leaps of faith.
Nate made his way along the corridor to his rooms. Even though he, Sterling, and Quint had spent a long evening at Sterling's club, he wasn't the least bit inebriated. Well, perhaps he was the least bit inebriated, but certainly not extensively.
He glanced at Gabriella's door and pulled up short. It was slightly open and a light still burned inside her room. Was she waiting for him? He grinned. What a delicious idea. Even before last night he knew he wished to share her bed every night for the rest of his life, but he hadn't expected to do so quite so soon. Not that it wasn't an excellent idea.
"Gabriella," he said softly, pushed open her door and slipped inside the room.
She lay curled on her side, one arm flung off the bed, a book she had been reading on the floor beneath her hand. She was obviously asleep. Disappointment stabbed him. As much as he would like to do so, he wasn't going to wake her up. He moved quietly to the side of the bed, picked up the book, glanced at the title and smiled. Fiction. And romantic fiction at that. She had certainly come a long way since they first met. It was such a short time ago and yet it seemed he had known her forever. In his dreams perhaps or in his heart.
He set the book down on the table by the side of the bed, started to extinguish the lamp, then paused to look at her. He would never tire of looking at her. Not if they lived to be as old as the relics he hunted. And one day there would be children and...
And he certainly couldn't continue hunting for antiquities if he had a wife. The thought pulled him up short. How could he leave her? Regardless of what Gabriella had hoped for her life, he couldn't possibly drag her around with him. Not now that he knew how she had grown up. It wouldn't be right, and it wouldn't be what she deserved. She deserved something...well, better. If he wanted Gabriella in his life, his life was going to have to change.
He looked at her once more, then extinguished the lamp. It was a small enough price to pay. Indeed, it was well worth it.
## Twenty-four
Would you be so good as to tell Lord Rathbourne I should like to see him," Gabriella said to his lordship's butler. Today as yesterday, Franks had greeted them with as few words as possible, and then escorted Gabriella and Xerxes into the library.
"As you wish, miss." The butler hesitated. "I have not yet spoken with his lordship this morning but, as the treasure room has been opened, I assume Lord Rathbourne is occupied elsewhere in the house. I shall inform him of your request as soon as I see him."
"There's nothing to be done about it, I suppose," Gabriella said to Xerxes when the servant had left the room. "I might as well finish the list I began yesterday. It's the least I can do."
Xerxes frowned. "You don't owe this man anything, girl."
"Aside from an apology, you mean?" She shook her head. "I agreed to take this position and now I am about to renege on that agreement. It does not sit well with me but..." She glanced around the library and shivered. "I think it's best."
"We all think it's best," Xerxes said firmly.
No doubt Nathanial would agree, if she ever again found the opportunity to speak with him. Today, she thought. She would definitely tell him everything today.
She took up where she'd left off yesterday, filing page after page with notes on the various collections. By late morning Lord Rathbourne had still not made an appearance.
"We could simply write him a note and leave," Xerxes suggested hopefully. He didn't wish to remain there any longer than she did.
"No," she said firmly. "However, I am going to step outside for a moment."
She pulled open the drapes that covered the French doors to the garden, sunlight flooding the dark room. What a pity, this could be such a pleasant house. She opened the doors and stepped down two steps onto the pathway. As he did yesterday, Xerxes stood in the doorway, leaning against the doorjamb, her guardian. And like yesterday, his presence was reassuring.
She walked the few steps around the hedge and stopped short. Lord Rathbourne sat on the end of the bench closest to her, his back to her, his head tilted to one side as if he were listening to something. In all honesty, lord or not, it was insufferably rude to have ignored her request to speak to him. She had no idea how he might have slipped out there without either she or Xerxes seeing him, and yet obviously he had.
"My lord," she said firmly and stepped closer. "I should like to have a word with you." It would be best just to say it quickly and get it over with. She drew a deep breath and released the words in a rush. "As much as I appreciate the opportunity the position as your curator affords me, I do regret that, after further and due consideration, I cannot accept the position."
Surely he'd want a reason? Surely she had one that would sound better than she couldn't work with the unease that surrounded her in this house, in his presence? She stepped closer. "My plans for the future—my future, that is—have changed." That wasn't exactly the truth but it wasn't a complete lie either. Her plans had changed, or at least, if she were lucky, they would. "Well then, my lord, you do have my eternal gratitude for your faith in me, and my apologies, of course. So, that said, I shall take my leave." She turned on her heel and started off. And stopped in mid-step.
This was certainly the height of cowardice. Why, she hadn't even allowed the man to get a single word in. She at least owed him the courtesy of listening to his response. She braced herself and turned back. "My lord?"
He still didn't respond. He was no doubt furious.
"Lord Rathbourne." She squared her shoulders and approached him. "I fully understand—" And froze.
Shock clutched at her throat. It was obvious the man was dead. The glassy unfocused look in his eyes would have told her that even if it weren't for the slash that stretched from one side of his throat to the other and the drying reddish-brown blood that had soaked his clothes and settled in a thickened puddle at his feet.
Gabriella couldn't pull her gaze away. His face was ashen, drained of color. The thought occurred to her that any other woman would have swooned or at least screamed. She, however, was made of sterner stuff. She had seen and studied more than a few mummies over the years. Still, it was one thing to look at a three-thousand-year-old Egyptian and quite another to look at a newly dead British lord. Her stomach heaved. She turned on her heel, stumbled a few steps to the nearest flower bed, bent over and retched.
Almost at once she heard Xerxes hurry up behind her. "Gabriella, are you—"
"I'm fine." A slight touch of queasiness lingered but she did feel much better. She wiped her mouth with the back of her hand, straightened, and turned back toward the grisly scene.
"You shouldn't look at him, girl," Xerxes said grimly, handing her his handkerchief.
"I've already seen him." She dabbed at her mouth. "I can't imagine he looks any more dreadful now than he did a moment ago." She studied the dead man. "His throat's been cut, hasn't it?"
"It would appear so." He grabbed her arm and steered her back along the path and into the house. "If you've seen enough..."
"More than enough." She shook her head. "It's not the type of thing one expects to see in the middle of London. In more uncivilized parts of the world, of course, this sort of thing is not unexpected. I daresay it happens all the time in places like Asia Minor or Egypt. Why, one might not even be surprised to see a man with his throat slit sitting in a garden on a pleasant spring day," she said brightly.
Xerxes stared at her. "You're babbling, girl." He set her down on the sofa, strode to the doorway and called to the butler.
"Utter nonsense," she said under her breath. "I don't babble. I'm not the type of woman who babbles." She was simply keeping up a steady stream of observation. After all, she'd never seen a dead man before, and never imagined she'd see one with his throat cut. In the back of her mind she had the oddest conviction that if she stopped making relatively rational comments, she'd start screaming and never stop. Regardless, at the moment she couldn't stop talking.
Franks sent for the authorities, and it seemed as though the library was filled with people in no time at all, although it could have been hours. She'd lost all track of time. She wished Xerxes would stop looking at her as if she were mere seconds away from insanity. She was fine. Perfectly fine. Why, even her stomach had settled down. And that nice constable who was the first to arrive hadn't seemed the least bit annoyed by her observations as to his lordship's nature and her opinion that very nearly anyone who had ever met him might have a certain desire to slit his throat. Not her, of course, she had no reason to wish him dead. After all, he had offered her an opportunity few women would have imagined. And an exceptionally fine salary as well. No, no, it wouldn't have made any sense for her to have killed him.
The constable had asked where they could be found if it was necessary to speak with them further, then sent them on their way, but not until he and Xerxes exchanged knowing glances. The kind of looks men traded when dealing with an irrational woman. It was most annoying. She might well be babbling, although it did seem to her that every word was significant, but she was certainly not irrational.
She kept up a steady stream of chatter in the carriage on the way home. Had Xerxes noticed the expression on Rathbourne's face? Admittedly it might be in her own mind but she thought his lordship looked somewhat surprised. Although she supposed surprise was to be expected unless one had known one's throat was about to be slit like a pig's. In which case one would certainly take steps to prevent such a thing. Didn't he think so? And didn't Xerxes think as well that the viscount had been dead for some time? After all, she and he had been in the house for a good three hours and they certainly would have seen his lordship and whomever might have been with him go into the garden. Even if they hadn't, surely they would have heard something. A man probably couldn't have his throat cut without making some sort of sound. Gurgling or something of that nature.
The moment they crossed the threshold of Harrington House, Xerxes ordered a footman to send for Nathanial, then took her into the parlor. Apparently, the older man's tone left no room for hesitation, as Nathanial arrived within minutes. Xerxes joined him outside the parlor doors, obviously warning him about her state of mind. Which was absurd. Her state of mind was perfectly fine, even excellent if one considered it had only been a short time ago that she had found a blood-soaked, surprised-looking, very dead viscount in a garden.
"Gabriella?" Nathanial stepped into the room, Andrews right behind, bearing a tray with a decanter of brandy and glasses. Probably the very one he had brought on the very first night she was there. How appropriate, or perhaps ironic; she wasn't sure. Nathanial nodded at Andrews, and the butler left the room. Nathanial looked as if he wasn't sure what he should do now. "Brandy?" he asked her.
"I would think tea would be more appropriate at this time of day." However, in spite of the pleasantness of the day, the tips of her fingers were icy. "But I am a bit chilled. I find brandy to be excellent when one is chilled. Or nervous. Don't you? It does seem to soothe the nerves."
He filled two glasses and handed her one. She took it and noticed that her hand shook. He raised a brow.
"You needn't look at me that way." She took a long sip of the brandy, its warmth comforting and welcome. "I am fine. Perfectly fine. Admittedly, my hand is shaking, but then it's been that sort of day. I suspect anyone would shake a bit upon finding a dead viscount in a garden."
"Yes of course."
"Lovely garden," she murmured. "Quite peaceful." Except of course for the dead, blood-soaked man with the staring eyes and the surprised expression.
He sipped his own drink and studied her warily.
"I am not a delicate, fragile flower, you know."
"I know." He moved closer. "You are not like most women."
"I most certainly am not." She shrugged. "Most women, at the very least, would have screamed at coming upon a scene like that. I simply..."
He nodded. "John told me."
"John?" She pulled her brows together. Xerxes. "Yes, of course, John."
"He sent word as to what had happened but it only arrived a few minutes ago. I was about to come after you."
"It wasn't necessary." She cast him a bright smile. "I am fine."
"Are you?"
She laughed, and even to her own ears it had a strange, vaguely hysterical sound to it. "Perfectly fine. And the brandy is much better than tea."
"Do you feel better now?"
"Much."
He studied her cautiously. "It is understandable, you know...your reaction, that is."
"I would think so," she huffed. "Why, the contrast alone between the serenity of the garden and the—" She shivered in spite of herself. "—violence of what must have happened was enough to make anyone ill."
"Are you sure you're all right?"
"Perfectly." She took another long drink. "He must have been dead for some time, you know. We'd been there all morning so it must have happened last night." She nodded. "He looked quite...dead."
"Gabriella."
"Not recently dead." She shook her head. "Not that I know what someone recently dead would look like, but he looked, well, rigid. Quite, quite dead I would say."
Nathanial's brow furrowed in concern. "Gabriella."
"Am I babbling?"
"Yes."
"Don't be absurd." She took another bracing swallow. "I never babble."
"And yet—"
"I don't think it was a robbery," she continued, as if he hadn't said a word. "There was nothing disturbed in the treasure room. At least not that I noticed. And I would have noticed. I notice such things. I am nothing if not observant."
He stared at her.
She ignored him. "And if one were to rob the viscount, there are any number of things—priceless things—one could quite easily abscond with simply by slipping them in a pocket." She shook her head in an earnest manner. "No, it wasn't robbery, but then he wasn't a very good man, was he? I imagine there are all sorts of people who would have gladly slit his..." She tossed back the rest of her brandy and held the glass out to him. Her hand shook uncontrollably and she noted it with a strange sort of detachment. As though she wasn't the one looking at her hand and it wasn't her hand.
"But you are still shaking." He took the glass from her and set both their glasses on a table.
"It's nothing."
"And your hands..." He took them in his. "Are very cold."
"They are, aren't they?" Her voice was oddly detached, as if it were someone else's voice. "How very unusual."
"You are not fine at all." He pulled her into his arms, and she rested her head against the solid protection of his chest. "You have been through an ordeal."
"I am fine." A sob rose up inside her. Where on earth had that come from? She didn't sob. Or weep. Or cry. She couldn't remember the last time she had. She swallowed hard. "Perfectly..."
"Yes, of course, perfectly fine." He chuckled. "And have you at last run out of things to say?"
"No," she muttered against his chest, but it did sound somewhat like a sob and his arms tightened around her. She closed her eyes and listened to his heart beat. The warmth of his body coupled with the heat of the brandy washed through her, and with it the most remarkable understanding and acceptance. Nothing in the world could hurt her if she was in his arms. She closed her eyes and sagged against him, abruptly too weary to stand.
He scooped her into his arms and carried her out of the room.
"What are you doing? Where are we going?" she murmured, but she didn't really care. He could take her anywhere and she would go willingly.
"I'm taking you to your room." His voice sounded very far away.
"Mmm." She snuggled against him. "How lovely."
He said something she didn't quite hear, and scarcely cared. In some distant still functioning part of her mind she realized she'd be asleep long before they reached her bed. And wasn't that a pity? She had so very much to tell him. About who she was and what she wanted. And that what she wanted most of all was him.
## Twenty-five
Well?" Nate demanded the moment Quint stepped into the library. It was already evening, and his brother had been gone for hours. Sterling and Mr. Dennison had been waiting with him for Quint. They each sat behind their respective desks.
"Did you learn anything of value?" Sterling asked.
Sterling had sent Quint to find out what was known thus far about Rathbourne's death. And only did so because of the possibility that the murder might be connected to Gabriella's search for the Montini seal, not because Sterling still harbored some feelings for Lady Rathbourne. He hadn't actually said that, and no one had dared to ask.
Quint grinned. "It's amazing what one can learn when bandying about the name of the Earl of Wyldewood."
Sterling shrugged. "It can be useful on occasion."
"The inspector in charge practically fell all over himself to help me."
Sterling raised a brow.
"Well, perhaps not all over himself but he was helpful." Quint plopped into a chair. "They don't know much at the moment." He glanced at Nate. "I don't have much more than what John told you."
Nate clenched his jaw impatiently. "Blast it all, Quint, just tell me what you do know, then." If Gabriella was in danger, he needed to know. "And his name is really Xerxes Muldoon. He is employed by Gabriella."
"My, she is full of surprises," Quint said under his breath.
"Quint!"
"Very well." Quint thought for a moment. "Nothing appears to have been taken. The servants said there was nothing missing in the rest of the house, and Miss Montini told the police nothing appeared to have been disturbed in his treasure room." He cast Nate an incredulous expression. "Did you know he had a treasure room? The inspector said it was on the order of a vault."
"You knew he was a collector of rare antiquities," Nate replied. "He was also very protective. He didn't display them as most collectors do but kept them locked away, for his enjoyment alone." He shrugged. "Most of his collections were antiquities, but there were gems and paintings as well. The sum worth must be in the millions." He glanced at Sterling, who displayed no particular reaction. Rathbourne's death would leave his wife a very wealthy woman, which might well come as a relief to anyone who might be concerned about her welfare.
"The police are using lists Miss Montini compiled to make certain nothing is missing, but at this point they are fairly certain robbery was not a motive." Quint paused. "It's believed from the state of the body that he had been dead ten to twelve hours." He met Nate's gaze. "Which means he was killed last night. It also means..." Quint winced. "When Miss Montini found him it was not a pretty sight."
"I know." Nate had talked to Xerxes again after Gabriella fell asleep. The older man had been quite detailed in his description of the morning's discovery.
"Excellent." Sterling nodded. "Then there is nothing to indicate any connection between Miss Montini's search and Rathbourne's murder." His younger brothers traded glances. Sterling narrowed his eyes. "Is there?"
"Rathbourne's throat was cut." Nate drew a deep breath. "As was Montini's."
"What?" Surprise crossed Sterling's face. "I thought Montini died of a fever."
"Not according to what Quint heard in Crete," Nate said. "I suspect Gabriella was only told that to protect her."
For the first time, Dennison spoke. "It is somewhat awkward for foreign officers to inform a relative, particularly a young lady, of a loved one's death, when that death has been violent, sir," he said. "It's often believed kinder to conceal the fact of a violent death, as nothing can be done about it. I have heard about such things happening before."
"And she still doesn't know?" Sterling asked.
"I don't see any reason why she needs to know," Nate said simply.
Quint glanced at him. "Have you told her the rest of it yet?"
Sterling frowned. "The rest of what?"
"About the seals," Quint said.
Nate shook his head. "I haven't had the chance. I fully intend to tell her, but the opportunity has not yet presented itself. I need to find the right time."
"Perhaps," Gabriella's voice sounded from the doorway, "that time is now."
For a long moment none of the men said a word. Then Quint jumped to his feet. "If you will excuse me, I have an errand to attend to."
Sterling stood. "Mr. Dennison and I were just on our way out as well."
His brothers and the secretary hurried out of the room, murmuring polite greetings to Gabriella as they passed by. So much for brotherly support, Nate thought. Gabriella stared at him, stone-faced. They were rats deserting a ship that was not only sinking fast but on fire.
"Do you feel better?" he asked cautiously. Just how much had she heard?
"I'm fine."
He smiled. "Perfectly fine?"
She ignored him. "What haven't you told me about the seals?"
"Perhaps you should sit down."
Her jaw clenched. "I prefer to stand."
"Brandy, then?"
"No, thank you."
"Whisky?" Nate strolled to the whisky decanter and poured himself a glass.
"No." She crossed her arms over her chest. "What haven't you told me about the seals?"
"Quite a lot really." He took a long sip of his drink. "Are you sure you wouldn't rather sit down?"
Her eyes narrowed.
"Very well, then." He wasn't exactly sure how to say this. It did not cast Quint in a good light. Although, on the other hand, at least Quint hadn't stolen the seal. He drew a deep breath. "Quint saw Gutierrez steal the seal from your brother. Some months later, he managed to wrest it from Gutierrez in a game of cards."
"Then Quinton has the seal?"
"Not exactly." He shifted uneasily. It had been awkward enough to tell her of the path the seal had taken, but to tell her now that its whereabouts were still unknown, might always be unknown, was even more difficult. "When Quint unwrapped the seal the other day, right here in the house, he discovered someone had taken it and substituted a different seal."
She stared at him. "That's exactly what happened to my brother."
"Ironic, isn't it?" Nate pulled the seal Quint had found in the attic from his waistcoat pocket and handed it to her. "This is the seal Quint had."
She turned it over in her hand, glancing at it briefly. "This is chalcedony. My brother's was greenstone." She met his gaze firmly. "Where is it?"
He shook his head. "I don't know."
"Does your brother know?"
"No."
Suspicion flashed in her eyes. "Are you sure?"
He drew his brows together. "Yes. Quint wouldn't lie to me."
She shrugged. "He said he didn't steal the seal."
"And he didn't." Nate frowned. "He came by it in a relatively legitimate manner."
She snorted. "Relatively."
"Nonetheless, he didn't steal it."
"He could have told us all this at the beginning." She put the chalcedony seal on the desk. "It would have saved us a great deal of trouble."
"It would have made no difference at all. And yes, he could have told us." Nate paused. "He should have."
"And you trust him now?"
"Yes, I trust him. He would never lie to me."
"Do you know that?"
"I haven't a doubt in my mind."
"Well, I don't trust him."
"You don't have to."
"His reputation is not one that engenders trust."
"His reputation is not nearly as bad as—" Nate caught himself.
"Not nearly as bad as what?" Challenge flashed in her blue eyes.
"Nothing," Nate muttered.
"Not nearly as bad as my brother's?" Her voice was hard. "Is that what you were going to say?"
He tried to deny it. "No, of course not."
"Come now, Nathanial, there's no need to protect me. I know exactly the kind of man my brother was."
Nate stared. "I didn't—"
"My brother, Nathanial..." She met his gaze directly. In spite of the cool tone of her voice, there lingered an undercurrent, the merest hint, of despair. "Was the kind of man one would expect to end his days with his throat cut in some foreign land."
"You heard that as well?"
"Yes." She sighed and brushed a stray stand of dark hair away from her face. "I heard that."
"I am so sorry."
"What? Sorry that I heard it? Or sorry that it happened?"
"Both."
"You needn't be." She studied him for a moment. "I didn't know how he died, but it comes as no surprise."
"Still, surely it's upsetting."
"Upsetting?" She scoffed. "Of course it's upsetting, especially after finding Lord Rathbourne." She shuddered. "Although I suspect it's a rather quick way to die."
"Gabriella, I—"
"But I did know what kind of man Enrico was. I have always known. It was simply difficult to acknowledge aloud. He was all I had, all the family I've ever known." She paused for a long moment. "I watch you with your brothers and your mother and sister. There is a bond between you all that is...quite remarkable. And I confess I am envious.
"My brother and I," she shook her head, "we did not share what I see between you and your family. I was an obligation for Enrico. Not that he treated me badly," she added quickly. "He saw to it that my needs were met."
"You needn't say anything more."
She ignored him and paced the room. Her words seem to come of their own accord. "It wasn't until he died that I learned we weren't in the financial straits he had always led me to believe we were. I discovered I had a significant fortune that had supported me and my brother's work. Enrico had never mentioned it."
He nodded. "I know."
She continued as if he hadn't said a word. "When Enrico found me, a few years after my father's death, my circumstances were dismal. He rescued me, Nathanial. He was my savior and I adored him.
"He dressed me as a boy and took me with him on his excavations and his hunts for treasure." She paused in mid-step and met Nate's gaze defiantly. "I loved every minute of it. It wasn't until he left me in England that I came to realize that what I had thought was his brotherly desire to keep me by his side was really no more than a convenience. It was easer for him to keep me with him than to arrange for my upbringing elsewhere. As much as I loved it, it was not in my best interests. Somewhere in the back of my mind, I think I have always known my brother's concerns were always and only for himself, for his work, even if I have not dared admit it even to myself until recently."
"I know."
"You know?" She jerked her gaze to his. "You know about my childhood? My fortune?"
He nodded.
"How could you possibly know any of this?"
He winced. "I had some inquires made."
"Inquiries?" Her eyes widened. "You had me investigated?"
"Can you blame me?" He glared at her. "You lied to me, to my family. Bloody hell, Gabriella, you broke into my house!"
"How long have you known?"
"Only since yesterday."
She studied him carefully. "And when did you know of Quinton's part in all this?"
He hesitated. She would not take this well. "A few days ago."
"Before I came to your room?" she said slowly.
He braced himself. "Yes."
"And you did not think to tell me then?"
"I had other things on my mind?" he said weakly.
"Yes," she said simply. "I suppose you did."
Long moments ticked by in silence.
"So." She met his gaze. "It's at an end, isn't it?"
Panic flickered through him. "What's at an end?"
"The search for the seal. We are no closer now than when we started." She cast him a wry smile. "So ends my efforts to give my brother the acknowledgment in death that he didn't have in life. That he might not have deserved. I owed him that much."
"You didn't owe him anything," Nate said in a harsher tone than he'd intended.
"Oh but I did." She shook her head. "My brother gave me a purpose and a love for all things ancient. He did not prevent me from achieving far more in terms of my studies than most women ever dream of. He gave me as well Florence and Miriam and Xerxes, an odd sort of family admittedly, but a family nonetheless. My life has not been a bad one, Nathanial, and for that I do indeed owe him. Besides, he was my brother and I loved him. No matter how he regarded me." She blew a long breath. "I was told recently that no matter what I did, how hard I worked, he would never love me. I hadn't even realized that was what I'd wanted."
His heart twisted for her. He should tell her that he loved her, that he wanted her in his life for the rest of his days. That he would never leave her. But telling her now might seem as though he was simply trying to ease her pain. No, his feelings could wait.
"Gabriella." He started toward her.
She stepped back. "Please don't." She drew a deep breath. "If you take me in your arms, I will let you. If you take me to your bed, I will allow that as well." Her blue eyes were shadowed with emotion. "I have a great deal to think about. The rest of my life. You."
"Me?" Hope rose within him.
"Yes, you." She shook her head and started out of the room. She paused at the doorway and looked back at him. "I trusted you."
He shook his head. "I never betrayed your trust. I didn't tell you everything but I fully intended to. I never lied to you." He paused. "And trust, Gabriella, as well as honesty, has to go both ways."
"I know, Nathanial." She stared at him for a long moment. "As do leaps of faith."
When had finding the seal become less important than what she might have found with Nathanial?
She'd spent the rest of the evening alone in her rooms trying to sort out her thoughts and emotions. Now she tossed and turned and tried to sleep, with no success. Not surprising, really. She had entirely too much on her mind to sleep.
It was most annoying that he'd had her investigated, but she couldn't blame him. She had lied to him from the beginning. She had expected him to trust her but gave him little reason to do so. She wanted to be angry with him for keeping the information about Quinton's involvement from her, but didn't doubt that he would have told her. She couldn't fault him either for loyalty to his brother, loyalty that was no doubt returned. Nathanial was that sort of man. Honorable and honest. And she was a fool not to have realized that before now.
She had indeed trusted him. She groaned and punched the pillow. Obviously, she still did. What other reason could there be for confessing to him all the things she had? All those feelings about Enrico she had never admitted to herself, let alone to anyone one else.
And she loved him. She should have told him, but after talking about Enrico it would have sounded...well, pathetic. Obviously she was more of a coward than she had ever imagined. Now, she feared telling him. Feared he wouldn't feel the same, and feared as well the look of pity in his eyes. She could accept almost anything but that.
It was past time to face the reality that the search for the seal was over. And hadn't she told Nathanial yet another lie? Deep in her heart, didn't she want to find the seal as much to justify her own life as well as her brother's? Wouldn't it have been her own triumph as well as his? Wouldn't his moment of glory been hers too?
Still, it scarcely mattered now. There was nothing to show for her efforts but the seal Quinton had had. She'd paid it scant attention. Carved of chalcedony, it appeared Assyrian...
Good God! She bolted upright in bed. How could she have been such an idiot?
Gabriella flung off the covers, leapt out of bed, and headed for Nathanial's room. She forced herself to knock quietly on his door; it wouldn't do to wake up the entire household. She tried the handle and grinned. The man didn't lock his door. Stepping into his room, she snapped the door closed behind her.
She crossed the dark sitting room, grateful for the faint light from the window. Silly man didn't even close his drapes. She felt her way to his bed and reached out to shake him.
"Nathanial," she whispered.
"Wha..." he groaned.
Laughter bubbled up inside her. "Gracious, Nathanial, how can you sleep when I can't?"
"Gabriella?" he said, his voice rough with sleep.
"Were you expecting some other visitor?"
"I wasn't expecting anyone." He grabbed her and pulled her onto the bed. "But I am nothing if not a grateful host."
Before she could protest, he pressed his lips to hers and kissed her long and hard, a kiss to melt even the sternest resolve. A kiss to steal a heart or seal a promise.
He pulled away. "Am I to take this to mean you are no longer angry with me?"
"I wasn't angry with you."
He chuckled. "Then I would hate to see you when you are." He paused. "Why are you here?"
"First..." She kissed him again. She couldn't see his eyes in the dark, and that was for the best. For good or ill...she drew a deep breath. "I love you, Nathanial, and I thought you should know."
"Gabriella—"
"No, you needn't say anything right now. But that's not all I have to tell you." She kissed him once more, then grinned. "I know where the seal is."
## Twenty-six
You what?" At once he was wide-awake.
"I know where the seal is." She laughed and rolled off the bed. "Get dressed and I'll meet you in the hallway."
"Why am I getting dressed?"
"We have to go find the seal." Her shadow moved away and disappeared into the sitting room.
"Now?" he called after her in a stage whisper. Excitement sparred with disappointment. Couldn't they go in the morning? After a long night in his bed?
"Yes, now."
He heard the door close behind her. She knew where the seal was? He grinned. Wasn't that incredible? Wasn't she incredible? And she loved him. Damnation, he hadn't expected that. Not tonight anyway. He had hoped, of course. He quickly threw on clothes and stepped into the hallway. She was already waiting for him.
He raised a brow. "I see you're wearing your housebreaking clothes."
Even in the scant light shed by the lone lamp in the hallway he could see her blush. He loved making her blush.
"They were...expedient." She shrugged. "And they seemed appropriate."
"For treasure hunting perhaps." He grabbed her and pulled her into his arms. "Does this mean we will be breaking into any houses?"
"Absolutely not," she said staunchly.
"Good." He rested one hand on the small of her back, his other hand drifting lower to caress her buttocks. "I love it when you wear men's clothes."
"Stop it, Nathanial." Her smile belied her words. She pulled out of his arms and started toward the stairs. "Come along."
"Why can't this wait until morning?"
"It can, I suppose, but I can't." She hurried down the stairs.
"Where are we going?"
"I have a small house." She glanced back at him. "I assume you know that?"
He grinned. "Will we be taking Mr. Muldoon with us?"
"I see no need to wake..." She heaved a resigned sigh. "You know about him as well."
"Gabriella, my love." They reached the front door. He opened it and they stepped outside. "I know everything."
She snorted. "We'll need a carriage, it's too far to walk. Can you harness one yourself?"
"Yes, but I know where we can get a cab." He took her elbow and they started off at a brisk pace.
"At this hour?"
He nodded. "It's not far from here." He paused. "It's a...business establishment."
"What kind of business has cabs waiting at this hour?"
"A very discreet business," he said firmly. "Which is all you need to know."
"Oh," she murmured. "I see"
A quarter of an hour later they reached their destination, and as he had thought, there were several cabs waiting for fares. He held his questions until they settled in one and were on their way.
"You think the seal is in your house?"
She nodded. "And has probably been there all along."
"Why?"
"The seal you showed me last night, the one Quinton had." She leaned toward him, excitement underlying her words. "It's the same one Enrico showed to the committee. The one that was substituted for his."
"You're certain it's the same one?"
She hesitated. "Yes. Yes, I am."
"So you're saying—"
"Enrico must have been the one to take it from Quinton." A smile sounded in her voice. "He would have liked the irony of that, of replacing his seal with the one that had been traded for his. As though it had all come full circle."
"But why didn't he tell you that he had recovered the Ambropia seal?
"I'm not sure he didn't." She thought for a moment. "You read his last letter. It was rambling and convoluted but there was an odd sense of victory to it. I paid no real attention to it at the time. But now..." She nodded. "Now, it makes sense.
They pulled up in front of her house and he asked the driver to wait. She hurried up to the front door, inserted a key and turned the lock easily. Too easily perhaps. Unease drifted through him and he ignored it. It was probably nothing more than the circumstance of being at a house unknown to him late in the night.
She stepped inside and glanced back at him. "Florence and Miriam were called away so there is no one here." She lit a lamp in the front entry, stepped into what he assumed was a parlor, and came back with a second lamp. She lit that one and started up the stairs. He trailed after her, unable to shake a sense of unease.
She turned at the top of the second flight of stairs, led him down a narrow hallway and pushed open a door.
"Hold this, please." She thrust the lamp at him.
He glanced around. They were in a bedroom sparsely furnished with only a bed and a dresser. A small wooden crate sat in one corner. Gabriella moved to the crate and shoved off the top. It clattered to the floor, the noise resounding in the dark house.
He drew his brows together. "Did you hear something?"
"No." She stared at the crate.
"I thought I heard something," he muttered, and strained to listen. It was probably nothing more than the sound of the top of the crate echoing in the house. Still, he could have sworn he had heard something else.
"This was sent to me after Enrico died. We opened it but..." She paused for a moment. Regardless of Enrico's character, he was still her brother and she had obviously cared for him. "But I haven't yet looked in it."
She drew a deep breath and knelt down in front of the crate. He positioned the lamp to give her better light, moving it farther away from his nose. It had an odd scent of smoke, and he wondered if it had been some time since it was used.
Gabriella reached into the container and began pulling out the odds and ends of Enrico Montini's life. A few articles of clothing, several books and bound notebooks, some interesting but not particularly important artifacts, a well-worn pair of boots. She started to set the boots down then hefted one in her hand and glanced at Nate.
"What?" He moved the lamp closer.
She reached into the boot and pulled out a cloth-wrapped bundle. She dropped the boot and started to unwrap the bundle but her hands shook. She grimaced. "I don't recall my hands ever shaking before yesterday."
"Here." He grabbed her free hand and pulled her to her feet, then traded her the lamp for the bundle. Quickly, he unrolled the wrappings to reveal an ancient cylinder seal. She moved the lamp closer. It was greenstone. He met Gabriella's gaze. "I think we've found your seal."
She grinned. "Thank you, Nathanial."
"No, it's I who should thank you." He leaned closer and kissed her. "I can think of nothing I'd rather do than hunt for treasure with you."
She laughed, and for the first time since they'd met it struck him as carefree and happy. The sound of it wrapped around his soul.
"Do you want this?" He held out the seal.
"No, you keep it." She brushed her lips across his. "I trust you."
He smiled, slipped the artifact into his pocket, and took the lamp from her. "As much as I would like to linger here, we should be on our way. We have a driver waiting and it is nearly morning. Besides, now that you have the seal, you obviously need to prepare your presentation to the Verification Committee. You haven't much time left."
Her eyes widened. "My presentation?"
He raised a brow. "Who else?"
"I hadn't considered that. I really hadn't thought beyond recovering the seal. But of course." She lifted her chin and cast him a satisfied grin. "It will be my presentation."
"You are going to be a handful," he said under his breath.
"What?"
"I said we should probably get the impression while we're here."
"That's not what you said."
"Nonetheless, we should still get the impression."
"I'm afraid that's a bit of a problem," she said slowly.
"Oh?"
"I don't actually have the impression."
"But you said—"
"Yes, well." She winced. "It sounded so good at the time. But I am confident it's in the house. Somewhere."
He narrowed his eyes. "Are there any other lies you wish to confess?"
"Not at the moment," she said in a somber manner, but her eyes twinkled with amusement. "But if something comes to mind I shall be sure to tell you. Although didn't you say you knew everything?"
"Hmph."
He led the way into the hall and stopped. The hairs on the back of his neck prickled. There was a distinct odor of smoke. "Do you smell that?"
"What?" She stopped and her eyes grew wide. "Smoke? But how?"
"It doesn't matter. We have to go." He grabbed her hand and started for the stairs. Smoke drifted up from the lower floors. "Is there a back stairway?"
She stared at the smoke now billowing up the stairs. "My house is on fire!"
"A back stair, Gabriella?"
"My house..." She shook her head as if to clear it. "This way."
She dashed toward a door at the opposite end of the hall and reached for it. He yanked her back and carefully placed his hand on the door. It was cool. He pushed it open. "I have the lamp, I'll lead the way. Grab onto my coat."
"Why?"
"Because I said so!" he snapped, and struggled for calm. It wouldn't do either of them any good if he did not keep his wits about him. "Because if I can feel you holding on, I'll know you're still with me."
"Very well." Fear sounded in her voice.
He quickly led the way down the two flights to the ground floor. There was some smoke but not much. The fire was obviously in the front of the house. Again he paused to feel the door before pushing it open. The moment he did, acrid smoke billowed around him and he choked. "Is there a back way?"
She coughed. "Yes." She grabbed his hand and pulled him toward the back of the house, into a room that appeared to be a scullery, to a door on the outside wall. She fumbled with the key. Panic rang in her voice. "I can't get it open!"
"Move!" He shoved her aside, set the lamp down, took aim and kicked the door with the flat of his foot. He tried again and the door gave somewhat. Smoke was filling the room, the lamp no longer any use. He braced himself and tried again. The door splintered. He grasped the door and yanked it free. Grabbing Gabriella around the waist, he hauled her out of the building. They stumbled a few steps away from the house, both struggling to catch their breath.
She gasped for air. "Come on."
She sprinted across a small garden to a back gate with him right behind her. The gate opened to a narrow lane. She ran down it, turned into the street and circled to the front of the house, Nate close on her heels. She skidded to a stop and stared.
The front door was wide open and smoke poured out. Flames could be seen in one window. Gabriella stared in shocked disbelief.
"Bloody hell." Quint's voice sounded behind him.
"What are you doing here?" Nate snapped.
Quint stared at the burning building. "I heard you in the hall. I thought you might need help. So I woke Muldoon. We just got here."
Gabriella gasped. "My letters!"
Nate peered around his brother. "Where is Muldoon?"
"He ran for the fire brigade as soon as we realized where the smoke was coming from." Quint's gaze shifted to a point behind his brother. "You might want to stop her."
Nate swiveled to see Gabriella dashing back to the house and through the open door.
"Damnation." He raced for the door, Quint no more than a step behind.
"Here." Quint thrust a handkerchief at him. "Cover your nose and mouth and for God's sake be careful."
Nate nodded, clutched the handkerchief to his face and stepped inside. Flames roared at the foot of the stairs, a good six feet or so from the front door. The fire had spread into a parlor to the right. Dear God, where was she?
"Gabriella!" he screamed, praying to be heard above the din of the flames. Fear for her clutched at his heart. The smoke was so thick, if not for the light of the fire itself he wouldn't have been able to see his hand in front of his face.
"Here." She staggered out of parlor, choking and gasping for breath. He started toward her. A loud crack ripped through the air. He reached for her. Her terror-filled gaze met his. A huge chunk of the ceiling collapsed, missing him by mere inches. Gabriella crumpled to the floor beneath it.
His heart lodged in his throat and he grabbed pieces of the ceiling and threw them aside. At once Quint was by his side. It took a moment or perhaps a lifetime to free her. Nate pulled her from the wreckage with his brother's help, scooped her into his arms, and the two men made their escape. A split second after they stepped out of the house, the rest of the ceiling collapsed and flames licked at their backs. They staggered away from the building and Nate noted the fire brigade had arrived.
Muldoon rushed up to them. "Is she..." Fear flickered in the other man's eyes.
"No." Nate could see the faint rise and fall of Gabriella's chest with her labored breathing. "But we have to get her home."
Quint nodded. "We have a carriage."
They settled in the carriage. Quint sent Muldoon to fetch their family physician to meet them at Harrington House. Nate had no doubt that despite the lateness of the hour, the big man would not take no for an answer.
He cradled Gabriella in his arms on the endless ride that probably took no time at all. There was a cut high on her temple but she didn't appear to have suffered any burns. She still clutched what looked like a packet of letters in her hand. Even though he knew head wounds tended to bleed a lot, there still seemed to be an inordinate amount of blood. And he did the only thing it was possible to do.
"Dear Lord," he prayed. "Don't let me lose her."
He had lost all sense of time.
Vaguely, Nate noted that the sun had risen. He wasn't sure how long he had sat here in the corridor in his mother's wing of the house. She had insisted they put Gabriella in one of the larger suites. Now she was in Gabriella's room with the physician while he waited. Fear had settled in a heavy lump in the pit of his stomach.
But as long as he had been there, Quint and Sterling remained by his side. As had Muldoon. Not appropriate perhaps for a footman, but Muldoon was as much a part of Gabriella's family as any blood relation.
"Is that what she went back in the house for?" Quint glanced at the packet of letters on Nate's lap.
He nodded.
"What are they?" Sterling asked.
"I have no idea." Nate shook his head. "But not worth her life."
"They are to her," Muldoon said quietly. "Although she'd probably deny it. They're letters to her mother. She found them after her brother's death." His expression hardened. "He never gave them to her."
Nate had never hated anyone in his life before, but at the moment he would have gladly slit Enrico Montini's throat himself.
"Do you have any idea what happened?" Sterling asked. "How the fire started?"
"There was someone else in the house. I thought I heard something but I wasn't sure." He blew a long breath. "I should have paid more attention. I should have..." I should have stopped her from going back into that house. "I don't know if the fire was deliberate or an accident, and I suppose it doesn't really matter."
"Do you think it was someone looking for the seal?" Quint said.
"Or possibly the impression that would prove the veracity of the seal. Without the impression, anyone could claim discovery of the seal." Nate shook his head. "But I don't know. Frankly, I don't care."
Quint paused. "Did you find it?"
"Find what?" Sterling looked from one brother to the other, then realization dawned on his face. "The seal? The Montini seal?"
"We did." Nate could still feel it in his pocket.
"Well that's..." Sterling searched for the right word. "Good?"
Nate's gaze met his older brother's.
Sterling's voice softened. "She'll be fine, Nate."
"Of course she will," Quint added staunchly.
The door to Gabriella's room opened and Dr. Crenshaw stepped out, a grim look on his face, followed by Mother. Nate jumped to this feet, the letters falling to the floor. "How is she?"
"The cut on her head was superficial. I doubt it will even leave a scar. However..." The doctor paused. "Her lungs are somewhat congested, as would be expected, given what she's been through. And she has a nasty bump on her head that concerns me."
Nate resisted the urge to snap at the man. "But will she be all right?"
"Quite honestly, Nathanial, it's too early to say for certain. I shall be back tomorrow morning. We should know a great deal more then, one way or the other." He turned to Mother. "Someone should stay with her. Send for me if there are any changes."
"Yes, of course. Thank you." Mother signaled to Andrews to escort the doctor out. "I shall stay with her."
"No." Panic gripped him at the thought of her waking up without him there. Or him not being there and Gabriella never waking up at all. "I'll stay with her."
"Darling." Sympathy shone in his mother's eyes. "I don't think that's wise."
"Let him, Mother," Sterling said.
Mother cast a quelling glance at Sterling then turned back to her youngest son. "Nathanial, you smell strongly of smoke. It's quite overpowering. As her house has just burned, I daresay that's not the aroma she should awaken to. In addition, you are exhausted. You cannot be of any help to her in this state. I want you to bathe and sleep, then you may sit with her for as long as you wish."
He didn't like it but knew she was probably right. "Very well."
She aimed a pointed glance at Quinton. "You smell no better than he does."
"Yes, Mother," Quinton murmured.
"As for you." She addressed Muldoon. "I am aware of your connection to Miss Montini. You are welcome to stay outside her room as long as you like, and we will inform you at once if there's any change at all."
"I do appreciate it, ma'am." Muldoon nodded.
"There is nothing any of us can do at the moment but wait. Nathanial." Mother laid her hand on his arm. "We have known Dr. Crenshaw for many years. He has cared for everyone in this family at one time or another. I trust him completely and I can tell you he is most optimistic."
Nate smiled wryly. "That was optimistic?"
"Yes," she said firmly. "Gabriella will be fine." She turned to go back into Gabriella's room then stopped and knelt down to pick up the packet of letters. "What's this?"
"It's why she went back into the house," Nate said. "They're letters written to her mother. Gabriella found them after her brother's death." His tone was grim. "He kept them from her."
"One can only hope he is burning in Hell." Mother turned the packet over in her hands. "One should do something about this."
"As Montini is dead, I believe someone has," Quint said wryly.
"Yes, of course," Mother murmured, and glanced pointedly at Nate. "Now, go."
Nate quickly bathed and donned fresh clothes, but he had no intention of staying away from Gabriella. He couldn't leave her, couldn't abandon her. He returned within an hour to sit by the side of her bed. Mother joined him for a while, and Muldoon stayed outside her door as well. He was grateful to know the older man was there.
The food his mother sent up for him later, as he kept his vigil, sat uneaten through the long hours of the day and into the night. Occasionally, in spite of himself, he dozed, but those brief moments of sleep were filled with images of Gabriella dashing back into the house, flames licking at the stairs, and the terror in her blue eyes.
So he watched her too still form in the bed. And he listened to her labored breathing and thought it eased somewhat through the night, but he was too weary to know if he had heard some improvement or just hoped he had.
And he thought of all the things he loved about her and how she'd said that she loved him and hadn't asked for anything in return. And he thought of all the things he hadn't said to her, all the things he'd hadn't had a chance to say.
And prayed it wasn't too late to say them.
## Twenty-seven
Dr. Crenshaw stepped out of Gabriella's room the next morning, his expression as grim as ever.
Muldoon and Nate jumped to their feet. "Well?"
The physician met Nate's gaze directly. "She is much improved. Her lungs sound almost clear, although she will cough a bit for the next few days. Fortunately, Miss Montini is a very strong and healthy young woman."
"And her head?" Nate asked.
"She will have a dreadful headache but her eyes look good. I am confident she will be fine within a few days."
Relief washed through Nate, and Muldoon heaved an audible sigh.
"I have given her something to ease the pain and instructions to continue its use. It will also help her sleep." The doctor's firm gaze pinned Nate's. "What she needs now is rest. No excitement of any kind, and the fewer visitors the better." Dr. Crenshaw narrowed his eyes. "I have known you all of your life, Nathanial Harrington, and it is obvious to me that you care for this young woman. The body has a way of healing itself. Rest and sleep are the best things for her. Your mother or the presence of another female would be acceptable, but I would strongly advise that you limit your visits."
"But I—"
"I suspect you would provide excitement she does not need. Leave her be, Nathanial, for now. However," the doctor's expression softened, "she is awake now, although probably only for a few minutes. I should warn you, the medication will make it difficult for her to concentrate and she may not make much sense. You may see her now, but for no more than a minute," he added firmly.
"Thank you." Nate started into Gabriella's room but Muldoon stopped him.
"My wife and Miss Henry are in the country. I didn't want to fetch them until we knew Miss Montini would recover. Now, I must go, but we'll return as quickly as possible." The big man's gaze locked with Nate's. "Take care of her."
"Always," he said, the conviction in his voice straight from his heart.
Mother stepped out of Gabriella's room. "Only a minute, remember."
Nate nodded, walked into her room and directly to her bedside. Gauze covered the cut on her head, her face was deathly pale, and her blue eyes seemed enormous. Still, she was the most beautiful thing he'd ever seen.
She cast him a weak smile. "Are you going to chastise me?"
"No." He sat down beside her bed and took her hand in his. "Not today."
"Oh dear." She sighed. "Then I must be dying."
"You're not dying." His voice was rough with emotion. "You shall live a very long time. With me."
"How very nice." Her eyelids were heavy, her voice soft. "I'm so tired. And so sorry."
He smiled. "Nothing to be sorry about."
"You could have been hurt. I couldn't bear it if you were hurt." Her eyes drifted closed, then opened, but it was obviously a struggle for her. "But I found it, didn't I?"
"Yes, you did." He leaned closer and gently kissed her forehead. "You found the seal."
"No." The word was no more than a sigh, and her eyes drifted closed again. "I found the letters," she murmured, and was again asleep.
He watched her a few moments, then quietly took his leave. He paused outside her door and blew a long breath. Staying away from her now might be the hardest part of any of this. But if that's what was necessary, that's what he would do.
His mother hooked her arm through his and walked him down the hall toward the stairs. "I have an idea, Nathanial, and I'm not certain you're going to like it."
He ran a weary hand through his hair. "What is it, Mother?"
"I think it would be best—"
"Where is she?" A short blond woman with a determined air and fire in her eyes appeared at the top of the stairs, Mr. Dennison a step behind her.
"I tried to stop her," Mr. Dennison said in a helpless tone. He'd never known Dennison to sound the least bit helpless before.
"Where is she? What have you done with her?" the woman demanded.
"Lady Wyldewood, Mr. Harrington, this is Miss Henry," Dennison said.
"Of course." Mother smiled. "I have been expecting you. Miss Montini is asleep now but the doctor assures us she will be quite all right in no time."
"Thank God." Miss Henry's expression crumbled. "That's what Mr. Muldoon said, but regardless, I have to see for myself. We met him a few minutes ago right outside the house. We came here as soon as we arrived home and discovered—" She drew a deep breath and squared her shoulders. "As soon as we saw our house was in ruins."
"You poor dear." Mother took her arm and steered her into a sitting room off the hall. "We have a great deal to tell you, not all of it pleasant, and the hallway is not the place to do it." She glanced at Nate. "Coming?" It was a command more than a question.
Dennison leaned toward Nate. "I shall be in the library if she—if you—need me."
Nate nodded and followed the two women.
"I wish to see Gabriella," Miss Henry said staunchly.
"And you will." Mother settled her on a love seat and rang for tea. "Nathanial, if you would be so kind as to begin."
Nate quickly recounted the events of yesterday and last night, from Gabriella's finding Lord Rathbourne's body to the fire. Miss Henry listened, eyes wide with horror and concern. "This is all quite unbelievable." She twisted her hands in her lap. "You should as well know we were sent on a fool's errand. We received a note saying Mrs. Muldoon's mother was gravely ill. We arrived to find her mother is fine so we saw no need not to return."
"Then someone wanted your house empty," Nate said slowly. "To search it, no doubt."
"So it would appear." Miss Henry shook her head. "This is my fault. I should have expected something like this. I never should have condoned her search for that blasted seal in the first place."
"My dear, I haven't known her very long," Mother said in a kindly manner, "but I sincerely doubt that you could have stopped her."
"No, you're right there. But I have been concerned from the beginning that this might prove dangerous. Her brother was..." She shot Nate a questioning glance.
Nate nodded. "We are all aware of what kind of man Enrico Montini was." He paused. "As is Gabriella."
Miss Henry heaved a resigned sigh. "I have long thought so but she never said anything. And I never brought it up."
Nate studied her. "We found the seal."
Miss Henry's eyes widened.
"It was among her brother's things." He paused. "But we don't have the impression."
"Oh." She waved in an offhand manner. "I have that."
"You do?" Nate stared.
"I thought it would be safer for Gabriella if she didn't have it. After all, no one would expect me to have it."
"Safer?" His voice rose. "Whoever was in that house last night might well have been looking for it. And as Gabriella had claimed to have it, she was scarcely safer."
"Yes, well, that might have been a mistake on my part," Miss Henry snapped.
"I think," Mother cut in, "what Gabriella needs now, in addition to rest," she shot her son a pointed glance, "is her family."
Miss Henry shook her head. "Aside from myself and the Muldoons, she has no family to speak of."
"No, Miss Henry," Mother said firmly. "In truth, Gabriella has a very large and extensive family."
"Her mother's family, you mean?"
Mother nodded.
"They didn't want her."
"Quite the contrary. They tried to find her for years." Mother paused. "They were told she was dead."
Miss Henry stared for a moment, then her eyes widened with understanding. "Enrico." She looked at Nate. "To keep control of her money?"
Nate shrugged. "Probably."
"Gabriella's aunt is one of my oldest friends," his mother said. "She, along with her sister and her daughter, have been in Paris. According to the last letter I received from Caroline, they are to return the day after tomorrow and should arrive in Dover by afternoon. From there, they have planned to go to Caroline's estate in the country rather than return to London." She looked at Nate. "We can leave on a morning train. That will get us to Dover in time to catch them, and we can bring them directly here."
"We?" Nate scoffed. "I'm not going anywhere. Can't we telegraph them?"
Mother's brows drew together. "This is not the kind of news one imparts in a telegram."
"I'm not leaving Gabriella."
"You are not supposed to bother her, remember?" Mother turned to Miss Henry. "The doctor says she is to have as few visitors as possible and she is not to be overly excited."
"I have no intention of overexciting her." Nate glared at his mother.
"No doubt." Miss Henry sniffed.
"Well, you certainly can't if you aren't here. Besides, it's likely she will sleep for at least another few days. We'll be back probably before she knows we're gone," Mother said firmly. "She needs her family, Nathanial. She needs to know there are people who care about her."
"I care about her!"
"She has lost her brother, the only family she thought she had. Regardless of the kind of man he was, she obviously loved him. Now she has lost her home as well. This will help her. Not her body as much as her heart, I think. And I know you want that." She leaned toward him, laid her hand on his arm and gazed into his eyes. "Do this for her." She straightened. "Besides, I am certain Miss Henry will wish to stay with Gabriella."
"Without question." Miss Henry nodded.
"And as you and Mrs. Muldoon have lost your home as well, I suggest you both stay here for as long as is necessary."
"That's most kind of you." Miss Henry thought for a moment. "If we are to avoid excitement, I would suggest, if she awakens, she not be told what you are up to." She shook her head. "The truth will come as a shock to her. She has long said she has no interest in her mother's family, although I have never quite believed it."
"Miss Henry," Mother said gently, "she went back into that burning house to retrieve her mother's letters. I think that says a great deal about her true feelings."
Miss Henry nodded. "I agree."
"Then we are all agreed." Mother beamed.
"No." Nate's gaze skipped from his mother to Miss Henry and back. "We are most certainly not agreed. The Verification Committee ends its meeting at noon two days from now. If we miss that, the opportunity to present the seal will be gone forever. It's what she's worked for. What she wanted."
"She wanted her mother's letters. Which makes this—" His mother set her chin in a stubborn manner. "—more important."
Nate clenched his jaw. "Miss Henry?"
Miss Henry thought for a long moment. "Discovery of the seal would have made her brother's reputation. She very much wanted that. But she has always wanted to...well, belong somewhere, and she has never felt that she did. So yes." She nodded. "I agree."
"I don't," Nate said. "However," he rolled his gaze at the ceiling, "I can see how knowing there were people who wanted her might be the best thing for her. And I certainly can't let you go alone."
"Excellent." Again his Mother beamed.
"And, as this might well turn out to be quite disastrous..." He blew a long breath. "I have an idea of my own."
"You want me to what?" Quint stared at him as if he'd lost his mind.
Nate had asked his brothers to join him in the library. If he was going to fetch Gabriella's lost family, he was going to need help.
"Listen to him, Quint," Sterling said from behind his desk.
"As I was saying..." Nate drew a deep breath. "The Verification Committee ends its meeting in two days. Even if I were to get back in time to make it, and in spite of Mother's assurances, I am not confident of that, there will be no chance to prepare a presentation. With luck, Gabriella will be well enough to present the seal herself, but she will not be able to prepare anything either. I can begin tomorrow but—" Nate met Quint's gaze firmly. "I need you to help me put a presentation in order. Indeed, I need you to do most of it."
Quint snorted. "I think not."
"You're more than qualified," Sterling said mildly. "Aside from Professor Ashworth, you probably know as much about Ambropia and the legend of the Virgin's Secret as anyone."
"I don't care," Quint scoffed. "I am not about to do anything to legitimize Enrico Montini's claim."
"You wouldn't be doing this for Enrico Montini," Nate said, "you'd be doing it for Gabriella. And me," he added pointedly.
"All things considered, it seems to me this is the least you can do." Sterling had been told of Quint's involvement with the seals and was not pleased with his brother's actions.
Quint glared at his younger brother. "If you're not back, I have no intention of presenting..." He gritted his teeth. "The Montini seal."
"If I'm not back, you'll have no seal to present." Nate shook his head. "I'm not letting the seal out of my sight."
Quint crossed his arms over his chest. "How am I expected to prepare an argument for the validity of an artifact without the artifact in question?"
"You'll have it until I leave for Dover, and I can give you the impression."
Quint narrowed his eyes. "You don't trust me."
"I trust you with my life." He met his brother's gaze directly. "I always have."
Quint studied him for a long moment, then shrugged. "I'll do what I can."
"Nate, is there any other way we can be of assistance?" Sterling asked.
"We?" Quint muttered.
"Yes," Nate said. "I don't like leaving her, but apparently I'm not permitted to see her. The doctor says she will be fine physically. But even though we have the seal, I'm not certain she isn't still in danger. Muldoon will be here and he has protected her for much of her life. Still, I would feel better—"
"Of course." Sterling nodded. "We will make certain she is kept safe. Anything else?"
"I don't know." Nate absently paced in front of the desk. This wasn't enough. Regardless of what his mother and Miss Henry thought, if he missed that meeting, Gabriella would never forgive him. No reunion could ever make up for it. He would need to do something to make it right. Some sort of grand gesture. Something unexpected to absolve him and capture her heart. The vaguest of plans took shape with every step.
"Perhaps," he nodded thoughtfully. "But I shall need the help of...of an intrepid earl and a daring smuggler king."
Sterling grinned. "As ever we ever have and ever will be."
Quint looked from one brother to the other, then smiled reluctantly. "Brothers, one for the other."
"One for the other." Nate grinned.
It was still a very good pact.
## Twenty-eight
Caroline," Mother called with an enthusiastic wave.
Lady Danworthy stared for a moment, then smiled, and made her way toward them through the crowd on the Dover pier. "Millicent!" The two women embraced. "What on earth are you doing here? Are you on your way to Paris? Wait until you see the gowns we purchased. They are magnificent. But then if you're going to Paris, you'll be visiting Mr. Worth's establishment yourself."
"I'm not going to Paris," Mother said. "I have come here to meet you."
"Why how very kind of you, and most unexpected." Lady Danworthy's brow furrowed. "My apologies, Millicent, but I don't understand why you're here."
"Get on with it, Mother," Nate said under his breath.
The boat from France had, of course, been late. Not that he hadn't expected exactly that. Regardless of the arguments made by his mother and Miss Henry, he knew this was a bad idea. Still, he and Quint had the presentation well in hand, and he'd made other arrangements as a precautionary measure. Mother was right about one thing, though. While Gabriella had awakened briefly a few times yesterday, barely long enough to take the medication the doctor left for her, she had gone right back to sleep. They might indeed return before she realized he was gone.
"I have something of great importance to tell you that simply could not wait," Mother said. "There's a café I noted at the end of the pier. We can talk there."
"Oh dear." Lady Danworthy's eyes widened. "It's something dreadful, isn't it?"
"No, dearest, it's something quite wonderful."
Lady Danworthy studied Mother for a long moment. "Millicent, I have known you for much of my life. If you say this is important, then it is." She turned and gestured at two women who stood some distance away, surrounded by several servants amidst a virtual sea of baggage. Nate winced at the sight. Oh, wouldn't that make all this easier?
"Nathanial," Mother said when the two women drew near. "You remember Lady Danworthy's sister, Mrs. Delong? And of course you know Emma."
"Although we haven't seen one another for years." Emma Carpenter held out her hand to Nate. "How are you, Nathanial?"
"Very well, thank you." He could barely choke out the words. He took her hand as much to give himself a moment to regain his composure as anything else. No wonder Gabriella seemed familiar to him when they first met. It wasn't merely that he had met her brother in Egypt, but aside from differences in the shade of their hair and eyes, and a slight difference in the shape of their mouths, Emma and Gabriella could have passed for twins. "You are as lovely as ever."
She laughed. "And you are more charming than ever, I see."
"Nathanial." His mother raised a brow. "Time is of the essence, remember?"
"Of course." Within a quarter of an hour he had arranged for the servants to stay with the luggage, settled the ladies in a café with an excellent view of the channel, and resisted the urge to check his watch more than twice.
"Well?" Lady Danworthy said. "I am dying of curiosity. What is this matter of great importance?"
"Caroline." Mother took her hand. "We have some news for you about Gabriella."
"Gabriella?" Confusion crossed Lady Danworthy's face, then she sucked in a sharp breath and her free hand reached for her sister's. "Gabriella, our niece?"
"Yes, Gabriella Montini." Mother paused, and Nate wasn't sure if it was to find the right words or prolong the drama of the moment, although it did seem to him dramatic enough. "Caroline, she's alive."
Mrs. Delong gasped. "What do you mean, she's alive?"
"I mean she's not dead. She's never been dead." Mother huffed. "Goodness, of all the things I have to explain I didn't think alive would be among them."
Lady Danworthy stared. "But we were told—"
"Yes, well that was a lie." Mother's expression hardened. "Gabriella's life up to now has been somewhat unusual, but I can tell you she is a lovely young woman. A bit headstrong and prone to impulsive behavior perhaps—"
Nate snorted to himself.
"—but brilliant and really quite delightful in her own, independent way."
Mrs. Delong's brows drew together. "Are you certain of this?"
"We have confirmed her identity, and you will have no doubts yourselves the moment you lay eyes on her." Mother smiled at Emma. "She looks very much like Emma."
"Who strongly resembles Helene," Mrs. Delong said under her breath, a stunned expression on her face.
"Helene's daughter," Lady Danworthy murmured, unshed tears glistening in her eyes. "But how?"
"It's a very long story and somewhat complicated. I shall tell it all to you on the way to London. You should know as well that Gabriella has been injured, although she is expected to be fine," Mother added quickly. "But she needs her family."
"Ladies, we should be on our way," Nate said, trying and failing to hide his impatience.
"No," Mrs. Delong said. "We can't go to London."
Nate groaned to himself. "Why not?"
Caroline looked at her sister. "Why not indeed?"
Mrs. Delong met her younger sister's gaze firmly. "We cannot meet Helene's daughter without her necklace."
"Yes, of course," Caroline murmured.
"I had forgotten all about the necklace." Mother shook her head. "I should have thought of that."
Nate clenched his jaw. "What necklace?"
"Nathanial," Emma began in a soothing manner. "One of our ancestors made his fortune working for the East India Company. He gave his wife a Chinese gaming chip set in gold to wear as a pendant, for luck I believe. She passed it down to her daughter, who passed it to hers. My grandmother had two more made, as she had three daughters, and never revealed which was the original."
"When Helene left England, hers was somehow left behind." Mrs. Delong set her jaw in a stubborn manner. "I will not meet her daughter without her necklace. She would have wanted her to have it. It means, more than anything else could, that we welcome her as a part of our family."
"Can't you give it to her later?" Nate said hopefully.
All four women stared at him as if, being a man, he couldn't possibly understand, and indeed he didn't. And the look in each and every eye told him this was not open for debate. He groaned to himself. "Where is this necklace?"
"It's at my country house," Lady Danworthy said.
He shook his head. "We can't—"
"Or course we can, Nathanial," Mother said firmly. "And can still make it back in time."
"In time for what?" Emma asked.
"As I said, my dear, this is a very long story. I shall explain it all to you on the way." She glanced at her son. "Shouldn't we be on our way?"
"Yes," he said sharply. "Let's be on our way."
They still might be able to make it back to London before the committee adjourned. If not, he had made plans for that as well. He only hoped his plan worked better than any of Gabriella's.
Gabriella struggled to open her eyes.
She was lost in the thickest of London fogs. Tendrils of haze, like incessant fingers, plucked at her, wrapped around her, reached into her soul. Voices sounded far in the distance, fading and growing more distinct and fading again. She tried to go toward them but couldn't seem to progress, couldn't seem to move at all. The fog grew deeper, darker, nearly black. So thick she could feel it envelop her, press against her skin, push into her mouth, her nose.
She couldn't see anything at all save for an orange glow off to her right. Fire, of course, the house was on fire. She turned to flee and realized she couldn't, she had to go back. She held out her hands. They shook and were empty. Shouldn't she have something? But what? And why couldn't she remember? She turned again and Lord Rathbourne stepped out of the blackness, a vaguely surprised look on his face, his shirt crimson and dripping. Somewhere in the distance a woman screamed. A high-pitched, rasping, hysterical sound of terror and panic, and...it was her voice!
She bolted upright in the bed. Pain shot through her head. She doubled over and pressed the heels of her hands to her temples and groaned.
"Gabriella?" A comforting hand rested on her shoulder.
She turned her head and peered through half-opened eyes. "Florence?"
Florence sat by her side. "Yes dear, I'm here. How do you feel?"
"I'd have to feel better to die." She groaned again. "Am I dying?"
"No, darling, you're going to be fine." Florence shook her head in a chastising manner. "You were really very lucky."
"Yes, well I don't feel lucky." She gingerly lifted her head and sat up slowly. "What happened to me?"
"You don't remember?"
"I don't seem to remember anything at the moment." Save the insidious fog and flames and Rathbourne. "What..."
"There was a fire at the house. You and Mr. Harrington—"
"The seal." She groped for the memory. "We found the seal."
"Indeed you did."
"And the fire." She remembered the heat and the smoke and the fear. Her throat ached almost as much as her head. "And Nathanial." She caught her breath. "Is Nathanial—"
"He's fine," Florence said. "There isn't a scratch on him."
Relief washed through her.
"Do you remember going back in the house?"
"Going back..." She drew her brows together. She remembered a sense of urgency...She shook her head carefully. "No."
"You went back in the house to get your mother's letters."
"Did I?" Gabriella murmured. "How very foolish of me."
"Yes, it was," Florence said firmly.
She remembered now, some of it, most of it. "And did I? Find the letters?"
"Yes, you did."
"I can't imagine why I would," she said under her breath. It made no sense to her now. Still, she did recall the feeling of urgency. "When..."
"The fire was three days ago. You have been asleep since then. You have needed it and you continue to need it." Florence nodded. "Complete rest is what the doctor said and no excitement."
Excitement was the last thing she wanted, although the throbbing in her head had subsided somewhat. "Where is Nathanial?"
"He is not here right now and that's none of your concern at the moment. No excitement, remember." Florence's voice softened. "He has been quite concerned about you."
Gabriella settled back against the pillows and managed a slight smile. "Has he?"
"Indeed he has," she said with a smile of her own, "and you shall see him as soon as you are up to it."
Gabriella plucked at the covers. "I feel up to it now."
"Nonetheless, it's not advisable," Florence said in the no nonsense manner Gabriella recognized. There would be no getting around her on this point. Probably. Florence rose to her feet. "What you need now is something to eat. Broth and tea and toast, I should think."
"I am hungry," Gabriella murmured. A thought struck and she widened her eyes. "Three days did you say?"
Florence nodded warily.
"Then the Verification Committee ends its meeting tomorrow. I have to—"
"You have to do nothing at the moment but rest." Florence's look left no room for argument. She paused. "The doctor left something for the pain in your head and to help you sleep."
"I don't want it." She shook her head gently. "The dreams..." She shuddered. "The pain in my head is nothing compared to the dreadful dreams. No, I shall do without it." She forced a smile. "I do feel much better."
Florence considered her carefully. "Very well, then. I shall be back in a minute."
She took her leave, and Gabriella rested against the pillows. It was all coming back to her. Discovering the dead viscount, hearing how her brother had died, finding the seal, the fire, the letters...
Nathanial was safe and he had the seal. She glanced at the window. It was already afternoon. Still, she needn't worry, there was time. She closed her eyes and blew a long breath. Nathanial had the seal and all would be well. He would make certain of it.
She dozed on and off through the day but by early evening her mind had cleared considerably. Where was Nathanial? Florence seemed particularly evasive and finally refused to discuss Nathanial at all save to tell her that she needed to avoid excitement, which meant avoiding Mr. Harrington. It was a most convenient excuse, and Gabriella had stopped asking. Once, when Florence was out of the room, a maid came in to bring fresh linens. Gabriella had asked her to fetch Nathanial, but the maid said that Master Nathanial had left the city. She'd had no more information that that.
Where was he? Where had he gone? He had the seal. Surely he realized the meeting was to end tomorrow at noon. What if he didn't make it? What if he didn't come back at all? What if he had gone off to find the lost city himself?
No, she told herself, firmly trying to thrust the disquieting thoughts aside. She trusted him completely. Nathanial would never betray her like that. He would never betray her at all. It was simply the circumstances and her own suspicious nature that made her wonder otherwise.
But as the day wore on into evening and night, dread curled inside her. She wanted to trust him. No, she did trust him. Surely wherever he was, whatever he was doing, it was a matter of importance. He would never abandon her. She knew that, not merely in her mind but in her heart.
Nathanial Harrington was the one person in the world she could count on.
Still, as she fell into a restless sleep that night a voice in the back of her head nagged at her.
What if she was wrong?
## Twenty-nine
This was the final day.
The thought struck Gabriella the moment her eyes opened. She threw off the covers and slid out of bed. The ache in her head was nearly gone, and certainly no reason to stay abed. Was Nathanial back? A glance at the window told her it was already late morning. Damnation! She pulled on her wrapper, swept out of her room and stepped across the hall to his door. She paused, then turned the handle and stepped inside.
"Nathanial?" She crossed the sitting room to his bedchamber. His bed was untouched. Surely it had already been made up. Unless, of course, he hadn't slept in it last night. Where was he? Not that it mattered. She trusted him.
She hurried out of his room. She should dress, it was most improper to wander about the house in her nightclothes, but it couldn't be helped. At this particular moment propriety was not uppermost in her mind. She sped down the corridor to the stairway and fairly flew down the stairs, where she met the butler's startled gaze.
"Andrews," she said without preamble. "Have you seen Mr. Harrington? Nathanial?"
"Not today, miss."
"Do you know where he is?"
"No, miss." Andrews shook his head. "I have no idea where he is at the moment."
She huffed. "What about his brother?"
"Which brother, miss?"
"Any brother," she snapped.
"Neither Master Quinton nor his lordship are at home, miss."
"And Lady Wyldewood." She raised a brow. "Has she vanished as well?"
"I would not say she has vanished, miss. But no, she is not at home. Lady Regina, however, is still abed," he added in a helpful manner.
She gritted her teeth. "So is no one else home?"
"Miss Henry and Mr. Dennison are in the library, miss."
"That's something, at any rate," she muttered, and headed toward the library. "Thank you, Andrews," she tossed back over her shoulder.
"You are quite welcome, miss."
She flung open the library door and stormed into the room, interrupting what looked to be a discussion of a somewhat intense nature between Florence and Mr. Dennison. "Where is he?"
Florence rose to her feet, Mr. Dennison a scant beat behind her. "What are you doing out of bed?"
"I feel fine, perfectly fine," she snapped. "The only thing that would make me feel better is knowing where Nathanial is."
"Quite honestly, Gabriella." Florence met her gaze directly. "I don't know."
Gabriella's jaw tightened. "Mr. Dennison?"
He shook his head. "I can't say, miss."
"Can't say or won't say?"
"At this moment, Miss Montini, I have no idea where Mr. Harrington might be."
Gabriella's gaze shifted back and forth between Mr. Dennison and Florence. "I don't believe either of you."
"Nonetheless, we are not lying to you." Florence's lips pressed together in a disapproving manner. "You do realize you are not appropriately dressed?"
"I had other things on my mind," Gabriella said sharply. She paused and drew a deep breath. "I am going to my room now to dress appropriately and then I am going to the Antiquities Society in hopes that Nathanial has brought the seal to the committee."
Florence and Mr. Dennison exchanged glances.
"And you are not going to stop me."
"We wouldn't think of doing such a thing," Florence said. "By all means, go to the Antiquities Society. I think it's an excellent idea." She nodded. "In fact, Mr. Dennison and I will be happy to accompany you."
"You will?" Gabriella narrowed her eyes. "Why?"
"Goodness, Gabriella, will you ever stop being such a suspicious creature?" Florence huffed. "First of all, it would be entirely improper for you to be unaccompanied. Secondly, as perfectly fine as you may feel, I am not especially confident that you will not collapse at any moment. And third..." She paused. "I've been with you at the beginning of all this and I'd rather like to be with you at the end. Now, do put on some proper clothing and we shall wait for you here."
"Excellent." Gabriella nodded, turned, and started back to her rooms. She knew she shouldn't be annoyed at Florence, and in truth she wasn't. She could trust Florence. As she could trust Nathanial. And as long as she kept saying that to herself, she could keep this mounting sense of doom at bay. After all, he'd done nothing to earn her distrust. Not yet. She dashed the thought from her mind. Leaps of faith, Gabriella, she told herself firmly. Leaps of faith.
It was well past noon when they finally arrived at the Antiquities Society building. The Verification Committee had adjourned and the annual general meeting would begin in a few minutes.
The moment she realized they would be too late, a heavy weight had settled in the pit of Gabriella's stomach. Still, it was not yet time to give up. She spotted Mr. Beckworth amidst the crowd milling in the corridors and hurried toward him, Florence and Mr. Dennison hard on her heels.
"Mr. Beckworth," she called.
"Gabriella." The director addressed her with a concerned smile. "I heard about the fire. Nasty business. Are you all right?"
"Perfectly fine, thank you. Mr. Beckworth..." She held her breath. "At the Verification Committee meeting, did Mr. Harrington present my brother's seal?"
"I am sorry, my dear." Sympathy shone in the older man's eyes. "I haven't seen Mr. Harrington since I met with the two of you in my office."
"I see," she said slowly. A terrible sense of defeat and disappointment washed through her. Despair caught at her throat. Still, she preferred not to let it show. She managed a polite smile. "Thank you, Mr. Beckworth."
"As you are already here, I assume you will be joining us for the general meeting."
She shook her head. "I'm not actually a member of the society."
"I know that, my dear, but it seems to me you usually attend the meeting." Mr. Beckworth smiled. "In the upstairs gallery, of course, with the other ladies."
"I don't think—"
"Of course she will," Florence said in a gracious manner. "She wouldn't think of missing it."
"Excellent." Mr. Beckworth nodded and took his leave.
"Come along." Florence hooked her arm through Gabriella's and steered her toward the stairs that led to the observation galleries. "If we don't go now, we won't find good seats."
"I have no desire to observe the meeting," Gabriella said, but allowed Florence to lead her up the stairs nonetheless.
Mr. Dennison had disappeared, but then why should he be any different from anyone else today? In truth, she had no desire to do much of anything at the moment. It was as if she had stood out in the cold for a very long time and was now numb to the touch. She dimly understood that this feeling would fade and leave in its stead despair and anger. He had failed her. She'd trusted him and he had failed her. And in an odd way, she still did trust him. Perhaps because when she fully realized that she was wrong to do so, her devastation would be complete. And she was not yet prepared for that.
They managed to procure seats in the front row, right behind the railing. The meeting would begin any minute. There was some sort of activity, apparently outside of the doorway near the dais. Florence leaned closer to the railing to see what was happening. Gabriella stared ahead unseeing. It simply didn't matter. Nothing mattered save this awful ache that was growing inside her, somewhere near her heart.
"Well?" Nate stared at his brother.
Sterling chuckled. "It's hard to turn down a request from the Earl of Wyldewood."
"Excellent." Nate breathed a sigh of relief.
Mr. Dennison hurried up to them and nodded. "She's in the gallery. It was an excellent idea to put this part in the hands of Miss Henry, sir."
"I shall have to thank her later, and you as well." Nate turned to Quinton. "And?"
"Here." Quinton thrust a thin open book at him. "These are the rules of the society. I've marked the one you want."
"Thank you."
"You should thank me," Quinton replied. "It goes against everything I believe in to look at rules of any kind." He cast his brother a reluctant grin. "One for the other."
Nate returned his grin, acknowledged Sterling's nod of support, and stepped into the room.
The director took his place behind the podium, banged his gavel, and called the meeting to order. The room quieted. Mr. Beckworth began the way he always began, welcoming the members, and then droned on, the way he always droned on, at the beginning of any meeting. Gabriella had rather enjoyed it in the past. Even in the gallery it was as if she were truly a part of it all. As if she belonged.
She tried to focus on his words, tedious though they might be. Anything to keep from thinking her own thoughts.
Beckworth paused. "This year we have had a most unusual request, but as it comes from the Earl of Wyldewood..."
The announcement caught her attention. Florence nudged her. "Are you listening to this?"
"Yes," she murmured, and stared at the dais. What on earth was going on?
"The board has agreed to allow Mr. Nathanial Harrington to address the assembly. Mr. Harrington."
The director stepped aside and Nathanial took his place, a sheaf of papers and a small book in his hand.
Nathanial?
"Good day, gentlemen. I am most grateful to be allowed the opportunity to speak to you this afternoon," Nathanial began.
Gabriella stared in shock. What was he doing? She leaned closer.
"According to the rules of the Verification Committee, an artifact presented and rejected can only be presented again before the end of the next year's meeting. Exceptions can be made only under extraordinary circumstances." Nathanial's voice rang out over the gathering, strong and confident. "I believe this particular case meets that criteria.
"Last year, Enrico Montini had in his possession an Akkadian cylinder seal. The carvings on that seal included symbols for the lost city of Ambropia and the Virgin's Secret."
Murmurs of interest washed through the crowd. Gabriella's heart lodged in her throat.
"Unfortunately, it was stolen from him and he lost his life attempting to reclaim it." Nathanial's gaze rose to the gallery and found hers. "It has now been recovered thanks to the courageous efforts of Miss Gabriella Montini, but unfortunately too late for this year's committee meeting. According to the rules of this august institution..." He paused and glanced down at the book on the podium. "The general assembly may, through a simple majority vote, call for the reconvening of the Verification Committee. I urge you to do so now.
"Aside from the unparalleled historical importance of the seal, it is to be donated to the society, thus its validation is especially important. In addition," his gaze again met hers, "I propose it be known from this point forward as the Montini seal, in recognition of the man who brought it into the light of public knowledge and the woman who risked all to return it to us. Thank you." Nathanial nodded and left the dais.
"Most unusual," Mr. Beckworth said, retaking his position. "We shall vote on that proposal at the end of the meeting when we vote on other matters. And now turning to the..."
She stared in shocked disbelief. Blood roared in her ears. Her heart thudded in her chest. Nathanial hadn't betrayed her. He hadn't abandoned her. How could she have doubted him? Even for a minute?
"Come along." Florence got to her feet. "We must be going."
Gabriella stared at her. "What?"
"Come along," Florence said firmly, took her arm and urged her to her feet. "Now."
Gabriella was fairly certain she was putting one foot in front of the other but had no knowledge of doing so. One minute they were in the gallery, and the next in the downstairs corridor.
"Miss Montini? Gabriella?"
Gabriella turned to find the Earl of Wyldewood standing behind her, a genuine smile on his face. She wasn't at all sure she'd ever seen him truly smile before. "Yes?"
"If all went as Nathanial had planned, I was to give you this." He handed her a folded note. "And then escort you to the director's office."
She stared at the note in her hand.
He leaned closer to her and lowered his voice. "You should read it now, Gabriella."
She nodded and unfolded the note. Why did her hands always seem to shake of late?
The note read:
My Dearest Gabriella,
A much as surprise might be nice, it seems to me it might be wise to introduce something of this magnitude in private. You should know that after your father's death, your mother's family tried to find you. They ended their efforts only when they were told you were dead. They have always wanted you and have loved you as they loved your mother. My mother has arranged for your reunion.
You should know as well that I too love you.
Always,
Nathanial
She stared at the words on the page. A lump rose to her throat and her eyes fogged.
"Gabriella?" the earl said softly.
She glanced at him. "Do you know what this says?"
He grinned. "I have an idea."
She smiled at him. "Why, you're not at all stiff and stodgy, are you?"
"Don't tell anyone." He offered his arm. "Shall we, then?" She nodded and he escorted her along the corridor.
"Where..."
They paused before the door to the director's office and he stepped aside. "They're expecting you."
"They? Who?"
"Go on, my dear," Florence said behind her.
Gabriella looked at the earl. "Aren't you coming in?"
He shook his head. "I think not."
She glanced around. "Where is Nathanial?"
"I'm not exactly sure." He smiled in a brotherly sort of way. "Go on now. I suspect your future and some of your past awaits you."
"Very well." This was entirely too much to comprehend all at once. Nathanial had proven to truly be her hero. The news about her family was something she'd never dared dream of. And he had declared his love. It was all she'd ever wanted. No, it was so much more. Why, then, was she hesitating? She drew a deep breath and opened the door.
The door swung open and Gabriella stepped into the room. She was as lovely as always, if still a bit pale. She appeared cool and serene and confident, and he knew her well enough by now to know she probably wasn't the least bit serene.
Mother bustled over to her, took her arm and brought her to her family. "Gabriella, I want you to meet your aunts, Caroline and Beatrice, and your cousin Emma."
For a long awkward second, no one said a word. Then Lady Danworthy burst into tears and threw her arms around Gabriella. Followed a scant moment later by Mrs. Delong and Emma, all of them crying and talking at once. In the midst of it, one of her aunts fastened her mother's necklace around Gabriella's neck. The only one not weeping was Gabriella herself, who looked stunned. But not unhappy. He had been right to warn her.
Even Mother sniffed back a tear. "Isn't it wonderful. See what you've done, darling?"
"I didn't do it, Mother, you did."
"Nonsense, Nathanial." Mother shrugged. "You could have refused to accompany me. You could have put your foot down and stopped me. I am, after all, a mere woman."
He laughed. "There is nothing mere about you."
"Nor is there anything mere about her," Mother said pointedly.
"Nothing mere about her at all," he murmured. Gabriella had family now, and her brother's reputation had been assured in death regardless of what it might have been in life. She had everything she'd ever wanted. Would she still want him?
After a few more minutes she turned away from her family and stepped toward him.
"Thank you." Her voice was barely more than a whisper, whether from emotion or the lingering effects of the fire, he didn't know. "For what you did today and this." She drew a deep breath. "I read your note. Without it I would have said the wrong things entirely. I might have been, well, less than pleasant. I had no idea."
"I suspected that." He stared down at her. "I thought you should know."
"And the rest of it?" Her gaze searched his. "The part that had nothing to do with my family?"
"Oh that, yes, well..." He studied her for a moment. "I cannot ask you to join with me in my work, my travels, because there would always be a risk to you, to your safety. I cannot put you in the kind of jeopardy you experienced in your youth, and I cannot take you away from the family you have just found."
Disbelief sparked in her blue eyes. "Nathanial, I—"
"Therefore, I will stay here in England if you will stay with me as my wife."
"I—"
He took her hands. "You once told me you wanted to become indispensable to your brother. I find that you have become indispensable to me."
She swallowed hard.
"Not to my work, although you are brilliant and clever, but you have become indispensable to my life because I cannot imagine another day without you. And you have become indispensable to my heart because it would surely shatter if you were not by my side."
She stared up at him, tears glittering in her eyes.
An audible sob sounded from her aunts.
"Are you going to say anything?" He raised her hands to his lips. "Leaps of faith, Gabriella. Will you leap? I promise always to catch you."
"No." She shook her head. "I can't."
His heart twisted. "I see."
"No you don't." She smiled up at him with a radiance that caught at his soul. "I can't leap now, Nathanial—" Her voice faltered. "I already have."
At once she was in his arms, and he pressed his lips to hers and didn't care where they were or who was watching. He knew only that he would keep her by his side for as long as they lived. And knew as well that this love they shared was as timeless, as priceless, and as rare as anything left by the ancients.
And further knew that she was the greatest treasure of all.
## Epilogue
You should have taken the seal back from Montini when you had the chance. It would have made everything much easier. I assume there was a moment after you killed him."
"No." Javier Gutierrez's gaze slid from one of his employers to the other. "I was interrupted. And I was not about to risk a hangman's noose for the seal or for you." He settled back in his chair and narrowed his eyes. "You pay well but not well enough."
"It was a pity to have burned down the girl's house."
"An unfortunate accident." Gutierrez blew an annoyed breath. "No one was supposed to be there. In my haste to leave, I knocked over the lamp."
"And Rathbourne's death? Rather unnecessary, I would think, although I can't imagine he'll be missed. Another accident?"
"Let us say...a moment of passion. He owed me money." Gutierrez shrugged. "He refused to pay. What could I do?"
His employers exchanged glances. They were an unusual pair, in his experience. Cold, ruthless, and well matched. If he was the type of man to be easily scared, these two would have done it. As it was, merely being in their presence sent a chill of unease along his spine. Gutierrez shifted in his seat. He was not used to unease.
"Still, obtaining Rathbourne's seal from his wife will be far easier than wresting it from him."
The other nodded. "Then we will have three."
"And they will reveal the location of the lost city, the final resting place of Ambropia. And we alone will at long last know the Virgin's Secret."
## About the Author
VICTORIA ALEXANDER was an award-winning television reporter until she discovered fiction was much more fun than real life. She turned to writing full time and has never looked back.
Victoria grew up traveling the country as an Air Force brat and is now settled in Omaha, Nebraska, with her husband, two kids in college (buy her books!), and two bearded collies named Sam and Louie. She firmly believes housework is a four-letter word, there are no calories in anything eaten standing up, procrastination is an art form, and it's never too soon to panic.
And she loves getting mail that doesn't require a return payment. Write to her at P.O. Box 31544, Omaha, NE 68131.
www.eclectics.com/victoria
Visit www.AuthorTracker.com for exclusive information on your favorite HarperCollins author.
## By Victoria Alexander
THE VIRGIN'S SECRET
SEDUCTION OF A PROPER GENTLEMAN
THE PERFECT WIFE
SECRETS OF A PROPER LADY
WHAT A LADY WANTS
A LITTLE BIT WICKED
LET IT BE LOVE
WHEN WE MEET AGAIN
A VISIT FROM SIR NICHOLAS
THE PURSUIT OF MARRIAGE
THE LADY IN QUESTION
LOVE WITH THE PROPER HUSBAND
HER HIGHNESS, MY WIFE
THE PRINCE'S BRIDE
THE MARRIAGE LESSON
THE HUSBAND LIST
THE WEDDING BARGAIN
Coming Soon
BELIEVE
## Copyright
This is a work of fiction. Names, characters, places, and incidents are products of the author's imagination or are used fictitiously and are not to be construed as real. Any resemblance to actual events, locales, organizations, or persons, living or dead, is entirely coincidental.
THE VIRGIN'S SECRET. Copyright © 2009 by Cheryl Griffin. All rights reserved under International and Pan-American Copyright Conventions. By payment of the required fees, you have been granted the non-exclusive, non-transferable right to access and read the text of this e-book on-screen. No part of this text may be reproduced, transmitted, down-loaded, decompiled, reverse engineered, or stored in or introduced into any information storage and retrieval system, in any form or by any means, whether electronic or mechanical, now known or hereinafter invented, without the express written permission of HarperCollins e-books.
Adobe Digital Edition March 2009 ISBN 978-0-06-186635-7
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{"url":"http:\/\/mathhelpforum.com\/calculus\/119470-root-test-convergence-divergence.html","text":"# Math Help - Root test for convergence\/Divergence\n\n1. ## Root test for convergence\/Divergence\n\nsum 1-->infinity of (arctan(1\/n)^n\n\nHere's what I've done:\n\nlim 1->infinity of: nth root (above expression)\n\n= lim 1-> infinity of arctan(1\/n)\n\nNow what?\n\n2. Originally Posted by jzellt\nsum 1-->infinity of (arctan(1\/n)^n\n\nHere's what I've done:\n\nlim 1->infinity of: nth root (above expression)\n\n= lim 1-> infinity of arctan(1\/n)\n\nNow what?\n\nWell the properties of limits for composition functions, we can notice that the inside limit goes to 0 as we got to infinity, now we are approach the value of the arctan of 0 which is?\n\nCorrect me if wrong.\n\n3. Any other advice on this one...\n\n4. yes, brush up your notes from calculus I, since RockHard told ya what you need to do.\n\n5. At first the problem seems an easy application of a convergence criterion... may be however that it covers up an insidious trap ...\n\nFrom...\n\nhttp:\/\/en.wikipedia.org\/wiki\/Convergent_series\n\nRoot test: suppose that the terms of the sequence are non negative and that there exists r such that...\n\n$\\lim_{n \\rightarrow \\infty} \\sqrt[n] {a_{n}} = \\frac{1}{r}$ (1)\n\nIf $r>1$ the series converges, if $r<1$ the series diverges, if $r=1$ the root test is inconclusive...\n\nIt is not quite obvious the reason to have in (1) $\\frac{1}{r}$ instead of $r$. Anyway in our case is...\n\n$\\lim_{n \\rightarrow \\infty} \\sqrt[n] {a_{n}} = \\lim_{n \\rightarrow \\infty} \\tan^{-1} \\frac{1}{n} = 0$ (2)\n\nNow, if we want to apply the root test as explained by Wiky, the problem is that an $r$ such that $\\frac{1}{r}=0$ doesn't exists ...\n\nMerry Christmas from Italy\n\n6. Well...I'd find wolfram to be more accurate than wiki. Wolfram states this converges by root test, if I entered the equation right","date":"2014-07-11 00:28:26","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 9, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.851262092590332, \"perplexity\": 1897.6117440185421}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2014-23\/segments\/1404776423684.4\/warc\/CC-MAIN-20140707234023-00072-ip-10-180-212-248.ec2.internal.warc.gz\"}"}
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\section{Introduction}
Galton-Watson branching processes are a classical model for the evolution of population sizes, see e.g. \cite{AtN,Har-55}.
More specifically, there is an interest in the underlying genealogical trees. In the most basic
model, there is a single \em progenitor \em that produces $i$ children with probability $q(i)$ for
some \em offspring distribution \em $q$ on $\bN_0=\{0,1,2,\ldots\}$; recursively, each individual
in the population produces children independently and according to the same distribution $q$. We
represent this model by a \em graph-theoretic tree \em rooted at the progenitor, where each individual is a vertex and the \em parent-child relation \em specifies edges $v\rightarrow w$ between parent $v$ and child $w$. Vertices related to just their parent vertex and to no child
vertices are called
\em leaves\em. More precisely, we will follow
Neveu \cite{Nev-GW} to distinguish individuals (see Section \ref{trees}). We
will consider in this paper the following well-known and/or natural variants of Galton-Watson
trees (see e.g. Jagers \cite{Jag-68}):
\begin{itemize}\item ${\rm GW}(q)$-trees as the most basic model just described;
\item ${\rm GW}(q,\kappa)$-trees as ${\rm GW}(q)$-trees, where each individual is marked by an independent
\em lifetime \em with distribution $\kappa$ on $(0,\infty)$; this includes the case
$\kappa={\rm Exp}(c)$ of the exponential lifetime distribution with rate parameter
$c\in(0,\infty)$;
\item ${\rm GW}(q,\kappa,\beta)$-bushes as \em bushes \em (finite sequences) of a random number $N$
of ${\rm GW}(q,\kappa)$-trees, where $N$ is Poisson distributed with parameter $\beta\in[0,\infty)$, in shorthand: $N\sim{\rm Poi}(\beta)$;
\item ${\rm GWI}(q,\kappa,\eta,\chi)$-forests as \em forests \em (point processes on the \em forest
floor \em $[0,\infty)$) of independent bushes of $N_i$ ${\rm GW}(q,\kappa)$-trees at the locations
$S_i$ of a renewal process with inter-renewal distribution $\chi$ on $(0,\infty)$, where
each $N_i$, $i\ge 1$, has distribution $\eta$ on $\bN=\{1,2,\ldots\}$.
\end{itemize}
With the common interpretation that individuals give birth only at the end of their life and that renewal locations are \em immigration times\em, all but the first model give rise to continuous-time processes counting the number $Y_t$ of individuals in the population at
time $t\ge 0$. In these continuous-time models it is natural to take $q(1)=0$, since an
individual producing a single child at its death time can be viewed as continuing to live instead of being replaced by its child.
Reduction by Bernoulli leaf colouring was studied in \cite{DuW} and reads as follows in our
setting:
\begin{itemize}\item independently mark each leaf of a tree $T$ (with lifetimes), or of a bush $B=(T_{(1)},\ldots,T_{(N)})$ or of a forest $F=(B(t),t\ge 0)$ by a
\em black \em colour mark with probability $1-p\in(0,1)$, by a \em red \em colour mark otherwise; for this to be non-trivial, let $q(0)>0$, as $T$ then has leaves;
\item if there are any black leaves, consider, as illustrated in Figure \ref{discperc},
\begin{itemize}\item the \em $p$-reduced subtree \em $T^{p-\rm rdc}_{\rm sub}$ of $T$ as the
subtree of $T$ spanned by the root and the black leaves (with lifetime marks inherited);
\item and the \em $p$-reduced tree \em $T^{p-\rm rdc}$ derived from $T^{p-\rm rdc}_{\rm sub}$ by
identifying vertices via the equivalence relation generated by $v\equiv w$ for vertices
in $T_{\rm sub}^{p-\rm rdc}$ if $v\rightarrow w$ and $w$ is the only child of $v$ in $T_{\rm sub}^{p-\rm rdc}$
(marked by the sum of lifetimes in each equivalence class);
\item or the \em $p$-reduced bush \em
$B^{p-\rm rdc}=(T_{(I_1)}^{p-\rm rdc},\ldots,T_{(I_{N^{p-\rm rdc}})}^{p-\rm rdc})$ as the
$p$-reduced trees associated with the subsequence $(I_1,\ldots,I_{N^{p-\rm rdc}})$ of trees
in $B$ that have black leaves;
\item or the \em $p$-reduced forest \em $F^{p-\rm rdc}=(B^{p-\rm rdc}(t),t\ge 0)$ of
$p$-reduced bushes.
\end{itemize}
\end{itemize}
\vspace{-0.3cm}
\begin{figure}[ht]
\psfrag{Tsub}{$T_{\rm sub}^{p-\rm rdc}$}
\psfrag{Tcol}{$T^{p-\rm col}$}
\psfrag{Tblue}{$T^{p-\rm rdc}$}
\psfrag{time}{\scalebox{.7}{$t$}}
\epsfxsize=7cm
\centerline{\epsfbox{reductiontrees3.eps}}
\caption{{Black vertices are represented by solid circles
and red ones by circle lines.
}}
\label{discperc}
\end{figure}
It is easily seen that if $T$ is a Galton-Watson tree, then given that there are
any black leaves, the $p$-reduced subtree $T_{\rm sub}^{p-\rm rdc}$ and the $p$-reduced tree $T^{p-\rm rdc}$ are also Galton-Watson trees \cite{DuW}. By standard thinning properties of Poisson point processes, the $p$-reduced bush $B^{p-\rm rdc}$ associated with a ${\rm GW}(q,\kappa,\beta)$-bush $B$ is also a
Galton-Watson bush. We refer to the offspring distribution $q^{p-\rm rdc}$, the lifetime
distribution $\kappa^{p-\rm rdc}$ and the Poisson parameter $\beta^{p-\rm rdc}$ of $B^{p-\rm rdc}$ as the \em
$p$-reduced triplet $(q^{p-\rm rdc},\kappa^{p-\rm rdc},\beta^{p-\rm rdc})$ associated with $(q,\kappa,\beta)$\em, similarly for forests etc. It is not hard to find offspring distributions $\widehat{q}$ that do \em not \em arise as $p$-reduced offspring distributions for any $q$ (e.g. $\widehat{q}(0)=\widehat{q}(3)=1/2$). More precisely, \cite{DuW} obtained the following characterisation.
\begin{theo}[Theorem 4.2 of \cite{DuW}]\label{thm1} For an offspring distribution $q$, the following are
equivalent:
\begin{enumerate}\item[\rm(i)] There is a family $(q_\lambda)_{\lambda\ge 0}$ of offspring
distributions with $q_1=q$ such that $q_\mu$ is the $(1-\mu/\lambda)$-reduced offspring
distribution associated with $q_\lambda$, for all $0\le\mu<\lambda<\infty$.
\item[\rm(ii)] The generating function $\varphi_q$ of $q$ satisfies
\begin{equation} \varphi_q(s)=\sum_{i=0}^\infty s^iq(i)=s+\widetilde{\psi}(1-s),\qquad 0\le s\le 1,\label{genfnbm}
\end{equation}
where for some $\widetilde{b}\in\bR$, $\widetilde{a}\ge 0$ and a measure $\widetilde{\Pi}$ on $(0,\infty)$ with $\int_{(0,\infty)}(1\wedge x^2)\widetilde{\Pi}(dx)<\infty$,
\begin{equation} \widetilde{\psi}(r)=\widetilde{b}r+\widetilde{a}r^2+\int_{(0,\infty)}(e^{-rx}-1+rx\mathbf{1}_{\{x<1\}})\widetilde{\Pi}(dx),\qquad r\ge 0.\label{lkbm}
\end{equation}
\end{enumerate}
In the setting of {\rm (i)} and {\rm (ii)}, a consistent family
$(B_\lambda)_{\lambda\ge 0}$ of
${\rm GW}(q_\lambda,{\rm Exp}(c_\lambda),\beta_\lambda)$-bushes can be constructed such that $(B_\mu,B_\lambda)\overset{{\rm (d)}}{=}(B_\lambda^{(1-\mu/\lambda)-\rm rdc},B_\lambda)$ for all
$0\le\mu<\lambda<\infty$. For each $c=c_1\in(0,\infty)$ and $\beta=\beta_1\in(0,\infty)$, the family $(q_\lambda,c_\lambda,\beta_\lambda)_{\lambda\ge 0}$ is unique, $(q_\lambda)_{\lambda\ge 0}$ does not depend on $(c,\beta)$, while $(c_\lambda)_{\lambda\ge 0}$ depends on $q$ but not on $\beta$ and $(\beta_\lambda)_{\lambda\ge 0}$ on $q$ but not on $c$.
\end{theo}
In \cite{DuW}, this result is a key step in the construction of L\'evy trees as
genealogies of Markovian continuous-state branching processes with branching mechanism $\psi$, where $\psi$ is a linear transformation of $\widetilde{\psi}$ that we recall in Section \ref{Consistent family}. In the present paper we establish characterisations analogous to Theorem \ref{thm1} for the other variants of
Galton-Watson trees, bushes and forests.
\begin{theo}\label{thm2} For a pair $(q,\kappa)$ of offspring and lifetime distributions, the
following are equivalent:
\begin{enumerate}\item[\rm(i)] There are families $(q_\lambda,\kappa_\lambda)_{\lambda\ge 0}$
with $q_1=q$ and $\kappa_1=\kappa$ such that
$(q_\mu,\kappa_\mu)$ is the $(1-\mu/\lambda)$-reduced pair associated with
$(q_\lambda,\kappa_\lambda)$, for all $0\le\mu<\lambda<\infty$.
\item[\rm(ii)] The generating function $\varphi_q$ of $q$ satisfies $\varphi_q(s)=s+\widetilde{\psi}(1-s)$,
where $\widetilde{\psi}$ is of the form {\rm(\ref{lkbm})}. Moreover, $\kappa$ is geometrically
divisible in that there is a family $(X_\alpha^{(j)},j\ge 1)$ of independent identically
distributed random variables and $G^{(\alpha)}\sim{\rm Geo}(\alpha)$ independent geometric with parameter
$\alpha$, i.e. $\bP(G^{(\alpha)}=k)=\alpha(1-\alpha)^{k-1}$, $k\in\bN$, such that $X_\alpha^{(1)}+\cdots+X_\alpha^{(G^{(\alpha)})}\sim\kappa$
\begin{itemize}\item for all $\alpha>1/\widetilde{\psi}^\prime(\infty)$ if $\widetilde{\psi}^\prime(\infty)<\infty$, where $\widetilde{\psi}^\prime(\infty)$ means $\lim_{r\rightarrow\infty}\widetilde{\psi}^\prime(r)$;
\item for all $\alpha>0$ if $\widetilde{\psi}^\prime(\infty)=\infty$.
\end{itemize}
\end{enumerate}
In the setting of {\rm (i)} and {\rm (ii)}, a consistent family $(B_\lambda)_{\lambda\ge 0}$ of
${\rm GW}(q_\lambda,\kappa_\lambda,\beta_\lambda)$-bushes can be constructed such that $(B_\mu,B_\lambda)\overset{{\rm (d)}}{=} (B_\lambda^{(1-\mu/\lambda)-\rm rdc},B_\lambda)$ for all
$0\le\mu<\lambda<\infty$. For each $\beta=\beta_1\in(0,\infty)$, the family $(q_\lambda,\kappa_\lambda,\beta_\lambda)_{\lambda\ge 0}$ is unique, $(q_\lambda)_{\lambda\ge 0}$ does not depend on $(\kappa,\beta)$ while $(\kappa_\lambda)_{\lambda\ge 0}$ depends on $q$ but not on $\beta$ and $(\beta_\lambda)_{\lambda\ge 0}$ on $q$ but not on $\kappa$.
\end{theo}
The requirement on $\kappa$ set in the second bullet point is referred to as geometric infinite
divisibility in the literature, see \cite{KMM}, also Section \ref{contstate} here. Since the distribution $\kappa={\rm Exp}(c)$ is geometrically infinitely
divisible, Theorem \ref{thm2} is an extension of Theorem \ref{thm1}.
\begin{theo}\label{thm3} For a pair $(q,\eta)$ of offspring and immigration distributions, the following are equivalent:
\begin{enumerate}\item[\rm(i)] There are families $(q_\lambda,\eta_\lambda)_{\lambda\ge 0}$
with $q_1=q$ and $\eta_1=\eta$ such that
$(q_\mu,\eta_\mu)$ is the $(1-\mu/\lambda)$-reduced pair associated with
$(q_\lambda,\eta_\lambda)$, for all $0\le\mu<\lambda<\infty$.
\item[\rm(ii)] The generating function $\varphi_q$ of $q$ satisfies $\varphi_q(s)=s+\widetilde{\psi}(1-s)$,
where $\widetilde{\psi}$ is of the form {\rm(\ref{lkbm})}. Moreover, the generating function $\varphi_\eta$ of $\eta$ satisfies
$$\varphi_\eta(s)=\sum_{i=1}^\infty s^i\eta(i)=1-\widetilde{\phi}(1-s),\qquad 0\le s\le 1,\label{genfnim}
$$
where for some $\widetilde{d}\in\bR$, and a measure $\widetilde{\Lambda}$ on $(0,\infty)$ with $\int_{(0,\infty)}(1\wedge x)\widetilde{\Lambda}(dx)<\infty$,
\begin{equation} \widetilde{\phi}(r)=\widetilde{d}r+\int_{(0,\infty)}(1-e^{-rx})\widetilde{\Lambda}(dx),\qquad r\ge 0.\label{lkim}
\end{equation}
\end{enumerate}
In the setting of {\rm (i)} and {\rm (ii)}, a consistent family $(F_\lambda)_{\lambda\ge 0}$ of
${\rm GWI}(q_\lambda,{\rm Exp}(c_\lambda),\eta_\lambda,{\rm Exp}(\paramgamma_\lambda))$-forests can
be constructed such that $(F_\mu,F_\lambda)\overset{{\rm (d)}}{=} (F_\lambda^{(1-\mu/\lambda)-\rm rdc},F_\lambda)$ for all
$0\le\mu<\lambda<\infty$. For each $c=c_1\in(0,\infty)$ and $\paramgamma=\paramgamma_1\in(0,\infty)$, the family $(q_\lambda,c_\lambda,\eta_\lambda,\paramgamma_\lambda)_{\lambda\ge 0}$ is unique, $(q_\lambda)_{\lambda\ge 0}$ does not depend on $(c,\eta,\paramgamma)$, while $(c_\lambda)_{\lambda\ge 0}$ depends on $q$ but not on $(\eta,\paramgamma)$, $(\eta_\lambda)_{\lambda\ge 0}$ depends on $q$ but not on $(c,\paramgamma)$ and $(\paramgamma_\lambda)_{\lambda\ge 0}$ depends on $(q,\eta)$ but not on $c$.
\end{theo}
\noindent The binary special case with single immigrants, where for some $\theta\ge 0$ and all $\lambda\ge 0$
$$q_\lambda(0)=\frac{1}{2}+\frac{1}{2\sqrt{\theta^2+2\lambda}},\quad q_\lambda(2)=\frac{1}{2}-\frac{1}{2\sqrt{\theta^2+2\lambda}},\qquad c_\lambda=\sqrt{\theta^2+2\lambda},$$
$$\eta_\lambda(1)=1,\qquad h_\lambda=\sqrt{\theta^2+2\lambda}-\theta,$$
leads to the setting of \cite{PiW-05}, where $(F_\lambda)_{\lambda\ge 0}$ was shown to have independent ``increments'' expressed by a composition rule, and to converge to the forest in Brownian motion with drift $-\theta$.
Theorems \ref{thm2} and \ref{thm3} describe in the same way the genealogy of associated continuous-state branching processes (CSBP) as Theorem \ref{thm1}. Specifically, for Theorem
\ref{thm2} the continuous-state processes are Sagitov's age-dependent ${\rm CSBP}(K,\psi)$ based on a branching mechanism $\psi$ and the distribution of a local time process $K$, i.e.\ either an inverse subordinator or an inverse increasing random walk, see \cite{KaS-98,Sag-91} and Section \ref{CSBPKpsi} here; for Theorem \ref{thm3}, they are CSBP with immigration,
${\rm CBI}(\psi,\phi)$, where $\phi$ is an immigration mechanism, see \cite{KaW-71,Lam-CBI} and Section \ref{contstate} here.
\begin{prop}\label{prop4a} Let $(Z_t^\lambda,t\ge 0)$ be the population size process in the setting of Theorem \ref{thm2}. If $\widetilde{\psi}^\prime(0)>-\infty$, then
$$ \frac{Z^\lambda_t}{\psi^{-1}(\lambda)}\rightarrow Z_t\qquad\mbox{almost surely as $\lambda\rightarrow\infty$, for all $t\ge 0$,}
$$
where $(Z_t,t\ge 0)$ is a ${\rm CSBP}(K,\psi)$ with $Z_0=\beta$, for some $\psi$ linear transformation of $\widetilde{\psi}$ and $K=(K_s,s\ge 0)$ such that $\inf\{s\ge 0\colon K_s>V_\lambda\}\sim\kappa_\lambda$ for $V_\lambda\sim{\rm Exp}(c_\lambda)$ with $c_\lambda$ as in Theorem \ref{thm1}.
\end{prop}
\begin{prop}\label{prop4} Let $(Y_t^\lambda,t\ge 0)$ be the population size process in the setting of Theorem \ref{thm3}. Then
$$ \frac{Y_t^\lambda}{\psi^{-1}(\lambda)}\rightarrow Y_t\qquad\mbox{in distribution as $\lambda\rightarrow\infty$, for all $t\ge 0$,}
$$
where $(Y_t,t\ge 0)$ is a ${\rm CBI}(\psi,\phi)$ with $Y_0\!=\!0$, for $\psi$ and $\phi$ linear transformations of $\widetilde{\psi}$ and $\widetilde{\phi}$. If furthermore $\psi^\prime(0)>-\infty$ and $\phi^\prime(0)<\infty$, then the convergence holds in the almost sure sense.
\end{prop}
\noindent These convergence results should be seen in the context of the large literature on space-time scaling limits of branching processes in discrete or continuous time, see
\cite{Duq-imm,DuW2,KaS-98,KaW-71,Lam-lim,Pin-lim,Sag-91}. Convergence in distribution holds under much weaker assumptions on the families $(q_\lambda,\kappa_\lambda,\beta_\lambda)_{\lambda\ge 0}$ or
$(q_\lambda,c_\lambda,\eta_\lambda,h_\lambda)_{\lambda\ge 0}$ and invariance principles in varying degrees of generality have been obtained. It is also well-known that the convergence in distribution at a fixed time for Markovian branching processes implies the convergence in distribution of the whole process in the Skorohod sense of convergence of right-continuous functions with left limits. In \cite{DuW2}, joint convergence of processes and their genealogical trees is shown, also for a wider class of families $(B_\lambda)_{\lambda\ge 0}$ suitably converging to bushes of L\'evy trees. The main contribution of the present work is to provide almost sure approximations of more general classes of continuous-state processes and consistent families of trees that contain full information about the genealogy of the population of the limiting continuous-state process, which is not contained in the limiting process itself nor in the approximating discrete-state branching processes.
The structure of this paper is as follows. In Section \ref{preliminaries} we formally set up the framework in which we represent trees, we recall preliminaries from Duquesne and Winkel \cite{DuW} and develop a bit further some aspects that readily transfer and serve in the more general context here. We also provide some background about continuous-state branching processes with immigration, and about geometric infinite divisibility. Section \ref{Consistend kappa bush} presents the theory around Theorem \ref{thm2} and Proposition \ref{prop4a}, while Section \ref{Section Immigration} deals with Theorem \ref{thm3} and Proposition \ref{prop4}. In each setting we provide explicit
formulas for offspring distributions, lifetime distributions and immigration distributions
as appropriate; we also provide explicit reconstruction procedures that reverse the reduction for the consistent families
of bushes and forests and establish connections with backbone decompositions (Theorem \ref{CSBPbackbone}) and L\'evy trees. We finally deduce generalisations combining lifetimes and immigration.
\section{Preliminaries}\label{preliminaries}
\subsection{Discrete trees with edge lengths and colour marks}\label{trees}
\subsubsection{Discrete trees and the Galton-Watson branching
property}\label{Discr Trees and GW BP}
Following Neveu \cite{Nev-GW}, Chauvin \cite{Cha-BH} and others, we let
$$
\bU=\bigcup_{n \geq 0}
\bN^n=\{\emptyset,1,2,\ldots,11,12,\ldots,21,22,\ldots,111,112,\ldots\}
$$
be the set of \em integer words\em, where $\mathbb{N}=\{1,2,3,\ldots \}$ and
where $\mathbb{N}^{0}=\{\emptyset\}$ has the \em empty word \em $\emptyset$ as its only element.
For $u=u_1u_2\cdots u_n\in\bU,$ and $v=v_1v_2\cdots v_m\in\bU$, we denote by
$|u|$ the \em length \em of $u$, e.g. $|u_1u_2\cdots u_n|=n$, and by
$uv=u_1u_2\cdots u_nv_1v_2\cdots v_m$ the \em concatenation \em of words
in $\bU$.
\begin{defi}\label{defn1}
\rm A subset $\mathbf{t}\subset\bU$ is called a \em tree \em if
\begin{itemize}
\item $\emptyset \in \mathbf{t}$; we refer to $\emptyset$ as the \em progenitor \em of $\ft$;
\item for all $u\in \mathbb{U}$ and $j \in \mathbb{N}$ with $uj\in \mathbf{t}$, we have $u \in \mathbf{t}$; we refer to $u$ as the \em parent \em of $uj$;
\item for all $u\in \mathbf{t}$, there exists $\nu_{u}(\mathbf{t})\in\bN_0=\{0,1,2,\ldots\}$ such that $uj\in \mathbf{t} \Leftrightarrow 1\leq j \leq
\nu_{u}(\mathbf{t})$; we refer to $\nu_{u}(\mathbf{t})$ as the \em number of children \em of $u$.
\end{itemize}
\end{defi}
We refer to the length $|u|$ of a word $u\in\ft$ as the \em generation \em of the \em individual \em $u$ in the \em genealogical tree \em $\ft$. An element $u\in \mathbf{t}$ is called a
\em leaf \em of $\mathbf{t}$ if and only if $\nu_{u}(\mathbf{t})=0$. We consider the lexicographical order $\le$ on $\bU$ and its restriction to $\mathbf{t}$ as the canonical \em total order\em. For $u,v\in\bU$, we write $u
\preceq uv$, defining a \em partial order \em on $\bU$, whose restriction to $\ft$ is the \em genealogical order \em on $\ft$. The partial order $\preceq$ is compatible with the total order $\le$ in that $u\preceq v\Rightarrow u\le q$. A tree $\ft$ in the sense of Definition \ref{defn1} can be represented graphically as in Figure \ref{fig2}.
\begin{figure}[b]
\begin{center}
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\put(22,10){\line(0,1){20}}
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\put(1,50){\line(1,0){14}} \put(15,50){\line(0,1){20}}
\put(1,50){\line(0,1){20}}\put(36,30){\line(0,1){20}}
\put(29,50){\line(1,0){7}}\put(29,50){\line(0,1){20}}
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\put(92,50){\line(1,0){7}}\put(99,50){\line(0,1){20}}
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\put(85,50){\line(0,1){20}}
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\put(78,31){\makebox(0,0)[b]{\scalebox{.7}{2}}}
\put(1,71){\makebox(0,0)[b]{\scalebox{.7}{111}}}
\put(36,51){\makebox(0,0)[b]{\scalebox{.7}{13}}}
\put(92,51){\makebox(0,0)[b]{\scalebox{.7}{22}}}
\put(99,71){\makebox(0,0)[b]{\scalebox{.7}{222}}}
\put(22,51){\makebox(0,0)[b]{\scalebox{.7}{12}}}
\put(43,71){\makebox(0,0)[b]{\scalebox{.7}{132}}}
\put(64,51){\makebox(0,0)[b]{\scalebox{.7}{21}}}
\put(85,71){\makebox(0,0)[b]{\scalebox{.7}{221}}}
\put(29,71){\makebox(0,0)[b]{\scalebox{.7}{131}}}
\put(51,9){\makebox(0,0)[tl]{\scalebox{.7}{$\emptyset$}}}
\put(99,-10){\makebox(0,0)[r]{$\mathbf{t}$}}
\end{picture}
\hspace{3cm}
\begin{picture}(160,140)(0,-20)
\put(0,0){\dashbox{5}(120,0)}
\put(0,30){\dashbox{5}(140,0)}\put(0,75){\dashbox{5}(140,0)}
\put(0,-20){\vector(0,1){135}} \thicklines
\put(80,-20){\line(0,1){20}}\put(40,0){\line(1,0){80}}
\put(0,-20){\dashbox{5}(80,0)}
\put(40,0){\line(0,1){90}}\put(20,40){\line(1,0){40}}\put(60,40){\line(0,1){40}}
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\put(-7,113){\makebox(0,0)[r]{\scalebox{.7}{$t$}}}\put(120,30){\circle*{4}}
\put(141,-20){\makebox(0,0)[r]{$\overline{\mathbf{t}}$}}
\put(121,29){\makebox(0,0)[tl]{\scalebox{.7}{$(2,\zeta_2)$}}}
\put(141,75){\makebox(0,0)[tl]{\scalebox{.7}{$(22,\zeta_{22})$}}}
\put(-1,0){\makebox(0,0)[r]{\scalebox{.7}{$\omega_\emptyset$}}}
\put(-1,-20){\makebox(0,0)[r]{\scalebox{.7}{0}}}
\put(-1,30){\makebox(0,0)[r]{\scalebox{.7}{$\omega_2$}}}
\put(-1,75){\makebox(0,0)[r]{\scalebox{.7}{$\omega_{22}$}}}
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\end{picture}
\end{center}
\vspace{-0.3cm}
\caption{{On the left, $\ft=\{\emptyset, 1,2,11,12,13,21,22,111,112,131,132,221,222\}$, and on the right
$\overline{\ft}=\{(\emptyset,\zeta_\emptyset),(1,\zeta_1),(11,\zeta_{11}),(12,\zeta_{12}),(13,\zeta_{13})
(2,\zeta_2),(21,\zeta_{21}),(211,\zeta_{211}),(212,\zeta_{212}),(22,\zeta_{22})\}$.}}
\label{fig3}
\label{fig2}
\end{figure}
Let $\mathbb{T}$ be the space of all such trees, and let
$\mathbb{T}_{u}=\{\mathbf{t} \in \mathbb{T}\colon
u \in \mathbf{t} \}$. Then $\nu_u$ is a map defined on $\bT_u$ taking values in
$\bN_0$. Note that $\mathbb{T}$ is
uncountable. A sigma-algebra on $\mathbb{T}$ is defined as $\mathcal{F}=\sigma\{\mathbb{T}_{u},u\in\bU\}$. We also specify the $n$th generation
$\pi_n(\ft)=\{u\in\ft\colon |u|=n\}=\ft\cap\bN^n$ and set
$\mathcal{F}_{n}=\sigma\{\mathbb{T}_{u}, |u|\leq n\}=\sigma\{\pi_m,m\le n\}$
We define the shift map/operator $\theta_u$ that assigns to a tree $\ft$ its \em subtree $\ft_u=\theta_u\ft$ above $u \in\ft$\em:
$$\theta_u\colon \mathbb{T}_{u} \rightarrow
\mathbb{T}, \qquad \mathbf{t} \mapsto \theta_u\ft=\ft_u=\{v \in
\mathbb{U}\colon uv \in \mathbf{t}\}.$$
Clearly $\bT_\emptyset=\bT$ and
$\{\nu_u \geq j\}\cap \mathbb{T}_{u}= \mathbb{T}_{uj}$, also
$\theta_u^{-1}(\mathbb{T}_{v})=\mathbb{T}_{uv}$ and
$$\mathbb{T}_{v}=\{\mathbf{t} \in \mathbb{T}\colon \nu_{v_1v_2\cdots v_k}(\mathbf{t})\geq
{v_{k+1}}\mbox{ for all $0\leq k < m$}\}\qquad\mbox{for $v=v_1v_2\cdots v_m\in\bU$.}$$
These relations allow us to consider a \em random \em tree $\tau$ whose distribution is a
probability measure $\bP_q$ on $\mathbb{T}$, under which the
numbers of children $\nu_u$ of the individuals $u$ in the random tree are independent random variables with distribution $q$. More formally, $\bP_q$ is
characterised as follows:
\paragraph{${\rm GW}(q)$-trees and their branching property (see e.g. Neveu \cite{Nev-GW})}
\begin{itemize}
\item[(a)] For any probability measure $q$ on $\mathbb{N}_0$, there exists a unique
probability measure $\mathbb{P}_q$ on $(\mathbb{T}, \mathcal{F})$ such that
$\mathbb{P}_q(\nu_{\emptyset}=j)=q(j)$, and conditionally on $\{\nu_{\emptyset}=j\}$ for
any $j\ge 1$ with $q(j)>0$, the subtrees $\theta_{i}$, $1\le i\leq j$, above the
first generation are independent with distribution $\mathbb{P}_q$. A random tree
$\tau$ with distribution $\mathbb{P}_q$ is called a ${\rm GW}(q)$-tree.
\item[(b)] Under $\bP_q(\,\cdot\,|\cF_n,\pi_n=A)$,
the subtrees $\theta_u$, $u\in A$, above the $n$th generation are
independent and with distribution $\mathbb{P}_q$, for all finite $A\subset\bN^n$ and $n\ge 1$ with $\bP(\pi_n=A)>0$.
\end{itemize}
For finite trees, in particular in the \em (sub)critical \em case $\bE_q(\nu_\emptyset)=\sum_{j\in\bN_0}jq(j)\le 1$, the measure $\bP_q$ can be expressed as
$$ \mathbb{P}_q(\{\mathbf{t}\})=\prod_{v \in \mathbf{t}} q(\nu_v(\mathbf{t})), \qquad \mbox{for all $\mathbf{t} \in {\mathbb{T}}$,}$$
but this does not specify the measure $\bP_q$ in the \em supercritical \em case $\bE_q(\nu_\emptyset)>1$, where $\bP_q$ assigns positive measure to infinite trees. Here, $\bE_q$ is the expectation operator associated with $\bP_q$.
For a ${\rm GW}(q)$-tree $\tau$, the process $G_n=\#\pi_n(\tau)$, $n\ge 0$, is known as a ${\rm GW}(q)$-branching process.
\subsubsection{Marked trees and discrete branching processes in continuous time}\label{Lifetime marks and discr BP in cont. time}
Let $(\bH,\cH)$ be a measurable space of marks. We can attach a mark $\xi_u\in\bH$ to each vertex $u$ of a given tree $\mathbf{t}$. Formally, a marked tree is a subset
\begin{equation}\overline{\mathbf{t}}\subset \mathbb{U}\times \mathbb{H} \quad
\mbox{such that} \quad \mathbf{t}= \{u\in\bU\colon (u,\xi_u)\in\overline{\mathbf{t}}\mbox{ for some $\xi_u\in\bH$}\}\in
\mathbb{T}\label{skel}\end{equation}
and where $\overline{\ft}\cap\{u\}\times\bH=\{(u,\xi_u)\}$, i.e. the map $\xi\colon \ft\rightarrow\bH$ is unique. So a marked tree has the form $\overline{\ft}=\{(u,\xi_u)\in\bU\times\bH\colon u\in \mathbf{t}\}$. We denote by
$\bT^\bH$ the set of marked trees. We consider the set of trees $\mathbb{T}_{u}^\bH=\{\overline{\mathbf{t}} \in\bT^\bH\colon u \in \mathbf{t}\}$ containing
individual $u$ and note that $\xi_u\colon \bT_u^\bH\rightarrow\bH$ is a map.
For marked trees, we set $\nu_u(\overline{\mathbf{t}})=\nu_u(\mathbf{t})$ and
$\overline{\ft}_u=\overline{\theta}_u\overline{\ft}=\{(v,\xi_v)\colon v
\in \mathbf{t}_u\}=\{(v,\xi_{v})\colon uv \in \mathbf{t}\}$. As sigma-algebra on $\bT^\bH$ we take
one that makes $\overline{\ft}\mapsto\ft$ in (\ref{skel}) and $\overline{\ft}\mapsto\xi_u(\overline{\ft})$ measurable:
$$\mathcal{F}^{\bH}
=\sigma\{\bT_{u,H}^\bH,u\in\bU,H\in\cH\},\mbox{ where $\bT_{u,H}^\bH=\{\overline{\ft}\in\mathbb{T}_u^\bH\colon \xi_u\in H\}$.}$$
For example, for $\bH=(0,\infty)$, the marks can represent the \em lifetimes \em of individuals. We will later use $\bH=(0,\infty)\times\{0,1\}$ so that $\xi_u=(\zeta_u,\gamma_u)$
consists of a lifetime mark $\zeta_u\in(0,\infty)$ and a colour mark $\gamma_u\in\{0,1\}$.
We consider a model when
$u \in \mathbf{t}$ will produce children at the moment of its death.
The birth and death times $\alpha_u$ and $\omega_u$ of each individual $u \in \mathbf{t},$ are then defined recursively by
$$\left\{ \begin{array}{ll}
\alpha_\emptyset=0,\ \omega_{\emptyset}=\zeta_\emptyset,\\
\alpha_{uj}=\omega_u,\ \omega_{uj}=\alpha_{uj}+\zeta_{uj},&1\le j\le\nu_u, u\in\ft.
\end{array}
\right. $$
We denote by $\overline{\pi}_t(\overline{\ft})=\{u\in\mathbf{t}\colon \alpha_u<t\le\omega_u\}$ the set of individuals
alive at time $t\ge 0$ and define $\cF_t^\bH=\sigma\{\overline{\pi}_s,s\le t\}$. For $u\in\overline{\pi}_t(\overline{\ft})$, we denote by
$$ \overline{\theta}_{u,t}(\overline{\ft})=\{(\emptyset,\omega_u-t)\}\cup\{(v,\zeta_{uv})\colon uv\in\ft\}
$$
the subtree of individual $u$ above $t$. Figure \ref{fig3} gives a graphical representation of a tree $\overline{\ft}$ where lifetimes are shown as edge lengths.
In this formalism, we can define and study ${\rm GW}(q,\kappa)$-trees as ${\rm GW}(q)$ trees with
independent lifetimes distributed according to a measure $\kappa$ on $(0,\infty)$:
\paragraph{${\rm GW}(q,\kappa)$-trees and their branching property (see Neveu \cite{Nev-GW}, Chauvin \cite{Cha-BH})} \label{bran2}
\begin{itemize}
\item[(a)] For any probability measure $\mathbb{Q}$ on $\mathbb{N}_0 \times \bH$,
there exists a unique probability measure $\mathbb{P}_{\mathbb{Q}}$ on
$(\mathbb{T}^\bH,\mathcal{F}^\bH)$, such that
$(\nu_\emptyset,\xi_\emptyset)\sim {\mathbb{Q}}$ and conditionally on
$\{\nu_\emptyset = j,\xi_\emptyset\in H\}$ for any $j\ge 1$, $H\in\cH$, with $\mathbb{Q}(\{j\}\times H)>0$, the subtrees
$\overline{\theta}_i$, $1\le i \le j$, are independent
with distribution $\mathbb{P}_{\mathbb{Q}}$. For $\bH=(0,\infty)$ and
$\bQ=q\otimes\kappa$ a random tree $T$ with distribution $\bP_\bQ$ is called a
${\rm GW}(q,\kappa)$-tree.
\item[(b)] Under
$\bP_{q\otimes\kappa}(\,\cdot\,|\cF_t^\bH,\overline{\pi}_t=A)$, the subtrees
$\overline{\theta}_{u,t}$, $u\in A$, above time $t$
are independent and distributed like $\overline{\theta}_{\emptyset,s}$ under
$\bP_{q\otimes\kappa}(\,\cdot\,|\zeta_\emptyset>s)$, where $s=t-\alpha_u$ is the
($\cF_t^\bH$-measurable) age of $u$ at time $t$, for all finite $A\subset\bU$ with
$\bP(\overline{\pi}_t=A)>0$.
\end{itemize}
For further details including strong branching properties, we refer to \cite{CeG,CGM}.
For a ${\rm GW}(q,\kappa)$-tree $T$, the process $Z_t=\#\overline{\pi}_t(T)$, $t\ge 0$, is known as a Bellman-Harris branching process \cite{BeH2,BeH1}. The Markovian special case for $\kappa={\rm Exp}(c)$ is also
called a continuous-time Galton-Watson process.
\subsubsection{Coloured leaves, coloured trees and a two-colours branching
property}\label{coloured leaves and 2-colour BP}
On a suitable probability space $(\Omega,\mathcal{A},\mathbb{P})$, let $T\colon (\Omega,
\mathcal{A}, \mathbb{P})\!\rightarrow\!(\mathbb{T}^{(0,\infty)},
\mathcal{F}^{(0,\infty)}, \mathbb{P}_{q\otimes\kappa})$ be a ${\rm GW}(q,\kappa)$-tree.
We assume $q(0)>0$, i.e.\ $T$ has leaves, and $q(1)=0$, as an individual producing a
single child can be viewed as continuing to live instead of being replaced by its child.
Following \cite{DuW}, we independently mark the \em leaves \em $u$ of $T$ with one of two colours, say \em red\em, $\bP(\gamma_u(T)=1|\nu_u(T)=0)=p$, or \em black\em, $\bP(\gamma_u(T)=0|\nu_u(T)=0)=1-p$,
for some given $p\in(0,1)$. It will be convenient to also mark each non-leaf individual in black if the subtree above it has at least one black leaf, red otherwise. Such a marked tree $T^{p\rm-col}$ is a random element of $\bT^\bH$ for $\bH=(0,\infty)\times\{0,1\}$. We denote its distribution by $\bP_{q\otimes\kappa}^{p\rm-col}$. Note that it is \em not \em of the form
$\bP_\bQ$ introduced in the previous section, because marks for non-leaf individuals will not be independent. We set
\begin{equation}\label{gp}
g(p)=\mathbb{P}_{q\otimes\kappa}^{p\rm-col}(\gamma_{\emptyset}=1)=\mathbb{P}(\mbox{$T^{p\rm-col}$ has only red colour marks})
=\mathbb{E}[p^{\#\{u\in T\colon \nu_u(T)=0\}}].
\end{equation}
\paragraph{Branching properties of coloured ${\rm GW}(q,\kappa)$-trees (cf. Duquesne and Winkel \cite{DuW})}
\begin{itemize}
\item[(a)]
For all Borel-measurable $k\colon (0,\infty)\rightarrow[0,\infty)$, $j\ge 2$, $\varepsilon_i\in\{0,1\}$ and
$\cF^\bH$-measurable $f_i\colon \bT^\bH\rightarrow[0,\infty)$, $i=1,\ldots,j$, we have
\begin{eqnarray*}\label{branching prop expec}
\lefteqn{\mathbb{E}_{q\otimes\kappa}^{p\rm-col}\left[k(\zeta_\emptyset)\prod_{i=1}^{j}f_i
(\overline{\theta}_i); \nu_\emptyset=j;(\gamma_1,\ldots,\gamma_j)=(\varepsilon_1,\ldots,\varepsilon_j)\right]}\nonumber\\
&
&=\int_{(0,\infty)}k(z)\kappa(dz)\ q(j)\ g(p)^{j_r}(1-g(p))^{j_b}\ \prod^{j}_{i=1}\mathbb{E}_{q\otimes\kappa}^{p\rm-col}[f_i|\gamma_{\emptyset}=\varepsilon_i]
\end{eqnarray*}
where $j_r=\varepsilon_1+\cdots+\varepsilon_j$ and $j_b=j-j_r$ are the numbers of red and black colour marks.
\item[(b)]
For all $t\ge 0$ and $\cF^\bH$-measurable
$f_u\colon \bT^\bH\rightarrow[0,\infty)$ $$\bE_{q\otimes\kappa}^{p\rm-col}\left[\left.\prod_{u\in\overline{\pi}_t}f_u(\overline{\theta}_{u,t})\right|\cF_t^\bH\right]=\prod_{u\in\overline{\pi}_t}\left.\bE_{q\otimes\kappa}^{p\rm-col}[f_u(\overline{\theta}_{\emptyset,s})|\zeta_\emptyset>s]\right|_{s=t-\alpha_u}$$
In the exponential case $\kappa={\rm Exp}(c)$, this simplifies to
\begin{equation}\bE_{q\otimes {\rm Exp}(c)}^{p\rm-col}\left[\left.\prod_{u\in\overline{\pi}_t}f_u(\overline{\theta}_{u,t})\right|\cF_t^\bH\right]=\prod_{u\in\overline{\pi}_t}\bE_{q\otimes{\rm Exp}(c)}^{p\rm-col}[f_u]
\label{bpexp}
\end{equation}
\end{itemize}
\paragraph{Reduction procedure to identify the ``black tree'' in a two-colours tree}
\begin{itemize}\item We can extract $\widetilde{T}^{p \rm -rdc}_{\rm sub}=\{(u,\zeta_u)\in\bU\times(0,\infty)\colon (u,\zeta_u,1)\in T^{p\rm-col}\}$, the individuals of $T^{p\rm-col}$ with black colour marks. If $\widetilde{T}^{p \rm -rdc}_{\rm sub}\neq\varnothing$,
we rename the individuals of $\widetilde{T}^{p\rm-rdc}_{\rm sub}$ by the unique injection
$$\iota\colon \widetilde{\tau}_{\rm sub}^{p \rm -rdc}=\{u\in\bU\colon (u,\zeta_u)\in\widetilde{T}_u^{p\rm-rdc}\}\rightarrow\bU,$$
that is increasing for the lexicographical total order on $\bU$, maps onto an element
$\tau^{p \rm -rdc}_{\rm sub}$ of $\bT$ and is compatible with the genealogical partial orders. We refer to the image tree $T^{p\rm-rdc}_{\rm sub}=\{(\iota(u),\zeta_{u})\colon u\in\widetilde{\tau}_{\rm sub}^{p\rm-rdc}\}$ as the \em $p$-reduced subtree of $T$\em.
\item As a further reduction, we remove single-child individuals and add their lifetimes to the child's lifetime. Formally, we define
$\widetilde{\tau}^{p\rm-rdc}=\{v \in \tau^{p \rm -rdc}_{\rm sub}\colon {\nu_{v}}(\tau^{p \rm -rdc}_{\rm sub}) \neq 1\}$,
and $$\widetilde{\zeta}_u=\sum_{i=J_u}^n\zeta_{u_1\cdots u_i}(\tau^{p\rm-rdc}_{\rm sub}),
\mbox{ where }J_u=\sup\{j\colon \nu_{u_1\cdots u_i}(\tau^{p\rm-rdc}_{\rm sub})\!=\!1\mbox{ for all $i\in\{j,\ldots,n-1\}$}\},$$
for all $u=u_1\cdots u_n\in\widetilde{\tau}^{p\rm-rdc}$.
Again, there is a unique injection $\iota^\prime\colon \widetilde{\tau}^{p\rm-rdc}\rightarrow\bU$ that is
increasing for the lexicographical total order on $\bU$, maps onto an element
$\tau^{p\rm-rdc}$ of $\bT$ and is compatible with the genealogical partial orders. We refer to the image tree $T^{p\rm-rdc}=\{(\iota^\prime(u),\widetilde{\zeta}_u)\colon u\in\widetilde{\tau}^{p\rm-rdc}\}$ as the \em $p$-reduced tree \em (or as the \em black tree\em).
\end{itemize}
Figure \ref{discperc} in the Introduction illustrates the reduction procedure.
\begin{rem}\label{remdet}\rm \begin{enumerate}\item[(a)] The reduction procedure is transitive in that for independent colouring, we have $(T^{(1-\overline{p}_1)\rm-rdc})^{(1-\overline{p}_2)\rm-rdc}\overset{{\rm (d)}}{=} T^{(1-\overline{p}_1\overline{p}_2)\rm-rdc}$. In particular, colouring for $T^{(1-\overline{p}_1)\rm-rdc}$ and
$T^{(1-\overline{p}_3)\rm-rdc}$ for $\overline{p}_3<\overline{p}_1$ can be coupled such that $T^{(1-\overline{p}_3)\rm-rdc}=(T^{(1-\overline{p}_1)\rm-rdc})^{(1-\overline{p}_3/\overline{p}_1)\rm-rdc}$.
\item[(b)] Although we have used notation for a random ${\rm GW}(q,\kappa)$-tree $T^{p\rm-col}$ with leaves
coloured independently, note that the reduction of $T^{p\rm-col}$ to a black tree is a purely deterministic procedure. Our focus here has been on the technical framework and how it is used to formulate relevant examples. We postpone to Section \ref{Growth of GW exponential edge lengths} the review of further developments, notably of the reconstruction/growth procedures that reverse the reduction.
\end{enumerate}
\end{rem}
\subsubsection{Bushes and forests -- models with several progenitors and immigration}
\label{discr BP with immigration section} Branching processes
with immigration have been studied widely (see e.g. Athreya and Ney
\cite{AtN} and Jagers \cite{Jag-68}). We consider the model, where
immigrants arrive at the times $S_i$, $i\ge 1$, of a renewal
process $J_t=\#\{i\ge 1\colon S_i\le t\}$, i.e. where $S_0=0$ and the
interarrival times $S_i-S_{i-1}$, $i\ge 1$, are
independent and identically distributed random variables with a common
distribution on $(0,\infty)$ that we denote by $\chi$. At each
immigration time $S_i$ the number $N_i$ of immigrants is independent and
has a common distribution $\eta$ on $\bN$.
Each immigrant produces offspring independently according to the rules
of ${\rm GW}(q,\kappa)$-trees.
Denoting by $Z^{(i)}_{t-S_i}$ the size at time $t$ of the population of
immigrants arriving at time $S_i$,
\begin{equation}\label{discrimm} Y_t=\sum^{J_t}_{i=1}Z^{(i)}_{t-S_i}\end{equation}
is the total population size at time $t\ge 0$. Here, $Z^{(i)}$ are independent sums of $N_i$
independent Bellman-Harris processes with offspring and lifetime distributions $q$ and $\kappa$ as in Section \ref{Lifetime marks and discr BP in cont. time}.
To capture the genealogical trees of the population, we use the notion of a \em bush \em as a
random sequence $B=(T_{(1)},\ldots,T_{(N)})$ of independent trees, and the notion of a \em
forest \em as a point process $F=(B(t),t\ge 0)$ of independent bushes
$$B(S_i)=B^{(i)},\quad i\ge 0,\qquad B(t)=\partial,\quad t\not\in\{S_i,i\ge 1\}\quad \mbox{for a cemetery point $\partial$.}$$
\paragraph{${\rm GW}(q,\kappa,\beta)$-bushes and ${\rm GWI}(q,\kappa,\eta,\chi)$-forests}
\begin{itemize}\item A ${\rm GW}(q,\kappa,\beta)$-bush is a bush $B=(T_{(1)},\ldots,T_{(N)})$ of
independent ${\rm GW}(q,\kappa)$-trees $T_{(j)}$, where $N\sim{\rm Poi}(\beta)$.
\item A ${\rm GWI}(q,\kappa,\eta,\chi)$-forest is a forest $F=(B(t),t\ge 0)$ where each
bush $B(S_i)=B^{(i)}$ is associated with immigration at the times $S_i$ of a renewal
process with inter-arrival distribution $\chi$ and consists of an independent
$\eta$-distributed number $N_i$ of trees $T_{(j)}^{(i)}$.
\end{itemize}
It is straightforward to transfer the notions of colouring and reduction to the setting of bushes
and forests, since they apply tree by tree. We will slightly abuse notation and write $u\in B$ to refer
to individuals in a bush, instead of writing formally $u=(i,u^{\prime\prime})$ with $u^{\prime\prime}\in T_{(i)}$. Similarly, $u\in F$ means $u=(t,u^{\prime})$ with $u^\prime\in B(t)$ in the sense just defined. We will also abuse notation
$\nu_u$, $\zeta_u$ and $\gamma_u$ accordingly for $u\in B$ and $u\in F$.
\subsection{Continuous-state branching processes and immigration}\label{contstate}
\label{CSBP and CBI} We have looked at branching processes with
immigration in the discrete state space $\bN_0=\{0,1,2,\ldots \}$ in
continuous time. In this section we will recall (Markovian) continuous-state
branching processes with immigration, where the state space
(population size) will no longer be $\mathbb{N}_0$ but
$[0,\infty)$, and time is also continuous.
\subsubsection{Subordinators and geometric infinite divisibility}
\label{sub and g.i.d section}
\begin{defi}\label{Subordinator}
\rm An increasing right-continuous stochastic process $\sigma=(\sigma(t), t \geq 0)$ in $[0,\infty)$ is called a \em subordinator \em if it has stationary independent increments, i.e. if for every
$u, t\ge 0$, the \em increment \em $\sigma(t+u)-\sigma(t)$ is independent of $(\sigma(s),s\le t)$ and $\sigma(t+u)-\sigma(t)\overset{{\rm (d)}}{=} \sigma(u)$.
\end{defi}
The distribution of a subordinator $\sigma$ on the space $\bD([0,\infty),[0,\infty))$ of functions $f\colon [0,\infty)\rightarrow[0,\infty)$ that are right-continuous and have left limits equipped with the Borel sigma-algebra generated by Skorohod's topology, see e.g. \cite{JaS}, is specified by the Laplace transforms of
its one-dimensional distributions. For every $t\ge 0$ and $r \ge 0$,
\begin{equation}\label{LaExponent}
\mathbb{E}(\exp\{-r \sigma(t)\})= \exp\{-t \phi (r)\}
\end{equation}
where the function $\phi \colon [0, \infty) \rightarrow [0, \infty)$ is
called the \it Laplace exponent \rm of $\sigma$.
There exist a unique real number $d\ge 0$ and a
unique measure $\Lambda$ on $(0, \infty)$ with $\int (1 \wedge x) \Lambda
(dx) < \infty$, such that for every $r \geq 0$
\begin{equation}\label{LK1}
\phi(r)= dr + \int_{(0,\infty)}(1-e^{-r x}) \Lambda
(dx).
\end{equation}
Conversely, any function $\phi$ of the form (\ref{LK1}) is the
Laplace exponent of a subordinator, which can be constructed as $(dt+\sum_{s\le t}\Delta\sigma_s,t\ge 0)$ for a Poisson point
process $(\Delta\sigma_s,s\ge 0)$ in $(0,\infty)$ with intensity measure $\Lambda$. Equation (\ref{LK1}) is
referred to as the L\'evy-Khintchine representation of $\phi$.
We refer to Bertoin
\cite{Ber-Sub} for an introduction to subordinators and their
applications.
\begin{defi}
\rm A random variable $X$ is
\em geometrically infinitely divisible \em (g.i.d.) if for all
$\alpha\in(0,1)$
\begin{equation}\label{GEO}
X \overset{{\rm (d)}}{=} \sum_{j=1}^{G^{(\alpha)}}X_{\alpha}^{(j)}, \qquad j\ge 1,
\end{equation}
for a sequence $X_\alpha^{(j)}$, $j\ge 1$, of independent identically distributed (i.i.d.) random variables and an independent $G^{(\alpha)}\sim{\rm Geo}(\alpha)$:
$$
\mathbb{P}(G^{(\alpha)}=k)=\alpha(1-\alpha)^{k-1},\qquad
k\ge 1.
$$
\end{defi}
For example, an exponential random variable $X \sim {\rm Exp}(c)$
is g.i.d.\ since (\ref{GEO}) holds for $X_{\alpha}^{(j)} \sim {\rm Exp}(c/\alpha )$. The class of g.i.d.\ distributions can be characterised as follows.
\begin{lemm}[\cite{KMM}] \label{sub and geo}
A random variable $X$ is g.i.d.\ if and
only if it can be expressed as $X \overset{{\rm (d)}}{=} \sigma(V)$, where $\sigma=(\sigma(t), t \geq 0)$ is
a subordinator and $V \sim {\rm Exp}
(c)$ independent of $\sigma$ for one equivalently all $c\in(0,\infty)$.
\end{lemm}
\noindent Indeed we can then express ($V$ and hence) $X$ as
$$V=\sum_{j=1}^{G^{(\alpha)}}V_{\alpha}^{(j)}\qquad\mbox{and}\qquad X=\sigma_V=\sum_{j=1}^{G^{(\alpha)}}\left(\sigma\left(V_\alpha^{(1)}+\cdots+V_\alpha^{(j)}\right)-\sigma\left(V_\alpha^{(1)}+\cdots+V_\alpha^{(j-1)}\right)\right).$$
\begin{lemm}\label{uniqgeodiv} If $X$ is such that {\rm(\ref{GEO})} holds for some $\alpha\in(0,1)$, the distribution of $X_\alpha^{(j)}$ is unique.
\end{lemm}
\begin{pf} $\displaystyle\bE(e^{-rX})=\frac{\bE(e^{-rX_\alpha^{(1)}})\alpha}{1-(1-\alpha)\bE(e^{-rX_\alpha^{(1)}})}\quad\Rightarrow\quad\bE(e^{-rX_{\alpha}^{(1)}})=\frac{\bE(e^{-rX})}{\alpha+(1-\alpha)\bE(e^{-rX})}.$
\end{pf}
\subsubsection{Continuous-state branching processes}\label{CSBP and CBI explanation
section} Continuous-state branching processes were
introduced by Jirina \cite{Jir} and Lamperti \cite{Lam-CSBP}.
They are the limiting processes of sequences of rescaled
Galton-Watson processes as the initial population size tends
to infinity and the mean lifetime tends to zero. In this section we follow Le Gall
\cite{LeG-snake}.
\begin{defi}\rm
A continuous-state branching process (CSBP) is a right-continuous Markov
process $(Z_t, t \geq 0)$ in $[0,\infty)$, whose transition
kernels $P_t(x, dz)$ are such that for every $t\ge 0$, $x \geq 0$ and $x^\prime\ge 0$ we have
$$P_t(x+x^\prime,\cdot)=P_t(x,\cdot)*P_t(x^\prime,\cdot),$$
where $*$ denotes convolution. In other words, if for a given transition kernel we denote, for each $x\ge 0$, by $Z^x$ a CSBP starting from $Z_0^x=x$, then for $\widetilde{Z}^{x^\prime}\overset{{\rm (d)}}{=}Z^{x^\prime}$ independent, we require $Z^x_t +\widetilde{Z}^{x^\prime}_t\overset{{\rm (d)}}{=} Z^{x+x^\prime}_t$.
\end{defi}
The transition kernel is specified by the Laplace transforms
\begin{equation}\label{CSBP and psi solution u}
\mathbb{E}(\exp\{-r Z_t\}|Z_0=x) =\exp\{-x u_t(r)\}
\end{equation}
where $u_t\colon [0,\infty)\rightarrow[0,\infty)$. In fact, $(t,r)\mapsto u_t(r)$
is necessarily the unique non-negative solution of
\begin{equation}\label{u solution of psi}
u_t(r)+\int^t_0 \psi (u_s (r)) ds = r \qquad
\mbox{or} \qquad \frac{\partial u_t(r)}{\partial t} = -
\psi (u_t (r))
\end{equation}
with $u_0 (r)=r$, for some $\psi\colon [0,\infty)\rightarrow\bR$ of the form
\begin{equation}\label{LK2}
\psi(r)=br+a r^{2} + \int_{(0,
\infty)}(e^{-r x}-1+ r x \mathbf{1}_{\{x<1\}})\Pi(dx)
\end{equation}
where $b \in \mathbb{R}$, $a \geq 0$ and $\Pi$ is a measure on $(0,\infty)$ with $\int(1 \wedge x^2)\Pi(dx)<\infty$ and where $\psi$ satisfies the non-explosion condition $\int_{0+}|\psi(r)|^{-1}dr=\infty$, see \cite{Grey}.
Equation (\ref{LK2}) is referred to as the L\'evy-Khintchine representation of $\psi$. The process $(Z_t,t\ge 0)$ is then called a CSBP \em with branching mechanism $\psi$\em, or a
${\rm CSBP}(\psi)$. We denote by $\bP^x_\psi$ the distribution of $(Z_t^x,t\ge 0)$ on $\bD([0,\infty),[0,\infty))$. There also exists a sigma-finite measure $\Theta_\psi$ on $\bD([0,\infty),[0,\infty))$, such
that
$$(Z_t^x,t\ge 0)\overset{{\rm (d)}}{=} \left(\sum_{0\le y\le x}E_t(y),t\ge 0\right),\qquad x\ge 0,$$
where $(E(y),y\ge 0)$ is a Poisson point process in $\bD([0,\infty),[0,\infty))$ with intensity measure $\Theta_\psi$.
\subsubsection{Continuous-state branching processes with immigration}
Similarly, a discrete-state branching process with immigration has a
continuous analogue, the continuous-state branching process with
immigration, CBI for short. Following Kawazu and Watanabe \cite{KaW-71}, see also
\cite{Duq-imm,Lam-CBI}, besides the branching mechanism
$\psi$ for a CSBP, we also have an immigration mechanism $\phi$ of the form (\ref{LK1}) for
the CBI, which we then refer to as ${\rm CBI}(\psi,\phi)$.
A ${\rm CBI}(\psi,\phi)$ is a Markov process $(Y_t, t \geq 0)$ on $[0,\infty)$ whose transition kernels are
characterized by their Laplace transform, which in terms of $\phi$ and $u_t(r)$ as in (\ref{u solution of psi}) satisfy
$$ \mathbb{E}(\exp \{-r Y_t \}|Y_0=x)=\exp\left\{-x u_t(r)-\int^t_0 \phi(u_s(r))ds\right\}.
$
In fact, a subordinator
$\sigma=(\sigma(t),t\ge 0)$ with Laplace exponent $\phi$ can be viewed as a pure immigration process ${\rm CBI}(\phi,0)$. Indeed, a general ${\rm CBI}(\psi,\phi)$ is such that by time $t\ge 0$ a population of total size $\sigma(t)$ has immigrated and evolved like a ${\rm CSBP}(\psi)$; specifically, consider a Poisson
point process $(E^s,s\ge 0)$ in $\bD([0,\infty),[0,\infty))$ with intensity measure $d\Theta_\psi+\int_{(0,\infty)}\bP^x_\psi\Lambda(dx)$,
then in analogy with (\ref{discrimm})
$$ Y_t=\sum_{s\le t}E_{t-s}^s,\qquad t\ge 0,
$$
is a ${\rm CBI}(\psi,\phi)$. This follows by the exponential formula and properties of Poisson point processes.
Examples of continuous-state branching processes with immigration include (sub)critical CSBP conditioned on survival, see \cite{Lam-07} and literature therein.
\subsection{Growth of Galton-Watson bushes with exponential edge
lengths}\label{Growth of GW exponential edge lengths} In Theorem \ref{thm1}, we consider
families of ${\rm GW}(q_\lambda,{\rm Exp}(c_\lambda),\beta_\lambda)$-bushes $B_\lambda$,
$\lambda\ge 0$, with the
consistency property that any two bushes, for parameters $\mu<\lambda$ say, are related by $p$-reduction as formally defined in Section \ref{coloured leaves and 2-colour BP}, for $p=1-\mu/\lambda$. The choice of $p$ is dictated (up to a positive power for the ratio) by the consistency requirement that the relation holds for \em all \em $\mu$ and \em all \em
$\lambda$ (cf.\ Remark \ref{remdet}(a)).
The equivalence of (i) and (ii) in Theorem \ref{thm1} is a statement purely about offspring distributions $(q_\lambda,\lambda\ge 0)$. The reason for including the other two parameter sequences $(c_\lambda,\lambda\ge 0)$ and $(\beta_\lambda,\lambda\ge 0)$ in the remainder of the statement is simplicity. Specifically, if we consider trees without lifetimes and
hence without embedding in time, the removal of single-child individuals will be more artificial as it reduces the height of the trees; if we look at trees instead of bushes, the reduced tree will only be GW if we condition on the existence of at least one black leaf, and this does not lead to consistent families of random trees on the same probability space.
In \cite{DuW}, the bushes of Theorem \ref{thm1} are used to construct
L\'evy trees as a strong representation of the genealogy of the limiting ${\rm CSBP}(\psi)$ under some extra conditions on $\psi$. In a weaker sense, the family $(B_\lambda,\lambda\ge 0)$
itself is already a representation of the genealogy of ${\rm CSBP}(\psi)$, under
no conditions on $\psi$ other than $\psi(\infty)=\infty$ to exclude the case of increasing CSBPs corresponding to ``no death'' i.e.\ ``no leaves''. Roughly, $B_\lambda$ is the genealogy of a Poisson sample chosen among
all individuals with intensity proportional to $\lambda$; as $\lambda\rightarrow\infty$, the set
of individuals included becomes dense.
\subsubsection{Consistent families of ${\rm GW}(q_\lambda, {\rm Exp}(c_\lambda), \beta_\lambda
)$-bushes}\label{Consistent family}
In \cite{DuW}, the parameters in a consistent family of
Galton-Watson bushes are represented in terms of a branching
mechanism $\psi$ that is just required to satisfy $\psi(\infty)=\infty$, so that $\psi$ is eventually increasing and has a right inverse $\psi^{-1}\colon[0,\infty)\rightarrow[\psi^{-1}(0),\infty)$:
\begin{equation}\label{offspr}\varphi_{q_\lambda}(s)=s+\frac{\psi(\psi^{-1}(\lambda)(1-s))}{\psi^{-1}(\lambda)\psi^\prime(\psi^{-1}(\lambda))},\qquad c_\lambda=\psi^\prime(\psi^{-1}(\lambda)),\qquad\beta_\lambda=\beta\psi^{-1}(\lambda).
\end{equation}
It follows from the derivation there that this $\psi$ and $(\widetilde{\psi},c)$ in the statement of Theorem \ref{thm1} are related in a linear way as
\begin{equation}\label{psipsitilde}\psi(r)=k_1\widetilde{\psi}(k_2 r),
\end{equation}
where $k_1=1/\widetilde{\psi}(1)=\psi^{-1}(1)\psi^\prime(\psi^{-1}(1))$ and $k_2=c\widetilde{\psi}(1)=1/\psi^{-1}(1)$. For the underlying CSBPs, the relationship (\ref{psipsitilde}) just means $Z_t=k_2\widetilde{Z}_{k_1k_2t}$, so $\psi$ and $\widetilde{\psi}$
essentially refer to the same CSBP. With this parameterisation, (\ref{gp}) can be expressed
more explicitly for $q=q_\lambda$ and $p=1-\mu/\lambda$ as
\begin{equation}\label{g in psi}
g_\lambda(1-\mu/\lambda)=\mathbb{P}_{q_\lambda\otimes{\rm Exp}(c_\lambda)}^{(1-\mu/\lambda)\rm-col}(\gamma_{\emptyset}=1)
=1-\frac{\widetilde{\psi}^{-1}(\mu)}{\widetilde{\psi}^{-1}(\lambda)}=1-\frac{\psi^{-1}(\mu)}{\psi^{-1}(\lambda)}.
\end{equation}
\subsubsection{Analysis of the reduction procedure for ${\rm GW}(q_\lambda,{\rm Exp}(c_\lambda),
\beta_\lambda)$-bushes}\label{Recon for GW expo bushes1} In this section
we will study some key points in the reduction procedure in the setting of Theorem \ref{thm1}.
These will be important for the proofs of Theorems \ref{thm2} and \ref{thm3}.
There are three steps in the reduction procedure from $\lambda$ to $\mu<\lambda$: colouring
with $p=1-\mu/\lambda$, passage to the $p$-reduced sub-bush and passage to the $p$-reduced bush.
In the last step, the lifetime $\zeta^{(1-\mu/\lambda) \rm -rdc}_{u}$ of individual $u\in T^{(1-\mu/\lambda) \rm -rdc}_\lambda$ is obtained combining a
number $G^{(\alpha)}_u$ of lifetimes of $T_\lambda$ when removing the
single-child individuals. By the two-colours branching property in Section \ref{coloured leaves and 2-colour BP}, the random numbers $G^{(\alpha)}_u$ are
${\rm Geo}(\alpha)$, where $\alpha$ is the probability that an individual in
$T^{(1-\mu/\lambda) \rm -rdc}_{\lambda, \rm sub}$ produce zero or more than two children. We can express $\alpha$ in terms of $\psi$:
\begin{lemm} \label{alpha and psi} Let $0\le\mu<\lambda<\infty$. Given that $u\in B^{(1-\mu/\lambda)\rm-rdc}_\lambda$, we have
$G^{(\alpha)}_u\sim{\rm Geo}(\alpha)$ with
\begin{equation}\label{imporesult}
\alpha= \mathbb{P}\left(\left.\nu_{\emptyset}\left(T_{\lambda,\rm sub}^{(1-\mu/\lambda)\rm-rdc}\right)\neq 1\;\right|\,
\gamma_{\emptyset}\left(T_\lambda^{(1-\mu/\lambda)\rm-col}\right)=0\right)=\frac{\psi'(\psi^{-1}(\mu))}{\psi'(\psi^{-1}(\lambda))}.
\end{equation}
\end{lemm}
\begin{pf} According to the definition of $\alpha$ and the two-colours branching property of Section \ref{coloured leaves and 2-colour BP},
\begin{eqnarray*}
1-\alpha&=&\mathbb{P}\left(\left.\nu_\emptyset\left(T_{\lambda,\rm sub}^{(1-\mu/\lambda)\rm-rdc}\right)=1\; \right|\, \gamma_\emptyset\left(T_\lambda^{(1-\mu/\lambda)\rm-col}\right)=0\right)\\
&=&\frac{1}{1-g_\lambda(1-\mu/\lambda)}\sum^{\infty}_{j=2}{j\choose 1}q_\lambda(j)g_\lambda(1-\mu/\lambda)^{j-1}(1-g_\lambda(1-\mu/\lambda))\\
&=&\varphi_{q_\lambda}^\prime(g_\lambda(1-\mu/\lambda))
\end{eqnarray*}
and by (\ref{offspr}) and (\ref{g in psi}) we obtain
$1-\alpha=1-\psi^\prime(\psi^{-1}(\mu))/\psi^\prime(\psi^{-1}(\lambda))$.
\end{pf}
In the reconstruction procedure reversing the reduction procedure, we will therefore subdivide each lifetime in the
${\rm GW}(q_\mu, {\rm Exp}(c_\mu),\beta_\mu)$-bush into a geometric random
number of $G^{(\alpha)}_u$ parts.
\begin{rem}\label{alphakappa}\rm By the branching property of coloured ${\rm GW}(q,\kappa)$-trees in Section \ref{coloured leaves and 2-colour BP}, lifetime marks are
independent of colour marks. Therefore, Lemma \ref{alpha and psi} also holds for general $B_\lambda\sim{\rm GW}(q_\lambda,\kappa_\lambda,\beta_\lambda)$-bushes, not just for ${\rm GW}(q_\lambda,{\rm Exp}(c_\lambda),\beta_\lambda)$-bushes.
\end{rem}
In the case of ${\rm Exp}(c_\mu)$ lifetimes, given the lifetime $\zeta_u(T_\mu)$ the
conditional distribution of the random variable $N_u=G^{(\alpha)}_u-1$ follows a Poisson distribution:
\begin{prop}\label{Poi&Exp}
Let $\zeta$ be a random variable having distribution $\zeta \sim
{\rm Exp}(\alpha c)$. Suppose that $\zeta$ is subdivided into $G^{(\alpha)}$
independent parts, $\zeta=\zeta_1+\cdots+\zeta_{G^{(\alpha)}}$, where
$\zeta_i \sim {\rm Exp}(c)$ and $G^{(\alpha)} \sim {\rm Geo}(\alpha)$ are independent. Then given $\zeta=z$ for $z \geq 0$, we have
\begin{equation}\label{Poi with exp condition}
\mathbb{P}(G^{(\alpha)}=k \ | \ \zeta=z )
=\frac{((1-\alpha)cz)^{k-1}e^{- (1-\alpha) c z}}{(k-1)!}.
\end{equation}
\end{prop}
\begin{pf} This is, of course, well-known in the context of Poisson processes, but let us
give a direct argument and write the left hand side as a conditional
expectation
$\mathbb{P}(G^{(\alpha)}=k \ | \ \zeta=z )=\mathbb{E}(\mathbf{1}_{\{G^{(\alpha)}=k\}} |
\zeta=z)$. We also set $g_k(z)={((1-\alpha)cz)^{k-1}e^{- (1-\alpha) c
z}}/{(k-1)!}$.
$$\mbox{\bf Claim: \rm $\mathbb{E}(f(\zeta)g_k(\zeta))=\mathbb{E}(f(\zeta)\mathbf{1}_{\{G^{(\alpha)}=k\}})$ for all measurable $f \geq 0$.}$$
As $\zeta \sim {\rm Exp}(\alpha c)$,
$\displaystyl
\mathbb{E}(f(\zeta)g_k(\zeta)) = \int_{0}^{\infty}f(z)g_k(z)\alpha c
e^{-
\alpha c z}dz
= \int_{0}^{\infty}f(z)\frac{\alpha c ((1-\alpha) c
z)^{k-1}e^{-cz}}{(k-1)!}dz
$.
On the other hand, since $\zeta=\zeta_1+\cdots+\zeta_{G^{(\alpha)}}$, the conditional
distribution of $\zeta$ given $G^{(\alpha)}=k$ is Gamma$(c, k)$. Therefore,
$
\mathbb{E}(f(\zeta)\mathbf{1}_{\{G^{(\alpha)}=k\}}) =
\mathbb{P}(G^{(\alpha)}=k)\int_{0}^{\infty}f(z)\frac{z^{k-1}c^k
e^{-cz}}{(k-1)!}dz
= \int_{0}^{\infty}f(z)\frac{\alpha c ((1-\alpha) c
z)^{k-1}e^{-cz}}{(k-1)!}dz,
$
and so $g_k(z)$ is a version of the conditional
probability $\mathbb{P}(G^{(\alpha)}=k \ | \ \zeta=z )$, as claimed.
\end{pf
Moreover, we can also find the conditional joint distribution of the
lifetimes $(\zeta_1,\ldots,\zeta_{G^{(\alpha)}-1})$ given $\zeta=z$ and $G^{(\alpha)}=k$ as follows:
\begin{equation}\label{joint exponential}
f_{\zeta_1,\zeta_2,\ldots,\zeta_{k-1}|G^{(\alpha)}=k,\zeta=z} (y_1,
y_2,\ldots,y_{k-1}
=\frac{(k-1)!}{z^{k-1}}
\end{equation}
for all $y_1>0,\ldots,y_{k-1}>0$ such that $y_1+\cdots+y_{k-1}<z$.
In the middle step of the reduction procedure, all red individuals are removed. \cite{DuW} noted that it is a
consequence of the two-colours branching property, see Section \ref{coloured leaves and 2-colour BP} here, that they form independent ${\rm GW}(q_{\lambda,\rm red}^{(1-\mu/\lambda)\rm-col},{\rm Exp}(c_\lambda))$-trees, ``red trees'', where
\begin{equation}\label{red offspring distribution}
\varphi_{q_{\lambda,\rm red}^{(1-\mu/\lambda)\rm-col}}(s)=s+\frac{\psi_\mu(\psi_\mu^{-1}(\lambda-\mu)(1-s))}{\psi_\mu(\lambda-\mu)\psi_\mu^\prime(\psi_\mu^{-1}(\lambda-\mu))},\qquad\mbox{with $\psi_\mu(r)=\psi(\psi^{-1}(\mu)+r)-\mu$.}
\end{equation}
Furthermore, the numbers of red trees removed at the branchpoints are conditionally independent, and
given that a branchpoint has $m \geq 1$ subtrees containing black leaves, the generating function of the number of red trees can be expressed in terms of the $m$th derivative $\psi_\mu^{(m)}$ of $\psi_\mu$:
\begin{equation}
\label{redgraft} \frac{\psi_\mu^{(m)}(\psi_\mu^{-1}(\lambda-\mu)(1-s))}{\psi_\mu^{(m)}(0)}\mbox{ for $m\ge 2$ or }\frac{\psi_\mu^\prime(\psi_\mu^{-1}(\lambda-\mu)(1-s))-\psi_\mu^\prime(\psi_\mu^{-1}(\lambda-\mu))}{\psi_\mu^\prime(\psi_\mu^{-1}(\lambda-\mu)-\psi_\mu^\prime(0)}\mbox{ for $m=1$.}
\end{equation}
\subsubsection{Reconstruction procedure for ${\rm GW}(q_\lambda,{\rm Exp}(c_\lambda),
\beta_\lambda)$-bushes}\label{Recon for GW expo bushes2}
Let $B_\mu=(T^{(1)}_\mu, T^{(2)}_\mu,\ldots ,T^{(N_\mu)}_\mu)$ be a ${\rm GW}(q_\mu,{\rm Exp}(c_\mu),\beta_\mu)$-bush. We will construct $B_\lambda$.
\begin{enumerate}\item[1.] For each individual $u$ in $T_\mu^{(i)}$, given the lifetime
$\zeta_u^{(i)}=z$, subdivide into a random number $G_u^{(\alpha,i)}$ of parts with
distribution (\ref{Poi with exp condition}), where $\alpha$ is as in (\ref{imporesult})
and the parts $(\zeta_{u,1}^{(i)},\ldots,\zeta_{u,G_u^{(\alpha,i)}}^{(i)})$ have joint
distribution (\ref{joint exponential}). Now for each $i\in\{1,\ldots,N_\mu\}$, there is a
unique injection
$$(\iota^\prime_i)^{-1}\colon \tau^{(i)}_\mu=\{u\in\bU\colon (u,\zeta_u^{(i)})\in T^{(i)}_\mu\}\rightarrow\bU$$
such that $(\iota^\prime_i)^{-1}(\emptyset)=1^k$ with $k=G_\emptyset^{(\alpha,i)}-1$ and
$(\iota^\prime_i)^{-1}(uj)=(\iota_i^\prime)^{-1}(u)j1^{k}$ with $k=G_{uj}^{(\alpha,i)}-1$, for all
$uj\in\tau_\mu^{(i)}$, where $1^k$ is a string of $k$ letters $1$. We define
$$\widehat{T}_{\mu}^{(i)}=\{(1^{n-1},\zeta_{\emptyset,n}^{(i)})\colon 1\!\le\! n\!\le\! G_\emptyset^{(\alpha,i)}\}\cup\{((\iota_i^\prime)^{-1}(u)j1^{n-1},\zeta_{u,n}^{(i)})\colon uj\in\tau_\mu^{(i)},1\!\le\! n\!\le\! G_u^{(\alpha,i)}\}.$$
\item[2.] For each individual $u$ in $\widehat{T}_{\mu}^{(i)}$, given
$\widehat{\nu}_u^{(i)}=m \geq 1$ children,
consider a random number $C_u^{(i)}$ of further children with distribution (\ref{redgraft}) and a
uniform random permutation $\varrho_u^{(i)}$ among the $(m+C_u^{(i)})!/m!$ permutations with
$\varrho_u^{(i)}(1)<\cdots<\varrho_u^{(i)}(m)$. Let $T^{(u,1,i)},\ldots,T^{(u,k,i)}$ with
$k=C_u^{(i)}$ be independent
${\rm GW}(q_{\lambda,\rm red}^{(1-\mu/\lambda)\rm-rdc},{\rm Exp}(c_\lambda))$-trees with offspring
distribution (\ref{red offspring distribution}). Then for each $i\in\{1,\ldots,N_\mu\}$, there
is a unique injection
$$(\iota_i)^{-1}\colon \widehat{\tau}^{(i)}_{\mu}=\{u\in\bU\colon (u,\widehat{\zeta}_u^{(i)})\in \widehat{T}_\mu^{(i)}\}\rightarrow\bU$$
such that $(\iota_i)^{-1}(\emptyset)=\emptyset$ and $(\iota_i)^{-1}(uj)=(\iota_i)^{-1}(u)\varrho_u^{(i)}(j)$
for all $uj\in\widehat{\tau}^{(i)}_\mu$. We define
\begin{eqnarray*}\widehat{T}_\lambda^{(i)}&=&\{((\iota_i)^{-1}(u),\widehat{\zeta}_u^{(i)})\colon u\in\widehat{\tau}_\mu^{(i)}\}\\ &&\cup\{(\iota_i)^{-1}(u)\varrho_u(\widehat{\nu}_u^{(i)}+j)v,\zeta_v^{(u,j,i)})\colon u\in\widehat{\tau}_\mu^{(i)},1\le j\le C_u^{(i)},v\in\tau^{(u,j,i)}\}.
\end{eqnarray*}
\item[3.] Given $N_\mu=n$, consider a random number
$N_{\lambda}^{\rm red}\sim{\rm Poi}(\beta_\lambda-\beta_\mu)$ of further progenitors and a
uniform random permutation $\varrho$ among the $(n+N_{\lambda}^{\rm red})!/n!$ permutations with
$\varrho(1)<\cdots<\varrho(n)$. Let $\widehat{T}^{(n+1)}_\lambda,\ldots,\widehat{T}^{(n+k)}$
with $k=N_{\lambda}^{\rm red}$ be independent
${\rm GW}(q_{\lambda,\rm red}^{(1-\mu/\lambda)\rm-rdc},{\rm Exp}(c_\lambda))$-trees with
offspring distribution (\ref{red offspring distribution}). Then we finally define
$$B_\lambda=(\widehat{T}_\lambda^{\varrho^{-1}(1)},\ldots,\widehat{T}_\lambda^{\varrho^{-1}(n+k)})\qquad\mbox{with $n=N_\mu$ and $k=N_\lambda^{\rm red}$.}$$
\end{enumerate}
\begin{rem}\label{Markov}\rm \begin{enumerate}\item[(a)] The constructed bush is indeed a ${\rm GW}(q_\lambda,{\rm Exp}(c_\lambda),\beta_\lambda)$-bush
and the pair obtained $(B_\mu,B_\lambda)$ has the same distribution as
$(B_\lambda^{(1-\mu/\lambda) \rm-rdc},B_\lambda)$. The intermediate trees
$\widehat{T}_\mu^{(i)}$ have the same distribution as the $(1-\mu/\lambda)$-reduced subtrees,
also jointly with the pair. If the split into two parts in the definition of $\widehat{T}_\lambda^{(i)}$ is used to assign black colour marks to the first part and red colour marks to the second part, then the resulting trees have the same distribution as the $(1-\mu/\lambda)$-coloured trees $B^{(1-\mu/\lambda)\rm-col}_\lambda$. We refer to Figure \ref{discperc} as a graphical illustration of the reduction and hence the reconstruction procedure.
\item[(b)] In \cite{DuW}, this reconstruction procedure is also formulated for representations of the trees
in a space of $\bR$-trees, tree-like metric spaces that we briefly address in Section \ref{realtree}.
\item[(c)] It is a simple consequence of the reduction procedure and/or the reconstruction
procedure that $(B_\lambda,\lambda\ge 0)$ is an inhomogeneous \em Markov \em process in a suitable space of finite sequences of $(0,\infty)$-marked trees. Similarly, (\ref{bpexp}) can be strengthened in the present context to
$$\bE\left(\left.\prod_{u\in\overline{\pi}_t(B_\lambda)}f_u\left(\overline{\theta}_{u,t}\left(B_\lambda^{(1-\mu/\lambda)\rm-col}\right)\right)\right|\cG_{\lambda,t}\right)=\prod_{u\in\overline{\pi}_t(B_\lambda)}\bE\left(f_u\left(B_\lambda^{(1-\mu/\lambda)\rm-col}\right)\right),$$
where $\cG_{\lambda,t}=\sigma\left\{\overline{\pi}_s(B_{\lambda^\prime}),\lambda^\prime \geq \lambda, s \leq t \right\}$, $\lambda\ge 0$, $t\ge 0$, and $B^{(1-\mu/\lambda)\rm-col}_\lambda$ is as in (a).
\end{enumerate}
\end{rem}
\subsubsection{Limiting behaviour as $\lambda \rightarrow \infty$} \label{limiting of lambda for edge} In the context of limiting results as $\lambda\rightarrow\infty$, we record the following corollary of Lemma \ref{alpha
and psi}.
\begin{coro}\label{9} If we fix $\mu>0$ in the setting of Lemma \ref{alpha and psi}, we have
$$\alpha=\alpha(\mu,\lambda)=\frac{\psi^\prime(\psi^{-1}(\mu))}{\psi^\prime(\psi^{-1}(\lambda))}\rightarrow\frac{\psi^\prime(\psi^{-1}(\mu))}{\psi^\prime(\infty)}\qquad\mbox{as $\lambda\rightarrow\infty$,}$$
where $\psi^\prime(\infty)$ means $\lim_{\lambda\rightarrow\infty}\psi^\prime(\lambda)$, in the following sense:
\begin{itemize}
\item If $\psi^\prime(\infty)< \infty,$ then $\alpha(\mu,\lambda)\rightarrow\alpha_0(\mu)=\psi^\prime(\psi^{-1}(\mu))/\psi^\prime(\infty)$.
\item If $\psi^\prime(\infty)=\infty$, then $\alpha(\mu,\lambda)\rightarrow\alpha_0(\mu)=0$.
\end{itemize}
\end{coro}
Note that this means that as $\lambda\rightarrow\infty$, lifetimes are cut into finite
${\rm Geo}(\alpha_0(\mu))$-distributed numbers of pieces in the first case, but into infinitely many
pieces in the second case.
\begin{lemm} \label{CSBP Limit Lemma x}
Let $(Z_t^{\lambda},t\ge 0)$, $\lambda\ge 0$, be continuous-time Galton-Watson processes associated with a consistent family $(B_\lambda)_{\lambda\ge 0}$, of ${\rm GW}(q_\lambda,{\rm Exp}(c_\lambda),\beta_\lambda)$-bushes as in Theorem \ref{thm1}. Then
$$\frac{1}{\psi^{-1}(\lambda)}Z^{\lambda}_t \rightarrow Z_t\qquad\mbox{in distribution, as $\lambda\rightarrow\infty$, for all $t\ge 0$,}$$
where $(Z_t,t\ge 0)$ is a ${\rm CSBP}(\psi)$ starting from $Z_0=\beta$, with $\psi$ as in {\rm(\ref{offspr})}. If furthermore
$\psi^\prime(0)>-\infty$, then the convergence holds in the almost sure sense.
\end{lemm}
\begin{pf} First consider $t=0$. The initial population sizes are Poisson distributed with
$$\bE(\exp\{-rZ_0^\lambda/\psi^{-1}(\lambda)\})=\exp\{\beta\psi^{-1}(\lambda)(e^{-r/\psi^{-1}(\lambda)}-1)\}\rightarrow e^{-\beta r},$$
as $\lambda\rightarrow\infty$, since $\psi^{-1}(\lambda)\rightarrow\infty$.
For $t>0$, the desired limiting distribution is characterised by (\ref{CSBP and psi solution u}) in terms of $u_t(r)$ and for $x=\beta$. If we integrate (\ref{u solution of psi}), we can identify $u_t(r)$ as the unique solution of
\begin{equation}\label{CSBP u bounds}
\int^{r}_{u_t(r)}\frac{dv}{{\psi}(v)} = t.
\end{equation}
Consider $Z^{\lambda}_t/{\psi}^{-1}(\lambda)$ as a sum of a Poisson number
of independent ${\rm GW}(q_\lambda,{\rm Exp}(c_\lambda))$-processes. Set
$s=e^{-r/{\psi}^{-1}(\lambda)}$ and apply the branching property at
the first branching time of a single ${\rm GW}(q_\lambda,
{\rm Exp}(c_\lambda))$-tree to obtain for its population size $X_t^\lambda$ at time $t$ with
$w^{\lambda}_t(s) = \bE(s^{X^{\lambda}_t})$
$
w^{\lambda}_t(s) \label{w solution}
=s e^{-c_\lambda t} + \int_0^t\sum_{k=0}^\infty q_\lambda(k)(\bE(s^{X_{t-y}^\lambda}))^kc_\lambda e^{-c_\lambda y}dy
=s e^{-c_\lambda t} + \int^{t}_{0} \varphi_{q_\lambda}(w^\lambda_z(s))c_\lambda e^{-c_\lambda (t-z)} dz.
$
Now apply (\ref{offspr}), multiply by $e^{c_\lambda t}$, differentiate with respect to $t$ and rearrange to get
\begin{equation}\label{wlambda}1=\frac{\psi^{-1}(\lambda)\frac{\partial}{\partial t}w_t^\lambda(s)}{\psi(\psi^{-1}(\lambda)(1-w_t^\lambda(s)))}\quad\Rightarrow\quad t=\int_{\psi^{-1}(\lambda)(1-w_t^\lambda(s))}^{\psi^{-1}(\lambda)(1-s)}\frac{dv}{\psi(v)}.\end{equation}
For $s=e^{-r/\psi^{-1}(\lambda)}$, we have $\psi^{-1}(\lambda)(1-s)\rightarrow r$ and by (\ref{CSBP u bounds}) also $\psi^{-1}(\lambda)(1-w_t^\lambda(s))\rightarrow u_t(r)$ and then
$$ \bE(\exp\{-rZ_t^{\lambda}/\psi^{-1}(\lambda)\})=\exp\{-\beta\psi^{-1}(\lambda)(1-w_t^\lambda(s))\}\rightarrow e^{-\beta u_t(r)}\qquad\mbox{as $\lambda\rightarrow\infty$,}
$$
as required.
For the proof of almost sure convergence, recall our notation $\overline{\pi}_t(T)\subset\bU$ for the population alive at time $t$ of a tree $T$, which we will slightly abuse and also apply to bushes.
Let $0\le\mu<\lambda$, $p=1-\mu/\lambda$ and $\mathcal{G}_{\lambda, t}=\sigma\left\{\overline{\pi}_s(B_{\lambda^\prime}),\lambda^\prime \geq \lambda, s \leq t \right\}$.
Now note that $Z^{\lambda}_t/{\psi}^{-1}(\lambda)$ is $\mathcal{G}_{\lambda, t}$-measurable, and that $\psi^\prime(0) > -\infty$ ensures that $\mathbb{E}(Z^{\mu}_t) < \infty$. Then, a.s.
$
\mathbb{E}\left(\left.Z^{\mu}_{t} \right| \mathcal{G}_{\lambda, t} \right)
=\mathbb{E}\left(\left.\sum_{u \in
\overline{\pi}_t(B_\lambda)}\mathbf{1}_{\{\gamma_u(B_\lambda^{p-\rm col})=0\}} \right|
\mathcal{G}_{\lambda, t} \right)
= \sum_{u \in
\overline{\pi}_t(B_\lambda)}\mathbb{E}\left(\left.\mathbf{1}_{\{\gamma_u(B_\lambda^{p-\rm col})=0\}}\right|\mathcal{G}_{\lambda, t} \right),
$
since $Z_t^\mu=\#\overline{\pi}_t(B_\mu)$. By the branching property in Remark \ref{Markov}(c), applied to functions
$f_u(\cdot)=\mathbf{1}_{\{\gamma_\emptyset(\cdot)=0\}},$ and $f_v(\cdot)\equiv 1$ for $v \neq u,$ we obtain, a.s.,
$$
\sum_{u \in
\overline{\pi}_t(B_\lambda)}\mathbb{E}\left(\left.\mathbf{1}_{\{\gamma_u(B_\lambda^{p-\rm col})=0\}}\right| \mathcal{G}_{\lambda, t} \right)
=Z^{\lambda,\beta}_t \mathbb{P}\left(\gamma_\emptyset(B_\lambda^{p-\rm col})=0\right)
=(1-g_\lambda(p))Z^{\lambda,\beta}_t.
$
According to (\ref{g in psi}), this shows that
$$\mathbb{E}\left(\left.\frac{Z^{\mu}_t}{\psi^{-1}(\mu)} \right| \mathcal{G}_{\lambda, t} \right) = \frac{Z^{\lambda}_t}{{\psi}^{-1}(\lambda)}\qquad\mbox{a.s.,}$$
which is the martingale property in the (decreasing) filtration $(\mathcal{G}_{\lambda,t},\lambda\ge 0)$ that implies that $Z^{\lambda}_{t}/\psi^{-1}(\lambda) \rightarrow Z_t$ almost surely as $\lambda \rightarrow
\infty$, see e.g. \cite{Wil-mart}.
\end{pf}
\section{Growth of ${\rm GW}(q_\lambda, \kappa_\lambda, \beta_\lambda)$-bushes: lifetimes}\label{Consistend kappa bush}
Theorem \ref{thm2} considers Galton-Watson bushes $B_\lambda$ with general
independent $\kappa_\lambda$-distributed lifetimes. The main statement beyond the exponential case of Theorem \ref{thm1} is that consistency of a full
family $(B_\lambda,\lambda\ge 0)$ under Bernoulli leaf colouring
requires $\kappa_\lambda$ to be geometrically divisible.
\subsection{Proof of Theorem \ref{thm2}}\label{Growth for general edge trees}
${\rm (i)}\Rightarrow{\rm (ii)}$: Suppose, (i) holds. In particular, $q_\mu$ is then the $(1-\mu/\lambda)$-reduced
offspring distribution associated with $q_\lambda$, for all $\mu<\lambda$. By Theorem \ref{thm1},
$\varphi_q(s)=s+\widetilde{\psi}(1-s)$, where $\widetilde{\psi}$ has the form (\ref{lkbm}). To be specific, let $c=1$,
$\beta\in(0,\infty)$ and parametrise $(q_\lambda,\beta_\lambda)$ using $\psi$ as in Section \ref{Consistent family}.
Consider a ${\rm GW}(q_\lambda,\kappa_\lambda)$-tree $T_\lambda$. By Remark \ref{alphakappa}, Lemma \ref{alpha and psi} applies. Using notation
similar to the reduction procedure of Section \ref{coloured leaves and 2-colour BP} for $T=T_\lambda$, $p=1-1/\lambda$, $\iota=\iota_\lambda$, $\lambda\ge 1$,
we can write $\zeta_\emptyset\sim \kappa_1$ as
\begin{equation}\label{writeas}\zeta_\emptyset=\sum_{v\in \iota_\lambda^{-1}(\emptyset)}\zeta_v^\lambda\overset{{\rm (d)}}{=} \sum_{j=1}^{G^{(\alpha)}}X_\alpha^{(j)},
\end{equation}
for $X_\alpha^{(j)}\sim\kappa_\lambda$, $j\ge 1$, and $G^{(\alpha)} \sim {\rm Geo}(\alpha)$ with
$\alpha=\psi^\prime(\psi^{-1}(1))/\psi^\prime(\psi^{-1}(\lambda))$ independent. Specifically, consider $\Gamma_n=\{G_\emptyset^{(\alpha)}=n\}$, where $G_\emptyset^{(\alpha)}=\iota^{-1}_\lambda(\emptyset)$ in the notation of Section \ref{Recon for GW expo bushes1}. On $\Gamma_n$, write $v_0=\emptyset$ and for
$1\le j\le n-1$, denote by $v_j$ the unique black child of $v_{j-1}$, then an $n$-fold inductive application of
the two-colours branching property of Section \ref{coloured leaves and 2-colour BP}, also summing over all offspring numbers and colour combinations
as in Lemma \ref{alpha and psi}, yields for $p=1-1/\lambda$
\beq \bE_{q_\lambda\otimes\kappa_\lambda}^{p-\rm col}\!\!\left[\left.\prod_{j=0}^{n-1} k_j(\zeta_{v_j});\Gamma_n\right|\gamma_\emptyset\!=\!0\right]
&\!\!\!\!\!=\!\!\!\!\!&\int_{(0,\infty)}\!\!\!\!k_0(z)\kappa_\lambda(dz)\varphi_{q_\lambda}^\prime(g_\lambda(p))\bE_{q_\lambda\otimes\kappa_\lambda}^{p-\rm col}\!\!\left[\left.\prod_{j=0}^{n-2}k_{j+1}(\zeta_{v_{j}});\Gamma_{n-1}\right|\gamma_\emptyset\!=\!0\right]\\
&\!\!\!\!\!=\!\!\!\!\!&\left(\prod_{j=0}^{n-1}\int_{(0,\infty)}k_j(z)\kappa_\lambda(dz)\right)(1-\alpha)^{n-1}\alpha.
\eeq
By Corollary \ref{9}, $\alpha\downarrow\psi^\prime(\psi^{-1}(1))/\psi^\prime(\infty)=1/\widetilde{\psi}^\prime(\infty)$ as $\lambda\rightarrow\infty$.
Therefore, we can write $\zeta_\emptyset\sim\kappa_1$ as in (\ref{writeas}) for all $\alpha>1/\widetilde{\psi}^\prime(\infty)$, with the convention $1/\widetilde{\psi}^\prime(0)=0$ if $\widetilde{\psi}^\prime(0)=\infty$. This yields (ii).
${\rm (ii)}\Rightarrow{\rm (i)}$: Suppose, (ii) holds. By Theorem \ref{thm1}, the family $(q_\lambda,\lambda\ge 0)$
exists as required. By Theorem \ref{thm1} and Remark \ref{alphakappa}, we can express $\varphi_{q_\lambda}$ and
$\alpha$ in terms of $\psi$ as in Section \ref{Consistent family} and Lemma \ref{alpha and psi}, choosing $c=1$.
For $\lambda>1$, geometric divisibility of $\kappa$ permits us to define $\kappa_\lambda$ as the
distribution of
$X_\alpha^{(j)}$
for $\alpha=\psi^\prime(\psi^{-1}(1))/\psi^\prime(\psi^{-1}(\lambda))=1/\widetilde{\psi}^\prime(\widetilde{\psi}^{-1}(\lambda\widetilde{\psi}(1)))>1/\widetilde{\psi}^\prime(\infty)$.
Now consider a ${\rm GW}(q_\lambda,\kappa_\lambda)$-tree and $p=1-1/\lambda$. Use notation from ${\rm (i)}\Rightarrow{\rm (ii)}$ and also set
$v^*=v_{G^{(\alpha)}_\emptyset-1}$. On $\Upsilon_j=\{\nu_{v^*}-\gamma_{v^*1}-\cdots-\gamma_{v^*\nu_{v^*}}=j\}$ for $j\ge 2$,
denote by $w_1,\ldots,w_j$ the black children of $v^*$. Then, by Remark \ref{alphakappa} and repeated application of the two-colours branching property,
we obtain
\beq &&\hspace{-0.5cm}\bE_{q_\lambda\otimes\kappa_\lambda}^{p-\rm col}\left[\left.\exp\left\{-r\sum_{m=0}^{G_\emptyset^{(\alpha)}-1}\zeta_{v_m}\right\}\prod_{i=1}^jf_i(\overline{\theta}_{w_i});\Upsilon_j\right|\gamma_\emptyset=0\right]\\
&&=\sum_{n=1}^\infty\left(\int_{(0,\infty)}e^{-rz}\kappa_\lambda(dz)\right)^n\ (1-\alpha)^{n-1}\ \varphi_{q_\lambda}^{(j)}(g_\lambda(p))\frac{(1-g_\lambda(p))^{j-1}}{j!}\ \prod_{i=1}^j\bE_{q_\lambda\otimes\kappa_\lambda}^{p-\rm col}\left[\left.f_i\right|\gamma_\emptyset=0\right].
\eeq This is the branching property characterizing ${\rm
GW}(q,\kappa)$, because the first term is the Laplace transform of a
geometric sum with distribution $\kappa$, up to a factor of
$\alpha$, and for the middle term we identify the offspring
distribution $q$
using all the cancellations due to
(\ref{offspr}), (\ref{g in psi}) and (\ref{imporesult})
\beq &&\hspace{-0.7cm}\sum_{j=2}^\infty s^j\frac{1}{\alpha}\frac{\varphi_{q_\lambda}^{(j)}(g_\lambda(p))}{j!}(1-g_\lambda(p))^{j-1}+\left(1-\sum_{j=2}^\infty\frac{1}{\alpha}\frac{\varphi_{q_\lambda}^{(j)}(g_\lambda(p))}{j!}(1-g_\lambda(p))^{j-1}\right)\\
&&\hspace{-0.2cm}=1+\frac{\varphi_{q_{\lambda}}(g_\lambda(p)+s(1-g_\lambda(p)))-1-(1-s)\varphi_{q_\lambda}^\prime(g_\lambda(p))(1-g_\lambda(p))}{\alpha(1-g_\lambda(p))}=s+\frac{\psi(\psi^{-1}(1)(1-s))}{\psi^{-1}(1)\psi^\prime(\psi^{-1}(1))}
\eeq
confirming that $(q,\kappa)$ is the $(1-1/\lambda)$-reduced pair associated with $(q_\lambda,\kappa_\lambda)$. For $\mu<1$, set $\alpha=\psi^\prime(\psi^{-1}(\mu))/\psi^\prime(\psi^{-1}(1))$ and define $\kappa_\mu$
to be the distribution of
$$\sum_{j=1}^{G^{(\alpha)}}X^{(j)},\qquad\mbox{for independent $X^{(j)}\sim\kappa$, $j\ge 1$, independent of $G^{(\alpha)}\sim{\rm Geo}(\alpha)$.}$$
As above, $(q_\mu,\kappa_\mu)$ is the
$(1-\mu)$-reduced pair associated with $(q,\kappa)$. The reduction relation for $0\le\mu<\lambda<\infty$ follows
using transitivity of colouring reduction (see Remark \ref{remdet}(a)) for $0\le\nu<\mu<\lambda<\infty$. Specifically, for $\mu=1$, this yields that $(B_\lambda^{(1-\nu/\lambda)-{\rm rdc}},B_\lambda)\overset{{\rm (d)}}{=} (B_\nu,B_\lambda)$. For $\nu=1<\mu<\lambda$ and $\nu<\mu<\lambda=1$, this argument can be combined with the uniqueness of the divisor distribution (Lemma \ref{uniqgeodiv}). This completes the proof of ${\rm (ii)}\Rightarrow{\rm (i)}$.
We identified $(\beta_\lambda,\lambda\ge 0)$ in ${\rm(i)}\Rightarrow{\rm (ii)}$. The same reasoning as in Remark \ref{alphakappa} allows us to combine (i) here and Theorem \ref{thm1} to see that $(q,\kappa,\beta)$ is the $(1-1/\lambda)$-reduced triplet associated with $(q_\lambda,\kappa_\lambda,\beta_\lambda)$. The existence of $(B_\lambda,\lambda\ge 0)$ now follows from Kolmogorov's consistency theorem. Uniqueness of the families $(q_\lambda,\lambda\ge 0)$, $(\kappa_\lambda,\lambda\ge 0)$ and $(\beta_\lambda,\lambda\ge 0)$ for each
$\beta=\beta_1\in(0,\infty)$ follows from the uniqueness results in Theorem \ref{thm1} and as shown in ${\rm (ii)}\Rightarrow{\rm (i)}$.\hspace*{\fill} $\square$
\subsection{Reconstruction procedures and backbone decomposition}\label{Reconstruction General Edge}
If $\kappa_\lambda(dz)=f_\lambda(z)dz$ is absolutely continuous for all $\lambda\ge 0$ and $\zeta=\zeta_1+\cdots+\zeta_{G^{(\alpha)}}\sim\kappa_\mu$ for independent $\zeta_j\sim\kappa_\lambda$ and $G^{(\alpha)}\sim{\rm Geo}(\alpha)$ for $\alpha$ as in Section \ref{Recon for GW expo bushes1}, we find conditional joint
distributions as in (\ref{joint exponential})
\begin{equation}\label{joint general edge}
f_{\zeta_{1},\ldots,\zeta_{n-1}|G^{(\alpha)}=n,\zeta=z}(y_1,\ldots,y_{n-1})=
\frac{f_\lambda(z-\sum^{n-1}_{j=1}y_j)\prod^{n-1}_{j=1}f_\lambda(y_j)}{f_\lambda^{*(n)}(z)}
\end{equation}
for $y_j>0$ with $y_1+\cdots+y_{n-1}<z$, $n\ge 1$, where
$f_\lambda^{*(n)}$ is the $n$th convolution power of $f_\lambda$.
\subsubsection{Reconstruction procedure for ${\rm GW}(q_\lambda, \kappa_\lambda,\beta_\lambda)$-bushes}
For $B_\mu\sim{\rm GW}(q_\mu,\kappa_\mu,\beta_\mu)$ the
procedure in Section \ref{Recon for GW expo bushes2} with ${\rm Exp}(c_\lambda)$ replaced by $\kappa_\lambda$ and
(\ref{joint exponential}) replaced by (\ref{joint general edge}) constructs $B_\lambda\sim{\rm GW}(q_\lambda,\kappa_\lambda,\beta_\lambda)$.
In the general case, one could use regular conditional distributions. Alternatively, we can adapt the procedure in
Section \ref{Recon for GW expo bushes2} using the subordinator (random walk) representation of geometrically infinitely
(finitely) divisible distributions as explained below.
\subsubsection{Reconstruction procedure with subordinators in the case where $\kappa$ is g.i.d.}\label{recprocgid}
In the g.i.d.\ case, let $\bH=\bigcup_{\zeta\in(0,\infty)}\{\zeta\}\times\bD([0,\zeta],[0,\infty))$,
where $\bD([0,\zeta],[0,\infty))$ is the set of functions $f\colon[0,\zeta]\rightarrow[0,\infty)$ that are right-continuous with left limits, equipped with the Borel sigma-algebra generated by the metric topology induced by $$d((\zeta_1,f_1),(\zeta_2,f_2))=|\zeta_1-\zeta_2|+d_{\rm Sk}(f_1(\cdot\wedge\zeta_1),f_2(\cdot\wedge\zeta_2)),$$ where $d_{\rm Sk}$ is a
metric that generates Skorohod's topology on $\bD([0,\infty),[0,\infty))$, see e.g. \cite[Section IV.1]{LeG-snake}. Consider a subordinator $(\sigma(t),t\ge 0)$, under $\bP$, such that $\sigma(V)\sim\kappa_1$ for an
independent $V\sim{\rm Exp}(1)$. Define the following measure on $\bN_0\times\bH$
\begin{equation}\label{Qmu}\bQ_\mu(\{j\}\times H)=q_\mu(j)\int_{(0,\infty)}\bP((z,(\sigma(t),0\le t\le z))\in H)c_\mu e^{-c_\mu z}dz.\end{equation}
Now consider a bush $\overline{B}_\mu$ of $N_\mu\sim{\rm Poi}(\beta_\mu)$ random trees with distribution $\bP_{\bQ_\mu}$ as defined in the branching property of Section \ref{Lifetime marks and discr BP in cont. time} for the measure $\bQ_\mu$ just defined.
Then the reconstruction procedure of Section \ref{Recon for GW expo bushes2} can be applied subdividing subordinator
lifetimes $\zeta_u^{(i)}\sim{\rm Exp}(c_\mu)$ rather than directly the population lifetimes $\sigma_u^{(i)}(\zeta_u^{(i)})\sim\kappa_\mu$. Also define
$$\sigma_{u,m}^{(i)}(t)=\sigma_u^{(i)}(\zeta_{u,1}^{(i)}+\cdots+\zeta_{u,m-1}^{(i)}+t)-\sigma_u^{(i)}(\zeta_{u,1}^{(i)}+\cdots+\zeta_{u,m-1}^{(i)}),\quad 0\le t\le\zeta_{u,m}^{(i)}, 1\le m\le G_u^{(\alpha,i)}.$$
The remainder of the procedure is easily adapted. The resulting $\overline{B}_\lambda$ is a bush of $N_\lambda\sim{\rm Poi}(\beta_\lambda)$ random trees with distribution $\bP_{\bQ_\lambda}$.
\subsubsection{Reconstruction procedure with random walks in the case where $\psi^\prime(\infty)<\infty$.}
In the case where $\kappa$ is geometrically divisible up to $\alpha_0(1)=\psi^\prime(\psi^{-1}(1))/\psi^\prime(\infty)>0$, let $\bH=\bigcup_{n\in\mathbb{N}}\{n\}\times[0,\infty)^{n+1}$ be the space of random walk paths. On $\bN_0\times\bH$, consider
$$\bQ^{\rm RW}_\mu(\{j\}\times H)=q_\mu(j)\sum_{n=1}^\infty\bP((n,(\sigma(k),0\le k\le n))\in H)(1-\alpha_0(\mu))^{n-1}\alpha_0(\mu),$$
where, under $\bP$, $(\sigma(k),k\ge 0)$ is a random walk with $\sigma(k+1)-\sigma(k)\sim\kappa_\infty$ i.i.d.
In fact, in this case, the distribution $\kappa_\infty$ exists since the Laplace transforms of $\kappa_\lambda$
converge as $\lambda\rightarrow\infty$ to a completely monotone function continuous at zero. Then the reconstruction
procedure of Section \ref{Recon for GW expo bushes2} can be applied subdividing geometric ``random walk lifetimes'' $G_u^{(i)}\sim{\rm Geo}(\alpha_0(\mu))$ into
$G_u^{(i)}=G_{u,1}^{(i)}+\cdots+G_{u,G_u^{(\alpha,i)}}$, for independent $G_{u,m}^{(i)}\sim{\rm Geo}(\alpha_0(\lambda))$, $m\ge 1$, and $G_u^{(\alpha,i)}\sim{\rm Geo}(\alpha)$.
Also define
$$\sigma_{u,m}^{(i)}(k)=\sigma_u^{(i)}(G_{u,1}^{(i)}+\cdots+G_{u,m-1}^{(i)}+k)-\sigma_u^{(i)}(G_{u,1}^{(i)}+\cdots+G_{u,m-1}^{(i)}),\qquad 0\le k\le G_{u,m}^{(i)}.$$
The remainder of the procedure is easily adapted. The resulting $\overline{B}_\lambda$ is a bush of $N_\lambda\sim{\rm Poi}(\beta_\lambda)$ random trees with distribution $\bP_{\bQ^{\rm RW}_\lambda}$.
\subsubsection{Backbone decomposition of supercritical Bellman-Harris processes}
The reconstruction procedures that build a ${\rm GW}(q_\lambda,\kappa_\lambda,\beta_\lambda)$-bush $B_\lambda$ from a ${\rm GW}(q_\mu,\kappa_\mu,\beta_\mu)$-bush give rise to decompositions of the associated Bellman-Harris process $Z_t^\lambda=\#\overline{\pi}_t(B_\lambda)$ along the ${\rm GW}(q_\mu,\kappa_\mu,\beta_\mu)$-bush $B_\mu$. In the sequel, we will write ${\rm BH}(q_\lambda,\kappa_\lambda)$ for such a Bellman-Harris process and when we specify its initial distribution, all these individuals are taken with zero age.
In the supercritical case $\psi^\prime(0)<0$, note that the ${\rm GW}(q_{\lambda,\rm red}^{(1-\mu/\lambda)-\rm rdc},\kappa_\lambda)$-trees
with offspring distribution as in (\ref{red offspring distribution}) that are grafted onto $B_\mu$ are subcritical for all $0\le\mu<\lambda$. The case $\mu=0$ in is at the heart of many decompositions in various settings, mainly continuous analogues with and without spatial motion, see \cite{BKM-10,DuW,EtW-99,EvO-91}. As an immediate consequence of our reconstruction procedures, we obtain a version of the backbone decomposition for Bellman-Harris processes.
\begin{coro}\label{cor14} Let $\psi$ be a supercritical
branching mechanism, $B_0$ a bush of $N_0\sim{\rm Poi}(\beta\psi^{-1}(0))$ random trees with distribution
$\bP_{\bQ_0}$ as in {\rm (\ref{Qmu})}. Subdivide each subordinator lifetime as in Section \ref{recprocgid} to get a bush $\widehat{B}_0$. For each $u\in\widehat{B}_0$
independently, given $\widehat{\nu}_u=m$ children, consider a ${\rm BH}(q_{\lambda,\rm red}^{1-\rm rdc},\kappa_\lambda)$-process $Z^{(u)}$ with $Z^{(u)}_0$
of distribution {\rm(\ref{redgraft})}. Also consider a ${\rm BH}(q_{\lambda,\rm red}^{1-\rm rdc},\kappa_\lambda)$-process $Z^{\rm root}$ with $Z^{\rm root}_0\sim{\rm Poi}(\beta(\psi^{-1}(\lambda)-\psi^{-1}(0)))$. Then the process
$$Z_t=\#\overline{\pi}_t(B_0)+Z^{\rm root}_t+\sum_{u\in\widehat{B}_0\colon \omega_u\le t}Z^{(u)}_{t-\omega_u}\vspace{-0.2cm}$$
is a ${\rm BH}(q_\lambda,\kappa_\lambda)$-process with $Z_0\sim{\rm Poi}(\beta\psi^{-1}(\lambda))$.
\end{coro}
\subsection{Limiting trees and branching processes as $\lambda\rightarrow\infty$}\label{realtree}
\subsubsection{Convergence of trees: $\bR$-tree representations, L\'evy trees and snakes}\label{secsnake}
A random marked tree $\overline{T}_\lambda$ with distribution $\bP_{\bQ_\lambda}$ as in Section \ref{recprocgid} specifies marks $\zeta_u\sim{\rm Exp}(c_\lambda)$ as well as $\sigma_u(\zeta_u)\sim\kappa_\lambda$. Therefore, we can associate coupled
trees
\begin{eqnarray*}&&T_\lambda^\circ=\{(u,\zeta_u)\colon (u,\zeta_u,\sigma_u)\in\overline{T}_\lambda\}\sim{\rm GW}(q_\lambda,{\rm Exp}(c_\lambda),\beta_\lambda)\\
\mbox{and}&&T_\lambda^\bullet=\{(u,\sigma_u(\zeta_u))\colon (u,\zeta_u,\sigma_u)\in\overline{T}_\lambda\}\sim{\rm GW}(q_\lambda,\kappa_\lambda,\beta_\lambda),
\end{eqnarray*}
that only differ in their lifetimes. In the same way, we obtain coupled bushes and consistent families $(B_\lambda^\circ,\lambda\ge 0)$ and $(B_\lambda^\bullet,\lambda\ge 0)$. Several other representations of $(T_\lambda^\circ,T_\lambda^\bullet)$ are natural. For an $\bR$-tree representation
$$\cT_\lambda^\circ=\{\rho\}\cup\bigcup_{u\in\tau_\lambda}\{u\}\times(\alpha_u^\circ,\omega_u^\circ]$$
of $T_\lambda^\circ$, with root $\rho=(\emptyset,0)$ and metric $\mathrm{d}$ given by $\mathrm{d}((v,s),(w,t))=|t-s|$ for $v\preceq w$ or $w\preceq v$,
$$\mathrm{d}((uiv,s),(ujw,t))=s+t-2\omega_u^\circ\qquad\mbox{for $u,v,w\in\bU$ and $i,j\in\bN$, $i\neq j$,}$$
see \cite[Sect. 3.3]{DuW}, we can consider the measure
$$\cW_\lambda(\{u\}\times(a,b])=\sigma_u(b-\alpha_u^\circ)-\sigma_u(a-\alpha_u^\circ),\qquad\alpha_u^\circ\le a<b\le\omega_u^\circ,u\in\tau_\lambda,$$
which for $\sigma_u\overset{{\rm (d)}}{=} \sigma$ as in (\ref{LaExponent})-(\ref{LK1}), $u\in\tau_\lambda$, is of the form $\cW_\lambda=d{\rm Leb}+\cR_\lambda$, where $\cR_\lambda$ is an infinitely divisible independently scattered random measure on $\cT_\lambda^\circ$, in the sense of \cite{Kal,RaR,WoU}, with $F$-measure ${\rm Leb}\otimes\Lambda$, where ${\rm Leb}$ denotes Lebesgue measure on $\cT_\lambda^\circ$.
For technical convenience, we can embed $(\cT_\lambda^\circ,\mathrm{d},\rho)$ as a metric subspace of $\ell_1(\bN)=\{x=(x_n)_{n\ge 1}\colon x_n\in[0,\infty),n\ge 1;||x||_1<\infty\}$, where $||x||_1=\sum|x_n|$ is the $\ell_1$-norm,
with root $0\in\ell_1(\bN)$. We can embed bushes $(\cB_\lambda^\circ,\lambda\ge 0)$ consistently following \cite[Remarks 4.9-4.10]{DuW}: in the (sub)critical case, this family starts from $\varnothing$, is piecewise constant and evolves by adding single branches, so we can represent the $j$th branch, of length $L_j^\circ$ say, as $[[x(j),x(j)+L_j^\circ e_j]]$ for some $x(j)=(x_1(j),\ldots,x_{j-1}(j),0,\ldots)\in\cB_{\lambda_{j-1}}^\circ$ and $e_j=(0,\ldots,0,1,0,\ldots)$ the $j$th coordinate vector in $\ell_1(\bN)$. Similarly, we can embed $(\cB_\lambda^\bullet,\lambda\ge 0)$ for certain $L_j^\bullet=\sigma_j(L_j^\circ)$. The supercritical case can be handled
using \cite[Proposition 3.7]{DuW}.
For the $\ell_1(\bN)$-embedding, we can consider $\cW_\lambda$ as a random measure on $\ell_1(\bN)$ (with support included in the embedded $\cB_\lambda^\circ$).
Then $\cB_\lambda^{\circ,\rm wt}=(\cB_\lambda^\circ,\cW_\lambda)$ is a weighted $\bR$-tree in the sense of \cite{EvW-07,GPW-09}, in the (sub)critical case, but weak or vague convergence as $\lambda\rightarrow\infty$ are not appropriate since in the limit the measures become infinite on any ball around a point in some $\cB_\lambda^\circ$. Nevertheless, consistency implies that in the case where convergence to a locally compact and separable L\'evy bush $\cB^\circ$ occurs in the Gromov-Hausdorff sense \cite[Theorem 5.1]{DuW}, the random measures $\cW_\lambda$ consistently build an infinitely divisible independently scattered random measure $\cW$ on $\cB^\circ\subset\ell_1(\bN)$ whose Poissonian component still has $F$-measure ${\rm Leb}|_{\cB^\circ}\otimes\Lambda$, while the continuous component is still $d{\rm Leb}|_{\cB^\circ}$ and where ${\rm Leb}$ is one-dimensional Lebesgue measure on $\ell_1(\bN)$.
For $x\in\cB^\circ$ consider the path $[[0,x]]=\{y\in\cB^\circ\colon y_n\le x_n,n\ge 1\}=\{f_x(t)\colon 0\le t\le||x||_1\}$, where $||f_x(t)||_1=t$, $0\le t\le||x||_1$. Then $\cS\colon\cB^\circ\rightarrow\bH$ with $\bH$ as in Section \ref{recprocgid} given by $\cS(x)=(||x||_1,(\cW([[0,f_x(t)]]),0\le t\le||x||_1))$ can be seen as a snake in the sense of \cite[Section I.3.2]{LeG-snake}, where the
spatial motion here is a subordinator with characteristic pair $(d,\Lambda)$ as in (\ref{LK1}). See \cite{JaM} for another setting where discontinuous snakes appear naturally.
\begin{prop} In the setting of Theorem \ref{thm2}, if $\cB^\circ$ is separable and locally compact, then
$\cB^\bullet$, the closure of $\bigcup\cB^\bullet_\lambda$ in $\ell_1(\bN)$, is separable, but locally compact only if $\Lambda=0$. Also, if $\cB^\circ$ is furthermore bounded, then $\cB^\bullet$ is bounded if and only if $\Lambda=0$.
\end{prop}
\begin{pf} Separability of $\cB^\bullet$ is trivial as $\cB^\bullet_\lambda$ is separable. Now argue conditionally given $(\cB^\circ_\lambda)_{\lambda\ge 0}$.
To show that boundedness and local compactness fail unless $\Lambda=0$, first consider $\Lambda=\delta_h$, the Dirac measure in $h>0$. It is not hard to
show that the subtree of $\cB^\circ$ above every vertex, except leaves, has infinite total length. As a consequence, there will be
$\lambda_1>0$ for which $\cW_{\lambda_1}$ has an atom $x_1$, and the subtree $\{y\in\cB^\circ\colon x_1\in[[0,y[[\}$ above $x_1$ has
infinite length. Inductively, we find an increasing sequence $x_1\prec x_2\prec\cdots$ of atoms of $\cW$ showing that
$\cB^\bullet$ is not bounded.
Now assume that $\cB^\bullet$ is locally compact. By the Hopf-Rinow theorem \cite{Gro}, closed balls are then compact. Since $\cB^\circ$ is locally compact, \cite[Remark 5.1]{DuW} implies that \cite[Theorem 5.1]{DuW} applies. In particular, $\psi^\prime(\infty)=\infty$, so that $\cB^\circ\setminus[[0,L^\circ_1e_1]]$ has infinitely many connected
components, each of infinite total length, so that they all contain atoms of $\cW$. However, $L_1^\bullet=\sigma_1(L_1^\circ)$ is bounded and for a large enough ball with $B(r)=\{x\in\cB^\bullet\colon ||x||_1\le r\}\supset[[0,L^\bullet_1e_1]]$, the set $B(r+h)\setminus[[0,L^\bullet_1e_1]]$ also has infinitely many connected components each exceeding diameter $h$. Considering a cover of $B(r+h)$ by open balls of radii less than $h/2$, there is no finite subcover, as each connected component needs at least one ball that does not intersect any other connected component, which contradicts the compactness of $B(r+h)$. So $\cB^\bullet$ is not locally compact.
For any $\Lambda\neq 0$ and $h>0$ with $\Lambda((h,\infty))>0$, let
$\widetilde{\Lambda}=\Lambda((h,\infty))\delta_h$, couple $\cB^\bullet$ and
$\widetilde{\cB}^\bullet$ and argue as above that $\widetilde{\cB}^\bullet$ is not bounded nor locally
compact, then deduce the same for $\cB^\bullet$.
\end{pf}
Since $\cB^\bullet$ is separable for a large class of branching mechanisms and lifetime subordinators, the framework of
\cite{GPW-09} can be used to further study these trees.
\subsubsection{Superprocesses, backbones and convergence of Bellman-Harris processes}\label{CSBPKpsi}
Sagitov \cite{Sag-91} studied convergence of Bellman-Harris processes to certain non-Markovian CSBP whose distribution is best described via Markovian superprocesses that record residual lifetimes, see also \cite{KaS-98}. Specifically, a $(\xi,K,\psi)$-superprocess \cite{Dyn-91} is a Markov process $M=(M_t,t\ge 0)$ on the space of finite Borel measures $\bM([0,\infty))$ with transition semigroup characterised by
\begin{equation}\label{meassg}\bE\left(\left.\!\exp\left\{\!-\!\!\int_{[0,\infty)}\!\!f(z)M_t(dz)\!\right\}\right|M_0\!=\!m\!\right)\!=\exp\left\{\!-\!\!\int_{[0,t]}\!\!u_{t-z}(f)m(dz)\!-\!\!\int_{(t,\infty)}\!\!f(z\!-\!t)m(dz)\!\right\},\hspace{-0.0cm}\end{equation
for all $f\colon[0,\infty)\rightarrow[0,\infty)$ bounded continuous, where $u_t(f)$ is the unique (at least if $\psi^\prime(0)>-\infty$) nonnegative solution of
\begin{equation}\label{measint}u_t(f)+\int_{[0,t]}\psi(u_{t-s}(f))dH_s=\bE(f(\xi_t))\end{equation}
and $H_s=\bE(K_s)$ is the renewal function of a strictly increasing subordinator or random walk $\sigma$,
here in terms of the local time process $K_s=\inf\{t\ge 0\colon \sigma(t)>s\}$, which is an additive functional of our particular choice $\xi_t=\sigma(K_t)-t$ of Markovian spatial motion. Since branching only occurs at the origin, $M$ is a catalytic superprocess \cite{DaF-94,Dyn-95}. We call the associated population size process $Z_t=M_t([0,\infty))$ a ${\rm CSBP}(K,\psi)$. Such processes appear as limits of Bellman-Harris processes, also in our setting. Unless stated otherwise, we understand $Z_0=\beta$ as $M_0=\beta\delta_0$.
\begin{prop}\label{Sagconv} Let $Z_t^{\lambda}=\#\overline{\pi}_t(B_\lambda)$ for a consistent family $(B_\lambda)_{\lambda\ge 0}$ of
${\rm GW}(q_\lambda,\kappa_\lambda,\beta_\lambda)$-bushes with branching mechanism $\psi$ as in Theorem \ref{thm2}. Suppose that $\psi$ is subcritical or critical, i.e. $\psi^\prime(0)\ge 0$, and that $\psi^\prime(\infty)=\infty$. Let $\sigma$ be a subordinator as in Section \ref{recprocgid}. Then, for all $t \geq 0$
$$\frac{1}{\psi^{-1}(\lambda)}Z^{\lambda}_t \longrightarrow Z_t\qquad\mbox{almost surely, as $\lambda\rightarrow\infty$,}$$
where $(Z_t,t\ge 0)$ is a ${\rm CSBP}(K,\psi)$ starting from $Z_0=\beta$.
\end{prop}
\begin{pf} The almost sure convergence follows from the same martingale argument as in Lemma \ref{CSBP Limit Lemma x}. The identification of the limiting process follows from
\cite[Theorem 1]{Sag-91}, see \cite{KaS-98} for a proof. Specifically, we consider the limit along the subsequence $(\lambda_n,n\ge 1)$ for which
$\psi^{-1}(\lambda_n)=n$.
\end{pf}
Proposition \ref{prop4a} states that the conclusion of Proposition \ref{Sagconv} holds in the supercritical as well as $\psi^\prime(\infty)<\infty$ cases. Sagitov \cite{Sag-91} announces his convergence result to include the (finite-mean) supercritical case, but the proof in Kaj and Sagitov \cite{KaS-98} only
treats the subcritical and critical cases. They say that the supercritical case would require some further assumptions and additional work. In our less general setting, this is not difficult -- our proof does not rely on \cite{KaS-98}.
\begin{pfofprop4a} In the g.i.d.\ case $\psi^\prime(\infty)=\infty$, we use Lemma \ref{sub and geo} to represent $\kappa_\lambda$ in terms of a strictly increasing subordinator $\sigma$ as $\kappa_\lambda=\bP(\sigma(V_\lambda)\in\cdot)$ for
$V_\lambda\sim{\rm Exp}(\psi^\prime(\psi^{-1}(\lambda)))$. We introduce Markovian measure-valued branching processes $M^\lambda$ that do not record residual
lifetimes of $Z^\lambda$, but what we may think of as limiting residual lifetimes (in the limit $\lambda\rightarrow\infty$),
$$M^\lambda_t=\sum_{u\in\overline{\pi}_t(B_\lambda^\bullet)}\delta_{\sigma_u(K^u_{t-\alpha_u^\bullet})-(t-\alpha_u^\bullet)}=\sum_{x\in\cB^\circ_\lambda\colon\sigma(x-)\le t\le\sigma(x)}\delta_{\sigma(x)-t},\qquad\mbox{where $\sigma(x-)=\cW_\lambda([[0,x[[)$,}$$
using notation as in Section \ref{secsnake} and also $K^u_s=\inf\{t\ge 0\colon\sigma_u(t)>s\}$.
By \cite[Lemma 3]{KaS-98} or \cite[Formula (1.5)]{Dyn-91}, for the Markov process $\xi_t=\sigma(K_t)-t$, the
semigroup of $M^\lambda$ is such that
$$\bE\left(\!\left.\exp\left\{\!-\!\int_{[0,\infty)}\!f(z)M_t^\lambda(dz)\right\}\right|M_0^\lambda\!=\!m\right)=\exp\left\{\!-\!\int_{[0,t]}\!v^\lambda_{t-z}(f)m(dz)+\int_{(t,\infty)}\!f(z\!-\!t)m(dz)\right\},$$
for all $f\colon[0,\infty)\rightarrow[0,\infty)$ bounded continuous, where $v_t^\lambda(f)$ satisfies
$$1-e^{-v_t^\lambda(f)}=\bE\left(1-e^{-f(\sigma(K_t)-t)}\right)-\int_{[0,t]}\left(\varphi_\lambda(e^{-v_{t-s}^\lambda(f)})-e^{-v_{t-s}^\lambda(f)}\right)d\bE(N_s^\lambda),$$
and where $N_s^\lambda=\#\{k\ge 1\colon R_k^\lambda\le s\}$ is the renewal process associated with a random walk with $R_k^\lambda-R_{k-1}^\lambda\sim\kappa_\lambda$. It
is easily checked that $\bE(N_s^\lambda)=H_s$. The remainder is straightforward (cf. \cite[Section 1.2]{Dyn-91}). We apply (\ref{psipsitilde}) to see that $u_t^\lambda(f)=\psi^{-1}(\lambda)(1-\exp\{-v_t^\lambda(f/\psi^{-1}(\lambda))\})$ satisfies
$$u_t^\lambda(f)+\int_{[0,t]}\psi(u_{t-s}^\lambda(f))dH_s=\bE\left(\psi^{-1}(\lambda)\left(1-\exp\left\{-\frac{f(\sigma(K_t)-t)}{\psi^{-1}(\lambda)}\right\}\right)\right).$$
Uniqueness in (\ref{measint}) means that $u_t^\lambda(f)=u_t(f_\lambda)$, where $f_\lambda=\psi^{-1}(\lambda)(1-e^{-f/\psi^{-1}(\lambda)})\uparrow f$. By
the Monotone Convergence Theorem, this implies for $N_\lambda\sim{\rm Poi}(\beta\psi^{-1}(\lambda))$ and $M_0^\lambda=N_\lambda\delta_0$ that
\beq\bE\left(\!\exp\left\{\!-\!\!\int_{[0,\infty)}\!\!f(z)\frac{M_t^\lambda(dz)}{\psi^{-1}(\lambda)}\right\}\right)&\!\!\!=\!\!\!&\exp\left\{-\beta\psi^{-1}(\lambda)(1-v_t^\lambda(f/\psi^{-1}(\lambda)))\right\}=\exp\{-\beta u_t(f_\lambda)\}\\
&\!\!\!=\!\!\!&\bE\left(\left.\!\exp\left\{\!-\!\!\int_{[0,\infty)}\!\!f_\lambda(z)M_t(dz)\!\right\}\right|M_0=\beta\delta_0\right)\\
&\!\!\!\rightarrow\!\!\!&\bE\left(\left.\!\exp\left\{\!-\!\!\int_{[0,\infty)}\!\!f(z)M_t(dz)\!\right\}\right|M_0=\beta\delta_0\right)=\exp\{-\beta u_t(f)\}.
\eeq
In particular, for $Z_t^\lambda=M_t^\lambda([0,\infty)$, we obtain $Z_t^\lambda/\psi^{-1}(\lambda)\rightarrow Z_t$, where $Z$ is a ${\rm CSBP}(K,\psi)$. The martingale argument of Lemma \ref{CSBP Limit Lemma x} establishes almost sure convergence in the case $\psi^\prime(0)>-\infty$.
In the finitely geometrically divisible case, we use bushes $\overline{B}_\lambda$ based on measures $\bQ^{\rm RW}_\lambda$ and
$$M_t^\lambda=\sum_{u\in\overline{\pi}({B}_\lambda^\bullet)}\delta_{\sigma_u(K^u_{t-\alpha_u^\bullet})-(t-\alpha_u^\bullet)},\qquad\mbox{where $K^u_s=\inf\{k\ge 1\colon \sigma_u(k)>s\}$.}$$
Then the argument above is easily adapted.
\end{pfofprop4a}
As an application, let us derive a backbone decomposition. This should be useful to deduce more general supercritical Bellman-Harris convergence results from subcritical results. Our present paper is not about superprocesses nor convergence of general triangular arrays, so we do not push for highest generality nor assumptions as in \cite{KaS-98}, but we would like to mention a now natural approach -- in a sense to be made precise, convergence of supercritical processes is equivalent to convergence of backbones and convergence of associated subcritical processes.
We write $\bP_{K,\psi}^r$ for the distribution of a ${\rm CSBP}(K,\psi)$ starting from $r\ge 0$. Just as for ${\rm CSBP}(\psi)$ in Section \ref{CSBP and CBI explanation
section}, we consider the sigma-finite measure $\Theta_{K,\psi}$ such that $\bP_{K,\psi}^r$ is the distribution of a sum over a Poisson point process with intensity measure $r\Theta_{K,\psi}$.
\begin{theo}[Backbone decomposition for ${\rm CSBP}(K,\psi)$]\label{CSBPbackbone} Let $\psi$ be a (non-explosive) supercritical, $\psi_0(r)=\psi(r+\psi^{-1}(0))$ the associated subcritical branching mechanism. Let $\overline{B}_0$ be a bush of
$N_0\sim{\rm Poi}(\beta\psi^{-1}(0))$ trees with distribution
$\bP_{\bQ_0}$ as in {\rm (\ref{Qmu})}, and, as in Section \ref{secsnake}, $(\cB_0^\circ,\cW_0)$ a representation as a weighted $\bR$-tree, $\sigma(x)=\cW_0([[0,x]])$. Given $(\cB_0^\circ,\cW_0)$, consider
\begin{itemize}\item points $(Z^x,x\in\cP)$ of a Poisson point process in $\bD([0,\infty),[0,\infty))$ with intensity measure
$$Q(df,dx)=\left(2a\Theta_{K,\psi_0}(df)+\int_{(0,\infty)}\bP^r_{K,\psi_0}(df)re^{-r\psi^{-1}(0)}\Pi(dr)\right){\rm Leb}|_{\cB_0^\circ}(dx),$$
\item extra points $(Z^x,x\in{\rm Br}(\cB_0^\circ))$ independent of $(Z^x,x\in\cP)$ with distribution
$$Q^{(l(x))}(df)=\frac{2a1_{\{l(x)=2\}}}{|\psi_0^{(l(x))}(0)|}\delta_0(df)+\int_{(0,\infty)}\bP^r_{K,\psi_0}(df)\frac{r^{l(x)}e^{-r\psi^{-1}(0)}}{|\psi^{(l(x))}_0(0)|}\Pi(dr),$$
where $l(x)+1$ is the number of connected components of $\cB_0^\circ\setminus\{x\}$ and ${\rm Br}(\cB_0^\circ)$ is the set of branchpoints
$\{x\in\cB_0^\circ\colon l(x)\ge 2\}\setminus\{0\}$,
\item and an extra point $Z^{0}$ independent of $(Z^x,x\in\cP\cup{\rm Br}(\cB_0^\circ))$ with distribution $\bP^\beta_{K,\psi_0}$.
\end{itemize}
Then the process $\displaystyle Z_t=\sum_{x\in\cP\cup{\rm Br}(\cB_0^\circ)\cup\{0\}}Z^x_{t-\sigma(x)}$ is a ${\rm CSBP}(K,\psi)$ starting from $Z_0=\beta$.
\end{theo}
\begin{pf} Since $Z$ is not a Markov process, we will deduce the theorem from the richer structure of a Markovian $(\xi,K,\psi)$-superprocesses $M$ starting from
$M_0=m$. Slightly abusing notation, we consider $M$ also under $\Theta_{K,\psi}$ and $\bP^r_{K,\psi}$ and $Q$. Then the intensity measures and distributions in the
bullet points specify a point process $(M^x,x\in\cP\cup{\rm Br}(\cB_0^\circ)\cup\{0\})$. From the exponential formula for
Poisson point processes and from (\ref{meassg}) for the subcritical branching mechanism $\psi_0$ with $u^{\rm sub}_t(f)$ associated via the analogue of (\ref{measint}), it
is not hard to calculate
\begin{eqnarray*}&&\hspace{-0.68cm}\bE\left(\!\left.\exp\left\{\!-\!\sum_{x\in\cP}\int_{[0,\infty)}\!\!f(z)M^x_{t-\sigma(x)}(dz)\!\right\}\right|\cB_0^\circ,\cW_0\!\right)\!=\exp\left\{\!-\!\int_{\cB_0^\circ}\!(\psi_0^\prime(u_{t-\sigma(x)}^{\rm sub}(f))\!-\!\psi_0^\prime(0)){\rm Leb}(dx)\!\right\},\\
&&\hspace{-0.68cm}\bE\left(\left.\exp\left\{-\sum_{x\in{\rm Br}(\cB_0^\circ)}\int_{[0,\infty)}f(z)M^x_{t-\sigma(x)}(dz)\right\}\right|\cB_0^\circ,\cW_0\right)=\prod_{x\in{\rm Br}(\cB_0^\circ)}\frac{\psi_0^{(l)}(u_{t-\sigma(x)}^{\rm sub}(f))}{\psi_0^{(l)}(0)},\\
&&\hspace{-0.68cm}\bE\left(\!\left.\exp\left\{\!-\!\int_{[0,\infty)}\!f(z)M^0_t(dz)\!\right\}\right|\cB_0^\circ,\cW_0\!\right)\!=\exp\left\{\!-\!\int_{[0,t]}\!u_{t-z}^{\rm sub}(f)m(dz)\!-\!\int_{(t,\infty)}\!f(z\!-\!t)m(dz)\!\right\},
\end{eqnarray*}
using the convention $u^{\rm sub}_s(f)=0$ for $s<0$. It now suffices to show that the backbone decomposition of approximations $M^\lambda$ of $M$, cf. Corollary
\ref{cor14}, appropriately converges to these quantities. Let us formulate that backbone decomposition in the current setting. Let
$\bP^{r,\lambda}_{\kappa,\psi_0}$ be the distribution of $M^\lambda$ given $M_0^\lambda=N_\lambda^r\delta_0$ with $N_\lambda^r\sim{\rm Poi}(\psi^{-1}(\lambda)r)$. Then, given $(\cB_0^\circ,\cW_0)$, consider points
$(M^{x,\lambda},x\in\cP^\lambda)$ of a Poisson point process with intensity measure
$$2a(\psi^{-1}(\lambda)-\psi^{-1}(0))\bP(M^\lambda\in\cdot|M^\lambda_0=\delta_0)+\int_{(0,\infty)}\bP^{r,\lambda}_{\kappa,\psi_0}re^{-r\psi^{-1}(0)}\Pi(dr),$$
and
$$M^{x,\lambda}\sim\frac{2a1_{\{l(x)=2\}}}{|\psi_0^{(l(x))}(0)|}\delta_0+\int_{(0,\infty)}\bP^{r,\lambda}_{K,\psi_0}\frac{r^{l(x)}e^{-r\psi^{-1}(0)}}{|\psi^{(l(x))}_0(0)|}\Pi(dr)\quad\mbox{for $x\in{\rm Br}(\cB_0^\circ)$ and } M^{0,\lambda}\sim\bP^{\beta,\lambda}_{K,\psi_0}.$$
Then the analogous calculations yield for $u^{\lambda,\rm sub}_t(f)=\psi_0^{-1}(\lambda)(1-e^{-v_t^{\lambda,\rm sub}(f/\psi^{-1}(\lambda))})$
\begin{eqnarray*}&&\hspace{-0.68cm}\bE\left(\!\left.\exp\left\{\!-\!\!\sum_{x\in\cP^\lambda}\int_{[0,\infty)}\!\!\!\!\!f(z)\frac{M^{x,\lambda}_{t-\sigma(x)}(dz)}{\psi^{-1}(\lambda)}\!\right\}\right|\cB_0^\circ,\cW_0\!\right)\!=\exp\left\{\!-\!\int_{\cB_0^\circ}\!\!(\psi_0^\prime(u^{\lambda,\rm sub}_{t-\sigma(x)}(f))\!-\!\psi_0^\prime(0)){\rm Leb}(dx)\!\right\}\!,\\
&&\hspace{-0.68cm}\bE\left(\left.\exp\left\{-\sum_{x\in{\rm Br}(\cB_0^\circ)}\int_{[0,\infty)}f(z)\frac{M^x_{t-\sigma(x)}(dz)}{\psi^{-1}(\lambda)}\right\}\right|\cB_0^\circ,\cW_0\right)=\prod_{x\in{\rm Br}(\cB_0^\circ)}\frac{\psi_0^{(l)}(u_{t-\sigma(x)}^{\lambda,\rm sub}(f))}{\psi_0^{(l)}(0)},\\
&&\hspace{-0.68cm}\bE\left(\!\left.\exp\left\{\!-\!\int_{[0,\infty)}\!f(z)\frac{M^0_t(dz)}{\psi^{-1}(\lambda)}\!\right\}\right|\cB_0^\circ,\cW_0\!\right)\!=\exp\left\{\!-\!\int_{[0,t]}\!u_{t-z}^{\lambda,\rm sub}(f)m(dz)\!-\!\int_{(t,\infty)}\!f(z\!-\!t)m(dz)\!\right\},
\end{eqnarray*}
But from the proof of Proposition \ref{prop4a}, we know that $u_t^{\lambda,\rm sub}(f)\rightarrow u_t^{\rm sub}(f)$ as $\lambda\rightarrow\infty$, and this completes the proof.
\end{pf}
We can specialise this backbone decomposition to the case $K_t=t$, when a ${\rm CSBP}(K,\psi)$ is simply a Markovian ${\rm CSBP}(\psi)$ and $\sigma(x)=d(0,x)$ is just the height of $x\in\cB_0^\circ$. In this framework, and even with a spatial motion added, this
decomposition was obtained recently by Berestycki et al. \cite{BKM-10}, generalising an analogous result of \cite[Theorem 5.6]{DuW} formulated in a context of L\'evy trees.
\section{Growth of ${\rm GWI}(q_\lambda,\kappa_\lambda,\eta_\lambda,\chi_\lambda)$-forests: immigration}\label{Section Immigration}
Theorem \ref{thm3} is about forests $F_\lambda$ of ${\rm GW}(q_\lambda,{\rm Exp}(c_\lambda))$-trees arising from immigration of independent $\eta_\lambda$-distributed
numbers of immigrants at ${\rm Exp}(\paramgamma_\lambda)$-spaced times. The main statement beyond the no-immigration case of Theorem \ref{thm1} is that consistency of a
family $(F_\lambda,\lambda\ge 0)$ under Bernoulli leaf colouring relates $\eta_\lambda$ to a continuous-state immigration
mechanism $\phi$. After some more general remarks, we focus on the Markovian case of exponential lifetimes and inter-immigration times.
\subsection{A two-colours regenerative property and associated forest reduction}\label{regprop}
Let $F=(B(t),t\ge 0)$ be a ${\rm GWI}(q,\kappa,\eta,\chi)$-forest as defined in Section \ref{discr BP with immigration section}, specifically denote by $S_1=\inf\{t\ge 0\colon B(t)\neq\partial\}\sim\chi$ the first immigration time, by $B(S_1)=(T_{(1)}^{(1)},\ldots,T_{(N_1)}^{(1)})$ the
bush of independent genealogical trees $T^{(1)}_{(j)}\sim{\rm GW}(q,\kappa)$, $j\ge 1$, of the $N_1\sim\eta$ time-$S_1$ immigrants and by $F_{\rm post}=(B^{\rm post}(t),t\ge 0)$ the post-$S_1$ forest
given by $B^{\rm post}(0)=\partial$ and $B^{\rm post}(t)=B(S_1+t)$ for $t>0$. It is immediate from the definition that $F$ satisfies a
\em regenerative property \em at $S_1$ in that $(S_1,B(S_1))$ is independent of $F_{\rm post}$ and $F_{\rm post}\overset{{\rm (d)}}{=} F$, and that the distribution of $(S_1,B(S_1))$ as above together with this regenerative
property characterises the distribution of $F$. Since colouring and reduction apply tree by tree, we obtain for the associated
forest $F^{p-\rm col}$ of coloured trees $T^{(i),p-\rm col}_{(j)}\sim\bP^{p-\rm col}_{q\otimes\kappa}$,
$1\le j\le N_i$, $i\ge 1$, a
\paragraph{Regenerative property of coloured ${\rm GWI}(q,\kappa,\eta,\chi)$-forests}
\begin{enumerate}\item[(a)] For all $n\ge 1$, $\varepsilon_j\in\{0,1\}$, and measurable functions $k$, $f_j$ and $G$, $1\le j\le n$,
we have
\begin{eqnarray*}&&\hspace{-1.5cm}\bE\left(k(S_1)G(F^{p-\rm col}_{\rm post})\prod_{j=1}^nf_j(T^{p-\rm col}_{(j)});N_1=n;\left(\gamma_\emptyset(T^{p-\rm col}_{(1)}),\ldots,\gamma_\emptyset(T^{p-\rm col}_{(n)})\right)=(\varepsilon_1,\ldots,\varepsilon_n)\right)\\
&&=\int_{(0,\infty)}k(z)\chi(dz)\eta(n)g(p)^{n_r}(1-g(p))^{n_b}\bE(G(F^{p-\rm col}))\prod_{j=1}^n\bE_{q\otimes\kappa}^{p-\rm col}[f_j|\gamma_\emptyset=\varepsilon_j],
\end{eqnarray*}
where $n_r=\varepsilon_1+\cdots+\varepsilon_n$ and $n_b=n-n_r$ are the numbers of red and black colours.
\item[(b)] For $t\ge 0$, consider the post-$t$ forest $F_{{\rm post}-t}^{p-\rm col}=(B^{p-\rm col}(t+s),s\ge 0)$ and
the pre-$t$ sigma-algebra $\cF_t=\sigma\{B^{p-\rm col}(r),r\le t\}$. Then for all measurable functions $f_u$, $u\in\bU$, and \nolinebreak $G$,
$$\hspace{-0.5cm}\bE\!\left(\!\left.G(F_{{\rm post}-t}^{p-\rm col})\!\!\!\prod_{u\in\overline{\pi}_t(F)}\!\!f_u(\overline{\theta}_{u,t}(F^{p-\rm col}))\right|\cF_t\right)\!
=\bE(G(F^{p-\rm col}))\!\!\!\prod_{u\in\overline{\pi}_t(F)}\!\!\left.\bE^{p-\rm col}_{q\otimes\kappa}[f_u(\overline{\theta}_{\emptyset,s})|\zeta_\emptyset>s]\right|_{s=t-\alpha_u}\!.$$
\end{enumerate}
As a trivial application of (a), we can calculate the probability that all immigrants are red
$$\bP\left(\left(\gamma_\emptyset(T^{p-\rm col}_{(1)}),\ldots,\gamma_\emptyset(T^{p-\rm col}_{(N_1)})\right)=(1,\ldots,1)\right)
=\sum_{n\ge 1}\eta(n)g(p)^n=\varphi_{\eta}(g(p)).
$$
We deduce the distribution of the number of red immigrants given that all immigrants are red
\begin{equation}\label{red disbribution}
\eta_{\rm red}^{p-\rm col}(m)=\left\{
\begin{array}{ll}
\frac{\eta(m)g(p)^{m}}{\varphi_{\eta}(g(p))}, & \hbox{if $m \geq 1$,} \\
0, & \hbox{if $m=0$,}
\end{array}
\right.\quad\mbox{with generating function }\varphi_{\eta_{\rm red}^{p-\rm col}}(s)=\frac{\varphi_\eta(sg(p))}{\varphi_\eta(g(p))}.
\end{equation}
Also by the regenerative property (a), the number $\widetilde{G}$ of immigrations until we see the first black immigrant is geometrically distributed with parameter $1-\varphi_\eta(g(p))$, i.e.
\begin{equation}\label{Immigration Geometric}
\bP(\widetilde{G}=j)=\varphi_{\eta}(g(p))^{j-1}(1-\varphi_{\eta}(g(p))),\qquad j\ge 1.
\end{equation}
Conditioning on having at least one black immigrant (probability $1-\varphi_\eta(g(p))$), we get for $\ell\ge 1$
$$\eta^{p-\rm rdc}(\ell)=\frac{1}{1-\varphi_\eta}(g(p))\sum_{m\ge 0}{m+\ell\choose m}\eta(m+\ell)(g(p))^m(1-g(p))^\ell=\frac{(1-g(p))^\ell\varphi_\eta^{(\ell)}(g(p))}{\ell!(1-\varphi_\eta(g(p))}.$$
Similarly, conditioning on having $\ell$ black immigrants, $\ell\ge 1$, we get for the number of red ones
\begin{equation}\label{red conditional distribution}\widetilde{\eta}_\ell(m)=\eta(m+\ell)\frac{(m+\ell)!}{m!}(g(p))^m\frac{1}{\varphi_\eta^{(\ell)}(g(p))},\qquad\mbox{for $m\ge 0$.}\end{equation}
These distributions have generating functions that we can express in terms of $\varphi_\eta$, for $s\in[0,1]$
$$\varphi_{\eta^{p-\rm rdc}}(s)=\frac{\varphi_\eta(g(p)+s(1-g(p)))-\varphi_\eta(g(p))}{1-\varphi_\eta(g(p))}\qquad\mbox{and}\qquad\varphi_{\widetilde{\eta}_\ell}(s)=\frac{\varphi_\eta^{(\ell)}(sg(p))}{\varphi_\eta^{(\ell)}(g(p))},\quad\ell\ge 1,$$
and as $\varphi_{\eta^{p-\rm rdc}}$ and $\varphi_\eta$ are analytic, we can extend $\varphi_\eta$ analytically to $[-g(p)/(1-g(p)),1]$. Evaluating at $s=v_p=-g(p)/(1-g(p))$, we get
$\varphi_\eta(g(p))=-\varphi_{\eta^{p-\rm rdc}}(v_p)/(1-\varphi_{\eta^{p-\rm rdc}}(v_p))$, so
\begin{equation}\label{exteta}\varphi_\eta(r)=\frac{\varphi_{\eta^{p-\rm rdc}}(v_p+r(1-v_p))-\varphi_{\eta^{p-\rm rdc}}(v_p)}{1-\varphi_{\eta^{p-\rm rdc}}(v_p)},\qquad r\in[0,1].
\end{equation}
As reduction preserves the Galton-Watson property for trees \cite{DuW}, we now see that a $p$-reduced ${\rm GWI}(q,\kappa,\eta,\chi)$-forest is a ${\rm GWI}(q^{p-\rm rdc},\kappa^{p-\rm rdc},\eta^{p-\rm{rdc}}, \chi^{p-\rm{rdc}})$-forest. Specifically, $\eta^{p-\rm rdc}$ is as above, $\chi^{p-\rm{rdc}}$ the distribution of a geom($1-\varphi_\eta(g(p))$) sum of independent
$\chi$-distributed variables.
\subsection{Growth of ${\rm GWI}(q_\lambda,{\rm Exp}(c_\lambda),\eta_\lambda,{\rm Exp}(\paramgamma_\lambda))$-forests}
\subsubsection{Proof of Theorem \ref{thm3}}
${\rm (i)}\Rightarrow{\rm (ii)}$: Suppose, (i) holds. In particular, $q_\mu$ is then the $(1-\mu/\lambda)$-reduced
offspring distribution associated with $q_\lambda$, for all $0\le\mu<\lambda<\infty$. By Theorem \ref{thm1},
$\varphi_q(s)=s+\widetilde{\psi}(1-s)$, where $\widetilde{\psi}$ has the form (\ref{lkbm}). To be specific, let $c=1$ and parametrise $(q_\lambda,c_\lambda)$ using $\psi$ as in Section \ref{Consistent family}.
Now consider the relationship between $\eta=\eta_\mu$ and $\eta_\lambda$ for $\lambda>\mu=1$. By the discussion above (\ref{exteta}), we can extend $\varphi_\eta$ analytically to $[-g_\lambda(1-1/\lambda)/(1-g_\lambda(1-1/\lambda)),1]$, where, expressing as in (\ref{g in psi}), we have $g_\lambda(1-1/\lambda)=1-\psi^{-1}(1)/\psi^{-1}(\lambda)\rightarrow 1$ as $\lambda\rightarrow\infty$. Differentiating (\ref{exteta}),
we see that $\varphi_\eta^\prime$ has positive derivatives on $(-\infty,1)$. Setting $\widetilde{\phi}(r)=1-\varphi_\eta(1-r)$,
$r\ge 0$, the derivative $\widetilde{\phi}^\prime$ is completely monotone on $(0,\infty)$ and, by Bernstein's theorem (see e.g. \cite{Fel}), there exists a
Radon measure $\widetilde{\Lambda}^*$ on $[0, \infty)$ such that
$$\widetilde{\phi}^{\prime}(r)=\int_{[0,\infty)}e^{-r x} \widetilde{\Lambda}^*(d x)<\infty,\qquad r>0.$$
From $\widetilde{\phi}(0)=0$, we get integrability $\int_{(1,\infty)}x^{-1}\widetilde{\Lambda}^*(dx)<\infty$ and
$$
\widetilde{\phi}(u) = \widetilde{\phi}(0) + \int^{u}_{0} \widetilde{\phi}^{\prime}(r) d r
= \widetilde{\Lambda}^*(\{0\})u + \int_{(0,\infty)}\frac{1-e^{-ux}}{x}\widetilde{\Lambda}^*(d x),
$$
and, in particular, setting $\widetilde{d}=\widetilde{\Lambda}^*(\{0\})$ and $\widetilde{\Lambda}(dx)=x^{-1}\widetilde{\Lambda}^*|_{(0,\infty)}(dx)$ yields (ii).
${\rm (ii)}\Rightarrow{\rm (i)}$: Now suppose that (ii) holds. According to Theorem \ref{thm1}, the family of offspring distribution $(q_\lambda,\lambda\ge 0)$ exists as required; furthermore, we can express $\varphi_{q_\lambda}$ in terms of $\psi$ as in Section \ref{Consistent family}, choosing $c=1$.
By (\ref{g in psi}), we have $g_\lambda(1-\mu/\lambda)=1-\psi^{-1}(\mu)/\psi^{-1}(\lambda)$, so $v_{\mu,\lambda}=-g_\lambda(1-\mu/\lambda)/(1-g_\lambda(1-\mu/\lambda))=1-\psi^{-1}(\lambda)/\psi^{-1}(\mu)$ for all $0\le\mu<\lambda<\infty$.
By (\ref{exteta}) and the discussion above (\ref{exteta}), the required immigration distributions must be of the following form, respectively for $\mu<1$ and $\lambda>1$
\beq\varphi_{\eta_\mu}(s)=\frac{\varphi_\eta(g_1(1-\mu)+s(1-g_1(1-\mu)))-\varphi_\eta(g_1(1-\mu))}{1-\varphi_\eta(g_1(1-\mu))}\!\!&=&\!\!1-\frac{\widetilde{\phi}((1-s)\psi^{-1}(\mu)/\psi^{-1}(1))}{\widetilde{\phi}(\psi^{-1}(\mu)/\psi^{-1}(1))},\\
\varphi_{\eta_\lambda}(r)=\frac{\varphi_\eta(v_{1,\lambda}+r(1-v_{1,\lambda}))-\varphi_\eta(v_{1,\lambda})}{1-\varphi_\eta(v_{1,\lambda})}\!\!&=&\!\!1-\frac{\widetilde{\phi}((1-r)\psi^{-1}(\lambda)/\psi^{-1}(1)}{\widetilde{\phi}(\psi^{-1}(\lambda)/\psi^{-1}(1))}.\eeq
Since $\widetilde{\phi}^\prime$ is completely monotone, simple differentiation yields that these functions are indeed generating functions of immigration distributions. Furthermore, $\eta_\mu$ is the $(1-\mu/\lambda)$-reduced immigration
distribution of $\eta_\lambda$ for all $0\le\mu<\lambda<\infty$, by the transitivity of colouring reduction noted in
Remark \ref{remdet}(a), which also applies to forests, since colouring and reduction are defined tree by tree. The full
statement of (i) can now be obtained formally as in the proof of Theorem \ref{thm2}, with the simpler regenerative property here taking the role of the branching property there.
In the setting of (i) and (ii) for $c=c_1\in(0,\infty)$, $\paramgamma=\paramgamma_1\in(0,\infty)$, Kolmogorov's consistency theorem allows us to set up a consistent family $(F_\lambda,\lambda\ge 0)$ of ${\rm GWI}(q_\lambda,{\rm Exp}(c_\lambda),\eta_\lambda,{\rm Exp}(\paramgamma_\lambda))$-forests. Uniqueness of ($q_\lambda,c_\lambda$), $\lambda \geq 0$,
follows from Theorem \ref{thm1}. Uniqueness of ($\eta_\lambda,\lambda\ge 0$) was noted in ${\rm (ii)}\Rightarrow{\rm (i)}$.
Uniqueness of $(\paramgamma_\lambda,\lambda\ge 0)$ follows from the relationship between inter-immigration times as geometric sums,
where we calculate $\paramgamma_\lambda$ from (\ref{Immigration Geometric}) for $\lambda>1>\mu$ as
$$\paramgamma_\lambda=(1-\varphi_{\eta_\lambda}(g_\lambda(1-1/\lambda)))\paramgamma=\frac{\paramgamma}{\widetilde{\phi}(\psi^{-1}(\lambda)/\psi^{-1}(1))}\quad\mbox{and}\quad \paramgamma_\mu
=\frac{\paramgamma}{\widetilde{\phi}(\psi^{-1}(\mu)/\psi^{-1}(1))}.$$
\begin{flushright}$\Box$\end{flushright}
\subsubsection{Freedom in parameterisation and standard choice}
In analogy to Section \ref{Consistent family}, we can use a single function $\phi$ to replace $(\widetilde{\phi},\paramgamma)$ of Theorem \ref{thm3} and parametrise $(\eta_\lambda,{\rm Exp}(\paramgamma_\lambda))$, $\lambda\ge 0$, such that
\begin{equation}\label{full immi gen by phi psi}
\varphi_{\eta_\lambda}(v)=1-\frac{{\phi}(\psi^{-1}(\lambda)(1-v))}{{\phi}(\psi^{-1}(\lambda))}\qquad\mbox{and}\qquad\paramgamma_\lambda=\phi(\psi^{-1}(\lambda)),
\end{equation}
where $\phi$ is a linear transformation $\phi(s)=k_3\widetilde{\phi}(k_4s)$ of $\widetilde{\phi}$. Specifically, we choose $k_3=\paramgamma$ and $k_4=1/\psi^{-1}(1)$. It is easy to check that this works, using $\widetilde{\phi}(1)=1-\eta(0)=1$.
In this parameterisation, we can also express in terms of $\psi$ and $\phi$ the remaining quantities studied in Section \ref{regprop}. E.g. (\ref{full immi gen by phi psi}) and (\ref{red disbribution}) now yield the generating function of the pure-red
immigration distribution
\begin{equation}\label{red immi gen}
\varphi_{\eta_{\lambda,\rm red}^{(1-\mu/\lambda)-\rm col}}(s)=\frac{\phi(\psi^{-1}(\lambda))-\phi( \psi^{-1}(\lambda)-s(\psi^{-1}(\lambda)-\psi^{-1}(\mu)))}{\psi^{-1}(\lambda)-\psi^{-1}(\mu)}.
\end{equation}
The parameters of the geometric distributions (\ref{Immigration Geometric}) take a simple form that leads to a distinction of finite/infinite immigration rate $\alpha^{\rm imm}_{\mu,\lambda}=1-\varphi_{\eta_\lambda}(g_\lambda(1-\mu/\lambda))=\phi(\psi^{-1}(\mu))/\phi(\psi^{-1}(\lambda))$, and
\begin{itemize}
\item $\alpha_{\mu,\lambda}^{\rm imm} \rightarrow\frac{\phi(\psi^{-1}(\mu))}{\phi(\infty)}>0$ as $\lambda \rightarrow \infty$, if $\phi(\infty) < \infty;$
\item $\alpha_{\mu,\lambda}^{\rm imm} \rightarrow 0$ as $\lambda \rightarrow \infty$, if $\phi(\infty) = \infty$.
\end{itemize}
With the formulas above we can formulate explicitly a reconstruction procedure.
\subsubsection{Reconstruction procedure for ${\rm GWI}(q_\lambda,{\rm Exp}(c_\lambda),\eta_\lambda,{\rm Exp}(\paramgamma_\lambda))$-forests}
For $F_\mu\sim{\rm GWI}(q_\mu,{\rm Exp}(c_\mu),\eta_\mu,{\rm Exp}(\paramgamma_\mu))$, we modify the steps of Section
\ref{Recon for GW expo bushes2} to construct $F_\lambda$.
\begin{enumerate}\item[1.] In every tree of $F_\mu$, subdivide lifetimes as in Section \ref{Recon for GW expo bushes2} and hence
construct a forest $\widehat{F}_\mu$.
\item[2.] In every tree of $\widehat{F}_\mu$, add further children and independent red trees as in Section
\ref{Recon for GW expo bushes2} and hence construct a forest $\widehat{F}_\lambda$.
\item[3.] At every immigration time, given that there are $N_\mu^{(i)}=\ell$ immigrants in $\widehat{F}_\mu$, consider a
random number $N_\lambda^{(i),\rm red}\sim\widetilde{\eta}_\ell$ of further immigrants as in (\ref{red conditional distribution}),
proceed as in Section \ref{Recon for GW expo bushes2} and superpose a further independent ${\rm GWI}(q_{\lambda,{\rm red}}^{(1-\mu/\lambda)-\rm col},{\rm Exp}(c_\lambda),\eta_{\lambda,\rm red}^{(1-\mu/\lambda)-\rm col},{\rm Exp}(\paramgamma_\lambda-\paramgamma_\mu))$-forest with distributions as in (\ref{red offspring distribution}) and (\ref{red immi gen}) to finally obtain $F_\lambda$.
\end{enumerate}
\subsubsection{Convergence of the population sizes: proof of Proposition \ref{prop4}}\label{section convergence of the population size}
For convergence in distribution, we calculate the Laplace transform of $Y_t^\lambda/\psi^{-1}(\lambda)$, where
$$Y_t^\lambda=\overline{\pi}_t\left(F_\lambda\right)=\sum_{i=1}^{J_t^\lambda}\overline{\pi}_{t-S_i^\lambda}\left(B_\lambda(S_i^\lambda)\right)=\sum_{i=1}^{J_t^\lambda}\sum_{j=1}^{N_i^\lambda}\overline{\pi}_{t-S_i^\lambda}\left(T_{(j)}^{(i),\lambda}\right)$$
with notation as in and around (\ref{discrimm}), but with all quantities $\lambda$-dependent.
We exploit that
$$E^\lambda(S_i^\lambda)=Z^{(i),\lambda}=\left(\overline{\pi}_t(B_\lambda(S_i^\lambda)),t\ge 0\right),\qquad E^\lambda(s)=0,\quad s\not\in\left\{S_i^\lambda,i\ge 1\right\},$$
is a Poisson point process with intensity measure $\paramgamma_\lambda$ times
the distribution of a ${\rm GW}(q_\lambda,{\rm Exp}(c_\lambda))$-process $Z^{(1),\lambda}$ starting from $Z^{(1),\lambda}_0\sim\eta_\lambda$. By the exponential formula for Poisson point processes,
$$\bE\left(\exp\left\{-rY_t^\lambda/\psi^{-1}(\lambda)\right\}\right)=\exp\left\{-\paramgamma_\lambda\int_0^t\sum_{m=1}^\infty\eta_\lambda(m)\left(1-\left(\bE\left(s^{Z_{t-v}^\lambda}\right)\right)^m\right)dv\right\},$$
where $s=e^{-r/\psi^{-1}(\lambda)}$ and $Z^\lambda$ is the population size of a single ${\rm GW}(q_\lambda,{\rm Exp}(c_\lambda))$-tree as in the proof of Lemma \ref{CSBP Limit Lemma x}. Using notation and asymptotics from there, as well as (\ref{full immi gen by phi psi}), this equals
$$\exp\left\{-\paramgamma_\lambda\int_0^t\left(1-\varphi_{\eta_\lambda}\left(w_{t-v}^\lambda(s)\right)\right)dv\right\}=\exp\left\{-\int_0^t\phi\left(\psi^{-1}(\lambda)\left(1-w_{t-v}^\lambda(s)\right)\right)dv\right\}.$$
Since $\psi^{-1}(\lambda)\left(1-w_{t-v}^\lambda(s)\right)\rightarrow u_{t-v}(r)$, and, by (\ref{CSBP u bounds}) and (\ref{wlambda}), all these quantities are bounded by
$\max\{r,\psi^{-1}(0)\}$, dominated convergence completes the proof of convergence in distribution.
Almost sure convergence follows by martingale arguments as in Lemma \ref{CSBP Limit Lemma x}, using a version of the regenerative property (ii) of Section \ref{regprop} rather than the version of the branching property (\ref{bpexp}) that we presented in Remark \ref{Markov}(c). {\hspace*{\fill} $\square$}
\bigskip
From Proposition \ref{prop4} and Lemma \ref{CSBP Limit Lemma x} we deduce the analogous convergence result for ${\rm GWI}$-processes starting from initial
population sizes $Y_0^\lambda\sim{\rm Poi}(\beta_\lambda)$. The limiting CBI then has $Y_0=\beta$.
\subsection{Analogous results for ${\rm GWI}(q_\lambda,\kappa_\lambda,\eta_\lambda,\chi_\lambda)$-forests}
Finally, let us combine Theorems \ref{thm2} and \ref{thm3} into a single statement and also deduce the analogous pattern for general inter-immigration distributions that now emerges naturally.
\begin{coro}\label{corolast} For a tuple $(q,\kappa,\eta,\chi)$ of offspring, lifetime, immigration and inter-immigration distributions, the
following are equivalent:
\begin{enumerate}\item[\rm(i)] There are $(q_\lambda,\kappa_\lambda,\eta_\lambda,\chi_\lambda)_{\lambda\ge 0}$
with $(q_1,\kappa_1,\eta_1,\chi_1)=(q,\kappa,\eta,\chi)$ such that
$(q_\mu,\kappa_\mu,\eta_\mu,\chi_\mu)$ is the $(1-\mu/\lambda)$-reduced tuple associated with
$(q_\lambda,\kappa_\lambda,\eta_\lambda,\chi_\lambda)$, for all $0\le\mu<\lambda<\infty$.
\item[\rm(ii)] The generating functions $\varphi_q$ of $q$ and $\varphi_\eta$ of $\eta$ satisfy $\varphi_q(s)=s+\widetilde{\psi}(1-s)$ for some
$\widetilde{\psi}$ of the form {\rm(\ref{lkbm})} and $\varphi_\eta(s)=1-\widetilde{\phi}(1-s)$ for some $\widetilde{\phi}$ of the form
{\rm(\ref{lkim})}; $\kappa$ is geometrically divisible for all $\alpha>1/\widetilde{\psi}^\prime(\infty)$ if $\widetilde{\psi}^\prime(\infty)<\infty$,
or for all $\alpha>0$ if $\widetilde{\psi}^\prime(\infty)=\infty$.
Moreover, $\chi$ is also geometrically divisible
\begin{itemize}\item for all $\alpha>1/\widetilde{\phi}(\infty)$ if $\widetilde{\phi}(\infty)<\infty$;
\item for all $\alpha>0$ if $\widetilde{\phi}(\infty)=\infty$.
\end{itemize}
\end{enumerate}
In the setting of {\rm (i)} and {\rm (ii)}, a consistent family $(F_\lambda)_{\lambda\ge 0}$ of
${\rm GWI}(q_\lambda,\kappa_\lambda,\eta_\lambda,\chi_\lambda)$-forests can
be constructed such that $(F_\mu,F_\lambda)\overset{{\rm (d)}}{=} (F_\lambda^{(1-\mu/\lambda)-\rm rdc},F_\lambda)$ for all
$0\le\mu<\lambda<\infty$.
\end{coro}
\noindent We omit the proof which is a straightforward combination of the proofs of Theorems \ref{thm2} and \ref{thm3}.
Similarly, the convergence results of Proposition \ref{prop4a} and \ref{prop4} find their analogue in this setting:
\begin{coro}\label{corolast2} Let $(Y_t^\lambda,t\ge 0)$ be the population size process in the setting of Corollary \ref{corolast}. Then
$$ \frac{Y_t^\lambda}{\psi^{-1}(\lambda)}\rightarrow Y_t\qquad\mbox{in distribution as $\lambda\rightarrow\infty$, for all $t\ge 0$,}
$$
where $(Y_t,t\ge 0)$ is a ${\rm CBI}(K,\psi,\widehat{K},\phi)$ with $Y_0\!=\!0$, for
\begin{itemize}\item branching mechanism $\psi$ a linear transformations of $\widetilde{\psi}$ as in (\ref{psipsitilde})
\item $K=(K_s,s\ge 0)$ with $\inf\{s\ge 0\colon K_s>V_\lambda\}\sim\kappa_\lambda$ for
$V_\lambda\sim{\rm Exp}(c_\lambda)$ with $c_\lambda$ of Theorem \ref{thm3},
\item immigration mechanism $\phi$ a linear transformation of $\widetilde{\phi}$ as in (\ref{full immi gen by phi psi})
\item $\widehat{K}=(\widehat{K}_s,s\ge 0)$ with $\inf\{s\ge 0\colon \widehat{K}_s>\widehat{V}_\lambda\}\sim\chi_\lambda$ for $\widehat{V}_\lambda\sim{\rm Exp}(\paramgamma_\lambda)$ with $\paramgamma_\lambda$ of Theorem \ref{thm3}.
\end{itemize}
If furthermore $\psi^\prime(0)>-\infty$ and $\phi^\prime(0)<\infty$, then the convergence holds in the almost sure sense.
\end{coro}
\noindent Like a ${\rm CBI}(\psi,\phi)$, a ${\rm CBI}(K,\psi,\widehat{K},\phi)$ can be constructed from a Poisson point process $(E^s,s\ge 0)$ in $\bD([0,\infty),[0,\infty))$
with intensity measure $d\Theta_{K,\psi}+\int_{(0,\infty)}\bP^x_{K,\psi}\Lambda(dx)$, where $(d,\Lambda)$ are the characteristics of $\phi$ in (\ref{lkim}). Also
consider an independent subordinator $\widehat{\sigma}$ or increasing random walk, in the $\phi(\infty)<\infty$ case, in fact $\widehat{K}_s=\inf\{t\ge 0\colon \widehat{\sigma}(t)>s\}$ clarifies the notation and the meaning of $\widehat{K}$. Then
$$Y_t=\sum_{s\le\widehat{K}_t}E_{t-\widehat{\sigma}(s)}^s$$
is a ${\rm CBI}(K,\psi,\hat{K},\phi)$ with $Y_0=0$. Like a ${\rm CSBP}(K,\psi)$, a ${\rm CBI}(K,\psi,\widehat{K},\phi)$ is non-Markovian, but admits a Markovian representation $(M,\vartheta)$
that records residual lifetimes as well as residual times to the next immigration, with values in $\bM([0,\infty))\times[0,\infty)$, where
$\vartheta_t=\widehat{\sigma}(\widehat{K}_t)-t$, so that
\beq&&\hspace{-0.5cm}\bE\left(\left.\!\exp\left\{\!-\!\!\int_{[0,\infty)}\!\!f(z)M_t(dz)-r\vartheta_t\!\right\}\right|M_0\!=\!m\!,\vartheta_0=s\right)\\
&&=\bE\Bigg(\exp\Bigg\{-\int_{[0,t]}u_{t-z}(f)m(dz)-\int_{(t,\infty)}f(z-t)m(dz)\\
&&\hspace{2.1cm}-\int_{[0,t-s]}\phi(u_{t-s-z}(f))d\widehat{K}_z-r\left(\widehat{\sigma}(\widehat{K}_{(t-s)^+})-(t-s)\right)\Bigg\}\Bigg),\eeq
where $u_t(f)$ is the unique nonnegative solution of (\ref{measint}). Then the process $Y_t=M_t([0,\infty))$ is a ${\rm CBI}(K,\psi,\hat{K},\phi)$. We leave any
further details including the proof of Corollary \ref{corolast2} to the reader.
\bibliographystyle{abbrv}
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\section{Introduction}\label{sec:intro}
Benchmarking quantum computers aims to determine the performance of a quantum computing system under an appropriate set of metrics. Within the Noisy Intermediate-Scale Quantum (NISQ) era \cite{Preskill-2018}, benchmarking the capabilities and performance of quantum computers when executing quantum programs is of paramount importance, especially for assessing their scalability.
An intuitive approach to benchmarking quantum computers is establishing a set of quantum programs and measuring the performance of a quantum computer when executing each one. Such a work gains more merit as bigger quantum computers are built. There are various advantages to this approach, for example benchmarking the limits and behaviour of a quantum machine within a scaling, computationally intensive environment and particularly testing the system when performing a ``real-world" task. Various companies appear to favor such an approach. IonQ for example tested their quantum computer using the Bernstein-Vazirani \cite{BernsteinVazirani} and the Hidden Shift \cite{vanDam-2006,Rotteler-2010} algorithms. The metric for performance was the likelihood of measuring the correct output \cite{Wright-2019}. On the other hand, Google focused on the problem of quantum sampling and achieved results they claim demonstrated quantum supremacy \cite{GoogleSuprem}. While the chosen application was not particularly useful in a ``real-world" scenario, it excelled at demonstrating the computing power of the system. So a question arises, highlighting the difficulty of creating quantum program benchmarks: which benchmarks are more insightful?
The different competing quantum technologies pose a major challenge. The technologies have different topologies and thus have unique strengths and weaknesses. For example, the connectivity of an ion-trap computer provided a large advantage on some benchmarks over a superconducting quantum computer \cite{Linke-2017}.
Clearly, systems can behave differently on different benchmarks, which introduces the issue of invested interests, i.e., using benchmarks that are expected to perform well on a current system \cite{Lilja-2000}. While impressive, current quantum computers are very small compared to the computers we hope to build in the coming years. Hence, current benchmarks are also relatively simple compared to truly useful programs. While running smaller versions of real-world applications introduces error, and is accepted in classical benchmarking, this is exacerbated for quantum computers. Entirely new issues may be introduced when scaling up and it is difficult to say whether performances measured today are good indicators of future performance. For example, IonQ's computer \cite{Wright-2019} has all $11$ qubits fully-connected. This configuration is possible at this scale, but this might not be true for a system with hundred or a thousand qubits. Such a system will likely require multiple fully-connected groups of qubits and communication will need to be orchestrated between them \cite{Martonosi-2019}. This introduces additional complexity which is not found in small-scale benchmarks.
Our previous question can be refined as: which quantum algorithms would be useful for program benchmarking quantum computers in the future? Algorithms such as quantum Markov chains \cite{Gudder-2008}, Shor's \cite{Shor-1997}, Grover's \cite{Grover-1996} and quantum chemistry \cite{Linke-2017,Olson-2017,McArdle-2020} are some obvious examples. Even if, for the most part, these algorithms will remain out of reach for near-term quantum computers, there is much to be gained from analyzing their scalability and reaction to noise. Currently, classical-quantum hybrid algorithms \cite{Yudong-2019,Schuld-2020,Wecker-2015,Fahri-2014,Perruzo-2014} are popular due to their ability to make use of the limited resources of NISQ computers. Another example of an algorithm that holds great potential for benchmarking are quantum walks, due to their susceptibility to noise and clear quadratic advantage over classical random walks \cite{Szegedy}.
Within this paper we use quantum algorithms to benchmark three of the newer IBM superconducting quantum computers: the $5$-qubit Bogota and Santiago and the $7$-qubit Casablanca machines. The choice of quantum algorithms is crucial, as we want them to be (i) scalable, in order to be able to face the challenge of the growing number of qubits in quantum computers, (ii) predictable in a way that allows us to recognize its noisy behaviour, and (iii) demonstrate clear quantum advantage. With these criteria in mind, we choose five algorithms: discrete-time quantum walks \cite{Aharonov-2000}, continuous-time quantum walks \cite{Fahri-1998-ctqw}, a circuit simulating the continuous-time quantum walk by decomposing its Hamiltonian to a sequence of Pauli gates, quantum phase estimation \cite{qpe-2019} and Grover's algorithm \cite{Grover-1996}. Finally, we note that using a continuous-time quantum algorithm and its Hamiltonian decomposition for benchmarking a (digital) quantum computer has not been carried out before, to the best of our knowledge. For the experiments we make use of the IBM Qiskit development kit \cite{Qiskit,IBMQExp} to simulate and execute the quantum circuits.
The paper is organized as follows. Section \ref{sec:prelim} introduces the preliminary methods necessary for the benchmarking process. Section \ref{sec:framework} defines the three benchmark metrics that result from the benchmarking process as well as presents a concrete framework for program benchmarking quantum computers. Moving on, the experimental process and the benchmark results for the chosen quantum computers are layed out in Section \ref{sec:results} before finally concluding the paper in Section \ref{sec:concl}.
\section{Preliminary Methods}\label{sec:prelim}
This section introduces the noise model used to simulate the behaviour of quantum computers. Additionally, it offers a brief discussion on the five quantum algorithms we use for benchmarking. Finally, it presents the quantum computers we are interested in benchmarking and the mathematical foundation of the final benchmark metrics.
\subsection{Unified Noise Model}\label{subsec:unm}
To approximate the noisy behaviour of quantum computers we use the \textit{unified noise model} (UNM) we have recently developed \cite{KGeorgo-2020-UNM}. This model combines three sources of error: (i) hardware infidelities in the form of gate, state preparation and measurement errors, (ii) decoherence in the form of thermal relaxation and (iii) dephasing of the qubits. The experiments in \cite{KGeorgo-2020-UNM} show that the UNM performs very well at approximating the behaviour of the IBMQ $15$-qubit Melbourne computer and better than other state of the art noise models.
The main characteristic of the UNM is its architecture awareness: the architectural graph that encompasses all the information regarding the connectivity of the qubits within the quantum processing unit (QPU) gets encoded within the model itself. Additionally, the model uses a number of noise parameters (see Table \ref{table:noiseparams}) calibrated from the machine itself, i.e., parameters that express the error rates of the gates, state preparations and measurements as well as the time it takes for the qubits within the QPU to decohere and dephase. Each noise parameter is unique and corresponds to each qubit individually or pair of qubits.
\bgroup
\def\arraystretch{1.2}%
\setlength\tabcolsep{0.1cm}
\begin{table}[!t]
\centering
\begin{tabular}{c|c|c}
\hline
Parameter & Error Type & No. Parameters \\ \hline \hline
$p_{r}$ & Gate error rates & $r$ \\
$p_{m}$ & State preparation error rates & $m$ \\
$p_{s}$ & Measurement error rates & $s$ \\
$T_{1}$ & Thermal relaxation times & $n$ \\
$T_{2}$ & Dephasing times & $n$ \\ \hline
\end{tabular}
\caption{The noise parameters and number of noise parameters for each type of error within the UNM; $n$ is the number of qubits in the system, $m$ is the number of qubits that are measured, $s$ is the number of state preparations that occur and $r$ is the number of distinct types of gates implemented in the architecture, each considered once per qubit or pair of qubits.}
\label{table:noiseparams}
\end{table}
\egroup
\subsection{The Quantum Algorithms}\label{subsec:qalg}
Here we give a quick overview of the quantum algorithms we use for benchmarking, i.e., (i) discrete-time quantum walks (DTQW), (ii) continuous-time quantum walks (CTQW), (iii) Pauli decomposition of the CTQW Hamiltonian (PD) (iv) quantum phase estimation (QPE) and (v) quantum search (QS). These algorithms are chosen according to three major characteristics that are interesting for benchmarking:
\begin{itemize}
\item \textbf{Scalability.} The algorithm should be able to scale up (or down) and run on increasingly larger quantum systems.
\item \textbf{Predictability.} The algorithm should produce a result that is easily predictable. An important addition to predictability is \textit{noise susceptibility}: the algorithm should provide a result whose distortion under the effects of noise is easily distinguishable from the ideal evolution.
\item \textbf{Quantum advantage.} The algorithm should provide a computational speed-up over its classical counterpart or, in other words, represent a possibly relevant real-world application.
\end{itemize}
\paragraph{Discrete-time Quantum Walks.} Quantum walks (DTQW) are the quantum mechanical analogue of a classical random walk on a graph or a lattice \cite{Aharonov-2000,KGeorgo-2020-rots,Kempe-2003}. They exhibit intrinsic properties that render their evolution easily predictable and highly susceptible to noise, making them an ideal candidate for benchmarking. First of all, discrete-time quantum walks exhibit modular behaviour \cite{KGeorgo-2020-rots,Reitzner-2011-mod}. This characteristic describes the modular relationship between the parity of the number of coin-flips of the walk, the initial state and the current position of the walker, a property that gets violated in a noisy environment \cite{KGeorgo-2020-rots}. Secondly, quantum walks propagate quadratically further than classical random walks \cite{Aharonov-2000,Szegedy} thus showing clear quantum advantage over their classical counterparts.
Finally, quantum walks are a highly scalable process. The size of the state-space of a quantum walk (i.e., the number of states that the walk traverses, represented by the number of qubits in the relevant register) can easily increase to match the size of the quantum computer we are interested in benchmarking.
\paragraph{Continuous-time Quantum Walks.} Continuous-time quantum walks (CTQW) were first introduced by Fahri and Gutman in \cite{Fahri-1998-ctqw}. This algorithm, much like the DTQWs, have an easily predictable quantum evolution that is highly susceptible to quantum noise, but exhibit very different characteristics to the discrete case. First of all, the CTQW evolution is determined by a Hamiltonian, $H$, instead of a coin-flip and is driven by a unitary of the form $e^{-iHt}$. Unlike the DTQWs, continuous-time quantum walks do not exhibit modular behaviour. Although like DTQWs they feature a quadratic increase in the walker's propagation \cite{Chakra-2020,Ambainis-2020}.
Finally, CTQWs are an easily scalable process as adding qubits to circuit can scale up the size of the state-space. To the best of our knowledge, this is the first work that uses a continuous-time quantum algorithm to benchmark the performance of a digital quantum computer. There are a few ways that a continuous-time quantum walk can be implemented on a gate-based (i.e., discrete) quantum computer, for example simulating the CTQW evolution using a DTQW \cite{Childs-2009-CTQW} or approximating the CTQW through a decomposition of the unitary $e^{-iHt}$ to a sequence of universal gates. Within this paper, the latter approach is utilized.
\paragraph{Pauli Decomposition of CTQW Hamiltonian.} Decomposition of quantum Hamiltonians to a set of universal gates is a well-studied area of research \cite{Lloyd-1996-dec,Childs-2011-dec,Suzuki-1992-dec,Aharonov-2003-dec,Hedge-2016-Pdec}. For this paper, we are interested in decomposing the Hamiltonian of the CTQW using the well-known set of Pauli matrices as the universal gate set. A further detailed analysis of this process is presented in the following Section \ref{subsec:paulidec}.
This procedure, often called \textit{Hamiltonian simulation}, adheres to the criteria for a good benchmarking process via the algorithm it decomposes. In other words, since the CTQW is suitable for benchmarking, so is the circuit that implements the decomposition of the CTQW Hamiltonian. Furthermore, it provides added value to this research since one can evaluate the performance of the quantum computer when executing the Hamiltonian simulation of a quantum process (i.e., CTQW), as well as compare the decomposition with the original algorithm.
\paragraph{Quantum Phase Estimation.} The quantum phase estimation (QPE) algorithm is used to estimate the phase (or eigenvalue) of an eigenvector of a unitary operator. More precisely, given an arbitrary quantum operator $U$ and a quantum state $\ket{\psi}$ such that $U\ket{\psi}=e^{2i\pi\theta}\ket{\psi}$, the algorithm estimates the value of $\theta$, given an approximation error \cite{Kitaev-qpe,qpe-2019,Oh-2019}.
Within this paper we exploit three characteristics of QPE that make it interesting for benchmarking. First of all, it is a scalable algorithm as further accuracy of the result (i.e., the estimated phase) can be obtained by increasing the number of qubits in the system. Furthermore, the result of the QPE is predictable and highly susceptible to quantum noise. Finally, QPE offers clear quantum advantage, achieving an exponential speed-up over known classical methods, rendering the algorithm one of the most important subroutines in quantum computing and serving as the building block of major quantum algorithms, like Shor's \cite{Shor-1997} or the HHL algorithms \cite{Harrow-2009}.
\paragraph{Quantum Search.} Grover's algorithm \cite{Grover-1996} describes a process for searching for a specific item within a database. Within this paper we use quantum search (QS) to look for a specific number $s$ within a set of numbers $\mathcal{S}=\{0,\dots,2^{n-1}\}$, where $n$ is the number of qubits within the quantum system that participate in the computation.
Quantum search represents an ideal algorithm for benchmarking quantum computers. The algorithm can scale up to search for an item within a larger database simply by adding qubits to the relevant quantum register. The result is easily predictable, as it is simply the item (or number, in our case) sought, as well as highly susceptible to noise (for example, the wrong result might appear due to noise). Additionally, QS can speed up an unstructured search problem quadratically, thus making it a very alluring application for quantum computers. Finally, Grover's algorithm can serve as a general trick or subroutine to obtain quadratic runtime improvements for a variety of other algorithms through what is called amplitude amplification \cite{Brassard-2002}.
\subsection{Pauli Decomposition of CTQW Hamiltonian}\label{subsec:paulidec}
In general, an arbitrary Hamiltonian $H$ of size $N\times N$, where $N=2^{n}$ and $n$ the number of qubits in the system, can be decomposed into a sequence of Pauli operators of the set $S=\{ \sigma_{I}, \sigma_{X}, \sigma_{Y}, \sigma_{Z} \}$ with well-known matrix representation as
\begin{equation}
H = \sum_{i_{1},\dots,i_{N}=I,x,y,z} \alpha_{i_{1},\dots,i_{N}} \left( \sigma_{i_{1}}\otimes\dots\otimes\sigma_{i_{N}} \right), \label{eq:pd}
\end{equation}
where
\begin{equation*}
\alpha_{i_{1},\dots,i_{N}} = \frac{1}{N}\operatorname{tr}\left[ \left( \sigma_{i_{1}}\otimes\dots\otimes\sigma_{i_{N}} \right)\cdot H \right].
\end{equation*}
For this paper we implement a CTQW on an $N$-cycle with Hamiltonian defined as $H_{\text{qw}}=\gamma A=\frac{1}{d}A=\frac{1}{2}A$, where $\gamma=1/d$ is the hopping rate between the two adjacent nodes in the cycle (i.e., node degree $d=2$) and $A$ is the adjacency matrix. Note that, for the physical purposes of this evolution, the rate $\gamma_{n|i}$ has dimensions $1/t$, where $t$ is the time duration of the CTQW evolution, the Hamiltonian has the appropriate energy dimensions and, for simplicity, $\hbar=1$.
The next step is to construct the unitary evolution operator from the Hamiltonian. This can be done by exponentiating the Hamiltonian as $e^{-iHt}$ where $H$ is a sum of terms of the form of equation \eqref{eq:pd}. It is important here to consider two things. First of all, during matrix exponentiation, a decomposition of the form $e^{-i(H_{1}+H_{2})t}=e^{-iH_{1}t}e^{-iH_{2}t}$, where $H_{1}$ and $H_{2}$ are Hermitian operators, is possible if $H_{1}$ and $H_{2}$ commute, i.e., $H_{1}H_{2}-H_{2}H_{1}=0$. This rule is naturally expanded for more than two matrices on the exponent.
Secondly, in the case that not all matrices in the exponent commute, the unitary operator resulting from the Hamiltonian exponentiation needs to be decomposed using the Lie-product formula \cite{Lloyd-1996-dec} as
\begin{equation}
e^{-i(H_{1} + H_{2} + \dots)t} \approx \left( e^{-iH_{1}t/r}e^{-iH_{2}t/r}\dots \right)^{r}
\end{equation}
where, for our case, $H=\sum_{i} H_{i}$ is the Pauli decomposition of the Hamiltonian to a sequence of Hermitian terms. This formula will create an approximation of the Hamiltonian with bounded error depending on $r$ \cite{Lloyd-1996-dec,Suzuki-1992-dec}. To ensure that the Pauli Hamiltonian decomposition exhibits error at most $\epsilon$, the bound $r$ can be taken as \cite{Lloyd-1996-dec}
\begin{equation*}
r = \mathcal{O}\left( (||H||t)^{2}/ \epsilon \right),
\end{equation*}
where $||H||$ is the norm of the Hamiltonian $H$ and $t$ is the continuous-time duration of the quantum evolution.
Thus, using the above methodology, we can now decompose the unitary that describes the continuous-time evolution of the CTQW (i.e., $e^{-iHt}$) in a sequence of Pauli operations that \textit{approximate} said evolution. As the Pauli gates form a universal gate set, the resulting quantum circuit can be implemented on the gate-based quantum processors.
\subsection{The Quantum Computers and Circuits}\label{subsec:qcs}
Within this paper we are interested in benchmarking three computers, the $5$-qubit Bogota, Santiago and the $7$-qubit Casablanca machine. They all exhibit a quantum volume $V_{Q}=32$ \cite{IBMQExp,Moll-2018}.
\paragraph{Quantum computer architectures.} Throughout our research we find that it is essential for the benchmarking procedure to be architecturally aware. This is also reflected by our choice of the UNM, an architecture aware noise model \cite{KGeorgo-2020-UNM}. Figure \ref{fig:architectures} shows the qubit connectivity of the quantum computers we benchmark in this paper.
\begin{figure}[!t]
\begin{tabular}{c}
\includegraphics[width=8.5cm]{ibmqBogotaSantiago.png} \\
(a) \\ [6pt]
\includegraphics[width=5.2cm]{ibmqCasablanca.png} \\
(b) \\[6pt]
\end{tabular}
\caption{Architectural graphs of the (a) IBMQ $5$-qubit Bogota and Santiago machines and (b) the IBMQ $7$-qubit Casablanca machine. A node in the graph represents a physical qubit whereas an edge represents a connection between a pair of qubits.}
\label{fig:architectures}
\end{figure}
\paragraph{Quantum circuits and characteristics.} For the implementation of quantum walks, we use a gate efficient approach that is based on inverter gates, as shown in \cite{Douglas-2009-EffWalk}. The QPE circuit is based on \cite{QPECircuit} and heavily relies on quantum Fourier transform (and its inverse) \cite{Nielsen_Chuang_2010} to estimate the relevant eigenvalue. For QS we make use of two approaches to the implementation, one with ancilla qubits (QSa) and one without (QSn) \cite{Karlsson_2018}.
For the continuous-time quantum walk, the Hamiltonian is automatically implemented by the Qiskit API when submitted for execution on the quantum computer. The Pauli decomposition (PD) of the Hamiltonian can be easily implemented on the quantum computers as it already maps the continuous-time Hamiltonian to a discrete basis gate set decomposition.
We identify four quantum circuit characteristics that are of interest for benchmarking:
\begin{enumerate}[i]
\item the number of gates in the circuit,
\item the number of active qubits, or qubits that are utilized by the quantum circuit (also called workspace),
\item the depth of the circuit, i.e., the longest path between the start of the circuit and a measurement gate and
\item the runtime of the circuit on the quantum computer. Table \ref{table:charactQASMcirc} shows those characteristics of the quantum circuits that implement each quantum algorithm.
\end{enumerate}
\begin{table*}[!t]
\begin{tabular}{c}
\bgroup
\def\arraystretch{1.2}%
\setlength\tabcolsep{0.36cm}
\centering
\begin{tabular}{c|c|c|c|c}
\hline
Machine & No. Gates & Size of Workspace & Depth & QC Runtime $\pm$ s.d. ($\text{ms}$) \\ \hline \hline
$5$-qubit Bogota & $47$ & $3$ & $35$ & $2.41\pm 0.03$ \\
$5$-qubit Santiago & $47$ & $3$ & $35$ & $2.34\pm 0.02$ \\
$7$-qubit Casablanca & $47$ & $3$ & $35$ & $2.54\pm 0.05$ \\ \hline
\end{tabular}
\egroup \\ [6pt] (a) Discrete-time quantum walk circuit characteristics. \\ [6pt]
\bgroup
\def\arraystretch{1.2}%
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\centering
\begin{tabular}{c|c|c|c|c}
\hline
Machine & No. Gates & Size of Workspace & Depth & QC Runtime $\pm$ s.d. ($\text{ms}$) \\ \hline \hline
$5$-qubit Bogota & $19$ & $2$ & $13$ & $2.34\pm 0.04$ \\
$5$-qubit Santiago & $19$ & $2$ & $13$ & $3.01\pm 0.07$ \\
$7$-qubit Casablanca & $19$ & $2$ & $13$ & $2.62\pm 0.01$ \\ \hline
\end{tabular}
\egroup \\ (b) Continuous-time quantum walk circuit characteristics. \\ [6pt]
\bgroup
\def\arraystretch{1.2}%
\setlength\tabcolsep{0.36cm}
\centering
\begin{tabular}{c|c|c|c|c}
\hline
Machine & No. Gates & Size of Workspace & Depth & QC Runtime $\pm$ s.d. ($\text{ms}$) \\ \hline \hline
$5$-qubit Bogota & $243$ & $2$ & $183$ & $2.42\pm 0.04$ \\
$5$-qubit Santiago & $243$ & $2$ & $183$ & $2.28\pm 0.06$ \\
$7$-qubit Casablanca & $243$ & $2$ & $183$ & $3.76\pm 0.04$ \\ \hline
\end{tabular}
\egroup \\ (c) Pauli decomposition of CTQW circuit characteristics. \\ [6pt]
\bgroup
\def\arraystretch{1.2}%
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\centering
\begin{tabular}{c|c|c|c|c}
\hline
Machine & No. Gates & Size of Workspace & Depth & QC Runtime $\pm$ s.d. ($\text{ms}$) \\ \hline \hline
$5$-qubit Bogota & $93$ & $4$ & $66$ & $2.54\pm 0.01$ \\
$5$-qubit Santiago & $97$ & $4$ & $72$ & $2.26\pm 0.03$ \\
$7$-qubit Casablanca & $100$ & $4$ & $75$ & $2.68\pm 0.06$ \\ \hline
\end{tabular}
\egroup \\ (d) Quantum phase estimation circuit characteristics. \\ [6pt]
\bgroup
\def\arraystretch{1.2}%
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\centering
\begin{tabular}{c|c|c|c|c}
\hline
Machine & No. Gates & Size of Workspace & Depth & QC Runtime $\pm$ s.d. ($\text{ms}$) \\ \hline \hline
$5$-qubit Bogota & $497$ & $4$ & $358$ & $2.70\pm 0.08$ \\
$5$-qubit Santiago & $479$ & $4$ & $343$ & $2.56\pm 0.04$ \\
$7$-qubit Casablanca (ancilla) & $788$ & $6$ & $503$ & $2.81\pm 0.04$ \\
$7$-qubit Casablanca (no ancilla) & $465$ & $4$ & $336$ & $2.74\pm 0.03$ \\ \hline
\end{tabular}
\egroup \\ (e) Quantum search circuit characteristics. \\
\end{tabular}
\caption{Quantum circuit characteristics for the five quantum circuits. No. gates and size of workspace are the number of gates and active qubits in the circuit respectively; depth of the circuit is the longest path between the start of the circuit and a measurement gate; QC runtime is the approximate average execution time of the circuit on the quantum computer along with standard deviation (s.d.).}
\label{table:charactQASMcirc}
\end{table*}
\subsection{Benchmark Indicators: Hellinger Distance}\label{subsec:hd}
The end results of the benchmarking process will be in the form of a comparison of the quantum computer output distribution with the distributions resulting from the unified noise model simulations of each machine and the ideal evolution.
To compare the probability distributions and generate the benchmarks, we use the Hellinger distance (HD) \cite{Jin_2018}.
\begin{definition}[Hellinger distance]\label{def:hd}
For probability distributions $P=\{p_{i}\}_{i\in[s]}$, $Q=\{q_{i}\}_{i\in[s]}$ supported on $[s]$, the Hellinger distance between them is defined as
\begin{equation}
h(P,Q) = \frac{1}{\sqrt{2}} \sqrt{ \sum_{i=1}^{k} \left( \sqrt{p_{i}} - \sqrt{q_{i}} \right)^{2} } \label{eq:hd}.
\end{equation}
\end{definition}
The Hellinger distance is a metric satisfying the triangle inequality. It takes values between $0$ and $1$ (i.e. $h(P,Q)\in[0,1]$) with $0$ meaning that the two distributions are equal. Additionally, it is easy to compute, easy to read and it does not depend on the probability distributions having the same support. The last property is particularly useful since in many ideal output distribution of quantum circuits the probability mass is concentrated on a few states.
\section{Framework for Program Benchmarking}\label{sec:framework}
The desired benchmarks correspond to comparisons (via the Hellinger distance) between the evolution of the quantum system on the computer, the noisy simulation and the ideal case. To achieve a more memorable notation, in the following analysis the symbol $q$ denotes the quantum computer, $i$ the ideal evolution and $n$ the noisy simulation. For example, the subscript $q|n$ denotes a value that corresponds to the difference between the quantum computer evolution ($q$) and the noisy evolution ($n$). We define three distances of interest, or otherwise, three benchmark metrics, as follows.
\begin{definition}[alpha benchmark]\label{def:alpha}
The Hellinger distance between the probability distribution of the quantum computer evolution, $Q$, and the ideal distribution, $D$, namely $h_{\text{id}}(Q,D)$, is the alpha benchmark, with notation $\alpha_{q|i}$:
\begin{equation}
\alpha_{q|i} \equiv h(Q,D). \label{eq:alpha}
\end{equation}
\end{definition}
\begin{definition}[beta benchmark]\label{def:beta}
The Hellinger distance between the probability distribution of the quantum computer evolution, $Q$, and the distribution resulting from the noisy simulations, $N$, namely $h_{\text{nm}}(Q,N)$, is the beta benchmark, with notation $\beta_{q|n}$:
\begin{equation}
\beta_{q|n} \equiv h(Q,N). \label{eq:beta}
\end{equation}
\end{definition}
\begin{definition}[gamma benchmark]\label{def:gamma}
The Hellinger distance between the distribution resulting from the noisy simulations, $N$, and the ideal distribution, $D$, namely $h_{\text{sm}}(N,D)$, is the gamma benchmark, with notation $\gamma_{n|i}$:
\begin{equation}
\gamma_{n|i} \equiv h(N,D). \label{eq:gamma}
\end{equation}
\end{definition}
As mentioned in Section \ref{subsec:hd}, the Hellinger distance is a metric that satisfies the triangle inequality. Hence, since the benchmarks established in the above definitions describe the pairwise Hellinger distances between three probability distributions (the quantum computer, $Q$, the simulated evolution, $N$, and the ideal evolution, $D$), it is possible to derive a relationship between $\alpha_{q|i}$, $\beta_{q|n}$ and $\gamma_{n|i}$ via the triangle inequality, through the following Lemma (whose proof follows simply from the triangle inequality for a metric).
\begin{lemma}[benchmarks triangle inequality]\label{lemma:triangineq}
Given probability distributions $Q$, $N$ and $D$, as established in the benchmark definitions, the pairwise Hellinger distances between those distributions, i.e., $h(Q,D)$, $h(Q,N)$ and $h(N,D)$ follow the triangle inequality:
\begin{equation*}
h(Q,D) \leq h(Q,N) + h(N,D).
\end{equation*}
Thus, the relevant benchmarks will also follow the triangle inequality as:
\begin{equation}
\alpha_{q|i} \leq \beta_{q|n} + \gamma_{n|i} \label{eq:trianineq}
\end{equation}
\end{lemma}
The above Lemma effectively means that, according to the benchmark definitions, the deviation of the quantum computer evolution from the ideal ($\alpha_{q|i}$) will never be greater than the sum of the expected (i.e., simulated) evolution derived by the levels of noise within the machine and the distance between the simulated and ideal distributions ($\beta_{q|n}+\gamma_{n|i}$). In other words, defining the benchmark metrics using the Hellinger distance allows us to quantify the confidence on the estimated level of noise during the execution of the quantum circuit.
Following the definitions and relationship between the benchmarks, we present a framework for program benchmarking quantum computers as the sequence of the six steps below:
\begin{description}
\item[Step 1] \textbf{Select benchmark method(s).} The selection of quantum algorithms that will be used for benchmarking should adhere to the three criteria described in Section \ref{subsec:qalg}: (i) scalability, (ii) predictability and noise susceptibility and (iii) quantum advantage.
\item[Step 2] \textbf{Quantum noise model and simulator.} Select or implement a noise model that approximates the noisy evolution within the quantum machine and a simulator that can execute the noise model.
\item[Step 3] \textbf{Run experiments.} Design and execute a suitable number of experiments of the benchmark method(s) on the quantum computer. This step also includes calibrating the noise parameters that encapsulate the level of noise within the quantum computer at the time of the experiments.
\item[Step 4] \textbf{Simulate the noisy evolution.} Use the noise model in order to simulate the noisy evolution of the quantum computer using the calibrated noise parameters from the time of the experiment, as described in Step 3.
\item[Step 5] \textbf{Simulate the ideal evolution.} This can be done either through simple noise-free simulations or by calculating the probabilities through the quantum statevector.
\item[Step 6] \textbf{Calculate the benchmarks.} The final benchmarks $\alpha_{q|i}$, $\beta_{q|n}$ and $\gamma_{n|i}$ are the Hellinger distances between the quantum computer, the UNM and the ideal evolution in the setting described in Definitions \ref{def:alpha}, \ref{def:beta} and \ref{def:gamma}.
\end{description}
Following the benchmarking framework we can extract meaningful results from a series of comparisons between the benchmark metrics, $\alpha_{q|i}$, $\beta_{q|n}$ and $\gamma_{n|i}$. The $\beta_{q|n}$ benchmark essentially describes how closely the noise model simulates the behaviour of the quantum computer. The $\alpha_{q|i}$ benchmark shows how far the behaviour of the quantum computer falls from the noise-free case thus giving an estimate of the overall computer performance under the effects of noise. From the comparison between the $\beta_{q|n}$ and $\alpha_{q|i}$ benchmarks we can extract valuable information: if $\beta_{q|n}<\alpha_{q|i}$ the noise levels in the quantum computer are closer to the estimated ones from the noise simulations; on the opposite case the computer operates closer to the ideal evolution. In the latter case the quantum computer behaves more efficiently with lower level of noise than expected, thus giving us more confidence regarding the computational result.
The $\gamma_{n|i}$ benchmark, even though it can be used as an estimate of the noise levels in the quantum computer, does not give any relevant information on its own. The value comes when considering the $\gamma_{n|i}$ benchmark together with the $\alpha_{q|i}$ benchmark. First of all, the closer the values of $\alpha_{q|i}$ and $\gamma_{n|i}$ are, the smaller the value of $\beta_{q|n}$ (if $\beta_{q|n}=0$ then $\alpha_{q|i}=\gamma_{n|i}$ and vice versa). Additionally, if $\alpha_{q|i}>\gamma_{n|i}$, then the noise model, and hence the noise parameters, underestimate the levels of noise during the quantum computer evolution, with the opposite being true if $\alpha_{q|i}<\gamma_{n|i}$. Moreover, the absolute difference $\abs{\alpha_{q|i}-\gamma_{n|i}}$ can quantify this noise over- or underestimation. Precisely, the smaller the absolute difference the smaller the error in estimation. This information is useful to the benchmarking process as it further highlights whether the machine is more or less noisy than estimated while also showing the efficiency of the machine calibration techniques.
Finally, using the triangle inequality from Lemma \ref{lemma:triangineq} we can make further interesting remarks. From equation \eqref{eq:trianineq} we find that $\beta_{q|n} \geq \alpha_{q|i} - \gamma_{n|i}$. Furthermore, from the above analysis we know that a comparison between $\alpha_{q|i}$ and $\gamma_{n|i}$ can tell us whether the calibrated noise parameters over- or underestimate the level of noise during the evolution on the quantum computer, and the absolute difference $\abs{\alpha_{q|i}-\gamma_{n|i}}$ can give an indication of the scale of the error in estimation. Importantly, the above inequality does not hold for absolute values when subtraction takes place (i.e., for an inequality of the form $\abs{\beta_{q|n}} \geq \abs{\alpha_{q|i} - \gamma_{n|i}}$ when $\gamma_{n|i}>\alpha_{q|i}$). Considering this, we can interpret the triangle inequality as a measure of the confidence on the calibrated parameters encapsulating a picture of the noise that is accurate enough to provide a good estimation of the quantum evolution, expressed as follows:
\begin{itemize}
\item If $\beta_{q|n}\geq\abs{\alpha_{q|i}-\gamma_{q|i}}$, then the over- or underestimation of the noise is small enough to provide estimates of the quantum evolution with \textit{high confidence}.
\item If $\beta_{q|n}<\abs{\alpha_{q|i}-\gamma_{q|i}}$, then the calibrated parameters generate \textit{low confidence} on the levels of noise.
\end{itemize}
\section{Experiments and Results}\label{sec:results}
\subsection{Experimental Setup}\label{subsec:setup}
For the benchmarking experiments we run the quantum circuits that implement the chosen quantum algorithms. In the case of the DTQW we run one step of the algorithm (i.e., one coin-flip) on a workspace (i.e., number of active qubits used by the circuit) of three qubits with initial state $\ket{0}$, as previous work shows that this configuration is satisfactory for errors to take place and the behaviour of the quantum walk to evolve in a predictable manner \cite{KGeorgo-2020-rots}. For the CTQW and its PD we use a small two-qubit state-space, as the Pauli decomposition gets excessively large for a three-qubit state space or bigger. The algorithms are implemented for a continuous time of $t=3$. Our QPE routine is tailored to estimate a phase of $\theta = 2\pi/3$ using a workspace of four qubits with theoretical probability of success estimated at $0.688$. Finally, the QS implementation performs an unstructured search for the element $\ket{s}=\ket{10}$ (decimal) within a four qubits state-space i.e., a dataset containing numbers $\ket{0}$ to $\ket{15}$. The QS algorithm is run for three iterations and shows a theoretical probability of success estimated at $0.96$. Table \ref{table:initconfig} shows more comprehensively the initial configuration for each algorithm.
\bgroup
\def\arraystretch{1.2}%
\setlength\tabcolsep{0.32cm}
\begin{table*}[!t]
\centering
\begin{tabular}{c|c|c|c|c}
\hline
Algorithm & No. Qubits & Size of Workspace & Iterations/Duration & Probability of Success \\ \hline \hline
DTQW & $2$ & $3$ & $1$ (one coin-flip) & $p_{\text{succ}}=0.5$ in states $\ket{1}$ and $\ket{3}$ \\
CTQW & $2$ & $2$ & $t=3$ & $p_{\ket{2}}=0.99$, $p_{\ket{1}}=p_{\ket{3}}=0.005$ \\
PD & $2$ & $2$ & $t=3$ & $p_{\ket{2}}=0.99$, $p_{\ket{1}}=p_{\ket{3}}=0.005$ \\
QPE & $3$ & $4$ & $1$ & $p_{\text{succ}}=0.688$ in state $\ket{3}$ \\
QSa & $4$ & $6$ & $3$ & $p_{\text{succ}}=0.96$ in state $\ket{10}$ \\
QSn & $4$ & $4$ & $3$ & $p_{\text{succ}}=0.96$ in state $\ket{10}$ \\ \hline
\end{tabular}
\caption{The initial configuration for the quantum walk (DTQW), continuous-time quantum walk (CTQW) and its Pauli decomposition (PD), quantum phase estimation (QPE) and quantum search algorithms with ancilla (QSa) and without ancilla (QSn) for the benchmarking experiments. No. qubits is the number of qubits in the state space, size of workspace is the number of active qubits utilized by the circuit, iterations is the number of repetitions of the quantum circuit and the probability of success is the theoretical probability of the algorithm to give us the correct (or expected) result.}
\label{table:initconfig}
\end{table*}
\egroup
We run each algorithm independently $100{,}000$ times (see Section \ref{subsec:qalg}) on the three chosen computers (see Section \ref{subsec:qcs}) and as a simulation using the UNM (see Section \ref{subsec:unm}) and the ideal simulator \cite{Qiskit} to reproduce the noisy and the ideal quantum evolutions respectively. Finally, we compute the $\beta_{q|n}$, $\alpha_{q|i}$ and $\gamma_{n|i}$ benchmarks, as defined in Section \ref{sec:framework}. The results are shown on Table \ref{table:benchmarks} for each machine and are also visualized in Figure \ref{fig:benchmarks}.
\begin{figure*}[!t]
\begin{center}
\includegraphics[width=12cm]{benchmarks.png} \\
\end{center}
\caption{Visualization of the three benchmarks, $\alpha_{q|i}$, $\beta_{q|n}$ and $\gamma_{n|i}$ for each of the three quantum computers (mapped to the left-hand side $y$-axis) when executing the five quantum algorithms.}
\label{fig:benchmarks}
\end{figure*}
\begin{table*}[!t]
\begin{tabular}{c}
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\centering
\begin{tabular}{c|c|c|c|c|c|c}
\hline
Machine & $\alpha_{q|i}$ & $\beta_{q|n}$ & $\gamma_{n|i}$ & $\abs{\alpha_{q|i}-\gamma_{n|i}}$ & QC Runtime ($\text{sec}$) & Sim. Runtime ($\text{sec}$) \\ \hline \hline
$5$-qubit Bogota & $0.287$ & $0.025$ & $0.291$ & $0.004$ & $240.7$ & $3{,}331.5$ \\
$5$-qubit Santiago & $0.321$ & $0.054$ & $0.362$ & $0.041$ & $234.2$ & $3{,}119.3$ \\
$7$-qubit Casablanca & $0.269$ & $0.076$ & $0.331$ & $0.062$ & $254.5$ & $4{,}626.6$ \\ \hline
\end{tabular}
\egroup \\ [6pt] (a) Quantum walk algorithm on two qubits. \\ [6pt]
\bgroup
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\setlength\tabcolsep{0.4cm}
\centering
\begin{tabular}{c|c|c|c|c|c|c}
\hline
Machine & $\alpha_{q|i}$ & $\beta_{q|n}$ & $\gamma_{n|i}$ & $\abs{\alpha_{q|i}-\gamma_{n|i}}$ & QC Runtime ($\text{sec}$) & Sim. Runtime ($\text{sec}$) \\ \hline \hline
$5$-qubit Bogota & $0.139$ & $0.096$ & $0.234$ & $0.095$ & $234.4$ & $3{,}026.9$ \\
$5$-qubit Santiago & $0.154$ & $0.035$ & $0.184$ & $0.030$ & $301.7$ & $2{,}905.7$ \\
$7$-qubit Casablanca & $0.247$ & $0.044$ & $0.239$ & $0.008$ & $262.1$ & $3{,}035.2$ \\ \hline
\end{tabular}
\egroup \\ (b) Continuous-time quantum walk algorithm on two qubits. \\ [6pt]
\bgroup
\def\arraystretch{1.2}%
\setlength\tabcolsep{0.4cm}
\centering
\begin{tabular}{c|c|c|c|c|c|c}
\hline
Machine & $\alpha_{q|i}$ & $\beta_{q|n}$ & $\gamma_{n|i}$ & $\abs{\alpha_{q|i}-\gamma_{n|i}}$ & QC Runtime ($\text{sec}$) & Sim. Runtime ($\text{sec}$) \\ \hline \hline
$5$-qubit Bogota & $0.436$ & $0.041$ & $0.467$ & $0.031$ & $242.7$ & $11{,}764.9$ \\
$5$-qubit Santiago & $0.514$ & $0.084$ & $0.465$ & $0.049$ & $228.5$ & $11{,}618.6$ \\
$7$-qubit Casablanca & $0.577$ & $0.039$ & $0.612$ & $0.035$ & $376.4$ & $12{,}861.3$ \\ \hline
\end{tabular}
\egroup \\ (c) Pauli decomposition of the CTQW algorithm on two qubits. \\ [6pt]
\bgroup
\def\arraystretch{1.2}%
\setlength\tabcolsep{0.4cm}
\centering
\begin{tabular}{c|c|c|c|c|c|c}
\hline
Machine & $\alpha_{q|i}$ & $\beta_{q|n}$ & $\gamma_{n|i}$ & $\abs{\alpha_{q|i}-\gamma_{n|i}}$ & QC Runtime ($\text{sec}$) & Sim. Runtime ($\text{sec}$) \\ \hline \hline
$5$-qubit Bogota & $0.469$ & $0.177$ & $0.351$ & $0.118$ & $254.2$ & $6{,}287.1$ \\
$5$-qubit Santiago & $0.488$ & $0.144$ & $0.369$ & $0.119$ & $226.8$ & $6{,}304.2$ \\
$7$-qubit Casablanca & $0.524$ & $0.267$ & $0.351$ & $0.173$ & $268.3$ & $7{,}741.1$ \\ \hline
\end{tabular}
\egroup \\ (d) Quantum phase estimation algorithm on three qubits. \\ [6pt]
\bgroup
\def\arraystretch{1.2}%
\setlength\tabcolsep{0.28cm}
\centering
\begin{tabular}{c|c|c|c|c|c|c}
\hline
Machine & $\alpha_{q|i}$ & $\beta_{q|n}$ & $\gamma_{n|i}$ & $\abs{\alpha_{q|i}-\gamma_{n|i}}$ & QC Runtime ($\text{sec}$) & Sim. Runtime ($\text{sec}$) \\ \hline \hline
$5$-qubit Bogota & $0.754$ & $0.014$ & $0.752$ & $0.002$ & $270.4$ & $22{,}834.3$ \\
$5$-qubit Santiago & $0.751$ & $0.051$ & $0.738$ & $0.013$ & $256.7$ & $21{,}923.2$ \\
$7$-qubit Casablanca (ancilla) & $0.760$ & $0.040$ & $0.752$ & $0.008$ & $280.1$ & $36{,}425.7$ \\
$7$-qubit Casablanca (no ancilla) & $0.763$ & $0.061$ & $0.738$ & $0.025$ & $274.1$ & $22{,}889.9$ \\ \hline
\end{tabular}
\egroup \\ (e) Quantum search algorithm on four qubits. \\
\end{tabular}
\caption{Results from benchmarking the three machines for each of the five algorithms. The benchmark indicators $\alpha_{q|i}$, $\beta_{q|n}$ and $\gamma_{n|i}$ are the HD between the quantum computer and the UNM, the quantum computer and the ideal distribution and the UNM and the ideal distribution respectively. QC runtime is the cumulative execution time in seconds for $100{,}000$ iterations of each algorithm on the respective machine; Sim. runtime is the time it takes to simulate for $100{,}000$ iterations the behaviour of each machine when executing the algorithms.}
\label{table:benchmarks}
\end{table*}
\subsection{Results}\label{subsec:res}
The results of the benchmark experiments are presented in Table \ref{table:benchmarks} and below for each of the algorithms used.
It is important at this point to emphasize an advantage of the structure presented for program benchmarking quantum computers. The use of the Hellinger distance to define the three benchmarks ($\alpha_{q|i}$, $\beta_{q|n}$ and $\gamma_{n|i}$) allows us to perform a comparison of the performance of different quantum algorithms on the same basis, i.e., a dimensionless quantity. In other words, through the proposed framework, apart from measuring the efficiency of a quantum processor, we can gather additional information and compare the efficiency of circuit implementations for the selected algorithms. This is done here for the two approaches on implementing the CTQW on a gate-based computer, one that is done through the IBMQ API and one that use the Pauli decomposition of the CTQW Hamiltonian (see Section \ref{subsec:paulidec}).
\subsubsection{Discrete-time Quantum Walk Benchmarking}\label{subsubsec:dtqwBench}
The DTQW implementation produces a circuit with a relatively small depth and number of gates. Our $\beta_{q|n}$ benchmark measures how closely a quantum computer operates to the expected noise levels (indicated by the noise parameters and simulated by the UNM). Thus, in the small circuit case of the DTQW, the Bogota machine operates much closer to the expected evolution than the other computers with the smallest $\beta_{q|n}$.
The $\alpha_{q|i}$ and $\gamma_{n|i}$ benchmarks lead to a coherent picture of the overall performance of the quantum computers. In the DTQW case, the $\alpha_{q|i}$ benchmark is relatively small compared to the larger algorithms. This means that the quantum computers are not as erroneous as in the larger circuit cases, an expected result. A comparison between the $\alpha_{q|i}$ and $\gamma_{n|i}$ benchmarks shows that $\alpha_{q|i} < \gamma_{n|i}$ for all the machines, indicating that the the calibrated parameters overestimate the levels of noise during the evolution of the circuit. Finally, $\beta_{q|n}\geq\abs{\alpha_{q|i}-\gamma_{q|i}}$ for all quantum computers, which implies high confidence of our noise level estimates.
\subsubsection{Continuous-time Quantum Walk Benchmarking}\label{subsubsec:ctqwBench}
The CTQW circuit is the smallest circuit we use for benchmarking, which leads us to expect that the machine evolution will be reasonably close to the ideal case for this circuit. Evidently, the $\alpha_{q|i}$ benchmark is by far the smallest for the CTQW algorithm on all three machines. As $\alpha_{q|i}<\gamma_{n|i}$ for the Bogota and Santiago machines, we can say that the noise parameters overestimate the noise for those two computers, whereas they slightly underestimate the noise for the Casablanca computer.
For the CTQW, the Santiago machine operates closer to the expected levels of noise according to its error rates, with the smallest $\beta_{q|n}$. Similarly to the DTQW, we are confident in the calibrated noise as $\beta_{q|n}\geq\abs{\alpha_{q|i}-\gamma_{q|i}}$.
\subsubsection{Pauli Decomposition Benchmarking}\label{subsubsec:pdBench}
Unlike the CTQW circuit, its Pauli decomposition is a large and quite deep circuit, which means that we should expect a very noisy evolution. The small $\beta_{q|n}$ values show that the quantum computers operate relatively close to expectations.
As shown by the large values of $\alpha_{q|i}$ for all quantum computers, the PD circuit is quite error-prone compared to the two previous cases of the quantum walk. Thus, an important conclusion we can safely draw is that, for a two-qubit continuous-time quantum walk, the Pauli decomposition of its Hamiltonian leads to a more complex and noisier circuit than the decomposition to base gates done automatically through the IBMQ API.
The $\gamma_{n|i}$ benchmarks show that the parameters overestimate the noise of the Bogota and Casablanca machines and underestimate the noise within the Santiago computer. Furthermore, $\beta_{q|n}\geq\abs{\alpha_{q|i}-\gamma_{q|i}}$ inspires high-confidence on our noise estimates. This result is important as, through our benchmarks, we produce evidence on the performance of two different techniques for implementing the same algorithm.
\subsubsection{Quantum Phase Estimation Benchmarking}\label{subsubsec:dqpeBench}
Moving on, the QPE circuit is slightly more complex than the DTQW circuit. An overall bigger $\beta_{q|n}$ benchmark on every machine indicates that the QPE circuit execution deviates slightly more from the expected evolution compared to the other algorithms. This could be the result of random fluctuations of the calibrated parameters.
Following the $\alpha_{q|i}$ benchmark, we can extract similar results for the quantum computers as the PD circuit benchmarks, finding that the larger values indicate that the quantum computer performance is hindered by the size of the circuit. Interestingly, we see that $\alpha_{q|i}>\gamma_{n|i}$ for all the quantum computers, showcasing that the calibrated parameters underestimate the noise and we can expect noisier results from all the quantum computers for computations of similar size. We have high-confidence of our calibrated parameters as $\beta_{q|n}\geq\abs{\alpha_{q|i}-\gamma_{q|i}}$ for all experiments.
\subsubsection{Quantum Search Benchmarking}\label{subsubsec:qsBench}
Finally, the quantum search circuit represents the largest implemented algorithm and thus, we expect it to also be the noisiest. Following the above methodology, we find that the $\beta_{q|n}$ benchmark shows the smallest values for each machine, indicating that the quantum computers operated close to the predicted evolutions. The $\alpha_{q|i}$ and $\gamma_{n|i}$ benchmarks are quite close, increasing our confidence on the estimated noise levels. Nevertheless, it is safe to conclude that the machines in the case of the QS exhibit intense levels of noise, as the $\alpha_{q|i}$ benchmarks are very large.
A quick comparison between the $\alpha_{q|i}$ and $\gamma_{n|i}$ shows that the calibrated parameters offer a very good picture of the noise within the quantum computer in this experiment.
\subsubsection{Analysis}\label{subsubsec:conclBenc}
A comparison of the $\alpha_{q|i}$ and $\gamma_{n|i}$ benchmarks shows that when executing the DTQW and CTQW circuits, the machines are less error-prone than the UNM and calibrated parameters imply, with the opposite being true for the PD, QS and QPE. This informs us that the machines are better at executing small circuits, an expected result. Additionally, a larger numerical difference between $\alpha_{q|i}$ and $\gamma_{n|i}$ indicates a more efficient computer behaviour. For example, Casablanca is the most efficient in terms of noise for the DTQW case as $\alpha_{q|i}<\gamma_{n|i}$ with the biggest difference. Similarly, in the QS case the Casablanca machine is the least efficient with $\alpha_{q|i}>\gamma_{n|i}$. Here, as we use two different implementations of the QS algorithm, we can also compare them with each other. The benchmarks show that the QS circuit with ancilla is closer to the expected evolution by the UNM, shown by $\beta_{q|n}$, and to the ideal evolution, shown by $\alpha_{q|i}$, although not by much, shown by $\alpha_{q|i}$ versus $\gamma_{n|i}$.
The results from the above analysis are crucial as they can lead towards the selection of a machine appropriate to specific circuit needs. Additionally, we provide a comparison between the continuous-time quantum walk circuit and its Pauli decomposition. This fact is clearly reflected on the benchmarks, as the values of $\alpha_{q|i}$ are much lower for the CTQW circuit, while both circuits operate within expectations (small $\beta_{q|n}$ values) and are not massively over- or underestimated by their noise parameters (small $|\alpha_{q|i}-\gamma_{n|i}|$). Thus, we can easily conclude that, for the two-qubit case, a Pauli decomposition of the CTQW Hamiltonian leads to a less efficient circuit.
Overall, we can concentrate the general benchmarking results as follows. In the smallest circuits (i.e., CTQW and DTQW case) the $\beta_{q|n}$ benchmarks show how close the computers operate to the expected levels of noise. The Casablanca machine shows the best $\alpha_{q|i}$ benchmark, i.e., it is the closest to the ideal evolution, closely followed by Bogota and with Santiago being the furthest away. The UNM and the calibrated parameters always overestimate the noise in this case as $\alpha_{q|i}<\gamma_{n|i}$, thus showing that all the quantum computers exhibit relatively low levels of noise when executing small circuits. In the slightly deeper circuit of the QPE, the Bogota machine outperforms the others, followed by Santiago and Casablanca. In this case though the $\beta_{q|n}$ benchmark indicates slightly bigger deviation from the expected level of noise which is biggest on the Casablanca machine. The UNM and calibrated parameters always underestimate the noise in the quantum computer as $\alpha_{q|i}>\gamma_{n|i}$, meaning that the quantum computers are more error-prone. Lastly, in the largest QS and PD circuits, the benchmarks indicate that all the machines operate close to the noise model with low $\beta_{q|n}$ benchmarks and exhibit very low performance as the $\alpha_{q|i}$ benchmarks are large, getting close to $1$ for the QS. The noise model slightly underestimates the noise, but in this case of very deep circuits, the benchmarks show that the machines will not produce any meaningful results, an expected outcome.
A visual representation of the comparison between the distributions resulting from the quantum computer executions and the ideal simulations for each quantum algorithm are shown in Appendix \ref{ap:benchcomp}. A further comparison between the quantum computer and the individual UNM distributions for each machine is given in Appendix \ref{ap:comp}.
\section{Discussion and Conclusions}\label{sec:concl}
In this paper we have presented an approach to benchmarking quantum computers using scaling, high-level quantum algorithms considered as attractive ``real-world" problems. We have defined three benchmark metrics, each highlighting different aspects of the machine's efficiency either as a standalone or through comparisons between them. Each benchmark metric describes the difference between two quantum evolutions and together they follow the triangle inequality.
In order to better present the main characteristics of our benchmarks, we streamline the discussion as a comparison with the vastly used metric of \textit{quantum volume} \cite{Moll-2018}, which quantifies the expected size of a circuit that can be reliably run on a quantum computer. In contrast, our program benchmarks describe the performance of the quantum computer when running a specific circuit itself. This approach has advantages and disadvantages over architecture-neutral benchmarks. First of all, our benchmark metrics show the exact performance of a QPU (and in extend, the quantum computer itself) when running the quantum circuit. Additionally, they highlight the difference with an expected (noisy) simulated evolution and an ideal (noise-free) simulated evolution, a result that better identifies the weaknesses of the machine in a more structural manner. More specifically, the metrics allow us to realise the manner and intensity that the computer deviates from the simulated evolutions. Finally, all three machines we benchmark use the same QPU technology and exhibit the same quantum volume of $32$. On the other hand, our benchmarking process exhibits different metrics and results for each quantum computer. Thus, our benchmarks capture the performance of each QPU more thoroughly and provide us with a more detailed representation of their performance.
On the con side, our benchmarking process is slower compared to the calculation of the quantum volume. This is an expected outcome as we need to run a number of experiments on the quantum computer as well as the noisy simulations. Moreover, the flip-side of the architecture-specific nature of our metrics dictates that each computer will exhibit different benchmarks when executing different algorithms.
Thus, we can conclude that our architecture-specific program benchmarks showcase the performance of a quantum computer in a ``real-world" environment, as well as highlight the performance when running a specific algorithm and carry out meaningful comparisons between related circuits (e.g., the CTQW vs the Pauli decomposition of its Hamiltonian). Additionally, we can gather information on whether the calibrated parameters over- or underestimate the levels of noise during the quantum evolution. On the other hand, architecture-neutral benchmarks like the quantum volume are more generic and excel at showcasing the limitations of QPUs when running arbitrary quantum circuits.
It is also noteworthy that the benchmarking experiments carried out during this work are not suitable for a comparative analysis regarding the efficiency of each qubit topology (see Figure \ref{fig:architectures}). One reason for this is the relatively small size and limited flexibility of the computers themselves. Nevertheless, when benchmarking larger quantum computers in the future, we believe that using our framework, one could focus, for example, on searching which architectures are more efficient for a specific task.
In conclusion, our work has shown that using quantum algorithms to benchmark quantum computers in a well-structured environment can stress different aspects and very informatively highlight the performance of a quantum computer. Additionally, it provides a way to compare the efficiency of different circuits used to implement the same task. To the best of our knowledge, this is the first work that uses a continuous-time quantum algorithm to benchmark the performance of a digital quantum machine. The results show that, for small state spaces of the continuous-time quantum walk, the Pauli decomposition does not produce an efficient circuit. This result is expected as the complexity of a Hamiltonian operating on two qubits is very small.
\section{Acknowledgements}
This work was supported by the Engineering and Physical Sciences Research Council, Centre for Doctoral Training in Cloud Computing for Big Data, United Kingdom [grant number EP/L015358/1].
\section{Introduction}\label{sec:intro}
Benchmarking quantum computers aims to determine the performance of a quantum computing system under an appropriate set of metrics. Within the Noisy Intermediate-Scale Quantum (NISQ) era \cite{Preskill-2018}, benchmarking the capabilities and performance of quantum computers when executing quantum programs is of paramount importance, especially for assessing their scalability.
An intuitive approach to benchmarking quantum computers is establishing a set of quantum programs and measuring the performance of a quantum computer when executing each one. Such a work gains more merit as bigger quantum computers are built. There are various advantages to this approach, for example benchmarking the limits and behaviour of a quantum machine within a scaling, computationally intensive environment and particularly testing the system when performing a ``real-world" task. Various companies appear to favor such an approach. IonQ for example tested their quantum computer using the Bernstein-Vazirani \cite{BernsteinVazirani} and the Hidden Shift \cite{vanDam-2006,Rotteler-2010} algorithms. The metric for performance was the likelihood of measuring the correct output \cite{Wright-2019}. On the other hand, Google focused on the problem of quantum sampling and achieved results they claim demonstrated quantum supremacy \cite{GoogleSuprem}. While the chosen application was not particularly useful in a ``real-world" scenario, it excelled at demonstrating the computing power of the system. So a question arises, highlighting the difficulty of creating quantum program benchmarks: which benchmarks are more insightful?
The different competing quantum technologies pose a major challenge. The technologies have different topologies and thus have unique strengths and weaknesses. For example, the connectivity of an ion-trap computer provided a large advantage on some benchmarks over a superconducting quantum computer \cite{Linke-2017}.
Clearly, systems can behave differently on different benchmarks, which introduces the issue of invested interests, i.e., using benchmarks that are expected to perform well on a current system \cite{Lilja-2000}. While impressive, current quantum computers are very small compared to the computers we hope to build in the coming years. Hence, current benchmarks are also relatively simple compared to truly useful programs. While running smaller versions of real-world applications introduces error, and is accepted in classical benchmarking, this is exacerbated for quantum computers. Entirely new issues may be introduced when scaling up and it is difficult to say whether performances measured today are good indicators of future performance. For example, IonQ's computer \cite{Wright-2019} has all $11$ qubits fully-connected. This configuration is possible at this scale, but this might not be true for a system with hundred or a thousand qubits. Such a system will likely require multiple fully-connected groups of qubits and communication will need to be orchestrated between them \cite{Martonosi-2019}. This introduces additional complexity which is not found in small-scale benchmarks.
Our previous question can be refined as: which quantum algorithms would be useful for program benchmarking quantum computers in the future? Algorithms such as quantum Markov chains \cite{Gudder-2008}, Shor's \cite{Shor-1997}, Grover's \cite{Grover-1996} and quantum chemistry \cite{Linke-2017,Olson-2017,McArdle-2020} are some obvious examples. Even if, for the most part, these algorithms will remain out of reach for near-term quantum computers, there is much to be gained from analyzing their scalability and reaction to noise. Currently, classical-quantum hybrid algorithms \cite{Yudong-2019,Schuld-2020,Wecker-2015,Fahri-2014,Perruzo-2014} are popular due to their ability to make use of the limited resources of NISQ computers. Another example of an algorithm that holds great potential for benchmarking are quantum walks, due to their susceptibility to noise and clear quadratic advantage over classical random walks \cite{Szegedy}.
Within this paper we use quantum algorithms to benchmark three of the newer IBM superconducting quantum computers: the $5$-qubit Bogota and Santiago and the $7$-qubit Casablanca machines. The choice of quantum algorithms is crucial, as we want them to be (i) scalable, in order to be able to face the challenge of the growing number of qubits in quantum computers, (ii) predictable in a way that allows us to recognize its noisy behaviour, and (iii) demonstrate clear quantum advantage. With these criteria in mind, we choose five algorithms: discrete-time quantum walks \cite{Aharonov-2000}, continuous-time quantum walks \cite{Fahri-1998-ctqw}, a circuit simulating the continuous-time quantum walk by decomposing its Hamiltonian to a sequence of Pauli gates, quantum phase estimation \cite{qpe-2019} and Grover's algorithm \cite{Grover-1996}. Finally, we note that using a continuous-time quantum algorithm and its Hamiltonian decomposition for benchmarking a (digital) quantum computer has not been carried out before, to the best of our knowledge. For the experiments we make use of the IBM Qiskit development kit \cite{Qiskit,IBMQExp} to simulate and execute the quantum circuits.
The paper is organized as follows. Section \ref{sec:prelim} introduces the preliminary methods necessary for the benchmarking process. Section \ref{sec:framework} defines the three benchmark metrics that result from the benchmarking process as well as presents a concrete framework for program benchmarking quantum computers. Moving on, the experimental process and the benchmark results for the chosen quantum computers are layed out in Section \ref{sec:results} before finally concluding the paper in Section \ref{sec:concl}.
\section{Preliminary Methods}\label{sec:prelim}
This section introduces the noise model used to simulate the behaviour of quantum computers. Additionally, it offers a brief discussion on the five quantum algorithms we use for benchmarking. Finally, it presents the quantum computers we are interested in benchmarking and the mathematical foundation of the final benchmark metrics.
\subsection{Unified Noise Model}\label{subsec:unm}
To approximate the noisy behaviour of quantum computers we use the \textit{unified noise model} (UNM) we have recently developed \cite{KGeorgo-2020-UNM}. This model combines three sources of error: (i) hardware infidelities in the form of gate, state preparation and measurement errors, (ii) decoherence in the form of thermal relaxation and (iii) dephasing of the qubits. The experiments in \cite{KGeorgo-2020-UNM} show that the UNM performs very well at approximating the behaviour of the IBMQ $15$-qubit Melbourne computer and better than other state of the art noise models.
The main characteristic of the UNM is its architecture awareness: the architectural graph that encompasses all the information regarding the connectivity of the qubits within the quantum processing unit (QPU) gets encoded within the model itself. Additionally, the model uses a number of noise parameters (see Table \ref{table:noiseparams}) calibrated from the machine itself, i.e., parameters that express the error rates of the gates, state preparations and measurements as well as the time it takes for the qubits within the QPU to decohere and dephase. Each noise parameter is unique and corresponds to each qubit individually or pair of qubits.
\bgroup
\def\arraystretch{1.2}%
\setlength\tabcolsep{0.1cm}
\begin{table}[!t]
\centering
\begin{tabular}{c|c|c}
\hline
Parameter & Error Type & No. Parameters \\ \hline \hline
$p_{r}$ & Gate error rates & $r$ \\
$p_{m}$ & State preparation error rates & $m$ \\
$p_{s}$ & Measurement error rates & $s$ \\
$T_{1}$ & Thermal relaxation times & $n$ \\
$T_{2}$ & Dephasing times & $n$ \\ \hline
\end{tabular}
\caption{The noise parameters and number of noise parameters for each type of error within the UNM; $n$ is the number of qubits in the system, $m$ is the number of qubits that are measured, $s$ is the number of state preparations that occur and $r$ is the number of distinct types of gates implemented in the architecture, each considered once per qubit or pair of qubits.}
\label{table:noiseparams}
\end{table}
\egroup
\subsection{The Quantum Algorithms}\label{subsec:qalg}
Here we give a quick overview of the quantum algorithms we use for benchmarking, i.e., (i) discrete-time quantum walks (DTQW), (ii) continuous-time quantum walks (CTQW), (iii) Pauli decomposition of the CTQW Hamiltonian (PD) (iv) quantum phase estimation (QPE) and (v) quantum search (QS). These algorithms are chosen according to three major characteristics that are interesting for benchmarking:
\begin{itemize}
\item \textbf{Scalability.} The algorithm should be able to scale up (or down) and run on increasingly larger quantum systems.
\item \textbf{Predictability.} The algorithm should produce a result that is easily predictable. An important addition to predictability is \textit{noise susceptibility}: the algorithm should provide a result whose distortion under the effects of noise is easily distinguishable from the ideal evolution.
\item \textbf{Quantum advantage.} The algorithm should provide a computational speed-up over its classical counterpart or, in other words, represent a possibly relevant real-world application.
\end{itemize}
\paragraph{Discrete-time Quantum Walks.} Quantum walks (DTQW) are the quantum mechanical analogue of a classical random walk on a graph or a lattice \cite{Aharonov-2000,KGeorgo-2020-rots,Kempe-2003}. They exhibit intrinsic properties that render their evolution easily predictable and highly susceptible to noise, making them an ideal candidate for benchmarking. First of all, discrete-time quantum walks exhibit modular behaviour \cite{KGeorgo-2020-rots,Reitzner-2011-mod}. This characteristic describes the modular relationship between the parity of the number of coin-flips of the walk, the initial state and the current position of the walker, a property that gets violated in a noisy environment \cite{KGeorgo-2020-rots}. Secondly, quantum walks propagate quadratically further than classical random walks \cite{Aharonov-2000,Szegedy} thus showing clear quantum advantage over their classical counterparts.
Finally, quantum walks are a highly scalable process. The size of the state-space of a quantum walk (i.e., the number of states that the walk traverses, represented by the number of qubits in the relevant register) can easily increase to match the size of the quantum computer we are interested in benchmarking.
\paragraph{Continuous-time Quantum Walks.} Continuous-time quantum walks (CTQW) were first introduced by Fahri and Gutman in \cite{Fahri-1998-ctqw}. This algorithm, much like the DTQWs, have an easily predictable quantum evolution that is highly susceptible to quantum noise, but exhibit very different characteristics to the discrete case. First of all, the CTQW evolution is determined by a Hamiltonian, $H$, instead of a coin-flip and is driven by a unitary of the form $e^{-iHt}$. Unlike the DTQWs, continuous-time quantum walks do not exhibit modular behaviour. Although like DTQWs they feature a quadratic increase in the walker's propagation \cite{Chakra-2020,Ambainis-2020}.
Finally, CTQWs are an easily scalable process as adding qubits to circuit can scale up the size of the state-space. To the best of our knowledge, this is the first work that uses a continuous-time quantum algorithm to benchmark the performance of a digital quantum computer. There are a few ways that a continuous-time quantum walk can be implemented on a gate-based (i.e., discrete) quantum computer, for example simulating the CTQW evolution using a DTQW \cite{Childs-2009-CTQW} or approximating the CTQW through a decomposition of the unitary $e^{-iHt}$ to a sequence of universal gates. Within this paper, the latter approach is utilized.
\paragraph{Pauli Decomposition of CTQW Hamiltonian.} Decomposition of quantum Hamiltonians to a set of universal gates is a well-studied area of research \cite{Lloyd-1996-dec,Childs-2011-dec,Suzuki-1992-dec,Aharonov-2003-dec,Hedge-2016-Pdec}. For this paper, we are interested in decomposing the Hamiltonian of the CTQW using the well-known set of Pauli matrices as the universal gate set. A further detailed analysis of this process is presented in the following Section \ref{subsec:paulidec}.
This procedure, often called \textit{Hamiltonian simulation}, adheres to the criteria for a good benchmarking process via the algorithm it decomposes. In other words, since the CTQW is suitable for benchmarking, so is the circuit that implements the decomposition of the CTQW Hamiltonian. Furthermore, it provides added value to this research since one can evaluate the performance of the quantum computer when executing the Hamiltonian simulation of a quantum process (i.e., CTQW), as well as compare the decomposition with the original algorithm.
\paragraph{Quantum Phase Estimation.} The quantum phase estimation (QPE) algorithm is used to estimate the phase (or eigenvalue) of an eigenvector of a unitary operator. More precisely, given an arbitrary quantum operator $U$ and a quantum state $\ket{\psi}$ such that $U\ket{\psi}=e^{2i\pi\theta}\ket{\psi}$, the algorithm estimates the value of $\theta$, given an approximation error \cite{Kitaev-qpe,qpe-2019,Oh-2019}.
Within this paper we exploit three characteristics of QPE that make it interesting for benchmarking. First of all, it is a scalable algorithm as further accuracy of the result (i.e., the estimated phase) can be obtained by increasing the number of qubits in the system. Furthermore, the result of the QPE is predictable and highly susceptible to quantum noise. Finally, QPE offers clear quantum advantage, achieving an exponential speed-up over known classical methods, rendering the algorithm one of the most important subroutines in quantum computing and serving as the building block of major quantum algorithms, like Shor's \cite{Shor-1997} or the HHL algorithms \cite{Harrow-2009}.
\paragraph{Quantum Search.} Grover's algorithm \cite{Grover-1996} describes a process for searching for a specific item within a database. Within this paper we use quantum search (QS) to look for a specific number $s$ within a set of numbers $\mathcal{S}=\{0,\dots,2^{n-1}\}$, where $n$ is the number of qubits within the quantum system that participate in the computation.
Quantum search represents an ideal algorithm for benchmarking quantum computers. The algorithm can scale up to search for an item within a larger database simply by adding qubits to the relevant quantum register. The result is easily predictable, as it is simply the item (or number, in our case) sought, as well as highly susceptible to noise (for example, the wrong result might appear due to noise). Additionally, QS can speed up an unstructured search problem quadratically, thus making it a very alluring application for quantum computers. Finally, Grover's algorithm can serve as a general trick or subroutine to obtain quadratic runtime improvements for a variety of other algorithms through what is called amplitude amplification \cite{Brassard-2002}.
\subsection{Pauli Decomposition of CTQW Hamiltonian}\label{subsec:paulidec}
In general, an arbitrary Hamiltonian $H$ of size $N\times N$, where $N=2^{n}$ and $n$ the number of qubits in the system, can be decomposed into a sequence of Pauli operators of the set $S=\{ \sigma_{I}, \sigma_{X}, \sigma_{Y}, \sigma_{Z} \}$ with well-known matrix representation as
\begin{equation}
H = \sum_{i_{1},\dots,i_{N}=I,x,y,z} \alpha_{i_{1},\dots,i_{N}} \left( \sigma_{i_{1}}\otimes\dots\otimes\sigma_{i_{N}} \right), \label{eq:pd}
\end{equation}
where
\begin{equation*}
\alpha_{i_{1},\dots,i_{N}} = \frac{1}{N}\operatorname{tr}\left[ \left( \sigma_{i_{1}}\otimes\dots\otimes\sigma_{i_{N}} \right)\cdot H \right].
\end{equation*}
For this paper we implement a CTQW on an $N$-cycle with Hamiltonian defined as $H_{\text{qw}}=\gamma A=\frac{1}{d}A=\frac{1}{2}A$, where $\gamma=1/d$ is the hopping rate between the two adjacent nodes in the cycle (i.e., node degree $d=2$) and $A$ is the adjacency matrix. Note that, for the physical purposes of this evolution, the rate $\gamma_{n|i}$ has dimensions $1/t$, where $t$ is the time duration of the CTQW evolution, the Hamiltonian has the appropriate energy dimensions and, for simplicity, $\hbar=1$.
The next step is to construct the unitary evolution operator from the Hamiltonian. This can be done by exponentiating the Hamiltonian as $e^{-iHt}$ where $H$ is a sum of terms of the form of equation \eqref{eq:pd}. It is important here to consider two things. First of all, during matrix exponentiation, a decomposition of the form $e^{-i(H_{1}+H_{2})t}=e^{-iH_{1}t}e^{-iH_{2}t}$, where $H_{1}$ and $H_{2}$ are Hermitian operators, is possible if $H_{1}$ and $H_{2}$ commute, i.e., $H_{1}H_{2}-H_{2}H_{1}=0$. This rule is naturally expanded for more than two matrices on the exponent.
Secondly, in the case that not all matrices in the exponent commute, the unitary operator resulting from the Hamiltonian exponentiation needs to be decomposed using the Lie-product formula \cite{Lloyd-1996-dec} as
\begin{equation}
e^{-i(H_{1} + H_{2} + \dots)t} \approx \left( e^{-iH_{1}t/r}e^{-iH_{2}t/r}\dots \right)^{r}
\end{equation}
where, for our case, $H=\sum_{i} H_{i}$ is the Pauli decomposition of the Hamiltonian to a sequence of Hermitian terms. This formula will create an approximation of the Hamiltonian with bounded error depending on $r$ \cite{Lloyd-1996-dec,Suzuki-1992-dec}. To ensure that the Pauli Hamiltonian decomposition exhibits error at most $\epsilon$, the bound $r$ can be taken as \cite{Lloyd-1996-dec}
\begin{equation*}
r = \mathcal{O}\left( (||H||t)^{2}/ \epsilon \right),
\end{equation*}
where $||H||$ is the norm of the Hamiltonian $H$ and $t$ is the continuous-time duration of the quantum evolution.
Thus, using the above methodology, we can now decompose the unitary that describes the continuous-time evolution of the CTQW (i.e., $e^{-iHt}$) in a sequence of Pauli operations that \textit{approximate} said evolution. As the Pauli gates form a universal gate set, the resulting quantum circuit can be implemented on the gate-based quantum processors.
\subsection{The Quantum Computers and Circuits}\label{subsec:qcs}
Within this paper we are interested in benchmarking three computers, the $5$-qubit Bogota, Santiago and the $7$-qubit Casablanca machine. They all exhibit a quantum volume $V_{Q}=32$ \cite{IBMQExp,Moll-2018}.
\paragraph{Quantum computer architectures.} Throughout our research we find that it is essential for the benchmarking procedure to be architecturally aware. This is also reflected by our choice of the UNM, an architecture aware noise model \cite{KGeorgo-2020-UNM}. Figure \ref{fig:architectures} shows the qubit connectivity of the quantum computers we benchmark in this paper.
\begin{figure}[!t]
\begin{tabular}{c}
\includegraphics[width=8.5cm]{ibmqBogotaSantiago.png} \\
(a) \\ [6pt]
\includegraphics[width=5.2cm]{ibmqCasablanca.png} \\
(b) \\[6pt]
\end{tabular}
\caption{Architectural graphs of the (a) IBMQ $5$-qubit Bogota and Santiago machines and (b) the IBMQ $7$-qubit Casablanca machine. A node in the graph represents a physical qubit whereas an edge represents a connection between a pair of qubits.}
\label{fig:architectures}
\end{figure}
\paragraph{Quantum circuits and characteristics.} For the implementation of quantum walks, we use a gate efficient approach that is based on inverter gates, as shown in \cite{Douglas-2009-EffWalk}. The QPE circuit is based on \cite{QPECircuit} and heavily relies on quantum Fourier transform (and its inverse) \cite{Nielsen_Chuang_2010} to estimate the relevant eigenvalue. For QS we make use of two approaches to the implementation, one with ancilla qubits (QSa) and one without (QSn) \cite{Karlsson_2018}.
For the continuous-time quantum walk, the Hamiltonian is automatically implemented by the Qiskit API when submitted for execution on the quantum computer. The Pauli decomposition (PD) of the Hamiltonian can be easily implemented on the quantum computers as it already maps the continuous-time Hamiltonian to a discrete basis gate set decomposition.
We identify four quantum circuit characteristics that are of interest for benchmarking:
\begin{enumerate}[i]
\item the number of gates in the circuit,
\item the number of active qubits, or qubits that are utilized by the quantum circuit (also called workspace),
\item the depth of the circuit, i.e., the longest path between the start of the circuit and a measurement gate and
\item the runtime of the circuit on the quantum computer. Table \ref{table:charactQASMcirc} shows those characteristics of the quantum circuits that implement each quantum algorithm.
\end{enumerate}
\begin{table*}[!t]
\begin{tabular}{c}
\bgroup
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\centering
\begin{tabular}{c|c|c|c|c}
\hline
Machine & No. Gates & Size of Workspace & Depth & QC Runtime $\pm$ s.d. ($\text{ms}$) \\ \hline \hline
$5$-qubit Bogota & $47$ & $3$ & $35$ & $2.41\pm 0.03$ \\
$5$-qubit Santiago & $47$ & $3$ & $35$ & $2.34\pm 0.02$ \\
$7$-qubit Casablanca & $47$ & $3$ & $35$ & $2.54\pm 0.05$ \\ \hline
\end{tabular}
\egroup \\ [6pt] (a) Discrete-time quantum walk circuit characteristics. \\ [6pt]
\bgroup
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\centering
\begin{tabular}{c|c|c|c|c}
\hline
Machine & No. Gates & Size of Workspace & Depth & QC Runtime $\pm$ s.d. ($\text{ms}$) \\ \hline \hline
$5$-qubit Bogota & $19$ & $2$ & $13$ & $2.34\pm 0.04$ \\
$5$-qubit Santiago & $19$ & $2$ & $13$ & $3.01\pm 0.07$ \\
$7$-qubit Casablanca & $19$ & $2$ & $13$ & $2.62\pm 0.01$ \\ \hline
\end{tabular}
\egroup \\ (b) Continuous-time quantum walk circuit characteristics. \\ [6pt]
\bgroup
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\centering
\begin{tabular}{c|c|c|c|c}
\hline
Machine & No. Gates & Size of Workspace & Depth & QC Runtime $\pm$ s.d. ($\text{ms}$) \\ \hline \hline
$5$-qubit Bogota & $243$ & $2$ & $183$ & $2.42\pm 0.04$ \\
$5$-qubit Santiago & $243$ & $2$ & $183$ & $2.28\pm 0.06$ \\
$7$-qubit Casablanca & $243$ & $2$ & $183$ & $3.76\pm 0.04$ \\ \hline
\end{tabular}
\egroup \\ (c) Pauli decomposition of CTQW circuit characteristics. \\ [6pt]
\bgroup
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\centering
\begin{tabular}{c|c|c|c|c}
\hline
Machine & No. Gates & Size of Workspace & Depth & QC Runtime $\pm$ s.d. ($\text{ms}$) \\ \hline \hline
$5$-qubit Bogota & $93$ & $4$ & $66$ & $2.54\pm 0.01$ \\
$5$-qubit Santiago & $97$ & $4$ & $72$ & $2.26\pm 0.03$ \\
$7$-qubit Casablanca & $100$ & $4$ & $75$ & $2.68\pm 0.06$ \\ \hline
\end{tabular}
\egroup \\ (d) Quantum phase estimation circuit characteristics. \\ [6pt]
\bgroup
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\centering
\begin{tabular}{c|c|c|c|c}
\hline
Machine & No. Gates & Size of Workspace & Depth & QC Runtime $\pm$ s.d. ($\text{ms}$) \\ \hline \hline
$5$-qubit Bogota & $497$ & $4$ & $358$ & $2.70\pm 0.08$ \\
$5$-qubit Santiago & $479$ & $4$ & $343$ & $2.56\pm 0.04$ \\
$7$-qubit Casablanca (ancilla) & $788$ & $6$ & $503$ & $2.81\pm 0.04$ \\
$7$-qubit Casablanca (no ancilla) & $465$ & $4$ & $336$ & $2.74\pm 0.03$ \\ \hline
\end{tabular}
\egroup \\ (e) Quantum search circuit characteristics. \\
\end{tabular}
\caption{Quantum circuit characteristics for the five quantum circuits. No. gates and size of workspace are the number of gates and active qubits in the circuit respectively; depth of the circuit is the longest path between the start of the circuit and a measurement gate; QC runtime is the approximate average execution time of the circuit on the quantum computer along with standard deviation (s.d.).}
\label{table:charactQASMcirc}
\end{table*}
\subsection{Benchmark Indicators: Hellinger Distance}\label{subsec:hd}
The end results of the benchmarking process will be in the form of a comparison of the quantum computer output distribution with the distributions resulting from the unified noise model simulations of each machine and the ideal evolution.
To compare the probability distributions and generate the benchmarks, we use the Hellinger distance (HD) \cite{Jin_2018}.
\begin{definition}[Hellinger distance]\label{def:hd}
For probability distributions $P=\{p_{i}\}_{i\in[s]}$, $Q=\{q_{i}\}_{i\in[s]}$ supported on $[s]$, the Hellinger distance between them is defined as
\begin{equation}
h(P,Q) = \frac{1}{\sqrt{2}} \sqrt{ \sum_{i=1}^{k} \left( \sqrt{p_{i}} - \sqrt{q_{i}} \right)^{2} } \label{eq:hd}.
\end{equation}
\end{definition}
The Hellinger distance is a metric satisfying the triangle inequality. It takes values between $0$ and $1$ (i.e. $h(P,Q)\in[0,1]$) with $0$ meaning that the two distributions are equal. Additionally, it is easy to compute, easy to read and it does not depend on the probability distributions having the same support. The last property is particularly useful since in many ideal output distribution of quantum circuits the probability mass is concentrated on a few states.
\section{Framework for Program Benchmarking}\label{sec:framework}
The desired benchmarks correspond to comparisons (via the Hellinger distance) between the evolution of the quantum system on the computer, the noisy simulation and the ideal case. To achieve a more memorable notation, in the following analysis the symbol $q$ denotes the quantum computer, $i$ the ideal evolution and $n$ the noisy simulation. For example, the subscript $q|n$ denotes a value that corresponds to the difference between the quantum computer evolution ($q$) and the noisy evolution ($n$). We define three distances of interest, or otherwise, three benchmark metrics, as follows.
\begin{definition}[alpha benchmark]\label{def:alpha}
The Hellinger distance between the probability distribution of the quantum computer evolution, $Q$, and the ideal distribution, $D$, namely $h_{\text{id}}(Q,D)$, is the alpha benchmark, with notation $\alpha_{q|i}$:
\begin{equation}
\alpha_{q|i} \equiv h(Q,D). \label{eq:alpha}
\end{equation}
\end{definition}
\begin{definition}[beta benchmark]\label{def:beta}
The Hellinger distance between the probability distribution of the quantum computer evolution, $Q$, and the distribution resulting from the noisy simulations, $N$, namely $h_{\text{nm}}(Q,N)$, is the beta benchmark, with notation $\beta_{q|n}$:
\begin{equation}
\beta_{q|n} \equiv h(Q,N). \label{eq:beta}
\end{equation}
\end{definition}
\begin{definition}[gamma benchmark]\label{def:gamma}
The Hellinger distance between the distribution resulting from the noisy simulations, $N$, and the ideal distribution, $D$, namely $h_{\text{sm}}(N,D)$, is the gamma benchmark, with notation $\gamma_{n|i}$:
\begin{equation}
\gamma_{n|i} \equiv h(N,D). \label{eq:gamma}
\end{equation}
\end{definition}
As mentioned in Section \ref{subsec:hd}, the Hellinger distance is a metric that satisfies the triangle inequality. Hence, since the benchmarks established in the above definitions describe the pairwise Hellinger distances between three probability distributions (the quantum computer, $Q$, the simulated evolution, $N$, and the ideal evolution, $D$), it is possible to derive a relationship between $\alpha_{q|i}$, $\beta_{q|n}$ and $\gamma_{n|i}$ via the triangle inequality, through the following Lemma (whose proof follows simply from the triangle inequality for a metric).
\begin{lemma}[benchmarks triangle inequality]\label{lemma:triangineq}
Given probability distributions $Q$, $N$ and $D$, as established in the benchmark definitions, the pairwise Hellinger distances between those distributions, i.e., $h(Q,D)$, $h(Q,N)$ and $h(N,D)$ follow the triangle inequality:
\begin{equation*}
h(Q,D) \leq h(Q,N) + h(N,D).
\end{equation*}
Thus, the relevant benchmarks will also follow the triangle inequality as:
\begin{equation}
\alpha_{q|i} \leq \beta_{q|n} + \gamma_{n|i} \label{eq:trianineq}
\end{equation}
\end{lemma}
The above Lemma effectively means that, according to the benchmark definitions, the deviation of the quantum computer evolution from the ideal ($\alpha_{q|i}$) will never be greater than the sum of the expected (i.e., simulated) evolution derived by the levels of noise within the machine and the distance between the simulated and ideal distributions ($\beta_{q|n}+\gamma_{n|i}$). In other words, defining the benchmark metrics using the Hellinger distance allows us to quantify the confidence on the estimated level of noise during the execution of the quantum circuit.
Following the definitions and relationship between the benchmarks, we present a framework for program benchmarking quantum computers as the sequence of the six steps below:
\begin{description}
\item[Step 1] \textbf{Select benchmark method(s).} The selection of quantum algorithms that will be used for benchmarking should adhere to the three criteria described in Section \ref{subsec:qalg}: (i) scalability, (ii) predictability and noise susceptibility and (iii) quantum advantage.
\item[Step 2] \textbf{Quantum noise model and simulator.} Select or implement a noise model that approximates the noisy evolution within the quantum machine and a simulator that can execute the noise model.
\item[Step 3] \textbf{Run experiments.} Design and execute a suitable number of experiments of the benchmark method(s) on the quantum computer. This step also includes calibrating the noise parameters that encapsulate the level of noise within the quantum computer at the time of the experiments.
\item[Step 4] \textbf{Simulate the noisy evolution.} Use the noise model in order to simulate the noisy evolution of the quantum computer using the calibrated noise parameters from the time of the experiment, as described in Step 3.
\item[Step 5] \textbf{Simulate the ideal evolution.} This can be done either through simple noise-free simulations or by calculating the probabilities through the quantum statevector.
\item[Step 6] \textbf{Calculate the benchmarks.} The final benchmarks $\alpha_{q|i}$, $\beta_{q|n}$ and $\gamma_{n|i}$ are the Hellinger distances between the quantum computer, the UNM and the ideal evolution in the setting described in Definitions \ref{def:alpha}, \ref{def:beta} and \ref{def:gamma}.
\end{description}
Following the benchmarking framework we can extract meaningful results from a series of comparisons between the benchmark metrics, $\alpha_{q|i}$, $\beta_{q|n}$ and $\gamma_{n|i}$. The $\beta_{q|n}$ benchmark essentially describes how closely the noise model simulates the behaviour of the quantum computer. The $\alpha_{q|i}$ benchmark shows how far the behaviour of the quantum computer falls from the noise-free case thus giving an estimate of the overall computer performance under the effects of noise. From the comparison between the $\beta_{q|n}$ and $\alpha_{q|i}$ benchmarks we can extract valuable information: if $\beta_{q|n}<\alpha_{q|i}$ the noise levels in the quantum computer are closer to the estimated ones from the noise simulations; on the opposite case the computer operates closer to the ideal evolution. In the latter case the quantum computer behaves more efficiently with lower level of noise than expected, thus giving us more confidence regarding the computational result.
The $\gamma_{n|i}$ benchmark, even though it can be used as an estimate of the noise levels in the quantum computer, does not give any relevant information on its own. The value comes when considering the $\gamma_{n|i}$ benchmark together with the $\alpha_{q|i}$ benchmark. First of all, the closer the values of $\alpha_{q|i}$ and $\gamma_{n|i}$ are, the smaller the value of $\beta_{q|n}$ (if $\beta_{q|n}=0$ then $\alpha_{q|i}=\gamma_{n|i}$ and vice versa). Additionally, if $\alpha_{q|i}>\gamma_{n|i}$, then the noise model, and hence the noise parameters, underestimate the levels of noise during the quantum computer evolution, with the opposite being true if $\alpha_{q|i}<\gamma_{n|i}$. Moreover, the absolute difference $\abs{\alpha_{q|i}-\gamma_{n|i}}$ can quantify this noise over- or underestimation. Precisely, the smaller the absolute difference the smaller the error in estimation. This information is useful to the benchmarking process as it further highlights whether the machine is more or less noisy than estimated while also showing the efficiency of the machine calibration techniques.
Finally, using the triangle inequality from Lemma \ref{lemma:triangineq} we can make further interesting remarks. From equation \eqref{eq:trianineq} we find that $\beta_{q|n} \geq \alpha_{q|i} - \gamma_{n|i}$. Furthermore, from the above analysis we know that a comparison between $\alpha_{q|i}$ and $\gamma_{n|i}$ can tell us whether the calibrated noise parameters over- or underestimate the level of noise during the evolution on the quantum computer, and the absolute difference $\abs{\alpha_{q|i}-\gamma_{n|i}}$ can give an indication of the scale of the error in estimation. Importantly, the above inequality does not hold for absolute values when subtraction takes place (i.e., for an inequality of the form $\abs{\beta_{q|n}} \geq \abs{\alpha_{q|i} - \gamma_{n|i}}$ when $\gamma_{n|i}>\alpha_{q|i}$). Considering this, we can interpret the triangle inequality as a measure of the confidence on the calibrated parameters encapsulating a picture of the noise that is accurate enough to provide a good estimation of the quantum evolution, expressed as follows:
\begin{itemize}
\item If $\beta_{q|n}\geq\abs{\alpha_{q|i}-\gamma_{q|i}}$, then the over- or underestimation of the noise is small enough to provide estimates of the quantum evolution with \textit{high confidence}.
\item If $\beta_{q|n}<\abs{\alpha_{q|i}-\gamma_{q|i}}$, then the calibrated parameters generate \textit{low confidence} on the levels of noise.
\end{itemize}
\section{Experiments and Results}\label{sec:results}
\subsection{Experimental Setup}\label{subsec:setup}
For the benchmarking experiments we run the quantum circuits that implement the chosen quantum algorithms. In the case of the DTQW we run one step of the algorithm (i.e., one coin-flip) on a workspace (i.e., number of active qubits used by the circuit) of three qubits with initial state $\ket{0}$, as previous work shows that this configuration is satisfactory for errors to take place and the behaviour of the quantum walk to evolve in a predictable manner \cite{KGeorgo-2020-rots}. For the CTQW and its PD we use a small two-qubit state-space, as the Pauli decomposition gets excessively large for a three-qubit state space or bigger. The algorithms are implemented for a continuous time of $t=3$. Our QPE routine is tailored to estimate a phase of $\theta = 2\pi/3$ using a workspace of four qubits with theoretical probability of success estimated at $0.688$. Finally, the QS implementation performs an unstructured search for the element $\ket{s}=\ket{10}$ (decimal) within a four qubits state-space i.e., a dataset containing numbers $\ket{0}$ to $\ket{15}$. The QS algorithm is run for three iterations and shows a theoretical probability of success estimated at $0.96$. Table \ref{table:initconfig} shows more comprehensively the initial configuration for each algorithm.
\bgroup
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\begin{table*}[!t]
\centering
\begin{tabular}{c|c|c|c|c}
\hline
Algorithm & No. Qubits & Size of Workspace & Iterations/Duration & Probability of Success \\ \hline \hline
DTQW & $2$ & $3$ & $1$ (one coin-flip) & $p_{\text{succ}}=0.5$ in states $\ket{1}$ and $\ket{3}$ \\
CTQW & $2$ & $2$ & $t=3$ & $p_{\ket{2}}=0.99$, $p_{\ket{1}}=p_{\ket{3}}=0.005$ \\
PD & $2$ & $2$ & $t=3$ & $p_{\ket{2}}=0.99$, $p_{\ket{1}}=p_{\ket{3}}=0.005$ \\
QPE & $3$ & $4$ & $1$ & $p_{\text{succ}}=0.688$ in state $\ket{3}$ \\
QSa & $4$ & $6$ & $3$ & $p_{\text{succ}}=0.96$ in state $\ket{10}$ \\
QSn & $4$ & $4$ & $3$ & $p_{\text{succ}}=0.96$ in state $\ket{10}$ \\ \hline
\end{tabular}
\caption{The initial configuration for the quantum walk (DTQW), continuous-time quantum walk (CTQW) and its Pauli decomposition (PD), quantum phase estimation (QPE) and quantum search algorithms with ancilla (QSa) and without ancilla (QSn) for the benchmarking experiments. No. qubits is the number of qubits in the state space, size of workspace is the number of active qubits utilized by the circuit, iterations is the number of repetitions of the quantum circuit and the probability of success is the theoretical probability of the algorithm to give us the correct (or expected) result.}
\label{table:initconfig}
\end{table*}
\egroup
We run each algorithm independently $100{,}000$ times (see Section \ref{subsec:qalg}) on the three chosen computers (see Section \ref{subsec:qcs}) and as a simulation using the UNM (see Section \ref{subsec:unm}) and the ideal simulator \cite{Qiskit} to reproduce the noisy and the ideal quantum evolutions respectively. Finally, we compute the $\beta_{q|n}$, $\alpha_{q|i}$ and $\gamma_{n|i}$ benchmarks, as defined in Section \ref{sec:framework}. The results are shown on Table \ref{table:benchmarks} for each machine and are also visualized in Figure \ref{fig:benchmarks}.
\begin{figure*}[!t]
\begin{center}
\includegraphics[width=12cm]{benchmarks.png} \\
\end{center}
\caption{Visualization of the three benchmarks, $\alpha_{q|i}$, $\beta_{q|n}$ and $\gamma_{n|i}$ for each of the three quantum computers (mapped to the left-hand side $y$-axis) when executing the five quantum algorithms.}
\label{fig:benchmarks}
\end{figure*}
\begin{table*}[!t]
\begin{tabular}{c}
\bgroup
\def\arraystretch{1.2}%
\setlength\tabcolsep{0.4cm}
\centering
\begin{tabular}{c|c|c|c|c|c|c}
\hline
Machine & $\alpha_{q|i}$ & $\beta_{q|n}$ & $\gamma_{n|i}$ & $\abs{\alpha_{q|i}-\gamma_{n|i}}$ & QC Runtime ($\text{sec}$) & Sim. Runtime ($\text{sec}$) \\ \hline \hline
$5$-qubit Bogota & $0.287$ & $0.025$ & $0.291$ & $0.004$ & $240.7$ & $3{,}331.5$ \\
$5$-qubit Santiago & $0.321$ & $0.054$ & $0.362$ & $0.041$ & $234.2$ & $3{,}119.3$ \\
$7$-qubit Casablanca & $0.269$ & $0.076$ & $0.331$ & $0.062$ & $254.5$ & $4{,}626.6$ \\ \hline
\end{tabular}
\egroup \\ [6pt] (a) Quantum walk algorithm on two qubits. \\ [6pt]
\bgroup
\def\arraystretch{1.2}%
\setlength\tabcolsep{0.4cm}
\centering
\begin{tabular}{c|c|c|c|c|c|c}
\hline
Machine & $\alpha_{q|i}$ & $\beta_{q|n}$ & $\gamma_{n|i}$ & $\abs{\alpha_{q|i}-\gamma_{n|i}}$ & QC Runtime ($\text{sec}$) & Sim. Runtime ($\text{sec}$) \\ \hline \hline
$5$-qubit Bogota & $0.139$ & $0.096$ & $0.234$ & $0.095$ & $234.4$ & $3{,}026.9$ \\
$5$-qubit Santiago & $0.154$ & $0.035$ & $0.184$ & $0.030$ & $301.7$ & $2{,}905.7$ \\
$7$-qubit Casablanca & $0.247$ & $0.044$ & $0.239$ & $0.008$ & $262.1$ & $3{,}035.2$ \\ \hline
\end{tabular}
\egroup \\ (b) Continuous-time quantum walk algorithm on two qubits. \\ [6pt]
\bgroup
\def\arraystretch{1.2}%
\setlength\tabcolsep{0.4cm}
\centering
\begin{tabular}{c|c|c|c|c|c|c}
\hline
Machine & $\alpha_{q|i}$ & $\beta_{q|n}$ & $\gamma_{n|i}$ & $\abs{\alpha_{q|i}-\gamma_{n|i}}$ & QC Runtime ($\text{sec}$) & Sim. Runtime ($\text{sec}$) \\ \hline \hline
$5$-qubit Bogota & $0.436$ & $0.041$ & $0.467$ & $0.031$ & $242.7$ & $11{,}764.9$ \\
$5$-qubit Santiago & $0.514$ & $0.084$ & $0.465$ & $0.049$ & $228.5$ & $11{,}618.6$ \\
$7$-qubit Casablanca & $0.577$ & $0.039$ & $0.612$ & $0.035$ & $376.4$ & $12{,}861.3$ \\ \hline
\end{tabular}
\egroup \\ (c) Pauli decomposition of the CTQW algorithm on two qubits. \\ [6pt]
\bgroup
\def\arraystretch{1.2}%
\setlength\tabcolsep{0.4cm}
\centering
\begin{tabular}{c|c|c|c|c|c|c}
\hline
Machine & $\alpha_{q|i}$ & $\beta_{q|n}$ & $\gamma_{n|i}$ & $\abs{\alpha_{q|i}-\gamma_{n|i}}$ & QC Runtime ($\text{sec}$) & Sim. Runtime ($\text{sec}$) \\ \hline \hline
$5$-qubit Bogota & $0.469$ & $0.177$ & $0.351$ & $0.118$ & $254.2$ & $6{,}287.1$ \\
$5$-qubit Santiago & $0.488$ & $0.144$ & $0.369$ & $0.119$ & $226.8$ & $6{,}304.2$ \\
$7$-qubit Casablanca & $0.524$ & $0.267$ & $0.351$ & $0.173$ & $268.3$ & $7{,}741.1$ \\ \hline
\end{tabular}
\egroup \\ (d) Quantum phase estimation algorithm on three qubits. \\ [6pt]
\bgroup
\def\arraystretch{1.2}%
\setlength\tabcolsep{0.28cm}
\centering
\begin{tabular}{c|c|c|c|c|c|c}
\hline
Machine & $\alpha_{q|i}$ & $\beta_{q|n}$ & $\gamma_{n|i}$ & $\abs{\alpha_{q|i}-\gamma_{n|i}}$ & QC Runtime ($\text{sec}$) & Sim. Runtime ($\text{sec}$) \\ \hline \hline
$5$-qubit Bogota & $0.754$ & $0.014$ & $0.752$ & $0.002$ & $270.4$ & $22{,}834.3$ \\
$5$-qubit Santiago & $0.751$ & $0.051$ & $0.738$ & $0.013$ & $256.7$ & $21{,}923.2$ \\
$7$-qubit Casablanca (ancilla) & $0.760$ & $0.040$ & $0.752$ & $0.008$ & $280.1$ & $36{,}425.7$ \\
$7$-qubit Casablanca (no ancilla) & $0.763$ & $0.061$ & $0.738$ & $0.025$ & $274.1$ & $22{,}889.9$ \\ \hline
\end{tabular}
\egroup \\ (e) Quantum search algorithm on four qubits. \\
\end{tabular}
\caption{Results from benchmarking the three machines for each of the five algorithms. The benchmark indicators $\alpha_{q|i}$, $\beta_{q|n}$ and $\gamma_{n|i}$ are the HD between the quantum computer and the UNM, the quantum computer and the ideal distribution and the UNM and the ideal distribution respectively. QC runtime is the cumulative execution time in seconds for $100{,}000$ iterations of each algorithm on the respective machine; Sim. runtime is the time it takes to simulate for $100{,}000$ iterations the behaviour of each machine when executing the algorithms.}
\label{table:benchmarks}
\end{table*}
\subsection{Results}\label{subsec:res}
The results of the benchmark experiments are presented in Table \ref{table:benchmarks} and below for each of the algorithms used.
It is important at this point to emphasize an advantage of the structure presented for program benchmarking quantum computers. The use of the Hellinger distance to define the three benchmarks ($\alpha_{q|i}$, $\beta_{q|n}$ and $\gamma_{n|i}$) allows us to perform a comparison of the performance of different quantum algorithms on the same basis, i.e., a dimensionless quantity. In other words, through the proposed framework, apart from measuring the efficiency of a quantum processor, we can gather additional information and compare the efficiency of circuit implementations for the selected algorithms. This is done here for the two approaches on implementing the CTQW on a gate-based computer, one that is done through the IBMQ API and one that use the Pauli decomposition of the CTQW Hamiltonian (see Section \ref{subsec:paulidec}).
\subsubsection{Discrete-time Quantum Walk Benchmarking}\label{subsubsec:dtqwBench}
The DTQW implementation produces a circuit with a relatively small depth and number of gates. Our $\beta_{q|n}$ benchmark measures how closely a quantum computer operates to the expected noise levels (indicated by the noise parameters and simulated by the UNM). Thus, in the small circuit case of the DTQW, the Bogota machine operates much closer to the expected evolution than the other computers with the smallest $\beta_{q|n}$.
The $\alpha_{q|i}$ and $\gamma_{n|i}$ benchmarks lead to a coherent picture of the overall performance of the quantum computers. In the DTQW case, the $\alpha_{q|i}$ benchmark is relatively small compared to the larger algorithms. This means that the quantum computers are not as erroneous as in the larger circuit cases, an expected result. A comparison between the $\alpha_{q|i}$ and $\gamma_{n|i}$ benchmarks shows that $\alpha_{q|i} < \gamma_{n|i}$ for all the machines, indicating that the the calibrated parameters overestimate the levels of noise during the evolution of the circuit. Finally, $\beta_{q|n}\geq\abs{\alpha_{q|i}-\gamma_{q|i}}$ for all quantum computers, which implies high confidence of our noise level estimates.
\subsubsection{Continuous-time Quantum Walk Benchmarking}\label{subsubsec:ctqwBench}
The CTQW circuit is the smallest circuit we use for benchmarking, which leads us to expect that the machine evolution will be reasonably close to the ideal case for this circuit. Evidently, the $\alpha_{q|i}$ benchmark is by far the smallest for the CTQW algorithm on all three machines. As $\alpha_{q|i}<\gamma_{n|i}$ for the Bogota and Santiago machines, we can say that the noise parameters overestimate the noise for those two computers, whereas they slightly underestimate the noise for the Casablanca computer.
For the CTQW, the Santiago machine operates closer to the expected levels of noise according to its error rates, with the smallest $\beta_{q|n}$. Similarly to the DTQW, we are confident in the calibrated noise as $\beta_{q|n}\geq\abs{\alpha_{q|i}-\gamma_{q|i}}$.
\subsubsection{Pauli Decomposition Benchmarking}\label{subsubsec:pdBench}
Unlike the CTQW circuit, its Pauli decomposition is a large and quite deep circuit, which means that we should expect a very noisy evolution. The small $\beta_{q|n}$ values show that the quantum computers operate relatively close to expectations.
As shown by the large values of $\alpha_{q|i}$ for all quantum computers, the PD circuit is quite error-prone compared to the two previous cases of the quantum walk. Thus, an important conclusion we can safely draw is that, for a two-qubit continuous-time quantum walk, the Pauli decomposition of its Hamiltonian leads to a more complex and noisier circuit than the decomposition to base gates done automatically through the IBMQ API.
The $\gamma_{n|i}$ benchmarks show that the parameters overestimate the noise of the Bogota and Casablanca machines and underestimate the noise within the Santiago computer. Furthermore, $\beta_{q|n}\geq\abs{\alpha_{q|i}-\gamma_{q|i}}$ inspires high-confidence on our noise estimates. This result is important as, through our benchmarks, we produce evidence on the performance of two different techniques for implementing the same algorithm.
\subsubsection{Quantum Phase Estimation Benchmarking}\label{subsubsec:dqpeBench}
Moving on, the QPE circuit is slightly more complex than the DTQW circuit. An overall bigger $\beta_{q|n}$ benchmark on every machine indicates that the QPE circuit execution deviates slightly more from the expected evolution compared to the other algorithms. This could be the result of random fluctuations of the calibrated parameters.
Following the $\alpha_{q|i}$ benchmark, we can extract similar results for the quantum computers as the PD circuit benchmarks, finding that the larger values indicate that the quantum computer performance is hindered by the size of the circuit. Interestingly, we see that $\alpha_{q|i}>\gamma_{n|i}$ for all the quantum computers, showcasing that the calibrated parameters underestimate the noise and we can expect noisier results from all the quantum computers for computations of similar size. We have high-confidence of our calibrated parameters as $\beta_{q|n}\geq\abs{\alpha_{q|i}-\gamma_{q|i}}$ for all experiments.
\subsubsection{Quantum Search Benchmarking}\label{subsubsec:qsBench}
Finally, the quantum search circuit represents the largest implemented algorithm and thus, we expect it to also be the noisiest. Following the above methodology, we find that the $\beta_{q|n}$ benchmark shows the smallest values for each machine, indicating that the quantum computers operated close to the predicted evolutions. The $\alpha_{q|i}$ and $\gamma_{n|i}$ benchmarks are quite close, increasing our confidence on the estimated noise levels. Nevertheless, it is safe to conclude that the machines in the case of the QS exhibit intense levels of noise, as the $\alpha_{q|i}$ benchmarks are very large.
A quick comparison between the $\alpha_{q|i}$ and $\gamma_{n|i}$ shows that the calibrated parameters offer a very good picture of the noise within the quantum computer in this experiment.
\subsubsection{Analysis}\label{subsubsec:conclBenc}
A comparison of the $\alpha_{q|i}$ and $\gamma_{n|i}$ benchmarks shows that when executing the DTQW and CTQW circuits, the machines are less error-prone than the UNM and calibrated parameters imply, with the opposite being true for the PD, QS and QPE. This informs us that the machines are better at executing small circuits, an expected result. Additionally, a larger numerical difference between $\alpha_{q|i}$ and $\gamma_{n|i}$ indicates a more efficient computer behaviour. For example, Casablanca is the most efficient in terms of noise for the DTQW case as $\alpha_{q|i}<\gamma_{n|i}$ with the biggest difference. Similarly, in the QS case the Casablanca machine is the least efficient with $\alpha_{q|i}>\gamma_{n|i}$. Here, as we use two different implementations of the QS algorithm, we can also compare them with each other. The benchmarks show that the QS circuit with ancilla is closer to the expected evolution by the UNM, shown by $\beta_{q|n}$, and to the ideal evolution, shown by $\alpha_{q|i}$, although not by much, shown by $\alpha_{q|i}$ versus $\gamma_{n|i}$.
The results from the above analysis are crucial as they can lead towards the selection of a machine appropriate to specific circuit needs. Additionally, we provide a comparison between the continuous-time quantum walk circuit and its Pauli decomposition. This fact is clearly reflected on the benchmarks, as the values of $\alpha_{q|i}$ are much lower for the CTQW circuit, while both circuits operate within expectations (small $\beta_{q|n}$ values) and are not massively over- or underestimated by their noise parameters (small $|\alpha_{q|i}-\gamma_{n|i}|$). Thus, we can easily conclude that, for the two-qubit case, a Pauli decomposition of the CTQW Hamiltonian leads to a less efficient circuit.
Overall, we can concentrate the general benchmarking results as follows. In the smallest circuits (i.e., CTQW and DTQW case) the $\beta_{q|n}$ benchmarks show how close the computers operate to the expected levels of noise. The Casablanca machine shows the best $\alpha_{q|i}$ benchmark, i.e., it is the closest to the ideal evolution, closely followed by Bogota and with Santiago being the furthest away. The UNM and the calibrated parameters always overestimate the noise in this case as $\alpha_{q|i}<\gamma_{n|i}$, thus showing that all the quantum computers exhibit relatively low levels of noise when executing small circuits. In the slightly deeper circuit of the QPE, the Bogota machine outperforms the others, followed by Santiago and Casablanca. In this case though the $\beta_{q|n}$ benchmark indicates slightly bigger deviation from the expected level of noise which is biggest on the Casablanca machine. The UNM and calibrated parameters always underestimate the noise in the quantum computer as $\alpha_{q|i}>\gamma_{n|i}$, meaning that the quantum computers are more error-prone. Lastly, in the largest QS and PD circuits, the benchmarks indicate that all the machines operate close to the noise model with low $\beta_{q|n}$ benchmarks and exhibit very low performance as the $\alpha_{q|i}$ benchmarks are large, getting close to $1$ for the QS. The noise model slightly underestimates the noise, but in this case of very deep circuits, the benchmarks show that the machines will not produce any meaningful results, an expected outcome.
A visual representation of the comparison between the distributions resulting from the quantum computer executions and the ideal simulations for each quantum algorithm are shown in Appendix \ref{ap:benchcomp}. A further comparison between the quantum computer and the individual UNM distributions for each machine is given in Appendix \ref{ap:comp}.
\section{Discussion and Conclusions}\label{sec:concl}
In this paper we have presented an approach to benchmarking quantum computers using scaling, high-level quantum algorithms considered as attractive ``real-world" problems. We have defined three benchmark metrics, each highlighting different aspects of the machine's efficiency either as a standalone or through comparisons between them. Each benchmark metric describes the difference between two quantum evolutions and together they follow the triangle inequality.
In order to better present the main characteristics of our benchmarks, we streamline the discussion as a comparison with the vastly used metric of \textit{quantum volume} \cite{Moll-2018}, which quantifies the expected size of a circuit that can be reliably run on a quantum computer. In contrast, our program benchmarks describe the performance of the quantum computer when running a specific circuit itself. This approach has advantages and disadvantages over architecture-neutral benchmarks. First of all, our benchmark metrics show the exact performance of a QPU (and in extend, the quantum computer itself) when running the quantum circuit. Additionally, they highlight the difference with an expected (noisy) simulated evolution and an ideal (noise-free) simulated evolution, a result that better identifies the weaknesses of the machine in a more structural manner. More specifically, the metrics allow us to realise the manner and intensity that the computer deviates from the simulated evolutions. Finally, all three machines we benchmark use the same QPU technology and exhibit the same quantum volume of $32$. On the other hand, our benchmarking process exhibits different metrics and results for each quantum computer. Thus, our benchmarks capture the performance of each QPU more thoroughly and provide us with a more detailed representation of their performance.
On the con side, our benchmarking process is slower compared to the calculation of the quantum volume. This is an expected outcome as we need to run a number of experiments on the quantum computer as well as the noisy simulations. Moreover, the flip-side of the architecture-specific nature of our metrics dictates that each computer will exhibit different benchmarks when executing different algorithms.
Thus, we can conclude that our architecture-specific program benchmarks showcase the performance of a quantum computer in a ``real-world" environment, as well as highlight the performance when running a specific algorithm and carry out meaningful comparisons between related circuits (e.g., the CTQW vs the Pauli decomposition of its Hamiltonian). Additionally, we can gather information on whether the calibrated parameters over- or underestimate the levels of noise during the quantum evolution. On the other hand, architecture-neutral benchmarks like the quantum volume are more generic and excel at showcasing the limitations of QPUs when running arbitrary quantum circuits.
It is also noteworthy that the benchmarking experiments carried out during this work are not suitable for a comparative analysis regarding the efficiency of each qubit topology (see Figure \ref{fig:architectures}). One reason for this is the relatively small size and limited flexibility of the computers themselves. Nevertheless, when benchmarking larger quantum computers in the future, we believe that using our framework, one could focus, for example, on searching which architectures are more efficient for a specific task.
In conclusion, our work has shown that using quantum algorithms to benchmark quantum computers in a well-structured environment can stress different aspects and very informatively highlight the performance of a quantum computer. Additionally, it provides a way to compare the efficiency of different circuits used to implement the same task. To the best of our knowledge, this is the first work that uses a continuous-time quantum algorithm to benchmark the performance of a digital quantum machine. The results show that, for small state spaces of the continuous-time quantum walk, the Pauli decomposition does not produce an efficient circuit. This result is expected as the complexity of a Hamiltonian operating on two qubits is very small.
\section{Acknowledgements}
This work was supported by the Engineering and Physical Sciences Research Council, Centre for Doctoral Training in Cloud Computing for Big Data, United Kingdom [grant number EP/L015358/1].
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{
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\section{Introduction}
This paper is centered on the area of \emph{distributed graph algorithms} and provides new methods and tools for developing improved \emph{deterministic} distributed algorithms.
It has been a central, well-known, and well-studied theme in this area that, for many of the graph problems of interest, known randomized algorithms outperform their deterministic counterparts. Concretely, the randomized variants have been much faster and/or achieved better output properties, e.g., approximation factors. As a prominent example, for several of the key problems of interest---including maximal independent set, maximal matching, $\Delta+1$ vertex coloring---we have known $O(\log n)$ round randomized algorithms since the 1986 work of Luby~\cite{luby86}. In contrast, developing even $\mathrm{poly}(\log n)$-time deterministic algorithms for many of these problems remained open for nearly four decades. See for instance the 2013 book of Barenboim and Elkin~\cite{barenboimelkin_book} which lists numerous such open questions. Only very recently, $\mathrm{poly}(\log n)$-time deterministic algorithms for these problems were developed~\cite{rozhonghaffari20, GGR20, chang2021strong, GhaffariK21, elkin2022deterministic}. However, currently, these deterministic algorithms are still quite far from their randomized counterparts.
In this paper, we focus on two of the most central tools in developing deterministic algorithms for local graph problems, namely \emph{network decompositions} and \emph{hitting sets}, and we present significantly improved deterministic distributed constructions of these tools. From a technical perspective, our novelty is in developing new randomized algorithms for these tools in such a way that we can analyze the algorithm by assuming only pairwise independence in the randomness it uses. We then describe how one can leverage this to derandomize the algorithms, i.e., to transform the randomized algorithm into an efficient deterministic algorithm. We next review the model and then state our contributions in the context of the recent progress.
\paragraph{Model.} We work with the standard distributed message-passing model for graph algorithms~\cite{peleg00}. The network is abstracted as an $n$-node graph $G=(V, E)$ where each node $v\in V$ corresponds to one processor in the network. Communications take place in synchronous rounds. Per round, each processor/node can send an $O(\log n)$-bit message to each of its neighbors in $G$. This model is called $\mathsf{CONGEST}\,$. The relaxed variant of the model where we allow unbounded message sizes is called $\mathsf{LOCAL}\,$. At the end of the round, each processor/node performs some computations on the data that it holds, before we proceed to the next communication round.
A graph problem in this model is captured as follows: Initially, the network topology is not known to the nodes of the graph, except that each node $v\in V$ knows its own unique $O(\log n)$-bit identifier and perhaps some of the global parameters of the network, e.g., the number $n$ of nodes in the network or a suitably tight upper bound on it. At the end of the computation, each node should know its own part of the output, e.g., in the graph coloring problem, each node should know its own color. When we discuss a particular graph problem, we will specify what part of the output should be known by each node.
\subsection{Network Decomposition}
Perhaps the most central object in the study of deterministic distributed algorithms for local graph problems has been the concept of \emph{network decomposition}, which was introduced by Awerbuch, Luby, Goldberg, and Plotkin~\cite{awerbuch89}. We next define this concept and explain its usefulness. Then, we discuss its existence and randomized distributed constructions. Afterward, we review the deterministic distributed constructions, especially the recent breakthroughs, and state our contributions.
Generally, the vertices of any $n$-node network can be colored using $O(\log n)$ colors such that in the subgraph induced by each color, each connected component has diameter $O(\log n)$. We call this an $O(\log n)$-color $O(\log n)$-diameter network decomposition (or sometimes $O(\log n)$-color $O(\log n)$-strong-diameter network decomposition, to contrast it with a weaker variant which we discuss later). This decomposition enables us to think of the entire graph as a collection of $O(\log n)$ node-disjoint graphs, each of which has a small $O(\log n)$-diameter per component; the latter facilitates distributed coordination and computation in the component.
As a prototypical example, given such a network decomposition, one easily gets an $O(\log^2 n)$-round deterministic algorithm for maximal independent set in the $\mathsf{LOCAL}\,$ model: we process the color classes one by one, and per color, in each $O(\log n)$-diameter component, we add to the output a maximal independent set of the nodes of the component that do not have a neighbor in the independent sets computed in the previous colors. Each color is processed in $O(\log n)$ rounds, as that is the component diameter, and thus the overall process takes $O(\log^2 n)$ rounds. See \cite{rozhonghaffari20, ghaffari2017complexity, ghaffari2018derandomizing} for how network decomposition leads to a general derandomization method in the $\mathsf{LOCAL}\,$ model, which transforms any $\mathrm{poly}(\log n)$-time randomized algorithm for any locally checkable problem~\cite{naor95} (roughly speaking, problems in which any proposed solution can be checked deterministically in $\mathrm{poly}(\log n)$-time, e.g., coloring, maximal independent set, maximal matching) into a $\mathrm{poly}(\log n)$-time deterministic algorithm.
The existence of such a $O(\log n)$-color $O(\log n)$-diameter network decomposition follows by a simple ball-growing process~\cite{Awerbuch-Peleg1990}.
We build the colors one by one, and each time, we color at least half of the remaining nodes with the next color. For one color $i$, start from an arbitrary node and grow its ball hop by hop, so long as the size is increasing by at least a $2$ factor per hop. This stops in at most $O(\log n)$ hops. Once stopped, color the inside of the ball with the current color $i$, and remove the boundary nodes, deferring them to the next colors. If we continue doing this from nodes that remain in the graph, in the end, at least half of the nodes of the graph (which remained after colors $1$ to $i-1$) are colored in this color $i$, each carved ball has diameter $O(\log n)$, and different balls are non-adjacent as we remove their boundaries.
Linial and Saks~\cite{linial92} gave a randomized distributed algorithm that computes almost such a network decomposition in $O(\log^2 n)$ rounds of the $\mathsf{CONGEST}\,$ model. The only weakness was in the diameter guarantee: the vertices of each color are partitioned into non-adjacent clusters so that per cluster, every two vertices of this cluster have a distance of at most $O(\log n)$ in the original graph. This is what we call $O(\log n)$ weak-diameter. In contrast, if the distance was measured in the subgraph induced by the nodes of this color, it is called a \emph{strong-diameter}. A $O(\log^2 n)$-round $\mathsf{CONGEST}\,$-model randomized algorithm for $O(\log n)$-color $O(\log n)$-strong-diameter network decomposition was provided much later, by Elkin and Neiman~\cite{elkin16_decomp}, building on a parallel algorithm of Miller, Peng, and Xu~\cite{miller2013parallel}.
In contrast, even after significant recent breakthroughs, deterministic constructions for network decomposition are still far from achieving similar measures, and this suboptimality spreads to essentially all applications of network decomposition in deterministic algorithms. The original work of Awerbuch et al.~\cite{awerbuch89} gave a $T$-round deterministic $\mathsf{LOCAL}\,$ algorithm for $c$-color and $d$-strong-diameter network decomposition where $c=d=T=2^{O(\sqrt{\log n\log\log n})}$. All these bounds were improved to $c=d=T=2^{O(\sqrt{\log n})}$ by Panconesi and Srinivasan~\cite{panconesi-srinivasan}. A transformation of Awerbuch et al.~\cite{awerbuch96} in the $\mathsf{LOCAL}\,$ model can transform these into a $O(\log n)$-color $O(\log n)$-strong-diameter network decomposition, but the time complexity remains $2^{O(\sqrt{\log n})}$ and this remained the state of the art for over nearly three decades.
Rozho\v{n} and Ghaffari~\cite{rozhonghaffari20} gave the first $\mathrm{poly}(\log n)$ time deterministic network decomposition with $\mathrm{poly}(\log n)$ parameters. Concretely, their algorithm computes a $O(\log n)$-color $O(\log^3 n)$-weak-diameter network decomposition in $O(\log^8 n)$ rounds of the $\mathsf{CONGEST}\,$ model. The construction was improved to a $O(\log n)$-color $O(\log^2 n)$-weak-diameter network decomposition in $O(\log^5 n)$ rounds of the $\mathsf{CONGEST}\,$ model, by Grunau, Ghaffari, and Rozho\v{n}~\cite{GGR20}. Both of these constructions were limited to only a weak-diameter guarantee. If one moves to the relaxed $\mathsf{LOCAL}\,$ model with unbounded message sizes, then by combining these with a known transformation of Awerbuch et al.~\cite{awerbuch96}, one gets $O(\log n)$-color $O(\log n)$-strong-diameter network decompositions, in a time complexity that is slower by a few logarithmic factors. However, such a transformation was not known for the $\mathsf{CONGEST}\,$ model, until a recent work of Chang and Ghaffari~\cite{chang2021strong}. They gave a $\mathsf{CONGEST}\,$-model reduction, which can transform the weak-diameter construction algorithm of Grunau et al.~\cite{GGR20} into a strong-diameter one, sacrificing some extra logarithmic factors. Concretely, they achieved a $O(\log n)$-color $O(\log^2 n)$-strong-diameter decomposition in $O(\log^{11} n)$ rounds. The time complexity of decomposition with these parameters was improved very recently by Elkin et al.~\cite{elkin2022deterministic}, obtaining a $O(\log n)$-color $O(\log^2 n)$-strong-diameter decomposition in $O(\log^5 n)$ rounds.
However, all these constructions are still far from building the arguably right object, i.e., an $O(\log n)$-color $O(\log n)$-strong-diameter decomposition, in the $\mathsf{CONGEST}\,$ model. This was true even if we significantly relax the time complexity, and as mentioned before, this sub-optimality spreads to all applications.
\paragraph{Our contribution.} In this paper, we present a novel deterministic construction of network decomposition which builds \emph{almost the right object}, achieving an $O(\log n)$-color $O(\log n \cdot \log\log\log n)$-strong-diameter decomposition, in $\mathrm{poly}(\log n)$ rounds. We note that all previous construction techniques seem to require diameter at least $\Omega(\log^2 n)$; see \cite{chang2021strong} for an informal discussion on this. Our algorithm breaks this barrier and reaches diameter $O(\log n \cdot \log\log\log n)$. The key novelty is in designing a new randomized algorithm that can be analyzed using only pairwise independence. We can thus derandomize this algorithm efficiently by using previously known network decompositions in a black-box manner, and in $\mathrm{poly}(\log n)$ time.
Furthermore, if we want faster algorithms, by a black-box combination of our new construction with the technically-independent recent work of Faour, Ghaffari, Grunau, Kuhn, and Rozho\v{n}~\cite{Faour2022} on locally derandomizing pairwise-analyzed randomized algorithms (roughly speaking, their approach works by a specialized weighted defective coloring, instead of using network decompositions), our construction becomes much faster than all the previous constructions, and therefore provides the new state-of-the-art:
\begin{theorem}\label{thm:NetDecomp} There is a deterministic algorithm that, in any $n$-node network, computes an $O(\log n)$-color $O(\log n \cdot \log\log\log n)$-strong-diameter decomposition in $\widetilde{O}(\log^3 n)$ rounds\footnote{We use the notation $\widetilde{O}(f(x)) = O(f(x) \cdot \mathrm{poly}\log f(x))$. } of the $\mathsf{CONGEST}\,$ model. The algorithm performs $\widetilde{O}(m)$ computations in total, where $m$ denotes the number of edges.
\end{theorem}
\subsection{Hitting Set}
While network decomposition is a generic tool for derandomization in the $\mathsf{LOCAL}\,$ model, and also a key tool for derandomization in the $\mathsf{CONGEST}\,$ model with extensive applications, a more basic tool that captures the usage of randomness in a range of distributed algorithms is \emph{hitting set}, as we describe next.
\paragraph{The Hitting Set Problem (basic case).} Given a collection of \lq\lq large\rq\rq\, sets in a ground set of elements, randomness gives us a very simple way of selecting a ``small\rq\rq\, portion of the elements such that we have at least one member of each set. The most basic variant is this: consider a bipartite graph $G=(A\sqcup B, E)$ where each node on one side $A$ has degree at least $k$. By using randomness, we can easily define a small subset $B'\subseteq B$ which, with high probability, has size $O(|B|\log n/k)$ and hits/dominates $A$, that is, each node $a\in A$ has a neighbor in $B'$. For that, simply include each element of $B$ in $B'$ with probability $p=O(\log n/k)$. This randomized selection in fact works in zero rounds. Finding such a small subset $B'$ in a deterministic manner is a key challenge in designing efficient deterministic distributed algorithms for many problems. For instance, Ghaffari and Kuhn~\cite{ghaffari2018congest-derandomizing} pointed out that this is the only use of randomness in some classic randomized algorithms for the construction of spanners and approximations of set cover. Indeed, a variant of this hitting set problem is a key ingredient even in our construction of the network decompositions mentioned above.
\paragraph{The Hitting Set Problem (general case).} Generalizing the problem allows us to capture a much wider range of applications. In some applications, we need to consider different sizes of the sets. Furthermore, we may not need to hit all sets, but instead, we would like to minimize the number, or the total cost, of those not hit. Following the bipartite graph terminology mentioned above, suppose each node $a\in A$ has a cost $c_a$, and its degree is denoted by $d_a$. Randomized selection with probability $p$ picks a subset $B'\subseteq B$ of size $p|B|$, in expectation, where the total cost of $A$-nodes that do not have a neighbor in $B'$ is $\sum_{a\in A} c_a (1-p)^{\deg(a)}$, in expectation. As a side comment, we note that in all applications that we are aware of, we may assume that $c_a\in [1, \mathrm{poly}(n)]$. Because of this, essentially without loss of generality, we can assume that for each node $a\in A$ we have $\deg(a)\leq O(\frac{1}{p} \cdot \log n)$. This is because the total expected cost of higher degree nodes is $1/\mathrm{poly}(n)$, which is negligible.
As an instructive example application, by defining $c_a:=\deg(a)$, we get that the total number of edges incident on $A$-nodes that are not hit is at most $O(|A|/p)$, in expectation. This particular guarantee is the sole application of randomness in some distributed constructions, e.g., the celebrated spanner construction of Baswana and Sen~\cite{baswana2007simple}.
\paragraph{Prior deterministic distributed algorithms for hitting set.} There are two known distributed constructions for hitting set~\cite{ghaffari2018congest-derandomizing, bezdrighin2022deterministic}, as we review next. Both of these algorithms are based on showing that a small collection of random bits are sufficient for the randomized algorithm and then using the conditional expectation method to derandomize this. However, both algorithms are computationally inefficient and use superpolynomial-time computations.
Ghaffari and Kuhn~\cite{ghaffari2018congest-derandomizing} observed that $O(\log n)$-wise independence is sufficient for the randomized algorithm in the basic hitting set problem, and thus $O(\log^2 n)$ bits of randomness are sufficient for the algorithm. Then, given a network decomposition with $c$ colors and strong diameter $d$, we can derandomize these bits one by one in a total of $O(cd \log^2 n)$ rounds, by processing the color classes one by one and fixing the bits in each color class in $O(d)$ time. However, the resulting algorithm is not computationally efficient: Each node has to perform $n^{O(\log n)}$-time local computations to calculate the conditional probabilities needed in the method of conditional expectation.
Parter and Yogev~\cite{parter2018congested} pointed out that one can replace the $O(\log n)$-wise independence with a pseudorandomness generator for read-once DNFs and this reduces the number of bits to $O(\log n (\log\log n)^3)$--this was presented in a different context of spanners in the congested clique model of distributed computing. More recently, Bezdrighin et al.~\cite{bezdrighin2022deterministic} further reduced that bound to $O(\log n \log\log n)$, by applying a pseudorandomness generator of Gopalan and Yeudayoff~\cite{gopalan2020concentration}, which is particularly designed for hitting events. This decreased the round complexity slightly to $O(cd\log n\log\log n)$. However, the conditional probability computations still remain quite inefficient: they are $n^{O(\log\log n)}$-time, which is still super-polynomial.
\paragraph{Our contribution.}
Instead of viewing the randomized hitting set algorithm as a one-shot process, we turn it into a more gradual procedure. Concretely, we show that one can turn the natural randomized algorithm for the general hitting set problem into a number of randomized algorithms (bounded by $O(\log n)$), in such a way that pairwise independence is sufficient for analyzing each step. Because of this, we can derandomize each step separately (using an overall potential function that ensures that the result after derandomizing all steps has the same guarantees as discussed above for the randomized algorithm). Thanks to this, in contrast to the prior algorithms~\cite{ghaffari2018congest-derandomizing, bezdrighin2022deterministic} which required super-polynomial computations, our algorithm uses only $\widetilde{O}(m)$ computations, summed up over the entire graph, where $m$ denotes the number of edges. Hence, our distributed algorithm directly provides a near-linear time low-depth deterministic parallel algorithm for the hitting set problem.
\begin{theorem} [Informal] There is a deterministic distributed algorithm that in $\mathrm{poly}(\log n)$ rounds and using $\widetilde{O}(m)$ total computations solves the generalized hitting set problem. That is, in the bipartite formulation mentioned above, the selected subset $B'$ has size $O(p \cdot |B|)$ and the total weight of nodes of $A$ not hit by $B'$ is $O(\sum_{a\in A} c_a (1-p)^{\deg(a)})$.
\end{theorem}
We present the formal version of this theorem in \Cref{thm:hittingset-congest} for the $\mathsf{CONGEST}\,$ model of distributed computing, and in \Cref{thm:hittingset-pram} for the $\mathsf{PRAM}$ model of parallel computation.
\paragraph{Applications of hitting set.} This efficiently derandomized hitting set has significant applications for a number of graph problems of interest. In this paper, as two examples, we discuss spanners and distance oracles. In the case of spanners, this deterministic hitting set leads to the first deterministic spanner algorithm with the best-known stretch-size trade-off, polylogarithmic round complexity, that has near-linear time computations. The best previously known deterministic constructions required superpolynomial computations~\cite{bezdrighin2022deterministic} (and in the \cite{ghaffari2018congest-derandomizing} case, had extra logarithmic factors in size). The formal statements are as follows, and the proofs are presented in \Cref{subsec:spanners}.
\begin{corollary} (\textbf{general stretch spanners, unweighted and weighted})
There is deterministic distributed algorithm that, in $\mathrm{poly}(\log n)$ rounds of the $\mathsf{CONGEST}\,$ model and with total computations $\widetilde{O}(m)$, for any integer $k\geq 1$, computes a $(2k-1)$-spanner with $O(nk + n^{1 + 1/k} \log k)$ and $O(nk + n^{1 + 1/k} k)$ edges for unweighted and weighted graphs, respectively.
\end{corollary}
\begin{corollary} (\textbf{ultra-sparse spanners})
There is deterministic distributed algorithm that, in $\mathrm{poly}(\log n)$ rounds of the $\mathsf{CONGEST}\,$ model and with total computations $\widetilde{O}(m)$, computes a spanner with size $(1+o(1)) n$ and with stretch $\log n \cdot 2^{O(\log^* n)}$ in weighted graphs.
\end{corollary}
\medskip
By slight generalizations of our hitting set, we also obtain an efficient parallel derandomization of approximate distance oracles constructions:
\begin{corollary} (\textbf{approximate distance oracle})
Given an undirected weighted graph $G = (V,E)$, a set of sources $S \subseteq V$ with $s=|S|$, and stretch and error parameters $k$ and $\varepsilon > 0$, there is a deterministic algorithm that solves the source-restricted distance oracle problem with $\widetilde{O}_{\varepsilon}(ms^{1/k})$ work and $\widetilde{O}_{\varepsilon}(\mathrm{poly}(\log n))$ depth in the $\mathsf{PRAM}\,$ model. The data structure has size $O(nk s^{1/k})$ and for each query $(u,v)$, the oracle can return a value $q$ in $O(k)$ time that satisfies
\begin{equation*}
d(u,v) \leq q \leq (2k-1)(1 + \varepsilon)d(u,v).
\end{equation*}
\end{corollary}
The proof is presented in \Cref{subsec:distance-oracles}. The corresponding centralized randomized construction was presented in the celebrated work of Thorup and Zwick~\cite{thorup2005approximate}. A centralized derandomization was given by Roddity, Thorup, and Zwick~\cite{roditty2005deterministic} but that approach does not appear to be applicable in parallel/distributed settings of computation.
\section{Preliminaries}
\label{sec:preliminaries}
We use standard graph theoretic notation throughout the paper.
All graphs are undirected and unweighted.
For a graph $G = (V, E)$, we use $d_G$ or just $d$ to denote the distance metric induced by its edges. For sets of nodes $U,W \subseteq V(G)$, we generalize $d$ by $d(U, W) = \min_{u \in U, w \in W} d(u,w)$.
\paragraph{Clustering}
Given a graph $G$, its cluster $C$ is simply a subset of nodes of $V(G)$.
The strong-diameter $\textrm{diam}(C)$ of a cluster $C$ is defined as $\textrm{diam}(C) = \max_{u,v \in C} d_{G[C]}(u,v)$.
We note that there is a related notion of \emph{weak-diameter} of a cluster $C$ which is defined as the smallest $D$ such that $\forall u,v \in C : d_G(u,v) \le D$. That is, $C$ can even be disconnected, but there has to be a short path between any two nodes if we are allowed to use all nodes of $G$, not just nodes of $C$.
Although a cluster is simply a subset of $V(G)$, during the construction of a clustering, we keep its \emph{center} node $v \in C$ and often we work with an arbitrary breadth first search tree of $C$ from $v$.
The basic object we construct in this paper is \emph{separated clusterings}, which we formally define next.
\begin{definition}[$s$-separated clustering]
\label{def:clustering}
Given an input graph $G$, a \emph{clustering} $\mathcal{C}$ is a collection of disjoint \emph{clusters} $C_1, \dots, C_t$, such that for each $i$ we have $C_i \subseteq V(G)$. We say that the clustering has \emph{(strong-)diameter} $D$ whenever the diameter of each graph $G[C_i]$, $1 \le i \le t$, is at most $D$. We say that the clustering is $s$-separated if for every $1 \le i < j \le t$ we have $d_G(C_i, C_j) \ge s$. We sometime refer to this by saying that the clustering has separation $s$.
\end{definition}
We will also need the following non-standard notion of $s$-hop degree of a cluster defined as follows.
\begin{definition}[$s$-hop degree]
\label{def:shop_degree}
Let $\mathcal{C}$ be some clustering and $C \in \mathcal{C}$ be a cluster with a fixed spanning tree $T_C$ rooted at $r \in C$.
The $s$-hop degree of $C$ in $\mathcal{C}$ is the minimum number $d$ such that for each $u \in C$ and the unique path $P_u$ from $u$ to $r$ in $T_C$ the following holds: The number of different clusters $C' \in \mathcal{C}$ such that $d(P_u, C') \le s$ is at most $d$.
\end{definition}
The $s$-hop degree of a clustering $\mathcal{C}$ is the maximum $s$-hop degree over all clusters $C \in \mathcal{C}$.
\section{Improved Network Decomposition, Outline}
\label{sec:outline}
To prove \Cref{thm:NetDecomp}, our core result is captured by the following low-diameter clustering statement, which clusters at least half of the vertices. \Cref{thm:NetDecomp} follows directly by repeating this clustering for $O(\log n)$ iterations, each time in the graph induced by the nodes that remain unclustered in the previous iterations.
\begin{theorem}\label{thm:Clustering-main} There is a deterministic $\mathsf{CONGEST}\,$ algorithm that runs in $\widetilde{O}(\log^2 n)$ rounds and computes a clustering of at least $\frac{n}{2}$ nodes, with strong diameter $O(\log n \cdot \log\log\log n)$, and separation $2$.
\end{theorem}
There are three ingredients in proving \Cref{thm:Clustering-main}, as we discuss next:
\paragraph{(A) Low-Degree Clustering.} The most important ingredient, captured by \Cref{thm:low_degree_main} and proven in \Cref{sec:low_degree}, is a clustering that manages to cluster half of the vertices but in which we have relaxed the separation/non-adjacency requirement of the clustering. Instead, we want each cluster to have $\text{s}$-hop degree of at most $\lceil 100 \log \log (n) \rceil$. See \cref{def:shop_degree} for the definition.
For this ingredient, we present a randomized algorithm with pairwise analysis and then we derandomize it.
\begin{restatable*}{theorem}{lowdegreemain}
\label{thm:low_degree_main}
Let $\text{s} \geq 2$ be arbitrary. There exists a deterministic $\mathsf{CONGEST}\,$ algorithm running in $\widetilde{O}(\text{s} \log^2(n))$ rounds which computes a clustering $\mathcal{C}$ with
\begin{enumerate}
\item strong diameter $O(\text{s} \log(n))$,
\item $\text{s}$-hop degree of at most $\lceil 100 \log \log (n) \rceil$, and
\item the number of clustered nodes is at least $n/2$.
\end{enumerate}
\end{restatable*}
\paragraph{(B) From Low-Degree to Isolation.} The second ingredient, captured by \Cref{thm:subsampling_main} and proven in \Cref{sec:LowDegtoIsolation} is able to receive the clustering algorithm of (A) and turn it into a true clustering with separation $s$, but at the expense of reducing the number of clustered nodes by an $O(\log\log n)$ factor. For this ingredient as well, we first present a simple randomized algorithm with pairwise analysis, and then we derandomize it.
\begin{restatable*}{theorem}{subsamplingmain}
\label{thm:subsampling_main}
Assume we are given a clustering $\mathcal{C}$ with
\begin{enumerate}
\item strong diameter $O(\text{s} \log (n))$ and
\item $\text{s}$-hop degree of at most $\lceil 100 \log \log (n) \rceil$.
\end{enumerate}
There exists a deterministic $\mathsf{CONGEST}\,$ algorithm running in $\widetilde{O}(\text{s} \log^2(n))$ rounds which computes a clustering $\mathcal{C}^{out}$ with
\begin{enumerate}
\item strong diameter $O(\text{s} \log (n))$,
\item separation of $\text{s}$ and
\item the number of clustered nodes is $\frac{|\mathcal{C}|}{1000 \log \log (n)}$.
\end{enumerate}
\end{restatable*}
\paragraph{(C) Improving Fraction of Clustering Nodes.} The third and last ingredient, captured by \Cref{thm:clusteringmorenodesmain} and proven in \Cref{sec:clusteringmorenodes}, receives the clustering algorithm of part (B) with a suitably high separation parameter (which is at least logarithmically related to the fraction of nodes clustered) and transforms it into a clustering of at least half of the nodes, at the expense of reducing the separation to simply $2$. This ingredient is a deterministic reduction and needs no derandomization and explains the final logarithm in the guarantees of \cref{thm:Clustering-main} (the first two logarithms are coming already from \cref{thm:low_degree_main}).
\begin{restatable*}{theorem}{clusteringmorenodesmain}
\label{thm:clusteringmorenodesmain}
Let $x \geq 2$ be arbitrary.
Assume there exists a deterministic $\mathsf{CONGEST}\,$ algorithm $\mathcal{A}$ running in $R$ rounds which computes a clustering $\mathcal{C}$ with
\begin{enumerate}
\item strong diameter $O(x \log n)$,
\item separation $10 \cdot x$ and
\item clustering at least $\frac{n}{2^x}$ nodes.
\end{enumerate}
Then, there exists a deterministic $\mathsf{CONGEST}\,$ algorithm $\mathcal{A}'$ running in $O(2^x (R + x \log n))$ rounds which computes a clustering $\mathcal{C}'$ with
\begin{enumerate}
\item strong diameter $O(x \log n)$,
\item separation $2$ and
\item clustering at least $\frac{n}{2}$ nodes.
\end{enumerate}
\end{restatable*}
\vspace{1em}
Having all three ingredients \cref{thm:low_degree_main,thm:subsampling_main,thm:clusteringmorenodesmain}, we simply put them all together to prove \cref{thm:Clustering-main}.
\begin{proof}[Proof of \Cref{thm:Clustering-main}]
Let $x=\lceil \log(2000 \log\log n)\rceil$. First, from \Cref{thm:low_degree_main}, we get a clustering of $n/2$ nodes with strong diameter $O(\log n\cdot \log\log\log n)$ and $10x$-hop degree at most $\lceil 100 \log \log (n) \rceil$, in $\widetilde{O}(\log^2 n)$ rounds. Feeding this clustering algorithm to \Cref{thm:subsampling_main} produces a clustering algorithm that clusters $\frac{n}{2000 \log\log n}$ nodes with strong diameter $O(\log n\cdot \log\log\log n)$ and separation $10 x$, in $\widetilde{O}(\log^2 n)$ rounds. Hence, this clustering can be put as input for \Cref{thm:clusteringmorenodesmain}, which as a result gives a clustering of at least $n/2$ nodes with strong diameter $O(\log n \cdot \log\log\log n)$, and separation $2$, in $\widetilde{O}(\log^2 n)$ rounds.
\end{proof}
\section{Low-Degree Clustering}
\label{sec:low_degree}
This section is devoted to proving the following theorem discussed in \cref{sec:outline}.
\lowdegreemain
\paragraph{Intuition Behind the Proof of \Cref{thm:low_degree_main}.}
In this paragraph, we give a brief intuition behind the proof of \cref{thm:low_degree_main}.
Our clustering algorithm can be viewed as derandomization of the randomized clustering algorithm of Miller, Peng, and Xu~\cite{miller2013parallel} (MPX). This is an algorithm that can cluster $n/2$ nodes with strong diameter $O(s \log n)$ such that the $s$-hop degree of the constructed clustering is in fact $1$, or in other words, the clustering is $s$-separated.
In the MPX algorithm, we simply run a breadth first search from all nodes of $V(G)$ at once, but every node starts the search only after a random delay computed as follows. Every node $v \in V(G)$ starts with the delay $\mathrm{del}(v) = O(s \log n)$. Next, each node starts flipping a coin and each time it comes up heads, it decreases its delay by $5s$. If it comes up tails, it stops the process. That is, the delays come from an exponential distribution; even more precisely, each node gets a head start coming from an exponential distribution, we talk about delays and add $O(s \log n)$ to make all numbers positive with high probability.
The guarantees of the MPX algorithm stem from the following observation. Let $u \in V(G)$ be arbitrary and let $\mathrm{wait}(u)$ be the first time $u$ is reached by above breadth first search with delays. Let $\mathrm{frontier}^{2s}(u)$ be the number of nodes $v \in V(G)$ such that $\mathrm{del}(v) + d(v,u) \le \mathrm{wait}(u) + 2s$. That is, $\mathrm{frontier}^{2s}(u)$ contains nodes who can reach $u$ after at most $2s$ additional steps after $u$ is reached for the first time.
We claim that with positive constant probability $\mathrm{frontier}^{2s}(u) = 1$, i.e., after the first node reaches $u$, it takes at least $2s$ additional steps until the next node reaches $u$.
To see this, replace each node $v \in V(G)$ by a runner on a real line who starts at position $d(u,v) + O(s \log n)$ (and may move toward left, as we soon discuss). Then, the exponential distribution that defines the delays corresponds to each runner flipping her coin until it comes up tails. For each heads, the runner runs distance $5s$ to the left.
We now let the runners flip the coins one by one. When a runner $r_j$ is flipping her coin, we consider the leftmost runner $r'_j$ out of the runners $r_1, \dots, r_{j-1}$ that already flipped their coins.
We observe that if $r_j$ at some point reaches a position at most $5s$ to the right from $r'_j$, we also have that $r_j$ runs to the distance $5s$ to the left of $r'_j$ with positive constant probability.
{\bf Derandomization: } Let us now explain the intuitive reason why we lose a factor of $O(\log \log n)$ in \cref{thm:low_degree_main}. Our derandomized algorithm simulates the coin flipping procedure step by step, for $O(\log n)$ steps, until every runner finally flips a tail and finishes.
In contrast to the previous simple algorithm, we now have to track our progress after every step.
So, our analysis is a derandomization of the following, different, randomized analysis of the same running process.
In this new randomized analysis, in each step $i$ and for each node $u$, we consider, very informally speaking, the event that the coin of all the runners that are currently at distance at most $2s$ from the leading runner comes up tail, where $t$ is a parameter we compute later.
The probability of this event is $2^{-t}$.
This means that the probability of this bad event happening in one of the $O(\log n)$ steps is at most $O(\log n) \cdot 2^{-t}$.
Choosing $t = O(\log \log n)$ makes this probability constant. Going back to the analysis of MPX, we get that at least half of nodes $u$ have $|\mathrm{frontier}^{2s}(u)| = O(\log \log n)$.
Although this new randomized analysis loses a factor of $O(\log \log n)$, we can derandomize it in this section by setting up suitable potentials and derandomizing the coin flips of each step. To do so, we in fact simulate one fully-independent coin flip of each node in $O(\log\log n)$ steps where in each step we only use pairwise-independent random bits.
The rest of the section is structured as follows. In \cref{lem:pairwise_del_clustering}, we show how computing suitable delays gives rise to the final clustering. This step is simple and does not rely on derandomization. \cref{thm:low_degree_delay_main} then shows how to compute the node delays that simulate the MPX analysis as discussed above.
To sample even one ``coin flip'' of MPX, we need to invoke $O(\log\log n)$ times the local derandomization lemma of \cite{Faour2022}. One call of this lemma corresponds to \cref{thm:deterministic-goodSet-selection}.
\paragraph{Basic Definitions}
To prove \Cref{thm:low_degree_main}, we first need to define the notions of delay, waiting time, and a frontier:
\begin{definition}[delay $\mathrm{del}$, waiting time $\mathrm{wait}_\mathrm{del}(u)$, and frontier $\mathrm{frontier}^{D}_{\mathrm{del}}(u)$]
\label{def:low_degree_del}
A delay function $\mathrm{del}$ is a function assigning each node $u \in V$ a \emph{delay} $\mathrm{del}(u) \in \{0,1,\ldots,O(\text{s} \log (n))\}$.
The waiting time of a node $u \in V$, with respect to a delay function $\mathrm{del}$, is defined as
%
\[\mathrm{wait}_{\mathrm{del}}(u) = \min_{v \in V} \left( \mathrm{del}(v) + d(v,u)\right).\]
The intuition behind $\mathrm{wait}_{\mathrm{del}}(u)$ is as follows: Assume that each node $v$ starts sending out a token at time $\mathrm{del}(v)$. Then, $\mathrm{wait}(u)$ is the time it takes until $u$ receives the first token.
Furthermore, for every parameter $D \geq 0$, the \emph{frontier of width $D$} of a node $u \in V$, with respect to a delay function $\mathrm{del}$, is defined as
\[\mathrm{frontier}^{D}_\mathrm{del}(u) = \{v \in V \colon \mathrm{del}(v) + d(v,u) \leq \mathrm{wait}_{\mathrm{del}}(u) + D\}.\] Informally, $\mathrm{frontier}^D(u)$ contains each node $v$ whose token arrives at $u$ at most $D$ time units after $u$ receives the first token.
\end{definition}
\smallskip
\paragraph{Clustering from given delays.} The delay of each vertex is computed by a procedure provided in \Cref{alg:low_degree_delay_alg}. Before discussing that, we first explain how each delay function $\mathrm{del}$, along with a separation parameter $s$, give rise to a clustering $\mathcal{C}^{\mathrm{del}}$: The clustering $\mathcal{C}^{\mathrm{del}}$ clusters all the nodes that have a small frontier of width $2 \text{s}$.
In particular, each node $u \in V(G)$ satisfying $|\mathrm{frontier}^{2\text{s}}(u)| \leq \lceil 100 \log \log(n)\rceil$ is included in some cluster of $\mathcal{C}^{\mathrm{del}}$.
More concretely, each clustered node $u$ gets clustered to the cluster corresponding to the node with the smallest identifier in the set $\mathrm{frontier}^0(u)$. In other words, $u$ gets clustered with the cluster of the minimizer of $\mathrm{wait}(u)$, where we use the smallest identifier to break ties. In the following text, we denote this node by $c_u$. See \cref{fig:mpx} for an illustration of this clustering.
\begin{figure}
\centering
\includegraphics[width = .6\textwidth]{img/mpx.eps}
\caption{The figure shows the run of the clustering algorithm constructing $\mathcal{C}^{\mathrm{del}}$.
The algorithm can be seen as starting a breadth first search from a single node $\sigma$ connected to every node $u \in V(G)$ with an edge of length $\mathrm{del}(u)$ (the $\mathsf{CONGEST}\,$ implementation of the algorithm does not need to simulate any such node $\sigma$).
The value $\mathrm{wait}(u)$ is the time until the search reaches the node $u$. The node that reaches $u$ the first is denote $c_u$.
Moreover, we cluster only nodes such that the size of their frontier of width $2s$ is at most $O(\log\log n)$. For example, the node $v$ is not clustered because after it is reached by the first node $c_v$, it is reached by $\Omega(\log \log n)$ other nodes in the following $2s$ steps.
It can be seen that for any $w$ on the path from $c_u$ to $u$, we have $\mathrm{frontier}^{2s}(w) \subseteq \mathrm{frontier}^{2s}(u)$, hence the constructed clusters are connected.
}
\label{fig:mpx}
\end{figure}
\begin{lemma}
\label{lem:pairwise_del_clustering}
Let $\mathrm{del}$ be a delay function and $\mathcal{C}^{\mathrm{del}}$ the corresponding clustering, as described above. Then, the clustering $\mathcal{C}^{\mathrm{del}}$ has
\begin{enumerate}
\item strong diameter $O(\text{s} \log n)$,
\item $\text{s}$-hop degree at most $\lceil 100 \log \log(n)\rceil$ and
\item the set of clustered nodes is equal to $|\{u \in V \colon \mathrm{frontier}^{2\text{s}}(u)| \leq \lceil 100 \log \log(n)\rceil\}|$.
\end{enumerate}
Moreover, the clustering $\mathcal{C}^{\mathrm{del}}$ can be computed in $O(\text{s} \log n \log\log n)$ $\mathsf{CONGEST}\,$ rounds.
\end{lemma}
To prove \Cref{lem:pairwise_del_clustering}, we first observe that the frontiers have the following property.
\begin{claim}
\label{cl:frontier}
Let $w$ be any node on a shortest path from $u$ to $c_u$ and let $D \ge 0$. Then, we have (I) $\mathrm{frontier}^D(w) \subseteq \mathrm{frontier}^D(u)$, and (II) $c_w = c_u$.
\end{claim}
\begin{proof}
First, we prove (I) $\mathrm{frontier}^D(w) \subseteq \mathrm{frontier}^D(u)$. Consider any $v \in \mathrm{frontier}^D(w)$. We prove that $v \in \mathrm{frontier}^D(u)$. Since $v \in \mathrm{frontier}^D(w)$, we have
$$\mathrm{del}(v) + d(v, w) \le \mathrm{del}(c_u) + d(c_u, w) + D.$$
Since $w$ lies on a shortest path from $c_u$ to $u$, we can add $d(w, u)$ to both sides of the equation to conclude that
$$\mathrm{del}(v) + d(v, w) + d(w, u) \le \mathrm{del}(c_u) + d(c_u, u) + D = \mathrm{wait}(u) + D,$$
and thus we have
$$\mathrm{del}(v) + d(v, u) \le \mathrm{wait}(u) + D.$$
Hence, we have $v \in \mathrm{frontier}^D(u)$ and (I) is proven.
Next, we prove (II) $c_w = c_u$. In view of the above proof of (I), it suffices to show that $c_u \in \mathrm{frontier}^0(w)$.
%
To prove $c_u \in \mathrm{frontier}^0(w)$, we use the fact that $c_u \in \mathrm{frontier}^0(u)$ and write
\[\mathrm{del}(c_u) + d(c_u,u) = \mathrm{wait}(u) \leq \mathrm{wait}(w) + d(w,u)\]
%
Subtracting $d(w, u)$ from both sides of the equation and using that $w$ lies on a shortest path from $u$ to $c_u$ gives
\[\mathrm{del}(c_u) + d(c_u,w) \leq \mathrm{wait}(w).\]
%
Thus $c_u \in \mathrm{frontier}^0(w)$ and we are done.
\end{proof}
\medskip
Having \Cref{cl:frontier}, we now go back to present a proof of \cref{lem:pairwise_del_clustering}.
\begin{proof}[Proof of \Cref{lem:pairwise_del_clustering}]
We start with the first property. Let $u$ be an arbitrary clustered node and recall that $c_u$ is its cluster center.
As $c_u \in \mathrm{frontier}^0(u)$, we have $d(c_u,u) = \mathrm{wait}(u)$.
Moreover,
%
\[\mathrm{wait}(u) = \min_{v \in V} \mathrm{del}(v) + d(v,u) \leq \mathrm{del}(u) + d(u,u) = \mathrm{del}(u) = O(\text{s} \log n).\]
%
Hence, we have $d(c_u,u) = O(\text{s} \log n)$.
Moreover, \cref{cl:frontier} gives that all nodes on a shortest path from $c_u$ to $u$ are also clustered to $c_u$, implying that the diameter of the cluster is $O(s \log n)$.
Next, we prove the second property.
Consider an arbitrary clustered node $w$. We first show that for an arbitrary clustered node $y$ with $d(w,y) \leq \text{s}$, it holds that $c_y \in \mathrm{frontier}^{2\text{s}}(w)$.
%
To see this, we first use the definition of $c_w$ to write
\begin{align*}
\mathrm{del}(c_y) + d(c_y,y)
\leq \mathrm{del}(c_w) + d(c_w,y)
\end{align*}
%
On one hand, we can use triangle inequality to lower bound the left-hand side by
%
\begin{align*}
\mathrm{del}(c_y) + d(c_y,y) \ge \mathrm{del}(c_y) + d(c_y, w) - d(w,y)
\end{align*}
On the other hand, we can use triangle inequality to upper bound the right hand side by
\begin{align*}
\mathrm{del}(c_w) + d(c_w,y)
\le \mathrm{del}(c_w) + d(c_w, w) + d(w, y).
\end{align*}
%
Putting the two bounds together, we conclude that
\begin{align*}
\mathrm{del}(c_y) + d(c_y, w) \le \mathrm{del}(c_w) + d(c_w, w) + 2d(w,y) \le \mathrm{wait}(w) + 2s,
\end{align*}
where we used our assumption $d(w,y) \le 2s$. Thus, $c_y \in \mathrm{frontier}^{2\text{s}}(w)$.
Now, let $u$ be an arbitrary clustered node and $P_u$ the unique path between $u$ and $c_u$ in the tree associated with the cluster. Furthermore, let $C \in \mathcal{C}^{\mathrm{del}}$ be a cluster with $d(P_u,C) \leq \text{s}$. Then there exists $w \in P_u$ and $y \in C$ with $d(w,y) \leq \text{s}$ and the discussion above implies $c_y \in \mathrm{frontier}^{2\text{s}} (w)$.
We now use \cref{cl:frontier}, which implies that $\mathrm{frontier}^{2\text{s}} (w) \subseteq \mathrm{frontier}^{2\text{s}}(u)$.
Hence, for each cluster $C$ with $d(P_u,C) \leq \text{s}$, the corresponding cluster center is contained in $\mathrm{frontier}^{2\text{s}}(u)$. As $u$ is clustered, we know that $|\mathrm{frontier}^{2\text{s}}(u)| \leq \lceil100 \log \log n\rceil$. Therefore, the $\text{s}$-hop degree of $\mathcal{C}^{\mathrm{del}}$ is at most $\lceil100 \log \log (n)\rceil$.
The third property follows directly from the definition.
To finish the proof, we need to show that the algorithm can be implemented in $O(s \log n \cdot \log\log n)$ rounds.
To see this, note that we can compute for each node $u$ whether $|\mathrm{frontier}^{2s}(u)| \le \lceil 100 \log \log(n)\rceil$ or not, as follows: We run a variant of breadth first search that takes into account the delays, where each node $v$ starts sending out a BFS token at time $\mathrm{del}(v)$. Recall that in classical breadth first search, after a node $u$ is reached for the first time by a token (sent from $c_u$), it broadcasts this token to all its neighbors and then it does not redirect any other tokens sent to it.
In our version of the search, each node stops redirecting only after at least $\lceil 100 \log \log(n)\rceil$ tokens arrived (we do not take into account tokens that have already arrived earlier) or after it counts $2s$ steps from the arrival of the first token.
It can be seen that this algorithm can be implemented in the desired number of rounds. Moreover, every node $u$ learns the value of $|\mathrm{frontier}^{2s}(u)|$ whenever the value is at most $\lceil 100 \log \log(n)\rceil$ and otherwise, it learns the value is larger than this threshold.
\end{proof}
In view of \cref{lem:pairwise_del_clustering}, to prove the randomized variant of \Cref{thm:low_degree_main} with pairwise analysis, it suffices to show that \cref{alg:low_degree_delay_alg} stated next computes a delay function $\mathrm{del}$ such that the expected number of nodes $u$ with $|\mathrm{frontier}^{2\text{s}}(u)| \leq \lceil 100 \log \log(n)\rceil$ is at least $n/2$. We later discuss how this is derandomized.
\subsection{Computing Delays}
This subsection is dedicated to proving the following theorem that asserts that we can compute a suitable delay function that can be plugged in \cref{lem:pairwise_del_clustering} that constructs a clustering from it.
\begin{theorem}
\label{thm:low_degree_delay_main}
\cref{alg:low_degree_delay_alg} runs in $\widetilde{O}(\text{s} \log^2(n))$ $\mathsf{CONGEST}\,$ rounds and computes a delay function $\mathrm{del}$ that satisfies
\begin{align}
\label{eq:houkacka}
|\{u \in V \colon |\mathrm{frontier}_{\mathrm{del}}^{2s}(u)| \leq \lceil 100 \log \log (n) \rceil\}| \geq n/2.
\end{align}
\end{theorem}
\begin{restatable}{algorithm}{delays}
\caption{Computing Delay Function $\mathrm{del}$}
\label{alg:low_degree_delay_alg}
Input: A parameter $s$, an algorithm $\mathcal{A}_{i,j}$ computing a \emph{good set} from \cref{def:low_degree_del_good_set} in $\widetilde{O}(s \log n)$ rounds\\
Output: A delay function $\mathrm{del}$ from \cref{def:low_degree_del} satisfying \cref{eq:houkacka}
\begin{algorithmic}[1]
\Procedure{Delays}{}
\State $V_0^ { \textrm{active}} \leftarrow V$
\State $R \leftarrow \lfloor 2 \log(n) \rfloor$
\State $k \leftarrow \lceil 100 \log \log (n) \rceil$
\State $\forall u \in V \colon \mathrm{del}_0(u) \leftarrow 5 \text{s} R$
\For{$i \leftarrow 1,2, \ldots, R$}
\State $W_{i,0} \leftarrow \emptyset$
\For{$j \leftarrow 1,2,\ldots,k$}
\State $S_{i,j} \leftarrow \mathcal{A}_{i,j}(\mathrm{del}_{i-1}, W_{i,j-1})$ \Comment{$S_{i,j} \subseteq V_{i-1}^ { \textrm{active}}$}
\State $W_{i,j} \leftarrow W_{i,j-1} \cup S_{i,j}$
\EndFor
\State $V_i^{ \textrm{active}} \leftarrow W_{i,k}$
\For{$\forall u \in V$}
\If{$u \in V^\textrm{active}_i$}
\State $\mathrm{del}_i(u) \leftarrow \mathrm{del}_{i-1}(u) - 5s$
\Else
\State $\mathrm{del}_i(u) \leftarrow \mathrm{del}_{i-1}(u)$
\EndIf
\EndFor
\EndFor
\State $\mathrm{del} \leftarrow \mathrm{del}_R$
\State \Return $\mathrm{del}$
\EndProcedure
\end{algorithmic}
\end{restatable}
\paragraph{Intuitive Description of \cref{alg:low_degree_delay_alg}.}
The algorithm runs in $R = \lfloor 2 \log(n)\rfloor$ phases and each phase consists of $k = \lceil 100 \log \log n\rceil$ iterations.
In iteration $j$ of phase $i$, algorithm $\mathcal{A}_{i,j}$ is a deterministic algorithm which computes a good set $\mathcal{S}_{i,j} \subseteq V^{\textrm{active}}_{i-1}$ as defined later in \cref{def:low_degree_del_good_set}. The algorithm description of $\mathcal{A}_{i,j}$ is deferred to \cref{sec:low_degree_local_derandomization}. The high-level intuition is that $\mathcal{A}_{i,j}$ derandomizes the randomized process which obtains $S_{i,j}$ from $V^{\textrm{active}}_{i-1}$ by including each vertex with probability $\frac{1}{4k}$, pairwise independently.
Repeating this pairwise independent sampling process $k$ times then simulates including each vertex from $V^{\textrm{active}}_{i-1}$ to $V^{\textrm{active}}_{i}$ with positive probability.
The derandomization of the pairwise independent process is done efficiently using a novel local derandomization procedure introduced in \cite{Faour2022} which essentially allows to efficiently derandomize algorithms that only rely on pairwise analysis.
Throughout the algorithm, each node is assigned a delay. At the beginning, each node $u$ is assigned a delay of $\mathrm{del}_0(u) = 5 \text{s} R = O(\text{s} \log n)$. In each subsequent phase, for each node $u$, we have two possibilities: if $u \in V^{\textrm{active}}_i$, the delay of node $u$ is decreased by $5 \text{s}$, i.e., $\mathrm{del}_i(u) = \mathrm{del}_{i-1}(u) - 5 \text{s}$ if $u \in V^{\textrm{active}}_i$; if $u \notin V^{\textrm{active}}_i$, then its delay stays the same, i.e., $\mathrm{del}_i(u) = \mathrm{del}_{i-1}(u)$ if $u \notin V^{\textrm{active}}_i$.
For every $u \in V$, we define the shorthand $\mathrm{wait}_i(u) = \mathrm{wait}_{\mathrm{del}_i}(u)$ and for every $D \geq 0$, we define $\mathrm{frontier}_i^D(u) = \mathrm{frontier}_{\mathrm{del}_i}^D(u)$.
\paragraph{Communication Primitives.}
%
For the deterministic algorithm $\mathcal{A}_{i,j}$ which computes the set $S_{i,j}$, it is important that each node $u$ can efficiently compute the set $\textrm{alive}_{i-1}(u)$ and $\textrm{dead}_{i-1}(u)$ that are defined next.
Let us give a brief intuition behind the definition. In the ``runner intuition'' from the beginning of the section, we want to know in every step all runners that are currently at distance at most $2s$ after the front runner. For these runners, we want to ensure that not all of them stop running in one step. In the reality of the distributed $\mathsf{CONGEST}\,$ model, we however cannot compute even the size of $\mathrm{frontier}^{2s}(u)$.
Fortunately, for our purposes if the number of ``runners'' that are distance at most $2s$ from the front runner is larger than $O(\log\log n)$, it roughly speaking suffices to work with the first $O(\log \log n)$ runners in the analysis. This is formalized by the following definition of alive and dead nodes (dead nodes are runners that stopped flipping coins).
\begin{restatable}{definition}{alivedead}[$\textrm{alive}_i(u)$/$\textrm{dead}_i(u)$]
For every vertex $u \in V$ and $i \in \{0,1,\ldots,R\}$, let $\textrm{dead}_i(u) \subseteq \mathrm{frontier}^{2 \text{s}}_i(u) \setminus V_i^{\textrm{active}}$ be an arbitrary subset of size $\min(k, |\mathrm{frontier}^{2 \text{s}}_i(u) \setminus V_i^{\textrm{active}}|)$ and $\textrm{alive}_i(u) \subseteq \mathrm{frontier}^{2 \text{s}}_i(u) \cap V_i^{\textrm{active}}$ be an arbitrary subset of size $\min(k - |\textrm{dead}_i(u)|, |\mathrm{frontier}^{2 \text{s}}_i(u) \cap V_i^{\textrm{active}}|)$.
\end{restatable}
Note that $|\textrm{alive}_i(u)| + |\textrm{dead}_i(u)| \leq \min(k, |\mathrm{frontier}^{2 \text{s}}_i(u)|)$ and $|\textrm{dead}_i(u)| \leq |\textrm{alive}_{i-1}(u)| + |\textrm{dead}_{i-1}(u)|$.
For each node $v \in V$, let $M_{i-1}(v) = \{u \in V \colon v \in \textrm{alive}_{i-1}(u)\}$. Then, we need some simultaneous and efficient communication, that allows each $v\in V$ to broadcast a message to all nodes in $M_{i-1}(v)$, and for $v$ to receive an aggregate of messages prepared for $v$ in nodes $M_{i-1}(v)$.
\begin{lemma}
\label{lem:low_degree_delays_communication}
Suppose that we are at the beginning of some phase $i \in [R]$. Given delay function $\mathrm{del}_{i-1}$, and given the set $V_{i-1}^{\textrm{active}}$, there exists a $\mathsf{CONGEST}\,$ algorithm running in $\widetilde{O}(\text{s} \log n)$ rounds which computes for each node $u \in V$ the sets $\textrm{alive}_{i-1}(u)$ and $\textrm{dead}_i(u)$. Moreover, let $M_{i-1}(v) = \{u \in V \colon v \in \textrm{alive}_{i-1}(u)\}.$ Then, there exists an $\widetilde{O}(\text{s} \log n)$ round $\mathsf{CONGEST}\,$ algorithm that allows each node $v$ to send one $O(\log n)$-bit message that is delivered to all nodes in $M_{i-1}(v)$. Similarly, there also exists an $\widetilde{O}(\text{s} \log n)$ round $\mathsf{CONGEST}\,$ algorithm that given $O(\log n)$-bit messages prepared at nodes in $M_{i-1}(v)$ specific for node $v$, it allows node $v$ to receive an aggregation of these messages, e.g., the summation of the values, in $\widetilde{O}(\text{s} \log n)$ rounds.
\end{lemma}
\begin{proof}[Proof of \Cref{lem:low_degree_delays_communication}] We run a variant of breadth first search (BFS) that takes into account the delays, and runs in $\widetilde{O}(s\log n)$ rounds: Each node $v$ starts sending out a BFS token at time $\mathrm{del}_{i-1}(v)$, where the token also includes the information whether $v\in V_{i}^{\textrm{active}}$ or not. During the entire process, each node $u$ forwards per time step at most $k= \lceil100\log\log n \rceil$ BFS tokens, breaking ties in favoring of tokens coming from nodes $v\notin V_{i}^{\textrm{active}}$. That is, all tokens that arrive at the same time step are forwarded in the next time step, except that the node forwards at most $k$ tokens in this time step, and moreover, the node first includes all tokens from nodes $v\notin V_{i}^{\textrm{active}}$ (up to $k$) before including tokens from nodes $v\in V_{i}^{\textrm{active}}$. Since per time step each node forwards at most $k$ tokens, each time step can be implemented in at most $k$ rounds of the $\mathsf{CONGEST}\,$ model. Furthermore, node $u$ starts counting time from the moment that it received the very first token (while forwarding any received tokens, up to $k$ per time step), and after $2s$ time steps have passed, node $u$ does not forward any other tokens.
Let us think of one tree for each node $v$, which is rooted at $v$ and includes all nodes $u$ that have received the token of node $v$. Every node $u$ receives the tokens of all nodes in $\mathrm{frontier}_{i-1}^{2s}(u)$, if there are at most $k$ of them. If there are more than $k$, node $u$ learns at least $k$ of them, with the following guarantee: The set of learned tokens includes all tokens from $v\notin V_{i}^{\textrm{active}}$, up to $k$ (if there were more).
Hence, given the received tokens, each node $u$ can form $\textrm{dead}_{i-1}(u) \subseteq \mathrm{frontier}^{2 \text{s}}_{i-1}(u) \setminus V_{i-1}^{\textrm{active}}$, which is subset of size $\min(k, |\mathrm{frontier}^{2 \text{s}}_{i-1}(u) \setminus V_{i-1}^{\textrm{active}}|)$. Furthermore, node $u$ can form $\textrm{alive}_{i-1}(u) \subseteq \mathrm{frontier}^{2 \text{s}}_{i-1}(u) \cap V_{i-1}^{\textrm{active}}$, which is a subset of size $\min(k - |\textrm{dead}_{i-1}(u)|, |\mathrm{frontier}^{2 \text{s}}_{i-1}(u) \cap V_{i-1}^{\textrm{active}}|)$.
By repeating the same communication, each node $v$ is able to send one message which is delivered to all nodes $M_{i-1}(v) = \{u \in V \colon v \in \textrm{alive}_{i-1}(u)\}$, all simultaneously in $\widetilde{O}(s\log n)$ rounds. Moreover, by repeating the same communication pattern but in the reverse direction of time, we can do an aggregation along each tree, again all simultaneously in $\widetilde{O}(s\log n)$ rounds, allowing each node $v$ to receive an aggregation of the messages prepared for $v$ in nodes $M_{i-1}(v)$.
\end{proof}
\paragraph{Potential Functions}
In this paragraph, we define an outer potential $\Phi_i$ for every phase $i$ and an inner potential $\phi_{i,j}$ for every
iteration $j$ within phase $i$. The inner potential satisfies that if $\phi_{i,j-1} \leq \phi_{i,j}$ in each iteration $j$, then $\Phi_i \leq \Phi_{i-1} + n$. The outer potential satisfies that $\Phi_0 = 2n$ and if $\Phi_R \leq 10n \log(n)$, then $|\{u \in V \colon |\mathrm{frontier}^{2 \text{s}}(u)|\}| \geq \frac{9n}{10}$.
\begin{definition}[Outer Potential]
\label{def:low_degree_delays_outer_potential}
For every $i \in \{0,1,\ldots,R\}$, the outer potential of a node $u$ after phase $i$ is defined as
\[\Phi_i(u) = e^{\frac{|\textrm{dead}_i(u)|}{10}}.\]
The outer potential after phase $i$ is defined as
\[\Phi_i = \sum_{u \in V} \Phi_i(u) + 2^i|V^{\textrm{active}}_i|.\]
\end{definition}
Here, "after phase $0$" should be read as "the beginning of phase $1$".
\cref{alg:low_degree_delay_alg} will make sure that the outer potential is sufficiently small. A small outer potential after phase $i$ implies on one hand that there are not too many nodes $u$ for which $|\textrm{dead}_i(u)|$ is large and on the other hand ensures that there are not too many nodes in $V^{\textrm{active}}_i$, i.e., $|V^{\textrm{active}}_i| \lesssim\frac{n}{2^i}$.
The following lemma captures the usefulness of the outer potential.
\begin{lemma}[Outer Potential Lemma]
\label{lem:low_degree_delays_outer_potential_lemma}
We have $\Phi_0 \leq 2n$. Moreover, if $\Phi_R \leq 10 n \log(n)$, then $|\{u \in V \colon |\mathrm{frontier}_{\mathrm{del}}^{2\text{s}}(u)| \leq 100 \log \log (n)\}| \geq \frac{9n}{10}$.
\end{lemma}
\begin{proof}
First, note that $\Phi_R \geq 2^R|V_R^ { \textrm{active}}| > 10n \log(n)|V_R^{ \textrm{active}}|$.
As we assume that $\Phi_R \leq 10 n \log(n)$, this directly implies $V_R^ { \textrm{active}} = \emptyset$.
In particular, every $u \in V$ with $|\mathrm{frontier}^{2\text{s}}_{\mathrm{del}}(u)| > 100 \log \log (n)$ contributes
%
\[\Phi_R(u) = e^{\frac{|\textrm{dead}_R(u)|}{10}}\geq e^{\frac{\min(k,|\mathrm{frontier}_{\mathrm{del}}^{2\text{s}}(u)|}{10})} \geq 100 \log(n) \]
%
to the potential. Hence, there can be at most $\Phi_R/(100 \log(n)) \leq n/10$ such nodes and therefore $\{u \in V \colon |\mathrm{frontier}_{\mathrm{del}}^{2\text{s}}(u)| \leq 100 \log \log (n)\} \geq \frac{9n}{10}$, as desired.
\end{proof}
\begin{restatable}{definition}{pessimisticprobability}[Pessimistic Estimator Probability $p_{i,j}(u)$]
For $i \in [R]$ and $j \in \{0,1,\ldots,k\}$, the pessimistic estimator probability of a node $u$ after iteration $j$ within phase $i$ is defined as
\[p_{i,j}(u) = I(\textrm{alive}_{i-1}(u) \cap W_{i,j} = \emptyset) \cdot \left(1 - \frac{|\textrm{alive}_{i-1}(u)|}{10k}\right)^{k-j}.\]
\end{restatable}
Here, "after iteration $0$", should be read as "the beginning of iteration $1$".
Let us briefly elaborate on the definition of $p_{i,j}(u)$. Assume that we would compute $S_{i,j}$ by sampling each vertex in $V^{\textrm{active}}_{i-1}$ with probability $\frac{1}{4k}$ pairwise independently.
By a simple pairwise analysis, one can show that this would imply $\Pr[S_{i,j} \cap \textrm{alive}_{i-1}(u) \neq \emptyset] \geq \frac{|\textrm{alive}_{i-1}(u)|}{10k}$. Hence, if we are currently at the beginning of iteration $j$ within phase $i$ just prior to sampling the set $S_{i,j}$, then $p_{i,j-1}(u)$ is an upper bound on the probability that no node in $\textrm{alive}_{i-1}(u)$ is contained in $ V^{\textrm{active}}_i$ (which one should think of as a bad event).
\begin{restatable}{definition}{innerpotential}[Inner Potential]
\label{def:low_degree_del_inner}
The inner potential of a node $u$ after iteration $j$ of phase $i$ is defined as
\[\phi_{i,j}(u) = p_{i,j}(u) e^{\frac{|\textrm{dead}_{i-1}(u)| + |\textrm{alive}_{i-1}(u)|}{10}}.\]
The inner potential after iteration $j$ of phase $i$ is defined as
\[\phi_{i,j} = \sum_{u \in V} \phi_{i,j}(u) + |W_{i,j}|2^i + \frac{k-j}{k}2^{i-1} |V_{i-1}^{\textrm{active}}|.\]
\end{restatable}
Again, assume for a moment that we would compute $S_{i,j}$ by sampling each vertex in $V^{\textrm{active}}_{i-1}$ with probability $\frac{1}{4k}$ pairwise independently. Assume we are at the beginning of iteration $j$ within phase $i$ just prior to sampling $S_{i,j}$. Then, using the fact that $\mathbb{E}[p_{i,j}(u)] \leq p_{i,j-1}(u)$, one directly gets that $\mathbb{E}[\phi_{i,j}(u)] \leq \phi_{i,j}(u)$ and it also follows that $\mathbb{E}[\phi_{i,j}] \leq \phi_{i,j-1}$. Moreover, one can also show that $\mathbb{E}[\Phi_i(u)] \leq \phi_{i,j-1}(u) + 1$ and $\mathbb{E}[\Phi_i] \leq \phi_{i,j-1} + n$.
In more detail, if at least one node in $\textrm{alive}_{i-1}(u)$ is contained in $V^{\textrm{active}}_i$, one can show that this implies $\textrm{dead}_i(u) = \emptyset$ and therefore $\Phi_i(u) = 1$. On the other hand, in the previous discussion we mentioned that with probability at most $p_{i,j-1}(u)$ no node in $\textrm{alive}_{i-1}(u)$ is included in $V^{\textrm{active}}_i$, and as $|\textrm{dead}_{i-1}(u)| + |\textrm{alive}_{i-1}(u)| \leq |\textrm{dead}_i(u)|$, we have $\ p_{i,j-1}(u)\Phi_i(u) \leq \phi_{i,j-1}$.
\begin{lemma}[Inner Potential Lemma]
\label{lem:low_degree_delays_inner}
For $i \in [R]$ and $j \in \{0,1,\ldots,k\}$,
Assume that in every iteration $j$ of phase $i$, $S_{i,j}$ is computed in such a way that $\phi_{i,j} \leq \phi_{i,j-1}$. Then, $\Phi_i \leq \Phi_{i-1} + n$ and $\Phi_R \leq 4n \log(n)$.
\end{lemma}
\begin{proof}
For each node $u \in V$, we have
\[\phi_{i,0}(u) = p_{i,0}(u)e^{\frac{|\textrm{dead}_{i-1}(u)| + |\textrm{alive}_{i-1}(u)|}{10}} = \left( 1 - \frac{|\textrm{alive}_{i-1}(u)|}{10k}\right)^{k-0} e^{\frac{|\textrm{dead}_{i-1}(u)| + |\textrm{alive}_{i-1}(u)|}{10}} \leq e^{\frac{|\textrm{dead}_{i-1}(u)|}{10}} = \Phi_{i-1}(u).\]
Therefore,
\[\phi_{i,0} = \sum_{u \in V} \phi_{i,0}(u) + |W_{i,0}|2^i + \frac{k-0}{k}2^{i-1} |V^{\textrm{active}}_{i-1}| \leq \sum_{u \in V} \Phi_{i-1}(u) + 2^{i-1}|V^{\textrm{active}}_{i-1}| = \Phi_{i-1}.\]
Consider an arbitrary $u \in V$. Next, we show that
\[e^{\frac{|\textrm{dead}_i(u)|}{10k}}=: \Phi_i(u) \leq \phi_{i,k}(u) + 1 = I(\textrm{alive}_{i-1}(u) \cap V_i^\textrm{active} = \emptyset)e^{\frac{|\textrm{dead}_{i-1}(u)| + |\textrm{alive}_{i-1}(u)|}{10k}} + 1.\]
It is easy to verify that the inequality is satisfied if $\textrm{alive}_{i-1}(u) \cap V_i^\textrm{active} = \emptyset$, as $|\textrm{dead}_i(u)| \leq |\textrm{dead}_{i-1}(u)| + |\textrm{alive}_{i-1}(u)|$.
Therefore, it remains to consider the case that there exists at least one node $v \in \textrm{alive}_{i-1}(u) \cap V_i^{\textrm{active}}$. The existence of such a node $v$ implies
%
\[\mathrm{wait}_i(u) \leq \mathrm{del}_i(v) + d(v,u) = \mathrm{del}_{i-1}(v) + d(v,u) - 5 \text{s} \leq \mathrm{wait}_{i-1}(u) + 2 \text{s} - 5 \text{s} = \mathrm{wait}_{i-1}(u) - 3 \text{s}.\]
%
For every node $w \notin V^{\textrm{active}}_i$, we have $ \mathrm{del}_i(w) = \mathrm{del}_{i-1}(w)$ and therefore
%
\[ \mathrm{del}_i(w) + d(w,u) = \mathrm{del}_{i-1}(w) + d(w,u) \geq \mathrm{wait}_{i-1}(u) \geq \mathrm{wait}_i(u) + 3 \text{s} > \mathrm{wait}_i(u) + 2 \text{s}\]
%
and thus $w \notin \mathrm{frontier}_i^{2 \text{s}}(u)$.
Hence, $\textrm{dead}_i(u) = \emptyset$ and the inequality is satisfied.
Therefore,
\[\phi_{i,k} = \sum_{u \in V} \phi_{i,k}(u) + |W_{i,k}|2^i + \frac{k-k}{k}2^{i-1} |V^{\textrm{active}}_{i-1}| \leq \left(\sum_{u \in V} \Phi_i(u) - 1 \right) + |V^{\textrm{active}}_i|2^i = \Phi_i - n.\]
A simple induction implies $\phi_{i,k} \leq \phi_{i,0}$.
Therefore,
\[\Phi_i \leq \phi_{i,k} + n \leq \phi_{i,0} + n = \Phi_{i-1} + n.\]
As $\Phi_0 \leq 2n$ according to \cref{lem:low_degree_delays_outer_potential_lemma}, a simple induction implies
\[\Phi_R \leq \Phi_0 + Rn \leq (2+R)n \leq 4n \log n.\]
\end{proof}
\paragraph{Good Set $S_{i,j}$:}
We are now going to define the good set of nodes $S_{i,j}$. Note that this is the part of \cref{alg:low_degree_delay_alg} whose definition we postponed.
\begin{definition}[Good Set $S_{i,j}$]
\label{def:low_degree_del_good_set}
For a set $S_{i,j} \subseteq V^{\textrm{active}}_{i-1}$ and $u \in V$, let
\[Y_{i,j}(u) = 1 - |\textrm{alive}_{i-1}(u) \cap S_{i,j}| + \binom{|\textrm{alive}_{i-1}(u) \cap S_{i,j}|}{2}.\]
We refer to the set $S_{i,j}$ as good if
\[ \sum_{u \in V} Y_{i,j}(u) \frac{\phi_{i,j-1}(u)}{1 - (|\textrm{alive}_{i-1}(u)|/(10k))} + |S_{i,j}|\cdot 2^i \leq \sum_{u \in V} \phi_{i,j-1}(u) + \frac{2^{i-1}}{k}|V^{\textrm{active}}_{i-1}|.\]
\end{definition}
\begin{lemma}
\label{lem:low_degree_del_potential_nonicreasing}
If $S_{i,j}$ is a good set, then $\phi_{i,j} \leq \phi_{i,j-1}$.
\end{lemma}
\begin{proof}
For each $u \in V$, we have
\begin{align*}
\frac{I(\textrm{alive}_{i-1}(u) \cap S_{i,j} = \emptyset)p_{i,j-1}(u)}{1 - (|\textrm{alive}_{i-1}(u)|/(10k))} &=
\frac{I(\textrm{alive}_{i-1}(u) \cap S_{i,j} = \emptyset)I(\textrm{alive}_{i-1}(u) \cap W_{i,j-1} = \emptyset) \cdot \left(1 - \frac{|\textrm{alive}_{i-1}(u)|}{10k}\right)^{k-(j-1)}}{1 - (|\textrm{alive}_{i-1}(u)|/(10k))} \\
&= I(\textrm{alive}_{i-1}(u) \cap W_{i,j} = \emptyset) \left(1 - \frac{|\textrm{alive}_{i-1}(u)|}{10k}\right)^{k-j} \\
&= p_{i,j}(u).
\end{align*}
It also holds that $I(\textrm{alive}_{i-1}(u) \cap S_{i,j} = \emptyset) \leq Y_{i,j}(u)$. Therefore,
\begin{align*}
Y_{i,j}(u) \frac{\phi_{i,j-1}(u)}{1 - (|\textrm{alive}_{i-1}(u)|/(10k))} &\geq I(\textrm{alive}_{i-1}(u) \cap S_{i,j} = \emptyset)\frac{p_{i,j-1}(u)e^{\frac{|\textrm{dead}_{i-1}(u)| + |\textrm{alive}_{i-1}(u)|}{10}}}{1 - (|\textrm{alive}_{i-1}(u)|/10k)} \\
&= p_{i,j}(u) e^{\frac{|\textrm{dead}_{i-1}(u)| + |\textrm{alive}_{i-1}(u)|}{10}} \\
&= \phi_{i,j}(u).
\end{align*}
Thus, we get
\begin{align*}
\phi_{i,j} &= \sum_{u \in V} \phi_{i,j}(u) + |W_{i,j}|2^i + \frac{k-j}{k}2^{i-1}|V_{i-1}^{\textrm{active}}| \\
&\leq \sum_{u \in V} Y_{i,j}(u) \frac{\phi_{i,j-1}(u)}{1 - (|\textrm{alive}_i(u)|/10k)} + (|S_{i,j}| + |W_{i,j-1}|)2^i + \frac{k-j}{k}2^i|V^{\textrm{active}}_{i-1}| \\
&\leq \sum_{u \in V} \phi_{i,j-1}(u) + \frac{2^{i-1}}{k}|V^{\textrm{active}}_{i-1}| + |W_{i,j-1}|2^i + \frac{k-j}{k}2^{i-1}|V^{\textrm{active}}_{i-1}| \\
&= \phi_{i,j-1}.
\end{align*}
\end{proof}
We now combine all the pieces to prove the main theorem of this subsection.
\paragraph{Proof of \cref{thm:low_degree_delay_main}}
We assume that in iteration $j$ of phase $i$, $\mathcal{A}_{i,j}$ computes a good set $S_{i,j}$. Therefore, \cref{lem:low_degree_del_potential_nonicreasing} implies that $\phi_{i,j} \leq \phi_{i,j-1}$. According to \cref{lem:low_degree_delays_inner}, this implies that $\Phi_R \leq 4n \log(n)$. Therefore, \cref{lem:low_degree_delays_outer_potential_lemma} implies that $|\{u \in V \colon |\mathrm{frontier}^{2 \text{s}}_{\mathrm{del}}(u)| \leq \lceil 100 \log \log (n)\rceil\}| \geq n/2$, as desired. It remains to discuss the $\mathsf{CONGEST}\,$ round complexity.
\cref{alg:low_degree_delay_alg} has $R \cdot k = \widetilde{O}(\log n)$ iterations in total. In iteration $i$ of phase $j$, algorithm $\mathcal{A}_{i,j}$ runs in $\widetilde{O}(\text{s} \log n)$ $\mathsf{CONGEST}\,$ rounds. Hence, the overall $\mathsf{CONGEST}\,$ complexity of \cref{alg:low_degree_delay_alg} is $\widetilde{O}(\text{s} \log^2 n)$.
\paragraph{Global Derandomization}
Here, we informally sketch a variant of \cref{alg:low_degree_delay_alg} which performs a global derandomization using the method of conditional expectation. A more formal discussion of this approach, though in a different context, is discussed in \cref{subsec:hittingset-impl} where we derandomize our algorithm for the hitting set problem in the $\mathsf{CONGEST}\,$ model. See in particular \cref{thm:hittingset-congest}.
\begin{definition}[Good Random Set $S_{i,j}$ (In Expectation)]
\label{def:low_degree_del_random_good_set}
For a set $S_{i,j} \subseteq V^{\textrm{active}}_{i-1}$ and $u \in V$, let
\[Y_{i,j}(u) = 1 - |\textrm{alive}_{i-1}(u) \cap S_{i,j}| + \binom{|\textrm{alive}_{i-1}(u) \cap S_{i,j}|}{2}.\]
%
We refer to a randomly computed subset $S_{i,j} \subseteq V^{\textrm{active}}_{i-1}$ as good in expectation if
%
\[ \mathbb{E} \left[\sum_{u \in V} Y_{i,j}(u) \frac{\phi_{i,j-1}(u)}{1 - (|\textrm{alive}_{i-1}(u)|/(10k))} + |S_{i,j}|\cdot 2^i \right] \leq \sum_{u \in V} \phi_{i,j-1}(u) + \frac{2^{i-1}}{k}|V^{\textrm{active}}_{i-1}|.\]
\end{definition}
Note that we can recover \cref{def:low_degree_del_good_set} if we drop the expectation. Assume we choose $S_{i,j}$ by including each node in $V^{\textrm{active}}_{i-1}$ with probability $\frac{1}{4k}$, pairwise independently. One can show that the resulting set $S_{i,j}$ is good in expectation. Moreover, the pairwise distribution over the random set $S_{i,j}$ can be realized with a random seed length of $\widetilde{O}(\log n)$ using the construction of~\cite{roditty2005deterministic, luby1993removing} that is described in \Cref{subsec:hittingset-impl}.
The goal is now to fix the random seed one by one in such a way that the resulting deterministic set $S_{i,j}$ is a good set.
For the following discussion, let $X = \sum_{u \in V} Y_{i,j}(u) \frac{\phi_{i,j-1}(u)}{1 - (|\textrm{alive}_{i-1}(u)|/(10k))} + |S_{i,j}|\cdot 2^i$.
The method of conditional expectation works by fixing the bits of the random seed one by one, each time fixing the $i$-th bit in such a way that
%
\[\mathbb{E}[X|\text{first $i$ bits are fixed to $b_0,\ldots,b_i$}] \leq \mathbb{E}[X|\text{first $i-1$ bits are fixed to $b_0,\ldots,b_{i-1}$}].\]
%
In particular, this ensures that
%
\[\mathbb{E}[X|\text{all bits are fixed}] \leq \mathbb{E}[X] \leq \sum_{u \in V} \phi_{i,j-1}(u) + \frac{2^{i-1}}{k}|V^{\textrm{active}}_{i-1}|\]
%
and hence the corresponding deterministic set $S_{i,j}$ is indeed a good set.
%
To find such a bit $b_i$, it suffices to compute two things:
\begin{itemize}
\item $\mathbb{E}[X|\text{first $i$ bits are fixed to $b_0,\ldots,b_{i-1},0$}]$, and
\item $\mathbb{E}[X|\text{first $i$ bits are fixed to $b_0,\ldots,b_{i-1}$}]$.
\end{itemize}
It is possible to decompose $X$ into $X = \sum_{u \in V} X_u$ such that each node $u$, when given $b_0,b_1,\ldots,b_i$, $\textrm{alive}_{i-1}(u)$ and $\phi_{i-1,j}(u)$, can efficiently compute $\mathbb{E}[X_u|b_0,b_1,\ldots,b_i]$, without any further communication. This in turn allows us to compute $\mathbb{E}[X|b_0,b_1,\ldots,b_i]$ in $O(D)$ rounds.
Hence, given that every node knows $\textrm{alive}_{i-1}(u)$ and $\phi_{i-1,j}(u)$, one can find a good set in $\widetilde{O}(D \log n)$ $\mathsf{CONGEST}\,$ rounds, where $D$ denotes the diameter of the network. Hence, computing $\textrm{alive}_{i-1}(u)$ and $\phi_{i-1,j}(u)$ can be done in $\widetilde{O}(\text{s} \log n)$ rounds at the beginning of phase $i$. Moreover, $\textrm{alive}_{i-1}(u)$ and $\phi_{i-1,j}(u)$ can be computed at the beginning of phase $i$ in $\widetilde{O}(\text{s} \log n)$ rounds according to \cref{lem:low_degree_delays_communication}. Hence, the overall resulting run-time of this variant of \cref{alg:low_degree_delay_alg} is $\widetilde{O}((D + \text{s}) \log^2(n)) = \widetilde{O}(D \log^2 n)$. This is the complexity for the setting where we have a low-diameter global tree of depth $D$. One can replace this by a standard application of network decomposition to reduce the round complexity to $\mathrm{poly}(\log n)$. In particular, given a $c$-color $d$-diameter network decomposition of $G^{O(s)}$, we can use independent randomness for the nodes of different colors, and for each color, we can perform the gathering and bit fixing in $(s+d)\mathrm{poly}(\log n)$ rounds. Hence, we can perform the same derandomization in $(s+d)\mathrm{poly}(\log n)$ rounds. With the algorithm of \cite{elkin2022deterministic} that computes a $O(\log n)$-color $\mathrm{poly}(\log n)$-strong-diameter network decomposition in $s \mathrm{poly}(\log n)$ rounds~\cite{elkin2022deterministic}, this becomes a complexity of $s \mathrm{poly}(\log n)$ rounds overall for the whole derandomization procedure. Please see the proof of \Cref{thm:hittingset-pram} where we perform such a global derandomization via network decomposition for the hitting set problem and provide more of the lower-order details. Instead of diving into those details here, in this section, we focus on the local derandomization which leads to a faster round complexity, as discussed in the next subsection.
\subsection{Algorithm $\mathcal{A}_{i,j}$ via Local Derandomization}
\label{sec:low_degree_local_derandomization}
This subsection is dedicated to providing the description of $\mathcal{A}_{i,j}$, that is proving \Cref{thm:deterministic-goodSet-selection} stated below. We note that this is the final missing piece in the proof of \cref{thm:low_degree_main}.
\begin{theorem}\label{thm:deterministic-goodSet-selection}
For every iteration $j$ of phase $i$, there exists a $\mathsf{CONGEST}\,$ algorithm $\mathcal{A}_{i,j}$ which computes a good set $S_{i,j} \subseteq V^{\textrm{active}}_{i-1}$ in $\widetilde{O}(\text{s} \log n)$ rounds.
\end{theorem}
The algorithm $\mathcal{A}_{i,j}$ makes use of the local rounding framework of Faour et al.~\cite{Faour2022} to compute a good set $S_{i,j}$. Their rounding framework works via computing a particular weighted defected coloring of the vertices, which allows the vertices of the same color to round their values simultaneously, with a limited loss in some objective functions that can be written as summation of functions each of which depend on only two nearby nodes. Next, we provide a related definition and then state their black-box local rounding lemma.
\begin{definition}[long-range d2-Multigraph]\label{def:long-range-d2multigraph}
A long-range \emph{d2-multigraph} is a multigraph $H=(V_H,E_H)$ that is simulated on top of an underlying communication graph $G=(V,E)$ by a distributed message-passing algorithm on $G$. The nodes of $H$ are a subset of the nodes of $G$, i.e., $V_H\subseteq V$. The edge set $E_H$ consists of two kinds of edges, \emph{physical edges} and \emph{virtual edges}. Physical edges in $E_H$ are edges between direct neighbors in $G$. For each physical edge in $e\in E_H$ with $V(e)=\set{u,v}$, both nodes $u$ and $v$ know about $e$. Virtual edges in $E_H$ are edges between two nodes $u,v\in V_H$, and for each such virtual edge, there is a manager node $w$ which knows about this edge.
We next describe the assumed communication primitives. Let $M(v)$ be the set of nodes $w$ who manage virtual edges that include $v$. We assume $T$-round primitives that provide the following: (1) each node $v$ can send one $O(\log n)$-bit message that is delivered to all nodes in $M(v)$ in $T$ rounds; (2) given $O(\log n)$-bit messages prepared at nodes $M(v)$ specific for node $v$, node $v$ can receive an aggregation of these messages, e.g., the summation of the values, in $T$ rounds.
\end{definition}
\begin{definition} (Pairwise Utility and Cost Functions) Let $H=(V_H,E_H)$ be a long-range d2-multigraph of an underlying communication graph $G=(V,E)$. For any label assignment $\vec{x}: V_H \rightarrow \Sigma$, a pairwise utility function is defined as $\sum_{u \ in V_H} \mathbf{u}(u, \vec{x}) + \sum_{e \in E_H} \mathbf{u}(e, \vec{x})$, where for a vertex $u$, the function $\mathbf{u}(u, \vec{x})$ is an arbitrary function that depends only on the label of $u$, and for each edge $e=\{u, v\}$, the function $\mathbf{u}(e, \vec{x})$ is an arbitrary function that depends only on the labels of $v$ and $u$. These functions can be different for different vertices $u$ and also for different edges $e$. A pairwise cost function is defined similarly. For a probabilistic/fractional assignment of labels to vertices $V_{H}$, where vertex $v$ assumes each label in $\Sigma$ with a given probability, the utility and costs are defined as the expected values of the utility and cost functions, if we randomly draw integral labels for the vertices from their corresponding distributions (and independently, though of course each term in the summation depends only on the labels of two vertices and thus pairwise independence suffices).
\end{definition}
\begin{lemma}\label{lemma:long-range-d2rounding}[Faour et al.~\cite{Faour2022}]
Let $H=(V_H,E_H)$ be a long-range d2-multigraph of an underlying communication graph $G=(V,E)$ of maximum degree $\Delta$, where the communication primitives have round complexity $T$. Assume that $H$ is equipped with pairwise utility and cost functions $\mathbf{u}(\cdot)$ and $\mathbf{c}(\cdot)$ (with label set $\Sigma$) and with a fractional label assignment $\lambda$. Further assume that the given rounding instance is polynomially bounded in a parameter $q \leq n$. Then for every constant $c>0$ and every $\varepsilon,\mu>\max\set{1/q^c, 2^{-c\sqrt{\log n}}}$, if $\mathbf{u}(\lambda)-\mathbf{c}(\lambda)>\mu\mathbf{u}(\lambda)$, there is a deterministic $\mathsf{CONGEST}\,$ algorithm on $G$ to compute an integral label assignment $\ell$ for which $\mathbf{u}(\ell)-\mathbf{c}(\ell)\geq (1-\varepsilon)\cdot\big(\mathbf{u}(\lambda)-\mathbf{c}(\lambda)\big)$ and such that the round complexity of the algorithm is
\[
T \cdot O\left(\frac{\log^2 q}{\varepsilon\cdot\mu}\cdot\left(\frac{|\Sigma| \log(q\Delta)}{\log n} + \log\log q \right)+\log q\cdot\log^* n \right).
\]
\end{lemma}
\paragraph{Our Local Derandomization.} In the following, for each node $u \in V$, we define $c_u = \frac{\phi_{i,j-1}(u)}{1- (|\textrm{alive}_{i-1} (u)|/(10k))}$.
The labeling space is whether each node in $V^{\textrm{active}}_{i-1}$ is contained in $S_{i,j}$ or not, i.e., each node in $V^{\textrm{active}}_{i-1}$ takes simply one of two possible labels $\Sigma=\{0,1\}$ where $1$ indicates that the node is in $S_{i,j}$. For a given label assignment $\vec{x} \in \{0,1\}^{V^{\textrm{active}}_{i-1}}$, we define the utility function
\[\mathbf{u}(\vec{x}) = \sum_{u \in V} c_u \sum_{v \in \textrm{alive}_{i-1}(u)} x_v + \frac{2^{i-1}}{k}|V_{i-1}^\textrm{active}| = \sum_{v \in V} \left( \sum_{u \in M_i(v)} c_u \right) x_v + \frac{2^{i-1}}{k}|V_{i-1}^{\textrm{active}}|,\]
and the cost function
\[\mathbf{c}(\vec{x}) = \sum_{u \in V} c_u \sum_{v \neq v' \in \textrm{alive}_{i-1}(u)} x_v x_{v'} + \sum_{v \in V^{\textrm{active}}_{i-1}} 2^i x_v.\]
If the label assignment is relaxed to be a fractional assignment $\vec{x} \in [0,1]^{V^{\textrm{active}}_{i-1}}$, where intuitively now $x_v$ is the probability of $v$ being contained in $S_{i,j}$, the same definitions apply for the utility and cost of this fractional assignment.
Note that the utility function is simply a summation of functions, each of which depends on the label of one vertex. Hence, it directly fits the rounding framework.
To capture the cost function as a summation of costs over edges, we next define an auxiliary multi-graph $H$ as follows: For each node $u \in V$ and every $v \neq v' \in \textrm{alive}_i(u)$, we add an auxiliary edge between $v$ and $v'$, with a cost function which is equal to $c_u$ when both $v$ and $v'$ are marked, and zero otherwise.
Note that $H$ is a long-range d2-Multigraph where the communication primitives have round complexity $\widetilde{O}(\text{s} \log n)$ as provided by \cref{lem:low_degree_delays_communication}.
We next argue that the natural fractional assignment where $x_v = \frac{1}{4k}$ for each $v\in V^{\textrm{active}}_{i-1}$ satisfies the conditions of \Cref{lemma:long-range-d2rounding}. First, note that these fractional assignments are clearly polynomially bounded in $q$ for $q=k=O(\log\log n)$. Next, we discuss that, for the given fractional assignment, utility minus cost is at least a constant factor of utility.
\begin{claim}
Let $\vec{x} \in [0,1]^{V^{\textrm{active}}_{i-1}}$ with $x_v = \frac{1}{4k}$ for every $v \in V^{\textrm{active}}_{i-1}$. Then, $\mathbf{u}(\vec{x}) - \mathbf{c}(\vec{x}) \geq \mathbf{u}(\vec{x})/2$.
\end{claim}
\begin{proof}
We have
\begin{align*}
\mathbf{u}(\vec{x}) &= \sum_{u \in V} c_u \sum_{v\in \textrm{alive}_{i-1}(u)} x_v + \frac{2^{i-1}}{k}|V_{i-1}^{\textrm{active}}| \\
&\geq 2 \left(\sum_{u \in V} c_u \sum_{v,v' \in \textrm{alive}_{i-1}(u)} x_v x_{v'} + \sum_{v \in V^{\textrm{active}}_{i-1}} 2^i \frac{1}{4k} \right) \\
&\geq 2 \mathbf{c}(\vec{x}).
\end{align*}
and therefore indeed $\mathbf{u}(\vec{x}) - \mathbf{c}(\vec{x}) \geq \mathbf{u}(\vec{x})/2$.
\end{proof}
Hence, we can apply \Cref{lemma:long-range-d2rounding} on these fractional assignments with $\mu=1/2$ and $\varepsilon=0.1$, which runs in $\widetilde{O}((\log\log \log n)^2)$ iterations of calling the communication primitives, each taking $\widetilde{O}(s\log^2 n)$ rounds. Hence, the entire procedure runs in $\widetilde{O}(s\log^2 n)$ rounds. As a result of applying
\Cref{lemma:long-range-d2rounding} with these parameters, we get an integral label assignment
$\vec{y} \in \{0,1\}^{V^\textrm{active}_{i-1}}$ which satisfies $\mathbf{u}(\vec{y}) - \mathbf{c}(\vec{y}) \geq 0.9 (\mathbf{u}(\vec{x}) - \mathbf{c}(\vec{x}))$. We can then conclude
%
\begin{align*}
\mathbf{u}(\vec{y}) - \mathbf{c}(\vec{y}) &\geq 0.9 (\mathbf{u}(\vec{x}) - \mathbf{c}(\vec{x})) \\
&\geq0.9 \left( \sum_{u \in V} c_u \frac{|\textrm{alive}_{i-1}(u)|}{4k} + \frac{2^{i-1}}{k}|V^{\textrm{active}}_{i-1}| - \left( \sum_{u \in V} c_u \frac{|\textrm{alive}_{i-1}(u)|}{16 k} + \frac{2^{i-2}}{k}|V^{\textrm{active}}_{i-1}|\right) \right)\\
&\geq \sum_{u \in V} c_u \frac{|\textrm{alive}_{i-1}(u)|}{10k}.
\end{align*}
%
This integral label assignment directly gives us $S_{i,j}$. In particular, let $S_{i,j} = \{v \in V^{\textrm{active}}_{i-1} \colon y_v = 1\}$. Note that
%
\begin{align*}
\mathbf{u}(\vec{y}) - \mathbf{c}(\vec{y}) = \sum_{u \in V} c_u \left( |\textrm{alive}_{i-1}(u) \cap S_{i,j}| - \binom{|\textrm{alive}_{i-1}(u) \cap S_{i,j}|}{2} \right) + \frac{2^{i-1}}{k}|V^{\textrm{active}}_{i-1}| - 2^i|S_{i,j}|,
\end{align*}
%
and therefore
%
\begin{align*}
\sum_{u \in V} Y_{i,j}(u)\frac{\phi_{i,j-1}(u)}{1 - (|\textrm{alive}_{i-1}(u)|/(10k))} + |S_{i,j}| \cdot 2^i
&= \sum_{u \in V} c_u - \mathbf{u}(\vec{y}) + \mathbf{c}(\vec{y}) + \frac{2^{i-1}}{k}|V^{\textrm{active}}_{i-1}| \\
&\leq \sum_{u \in V} c_u - \sum_{u \in V} c_u \frac{|\textrm{alive}_{i-1}(u)|}{10k} + \frac{2^{i-1}}{k}|V^\textrm{active}_{i-1}| \\
&\leq \sum_{u \in V} \phi_{i,j-1}(u) + \frac{2^{i-1}}{k}|V^{\textrm{active}}_{i-1}|, \end{align*}
which shows that $S_{i, j}$ is indeed a good set according to \Cref{def:low_degree_del_good_set}. This completes the description of our locally derandomized construction of good sets $S_{i, j}$, hence completing the proof of \Cref{thm:deterministic-goodSet-selection}.
\section{From Low-Degree Clusters to Isolated Clusters}
\label{sec:LowDegtoIsolation}
\subsamplingmain
Similar as in \cref{sec:low_degree}, we could get the same guarantees with a $\mathsf{CONGEST}\,$ algorithm with round complexity $O(s \mathrm{poly}(\log n))$ by performing a global derandomization with the help of a previously computed network decomposition.
\begin{proof}[Proof of \Cref{thm:subsampling_main}]
The clustering $\mathcal{C}^{out}$ is computed in two steps.
In the first step, we use the local rounding procedure to compute a clustering $\mathcal{C}'$ which one obtains from $\mathcal{C}$ by only keeping some of the clusters in $\mathcal{C}$ (any such cluster is kept in its entirety). Intuitively, the local rounding procedure derandomizes the random process which would include each cluster $C$ from $\mathcal{C}$ in the clustering $\mathcal{C}'$ with probability $\frac{1}{2k}$, $k = \lceil 100 \log \log n\rceil$, pairwise independently.
%
Given the clustering $\mathcal{C}'$, we keep each node $u \in \mathcal{C}'$ clustered in $\mathcal{C}^{out}$ if and only if the $\text{s}$-hop degree of $u$ in $\mathcal{C}'$ is $1$.
Note that given $\mathcal{C}'$, the output clustering $\mathcal{C}^{out}$ can be computed in $\widetilde{O}(\text{s} \log n)$ $\mathsf{CONGEST}\,$ rounds.
First, we discuss the first property, i.e., the strong diameter of the output clustering. The fact that the clustering $\mathcal{C}$ has strong diameter $O(\text{s} \log (n))$ directly implies that the clustering $\mathcal{C}'$ also has strong diameter $O(\text{s} \log(n))$, simply because each cluster of $\mathcal{C}'$ is exactly one of the clusters of $\mathcal{C}$. We next argue that $\mathcal{C}^{out}$ also has strong diameter $O(\text{s} \log n)$. Let $u$ be a node clustered in $\mathcal{C}'$ and $P_u$ the unique path from $u$ to its center in the tree associated with its cluster. Then, it directly follows from the definition that for every $w \in P_u$, the $\text{s}$-hop degree of $w$ in $\mathcal{C}'$ is at most the $\text{s}$-hop degree of $u$ in $\mathcal{C}'$. Therefore $u$ being clustered in $\mathcal{C}^{out}$ implies that $w$ is also clustered in $\mathcal{C}^{out}$. Hence, we conclude that $\mathcal{C}^{out}$ indeed has strong diameter $O(\text{s} \log n)$.
Next, we discuss the second property: The clustering $\mathcal{C}^{out}$ is $\text{s}$-hop separated. This directly follows from the fact that by definition every clustered node has a $\text{s}$-hop degree of $1$.
Finally, To prove \Cref{thm:subsampling_main}, the only remaining thing is to prove the third property, i.e., that we compute $\mathcal{C}'$ in such a way that $\mathcal{C}^{out}$ clusters at least $\frac{|\mathcal{C}|}{1000 \log \log (n)}$ nodes. The rest of this proof is dedicated to this property.
For each cluster $C \in \mathcal{C}$, we let $center(C)$ denote the cluster center of $C$ and define $Centers = \{center(C) \colon C \in \mathcal{C}\}$ as the set of cluster centers of $\mathcal{C}$.
%
Moreover, for each $u$ clustered in $\mathcal{C}$, recall that $c_u$ is the cluster center of the cluster of $u$ and let $P_u$ be the unique $u$-$c_u$ path in the tree associated with this cluster $C_u$. Now, let
%
\[S_u = \{C \in \mathcal{C} \colon d(P_u,C) \leq \text{s}\}.\]
%
Note that the size of $S_u$ is equal to the $\text{s}$-hop degree of $u$, which by assumption is at most $k$.
The labeling space is for each cluster center whether its cluster is contained in $\mathcal{C}'$ or not, i.e., each node in $Centers$ takes simply one of two possible labels $\{0,1\}$ where $1$ indicates that the corresponding cluster is in $\mathcal{C}'$. For a given label assignment $\vec{x} \in \{0,1\}^{Centers}$, we define
%
\[\mathbf{u}(\vec{x}) = \sum_{C \in \mathcal{C}} |C|x_{center(C)}\]
%
and
\[\mathbf{c}(\vec{x}) = \sum_{u \in V \colon \text{$u$ is clustered in $\mathcal{C}$}}\sum_{C \in S_u \setminus {C_u}} x_{c_u}x_{center(C)}.\]
If the label assignment is relaxed to be a fractional assignment $\vec{x} \in [0,1]^{Centers}$, where intuitively now $x_v$ is the probability of $v$'s cluster being contained in $\mathcal{C}'$, the same definitions apply for the utility and cost of this fractional assignment.
The utility function is simply a summation of functions, each of which depends on the label of one vertex in $Centers$. Hence, it directly fits the rounding framework.
To capture the cost function as a summation of costs over edges, we next define an auxiliary multi-graph $H$ as follows: For each node $u $ clustered in $\mathcal{C}$ and every $C_1 \neq C_2 \in S_u$, we add an auxiliary edge between $center(C_1)$ and $center(C_2)$ with a cost function which is equal to $1$ when both $C_1$ and $C_2$ are contained in $\mathcal{C}'$, and zero otherwise.
Note that $H$ is a long-range d2-Multigraph, according to \Cref{def:long-range-d2multigraph}. The communication primitives can be implemented in $\widetilde{O}(\text{s} \log n)$ rounds according to the lemma below.
\begin{lemma}
\label{lem:subsampling_delays_communication}
Let $\mathcal{C}$ be the input clustering of \cref{thm:subsampling_main}. There exists a $\mathsf{CONGEST}\,$ algorithm running in $\widetilde{O}(\text{s} \log n)$ rounds which computes for each node $u \in V$ the sets $\{center(C) \colon C \in S_u\}$. Moreover, for each $v \in Centers$, let $M(v) = \{u \in V \colon C_v \in S_u\}.$ Then, there exists an $\widetilde{O}(\text{s} \log n)$ round $\mathsf{CONGEST}\,$ algorithm that allows each node $v$ to send one $O(\log n)$-bit message that is delivered to all nodes in $M(v)$. Similarly, there also exists an $\widetilde{O}(\text{s} \log n)$ round $\mathsf{CONGEST}\,$ algorithm that given $O(\log n)$-bit messages prepared at nodes in $M(v)$ specific for node $v$, it allows node $v$ to receive an aggregation of these messages, e.g., the summation of the values, in $\widetilde{O}(\text{s} \log n)$ rounds.
\end{lemma}
\begin{proof}[Proof of \Cref{lem:subsampling_delays_communication}]
The proof follows roughly along the lines of the proof of \cref{lem:low_degree_delays_communication}. First, we run the following variant of breadth first search: At the beginning, each node clustered in $\mathcal{C}$ has a token which is equal to the identifier of its cluster.
Now, in each of the $\text{s}$ iterations, each node that has received at most $k = \lceil 100 \log \log n \rceil$ identifiers in the previous iteration forwards all the identifiers it has received to its neighbors. If a node has received more than $k$ identifiers, it selects $k$ of them to forward.
The first phase can be implemented in $O(k \text{s})$ $\mathsf{CONGEST}\,$ rounds.
It directly follows from the fact that the $\text{s}$-hop degree of $\mathcal{C}$ is at most $k$ that after the first phase each node $w$ learns the identifiers of all cluster centers such that the corresponding cluster $C$ satisfies $d(w,C) \leq \text{s}$. The next phase propagates this information up in the cluster tree, from the root toward the leaves, such that each descendant of $w$---i.e., any node whose cluster path to the root passes through $w$---learns about all those cluster centers as well.
The second phase consists of $O(\text{s} \log n)$ iterations. In each iteration, each
clustered node sends all the identifiers it learned about so far to each of its children in the corresponding cluster tree.
It again follows from the fact that the $\text{s}$-hop degree of $\mathcal{C}$ is at most $k$ that each of the $O(\text{s} \log n)$ iterations in the second phase can be implemented in $O(k)$ $\mathsf{CONGEST}\,$ rounds. Hence, the overall $\mathsf{CONGEST}\,$ runtime is $\widetilde{O}(\text{s} \log n)$.
For each $v \in Centers$, let $M(v) = \{u \in V \colon C_v \in S_u\}.$ By repeating the above communication, we have a $\widetilde{O}(\text{s} \log n)$-round procedure that delivers one message from each node $v$ to all nodes $M(v)$. By reversing the same communication in time, we can also provide the opposite direction: if each node in $M(v)$ starts with a message for $v$, then in $\widetilde{O}(\text{s} \log n)$ rounds, we can aggregate these messages and deliver the aggregate to $v$, simultaneously for all $v$.
\end{proof}
\begin{claim}
Let $k= \lceil 100 \log \log (n) \rceil$ and $\vec{x} \in [0,1]^{Centers}$ with $x_v = \frac{1}{2k}$ for every $v \in Centers$. Note that this fractional label assignment is polynomially bounded in $q=k=O(\log\log n)$. Furthermore, we have $\mathbf{u}(\vec{x}) - \mathbf{c}(\vec{x}) \geq \mathbf{u}(\vec{x})/2$.
\end{claim}
\begin{proof}
We have
%
\[\mathbf{u}(\vec{x}) = \sum_{C \in \mathcal{C}} |C|x_{center(C)} = \sum_{C \in \mathcal{C}} \frac{|C|}{2k} = \frac{|\mathcal{C}|}{2k},\]
%
and
%
\[\mathbf{c}(\vec{x}) = \sum_{u \in V \colon \text{$u$ is clustered in $\mathcal{C}$}}\sum_{C \in S_u \setminus \{C_u\}} x_{c_u}x_{center(C)} \leq \sum_{u \in V \colon \text{$u$ is clustered in $\mathcal{C}$}} \frac{1}{4k} \frac{|S_u|}{k} \leq \frac{|\mathcal{C}|}{4k}.\]
%
Therefore, indeed $\mathbf{u}(\vec{x}) - \mathbf{c}(\vec{x}) \geq \mathbf{u}(\vec{x})/2$.
\end{proof}
We now invoking the rounding of \Cref{lemma:long-range-d2rounding} with parameters $\mu=0.5$, $\varepsilon=0.5$, and $q=k=O(\log\log n)$ on the fractional label assignment of $\vec{x} \in [0,1]^{Centers}$ where $x_v = \frac{1}{2k}$ for every $v \in Centers$. The procedure runs in $\widetilde{O}(s\log n)$ rounds. As output, we get an integral label assignment $\vec{y} \in \{0,1\}^{Centers}$ which satisfies
%
\[\mathbf{u}(\vec{y}) - \mathbf{c}(\vec{y}) \geq 0.5 (\mathbf{u}(\vec{x}) - \mathbf{c}(\vec{x})) \geq \frac{|\mathcal{C}|}{8k}.\]
Let $C' = \{C \in \mathcal{C} \colon y_{center(C)} = 1\}$. Note that for every $u \in \mathcal{C}$,
%
\[I(\text{$u$ is clustered in $\mathcal{C}^{out}$}) \geq y_{c_u} - \sum_{C \in S_u \setminus \{C_u\}} y_{c_u} y_{center(C)}.\]
%
%
Therefore,
%
\begin{align*}
|\mathcal{C}^{out}|&\geq \sum_{u \in V \colon \text{$u$ is clustered in $\mathcal{C}$}} I(\text{$u$ is clustered in $\mathcal{C}^{out}$}) \\
&\geq \sum_{u \in V \colon \text{$u$ is clustered in $\mathcal{C}$}} \left(y_{c_u} - \sum_{C \in S_u \setminus \{C_u\}} y_{c_u} y_{center(C)}\right)\\
&= \mathbf{u}(\vec{y}) - \mathbf{c}(\vec{y}) \\
&\geq \frac{|\mathcal{C}|}{8k}
\end{align*}
%
and therefore $\mathcal{C}^{out}$ clusters enough vertices to prove \Cref{thm:subsampling_main}.
\end{proof}
\section{Clustering More Nodes}
\label{sec:clusteringmorenodes}
In this section, we prove the following result, which says that once we have access to a clustering algorithm that clusters a nontrivial proportion of nodes with sufficient separation, we can turn it into an algorithm that clusters a constant proportion of nodes. We are paying for this with a slight decrease in the separation guarantees.
\clusteringmorenodesmain
It follows from the analysis of \cref{alg:clustering_with_more_vertices} and its subroutine \cref{alg:expanding}.
To understand the pseudocode of the algorithms, we note that for a set of nodes $C \subseteq V(G)$ and $D \in \mathbb{N}_0$, we define $$C^{\leq D} = \{v \in V \colon d(C,v) \leq D\}.$$
Moreover, we say that a cluster $C$ is \emph{good} in \cref{alg:expanding} if $\textrm{cut}(C) < +\infty$. Otherwise, $C$ is \emph{bad}.
\begin{algorithm}
\caption{Making a clustering algorithm cluster half of the nodes}
\label{alg:clustering_with_more_vertices}
\begin{algorithmic}[1]
\Procedure{\textsc{ClusterHalfNodes($G$)}}{}
\State $\mathcal{C}_0 = \emptyset$
\State $N = \lceil 4 \cdot 2^x\rceil$
\For{$i = 1,2 \ldots, N$}
\State $G_i = G\left[V \setminus \left( \bigcup_{C \in \mathcal{C}_{i-1}} C \right)^{\le 1} \right]$
\State $\mathcal{C} \leftarrow \mathcal{A}(G_i)$
\State $\hat{\fC}_i \leftarrow \textsc{Expand}(G_i, \mathcal{C})$
\State $\mathcal{C}_i = \mathcal{C}_{i-1} \cup \hat{\mathcal{C}}_i$
\EndFor
\Return $\mathcal{C}_N$
\EndProcedure
\end{algorithmic}
\end{algorithm}
\begin{algorithm}
\caption{Expanding an input clustering}
\label{alg:expanding}
{\bf Input:} A graph $G$ and its $10x$-separated clustering $\mathcal{C}$\\
{\bf Output:} An expanded clustering $\hat{\fC}$ with small boundary
\begin{algorithmic}[1]
\Procedure{\textsc{Expand($G, \mathcal{C}$)}}{}
\For{$C \in \mathcal{C}$}
\State Define $\textrm{cut}(C) = \min \left\{0 \le i \le 3x \colon |C^{\leq i+1}| \leq 1.5 |C^{\leq i}| \right\}$ and $\textrm{cut}(C) = +\infty$ if no such $i$ exists.
\State If $\textrm{cut}(C) < +\infty$, define $\textrm{expand}(C) = C ^{\textrm{cut}(C)}$
\EndFor
\State \Return $\hat{\fC} = \{\textrm{expand}(C) : C \in \mathcal{C}, \textrm{cut}(C) < +\infty\}$
\EndProcedure
\end{algorithmic}
\end{algorithm}
We start by analyzing \cref{alg:expanding} in the following lemma. Importantly, the fourth condition for $\hat{\fC}$ in the statement below states that the total number of unclustered vertices neighboring one of the clusters in $\hat{\mathcal{C}}$ is at most half the total number of clustered vertices. This is the reason why we can
\begin{lemma}
\label{lem:clustering_transformed}
Let $x \geq 2$ be arbitrary and $\mathcal{C}$ a clustering with
\begin{enumerate}
\item strong diameter $O(x \log n)$,
\item separation $10 x$ and
\item clustering at least $\frac{n}{2^x}$ nodes.
\end{enumerate}
Then, $\hat{\mathcal{C}}$ constructed in \cref{alg:expanding} is a clustering with
\begin{enumerate}
\item strong diameter $O(x \log n)$,
\item separation $4 x$,
\item clustering at least $0.5\frac{n}{2^x}$ nodes and
\item $\left|\left(\bigcup_{C \in \hat{\fC}} C\right)^{\leq 1}\right| \leq 1.5 \left|\left(\bigcup_{C \in \hat{\fC}} C\right)^{\leq 1}\right|$.
\end{enumerate}
Moreover, the algorithm can be implemented in $O(x \log n)$ $\mathsf{CONGEST}\,$ rounds.
\end{lemma}
\begin{proof}
The first property follows from the fact that for a set $S$ and $D \in \mathbb{N}_0$, $\textrm{diam}(S^{\leq D}) \leq \textrm{diam}(S) + 2D$. Hence, for a good cluster $C$,
\[\textrm{diam}(\textrm{expand}(C)) \leq \textrm{diam}(C) + 2\textrm{cut}(C) = O(x \log n).\]
To prove the second property, let $C_1 \neq C_2 \in \mathcal{C}$ be two arbitrary good clusters. For $i \in \{1,2\}$, let $u_i \in \textrm{expand}(C_i)$ be arbitrary. By triangle inequality, we have:
\[d(u_1,u_2) \geq d(C_1,C_2) - d(C_1,u_1) - d(C_2,u_2) \geq 10 x - 2 \cdot 3x \geq 4x.\]
To prove the third property, it suffices to show that at most $0.5 \frac{n}{2^x}$ of the nodes are contained in bad clusters. For a bad cluster $C$, a simple induction implies $|C^{\leq 3x}| \geq 1.5^{3x} |C| \geq 2 \cdot 2^x |C|$. Therefore,
\[\sum_{C \in \mathcal{C}, \text{$C$ is a bad cluster}} |C| \leq \frac{1}{2^{x+1}} \sum_{C \in \mathcal{C}, \text{$C$ is a bad cluster}} |C^{\leq 3x}| \leq \frac{n}{2^{x+1}}, \]
where the last inequality follows from the fact that for two clusters $C_1 \neq C_2 \in \mathcal{C}$, $C_1^{\leq 3x} \cap C_2^{\leq 3x} = \emptyset$.
To prove the fourth property we write
\begin{align*}
\left|\left(\bigcup_{C \in \hat{\mathcal{C}}} C\right)^{\leq 1}\right| &\leq \sum_{\hat{C} \in \hat{\mathcal{C}}} |\hat{C}^{\leq 1}| \\
&= \sum_{C \in \mathcal{C} \colon \text{$C$ is a good cluster}} |(C^{\leq \textrm{cut}(C)})^{\leq 1}| \\
&= \sum_{C \in \mathcal{C} \colon \text{$C$ is a good cluster}} |C^{\leq \textrm{cut}(C) + 1}| \\
&\leq 1.5 \sum_{C \in \mathcal{C} \colon \text{$C$ is a good cluster}} |C^{\leq \textrm{cut}(C)}| \\
&= 1.5 \left|\left(\bigcup_{C \in \hat{\fC}} C\right)^{\leq 1}\right|.
\end{align*}
It remains to discuss the $\mathsf{CONGEST}\,$ computation. Since we have for any $C_1, C_2 \in \mathcal{C}$ that $C_1^{\le 3x} \cap C_2^{\le 3x} = \emptyset$, each cluster $C \in \mathcal{C}$ can compute the values of $C_1^{\le 0}, C_1^{\le 1}, \dots, C_1^{\le 3x}$ by running one breadth first search from $C_1$ up to distance of $3x$.
\end{proof}
We are now ready to prove \cref{thm:clusteringmorenodesmain}.
\begin{proof}[Proof of \cref{thm:clusteringmorenodesmain}]
We show that the algorithm satisfies the following invariants for $i \in \{0,1,\ldots,N\}$:
\begin{enumerate}
\item $\mathcal{C}_i$ is $2$-separated
\item $|V(\mathcal{C}_i)| \geq n \cdot \min(0.5,\frac{i}{8 \cdot 2^x})$
\item $|V(\mathcal{C}_i^{\leq 1})| \leq 1.5 |V(\mathcal{C}_i)| $
\end{enumerate}
The base case $i = 0$ trivially holds. Now, consider an arbitrary $i \in [N]$ and assume that the invariant is satisfied for $i-1$. To check the first invariant, let $C_1 \neq C_2 \in \mathcal{C}_i$ be arbitrary. If $C_1,C_2 \in \mathcal{C}_{i-1}$, then it follows by induction that $d(C_1,C_2) \geq 2$. If $C_1,C_2 \in \hat{\mathcal{C}}_i$, then it follows from \cref{lem:clustering_transformed} that $d_{G_i}(C_1,C_2) \geq 2$ which also directly implies $d_G(C_1,C_2) \geq 2$. It remains to consider the case that one cluster, let's say $C_1$, is in $\mathcal{C}_{i-1}$ and $C_2$ is in $\hat{\mathcal{C}}_i$.
We have
\[C_2 \subseteq V(G_i) = V \setminus V(\mathcal{C}_{i-1}^{\leq 1}) \subseteq V \setminus C_1^{\leq 1}\]
and therefore $d(C_1,C_2) \geq 2$, as desired.
Next, we show that the second invariant is preserved. If $|V(\mathcal{C}_{i-1})| \geq n/2$, then there is nothing to show. Otherwise, we have
\[|V(G_i)| \geq n - |V(\mathcal{C}_{i-1}^{\leq 1})| \geq n - 1.5 |V(\mathcal{C}_{i-1})| \geq n - 1.5\frac{n}{2} = \frac{n}{4}.\]
Therefore, according to \cref{lem:clustering_transformed}, $\hat{C}_i$ clusters at least $0.5 \frac{(n/4)}{2^x} = \frac{n}{8 \cdot 2^x}$ vertices, which together with $|V(\mathcal{C}_{i-1})| \geq n \cdot \min(0.5,\frac{i-1}{8 \cdot 2^x})$ directly implies $|V(\mathcal{C}_i)| \geq n \cdot \min(0.5,\frac{i}{8 \cdot 2^x})$.
It remains to verify the third property. According to \cref{lem:clustering_transformed}, we have
\[|V(\hat{\mathcal{C}}^{\leq 1}_i) \setminus V(\mathcal{C}^{\leq 1}_{i-1})| \leq 1.5 |V(\hat{\mathcal{C}}_i)|.\]
Therefore,
\[|V(\mathcal{C}^{\leq 1}_i)| = |V(C_{i-1}^{\leq 1})| + |V(\hat{\mathcal{C}}^{\leq 1}_i) \setminus V(\mathcal{C}^{\leq 1}_{i-1})| \leq 1.5 |V(\mathcal{C}_{i-1})| + 1.5 |V(\hat{\mathcal{C}}_i)| = 1.5 |V(\mathcal{C}_i)|.\]
This finishes the proof that the invariants are satisfied throughout the algorithm. Hence, $\mathcal{C}_N$ is a $2$-separated clustering that clusters at least half of the vertices. Moreover, it directly follows from the strong diameter guarantee of \cref{lem:clustering_transformed} that $\mathcal{C}_N$ has strong diameter $O(x \log n)$. Finally, as $\hat{\mathcal{A}}$ has a round complexity of $O(x \log n)$, it follows that $\mathcal{C}_N$ is computed in $O(2^x(R + x \log n))$ $\mathsf{CONGEST}\,$ rounds. This concludes the proof of \cref{lem:pairwise_del_clustering}.
\end{proof}
\iffalse
\section{Low-Diameter Clustering With Pairwise Analysis}
\label{sec:pairwise}
This section is devoted to proving the following theorem.
\begin{theorem}
\label{thm:pairwise}
Let $\text{s} \geq 2$ be arbitrary.
There exists a randomized $\mathsf{CONGEST}\,$ algorithm running in $\widetilde{O}(\text{s} \log n)$ rounds requiring access only to pairwise independent random variables that outputs a clustering $\mathcal{C}$ with
\begin{enumerate}
\item strong diameter $O(\text{s} \log (n))$,
\item separation of $\text{s}$, and
\item the expected number of clustered nodes, i.e., $\mathbb{E} \left[ \left| \bigcup_{C \in \mathcal{C}} C\right|\right]$, is at least $\Omega(n /(\log \log (n)))$.
\end{enumerate}
\end{theorem}
In the subsequent sections, we will strengthen \cref{thm:pairwise} in two ways.
First, in ..., we derandomize the algorithm of \cref{thm:pairwise} using the method of conditional expectation. This derandomization method requires the computation of global aggregates and therefore the resulting round complexity of O() has a dependency in terms of the diameter of the graph.
Second, in ..., we derandomize the algorithm of \cref{thm:pairwise} using a novel local derandomization technique developed in a concurrent paper, resulting in a deterministic algorithm with round complexity O().
Second, one shortcoming of \cref{thm:pairwise} is that the (expected) number of clustered nodes is $\Omega(n/\log \log (n))$, while existentially it is possible to satisfy the first two guarantees and at the same time clustering half of the nodes.
In \cref{sec:clustering_deterministic_transformation}, we show that we can compute a clustering with strong diameter $O(\log n \log \log \log n)$ clustering half of the nodes. This clustering is computed by invoking the algorithm of the deterministic version of \cref{thm:pairwise} $O(\log \log n)$ times, each time with separation parameter $\text{s} = O(\log \log \log n)$.
The algorithm proving \cref{thm:pairwise} consists of two parts. The first part computes a clustering $\mathcal{C}$ with strong diameter $O(\text{s} \log n)$ which, in expectation, clusters at least half of the nodes. However, the clustering only satisfies that the $\text{s}$-hop degree, defined below, is $O(\log \log n)$. This relaxes the separation property in the sense that a (non-empty) clustering is $\text{s}$-separated if and only if its $\text{s}$-hop degree is $1$.
In the second part of the algorithm, we make the clusters $\text{s}$-separated, at the cost of clustering only $\Omega(1/\log\log n)$ fraction of nodes. We define the notion of $s$-hop degree next.
\begin{definition}[$\text{s}$-hop degree of a clustering $\mathcal{C}$]
Let $\text{s} \geq 2$ be arbitrary.
For a clustered node $u$, let $P_u$ denote the path between $u$ and its cluster center in the tree associated with its cluster. The $\text{s}$-hop degree of $u$ is defined as the number of different clusters $C \in \mathcal{C}$ with $d(P_u,C) \leq \text{s}$.
The $\text{s}$-hop degree of a clustering $\mathcal{C}$ is defined as the largest $\text{s}$-hop degree of any node $u \in C \in \mathcal{C}$.
\end{definition}
\cref{sec:pairwise_low_degree} is dedicated to proving the following theorem.
\begin{theorem}
\label{thm:pairwise_low_degree}
Let $\text{s} \geq 2$ be arbitrary. There exists a randomized $\mathsf{CONGEST}\,$ algorithm running in $\widetilde{O}(\text{s} \log(n))$ rounds requiring only access to pairwise independent random variables and which computes a clustering $\mathcal{C}$ with
\begin{enumerate}
\item strong diameter $O(\text{s} \log(n))$,
\item $\text{s}$-hop degree of at most $10 \log \log (n)$, and
\item the expected number of clustered nodes is at least $n/2$.
\end{enumerate}
\end{theorem}
Given such a clustering, a simple subsampling idea can be used to compute a $\text{s}$-separated clustering at the expense that the expected number of clustered nodes goes down by an $O(\log \log n)$ factor. In particular, in \cref{sec:pairwise_subsampling} we prove the following theorem.
\begin{restatable}{theorem}{subsampling}
\label{thm:pairwise_subsampling}
Assume we are given a clustering $\mathcal{C}$ with
\begin{enumerate}
\item strong diameter $O(\text{s} \log (n))$ and
\item $\text{s}$-hop degree of at most $10 \log \log (n)$.
\end{enumerate}
There exists a randomized $\mathsf{CONGEST}\,$ algorithm running in $\widetilde{O}(\text{s} \log(n))$ rounds requiring only access to pairwise independent random variables which computes a clustering $\mathcal{C}^{out}$ with
\begin{enumerate}
\item strong diameter $O(\text{s} \log (n))$,
\item separation of $\text{s}$ and
\item the expected number of clustered nodes is at least $\frac{|\mathcal{C}|}{1000 \log \log (n)}$.
\end{enumerate}
\end{restatable}
Note that \cref{thm:pairwise} directly follows from \cref{thm:pairwise_low_degree} together with \cref{thm:pairwise_subsampling}.
\subsection{Computing a Low-Degree Clustering}
\label{sec:pairwise_low_degree}
This section is dedicated to the proof of \cref{thm:pairwise_low_degree}.
The algorithm proving \cref{thm:pairwise_low_degree} relies on the notion of a delay function $\mathrm{del}$ which assigns each vertex a value from the set $\{0,1,2,\ldots O(\text{s} \log(n))\}$.
\begin{definition}[Delay Function]
Given a graph $G$, a delay function $\mathrm{del}$ is a function assigning each node $u \in V(G)$ a its \emph{delay} $\mathrm{del}(u) \in \{0,1,\ldots,O(\text{s} \log (n))\}$.
\end{definition}
The delay of each vertex is computed by a simple subsampling procedure requiring only pairwise independence (\cref{alg:pairwise_delays}).
Before discussing and analyzing \cref{alg:pairwise_delays}, we first explain how each delay function $\mathrm{del}$ gives rise to a clustering $\mathcal{C}^{\mathrm{del}}$ with strong diameter $O(\text{s} \log n)$ and $\text{s}$-hop degree $O(\log \log n)$.
To this end, we need to introduce additional notation. Given a delay function $\mathrm{del}$, we define for every vertex $u \in V$ its \mathrm{wait} function $\mathrm{wait}(u)$ as
\[\mathrm{wait}(u) = \min_{v \in V} \mathrm{del}(v) + d(v,u).\]
The intuition behind $\mathrm{wait}(u)$ is as follows: Assume that each node $v$ starts sending out a token at time $\mathrm{del}(v)$. Then, $\mathrm{wait}(u)$ is the time it takes until $u$ receives the first token.
Finally, for every parameter $D \ge 0$ and any node $u \in V(G)$ we define its \emph{frontier of width $D$} as
\[\mathrm{frontier}^D(u) = \{v \in V \colon \mathrm{del}(v) + d(v,u) \leq \mathrm{wait}(u) + D\}.\]
Informally speaking, $\mathrm{frontier}^D(u)$ contains each node $v$ whose token arrives at $u$ at most $D$ time units after $u$ receives the first token.
We are now ready to define a clustering $\mathcal{C}^{\mathrm{del}}$ based on a delay function $\mathrm{del}$ and a separation parameter $s$.
The clustering $\mathcal{C}^{\mathrm{del}}$ clusters all the nodes which have a small frontier of width $2 \text{s}$.
In particular, each node $u \in V(G)$ satisfying $|\mathrm{frontier}^{2\text{s}}(u)| \leq 100 \log \log(n)$ is included in some cluster of $\mathcal{C}^{\mathrm{del}}$.
More concretely, each clustered node $u$ gets clustered to the cluster corresponding to the node with the smallest identifier in the set $\mathrm{frontier}^0(u)$. In other words, it gets clustered to the minimizer of $\mathrm{wait}(u)$ with the smallest identifier. In the following text, we denote this node by $c_u$ (see \cref{fig:mpx}).
\begin{figure}
\centering
\includegraphics[width = .6\textwidth]{img/mpx.eps}
\caption{The figure shows the run of the clustering algorithm constructing $\mathcal{C}^{\mathrm{del}}$.
The algorithm can be seen as starting a breadth first search from a single node $\sigma$ connected to every node $u \in V(G)$ with an edge of length $\mathrm{del}(u)$ (the $\mathsf{CONGEST}\,$ implementation of the algorithm does not need to simulate any such node $\sigma$).
The value $\mathrm{wait}(u)$ is the time until the search reaches the node $u$. The node that reaches $u$ the first is denote $c_u$.
Moreover, we cluster only nodes such that the size of their frontier of width $2s$ is at most $O(\log\log n)$. For example, the node $v$ is not clustered because after it is reached by the first node $c_v$, it is reached by $\Omega(\log \log n)$ other nodes in the following $2s$ steps.
It can be seen that for any $w$ on the path from $c_u$ to $u$, we have $\mathrm{frontier}^{2s}(w) \subseteq \mathrm{frontier}^{2s}(u)$, hence the constructed clusters are connected.
}
\label{fig:mpx}
\end{figure}
\begin{lemma}
\label{lem:pairwise_del_clustering}
Let $\mathrm{del}$ be a delay function and $\mathcal{C}^{\mathrm{del}}$ the corresponding clustering described above. Then, $\mathcal{C}^{\mathrm{del}}$ has
\begin{enumerate}
\item strong diameter $O(\text{s} \log n)$,
\item $\text{s}$-hop degree at most $100 \log \log (n)$ and
\item the total number of clustered nodes is equal to $|\{u \in V \colon \mathrm{frontier}^{2\text{s}}(u)| \leq 100 \log \log(n)\}|$.
\end{enumerate}
Moreover, the clustering $\mathcal{C}^{\mathrm{del}}$ can be computed in $O(\text{s} \log n \log\log n)$ $\mathsf{CONGEST}\,$ rounds.
\end{lemma}
To prove the lemma, we first observe that the frontiers have the following property.
\begin{claim}
\label{cl:frontier}
Let $w$ be any node on a shortest path from $u$ to $c_u$ and let $D \ge 0$. Then,
\[
\mathrm{frontier}^D(w) \subseteq \mathrm{frontier}^D(u).
\]
Moreover, $c_w = c_u$.
\end{claim}
\begin{proof}
Fix any such $w$ and consider any $v \in \mathrm{frontier}^D(w)$. The definition of $v$ implies that
$$\mathrm{del}(v) + d(v, w) \le \mathrm{del}(c_u) + d(c_u, w) + D.$$
Since $w$ lies on a shortest path from $c_u$ to $u$, we can add $d(w, u)$ to both sides of the equation to conclude that
$$\mathrm{del}(v) + d(v, w) + d(w, u) \le \mathrm{del}(c_u) + d(c_u, u) + D = \mathrm{wait}(u) + D$$
thus
$$\mathrm{del}(v) + d(v, u) \le \mathrm{wait}(u) + D.$$
Hence, we have $v \in \mathrm{frontier}^D(u)$ and we are getting $\mathrm{frontier}^D(w) \subseteq \mathrm{frontier}^D(u)$ as needed.
We continue with the second part of the statement. In view of the above proof that implies that $\mathrm{frontier}^0(w) \subseteq \mathrm{frontier}^0(u)$, it suffices to show that $c_u \in \mathrm{frontier}^0(w)$.
To see this, we use the fact that $c_u \in \mathrm{frontier}^0(u)$ to write
\[\mathrm{del}(c_u) + d(c_u,u) = \mathrm{wait}(u) \leq \mathrm{wait}(w) + d(w,u)\]
Subtracting $d(w, u)$ from both sides of the equation and using that $w$ lies on a shortest path from $u$ to $c_u$ gives
\[\mathrm{del}(c_u) + d(c_u,w) \leq \mathrm{wait}(w).\]
Thus $c_u \in \mathrm{frontier}^0(w)$ and we are done.
\end{proof}
We continue with the proof of \cref{lem:pairwise_del_clustering}.
\begin{proof}
We start with the first property. Let $u$ be an arbitrary clustered node and recall that $c_u$ is its cluster center.
As $c_u \in \mathrm{frontier}^0(u)$, we have $d(c_u,u) = \mathrm{wait}(u)$.
Moreover,
\[\mathrm{wait}(u) = \min_{v \in V} \mathrm{del}(v) + d(v,u) \leq \mathrm{del}(u) + d(u,u) = \mathrm{del}(u) = O(\text{s} \log n).\]
Hence, we have $d(c_u,u) = O(\text{s} \log n)$.
Moreover, \cref{cl:frontier} gives that all nodes on a shortest path from $c_u$ to $u$ are also clustered to $c_u$, implying that the diameter of the cluster is $O(s \log n)$.
Next, we prove the second property.
Consider an arbitrary clustered node $w$. We first show that for an arbitrary clustered node $y$ with $d(w,y) \leq \text{s}$, it holds that $c_y \in \mathrm{frontier}^{2\text{s}}(w)$.
To see this, we first use the definition of $c_w$ to write
\begin{align*}
\mathrm{del}(c_y) + d(c_y,y)
\leq \mathrm{del}(c_w) + d(c_w,y)
\end{align*}
On one hand, we can use triangle inequality to lower bound the left-hand side by
\begin{align*}
\mathrm{del}(c_y) + d(c_y,y) \ge \mathrm{del}(c_y) + d(c_y, w) - d(w,y)
\end{align*}
On the other hand, we can use triangle inequality to upper bound the right hand side by
\begin{align*}
\mathrm{del}(c_w) + d(c_w,y)
\le \mathrm{del}(c_w) + d(c_w, w) + d(w, y)
\end{align*}
Putting the two bounds together, we conclude that
\begin{align*}
\mathrm{del}(c_y) + d(c_y, w) \le \mathrm{del}(c_w) + d(c_w, w) + 2d(w,y) \le \mathrm{wait}(w) + 2s
\end{align*}
where we used our assumption $d(w,y) \le 2s$. Above inequality is in other words saying that $c_y \in \mathrm{frontier}^{2\text{s}}(w)$ as we wanted to prove.
Now, let $u$ be an arbitrary clustered node and $P_u$ the unique path between $u$ and $c_u$ in the tree associated with the cluster. Furthermore, let $C \in \mathcal{C}^{\mathrm{del}}$ be a cluster with $d(P_u,C) \leq \text{s}$. Then there exists $w \in P_u$ and $y \in C$ with $d(w,y) \leq \text{s}$ and the discussion above implies $c_y \in \mathrm{frontier}^{2\text{s}} (w)$.
We now use \cref{cl:frontier} that says that $\mathrm{frontier}^{2\text{s}} (w) \subseteq \mathrm{frontier}^{2\text{s}}(u)$.
Hence, for each cluster $C$ with $d(P_u,C) \leq \text{s}$, the corresponding cluster center is contained in $\mathrm{frontier}^{2\text{s}}(u)$. As $u$ is clustered, $|\mathrm{frontier}^{2\text{s}}(u)| \leq 100 \log \log n$, and therefore the $\text{s}$-hop degree of $\mathcal{C}^{\mathrm{del}}$ is at most $100 \log \log (n)$.
The third property follows directly from the definition.
To finish the proof, we need to show that the algorithm can be implemented in $O(s \log n \cdot \log\log n)$ rounds.
To see this, note that we can compute for each node $u$ whether $|\mathrm{frontier}^{2s}(u)| \le 100\log\log n$ as follows: We run a variant of breadth first search that takes into account the delays.
Moreover, recall that in classical breadth first search, after a node is $u$ is reached for the first time by a packet sent from $c_u$, it broadcasts it to all its neighbors and then it does not redirect any other packets sent to it.
In our version of the search, each node stops redirecting only after at least $100\log\log n$ packets arrived (we do not take into account packets that have already arrived earlier) or after it counts $2s$ steps from the arrival of the first packet.
It can be seen that this algorithm can be implemented in the desired number of rounds. Moreover, every node $u$ learns the value of $|\mathrm{frontier}^{2s}(u)|$ whenever the value is at most $100\log\log n$ and it learns the value is larger than this threshold otherwise.
\end{proof}
In view of \cref{lem:pairwise_del_clustering}, it suffices to show that \cref{alg:pairwise_delays} stated next computes a delay function $\mathrm{del}$ such that the expected number of nodes $u$ with $|\mathrm{frontier}^{2\text{s}}(u)| \leq 100 \log \log (n)$ is at least $n/2$.
\begin{restatable}{algorithm}{delays}
\caption{Computes Delay Function $\mathrm{del}$}
\label{alg:pairwise_delays}
\begin{algorithmic}[1]
\Procedure{Delays}{}
\State $V_0^ { \textrm{active}} = V$
\State $R = \lfloor 2 \log(n) \rfloor$
\State $k = \lceil 100 \log \log (n) \rceil$
\For{$i = 1,2, \ldots, R$}
\State $W_{i,0} = \emptyset$
\For{$j = 1,2,\ldots,k$}
\State $W_{i,j} = W_{i,j-1} \cup S_{i,j} = \mathcal{A}_{i,j}(V_0^ { \textrm{active}}, V_1^ { \textrm{active}}, \ldots, V_{i-1}^ { \textrm{active}}, W_{i,j-1})$ \Comment{In this section, we obtain the set $S_{i,j}$ by including each vertex with probability $\frac{1}{2k}$, assuming only pairwise independence.}
\EndFor
\State $V_{i+1}^{ \textrm{active}} = W_{i,k}$
\EndFor
\State For every $v \in V, i_v = \max \{i \in \{0,1,\ldots,R\} \colon v \in V_i^ { \textrm{active}}\}$
\State For every $v \in V$, $\mathrm{del}(v) = 5\text{s} (R - i_v)$ \\
\Return $\mathrm{del}$
\EndProcedure
\end{algorithmic}
\end{restatable}
The algorithm runs in $R = \lfloor 2 \log(n)\rfloor$ phases.
In the beginning, all of the nodes are active and in each phase, roughly half of the active nodes stop being active. Hence, after the last phase, all nodes are inactive, with high probability.
In the end, each node gets a delay of $\mathrm{del}(v) = 5 \text{s} (R - i_v) = O(\text{s} \log(n))$, where $i_v$ is the last phase where $v$ was still active.
For the sake of analysis, we define a delay function $\mathrm{del}_i$ for every phase $i \in \{0,1,2\ldots,R\}$ by defining
\[\mathrm{del}_i(v) = 5\text{s}(R - \min(i,i_v)).\]
Note that the delay of each node $v$ decreases by $5 \text{s}$ in each phase until $v$ stops being active, in which case the delay remains unchanged until the end. We also have $\mathrm{del}(v) = \mathrm{del}_R(v)$.
Similar as above, we now define for every vertex $u \in V$,
\[\mathrm{wait}_i(u) = \min_{v \in V} \mathrm{del}_i(v) + d(v,u)\]
and for every $D \in \mathbb{N}_0$,
\[\\frontierD_i(u) = \left\{v \in V \colon \mathrm{del}_i(v) + d(v,u) \leq \mathrm{wait}_i(u) + D \right\}.\]
\begin{restatable}{definition}{alivedead}[$\textrm{alive}_i(u)$/$\textrm{dead}_i(u)$]
For every vertex $u \in V$ and $i \in \{0,1,\ldots,R\}$, let $\textrm{dead}_i(u) \subseteq \mathrm{frontier}^{2 \text{s}}_i(u) \setminus V_i^{\textrm{active}}$ be an arbitrary subset of size $\min(k, |\mathrm{frontier}^{2 \text{s}}_i(u) \setminus V_i^{\textrm{active}}|)$ and $\textrm{alive}_i(u) \subseteq \mathrm{frontier}^{2 \text{s}}_i(u) \setminus V_i^{\textrm{active}}$ be an arbitrary subset of size $\min(k - |\textrm{dead}_i(u)|, |\mathrm{frontier}^{2 \text{s}}_i(u) \cap dead_i(u)|)$.
\end{restatable}
Recall that our goal is to show that the expected number of nodes in the set $\{u \in V \colon |\mathrm{frontier}^{2\text{s}}(u)| \leq 100 \log \log (n)\}$ is at least $n/2$. To that end, we define a suitable potential function $\Phi_i$ for every phase $i$.
\begin{definition}[Outer Potential]
\[\Phi_i(u) = e^{\frac{|\textrm{dead}_i(u)|}{10}}\]
\[\Phi_i = \sum_{u \in V} \Phi_i(u) + 2^i|V^{\textrm{active}}_i|.\]
\end{definition}
The following lemma states that the outer potential is at most $2n$ at the beginning and if the outer potential is sufficiently small at the end, then the corresponding clustering clusters a large fraction of the vertices.
\begin{lemma}[Outer Potential Lemma]
\label{lem:pairwise_outer_potential_lemma}
We have $\Phi_0 \leq 2n$. Moreover, if $\Phi_R \leq 10 n \log(n)$, then $\{u \in V \colon |\mathrm{frontier}^{2\text{s}}(u)| \leq 100 \log \log (n)\} \geq \frac{9n}{10}$.
\end{lemma}
\begin{proof}
First, note that $\Phi_R \geq 2^R|V_R^ { \textrm{active}}| > 10n \log(n)|V_R^{ \textrm{active}}|$.
As we assume that $\Phi_R \leq 10 n \log(n)$, this directly implies $V_R^ { \textrm{active}} = \emptyset$.
In particular, every $u \in V$ with $|\mathrm{frontier}^{2\text{s}}(u)| > 100 \log \log (n)$ contributes
\[\Phi_R(u) = e^{\frac{|\textrm{dead}_R(u)|}{10}}\geq e^{\frac{\min(k,|\mathrm{frontier}^{2\text{s}}(u)|}{10}} \geq 100 \log(n) \]
to the potential. Hence, there can be at most $\Phi_R/(100 \log(n)) \leq n/10$ such nodes and therefore $\{u \in V \colon |\mathrm{frontier}^{2\text{s}}(u)| \leq 100 \log \log (n)\} \geq \frac{9n}{10}$, as desired.
\end{proof}
\begin{restatable}{definition}{pessimisticprobability}[Pessimistic Estimator Probability $p_{i,j}(u)$]
\[p_{i,j}(u) = I(\textrm{alive}_i(u) \cap W_{i,j} = \emptyset) \cdot \left(1 - \frac{|\textrm{alive}_i(u)|}{10k}\right)^{k-j}\]
\end{restatable}
\todo{Briefly mention the intuition behind it}
\begin{restatable}{definition}{innerpotential}[Inner Potential]
Let .
\[\phi_{i,j}(u) = p_{i,j}(u) e^{\frac{|\textrm{dead}_i(u)| + |\textrm{alive}_i(u)|}{10}}\]
and
\[\phi_{i,j} = \sum_{u \in V} \phi_{i,j}(u) + |W_{i,j}|2^{i+1} + \frac{k-j}{k}2^i |V_i^{\textrm{active}}|.\]
\end{restatable}
\begin{restatable}{lemma}{Sijalg}
\label{lem:pairwise_Sij_alg}
Assume that for every $i \in [R], j \in [k]$, algorithm $\mathcal{A}_{i,j}$ is a deterministic algorithm which in $R$ rounds computes a set $S_{i,j}$ such that for
\[Y_{i,j}(u) = 1 - |\textrm{alive}_i(u) \cap S_{i,j}| + \binom{|\textrm{alive}_i(u) \cap S_{i,j}|}{2},\]
it holds that
\[ \sum_{u \in V} Y_{i,j}(u) \frac{\phi_{i,j-1}(u)}{1 - (|\textrm{alive}_i(u)|/10k)} + |S_{i,j}|\cdot 2^{i+1} \leq \sum_{u \in V} \phi_{i,j-1}(u) + \frac{2^i}{k}|V^{\textrm{active}}_i|.\]
Then, \cref{alg:pairwise_delays} runs in $O(R \log(n))$ rounds and computes a delay function $\mathrm{del}$ satisfying $|\mathcal{C}^{\mathrm{del}}| \geq \frac{n}{2}$.
Suppose $\mathcal{A}_{ij}$ is a randomized algorithm that in $R$ rounds computes a set $S_{i,j}$ such that
\[ \mathbb{E}[\sum_{u \in V} Y_{i,j}(u) \frac{\phi_{i,j-1}(u)}{1 - (|\textrm{alive}_i(u)|/10k)} + |S_{i,j}|\cdot 2^{i+1} ]\leq \sum_{u \in V} \phi_{i,j-1}(u) + \frac{2^i}{k}|V^{\textrm{active}}_i|\]
Then, \cref{alg:pairwise_delays} runs in $O(R \log(n))$ rounds and computes a delay function $\mathrm{del}$ satisfying $\mathbb{E}[|C^{\mathrm{del}}| \geq \frac{n}{2}]$.
\end{restatable}
\begin{proof}
Consider an arbitrary node $u \in V$. Note that $I(\textrm{alive}_i(u) \cap S_{i,j} = \emptyset) \leq Y_{i,j}(u)$. Therefore,
\begin{align*}
&Y_{i,j}(u) \frac{\phi_{i,j-1}(u)}{1 - (|\textrm{alive}_i(u)|/10k)} \\
&\geq I(\textrm{alive}_i(u) \cap S_{i,j} = \emptyset)\frac{p_{i,j-1}(u)e^{\frac{|\textrm{dead}_i(u)| + |\textrm{alive}_i(u)|}{10}}}{1 - (|alive_i(u)|/10k)} \\
&= I(\textrm{alive}_i(u) \cap S_{i,j} = \emptyset)\frac{I(\textrm{alive}_i(u) \cap W_{i,j-1} = \emptyset)\left( 1 - \frac{|\textrm{alive}_i(u)}{10k}\right)^{k-(j-1)}1.1^{|\textrm{dead}_i(u)| + |\textrm{alive}_i(u)|}}{1 - (|\textrm{alive}_i(u)|/10k)}\\
&= I(\textrm{alive}_i(u) \cap W_{i,j} = \emptyset)\left( 1 - \frac{|\textrm{alive}_i(u)|}{10k}\right)^{k-j}1.1^{|\textrm{dead}_i(u)| + |\textrm{alive}_i(u)|} \\
&= \phi_{i,j}(u).
\end{align*}
Therefore,
\begin{align*}
\phi_{i,j} &= \sum_{u \in V} \phi_{i,j}(u) + |W_{i,j}|2^{i+1} + \frac{k-j}{k}2^i|V_i^{\textrm{active}}| \\
&\leq \sum_{u \in V} Y_{i,j}(u) \frac{\phi_{i,j-1}(u)}{1 - (|\textrm{alive}_i(u)|/10k)} + (|S_{i,j}| + |W_{i,j-1}|)2^{i+1} + \frac{k-j}{k}2^i|V^{\textrm{active}}_i| \\
&\leq \sum_{u \in V} \phi_{i,j-1}(u) + \frac{2^i}{k}|V^{\textrm{active}}_i| + |W_{i,j-1}|2^{i+1} + \frac{k-j}{k}2^i|V^{\textrm{active}}_i| \\
&= \phi_{i,j-1}.
\end{align*}
As $\phi_{i,j} \leq \phi_{i,j-1}$, \cref{lem:inner_potential} implies that $\Phi_R \leq 4n \log n$. Therefore, \cref{lem:pairwise_outer_potential_lemma} gives that $\{u \in V \colon |\mathrm{frontier}^{2s}(u)| \leq 100 \log \log (n)\} \geq \frac{9n}{10}$. Hence, according to \cref{lem:pairwise_del_clustering}, $\mathcal{C}^{\mathrm{del}}$ clusters at least $\frac{9n}{10} \geq \frac{n}{2}$ nodes, as needed.
In the randomized case, the same set of calculations gives $\mathbb{E}[\phi_{i,j}] \leq \phi_{i,j-1}$ and therefore \cref{lem:inner_potential} implies that $\mathbb{E}[\Phi_R] \leq 4n \log(n)$. Hence, a Markov bound implies that $Pr[\Phi_R \leq 10n \log(n)] \geq 1 - \frac{4}{10} = \frac{3}{5}$. Hence, according to \cref{lem:pairwise_del_clustering},$\mathcal{C}^{\mathrm{del}}$ clusters at least $\frac{9n}{10}$ nodes with probability at least $\frac{3}{5}$ and hence the expected number of clustered nodes is at least $\frac{n}{2}$.
The $\mathsf{CONGEST}\,$ round complexity of $O(R \log n \log \log n)$ directly follows from the fact that for each $i \in [R]$ and $j \in [k]$, algorithm $\mathcal{A}_{i,j}$ is invoked exactly once.
\end{proof}
Hence, it suffices to show that the final potential is small in expectation. This directly follows from the following lemma.
\begin{restatable}{lemma}{pairwisesampling}[Pairwise Sampling Lemma]
Let $i \in [N]$ and $j \in [k]$ be arbitrary.
\label{lem:pairwise_analysis_sampling_lemma}
Let $\{X(v) \colon v \in V^{\textrm{active}}_i\}$ be a family of pairwise independent $0/1$-random variables with $Pr[X(v) = 1] = p$ for some $p \in [\frac{1}{4k},\frac{1}{2k}]$ for every $v \in V^{\textrm{active}}_i$. If one defines $S_{i,j} = \{v \in V^{\textrm{active}}_i \colon X(v) = 1\}$, then
\[\mathbb{E} \left[\sum_{u \in V} Y_{i,j}(u) \frac{\phi_{i,j-1}(u)}{1 - (|\textrm{alive}_i(u)|/10k)} + |S_{i,j}|\cdot 2^{i+1} \right] \leq \sum_{u \in V} \phi_{i,j-1}(u) + \frac{2^i}{k}|V^{\textrm{active}}_i|.\]
\end{restatable}
\begin{proof}
Let $u \in V$ be arbitrary. We have
\begin{align*}
\mathbb{E}[Y_{i,j}(u)] &= \mathbb{E} \left[ 1 - |\textrm{alive}_i(u) \cap S_{i,j}| + \binom{|\textrm{alive}_i(u) \cap S_{i,j}|}{2} \right] \\
&= \mathbb{E}[1 - \sum_{v \in \textrm{alive}_i(u)} X(v) + \sum_{v \neq v' \in \textrm{alive}_i(U)} X(v)X(v')] \\
&= 1 - \sum_{v \in \textrm{alive}_i(u)} \mathbb{E}[X(v)] + \sum_{v \neq v' \in \textrm{alive}_i(U)} \mathbb{E}[X(v)]\mathbb{E}[X(v')] \\
&\leq 1 - p|\textrm{alive}_i(u)| + p^2|\textrm{alive}_i(u)|^2 \\
&\leq 1 - (p/2)|\textrm{alive}_i(u)| \\
&\leq 1 - \frac{|\textrm{alive}_i(u)|}{10k}.
\end{align*}
Therefore,
\begin{align*}
\mathbb{E} \left[\sum_{u \in V} Y_{i,j}(u) \frac{\phi_{i,j-1}(u)}{1 - (|\textrm{alive}_i(u)|/10k)} + |S_{i,j}|\cdot 2^{i+1} \right] &= \sum_{u \in V} \mathbb{E}[Y_{i,j}(u)] \frac{\phi_{i,j-1}(u)}{1 - (|\textrm{alive}_i(u)|/10k)} + \mathbb{E}[|S_{i,j}|]\cdot 2^{i+1} \\
&\leq \sum_{u \in V} \phi_{i,j-1}(u) + \frac{2^i}{k}|V^{\textrm{active}}_i|.
\end{align*}
\end{proof}
We now have all the pieces together to prove \cref{thm:pairwise_low_degree}.
\begin{proof}[Proof of \cref{thm:pairwise_low_degree}]
\cref{lem:pairwise_analysis_sampling_lemma} implies that for every $i \in [R], j \in [k]$, there exists a $0$-round randomized $\mathsf{CONGEST}\,$ algorithm only using pairwise independence which computes a set $S_{i,j}$ such that
\[\mathbb{E} \left[\sum_{u \in V} Y_{i,j}(u) \frac{\phi_{i,j-1}(u)}{1 - (|\textrm{alive}_i(u)|/10k)} + |S_{i,j}|\cdot 2^{i+1} \right] \leq \sum_{u \in V} \phi_{i,j-1}(u) + \frac{2^i}{k}|V^{\textrm{active}}_i|.\]
Hence, according to
\cref{lem:pairwise_Sij_alg}, there exists a $0$-round $\mathsf{CONGEST}\,$ algorithm only using pairwise independence which computes a delay function $\mathrm{del}$ such that the expected number of clustered nodes in the clustering $\mathcal{C}^{\mathrm{del}}$ is at least $n/2$.
Thus, \cref{thm:pairwise_low_degree} now directly follows from applying \cref{lem:pairwise_del_clustering}.
\end{proof}
\paragraph{Global Derandomization}
In \cref{sec:pairwise_low_degree}, we have proven the following lemma.
\pairwisesampling*
Next, we explain how we can use \cref{lem:pairwise_analysis_sampling_lemma} together with the method of conditional expectation to prove the following Lemma.
\begin{lemma}[Global Derandomization]
Let $D$ be the diameter of the input graph.
For every $i \in [R]$, $j \in [k]$, there exists a deterministic algorithm $\mathcal{A}_{i,j}$ which in $\widetilde{O}((\text{s} + D) \log n)$ rounds computes a set $S_{i,j}$ such that
\[\sum_{u \in V} Y_{i,j}(u) \frac{\phi_{i,j-1}(u)}{1 - (|\textrm{alive}_i(u)|/10k)} + |S_{i,j}| \cdot 2^{i+1} \leq \sum_{u \in V} \phi_{i,j-1}(u) + \frac{2^i}{k}|V^{\textrm{active}}_i|.\]
\end{lemma}
\begin{proof}
Let $\{X(v) \colon v \in V_i^{\textrm{active}}\}$ be a set of $0/1$-random variables following some distribution $Q$ and let $S_{i,j} = \{v \in V^{\textrm{active}}_i \colon X(v) = 1\}$.
For each $u \in V$, we define
\[T_Q(u) = \mathbb{E} \left[Y_{i,j}(u)\frac{\phi_{i,j-1}(u)}{1 - (|\textrm{alive}_i(u)|/10k)} + I(u \in S_{i,j}) \cdot 2^{i+1}\right]\]
and $T_Q = \sum_{u \in V} T_Q(u)$.
The proof follows along the exact same lines as in \cref{subsec:hittingset-impl} where also the method of conditional expectation was used. We will therefore only discuss the necessary changes.
Let $p \in [\frac{1}{4k},\frac{1}{2k}]$ such that $p = 2^{-\ell}$ for some $\ell \in \mathbb{N}$.
The pairwise distribution of \cref{lem:pairwise_analysis_sampling_lemma} can be realized with a random seed length of $B = O(\log 1/p \cdot \log n) = \widetilde{O}(\log n)$ by using the pairwise distribution used in~\cite{luby1993removing, berger1989efficient}.
Now, algorithm $\mathcal{A}_{i,j}$ runs in $B$ rounds. At the beginning of round $i$, the first $i-1$ bits of the random seed are fixed and we deon and in round $i$ the $i$-th bit of the random seed is fixed in such a way .
\end{proof}
\paragraph{Local Derandomization}
\begin{lemma}[Global Derandomization]
\label{lem:global_derandomization}
For every $i \in [R]$, $j \in [k]$, there exists a deterministic algorithm $\mathcal{A}_{i,j}$ which in $\widetilde{O}(\text{s} \log^2 n)$ rounds computes a set $S_{i,j}$ such that
\[\sum_{u \in V} Y_{i,j}(u) \frac{\phi_{i,j-1}(u)}{1 - (|\textrm{alive}_i(u)|/10k)} + |S_{i,j}| \cdot 2^{i+1} \leq \sum_{u \in V} \phi_{i,j-1}(u) + \frac{2^i}{k}|V^{\textrm{active}}_i|.\]
\end{lemma}
\subsection{Derandomizing \cref{sec:pairwise_subsampling}}
\begin{restatable}{theorem}{subsampling}
\label{thm:pairwise_subsampling}
Assume we are given a clustering $\mathcal{C}$ with
\begin{enumerate}
\item strong diameter $O(\text{s} \log^2 (n))$ and
\item $\text{s}$-hop degree of at most $10 \log \log (n)$.
\end{enumerate}
There exists a deterministic $\mathsf{CONGEST}\,$ algorithm running in $\widetilde{O}(\text{s} \log(n))$ rounds which computes a clustering $\mathcal{C}^{out}$ with
\begin{enumerate}
\item strong diameter $O(\text{s} \log (n))$,
\item separation of $\text{s}$ and
\item the number of clustered nodes is at least $\frac{|\mathcal{C}|}{1000 \log \log (n)}$.
\end{enumerate}
Let $D$ be the diameter of the input graph. There also exists a deterministic $\mathsf{CONGEST}\,$ algorithm with the same guarantees running in $\widetilde{O}((D + \text{s}) \log(n))$ rounds.
\end{restatable}
\paragraph{Global Derandomization}
\paragraph{Local Derandomization}
\subsection{Global Derandomization}
\subsection{Local Derandomization}
\fi
\section{Hitting Set}
\label{sec:hittingset}
In this section, first, we introduce a variant of the hitting set problem. Next, we propose a simple randomized algorithm for this problem using only pairwise independence. In the end, we describe an efficient distributed/parallel derandomization of our randomized algorithm.
\subsection{Problem Definition}
Consider a collection $\mathcal{S} = \{S_1, \dots, S_N\}$ of $N$ subsets from the universe $\{1,\dots,n\}$ and let $w_i \geq 0$ be the weight that is assigned to $S_i$. We say a subset $H \subseteq [n]$ hits $S_i$ if $H \cap S_i \neq \emptyset$. Our goal is to find a small $H$ with a small cost. Cost of $H$ is total weights of $S_i$ that are not hit by $H$, i.e., $\sum_{i:S_i \cap H = \emptyset} w_i$. For a random subset $H$ that includes each element with probability $p$ independently, the expected size of $H$ is $\mathbb{E}[|H|] = np$ and its expected cost is
\begin{equation*}
\sum_{i=1}^{N} w_i (1-p)^{|S_i|} \approx \sum_{i=1}^{N} w_i e^{-|S_i|p} = \tau^p_{\mathcal{S}}.
\end{equation*}
For example, suppose the regular case where $|S_i| = \Delta$. For $p = 10\log N / \Delta$, a random subset hits all sets with high probability $1 - 1/\mathrm{poly}(N)$ and for $p = 1 / \Delta$, constant fraction of sets are hit. Two important examples for weights is when $w_i = 1$ and $w_i = |S_i|$. In the former, we simply count the number of not hit sets. The latter indeed appears in our applications for constructing spanners and distance oracles (see \Cref{sec:hittingset-app}). There, we get penalized for each not hit set by its size.
In many applications, the expected size and cost of a random subset are enough. The challenge is to find a subset deterministically. Based on this, we formulate the following problem where we combine our two objectives in one potential function.
\label{subsec:hittingset-define}
\begin{definition}[Hitting Set Problem]
\label{def:hittingset}
Given a collection $\mathcal{S} = \{S_1, \dots, S_N\}$ of $N$ subsets from the universe $\{1,\dots,n\}$, an integer weight $w_i \geq 0$ for each $S_i$, and a sampling parameter $p \in (0,1)$, find a subset $H$ that minimizes the potential function
\begin{equation}
\label{eq:hittingset-potential}
\Phi^{p}_{\mathcal{S}}(H) = \frac{\sum_{i=1}^{N} w_i \cdot \mathbb{1}[H\cap S_i = \emptyset]}{\tau^p_{\mathcal{S}}} + \frac{|H|}{np}.
\end{equation}
\end{definition}
So if $\Phi^{p}_{\mathcal{S}}(H) = O(1)$, then $H$ has size $O(np)$ and its cost is $O(\tau^{p}_{\mathcal{S}})$. Our goal is to find such a set with constant potential function deterministically and efficiently. In the rest, we assume that $N \geq n$ as we can add dummy sets with zero costs. We also assume that $p \leq 1/2$ to ensure that $1 - p = e^{-\Theta(p)}$. Note that the case $p \geq 1/2$ is trivial since we tolerate constant deviation from a random subset and for $p \geq 1/2$, the expected size of a random subset is at least $n/2$. So our hitting can include all the $n$ elements.
\paragraph{Hitting in Ordered Sets.}
There are applications where $H$ is partially penalized even if we hit $S_i$. The amount of cost depends on which element of $S_i$ is being hit. In \Cref{sec:hittingset-app}, we encounter a particular instance of this generalization which is described in the following.
For each $S_i$, there is no weight but there is an order $\pi_i(\cdot)$ on its elements where $\pi_i(j)$ denotes the $j$-th element of $S_i$ for $j=1,\dots,|S_i|$. Then, $H$ has to pay $k-1$ for $S_i$ if $\pi_i(k) \in H$ and
\begin{equation*}
H \cap \{\pi_i(1), \dots, \pi_i(k-1)\} = \emptyset.
\end{equation*}
If $H$ does not hit $S_i$ at all, it has to pay $|S_i|$. Cost of $H$ is the sum of the expenses incurred by each $S_i$. We call this problem \textit{hitting ordered set}. With this definition, the expected cost for a random $H$ is
\begin{equation*}
\sum_{i=1}^{N} \sum_{j=1}^{|S_i|} (1-p)^j
\end{equation*}
The hitting ordered set problem is related to the original setting of \Cref{def:hittingset} in the following sense.
\begin{lemma}
\label{lem:hittingset-ordered-reduction}
Given an instance $\mathcal{I}_1$ of the hitting ordered set problem with $N$ sets $S_1, \dots, S_N \subseteq [n]$, we can construct an instance $\mathcal{I}_2$ of the original hitting set problem (see \Cref{def:hittingset}) with $O(N \log n)$ sets in $O(\sum_{i=1}^{N} |S_i|)$ time such that the following holds: For any $H \subseteq [n]$, if $c_1$ is the cost of $H$ in $\mathcal{I}_1$ and $c_2$ is the cost of $H$ in $\mathcal{I}_2$, then $c_1 \leq c_2 \leq 3c_1$.
\end{lemma}
\begin{proof}
To construct $\mathcal{I}_2$, for each $S_i$ in $\mathcal{I}_1$, we add $O(\log n)$ sets to $\mathcal{I}_2$. Suppose $2^\ell \leq |S_i| < 2^{\ell + 1}$. For $j\in [\ell]$, let $S_i^{j} = \{\pi_i(1), \dots, \pi_i(2^j)\}$ and let $S_i^{\ell+1} = S_i$. This completes the construction of sets of $\mathcal{I}_2$. Weight of $S_i^j$ in $\mathcal{I}_2$ is its size $|S_i^j|$.
Consider a subset $H \subseteq [n]$ and let $k$ be the minimum index that $\pi_i(k) \in H$. Suppose $k$ is $|S_i|+1$ if there is no such index. So $H$ has to pay $k-1$ in $\mathcal{I}_1$. In $\mathcal{I}_2$, it has to pay $\sum_{j:|S_i^j| < k} |S_i^j|$ which lies in the range $[(k-1), 3(k-1)]$ and concludes the proof.
\end{proof}
\subsection{Iterative Sampling}
\label{subsec:hittingset-alg}
The goal of this section is to find $H$ with $\Phi^{p}_{\mathcal{S}} = O(1)$ for the hitting set problem \Cref{def:hittingset}. Let $\Delta = \max_{i \in [N]} |S_i|$. Our algorithm has $T = \lceil 8p\Delta\rceil$ iterations. We start with a randomized algorithm and then we derandomize it. For $t = 1,\dots, T$, let $\mathcal{P}^t$ be a pairwise-independent distribution over $n$ binary random variables $X^t_1, \dots, X^t_n \in \{0,1\}$ with bias $q = 4p/T$. That is:
\begin{align*}
\forall i \in [n], \forall b\in \{0,1\},&\quad \Pr[X^t_i = b] = q^b (1-q)^{(1-b)},\\
\forall i,j \in [n], i \neq j,\forall b,b' \in \{0,1\},&\quad \Pr[X^t_i = b, X^t_j = b'] = q^{b+b'} (1-q)^{2-(b+b')}.
\end{align*}
Let the random subset $G^t$ be $\{i \in [n] \mid X^t_i = 1\}$. We replace $G^t$s one by one with an explicit set $H^t$. The final output of the algorithm is $H = \cup_{t=1}^T H^t$. Suppose we are in iteration $t$. Our goal is to find $H^t$. Let
\begin{equation}
\label{eq:pessimistic-def}
Y^t_i = \sum_{j \in S_i} X^t_j - \sum_{j \in S_i} \sum_{k \in S_i: j < k} X^t_j X^t_k.
\end{equation}
If $G^t$ does not hit $S_i$, then $Y^t_i = 0$. Otherwise, $Y^t_i \leq 1$ (because $a \leq \binom{a}{2}+1$ for all positive integers $a$). So $1 - Y^t_i$ is always greater than or equal to $\mathbb{1}[G^t \cap S_i = \emptyset]$ and is a pessimistic estimator for the event that $G^t$ does not hit $S_i$. We have the following upper bound on $\mathbb{E}[1 - Y^t_i]$.
\begin{lemma}
\label{lem:hittingset-pessimistic-bound}
$\mathbb{E}[1-Y^t_i] \leq 1 - 3|S_i|p/T \leq e^{-|S_i|p/T}.$
\end{lemma}
\begin{proof}
Note that:
\begin{equation*}
\mathbb{E}[Y^t_i] = |S_i|q - \binom{|S_i|}{2}q^2 \geq |S_i|q - |S_i|^2q^2/2 \geq 3|S_i|q/4 = 3|S_i|p/T
\end{equation*}
where in the last inequality we use $q = 4p/T \leq 1/2\Delta \leq 1/2|S_i|$.
\end{proof}
For a subset $G \subseteq [n]$, we define the function $f^t(G)$ as
\begin{equation*}
f^t(G) = \frac{\sum_{i: S_i \cap (H^1 \cup \dots \cup H^{t-1}) = \emptyset} (1 - Y_i) \cdot w_i e^{-|S_i|(T - t)p/T}}{\tau^p_{\mathcal{S}}} + \frac{\sum_{i=1}^{n} X_i + \sum_{j=1}^{t-1} |H^j| + 4n(T - t)p/T}{4np}
\end{equation*}
where $X_i = \mathbb{1}[i \in G]$ and $Y_i$ is defined from $X_1, \dots, X_n$ similar to \Cref{eq:pessimistic-def}.
\begin{lemma}
\label{lem:hittingset-first-iteration}
$\mathbb{E}[f^1(G^1)] \leq 2.$
\end{lemma}
\begin{proof}
Note that $\mathbb{E}[\sum_{i=1}^{n} X^1_i] = nq = 4np/T$ and from \Cref{lem:hittingset-pessimistic-bound}, we have $\mathbb{E}[1 - Y_i^1] \leq e^{-|S_i|p/T}$. Plugging these two bounds completes the proof.
\end{proof}
\begin{lemma}
\label{lemma:hittingset-invariant}
For $t \geq 2$, we have:
\begin{equation*}
\mathbb{E}[f^t(G^t)] \leq f^{t-1}(H^{t-1}).
\end{equation*}
\end{lemma}
\begin{proof}
Consider a subset $S_i$. If one of $H^1, \dots, H^{t-2}$ hits $S_i$, then the contribution of $S_i$ to the both sides of the inequality is zero. Otherwise, if $H^{t-1}$ hits $S_i$, the contribution of $S_i$ to $\mathbb{E}[f^t(G^t)]$ is zero. Note that it may contribute a non-zero amount into the RHS since we use pessimistic estimator $1 - Y_i$. The only remaining case is when $S_i$ is not hit in any of the first $t-1$ iterations. Then, the contribution of $S_i$ to the LHS is \begin{equation*}
\mathbb{E}[1 - Y^t_i] \cdot w_i e^{-|S_i|(T - t)p/T} \leq w_i e^{-|S_i|(T - t + 1)p / T}
\end{equation*}
where we use \Cref{lem:hittingset-pessimistic-bound}. On the other hand, the contribution of $S_i$ to the RHS is exactly $w_i e^{-|S_i|(T - t + 1)p / T}$. So the contribution of each $S_i$ to the LHS is less than or equal to its contribution to the RHS. Since $\mathbb{E}[|G_t|] = nq = 4np/T$, the second term that controls the size in $f^t(\cdot)$ and $f^{t-1}(\cdot)$ are equal which completes the proof.
\end{proof}
\begin{theorem}
\label{thm:hittingset-sampling}
If $f^t(H^t) \leq \mathbb{E}[f^t(G^t)]$ for all $t = 1, \dots, T$, then
\begin{equation*}
\Phi^p_{\mathcal{S}}(H = H^1 \cup \dots \cup H^T) \leq 2.
\end{equation*}
\end{theorem}
\begin{proof}
From \Cref{lem:hittingset-first-iteration} and \Cref{lemma:hittingset-invariant}, we get that $f^T(H^T) \leq 2$. Comparing $f^T(H^T)$ and $\Phi^p_{\mathcal{S}}(H)$ term by term, we can easily see that $f^T(H^T) \geq \Phi^p_{\mathcal{S}}(H)$.
\end{proof}
If $\Delta \gg 1/p$, then the number of iterations can be quite large. However, we are mostly interested in the regime where the number of iterations is logarithmic. We can achieve this as stated in the following.
\begin{corollary}
\label{cor:hittingset-sampling-logn}
Let $\mathcal{S}^+ = \{\text{the first $10 \log N/p$ elements of $S_i$} \mid |S_i| \geq 10 \log N / p\}$ and $\mathcal{S}^{-} = \mathcal{S} \setminus \mathcal{S}^+$. Run the algorithm twice: once for $\mathcal{S}^-$ with the same set of weights as before and once on $\mathcal{S}^+$ by setting all weights to $N^2$. Let the output of these two runs be $H^-$ and $H^+$. Then:
\begin{equation*}
\Phi^p_{\mathcal{S}}(H = H^- \cup H^+) \leq 4.
\end{equation*}
Each run takes at most $O(\log N)$ iterations. Moreover, all sets in $\mathcal{S}$ with size at least $10\log N/p$ are hit by $H$.
\end{corollary}
\subsection{Implementation}
\label{subsec:hittingset-impl}
The remaining piece of \Cref{thm:hittingset-sampling} is to find $H^t$ such that $f^t(H^t) \leq \mathbb{E}[f^t(G^t)]$. We first start with the construction of a suitable pairwise distribution.
\paragraph{Construction of Pairwise Independent Distribution.} From the algorithm of the previous section, we need a pairwise distribution $\mathcal{P}$ on $n$ binary random variables $X_1, \dots, X_n \in \{0,1\}$ with bias $q$. Assume that $q = 2^{-\ell}$ for some $\ell \in \mathbb{N}$ and $n$ is a positive integer of the form $n = 2^{m}-1$ for $m \in \mathbb{N}$. We use the pairwise distribution that is used in~\cite{luby1993removing, berger1989efficient} which has a random seed of length $\ell m = O(\log 1/p \cdot \log n)$. Let us quickly recall the construction. We first assign an $\ell$-bit label $L_i$ to each $X_i$. Then, we set $X_i$ to one if and only if all the $\ell$ bits of $L_i$ is one. To construct the labels, we decompose the random seed $R$ into $\ell$ groups each containing $m$ bits as follows:
\begin{equation*}
R = r^0_0\dots r^0_{m-1}r^1_0\dots r^1_{m-1}\dots r^{\ell-1}_{0}r^{\ell-1}_{m-1}
\end{equation*}
The $j$-th group $r^{j}_0\dots r^{j}_{m-1}$ is for constructing the $j$-bit of $L_i$s. To define $L_i(j)$ (the $j$-th bit of $L_i$), we use the bit representation of $i$. Suppose $i = \sum_{k=0}^{m-1} b_k 2^k$. Then:
\begin{equation*}
L_i(j) = b_0 r^{j}_0 \oplus \dots \oplus b_{m-1} r^{j}_{m-1}
\end{equation*}
This completes the construction. In the course of derandomization, we fix the random seed bit by bit. Suppose we fix the first $B$ bits of $R$ to $b_0,\dots,b_{B-1} \in \{0,1\}$. This gives us a new distribution $\mathcal{Q}$. The following result by Berger, Rompel, and Shor~\cite{berger1989efficient} is an important tool to achieve work-efficient derandomization.
\begin{lemma}[\cite{berger1989efficient}, Section 3.2]
\label{lem:berger-etal}
For any given subset $A \subseteq [n]$, we can compute
\begin{equation*}
\sum_{i \in A} \mathbb{E}_{\mathcal{Q}}[X_i],\quad
\sum_{i \in A} \sum_{j \in A} \mathbb{E}_{\mathcal{Q}}[X_i X_j]
\end{equation*}
with $O(|A|)$ processors and in $O(\log n)$ depth in the $\mathsf{PRAM}\,$ model. In particular, we can compute these two quantities in $O(|A| \log n)$ time in the standard model.
\end{lemma}
\paragraph{Bit Fixing.}
Suppose we are in iteration $t$ and we want to find $H^t$ such that $f^t(H^t) \leq \mathbb{E}[f^t(G^t)]$. Suppose $\mathcal{P}^t$ is $\mathcal{P}$ as described above. If $q$ is not a power of two (which is needed for the pairwise construction), replace it with a power of two in the range $[q,2q)$. We can observe that for any $H$
\begin{equation}
\label{eq:enlarge-q}
\Phi^{2p}_{\mathcal{S}}(H) \geq \Phi^{p}_{\mathcal{S}}(H)/2
\end{equation}
So with this replacement, we lose at most a two factor in the final bound for the potential function. Now, we start to fix the bits of the random seed of $\mathcal{P}^t$. Suppose we already fixed the first $B$ bits of the random seed $R$ by $b_0,\dots,b_{B-1}$. Let $e_x = \mathbb{E}[f(G^{t+1}) \mid R(0)=b_0,\dots,R(B-1)=b_{B-1},R(B)=x]$ for $x \in \{0,1\}$. If $e_0 \leq e_1$, then we fix $b_B$ to zero. Otherwise, we fix it to one. Suppose all the $\ell m$ bits are fixed and suppose that the random variable $X_i$ is $v_i \in \{0,1\}$ when we set the random seed to $b_0 \dots b_{\ell m - 1}$. Then, we set $H^t$ to $\{ i \in [n] \mid v_i = 1\}$. We can easily observe that $f^t(H^t) \leq \mathbb{E}[f^{t}(G^t)]$.
\paragraph{$\mathsf{PRAM}\,$ Model.} We have all the ingredients for implementing the algorithm in the $\mathsf{PRAM}\,$ model. This leads to the following theorem.
\begin{theorem}
\label{thm:hittingset-pram}
There is a deterministic algorithm that solves the hitting set problem by finding a subset $H$ with $\Phi^{p}_{\mathcal{S}}(H) \leq 4$ and with $\widetilde{O}(\sum_{i=1}^{N} |S_i|)$ work and
\begin{equation*}
O(\lceil p\Delta\rceil \cdot \log 1/p \cdot \log^2 n)
\end{equation*}
depth in the $\mathsf{PRAM}\,$ model. Moreover, there is a deterministic algorithm that finds a subset $H$ with $\Phi^{p}_{\mathcal{S}}(H) \leq 8$ and such that $H$ hits all $S_i$s with size greater than $10 \log N / p$. This algorithm runs with $\widetilde{O}(\sum_{i=1}^{N} |S_i|)$ work and
\begin{equation*}
O(\log N \cdot \log 1/p \cdot \log^2 n)
\end{equation*}
depth in the $\mathsf{PRAM}\,$ model.
\end{theorem}
\begin{proof}
The first algorithm is based on \Cref{thm:hittingset-sampling} and the second algorithm is based on \Cref{cor:hittingset-sampling-logn}. In those two algorithms, the potential function is upper bounded by $2$ and $4$. Here, we can only guarantee $4$ and $8$. This is because $q$, the sampling probability of one iteration, may not be a power of two. As discussed before (see \Cref{eq:enlarge-q}), we can handle this issue by paying an extra factor two in the approximation factor. In one iteration, we have $O(\log 1/p \cdot \log n)$ bit fixing. For each bit, we need to compute two conditional expectation which takes $O(\log n)$ depth and $O(\log n \cdot \sum_{i=1}^{N} |S_i|)$ work using \Cref{lem:berger-etal}. Multiplying the number of iterations gives us the claimed bounds.
\end{proof}
\paragraph{$\mathsf{CONGEST}\,$ Model.}
First, let us describe how the hitting set problem is represented in the distributed model. Consider an $(N+n)$-node bipartite network $G = (A \sqcup B, E)$ where $A = [N]$ and $B = [n]$. A node $i \in A$ represents set $S_i$ and a node $j \in B$ represents element $j \in [n]$. There is an edge between $i \in A$ and $j \in B$ if and only of $j \in S_i$. We assume that $p$, $n$, and $\tau_{\mathcal{S}}^p$ (or an upper bound of it) is known to all nodes.
To simulate global decision making, we use $3$-separated network decomposition. We need to execute the following operation fast: For an arbitrary color $j$, let $C_1, \dots, C_d$ be the set of clusters with color $j$ in the given $3$-separated network decomposition. Suppose that each node $v$ in $C_1 \cup \dots \cup C_d$ knows a value $a_v$. For each cluster $C_i$, we want to broadcast the value $\sum_{v \in C_i} a_v$ to all nodes in $C_i$. We denote the round complexity of executing this operation for all clusters $C_1, \dots, C_d$ by $T^{\mathrm{agg}}_{\mathrm{ND}}$.
\begin{theorem}
\label{thm:hittingset-congest}
Given a $Q$-color $3$-separated network decomposition with aggregation time $T^{\mathrm{agg}}_{\mathrm{ND}}$ (as described above), there is a deterministic algorithm that solves the hitting set problem by finding a subset $H$ with $\Phi_{\mathcal{S}}^{p}(H) \leq 4$ in
\begin{equation*}
O(\lceil p\Delta \rceil \cdot Q \cdot \log 1/p \cdot \log n \cdot T_{\mathrm{ND}}^{\mathrm{agg}})
\end{equation*}
rounds of the $\mathsf{CONGEST}\,$ model. Moreover, there is a deterministic algorithm that finds a subset $H$ with $\Phi_{\mathcal{S}}^{p}(H) \leq 8$ and such that $H$ hits all $S_i$s with size greater than $10 \log N / p$. This algorithm runs in
\begin{equation*}
O(\log N \cdot Q \cdot \log 1/p \cdot \log n \cdot T_{\mathrm{ND}}^{\mathrm{agg}})
\end{equation*}
rounds of the $\mathsf{CONGEST}\,$ model.
\end{theorem}
\begin{proof}
We want to derandomize iteration $t$. In contrast to the $\mathsf{PRAM}\,$ model \Cref{thm:hittingset-pram}, in the $\mathsf{CONGEST}\,$ model, we do not have global communication and so we cannot decide which bit should be fixed in a straightforward way. However, we can simulate such global decision-making with network decomposition paying an extra factor $Q$ in the round complexity. For each cluster $C$, we independently draw a sample from the pairwise-independent distribution $\mathcal{P}$ with bias $q$. Recall that the input graph is a bipartite graph $G = (A \sqcup B, E)$. These samples assign a binary value to each node of $B$. Observe that the assigned values are also pairwise independent since the product of pairwise independent distributions is pairwise-independent. Now, to derandomize, we go through the colors one by one. Suppose we are working on color $j \in [Q]$ with $d$ clusters $C_1, \dots, C_d$. Moreover, suppose the first $b$ bits of random seeds of $C_1, \dots, C_d$ are fixed. We fix the $(b+1)$-th bit. Let us emphasize that each cluster has its own random seed and different clusters may fix the $(b+1)$-th bit differently. Consider cluster $C_i$ and a node $v \in A$ that is either in $C_i$ or is in the boundary of $C_i$ (i.e., $v$ is not in $C_i$ but has a neighbor in $C_i$). So this node represents a set $S_v$ in the corresponding hitting set problem. We assign two values $a^0_v$ and $a^1_v$ to $v$ where $a^x_v$ corresponds to the case when we fix the $(b+1)$-bit of the random seed of $C_i$ to $x$. Note that each neighbor of $v$ represents an element of $S_v$. If $v$ has a neighbor in the clusters with color $\{1,\dots,j-1\}$ that is already decided to be in $H$ (our final hitting set), then we set $a^x_v$ to zero. So suppose this is not the case and let $d$ be the number of neighbors of $v$ that are not in $C_i$ and are in a cluster with color in $\{j+1, \dots, Q\}$. Then, we set $a_v^x$ to $$\frac{(1 + \binom{d}{2}q^2 - dq) \cdot F_v^x \cdot w_v e^{-|S_i|(T - t) p /T}}{\tau_{\mathcal{S}}^{p}}$$ where $F_v^b$ is
\begin{equation*}
F_v^x = \mathbb{E}[1 + \sum_{u \in C_i \cap B: u\in S_v} \sum_{u \in C_i \cap B: u\in S_v \wedge u < u'} X_u X_{u'} - \sum_{u \in C_i \cap B: u\in S_v} X_u \mid \text{first $b$ bits and $(b+1)$-th bit is $x$}]
\end{equation*}
where $X_u$ represents the indicator random variable of element $u$. Note that the given network decomposition is $3$-separated and so all the boundaries of $C_1, \dots, C_d$ are disjoint. So $v$ can compute $F_v^b$ in $\widetilde{O}(|S_v|)$ according to \Cref{lem:berger-etal}. Also, note that that this gives us the contribution of $S_v$ to
\begin{equation*}
\mathbb{E}[f^{t}(\cdot) \mid \text{first $b$ bits and $(b+1)$-th bit is $x$}]
\end{equation*}
Next, for each element $u \in A_i$, set $a_u^x$ to
\begin{equation*}
\frac{\mathbb{E}[X_u \mid \text{first $b$ bits and $(b+1)$-th bit is $x$}]}{4np}.
\end{equation*}
In the end, for each cluster $C_i$, we compute two values $e_i^x$ for $x \in \{0,1\}$ which is $$\sum_{v \in C_i \cup (\partial(C_i) \cap B)} a_v^x$$ where $\partial(C_i)$ denotes the boundary of $C_i$. We broadcast $e^b_i$ to each nodes in $C_i$. This can be done in $O(T_{\mathrm{ND}}^{\mathrm{agg}})$ rounds for all $C_i$s simultaneously. Next, nodes of $C_i$ set the $(b+1)$-bit of the random seed to zero if $e^0_i \leq e^0_i$ and set it to one otherwise. This completes the bit fixing.
There are $T$ sampling iterations (if we apply \Cref{thm:hittingset-sampling}, $T = \lceil p\Delta\rceil$, and if we apply \Cref{cor:hittingset-sampling-logn}, $T = O(\log N)$), $Q$ colors, and $O(\log 1/p \cdot \log n)$ bits to fix for each color. Multiplying these numbers gives us the number of bit fixing. Taking into account that fixing each bit takes $O(T_{\mathrm{ND}}^{\mathrm{agg}})$ rounds of the $\mathsf{CONGEST}\,$ model concludes the proof.
\end{proof}
\begin{corollary}
\label{cor:hittingset-congest-logn}
There is a deterministic algorithm that solves the hitting set problem by finding a subset $H\subseteq [n]$ with $\Phi^p_{\mathcal{S}}(H) = O(1)$ in $\mathrm{poly}(\log n)$ rounds of the $\mathsf{CONGEST}\,$ model and with total computations $\widetilde{O}(m)$.
\end{corollary}
\begin{proof}
There is a work-efficient deterministic algorithm for finding a $3$-separated $O(\log n)$-color network decomposition in $\textrm{polylog}(n)$ rounds and with $T_{\mathrm{ND}}^{\mathrm{agg}} = \textrm{polylog}(n)$(see Theorem 2.12 of Rozho\v{n} and Ghaffari~\cite{rozhonghaffari20}). Plugging this bound in \Cref{thm:hittingset-congest} concludes the proof.
\end{proof}
\section{Applications of Hitting Set}
In this section, we discuss two applications of the hitting set problem. One is the distributed construction of multiplicative spanners and the other is the parallel construction of distance oracles. Let us quickly define these notions. A subgraph $H = (V,E') \subseteq G = (V,E)$ is an $\alpha$-spanner of $G$ if for all pairs of nodes $u, v \in V$, we have:
\begin{equation*}
d_G(u,v) \leq d_H(u,v) \leq \alpha \cdot d_G(u,v).
\end{equation*}
A distance oracle is a data structure that accepts a pair of nodes $(u,v)$ as a query and returns their distance in $G$. In \Cref{subsec:distance-oracles}, we discuss \textit{source-restricted approximate distance oracle} in which $s$ nodes of $G$ are marked as source and it is guaranteed that $u$ is always a source. The term ``approximate'' allows the oracle to return an approximation of $d_G(u,v)$ rather than its exact value.
\label{sec:hittingset-app}
\subsection{Spanners}
\label{subsec:spanners}
\iffalse
In the paper~\cite{bezdrighin2022deterministic}, Bezdrighin et al. show
the following results on the deterministic construction of spanners in the $\mathsf{CONGEST}\,$ model are presented:
\begin{itemize}
\item A $\textrm{polylog}(n)$ rounds algorithm for constructing $(2k-1)$-spanner with $O(nk + n^{1 + 1/k} \log k)$ and $O(nk + n^{1 + 1/k} k)$ edges for unweighted and weighted graphs, respectively.
\item A $\textrm{polylog}(n)$ rounds algorithm for constructing $O(\log n \cdot 2^{\log^* n})$-spanner and $O(\log n \cdot 4^{\log^* n})$-spanner with $O(n)$ edges for unweighted and weighted graphs, respectively.
\end{itemize}
While all of these algorithms have poly-logarithmic round complexity, the total amount of computations by nodes is slightly super-polynomial, i.e., $O(n^{\log \log n})$. Keeping the $\textrm{polylog}(n)$ round complexity, here we improve the amount of computations to near-linear time, i.e., $\widetilde{O}(|E(G)|)$.
\fi
\begin{theorem}
There is deterministic algorithm in $\mathrm{poly}(\log n)$ rounds of the $\mathsf{CONGEST}\,$ model and with total computations $\widetilde{O}(m)$ that finds a $(2k-1)$-spanner with $O(nk + n^{1 + 1/k} \log k)$ and $O(nk + n^{1 + 1/k} k)$ edges for unweighted and weighted graphs, respectively.
\end{theorem}
\begin{proof}
We derandomize Baswana-Sen algorithm~\cite{baswana2007simple}. Let us quickly recall this algorithm. It consists of $k$ steps. The input of step $i$ is a clustering denoted by $\mathcal{C}_i$. Each cluster has a center node known to all of its members. The input of the first step is the trivial clustering: there are $n$ clusters each containing a single node. During one step, we sample some of the clusters, and then based on that sampling, some nodes stay in their clusters, some get unclustered, and some join other clusters. After this, the current step $i$ terminates, and the new clustering $\mathcal{C}_{i+1}$ is passed to the next step. Here is what we do in step $i$ for $i \leq k-1$ (we discuss the last step, $i$ equals $k$, later):
\begin{enumerate}
\item Each cluster of $\mathcal{C}_i$ is sampled with probability $p = n^{-1/k}$.
\item A node that is in a sampled cluster, stays put in its own cluster.
\item For a node $v$ in an unsampled cluster, let $C_1, \dots, C_d$ be the set of clusters containing at least one neighbor of $v$. Let $e_i = \{u_i \in C_i, v\}$ be an edge with the minimum weight between $v$ and one of the nodes in $C_i$. If there are several edges with the minimum weight, $v$ selects one of them arbitrarily. Let $w_i$ be the weight of $e_i$. Without loss of generality, suppose $w_1 \leq \dots \leq w_d$. If all of $C_1, \dots, C_d$ are unsampled, $v$ adds all edges $e_1, \dots, e_d$ to the output spanner and gets unclustered. Otherwise, let $j$ be the minimum index for which $C_j$ is sampled. Then, $v$ adds $e_1, \dots, e_{j}$ to the output spanner and joins the sampled cluster $C_j$. Note that all such $v$ runs this step simultaneously.
\end{enumerate}
In the last step, we do the exact same thing except that we sample no cluster (each cluster is sampled with probability zero rather than $n^{-1/k}$).
The output of Baswana-Sen is always a $(2k-1)$-spanner and only the size of the output depends on the randomness. From the algorithm description, you can see that the only randomized part of the Baswana-Sen algorithm is the sampling of clusters. Our goal is to find the set of sampled clusters of each step deterministically. If we have the following properties on the set of sampled clusters, then we can guarantee the claimed bounds on the size of the output spanner (see~\cite{bezdrighin2022deterministic}, Lemma 3.3):
\begin{enumerate}[(a)]
\item For each $i$, the number of clusters in $\mathcal{C}_i$ is at most $n^{1-(i-1)/k}$.
\item The number of edges added to the output spanner is bounded as follows: For the unweighted case, the total number of edges added by nodes with at least $\gamma_1 n^{1/k} \log k$ neighboring clusters for a large enough constant $\gamma_1 > 0$ is at most $O(n^{1 + 1/k}/k)$. For the weighted case, all nodes add at most $O(n^{1 + 1/k})$ edges to the output.
\item A node that is clustered in $\mathcal{C}_i$, remains clustered if it has at least $\gamma_2 n^{1/k} \log n$ neighbouring clusters for a large enough constant $\gamma_2 > 0$.
\end{enumerate}
We can frame these properties as a hitting set problem. To avoid cluttering the notation, we refer to the universe size in the corresponding hitting set problem of step $i$ by $n^h_i$ and its number of sets by $N^h_i$. In step $i$, we have the following hitting set problem: There is an element in the universe for each cluster in $\mathcal{C}_i$. So $n^h_i = |\mathcal{C}_i| \leq n$. For each clustered node $v$ in $\mathcal{C}_i$, there is a set $S_v$ containing all of its neighboring clusters. So $N^h_i \leq n$. The parameter $p$ for the hitting set problem is set to the sampling probability of Baswana-Sen divided by a large enough constant $\gamma_3 > 0$, i.e., $p = n^{-1/k}/\gamma_3$ (note that the last step is already deterministic and no derandomization is needed there). For unweighted graphs, we set the weight of $S_v$ to its size $w_v = |S_v|$. For weighted graphs, we consider the hitting ordered set problem as discussed in \Cref{lem:hittingset-ordered-reduction}. For each clustered node $v$, we assign the order $\pi_v(\cdot)$ on $S_v$. Suppose that the neighboring clusters of $v$ are $C_1, \dots, C_d$ and the minimum weight of an edge between $C_i$ and $v$ is $w_i$. Then $C_i$ comes before $C_j$ in $\pi_v(\cdot)$ if $w_i < w_j$ or $w_i = w_j$ and $i < j$.
With straightforward calculations, we can see that all the three required properties are satisfied if we solve the presented hitting set problem with \Cref{cor:hittingset-congest-logn} (for the hitting ordered set problem, we first use the reduction \Cref{lem:hittingset-ordered-reduction}).
We have $k \leq \log n$ steps in total. As described above, each step can be derandomized by solving a hitting set problem. So the total round complexity is $\mathrm{poly}(\log n)$ by applying \Cref{cor:hittingset-congest-logn}. One issue here is that each element in the defined hitting set problem corresponds to a cluster. This issue can be handled by contracting each cluster to a node and using the fact that the network decomposition of~\cite{rozhonghaffari20} also works on contracted graphs. This slows down the round complexity only by a factor $k = O(\log n)$ as each cluster has diameter $k$.
\end{proof}
\begin{theorem}
For any $\varepsilon > 0$, there is deterministic distributed algorithm in $\mathrm{poly}(\log n) / \varepsilon$ rounds of the $\mathsf{CONGEST}\,$ model and with total computations $\widetilde{O}(m)$ that finds a spanner with size $n(1 + \varepsilon)$ and with stretch $O(\log n \cdot 2^{\log^* n} / \varepsilon)$ and $O(\log n \cdot 4^{\log^* n} / \varepsilon)$ stretch for unweighted and weighted graphs, respectively.
\end{theorem}
\begin{proof}
We derandomize the algorithm of Pettie~\cite{pettie2010distributed} to get a spanner with $O(n)$ edges and with stretch $O(\log n \cdot 2^{\log^* n})$ and $O(\log n \cdot 4^{\log^*} n)$ for unweighted and weighted graphs, respectively. Pettie's algorithm is combining $O(\log^* n)$ application of Baswana-Sen back to back and the hitting set problem we encounter in Pettie's algorithm, is exactly the same as the Baswna-Sen. So we do not repeat this here. We refer interested readers to Theorem 1.5 of ~\cite{bezdrighin2022deterministic} where the full algorithm and a slower derandomized version of it is discussed. Let us note that the original algorithm of Pettie only works for unweighted graphs, but with a simple modification which is proposed in~\cite{bezdrighin2022deterministic}, it can work on weighted graphs as well. To reduce the number of edges from $O(n)$ to $n(1 + \varepsilon)$, we apply the deterministic reduction of ~\cite{bezdrighin2022deterministic}, Theorem 1.2.
\end{proof}
\subsection{Approximate Distance Oracles}
\label{subsec:distance-oracles}
This section is devoted to the parallel implementation of the approximate distance oracle by Roditty, Thorup, and Zwick~\cite{roditty2005deterministic}. There, given a weighted graph $G = (V,E)$, a stretch parameter $k$, and a set of $s$ sources $S \subseteq V$, they deterministically construct a data structure of size $O(kns^{1/k})$ and in $\widetilde{O}(ms^{1/k})$ time. For a query $(u,v)$, the data structure can compute a value $q$ such that
\begin{equation*}
d(u,v) \leq q \leq (2k-1) d(u,v)
\end{equation*}
in $O(k)$ time. See \Cref{alg:distance-oracle} for their algorithm for constructing the data structure and \Cref{alg:distance-oracle-query} for how they evaluate a query.
\begin{algorithm}
\caption{Approximate Distance Oracle~\cite{roditty2005deterministic}}
\label{alg:distance-oracle}
\begin{algorithmic}[1]
\Procedure{DistOracle}{G, k}
\State $A_0 = S, A_k = \emptyset$.
\State $\ell = 10s^{1/k} \log n$.
\For{$i=1,\dots,k-1$}
\State For each $v \in V$, find $p_i(v) \in A_{i-1}$ such that $d(p_i(v), v) = d(A_{i-1}, v).$
\State For every $v \in V$, compute $N_{i-1}(v)$ which is the set of $\ell$ closest nodes to $v$ in $A_{i-1}$.
\State Find a set $A_i \subseteq A_{i-1}$ such that:
\Statex $\quad \quad \quad \quad$ (a) $|A_i| \leq s^{1-i/k}$.
\Statex $\quad \quad \quad \quad$ (b) $A_i$ hits $N_{i-1}(v)$ for all $v \in V.$
\Statex $\quad \quad \quad \quad$ (c) $\sum_{v \in V} |\{w \in A_{i-1} - A_{i} \mid d(w,v) < d(A_i, v)\}| = O(ns^{1/k}).$
\EndFor
\State For each $v \in V$, compute $p_{k-1}(v)$.
\State For every $v \in V$, set $B(v) = A_{k-1}$.
\For{$i=0,\dots,k-2$}
\State For every $v \in V$, set $B(v) = B(v) \cup \{w \in N_i(v) \mid d(w,v) < d(A_{i+1}, v)\}$.
\EndFor
\State For each $v \in V$, create a hash table $H(v)$ with an entry $(v,d(v,w))$ for each $w \in B(v)$.
\EndProcedure
\end{algorithmic}
\end{algorithm}
\begin{algorithm}
\caption{Evaluating a query~\cite{roditty2005deterministic}}
\label{alg:distance-oracle-query}
\begin{algorithmic}[1]
\Procedure{Query}{$u \in S$,$v$}
\State $w = u$, $i = 0$.
\While{$w \not \in B(v)$}
\State $i = i + 1$.
\State $(u,v) \leftarrow (v,u)$.
\State $w \leftarrow p_i(u)$
\EndWhile
\Return $d(w,u) + d(w,v)$
\EndProcedure
\end{algorithmic}
\end{algorithm}
\begin{theorem}
\label{thm:distance-oracle}
Given an undirected weighted graph $G = (V,E)$, a set of $s$ sources $S \subseteq V$, stretch parameter $k$, and error $\varepsilon > 0$, there is a deterministic algorithm that solves the source-restricted distance oracle problem with $\widetilde{O}_{\varepsilon}(ms^{1/k})$ work and $\widetilde{O}_{\varepsilon}(\mathrm{poly}(\log n))$ depth in the $\mathsf{PRAM}\,$ model. The data structure has size $O(nks^{1/k})$ and for each query $(u,v)$, the oracle can return a value $q$ in $O(k)$ time that satisfies
\begin{equation*}
d(u,v) \leq q \leq (2k-1)(1 + \varepsilon)d(u,v).
\end{equation*}
\end{theorem}
\begin{proof}
It is enough to provide a parallel algorithm with $\widetilde{O}_{\varepsilon}(\mathrm{poly}(\log n))$ depth for computing $A_i$, $N_i(\cdot)$, and the hash table. This gives us all the ingredients we need to run the algorithm.
Note that finding a suitable $A_i$ in \Cref{alg:distance-oracle} is just an instance of hitting ordered set problem and we can apply \Cref{lem:hittingset-ordered-reduction} and \Cref{cor:hittingset-sampling-logn}. The universe is $A_{i-1}$ and for each $v \in V$, we want to hit the set $N_{i-1}(v)$. We also need to determine $\pi_{i,v}(\cdot)$. An element $w$ comes before $w'$ in this order if $d(w,v) < d(w',v)$. If the distances are equal, we break the tie based on the identifier of $w$ and $w'$. If we set the sampling probability to $p = s^{-1/k}/\gamma$ for a large enough constant $\gamma > 0$ (indeed $\gamma = 24$ is enough), then we can compute a suitable $A_i$ satisfying all the three required properties with $\widetilde{O}(m)$ work and $O_{\varepsilon}(\mathrm{poly}(\log n))$ depth in the $\mathsf{PRAM}\,$ model using \Cref{thm:hittingset-pram} and the reduction \Cref{lem:hittingset-ordered-reduction}.
In~\cite{roditty2005deterministic}, they compute $N_i(\cdot)$ by running $\ell$ instances of Single Source Shortest Path problem (SSSP). There is no known parallel algorithm for SSSP with poly-logarithmic depth. However, recently, Rozhoň et al.~\cite{rozhovn2022undirected} proposed a work-efficient algorithm for computing $(1+\varepsilon)$-approximation of SSSP with poly-logarithmic depth. We can replace the exact computation with an approximation, losing $(1+\varepsilon)$ in the final stretch guarantee.
For computing the hash tables, we can apply the construction of Alon and Naor~\cite{alon1996derandomization}. There, they provide a deterministic hash table of $t$ elements into $O(t)$ space with read access of $O(1)$ time. While they did not discuss the parallel implementation of their construction, their algorithm can be implemented in $\mathrm{poly}(\log n)$ depth in a straightforward way. Their approach is derandomizing a randomized hash function using the method of conditional expectation on epsilon-biased spaces. They define a potential function (see section 3.1. of~\cite{alon1996derandomization}) which is a simple aggregation and can be parallelized. We do not discuss the full details as the implementation is straightforward.
\end{proof}
\newpage
\section*{Acknowledgments}
M.G., C.G., S.I., and V.R. were supported in part by the European Research Council (ERC) under the European Unions Horizon 2020 research and innovation program (grant agreement No.~853109) and the Swiss National Science Foundation (project grant 200021\_184735). B.H. was supported in part by NSF grants CCF-1814603, CCF-1910588, NSF CAREER award CCF-1750808, a Sloan Research Fellowship, funding from the European Research Council (ERC) under the European Union's Horizon 2020 research and innovation program (grant agreement 949272), and the Swiss National Science Foundation (project grant 200021\_184735).
\bibliographystyle{alpha}
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{
"redpajama_set_name": "RedPajamaArXiv"
}
| 7,786
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Eighties Icon Band Tears for Fears to play at The Colosseum!
Windsor, ON – Caesars Windsor is excited to announce the rescheduled date for Tears for Fears is Friday September 30 at 9pm.
The UK band formed in 1981 and has sold more than 30 million albums worldwide to date. They have released four albums as a group. Their platinum-selling 1983 debut The Hurting reached #1 on the UK Albums Chart, while their second album, 1985's Songs from the Big Chair reached #1 on the US Billboard Top 200 and spawned two massive hits, "Shout" and "Everybody Wants to Rule the World". The latter won the BRIT Award for Best British single in 1986. Roland Orzabal and Curt Smith released The Seeds of Love in 1989, and after a break, reunited to release Everybody Loves a Happy Ending in 2004.
Tears for Fears are currently working hard both in the US and the UK on their new record, which will be released in 2017 on Warner Bros. Records.
Great seats are still available. Tickets start at $30 Canadian and are on sale now through Ticketmaster and the Caesars Windsor Box Office.
Tickets for the July 17 show will be honoured for the rescheduled date.
Upcoming shows: Gavin DeGraw (June 19), Olivia Newton-John (July 21), Maks & Val (July 22), Kesha (July 28), Rain: A Tribute to the Beatles (July 29), Dolly Parton (Aug. 4), Jeff Dunham (Aug. 5), Toby Keith (Aug. 6) Paul Anka (Aug 21), Chubby Checker (Aug 25), Russell Peters (Aug 26), Willie Nelson (Sept 8) and Alice Cooper (Oct 2).
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{
"redpajama_set_name": "RedPajamaC4"
}
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CWE Rhode Island - Exploring U.S. Small Business Administration Resources
by Center for Women & Enterprise
Thu, June 27, 2019, 9:30 AM – 11:30 AM EDT
Center For Women & Enterprise
132 George M Cohan Boulevard
Do You Need Small Business Assistance?
Wherever You Are In Your Business Lifecycle,
The U.S. Small Business Administration (SBA) Can Help!
This free workshop, presented by a member of the local SBA team, is designed to help entrepreneurs understand the various SBA programs and services available to help them as they start, expand, grow, and recover their businesses. This overview presentation covers:
Introduction to the SBA
SBA Resource Partner Network -- Business Training, & One-on-One Mentoring
Access to Capital – Loan & Surety Bond Programs
International Trade/Exporting Assistance
Selling to the Federal Government – Contracting Certification Programs (such as 8(a) Business Development, HUBZone (Historically Underutilized Business Zone), and Women-Owned Small Business)
Emergency Preparedness, & Disaster Assistance Loans
Cancellation and Notification Policy
Sessions with low enrollment may be cancelled. CWE will notify you of class cancellations and changes.
If you can no longer attend a no-cost session for which you registered, please notify us at least 24 hours in advance. These classes are offered at no cost as a benefit to you, but there are per-registration costs to CWE associated with them. We reserve the right to request a $25 contribution to defray our costs for no shows, so that we may continue to offer free classes in the future.
United States Events Rhode Island Events Things to do in Providence, RI Providence Seminars Providence Business Seminars
CWE Rhode Island - Exploring U.S. Small Business Administration Resources at Center For Women & Enterprise
132 George M Cohan Boulevard, Providence, RI 02903
Thu, Aug 22 9:30 AM
Center For Women & Enterprise, Providence
Thu, Jul 25 9:30 AM
CWE Rhode Island - Women-Owned Small Business Federal Contracting Program
CWE Rhode Island - Is Entrepreneurship Right for You ?
CWE Rhode Island - Business Plan Basics
CWE Rhode Island, Providence
Wed, Aug 21 9:30 AM
CWE Rhode Island - Legal Considerations for New Business Owners
Browse Providence Events
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 7,532
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Q: Maven, switching to a different profile I have a problem with proper maven profile configuration of a project that is deployed to a continuous integration server.
In my project, there are some resources that needs to be included only during tests at the daily building phase and others that needs to be included during nightly builds, and they can never be included both at the same time, because building process will fail, I can achive this locally by activating one profile at the same time.
Continuous integration server runs following maven commands:
-during daily builds:
mvn clean package -Pci -Dci
-during nightly builds
mvn clean install -Dmaven.test.failure.ignore -Pci,nightly -Dci -Dnightly
As you see, nightly build command include maven variables and profiles defined in daily build command, which makes some troubles for me, becouse I want to have only one profile activated at the same time.
Specifically, what I want is having 3 separate profiles:
-my-pforile (activated by default, not used on CI server)
-ci-profile (activated only on daily builds, used on CI server)
-nightly-profile (activated only on nightly builds, used on CI server)
How can I achieve that? I tried almost everything. Reconfiguring CI server is not an option.
A: When I have to configure the same build with different profiles, using Jenkins as a CI,
I usually create as much builds as profiles, so each build uses the correct configuration.
If adding a new build is not an option probably you can try to create a workaround
using something like the exec plugin (http://mojo.codehaus.org/exec-maven-plugin/) to download
the resources from a ftp (or something else).
You will have also to create a cron job (or equivalent) to replace the correct resources between the builds:
in the evening you put there the resources for the night, in the morning the ones for the day.
But considering how cumbersome this process will be, probably it is better to try to add
a new build.
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{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 9,006
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Tagore, Debendranath
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Debendranath Tagore
Religious reformer
Sarada Devi
Debendranath Tagore (Bangla: দেবেন্দ্রনাথ ঠাকুর Debendronath Ţhakur) (May 15, 1817 - January 19, 1905) was an Indian Bengali philosopher and Hindu reformer from current-day West Bengal, in India. His son, Rabindrantah Tagore was a Nobel Prize winning poet. Debendranath was himself a leading contributor to the Bengali renaissance. A philanthropist and social activist, his interest in democracy and in education helped to produce a generation of Indians from whom the leaders of the nation emerged in its independence struggle against the British Empire. His concern to integrate Indian and Western ideas and to be guided by reason had a major impact on the public consciousness. The desire to modernize and to industrialize but not to become a cultural replica of the West can be traced back to Debendranath, and to his influential family.
2 Debendranath and the Samaj
3 Social Activism
4 Teaching
6 Legacy
He was born in Calcutta, India. His father, Dwarkanath Tagore, was a wealthy landowner and successful entrepreneur with interests in shipping and banking, among other ventures. Dwarkanath was a co-founder, with Ram Mohan Roy of the reformist Brahmo Samaj. He had also financed such projects as the Calcutta Medical College. From the age of nine, Debendranath received a classical Brahmin education which included the study of Sanksrit as well as Persian but he also studied English and Western philosophy. In 1827 he enrolled in the Anglo-Hindu College in Calcutta, which Roy had founded.
After graduating, he started managing the family estate but his interest in religion and philosophy soon began to take up most of his time. His grandmother's death in 1838 further stimulated this interest. In 1839 he founded a Society to promote discussion of religion and philosophy, then in 1842 he succeeded to the leadership of the Brahmo Samaj. Founded by Ram Mouhun Roy and his father in 1828, the Samaj promoted worship of one God, opposed image-veneration, such practices as Sati (widow suicide on their husband's funeral pyre), repudiated any need for a mediator (such as an Avatar) between people and God and also the authority of the Vedas. The Samaj stressed equality of all before God, regardless of gender, race or class.
Debendranath and the Samaj
Debendranath embraced all of the above but wanted to locate the Samaj more firmly within Hindu culture. Initially, he revived interest in the Vedas, starting a Bengali translation of the Rig Veda. He also began to work on a liturgy for the movement's worship, which was introduced in 1845. He composed many devotional songs. He replaced the pujas with the Magh Festival, in which images played no role. His use of the Vedas resulted in a dispute with Keshub Chunder Sen, a leading member of the Samaj and a close friend of Debendranath. Sen was attracted to Christianity and wanted the movement to be more eclectic. Although by 1850, Debendranath had ceased to use the Vedas, suggesting that no scripture, however ancient, is binding for all time in 1866 Sen led a break-away group, which took the name of the Brahmo Samaj of India. The original society became known as the Adi (original) Samaj. In 1878, Sen started his Church of the New Dispensation. He believed in a universal religion which in different contexts would have a different cultural color. In India, that color would be Hindu. In 1867, Debendranath was awarded the title of "Maharishi" by the Samaj.
Debendranth campaigned to reduce the tax burden on the poor. In 1859 he founded a Brahmo school. He also co-founded a charitable institution. From 1851, as Secretary of the British Indian Association, he also campaigned for India's political autonomy. The Association aimed to represent Indian interests to the British government. He also campaigned for universal primary education in India. Debendranath was a supporter of democracy and was against entrenched, traditional authority in both the religious and the political spheres.
Debendranath stressed reason and discrimination. He wanted to combine the best of what he found in the West with the best of what was found in Indian culture. He was deeply spiritual but until his retirement from business affairs in 1867 he continued to be involved in worldly matters. He did not renounce his material possessions as some Hindu traditions prescribe but continued to enjoy them in a spirit of detachment. He was praised by no less a spiritual master than Sri Ramakrishna who compared him to the Puranic king Janaka, father of Sita, the heroine of the epic Ramayana, extolled in the scriptures as an ideal man who perfectly synthesized material and spiritual accomplishments.
What is remarkable in this achievement is that he excelled his father, who received the title Prince from the British colonial government owing to his large fortune and yet retained his dignity before them, famously wearing an all-white outfit bereft of all jewelry in a party attended by the Queen, with only his shoes studded with two diamonds bettering the Koh-i-noor in the Queen's crown. This was a gesture symbolizing the mastery of wealth, as opposed to its slavish pursuit. In 1867, Debendranath retired to the hermitage he had established in 1863, later made world-famous as Santi Niketan by his son, Rabindranath. Dabendranath wrote several books. His Bengali commentary on scripture, the Brahmo-Dharma (1854) was widely acclaimed.
Debendranath played no small role in the education of his sons. Dwijendranath (1840-1926) was a great scholar, poet and music composer. He initiated shorthand and musical notations in Bengali. He wrote extensively and translated Kalidas's Meghdoot into Bengali. Satyendranath (1842-1923) was the first Indian to join the Indian Civil Service. At the same time he was a great scholar with a large reservoir of creative talents. Jyotirindranath (1849-1925) was a scholar, artist, music composer and theatre personality. Rabindranath (1861-1941) was his youngest son. His other sons Hemendranath (1844-1884), Birendranath (1845-1915) and Somendranath did not achieve that great fame but everybody was filled with creative talents. His daughters were Soudamini, Sukumari, Saratkumari, Swarnakumari (1855-1932) and Barnakumari. Soudamini was one of the first students of Bethune School and a gifted writer. Swarnakumari was a gifted writer, editor, song-composer, and social worker. All of them were famous for their beauty and education.
His part in creating the legacy of Thakurbari—the House of Tagore—in the cultural heritage of Bengal, centered in Kolkata, was not negligible. It was largely through the influence of the Tagore family, following that of the writer Bankim Chandra Chatterjee, that Bengal took a leading role on the cultural front as well as on the nationalistic one, in the Renaissance in India during the nineteenth century.
The house of the Tagore family in Jorasanko, popular as Jorasanko Thakur Bari in North-western Kolkata, was later converted into a campus of the Rabindra Bharati University, eponymously named after Rabindranath.
Furrell, James W. The Tagore Family: A memoir. New Delhi: Rupa, 2004. ISBN 978-8129104113
Sharma, Arvind. The Concept of Universal Religion in Modern Hindu Thought. New York, N.Y.: St. Martin's Press, 1998. ISBN 9780312216474
Ṭhākura, Debendranātha. Brahmo Dharma. Brahmo classics. Calcutta: Brahmo Mission Press, 1928.
Tagore, Satyendranath, and Indira Devi. Autobiography of Debendranath Tagore. Whitefish, MT: Kessinger, 2006. ISBN 978-1428614970
Tagore, Debendranath history
History of "Tagore, Debendranath"
Retrieved from https://www.newworldencyclopedia.org/p/index.php?title=Tagore,_Debendranath&oldid=683562
Politicians and reformers
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Q: How do I make it so an account's User Account Control dialog appears without a password prompt? On my primary account (the original one with which I installed Windows), when a User Account Control dialog appears, I have simply "Yes" and "No" buttons. However, when my partner (to whom I have given a separate Administrator account) sees one on this same machine, it asks for their password as well.
Is there any way I can make it so that they only have to press "Yes" or "No" like I do?
Below is a screenshot from my Settings app, showing that all accounts on the machine are set to "Administrator"
Below are examples of each (not mine; from others on the Internet whom have shared such dialogs)
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"redpajama_set_name": "RedPajamaStackExchange"
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There have been two Hudson Baronetcies, both of which are now extinct. A third, which began with Sir Charles Hudson, 1st Baronet of Wanlip Hall on 21 June 1791, changed name with the second Baronet becoming the Palmer baronets.
Hudson baronets of Melton Mowbray, Leics (1660)
The Hudson Baronetcy of Melton Mowbray, in the County of Leicester was created on 3 July 1660, in the Baronetage of England, for Henry Hudson. It became extinct on the death of the 7th Baronet in c. 1781
Sir Henry Hudson, 1st Baronet (c. 1609–1690)
Sir Edward Hudson, 2nd Baronet (c. 1637–1702)
Sir Benjamin Hudson, 3rd Baronet (c. 1665–1730)
Sir Charles Hudson, 4th Baronet (died 1752)
Sir Skeffington Hudson, 5th Baronet (1683–1760)
Sir Charles Hudson, 6th Baronet (died 1773)
Sir Charles Vallavine Hudson, 7th Baronet (1755–c. 1781)
Hudson baronets of North Hackney, Middlesex (1942)
The Hudson Baronetcy of North Hackney, in the County of Middlesex was created on 9 July 1942, in the Baronetage of the United Kingdom, for the Conservative Party politician Austin Hudson. It became extinct on his death in 1956.
Sir Austin Uvedale Morgan Hudson, 1st Baronet (1897–1956)
References
Extinct baronetcies in the Baronetage of England
Extinct baronetcies in the Baronetage of the United Kingdom
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"redpajama_set_name": "RedPajamaWikipedia"
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\section{Introduction}
\label{sec:intro}
Obtaining speaker discriminative features is an important step for many speaker recognition tasks, such as speaker verification and speaker diarization. In recent years, extracting speaker embeddings from the intermediate layer of a neural network has become the state-of-the-art method for both tasks~\cite{Snyder2018,Sell2018}, outperforming the historically successful i-vector method~\cite{Dehak2011}.
The ideal speaker embedding space should discriminatively capture the variations between speakers in manner that generalises well to unseen speakers. The properties of speech encapsulating a speaker's identity are many and varied, including factors related to the underlying physical properties of the vocal apparatus (including factors related to gender, age and some medical conditions), as well as properties related to accent, dialect, native language and sociolect (education level, which affects things like lexicon, syntax and stylistics). When humans hear speech from a unfamiliar speaker, we intuitively infer many of these factors in the process of forming a mental picture of the speaker. This technique is used in the field of forensic phonetics and acoustics~\cite{Jessen2007,Hansen2015}, where speaker classification is performed by human experts for suspects in criminal cases. It is reasonable to expect therefore that a good speaker embedding should encode the multiple contributing factors that constitute our notion of a speaker identity.
Indeed, previous work has shown that both deep speaker embeddings and i-vectors encode a wide variety of information and meta-information about speakers and utterances, such as speaker emotion~\cite{Williams2019,Pappagari2020}, accent and language~\cite{Maiti2020} or speaker gender, channel and transcription information \cite{Raj2019}.
This work hypothesises that explicitly encouraging deep speaker embeddings to encode multiple speaker attributes may result in a more descriptive embeddings space that is better able capture speaker variation, particularly between unseen speakers. This is particularly important for the task of diarization where typically all speakers are previously unseen. For this goal we propose to use multi-task learning (MTL)~\cite{Caruana1998} to leverage speaker attribute information when training deep speaker embeddings. A challenge of the use of attribute labels is that data with the relevant labels may be hard to find, particularly from the domain of interest. We solve this problem by designing a system that is able to make use of attribute labels from out-of-domain data. Furthermore, because the attribute labels are used only as a secondary task in MTL, and discarded at test time, our method is robust to noise in the labels, which allows us to obtain them by means of web scraping. We demonstrate the method by web scraping nationality labels for VoxCeleb 2 \cite{Nagraniy2017,Chung2018} and age information for speakers in recordings of the US Supreme Court, showing that training on nationality and age classification tasks in conjunction with the standard speaker classification can improve deep speaker embedding performance on both verification and diarization tasks. Experimental and web scraping code has been made available\footnote{\url{https://github.com/cvqluu/MTL-Speaker-Embeddings}}.
The concept of MTL revolves around the concept that machine learning models that may be used to solve different problems using the same data can benefit from sharing a common representation. In the field of automatic speech recognition (ASR), the work of \cite{Parveen2003} found improvement in increasing the robustness of a hybrid RNN/HMM system by performing speech enhancement as an additional task to classification, with both tasks relying on the same hidden representation, while the work of~\cite{bell2017multitask} found that training to simultaneously predict both context-dependent and context-independent targets regularly improved performance in an ASR setting.
Previous works implementing MTL for speaker recognition specifically have explored using the word spoken in the utterance as an additional task in training deep speaker embeddings to increase verification performance~\cite{Dey2018}. Likewise,~\cite{Liu2019} found that learning to additionally classify the phonetic information also improved performance. While these previous works managed to improve speaker embedding robustness by utilizing transcription information to inform the learned speech representation, there is a risk that these approaches result in embeddings that are more sensitive to phonetic variation, a nuisance factor when performing speaker diarization; this work instead proposes to utilize information that is explicitly connected to the speaker identity to establish a more descriptive and robust speaker embedding space.
\section{Multi-task learning}
\label{sec:mtl}
\begin{figure}[!tb]
\centering
\captionsetup{width=\linewidth}
\includegraphics[width=\columnwidth]{figures/mtl_new.pdf}
\caption{\label{fig:mtl_diag}{An example of a deep speaker embedding extractor trained with multiple tasks.}}
\vspace{-5mm}
\end{figure}
The concept of multi-task learning is that machine learning models may benefit from sharing the same representations when solving different tasks on the same data. In the context of deep learning, this typically means initial layers of a neural network are shared between tasks, after which task specific layers act upon these shared representation of the input data.
We apply MTL in training the speaker embedding extractor, such as the x-vector network \cite{Snyder2018}, which takes acoustic features as input and performs speaker classification. The speaker embeddings are extracted from an intermediate layer in this network. If the layers up until the extracted embedding are considered to be the embedding extractor, one can consider the remaining layers to be a task specific `head'. The standard x-vector network has a single task specific head, a feed forward network performing speaker classification. Multi task learning can be applied by adding additional, separate task-specific heads with their own loss functions which also act on the embedding.
All components of the system, meaning the embedding extractor $\mathcal{G}$, the speaker classification head $\mathcal{C}_{\text{speaker}}$, and each additional task head $\mathcal{C}_{\text{m}}$ for some $M$ number of additional tasks, are trained as a whole. This is illustrated in Figure \ref{fig:mtl_diag}. Each task head produces a loss based on the specific task. For example, the speaker classification loss is likely to be the standard cross entropy loss or angular penalty loss \cite{Wang2018}. An example for an additional task could be to predict the speaker's age. This could be a regression task, predicting the age and utilizing a mean squared error loss, or a classification task, predicting the age category out of some discretized bins.
Starting from the assumption that a speaker classification loss will always be applied, for $M$ additional tasks, their losses $\mathcal{L}_m$ can be combined in the following manner,
\begin{equation}
\mathcal{L_{\text{multi-task}}} = \mathcal{L}_{\text{speaker}} + \sum^M_{i=1} \lambda_{m} \mathcal{L}_{\text{m}}
\end{equation}
\noindent where each additional task loss $\mathcal{L}_{\text{m}}$ is weighted by some chosen loss weighting $\lambda_{m}$, relative to the speaker loss $\mathcal{L}_{\text{speaker}}$. With this formulation, it is possible to consider many additional tasks based on a variety meta-information, but in this work we typically consider only a single additional task.
Having the embedding space explicitly predict properties which we expect to be informative of identity should lead to robustness in that aspect. For example, if accent is used as a task, we encourage the space to structure speaker identity in a fashion we are confident will lead to an appropriate embedding for a new speaker with an accent seen during training. Training only on speaker label may lead to this structure appearing naturally, but previous work has shown that this training objective alone may lead to the embedding space capturing non-speaker related information, such as channel information \cite{Luu2020, huh_augmentation_2020} that we would prefer to model to be invariant to.
\section{Speaker Attributes}
In order to explore leveraging speaker attributes for MTL, we require the availability of data which has this meta-information. Our insight here is that we can opportunistically take advantage of such information when it is happens to be available, by applying MTL in a domain where additional attribute labels are available, followed by transfer learning to adapt the resulting embedding to be more suitable for the intended target domain.
We illustrate the method using two datasets, each with an additional attribute label: the Supreme Court of the United States (SCOTUS) corpus, for which we obtained information on speaker age; and VoxCeleb, which has obtainable information about speaker nationality. We find SCOTUS to be more challenging than VoxCeleb for diarization and hence adopt it as our primary task, although speaker verification results on VoxCeleb are also presented.
It should be noted that both sets of attribute labels used in this work were obtained via web scraping, and thus are more susceptible to noise and inaccuracy than labels obtained via other means. Nevertheless, we find that they can provide useful information about how speaker identity should be structured in the embedding space. In general, while noisy labels generally do degrade performance in most tasks, it is still possible to train under this paradigm and yield positive results \cite{xie_disturblabel_2016,song_learning_2020}.
\subsection{Age: SCOTUS Corpus}
\label{sec:scotus}
The Supreme Court of the United States is the highest level court in the United States. Audio recordings and transcriptions of SCOTUS oral arguments have been made available via the Oyez project\footnote{\url{https://www.oyez.org/about}}. This corpus is unusual in that it features certain speakers across many years. These speakers are the justices (judges) of the Supreme Court, who serve indefinitely on the court until their retirement. The average length of a supreme court justice's tenure at time of writing is approximately 17 years. The phenomenon of how a human voice can change due to the effects of age is fairly well known \cite{Mueller1997}, with degradation in the vocal folds sometimes leading to changes in the fundamental frequency of the voice, and other qualities in vocal delivery, such as jitter, shimmer and volume. Severe cases can lead to diagnoses of vocal disorders \cite{Rapoport2018}. These effects on a speaker's voice makes age a fascinating aspect to explore of what contributes to speaker identity.
For this work, 1022 digital recordings with 913 unique speakers across 1032 hours of audio were considered. For most speakers, approximate ages were obtained for speakers based on their date of qualification to practice law, which was obtained via web scraping. The interested reader can refer to the code repository for more information. One can see the distribution of ages of utterances in Figure \ref{fig:scotus_age_dist}, with an average age of around 56.
\begin{figure}[!tb]
\centering
\captionsetup{width=\linewidth}
\includegraphics[width=\columnwidth]{figures/fixed_scotus_allage_dist.pdf}
\vspace{-5mm}
\caption{\label{fig:scotus_age_dist}{The age distribution of utterances in the SCOTUS corpus, split into 10 bins.}}
\vspace{-5mm}
\end{figure}
\subsection{Nationality: VoxCeleb}
VoxCeleb 1 and 2 \cite{Nagraniy2017,Chung2018} are speaker recognition datasets featuring celebrity speakers, meaning certain speaker attributes can be found in the public domain. Speaker nationality labels can be web scraped from sources such as Wikipedia and are a proxy for speaker accent, an attribute which is clearly informative of speaker identity.
\begin{figure}[tb]
\centering
\captionsetup{width=\linewidth}
\includegraphics[width=0.7\columnwidth]{figures/vox2_nat_dist.pdf}
\vspace{-5mm}
\caption{\label{fig:vox_dist}{The nationality distribution of the 5994 speakers in VoxCeleb 2's training set. Only nationalities with more than 50 are shown individually in this figure, with the rest being grouped as `Other'.}}
\vspace{-5mm}
\end{figure}
\section{Experimental Setup}
\label{sec:exp}
Our primary verification and diarization experiments were performed on SCOTUS corpus, whilst embeddings were also trained on VoxCeleb.
For the training and evaluation on SCOTUS, utterances were split into train and test sets by recording, at an approximate $80\%$ train proportion, and ensuring that the train and test distributions for age using 10 uniformly spaced bins were approximately similar. In order to evaluate the SCOTUS data as a verification task, 15 positive and negative trials were selected for each speaker in the test recordings, excluding any speakers seen in the training set from these trials. Cosine similarity was used to score all embeddings.
The speaker embedding extractors for both VoxCeleb and SCOTUS followed the original x-vector architecture, up until the embedding layer which had 256 hidden units. For classification heads that utilized the standard cross entropy loss, these had a similar architecture to the x-vector network, having two hidden layers also of dimension 256 before being projected to the number of classes. The non-linearity used throughout was Leaky ReLU. For classification heads that used an angular penalty loss, these were simply a single affine matrix on top of the embedding layer that projected into the number of classes, using the CosFace~\cite{Wang2018} loss.
Embedding extractors were trained with different configurations of classification heads, and verification and diarization performance was evaluated.
We also performed contrastive experiments with randomly shuffling the labels of the additional tasks -- such as age -- to eliminate the possibility any kind of positive regularization effect that the additional task may have irrespective of the information in the labels.
For all embedding extractor training setups for SCOTUS, regardless of the number of classification heads, networks were trained for 50,000 iterations on 350 frames of 30-dimensional MFCCs, with batch size 500, using a small held out set of training utterances for validation. Stochastic gradient descent was used with learning rate 0.2 and momentum 0.5.
For diarization of the SCOTUS corpus, embeddings were extracted for every 1.5s with 0.75s overlap using reference speaker activity detection segmentation, using cosine similarity for scoring and using agglomerative hierarchical clustering to the oracle number of speakers. Due to the supreme court justices appearing in both train and test recordings, the diarization error rate was evaluated for two scenarios: The first evaluation scenario was standard in that all the speech segments were scored, including the speakers which appeared in the training set. The second evaluation scenario was to only score portions of the speech in which unseen speakers were talking.
Speaker embedding extractors for VoxCeleb were trained on the VoxCeleb 2 training set, with 5994 speakers, augmented in the standard Kaldi~\cite{Povey_ASRU2011} fashion with babble, music and background noises along with reverberation as in \cite{Snyder2018}. Speakers who were the only members of their nationality in the training set were grouped into the same class as the speakers who could not have their nationality scraped, yielding 102 nationality classes. Note that not all of these are shown in Figure \ref{fig:vox_dist}. Networks were trained for 100,000 iterations with the same batch size and optimization settings as the SCOTUS models.
When evaluating models trained on VoxCeleb on SCOTUS, the VoxCeleb models was were fine-tuned for 5000 iterations on the SCOTUS training set, varying whether or not the full network or only the last linear layer of the extractor was fine-tuned, along with whether or not age in addition to speaker labels were trained on during the fine-tuning. For the first 1000 iterations of fine-tuning, all embedding extractor parameters were frozen to allow the freshly initialized classification head(s) to fit to the new data.
\section{Results and discussions}
\label{sec:results}
\begin{table*}[t]
\centering
\begin{tabular}{|c| m{4cm} || c | c c c || c |}
\hline
\multirow{3}{*}{\shortstack{Training\\Set}} & \multirow{3}{*}{Model} & \multirow{3}{*}{\shortstack{SCOTUS\\Fine-tune label set}} & \multicolumn{3}{c||}{SCOTUS} & VoxCeleb \\
\cline{4-7}
& & & \multirow{2}{*}{EER} & \multicolumn{2}{c||}{DER} & \multirow{2}{*}{EER} \\
& & & & All & Unseen & \\
\hline
\multirow{8}{*}{\rotatebox[origin=c]{90}{SCOTUS}} & Only Speaker & - & 3.14\% & 27.58\% & 19.75\% & - \\
& Speaker + Random labels &- & 3.78\% & 27.87\% & 19.14\% & - \\
& Speaker + Age & - & 2.68\% & 26.14\% & 18.02\% & - \\
\cline{2-7}
& Only Speaker (CosFace) & - & 2.71\% & 26.51\% & 19.75\% & - \\
& Speaker (CosFace) + Age & - & \textbf{2.62\%} & \textbf{21.80\%} & \textbf{14.08\%} &- \\
\cline{2-7}
& Only Age & - & 3.99\% & 37.08\% & 26.18\% & - \\
\cline{2-7}
& Only Gender & - & 19.42\% & 58.02\% & 44.78\% & - \\
\cline{2-7}
& Only Random & - & 23.31\% & 69.13\% & 48.97\% & - \\
\hhline{|=|=#=|===#=|}
\multirow{10}{*}{\rotatebox[origin=c]{90}{VoxCeleb 2}} & Only Speaker (Baseline) & (Last Linear) Sp. & 2.26\% & 20.09\% & 14.07\% & 3.04\% \\
& Speaker + Random & (Last Linear) Sp. & 2.98\% & 22.18\% & 16.32\% & 4.32\% \\
& Speaker + Nationality & (Last Linear) Sp. & 1.99\% & 18.74 \% & 13.54\% & \textbf{2.95\%} \\
\cline{2-7}
& Only Speaker & (LL) Sp. + Age & 2.00\% & 17.81\% & 12.44\% & 3.04\% \\
& Speaker + Nationality & (LL) Sp. + Age & 1.89\% & \textbf{14.82\%} & \textbf{10.45\%} &\textbf{2.95\%} \\
\cline{2-7}
& Only Speaker (Baseline) & (Full) Sp. & 1.63\% & 29.56\% & 20.02\% & 3.04\% \\
& Speaker + Nationality & (Full) Sp. & 1.57\% & 25.04\% & 16.93\% & \textbf{2.95\%} \\
\cline{2-7}
& Only Speaker & (Full) Sp. + Age & 1.57\% & 25.98\% & 17.44\% & 3.04\% \\
& Speaker + Nationality & (Full) Sp. + Age & \textbf{1.52\%} & 19.77\% & 13.57\% & \textbf{2.95\%} \\
\cline{2-7}
& Only Nationality & - & 17.68\% & 57.97\% & 42.78\% & 13.38\%\\
\hline
\end{tabular}
\caption{\label{tab:scotus_multi} Verification and diarization performance on SCOTUS and VoxCeleb for various models with multitask learning objectives.}
\end{table*}
All results are shown in Table \ref{tab:scotus_multi}.
Results for adding a 10-class age classification task to speaker embedding extractors for the SCOTUS corpus can be seen in the first portion of the table, separated into the models trained with cross entropy loss on the speaker classification head, and the models trained with CosFace loss on the speaker classification head. The Diarization Error Rate (DER) results are split into two scoring scenarios, `All' indicates that all speech was scored and `Unseen' indicates that only speech segments from unseen speakers was scored. Adding gender classification was not found to have any positive effect, and thus these results have been omitted for brevity.
For the standard cross entropy loss, for configurations of the age loss with $\lambda_{\text{age}}$=0.5, the verification and diarization performance was improved over the baseline. (For brevity, not all values of $\lambda$ are shown in the table). By contrast, both the control experiment of randomly shuffled labels and the control experiment featuring completely random speaker labels yielded no improvement.
The SCOTUS trained models also feature an experiment in which the only classification head was the Age classification head, and this performs surprisingly well on the speaker verification and diarization tasks, despite only being trained to distinguish between 10 age categories. This suggests that in order to predict age, some knowledge of speaker identity is also required. Although it is not shown on the table, the age accuracy of the non-CosFace networks trained with both speaker and age outperformed that of training on age alone, suggesting that performing tasks in combination was able to improve both tasks. This is supported by previous MTL literature, which indicates training on multiple tasks may be helpful to each individually.
The addition of gender classification not improving results is unsurprising, considering male and female voices can largely be distinguished based on their fundamental frequency (F0) \cite{Jessen2007, Pernet2012}, and thus this does not add much discriminatory power to the primary goal of speaker recognition.
For SCOTUS trained models, it is clear from Table \ref{tab:scotus_multi} that CosFace improves the speaker verification and diarization performance over the standard cross entropy loss, with $\lambda_{\text{age}}$ being changed to $0.01$ to account for the change in the relative scale of $\mathcal{L}_{\text{speaker}}$. The addition of the age classification head similarly improves over only using speaker labels, producing the best results for verification and diarization for all the configurations shown, making a relative improvement of 17.8\% in DER and 3.3\% in EER over the `Only Speaker' CosFace baseline.
The performance of VoxCeleb trained models can also be seen in Table \ref{tab:scotus_multi}. Note all speaker classification heads for VoxCeleb models utilized the CosFace loss. The verification performance of these models on VoxCeleb can be seen in the rightmost column, with the addition of the nationality task ($\lambda_{\text{nat}}$=0.05) yielding a 3\% relative performance improvement.
When evaluated on SCOTUS, the fine-tuning of the VoxCeleb models was performed on either the Last Linear (LL) layer of the embedding network, or the whole (Full) network. Either Speaker (Sp.) labels alone were used, or Age was added as an auxiliary fine-tuning task, with $\lambda_{\text{age}}$=0.05. Models which only use speaker labels at all stages of training (marked as `Baseline' in the Table) are improved upon in both verification and diarization by utilizing either Nationality or Age tasks during the primary training stage or the fine-tuning stage respectively. Indeed, the best performance for verification and diarization is found when auxiliary tasks are employed at both stages.
For the best baseline verification model with no additional tasks, a 6.7\% relative improvement in EER is found by using nationality and age (1.63\% $\rightarrow$ 1.52\%), and similarly for diarization, using auxiliary tasks at both stages yields a 26.2\% relative improvement in DER scoring all regions (20.09\% $\rightarrow$ 14.82\%).
While using both additional tasks yields the best performance, improvements are still found when using a single auxiliary task at either stage of training, suggesting that this technique is still valuable for scenarios in which speaker attribute information is limited or missing from the desired domain.
Overall, these experiments demonstrated that training on auxiliary speaker attribute tasks in addition to speaker classification can yield more robust representations for verification and diarization.
\section{Conclusions}
\label{sec:conclusion}
In this work we showed that training on additional tasks relating to speaker attributes, specifically approximate age and nationality, alongside the standard speaker classification can improve the performance of deep speaker embeddings for both verification and diarization. Future work is planned to investigate training on multiple corpora and multiple tasks simultaneously to extend the improvement in robustness from multi-task training shown here.
\pagebreak
\bibliographystyle{IEEEtran}
\section{Introduction}
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\label{tab:example}
\centering
\begin{tabular}{ r@{}l r }
\toprule
\multicolumn{2}{c}{\textbf{Ratio}} &
\multicolumn{1}{c}{\textbf{Decibels}} \\
\midrule
$1$ & $/10$ & $-20$~~~ \\
$1$ & $/1$ & $0$~~~ \\
$2$ & $/1$ & $\approx 6$~~~ \\
$3.16$ & $/1$ & $10$~~~ \\
$10$ & $/1$ & $20$~~~ \\
$100$ & $/1$ & $40$~~~ \\
$1000$ & $/1$ & $60$~~~ \\
\bottomrule
\end{tabular}
\end{table}
\subsection{Equations}
Equations should be placed on separate lines and numbered. Examples of equations are given below. Particularly,
\begin{equation}
x(t) = s(f_\omega(t))
\label{eq1}
\end{equation}
where \(f_\omega(t)\) is a special warping function
\begin{equation}
f_\omega(t) = \frac{1}{2 \pi j} \oint_C
\frac{\nu^{-1k} \mathrm{d} \nu}
{(1-\beta\nu^{-1})(\nu^{-1}-\beta)}
\label{eq2}
\end{equation}
A residue theorem states that
\begin{equation}
\oint_C F(z)\,\mathrm{d}z = 2 \pi j \sum_k \mathrm{Res}[F(z),p_k]
\label{eq3}
\end{equation}
Applying (\ref{eq3}) to (\ref{eq1}), it is straightforward to see that
\begin{equation}
1 + 1 = \pi
\label{eq4}
\end{equation}
Finally we have proven the secret theorem of all speech sciences. No more math is needed to show how useful the result is!
\begin{figure}[t]
\centering
\includegraphics[width=\linewidth]{figure.pdf}
\caption{Schematic diagram of speech production.}
\label{fig:speech_production}
\end{figure}
\subsection{Information for Word users only}
For ease of formatting, please use the styles listed in Table 2. The styles are defined in this template file and are shown in the order in which they would be used when writing a paper. When the heading styles in Table 2 are used, section numbers are no longer required to be typed in because they will be automatically numbered by Word. Similarly, reference items will be automatically numbered by Word when the ``Reference'' style is used.
\begin{table}[t]
\caption{Main predefined styles in Word}
\label{tab:word_styles}
\centering
\begin{tabular}{ll}
\toprule
\textbf{Style Name} & \textbf{Entities in a Paper} \\
\midrule
Title & Title \\
Author & Author name \\
Affiliation & Author affiliation \\
Email & Email address \\
AbstractHeading & Abstract section heading \\
Body Text & First paragraph in abstract \\
Body Text Next & Following paragraphs in abstract \\
Index & Index terms \\
1. Heading 1 & 1\textsuperscript{st} level section heading \\
1.1 Heading 2 & 2\textsuperscript{nd} level section heading \\
1.1.1 Heading 3 & 3\textsuperscript{rd} level section heading \\
Body Text & First paragraph in section \\
Body Text Next & Following paragraphs in section \\
Figure Caption & Figure caption \\
Table Caption & Table caption \\
Equation & Equations \\
\textbullet\ List Bullet & Bulleted lists \\\relax
[1] Reference & References \\
\bottomrule
\end{tabular}
\end{table}
If your Word document contains equations, you must not save your Word document from ``.docx'' to ``.doc'' because when doing so, Word will convert all equations to images of unacceptably low resolution.
\subsection{Hyperlinks}
For technical reasons, the proceedings editor will strip all active links from the papers during processing. Hyperlinks can be included in your paper, if written in full, e.\,g.\ ``http://www.foo.com/index.html''. The link text must be all black.
Please make sure that they present no problems in printing to paper.
\subsection{Multimedia files}
The INTERSPEECH organizing committee offers the possibility to submit multimedia files. These files are meant for audio-visual illustrations that cannot be conveyed in text, tables and graphs. Just like you would when including graphics, make sure that you have sufficient author rights to the multimedia materials that you submit for publication. The proceedings media will NOT contain readers or players, so be sure to use widely accepted file formats, such as MPEG, Windows WAVE PCM (.wav) or Windows Media Video (.wmv) using standard codecs.
Your multimedia files must be submitted in a single ZIP file for each separate paper. Within the ZIP file you can use folders and filenames to help organize the multimedia files. In the ZIP file you should include a TEXT or HTML index file which describes the purpose and significance of each multimedia file. From within the manuscript, refer to a multimedia illustration by its filename. Use short file names without blanks for clarity.
The ZIP file you submit will be included as-is in the proceedings media and will be linked to your paper in the navigation interface of the proceedings. Causal Productions (the publisher) and the conference committee will not check or change the contents of your ZIP file.
Users of the proceedings who wish to access your multimedia files will click the link to the ZIP file which will then be opened by the operating system of their computer. Access to the contents of the ZIP file will be governed entirely by the operating system of the user's computer.
\subsection{Page numbering}
Final page numbers will be added later to the document electronically. \emph{Do not make any footers or headers!}
\subsection{References}
The reference format is the standard IEEE one. References should be numbered in order of appearance, for example \cite{Davis80-COP}, \cite{Rabiner89-ATO}, \cite[pp.\ 417--422]{Hastie09-TEO}, and \cite{YourName21-XXX}.
\subsection{Abstract}
The total length of the abstract is limited to 200 words. The abstract included in your paper and the one you enter during web-based submission must be identical. Avoid non-ASCII characters or symbols as they may not display correctly in the abstract book.
\subsection{Author affiliation}
Please list country names as part of the affiliation for each country.
\subsection{Number of authors in the author list}
The maximum number of authors in the author list is twenty. If the number of contributing authors is more than twenty, they should be listed in a footnote or in acknowledgement section, as appropriate.
\subsection{Submitted files}
Authors are requested to submit PDF files of their manuscripts. You can use commercially available tools or for instance http://www.pdfforge.org/products/pdfcreator. The PDF file should comply with the following requirements: (a) there must be no PASSWORD protection on the PDF file at all; (b) all fonts must be embedded; and (c) the file must be text searchable (do CTRL-F and try to find a common word such as ``the''). The proceedings editors (Causal Productions) will contact authors of non-complying files to obtain a replacement. In order not to endanger the preparation of the proceedings, papers for which a replacement is not provided in a timely manner will be withdrawn.
\section{Discussion}
This is the discussion. This is the discussion. This is the discussion. Is there any discussion?
Lorem ipsum dolor sit amet, consectetur adipiscing elit. Cras consequat mollis odio, nec venenatis enim auctor sed. Integer tincidunt fringilla lectus eget condimentum. In eget sapien id eros dapibus interdum vel ac quam. Aenean vitae rutrum erat. Aenean et risus pharetra, lacinia augue ut, fermentum ante. Integer dui arcu, interdum at ornare a, faucibus quis est. Mauris quis quam felis. Etiam pulvinar massa et turpis lacinia, eu posuere mi iaculis. Fusce at velit quis leo dignissim porttitor.
Fusce ut nunc eu sapien venenatis finibus a vel ligula. Pellentesque habitant morbi tristique senectus et netus et malesuada fames ac turpis egestas. Ut quam eros, volutpat at gravida consectetur, rutrum ut leo. Aenean cursus euismod feugiat. Cras hendrerit, ligula eu feugiat malesuada, neque turpis auctor lacus, sit amet accumsan neque orci a quam. Mauris suscipit ultrices mattis. Nulla at interdum metus, id pharetra diam. Curabitur at vestibulum sem, sed elementum massa. Donec iaculis et arcu ut rutrum. Fusce gravida, mauris porta volutpat eleifend, enim mauris eleifend orci, eu ultrices leo purus vitae metus. In pretium dolor ut magna dictum, at imperdiet lectus porta.
Quisque mollis lectus id risus pretium mattis. Morbi scelerisque posuere est, id efficitur urna luctus non. Praesent quam lacus, facilisis id ante eu, vehicula maximus ex. Nullam mollis in arcu vitae efficitur. Aliquam molestie eleifend ante, in pretium velit ultrices ac. Etiam laoreet nec sem non pulvinar. Integer ligula felis, interdum non lacus id, malesuada imperdiet turpis.
Aenean sit amet volutpat nisi. Aliquam eu erat quis tortor ultrices laoreet. Vivamus fermentum semper metus, non faucibus libero euismod vitae. Sed efficitur porta congue. Aenean in faucibus nisi. Donec suscipit augue vitae orci consequat, sit amet aliquet felis varius. Duis efficitur lacinia dolor sit amet lobortis. Curabitur erat sapien, molestie nec nisi eu, dignissim accumsan ipsum. Fusce id nibh nec risus dictum posuere in ac magna. Donec malesuada massa sed erat lacinia cursus. Suspendisse ornare augue nec volutpat consequat.
Vestibulum et vulputate nisi, a malesuada mi. Nam pellentesque arcu sapien, at placerat odio imperdiet ut. Curabitur nec venenatis tellus, vel aliquet nisi. Curabitur vel ligula sit amet metus auctor pretium. Nullam nulla mi, blandit a mattis id, vulputate sit amet enim. Proin mollis fringilla dictum. Proin lacinia orci purus.
Curabitur porttitor bibendum dolor, nec consectetur sapien pulvinar id. Donec eleifend, est vel dignissim pretium, tortor augue euismod nunc, id fermentum erat felis ac neque. Morbi id lectus ultricies, rutrum justo eu, sollicitudin risus. Suspendisse lobortis efficitur nisi sit amet pellentesque. Ut eget augue at mi aliquet mattis. Proin et feugiat erat, sit amet sodales eros. Integer sed elit quis est mattis ullamcorper. Pellentesque lectus nisi, vulputate a imperdiet tincidunt, auctor nec orci. Pellentesque sagittis nisl orci, vitae placerat massa lacinia nec. Sed egestas magna sed augue sollicitudin luctus. Praesent interdum bibendum tortor, eu porta purus. Aliquam convallis velit id mi fermentum, sed ornare eros cursus.
Quisque congue leo a fringilla pharetra. Phasellus sed tempor est, sed auctor purus. Morbi vel lacus ullamcorper, auctor mauris id, pulvinar lorem. Suspendisse potenti. Nam porta, purus non eleifend bibendum, erat metus pellentesque elit, non luctus nibh nunc ornare nisl. Sed rutrum lacinia nisi ac suscipit. Curabitur non blandit augue. Integer viverra, ipsum vel molestie euismod, sem quam tempus massa, eget efficitur ante turpis non metus. Quisque efficitur posuere velit in iaculis. Cras imperdiet varius urna vitae vestibulum. Donec accumsan eget nisi sed pellentesque. Vestibulum id quam ut urna volutpat ullamcorper gravida sit amet libero. Aliquam bibendum, ligula vitae porta malesuada, arcu diam congue erat, a pharetra diam sem vulputate tortor. Etiam luctus iaculis leo cursus tristique.
Mauris mattis sem dolor, sit amet ullamcorper arcu tincidunt ac. Vestibulum at blandit tortor. Quisque bibendum congue leo, vitae eleifend massa. Vestibulum vitae odio elit. Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut sagittis quam vel felis ornare, in gravida felis tempor. Donec molestie dui quis leo venenatis blandit. Nunc sit amet finibus metus. Cras ut lectus ex.
Suspendisse commodo libero vel leo tincidunt, a tempus mauris porta. Integer varius eros ac sapien lacinia vehicula. Donec porttitor, lacus faucibus rhoncus venenatis, neque quam imperdiet nunc, id consectetur metus purus quis sapien. Phasellus interdum nulla vel euismod posuere. Vestibulum finibus magna vel finibus mollis. Curabitur mollis turpis tortor, hendrerit vulputate justo egestas quis. Nam dignissim luctus leo non elementum. Phasellus a metus at leo malesuada bibendum. Mauris quis eleifend magna, nec vehicula ex. Donec venenatis urna fermentum commodo vehicula. Ut mattis scelerisque aliquam. Vivamus pulvinar erat metus, id tempus mi vulputate quis.
Fusce lobortis a urna eget blandit. Vivamus in eleifend neque, at sollicitudin lectus. Quisque faucibus egestas lorem, in commodo diam maximus eu. Morbi finibus ante ac felis porttitor euismod. Donec lobortis aliquam ipsum sit amet luctus. Cum sociis natoque penatibus et magnis dis parturient montes, nascetur ridiculus mus. Etiam rutrum neque sapien, eget luctus turpis iaculis pulvinar. Duis quis pulvinar nunc, nec bibendum ligula. Phasellus suscipit sagittis lacus molestie laoreet. Pellentesque lacus diam, tincidunt a aliquam vitae, aliquet non justo.
Etiam lectus lacus, commodo eget consectetur eget, auctor vitae leo. Praesent vitae erat in diam blandit semper vitae et eros. Maecenas auctor pharetra nibh eget egestas. Donec accumsan ut risus eget rhoncus. Nam placerat, erat sit amet gravida mollis, purus arcu accumsan diam, tempus pharetra risus mi ac sapien. Ut et tortor porta, pulvinar elit vitae, tempor mi. Nam interdum, nisl non pharetra molestie, turpis neque commodo ligula, sit amet pretium nisl nibh quis ante. Quisque et ex eget velit lobortis suscipit. Integer aliquam finibus molestie. Sed pellentesque neque eu turpis aliquet, mattis ornare enim finibus. In hac habitasse platea dictumst.
Integer congue quis justo a posuere. Quisque porta, ante et dignissim suscipit, arcu mauris ultrices libero, nec sollicitudin purus lacus a enim. Aliquam feugiat eget lacus molestie sodales. Duis blandit placerat nunc, et venenatis turpis dictum vel. Nulla facilisi. Nam ullamcorper, tellus eu posuere mattis, arcu lacus dictum nulla, vel mattis nisi sem posuere tellus. Etiam quis eros condimentum lectus lobortis eleifend. In ex lacus, sodales scelerisque egestas ac, aliquam nec purus. Nunc sit amet magna non libero ullamcorper dictum. Phasellus porta faucibus tempus. Praesent blandit tortor sed tellus ornare consectetur. Sed sed nisi id neque porta varius eu eu velit. Curabitur varius convallis justo id facilisis. Mauris auctor velit nec aliquam cursus.
Integer suscipit scelerisque leo sed faucibus. Ut commodo nulla luctus diam posuere egestas. Integer ut augue ac velit ullamcorper tempus. Pellentesque in mi rhoncus, sodales sem quis, commodo sem. Aenean dapibus euismod diam id rhoncus. Nullam vehicula placerat eros consectetur luctus. Aliquam auctor ipsum vitae egestas imperdiet. Ut nulla lacus, imperdiet quis urna vel, ornare imperdiet tortor. Mauris nec diam ac nunc laoreet volutpat at id turpis. Nulla eu neque a risus feugiat iaculis ac vel risus. Ut tempus elementum lorem eget porta. Nullam et ullamcorper urna. Cum sociis natoque penatibus et magnis dis parturient montes, nascetur ridiculus mus. Phasellus eget dui vitae nulla hendrerit ultrices quis rutrum leo.
Proin consectetur lacus sit amet eleifend varius. Etiam eu blandit risus. Curabitur pellentesque urna sed dolor congue mattis. Vestibulum ut velit posuere, feugiat leo a, dignissim massa. Proin eu nulla risus. Fusce luctus bibendum est, sit amet venenatis nisi finibus non. Donec ultricies ornare nunc at lobortis. Nam a auctor metus. Vestibulum pretium condimentum turpis ac mattis. Curabitur semper sagittis rhoncus. Duis molestie facilisis mattis. Sed pharetra lorem id tortor efficitur, sed maximus leo posuere. Quisque suscipit molestie convallis. Duis imperdiet placerat congue. Morbi placerat, velit ut tempor porta, ex nisi imperdiet purus, non feugiat ex velit nec nulla.
Phasellus mattis at erat eget lobortis. Vivamus sodales odio non erat luctus faucibus. Curabitur aliquam luctus nulla quis consectetur. Fusce vulputate finibus vulputate. Cras at condimentum massa. Duis vestibulum ipsum ac tortor lobortis fringilla. Cras non neque at nunc pellentesque mollis. Sed quis erat mauris. Ut dapibus sem lectus, quis imperdiet diam bibendum et. Maecenas quis venenatis ante. Nunc blandit a risus sed scelerisque. Praesent cursus est sit amet nisi tempus, quis placerat libero rutrum. Phasellus lacinia nisi quis consequat mollis. Phasellus sagittis aliquam lacus.
Sed dolor quam, posuere nec nunc eget, feugiat lobortis ligula. Fusce lacinia fermentum dolor, luctus dapibus ex venenatis feugiat. In hac habitasse platea dictumst. Nullam vitae ligula dignissim, interdum turpis quis, tincidunt metus. Nullam in nisl vitae mauris egestas porta. Nam fringilla aliquet sapien, non dapibus nunc sollicitudin id. Nunc hendrerit felis et vehicula consequat. Praesent varius libero id volutpat iaculis. Aliquam vel dui imperdiet, pharetra augue sed, iaculis nulla. Quisque mollis orci nec odio eleifend, eget laoreet nunc feugiat. Cras varius tortor a fringilla gravida.
Sed posuere erat eu dolor consequat euismod. Donec imperdiet, tellus nec convallis commodo, lorem sem lobortis purus, a lacinia massa ipsum vitae nisl. Vivamus auctor tellus in urna iaculis luctus. In dui nibh, posuere a erat a, lobortis finibus nulla. Sed vel suscipit nisi. Nunc eget nibh risus. Sed posuere tempus eleifend. Nullam ac lacinia ligula, ut blandit erat. In at est sed turpis consequat rutrum. Nunc eget lectus venenatis, convallis ligula non, mollis orci. Aliquam sit amet ligula turpis. Sed finibus laoreet elit nec molestie.
Mauris in nisi et neque euismod aliquam ut eget felis. Sed eget dictum tellus, finibus rutrum nibh. Fusce placerat augue a faucibus semper. Nam sed nisl ligula. Vivamus ante augue, faucibus at risus ut, hendrerit viverra risus. Etiam justo eros, dignissim a nunc sed, porta mollis erat. Fusce in sem accumsan, laoreet nibh porta, porttitor orci. Donec venenatis vehicula ante eget dictum.
Aliquam fermentum a metus pellentesque cursus. Vivamus eleifend ultricies tellus, vel scelerisque diam. Sed hendrerit est at elit suscipit placerat. Praesent nec pretium erat. Quisque blandit nunc in felis bibendum consequat.
\section{Conclusions}
Authors must proofread their PDF file prior to submission to ensure it is correct. Authors should not rely on proofreading the Word file. Please proofread the PDF file before it is submitted.
\section{Acknowledgements}
The ISCA Board would like to thank the organizing committees of the past INTERSPEECH conferences for their help and for kindly providing the template files. \\
Note to authors: Authors should not use logos in the acknowledgement section; rather authors should acknowledge corporations by naming them only.
\bibliographystyle{IEEEtran}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 8,664
|
<?php if ( ! defined('BASEPATH')) exit('No direct script access allowed');
class Store_model extends MY_Model
{
private $tableName;
public $Name;
public function __construct()
{
parent::__construct();
$this->tableName = "Stores";
$this->Name = "Store";
}
public function Items()
{
$query = $this->db->get($this->tableName);
return $query->result();
}
public function Exists($name){
$this->db->where("Name",$name);
$query = $this->db->get($this->tableName);
return $query->num_rows()>0;
}
public function AddItem($name,$phone,$manager,$address,$fax)
{
if ($this->Exists($name)){
return array("result"=>false,"message"=>$this->Name." ".$name." exists");
}
$item = array("Name" => $name,
"Phone" =>$phone,
"Manager" =>$manager,
"Address" => $phone,
"Fax" => $fax);
$this->db->insert($this->tableName,$item);
return $this->Items();
}
public function EditItem($id,$name,$phone,$manager,$address,$fax){
$item = array(
"Name" => $name,
"Phone" =>$phone,
"Manager" =>$manager,
"Address" => $phone,
"Fax" => $fax);
$this->db->update($this->tableName,$item,array("Id"=>$id));
return $this->Items();
}
public function RemoveItem($id)
{
$this->db->query("delete from Stores where Id=".$id);
return;
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 2,020
|
"""
Module with errors in doctest formatting.
>>> 1
'this is\n an error'
"""
def foo():
pass
if __name__ == '__main__':
import doctest
doctest.testmod()
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 119
|
Lapinč (nemško Lafnitz, madžarsko Lapincs) je avstrijsko-madžarska reka in 114 km dolg pritok reke Rabe.
Lapinč izvira na avstrijskem Štajerskem in teče skozi Gradiščansko. Pri Monoštru v Slovenskem Porabju na Madžarskem se izliva v Rabo. Na Madžarskem je Lapinč najkrajša reka v državi.
Zunanje povezave
Reke na Madžarskem
Reke v Avstriji
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 1,318
|
Freiburger Schule steht für:
Freiburger Schule (Politikwissenschaft), Schule der deutschen Politikwissenschaft
Freiburger Schule (Kameratechnik), Pionierleistung innovativer Kameraführung
Freiburger Schule der Nationalökonomie, siehe Ordoliberalismus
Freiburger Schule, Forschungsrichtung der Mediävistik, siehe Gerd Tellenbach
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 7,226
|
Ever since her engagement to Ashton Kutcher and the announcement of her pregnancy, Mila Kunis has remained mostly mum. But she went on The Ellen DeGeneres Show, which airs today, and opened up and told the audience all about these major life events. Well, mostly all. She's still not revealing the due date ("Sometime this year, I promise you") or the name (though that's picked out), but we do get to hear about her weird cravings ("I eat sauerkraut all day long") and how she and Ashton were able to keep their engagement secret for two whole months. Watch below: video platformvideo managementvideo solutionsvideo player I'm mostly just impressed that she was wearing the ring in public for two months and somehow Kim didn't notice. Trust me, that is hard to do.
Ever since her engagement to Ashton Kutcher and the announcement of her pregnancy, Mila Kunis has remained mostly mum. But she went on The Ellen DeGeneres Show, which airs today, and opened up and told the audience all about these major life events. Well, mostly all. She's still not revealing the due date ("Sometime this year, I promise you") or the name (though that's picked out), but we do get to hear about her weird cravings ("I eat sauerkraut all day long") and how she and Ashton were able to keep their engagement secret for two whole months.
I'm mostly just impressed that she was wearing the ring in public for two months and somehow Kim didn't notice. Trust me, that is hard to do.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 4,472
|
Cps Trial Process
Was the child placed with one of their parents, or with someone who is not a parent? A trial must. Collaborative Problem Solving (CPS) in a Nutshell Summary provided by Clayton R. CPS Chief Economic Officer Heather McArthur said this cut was not significant to the district. How does the CPS decide to prosecute?. Jan 26, 2018 · ""The CPS and police have a vital role in ensuring there is a fair trial process in place to protect the public. Indicated means that: CPS found enough evidence to support the claim that a child has been abused or neglected. We pride ourselves on staying in close communication with our clients and treating them like family. We need a comprehensive understanding of the CPS system if we want to effectively target strategies to enhance or improve the system. The Pretrial Process While many citizens think the real action in the criminal courts happens during trials, they are wrong in that assessment. Durango Phx, AZ 85009-6292 Southeast Facility 1810 S. If CPS is investigating you or a loved one, you should contact an attorney as soon as possible. This thematic inspection of the Crown Prosecution Service. Call an experienced attorney to learn more about the CPS process. You probably do not know for sure who made the report - it could have been a family member, a neighbor, a teacher, a doctor, a police officer, or even a stranger. The FDA requires a multi-phase clinical trials process to be completed before deciding if the medicine under investigation is safe and effective for a broader patient population. CPS workers should be under supervision with a supervisor present at each visit. A case plan is developed. The Judge works for them (paid by them) and never dismisses a Petition. 1 Clinical trials serve as the foundation for evidence-based medicine by addressing specific clinical questions that may lead to improvements in current clinical practice. Trial by jury is more involved, as the jury selection process is an active process requiring the attention of the prosecution and the defense. The Criminal Trial Process can be confusing. Due Process of law - that fundamental, constitutional guarantee that your court case will be handled fairly and that the judge will play by the rules laid down by the legislature - is often missing in CPS courts. Often, false reports of child abuse occur when one parent files for divorce and the other parent did not want a divorce or wants sole custody of the child or children. Texas Rules of Civil Procedure. Cyber-physical systems (CPS) are engineered systems that are built from, and depend upon, the seamless integration of computational algorithms and physical components. The UK Crown Prosecution Service is supposedly impartial, though several recent cases have disproved that, with three cases in particular demonstrating how the CPS not only concealed evidence, but interfered in the process of justice and even closed down a …. 7 Treatment Issues in Pretrial and Diversion Settings The pretrial period of criminal justice processing is unique in that for most people it is brief and the outcome is uncertain. The process in CPS court ("juvenile dependency") is rapid and structured. Child Custody. When rights are trampled and due process is subverted in cases involving CPS it can so often lead to crucial information being missed and a child being put into a precarious circumstance. If you do go to trial, you and CPS can give testimony and evidence. This allows for greater insight into a drug's potential early in the process and leads to a more focused program for promising compounds, including an optimized clinical trial design. It's about what can help you deal with all the stress of life so that you. All it takes is an unethical Social worker who fabricates Child Abuse charges usually against Fathers in Divorce and Child Custody situations. A Tennessee Court of Appeals vacated and remanded the trial court's termination of a father's parental rights and concluded that the trial court's finding that the child was in all reasonable probability subject to abuse or neglect by the father was against the great weight of the evidence. Start studying CPS. Program Administrator In support of CFSA's teaming principles and best practice standards, the Child Protective Services administrator is. However, this study was limited by sample size (28 children in CPS, 19 in PMT) and the lack of a waitlist control group. Other than judges or attorneys, most people find court proceedings intimidating and confusing. Both the Guidelines and the Protocol emphasise that defence engagement with the disclosure process should be "early and meaningful", and that the defence statement should be "clear and detailed". The CPS attorney presents evidence through the testimony of the CPS caseworker and law enforcement or other witnesses. Tables for Process Management - Values for Flag in Customizing, Text Table zu CPS_PRES, Result list for stock information for MSA, and more. Our adversarial system should support the defendants' right to a fair trial, it should allow for a case to be put as strongly as possible, evidence to be challenged robustly and inconsistencies exposed. (CPS) | HVAC Controls - About Builders Exchange of Kentucky. Collaborative Problem Solving (CPS) in a Nutshell Summary provided by Clayton R. 10 Verdict Guilty Plea? Mode of trial hearing (if either way offence) Step 5. Be sure to check the box on the cover page showing you are filing the "Appellant's Opening Brief. Problems & Corruption in Child Protective Services (CPS) The articles on this page expose some deep problems that shouldn't be ignored within Child Protective Services (CPS). Services – An enhanced clinical research Investigator service We provide complete clinical trial Investigator services, from protocol input to study close out. Navigating the CPS process and agreements with CPS is difficult and tricky. State child protective services agencies are required to maintain records of the reports of suspected child abuse and neglect that they receive. Emanuel named Janice Jackson, a fast-rising middle manager at CPS, as the district's new chief education officer. Child Protective Services (CPS) is the name of a governmental agency in many states of the United States responsible for providing child protection, which includes responding to reports of child abuse or neglect. If the defendant is absent, for instance, because of a refusal to turn up, or the plea was entered by post, the magistrates have the power to proceed in his or her absence. No matter how absurd or unbelievable the CPS/DCFS social worker's claim(s) may seem, please understand that the social worker is dead serious , and most likely presumes - no… most likely BELIEVES that you are guilty as accused. 2 SECTION 9 JURY TRIALS October 2015 JURY TRIALS Both the U. Cbd Oil Merrillville Indiana Cps And Parents Using Cbd Oil Is Cbd Oil Banned By Ncaa Do Hemp Hearts Contain Cbd Oil sort results by: best selling new to store a-z z-a customer rating low to high price high to low price savings dollars savings percent. Appeals generally take between 11 and 13 months. 7 Treatment Issues in Pretrial and Diversion Settings The pretrial period of criminal justice processing is unique in that for most people it is brief and the outcome is uncertain. If it needs a face to face discussion with cps ie if the witnesses have been video interviewed or the evidence is CCTV etc then cps will need to see the officer in charge with all the material. A federal judge is a federal judicial officer, paid by the federal government to act impartially and lawfully. Indicated means that: CPS found enough evidence to support the claim that a child has been abused or neglected. Pro Se Information (Individuals Representing Themselves) Bid Protest Information. CPS Research is the premier clinical research company in the UK, providing bespoke clinical research solutions for a global client base. To start with, CPS is only authorized by law (Chapter 262 of the Texas Family Code) to remove children from their home when facts exist that would "satisfy a person of ordinary prudence and caution to believe that there is an immediate danger to the physical health or safety of the child, or the child has been a victim of neglect or sexual. The opening of the application period is shared with judges across Texas who hear CPS cases and also announced to attorneys who practice child welfare law via a joint communique. Issue queries or Data Clarification Forms (DCFs) to investigative sites and review responses for database amendments as needed. Below is a graphical representation of the court process: (Click to Enlarge) Child is removed from the home. The investigation process is designed to be thorough. How to Write a Declaration of Facts to Submit to the Court When parents appear in Juvenile Court they are handed a report written by a CPS social worker. Tacoma CPS & Family Lawyer Torrone Law will help you fight an unjust system to get your life back on track and restore your family. Why the Child Protective Services Social Worker Isn't Helping You June 29, 2015 - By Linda Martin - 228 Comments I saw these words written in an intro on the FightCPS forum today: "My caseworker seems like no help. The Crown Prosecution Service advises the police on cases for possible prosecution and reviews cases submitted by the police. During an appeal the trial court has the authority to order visitation or other contact between the child and birth family (7B-1113). CPS Investigation Process in Michigan If you're a parent and you've been accused of a drug crime, such as manufacturing and cultivating, distributing, or selling drugs, contacting a defense attorney is a wise decision. Child Protective Services isn't the enemy. Take the first step now and talk to an experienced local cps attorney. If Child Protective Services (CPS) is involved in your case, you probably have to go to the juvenile court to find out what you can do. Program Administrator In support of CFSA's teaming principles and best practice standards, the Child Protective Services administrator is. Some recently asked State of Michigan Service Specialist interview questions were, "Describe a weakness you may be presented with in this position?" and "What to do if enter home with dirty dishes, 1 year old dirty diapers, how would you handle". For biological parents it is 72 hours…For foster parents it should be the same length of time but it rarely is. A trained, impartial hearing officer acts as a judge and makes a decision about the case. 4 Fact-finding hearing or adjudicatory trial. If CP&P disagrees, CP&P has 30 calendar days from the order to report in writing to the court its disagreement with the recommendation. geotrust® ssl trial certificate subscriber agreement you must read this geotrust ssl trial certificate subscriber agreement ("agreement") before applying for, accepting, or using a geotrust ssl trial certificate (each, a "certificate"). FIRST Robotics. Also managed the organisational design, to ensure alignment with the solution design, and to support the organisation's growth and expansion objectives. Subscriber may only access or use any Services provided under this Agreement on a Trial Basis for restricted use in a non-production, test environment, and solely for the purpose of Subscriber's internal, non-. If you have a disability and need a document on this website to be provided to you in another format, please send an email to DHS. Create a new Academic Account. Child Protective Services Investigation Process The CPS process varies from state to state, and it can change depending on the allegation. NATURE OF CASE: Mother, personally and as next friend for her minor daughters, brought 1983 action against, inter alia, child protective services caseworker and deputy sheriff, alleging violations of Fourth Amendment and their familial rights under Fourteenth Amendment's Due Process Clause. In many parts of the world, trials are being replaced by legal regimes that encourage suspects to admit guilt and waive their right to a full trial. Any annoymous calls made to CPS should be reported but not investagted without Due Process. CPS co-owner says allegedly frivolous lab testing was legal Any rule that has not followed that process is known as an interpretive rule — ensuring CMS cannot evade its notice-and-comment. The law requires CPS to provide written notice to the parents or other subjects of the report concerning the rights accorded to them by the New York State Social Services Law. Further information. The Michigan Child Protection Law provides the framework for what CPS must do. The actions of CPS depend on circumstances of each family. Indicated means that: CPS found enough evidence to support the claim that a child has been abused or neglected. formal partnership with CPS before you are eligible to begin the registration process or student teach with our district. CPS will implement Individual Quality Assessments and increase the use of Disclosure Gateway Reviews which would involve the dip-sampling of processes. ca - [email protected] Lovet One of the most significant developments in the field of civil rights litigation has been the emergence of damages as a remedy for the en-forcement of constitutional guarantees. Test Drive the #1 Low-Code Application Platform | Ultimus CPS. Facts CPS Juvenile Dependency Cases Superior Court of Arizona Maricopa County Juvenile Court Durango Facility 3125 W. Child First. We combine RPA, cognitive capture, process orchestration, mobility and engagement, and analytics to ease impleme. See the complete list of Tables for Cps. In Michigan, a parent has a right to be present at a jurisdiction trial, but the court may proceed without you if you fail to appear after having been given notice of the date and time. The officer listens to both sides and decides the case. Overview of Department of Child Safety (DCS) Decision Making Process - pg. There's power in the pen. out a prosecution will make the prosecutor also aware of potentially exculpatory evidence, as defined by the caselaw, so that the prosecution may disclose it to the defense. If CP&P disagrees, CP&P has 30 calendar days from the order to report in writing to the court its disagreement with the recommendation. Advances in CPS will enable capability, adaptability, scalability, resiliency, safety, security, and usability that will far exceed the simple embedded systems of today. family does not cooperate CPS can change the ranking. Prices and more details inside. The Supreme Court of Appeals shall give priority to appeals of child abuse and/or neglect proceedings and termination of parental rights cases and shall establish and administer an accelerated schedule in each case, to include the completion of the record, briefing, oral argument, and decision. Your child may also attend the trial. Within the CPS there are specialist units who deal specifically with Rape and Sexual Offences. Will not include trial home visits or runaway episodes in calculating 15 months in foster care. The Expert Concierge™ is available to ensure a smooth on-boarding process and provide ongoing assistance for the whole team. REQUEST A RECORD. The expectation is that if QbD is built into the process from the beginning it will provide a product that is more robust and reproducible. Does my child have a right to a jury trial? The California Supreme Court has held that children do not have a right to a jury trial in a juvenile court adjudication. The judge may decide to dismiss the case based on lack of. of Children & Families) Guide to confidentiality laws applicable to CHIPS proceedings (WI Court System) What is CHIPS? (WI Court System). In the Kentucky legal system, most cases involving child abuse are called "DNA" actions, which stands for "Dependent, Neglected or Abused. Trial dates often are delayed. Appeals People convicted by a magistrates' court can appeal to the Crown Court against their conviction and the sentence. The draft CPS guidance makes explicit mention of the importance of witnesses not being coached by prosecutors. Costs and the CPS. We handle cases throughout Michigan. For updated process serving legislation, please visit the Texas Courts website. CPS Research is the premier clinical research company in the UK, providing bespoke clinical research solutions for a global client base. CPS (Collaborative Problem Solving) by Ross Green consists of three steps: 1. Washington, D. Search our free database to find email addresses and direct dials for CPS Process Solutions employees. Available for PC, iOS and Android. 147: Case review panel — Creation — Duties. On very rare occasions, the juvenile court might even dismiss a CPS petition at this early stage, though case law (decisions by Courts of Appeal or the Supreme Court) states that a court has to allow CPS its day in court and may not dismiss at the A/D hearing (even though that can be the CPS's day in court too when the petition is based upon. Who is an "officer of the court?". Learn vocabulary, terms, and more with flashcards, games, and other study tools. § 35-3-34 (a) (2) provides that the GCIC is authorized to make criminal history records of the defendant or witnesses in a criminal action available to counsel for the defendant upon receipt of a written request from the defendant's counsel. Failure to comply may be a criminal (punishable by incarceration) or civil offense (punishable by fine), depending on the type of order that was disobeyed. Often, false reports of child abuse occur when one parent files for divorce and the other parent did not want a divorce or wants sole custody of the child or children. During an appeal the trial court has the authority to order visitation or other contact between the child and birth family (7B-1113). centennialcollege. Such meeting will be held before the trial judge or a magistrate, or a judicial officer who possesses fewer judicial powers than a judge. Th ese techniques are likely to lead to negative conse-quences, such as false allegations and reduced likelihood of conviction (e. This factsheet discusses laws that require child welfare agencies to make reasonable efforts to provide services that will help families remedy the conditions that brought the child and family into the child welfare system. F360-easel. If you're facing allegations of child neglect or abuse, call us at 616-425-5556. David Knight The CPS is using a fraudulent strategy to avoid trial by jury. You probably do not know for sure who made the report - it could have been a family member, a neighbor, a teacher, a doctor, a police officer, or even a stranger. Job frustrations and other social pressures can lead to some hard times. Two brothers arrested over a fatal stabbing in #Battersea in Feb were charged with murder and due to face trial this morning at the Old Bailey CPS offered no evidence against them. The other half of patients were randomly assigned to receive the IMPACT model of depression care, also known as collaborative care. Section 300 Petition Filed; Occurs within 48 hours of the child being taken into custody. There's power in the pen. The Crown Prosecution Service (CPS) is reportedly considering whether to prosecute BT. IHS Markit Technology is now a part of Informa Tech. Acknowledging the falling charges for rape, the CPS maintain that they apply the Code for Crown Prosecutors to assess which cases should be taken to trial. Program Administrator In support of CFSA's teaming principles and best practice standards, the Child Protective Services administrator is. Judge criticises CPS for accepting plea to lesser charge. In the event that your "attorney" does say something resembling an objection or two, do not count on him doing it properly. Teams can monitor their company website for domain integrity, security, and potential malicious activity with SiteSafe Monitor™. The CPS attorney presents evidence through the testimony of the CPS caseworker and law enforcement or other witnesses. During an appeal the trial court has the authority to order visitation or other contact between the child and birth family (7B-1113). The most secure digital platform to get legally binding, electronically signed documents in just a few seconds. 1 Clinical trials serve as the foundation for evidence-based medicine by addressing specific clinical questions that may lead to improvements in current clinical practice. The Prosecution Process Figure 1. Introduction. The meeting of parties to a case conducted before trial is called a pretrial conference. So let that one go…. The final step in the process of a CPS case is Trial. When you receive the issued subpoena, make enough copies of the stamped subpoena for yourself, the witness, and all parties. The court is required to follow the same rules of evidence and procedure at every trial. Solicitor for student in rape case criticises police and CPS of a collapsed rape trial has dismissed a police and prosecution review of the case that resulted in an apology to her client as. until the trial or determine a temporary out-of-home placement, such as with a family member or with a foster family. At the time, investigators were able to get a DNA profile from semen found on Hammerberg's body, but it didn't help them identify a suspect at the time. Comfort & Process Solutions, Inc. Pretrial and Pleas Proceedings of CPS Investigation During a CPS investigation in Michigan there will be pretrial and/or plea proceedings. Advances in CPS will enable capability, adaptability, scalability, resiliency, safety, security, and usability that will far exceed the simple embedded systems of today. CPS co-owner says allegedly frivolous lab testing was legal Any rule that has not followed that process is known as an interpretive rule — ensuring CMS cannot evade its notice-and-comment. For access to Automotive, Energy & Power, and Cost Benchmarking & Teardown Analysis, please visit benchmarking. His office is in the process of hiring additional staff to handle that work. " In its most recent opinion, the judges were so frustrated by "repeated violations" of the parent's rights and other failures of the trial court, they wrote,. If the attorney for the intervener does not plead or prove-up sufficient testimony to prove substantial past contact, they have created liability for themselves. Corrigan (1921) as follows: "The due process clause requires that every man shall have the protection of his day in court, and the benefit of the general law, a law which hears before it condemns, which proceeds not arbitrarily or capriciously, but upon inquiry, and renders judgment only after trial, so that every citizen shall hold his life. family does not cooperate CPS can change the ranking. This lists the names, job, titles, departments and salaries of all full-time CPS employees. Finding yourself or a loved one at the center of a CPS investigation can be daunting and. Baltimore, pilot and full-scale trials;. How to File a Complaint (please click here for information on filing a Vaccine Petition) Complaint Cover Sheet (Form 2, RCFC) Pro Se Complaint Form. 4 Fact-finding hearing or adjudicatory trial. timely disposition consistent with the circumstances of the individual case. In addition to the laws, regulations and policies that govern the general conduct of the CPS program, there are additional requirements for CPS to. CPS faces arresting issues. Why Regulatory Affairs?The Master of Science in Regulatory Affairs for Drugs, Biologics, and Medical Devices program is designed to produce graduates who are highly qualified to manage the regulatory process for companies innovating and developing cutting edge products in science and medicine. The Michigan Child Protection Law provides the framework for what CPS must do. appeals of findings of abuse and neglect can be found in. 15 th for May exam and August 15 th for November exam. The caseworker can also get records from CPS of past problems involving the same family. The expectation is that if QbD is built into the process from the beginning it will provide a product that is more robust and reproducible. African swine fever is a highly contagious and deadly viral disease affecting both domestic and wild pigs of all ages. Contact Us For Help With CPS. In 2012 the Trial of the Facts process was used to determine if former Labour MP Margaret Moran had committed £53,000 expenses fraud. "DigiCert offers excellent interaction with the customer, and an efficient and thorough order process. Remember to request an attorney of both CPS (child protective services) and the court. The prosecution organisations listed here will be involved in the investigation of offences and/or are the prosecuting authority. Assigns specific tasks (roles and responsibilities) to other members of the CT study team including study monitor updates relevant to clinical trial strategies, strategy changes and relevant study activities and/or changes. This is done by filing a Judicial Council Notice Designating the Record on Appeal (APP-003). A special thanks goes to Barry Macha, District Attorney in Wichita Falls, Texas, who undertook the task of editing the booklet and putting it in its final form for printing. Intelius provides phone numbers, previous addresses and background checks. 2 CSO-1123A (10-14) CPS-1053A (12-12) 13. Foster Care Review Board reviews Out 3, permanency options are considered services for adolescents to 12a. Let's imagine Mr. [email protected] Start writing. Start a free trial now to save yourself time and money!. Daily average fluxes are then derived on a 1° x1° grid in the next stage of processing. Certification is the perfect next step in your career, whether you have a college degree or not. Installation is free and the solution generates a significant monthly cash rebate. The Expert Concierge™ is available to ensure a smooth on-boarding process and provide ongoing assistance for the whole team. The second trial you are entitled to is the "Dispositional" hearing, also called the "Dispo" hearing in most counties. Ayo Berubah!. Learn how PTC is changing the game in digital transformation. If in reviewing the intake information or during the assessment process, you discover information that may affect the jurisdiction of the child abuse assessment, notify the jurisdiction that has responsibility for the assessment. charge and trial. Why the Child Protective Services Social Worker Isn't Helping You June 29, 2015 - By Linda Martin - 228 Comments I saw these words written in an intro on the FightCPS forum today: "My caseworker seems like no help. We have an extensive range of calibration equipment for the process market. (859) 294-4400 Comfort & Process Solutions, Inc. , asking what might have happened). This means you need copies of your records from every party involved in the case, including CPS. Reunification is considered achieved when both care and custody are returned to parents or guardians. Child Protective Services Investigation- What to expect and how to handle the situation, Part 2 July 25, 2017 This blog post is the second in a series of posts that detail CPS investigations : what they are, how CPS conducts them and how you can present yourself most favorably during one. And Denise Little, a longtime CPS educator, was picked to serve as Claypool's. Welcome to the Maricopa County Superior Court Protective Orders website. Find out more about guardianships in juvenile court. There is a required process, unless either of the following applies: The parent is convicted of murder or voluntary manslaughter or caused bodily harm to a child; The parent abandoned the child as an infant. Proper disclosure is a fundamental part of this. However, this study was limited by sample size (28 children in CPS, 19 in PMT) and the lack of a waitlist control group. Our Intelligent Automation software platform helps organizations transform information-intensive business processes, reduce manual work and errors, minimize costs, and improve customer engagement. • Help family through the approval process by helping them gather documents for the home study, fill out paperwork and get to appointments by transporting • Continue to work with the Department to achieve permanency for youth in CPS care. 620(f), the Superior Court of California, County of Nevada is publishing its proposed budget for fiscal year 2019-2020. The FDA requires a multi-phase clinical trials process to be completed before deciding if the medicine under investigation is safe and effective for a broader patient population. Process Development Engineer/Chemical Engineer/EngineeringAs a Process Development Engineer, job…See this and similar jobs on LinkedIn. A CPS investigation must begin within 24 hours and usually includes:. After DCFS/CPS/DSS barges in and takes your children…be they biological or foster…the court is required to give you a hearing within a reasonable length of time. The most secure digital platform to get legally binding, electronically signed documents in just a few seconds. This means they give up the right to: see, hear and question witnesses, bring their own witnesses, testify or stay silent. At Turning Technologies, we're dedicated to improving learning experiences in the classroom and workplace through interactive learning technology and real-time audience response systems. However, this study was limited by sample size (28 children in CPS, 19 in PMT) and the lack of a waitlist control group. If you're facing allegations of child neglect or abuse, call us at 616-425-5556. she broke her back and was airlifted to tje medical center. Independent Television News Limited 2011. " Imwinkelried, Exculpatory Evidence (1996) at 1. Trial Skills Training is offered once a year. Obama had initially said he would veto the bill which contains the draconian language authorizing the US military to seize and incarcerate US citizens without warrant, due process, trial, etc. The actions of CPS depend on circumstances of each family. A Study Of Adverse Symptoms In Users Of Over The Counter Pain Relief Medications Dg Health Menstrual Pain Relief. Notes from Gary Davis's Creativity is Forever - 1998 Kendall Hunt. Diversity Club. The Supreme Court is in Austin, immediately northwest of the state Capitol. In 1871, Congress created a. Here is everything you need to know about each and every step in the process. Only a portion of child maltreatment is reported to CPS, and not all referrals are investigated. 00 Craigslist is another way to advertise your daycare. All citizens in the community are required to report unsafe activity by calling 1-800-252-5400. The State Law Library supports the legal research and information needs of Montana's judges and court staff, attorneys and legal assistants, state employees, students and members of the public. request amendment of the record when the investigation into the complaint has. CPS's counsel (county counsel) will amend a Petition when parents are barking too loud and have solid evidence or witnesses to prove their innocence but, they will never dismiss it. CPS has 60 days after receiving the report to determine whether the report is "indicated" or "unfounded". Cancer will neither thrive nor take birth under such illnesses. We have them as demo units at CPS, ask to trial one out. " This week, Saunders was severely criticised by the Commons justice select committee after a string of rape trials collapsed. The CPS certification audit is administered directly by the CDSA and consists of over 300 distinct controls that help secure and manage physical datacenters, harden services, and protect storage facilities. The CPS worker provides evidence supporting the state's case, and you may do the same in support of your own case. Entering the Dependency Court System. Your attorney experienced in CPS matters and the court will consider matters that would facilitate a fair trial at a conference. CPS Abuses in Fairfax have increased with over the past few years. Convictions and the CPS• CPS rarely drops evidentially weakcases, research confirms weak cases continue tobe prosecuted, public interest cases• Violations of due process, Police cautioning• CPS as decision reverser, not a decisionmaker, "Prosecution momentum". 460 days ago. Middle School SO. Failure to Comply Law and Legal Definition. Prices and more details inside. And when the harm is serious, the government can physically remove the children from parental custody. A judicial examination and determination of facts and legal issues arising between parties to a civil or criminal action. D&B Hoovers provides sales leads and sales intelligence data on over 120 million companies like CPS SRL A CAPITALE RIDOTTO and Retail Sector contacts in LAMEZIA TERME, CATANZARO, Italy and around the world. The most secure digital platform to get legally binding, electronically signed documents in just a few seconds. The FDA requires a multi-phase clinical trials process to be completed before deciding if the medicine under investigation is safe and effective for a broader patient population. Mesa, AZ, 85210-6234 I f a child has been found to be dependent, it is the responsibility of the Arizona Department of Economic Security Child. This means they give up the right to: see, hear and question witnesses, bring their own witnesses, testify or stay silent. Decision-making process regarding the continued use or closure of the foster or trial adoptive home. Has the law of criminal evidence managed to strike an appropriate balance between protecting the victims of sexual assault, who are usually the prime witnesses in prosecution proceedings, and protecting the due process rights […]. solving process so that students can continue to progress through the task. Though, as recommended by the Creative Education Foundation, Stormz has chosen to focus on an evolution of the Osborn-Parnes CPS called the CPS Learner's Model. " If the appellate court grants the emergency orders the family could get immediate relief. The information that was on this website is now available on one of our new, topic focussed websites. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Cps Stance On Cbd Oil In Texas Pure Natural CBD Oil for Pain | What Is Cbd Oil Ny Times Buy Cbd Oil C4 Healthlabs Cbd Oil For Dogs 100mg. Great that you guys have interviews etc. There's power in the pen. IEEE membership offers access to technical innovation, cutting-edge information, networking opportunities, and exclusive member benefits. A federal judge is a federal judicial officer, paid by the federal government to act impartially and lawfully. The judge may decide to dismiss the case based on lack of. Cases are allocated a court date through a process called 'listing', which is a judicial function. Notes from Gary Davis's Creativity is Forever - 1998 Kendall Hunt. The court is required to follow the same rules of evidence and procedure at every trial. of Children & Families) Guide to confidentiality laws applicable to CHIPS proceedings (WI Court System) What is CHIPS? (WI Court System). For access to Automotive, Energy & Power, and Cost Benchmarking & Teardown Analysis, please visit benchmarking. CPS cannot remove a child from the home without a court order. Following is a general outline of the most commonly used sections in the Trial Notebook for a trial before the Court of a "typical" divorce, and the step-by-step process used in the authors' law firm to put it together: A. series you will be impressed. Typically, CPS investigates reports of abuse and neglect. How Does CPS Conduct its Investigations?. —Trial by jury in civil trials, unlike the case in criminal trials, has not been deemed essential to due process, and the Fourteenth Amendment has not been held to restrain the states in retaining or abolishing civil juries. If CP&P disagrees, CP&P has 30 calendar days from the order to report in writing to the court its disagreement with the recommendation. Child Abuse or Neglect Jurisdiction Trial. answer victim's questions about process. Who is an "officer of the court?". In many parts of the world, trials are being replaced by legal regimes that encourage suspects to admit guilt and waive their right to a full trial. 1 or Flash Builder for PHP 4. — A guardian ad litem should adhere to the Guidelines for Children's Guardians Ad Litem in Child Abuse and Neglect Proceedings set forth in Appendix A of these Rules and submit a written report to the court and provide a copy to all parties at least five (5) days prior to the disposition hearing that complies with the requirements set forth. The caseworker usually visits the child and family next. There are several things that we, the patriotic, self-sufficient defenders of liberty can do to counter this effort. The Nevada County Superior Court is looking for a Business Analyst and Court Services Assistant. Brady and its progeny have also firmly stated that the "good faith" or "bad faith" of a prosecutor in failing to disclose exculpatory evidence does not matter. When the LDSS learns that a criminal process has been initiated in either juvenile or circuit court, the LDSS must notify the appellant in writing that the CPS administrative appeal process is stayed and that his right to access his CPS record is suspended until the criminal process is completed in trial courtthe and the judge enters a final. This means you need copies of your records from every party involved in the case, including CPS. This process is recorded but for some reason not counted by anyone. 2 CSO-1123A (10-14) CPS-1053A (12-12) 13. Pretrial and Pleas Proceedings of CPS Investigation During a CPS investigation in Michigan there will be pretrial and/or plea proceedings. If you know of others who need this accomodation, please let them know it is available. All it takes is an unethical Social worker who fabricates Child Abuse charges usually against Fathers in Divorce and Child Custody situations. CPS Investigation Process 101 Upon receiving a report, CPS conducts an investigation, within a specified time frame (typically within 24 or 48 hours or up to 5 days, depending on the State). Trial dates often are delayed. If the defendant is absent, for instance, because of a refusal to turn up, or the plea was entered by post, the magistrates have the power to proceed in his or her absence. It is to be distinguished from malicious prosecution, another type of tort that involves misuse of the public right of access to the courts. CPS Instruments produces our in the USA and ships these around world. This Certification Practice Statement ("CPS") describes the Unipass Service provided by Origo Secure Internet Services Limited ("OSIS") to members of the Unipass Community. CPS has 60 days after receiving the report to determine whether the report is "indicated" or "unfounded". CPS and parents agree to settle most dependencies. CPS's counsel (county counsel) will amend a Petition when parents are barking too loud and have solid evidence or witnesses to prove their innocence but, they will never dismiss it. DHS/CPS can take your children without due process in their jurisdictional trials in secret courts where DHS/CPS actions,. Check case status sheet and calendar more frequent updatings right up to trial. Child Protective Services (CPS) is a part of the Texas Department of Family and Protective Services (DFPS), a State Agency set up by law to make sure children are safe and to help families create a safe environment for their children. Juvenile dependency court can have a significant impact on the lives and futures of children and families. Examples: disputes between a landlord and tenant. " This is called the "presumption of fitness" and is in the Criminal Code. Complex process flows,Business driven schedules,Workload routing and load balancing, Monitoring and managing present and future workload Download the Document. The process in CPS court ("juvenile dependency") is rapid and structured. BROWARD SHERIFF'S OFFICE. Original text.
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{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 259
|
import rospy
from geometry_msgs.msg import Twist, Quaternion
from nav_msgs.msg import Odometry
import tf
from math import radians, copysign
import PyKDL
from math import pi
def quat_to_angle(quat):
rot = PyKDL.Rotation.Quaternion(quat.x, quat.y, quat.z, quat.w)
return rot.GetRPY()[2]
def normalize_angle(angle):
res = angle
while res > pi:
res -= 2.0 * pi
while res < -pi:
res += 2.0 * pi
return res
class CalibrateAngular():
def __init__(self):
# Give the node a name
rospy.init_node('calibrate_angular', anonymous=False)
# Set rospy to execute a shutdown function when terminating the script
rospy.on_shutdown(self.shutdown)
# How fast will we check the odometry values?
self.rate = rospy.get_param('~rate', 20)
r = rospy.Rate(self.rate)
# The test angle is 360 degrees
self.test_angle = radians(rospy.get_param('~test_angle', 360.0))
self.speed = rospy.get_param('~speed', 0.5) # radians per second
self.tolerance = radians(rospy.get_param('tolerance', 1)) # degrees converted to radians
self.odom_angular_scale_correction = rospy.get_param('~odom_angular_scale_correction', 1.0)
self.start_test = rospy.get_param('~start_test', True)
# Publisher to control the robot's speed
self.cmd_vel = rospy.Publisher('/cmd_vel', Twist, queue_size=5)
# The base frame is usually base_link or base_footprint
self.base_frame = rospy.get_param('~base_frame', '/base_footprint')
# The odom frame is usually just /odom
self.odom_frame = rospy.get_param('~odom_frame', '/odom')
# Initialize the tf listener
self.tf_listener = tf.TransformListener()
# Give tf some time to fill its buffer
rospy.sleep(2)
# Make sure we see the odom and base frames
self.tf_listener.waitForTransform(self.odom_frame, self.base_frame, rospy.Time(), rospy.Duration(60.0))
rospy.loginfo("Bring up rqt_reconfigure to control the test.")
reverse = 1
while not rospy.is_shutdown():
if self.start_test:
# Get the current rotation angle from tf
self.odom_angle = self.get_odom_angle()
last_angle = self.odom_angle
turn_angle = 0
self.test_angle *= reverse
error = self.test_angle - turn_angle
# Alternate directions between tests
reverse = -reverse
while abs(error) > self.tolerance and self.start_test:
if rospy.is_shutdown():
return
# Rotate the robot to reduce the error
move_cmd = Twist()
move_cmd.angular.z = copysign(self.speed, error)
self.cmd_vel.publish(move_cmd)
r.sleep()
# Get the current rotation angle from tf
self.odom_angle = self.get_odom_angle()
# Compute how far we have gone since the last measurement
delta_angle = self.odom_angular_scale_correction * normalize_angle(self.odom_angle - last_angle)
# Add to our total angle so far
turn_angle += delta_angle
# Compute the new error
error = self.test_angle - turn_angle
# Store the current angle for the next comparison
last_angle = self.odom_angle
# Stop the robot
self.cmd_vel.publish(Twist())
# Update the status flag
self.start_test = False
params = {'start_test': False}
rospy.sleep(0.5)
# Stop the robot
self.cmd_vel.publish(Twist())
def get_odom_angle(self):
# Get the current transform between the odom and base frames
try:
(trans, rot) = self.tf_listener.lookupTransform(self.odom_frame, self.base_frame, rospy.Time(0))
except (tf.Exception, tf.ConnectivityException, tf.LookupException):
rospy.loginfo("TF Exception")
return
# Convert the rotation from a quaternion to an Euler angle
return quat_to_angle(Quaternion(*rot))
def shutdown(self):
# Always stop the robot when shutting down the node
rospy.loginfo("Stopping the robot...")
self.cmd_vel.publish(Twist())
rospy.sleep(1)
if __name__ == '__main__':
try:
CalibrateAngular()
except:
rospy.loginfo("Calibration terminated.")
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 367
|
\section{Introduction}
\IEEEPARstart{T}{his} demo file is intended to serve as a ``starter file''
for IEEE journal papers produced under \LaTeX\ using
IEEEtran.cls version 1.7 and later.
I wish you the best of success.
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\subsection{Subsection Heading Here}
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\subsubsection{Subsubsection Heading Here}
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\section{Conclusion}
The conclusion goes here.
\appendices
\section{Proof of the First Zonklar Equation}
Appendix one text goes here.
\section{}
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\section*{Acknowledgment}
The authors would like to thank...
\ifCLASSOPTIONcaptionsoff
\newpage
\fi
\section{Introduction}
\label{sec:Introduction}
The classical results from \cite{Shannon_IT_1956} show that feedback does not increase the capacity of discrete memoryless channels. However, Polyanskiy et al.\cite{Polyanskiy_IT_2011_NonAsym} and Chen et al. \cite{Chen_2011_ICC} show that capacity can be approached in a smaller number of channel uses using feedback. Polyanskiy et al. \cite{Polyanskiy_IT_2011_NonAsym} introduce random-coding lower bounds for variable-length feedback coding with termination (VLFT) and without termination (VLF), which approach capacity with average blocklengths of hundreds of bits. A communication system without feedback, on the other hand, requires thousands of bits to closely approach capacity \cite{Polyanskiy_CCR_2010}. This paper demonstrates practical systems using non-binary low-density parity-check (NB-LDPC) codes that match or exceed the lower bounds of \cite{Polyanskiy_IT_2011_NonAsym}. Most of the analysis in this paper is not exclusive for NB-LDPC codes, but NB-LDPC codes are used for demonstration because they perform well in the short-blocklength regime (150 to 600 bits) that is of interest.
In VLFT analysis of \cite{Polyanskiy_IT_2011_NonAsym}, the receiver provides full noiseless feedback to the transmitter. The transmitter sends additional incremental bits until it knows the receiver has decoded the message correctly, resulting in zero probability of error. The ``T" in VLFT stands for termination and corresponds to a noiseless transmitter confirmation (NTC) bit that the transmitter uses to terminate the transmission. The NTC is transmitted through a channel different from the main communication channel. In contrast, VLF (without the ``T") does not have the advantage of an NTC. All VLF forward transmissions go over the same noisy channel. Thus, there is always a nonzero probability of undetected error in VLF.
VLF and VLFT are examples of hybrid automatic repeat request (HARQ) schemes.
Prior to Polyanskiy et al. \cite{Polyanskiy_IT_2011_NonAsym} and Chen et al. \cite{Chen_2011_ICC}, HARQ feedback schemes had been studied in great detail in many papers including for example \cite{Shu_Lin_Yu1,Costello_HARQ_IT_1998,Visotsky_RBIR_TCOM_2005, Lott_Soljanin_ITW_2007,Fricke_Reliability_HARQ_TCOM_2009,EminaAllerton2014}.
These papers provide an overview of HARQ, discuss how error correcting codes can be combined with ARQ and demonstrate applications of HARQ. In particular, \cite{EminaAllerton2014} shows that hybrid ARQ is especially useful in point-to-point scenarios. The coding schemes that are most commonly explored in HARQ systems \cite{Roongta_HARQ_ICC_2003,Roongta_HARQ_WCNC_2004,BCJR_TransIT_1974} are based on convolutional codes (CCs) or a concatenation of turbo and block parity-check codes, where the Bahl-Cocke-Jelinek-Raviv (BCJR) algorithm is used to determine which bit is unreliable and needs to be transmitted in the subsequent transmissions. These works use a genie (equivalent to NTC in VLFT) to terminate transmissions.
In order to remove the genie and realize a more practical system (equivalent to VLF) \cite{Raghavan_ROVA_TransIT_1998,Fricke_HARQ_NAW_2006,Fricke_Approx_ROVA_VTC_2007,Fricke_Reliability_HARQ_TCOM_2009, Visotsky_RBIR_TCOM_2005,Adam_TB-ROVA} consider reliability-based HARQ using convolutional codes where the transmission terminates when the probability of having a correctly decoded message is high enough. For example, in \cite{Fricke_Reliability_HARQ_TCOM_2009} the reliability metric is based on the average magnitude of the log-likelihood ratios of the source symbols.
In \cite{Soljanin_LDPC_HARQ_2005,Soljanin_LDPC_HARQ_ITW_2006}, Soljanin et al. study VLFT HARQ using rate-compatible binary LDPC codes. They use maximum likelihood (ML) decoding analysis to determine the size of incremental transmissions in case of decoding failure. In \cite{Soljanin_LDPC_ITW_2009,Soljanin_LDPC_IT_2012} Soljanin et al. extend their analysis to time-varying binary erasure channels.
Some other high-throughput ARQ schemes use rateless spinal codes as in \cite{Perry_Spinal_Hotnets_2011,Perry_Spinal_Sigcomm_2012}, where hash functions are used for the subsequent coded symbols. In \cite{Romero_MS_Thesis_2014}, Romero uses cyclic redundancy check (CRC) codes to study the performance of spinal codes in VLF setting. Use of polar codes with HARQ is also studied in \cite{Chen_Polar_HARQ_TCOM_2013, Chen_Polar_HARQ_WCNC_2014}. These works present polar-code-based HARQ schemes over binary-input additive white Gaussian noise (BI-AWGN) and Rayleigh fading channels using Chase combining.
The closest work to the analysis presented here is by Pfletschinger et al. in \cite{Pfletschinger_LDPC_HARQ_TWC_2014} which uses rate-adaptive, non-binary LDPC codes in a HARQ scheme over Rayleigh fading channel in the VLFT setting. They present two algorithms that use channel statistics and mutual information to optimize the blocklengths for each transmission to maximize the throughput. Based on channel state information at transmitter, the code rates, modulations, and maximum number of retransmissions are all optimized prior to initial transmission.
Chen et al.~\cite{Chen_2011_ICC, Chen_Feedback_Journal_2013} and Williamson et al. \cite{Williamson_ISIT_2012} analyzed a VLFT
scheme based on rate-compatible sphere-packing with an ML
decoder (RCSP-ML) and simulated a VLFT scheme using convolutional codes.
The approximation based on RCSP-ML extends sphere-packing analysis from a single fixed-length code to a family of rate-compatible codes, where each code in the family achieves perfect packing and is decoded by an
ML decoder.
For the 2-dB BI-AWGN channel
with feedback, the convolutional codes achieve about 95\% of the idealized RCSP-ML throughput ($R_{RCSP}$) for average blocklengths up to 50 bits. In \cite{Adam_Tcom_New}, Williamson et al. also analyzed VLF systems for similar blocklengths of up to 100 bits.
However, for average blocklengths of 100 bits and larger,
the throughput of the convolutional code decreases because the frame-error rate performance of the convolutional code degrades as the length of code increases.
As Chen et al. mention in~\cite{Chen_Feedback_Journal_2013}, coding schemes with throughput performance close to
RCSP-ML in VLFT still remain to be identified
for expected latencies (average blocklengths) of 200 to 600 bits. This blocklength regime is important because it is still short enough that feedback provides a real advantage but also long enough that the system can be practical.
The primary purposes of this paper are to show how to optimize the lengths of incremental transmissions and to demonstrate that
NB-LDPC codes with optimized incremental transmissions
can achieve throughputs close to theoretical limits for expected latencies of 150 to 500 bits in the VLFT and VLF settings. Most of the following analysis is applicable to any coding scheme, but we use NB-LDPC codes to demonstrate the possible performance motivated by \cite{6404676}, which shows that NB-LDPC codes without feedback, perform well in this short-blocklength regime.
In our precursor conference papers \cite{Kasraisit2014,Kasraitw2014} we preliminarily analyzed the performance of NB-LDPC codes in VLFT for a BI-AWGN channel with an SNR of 2 dB with an unlimited number of transmissions and with the number of transmissions $m$ fixed to be five. We also considered two-phase VLF system with $m=5$. In VLFT, the non-binary LDPC codes of \cite{Kasraisit2014} attain 91\% to 93\% of the predicted RCSP-ML throughput for average blocklengths
of 150 to 450 bits. In a VLF scheme of \cite{Kasraitw2014} incorporating a confirmation phase after each communication phase (hence called ``two-phase"), 92\% of capacity is achieved in less than 500 bits with a maximum of five transmissions.
In this paper, we extend the results of the previous papers to consider a broader range of $m$, the number of possible transmissions. We also introduce a new VLF system that uses a stopping criterion that incorporates a cyclic redundancy check (CRC). This new system achieves better throughput performance than the schemes of \cite{Kasraisit2014,Kasraitw2014} for the example BI-AWGN channel with an SNR of 2 dB in the blocklength regime of 150 to 600 bits. For this channel, the CRC-based VLF scheme achieves about 94\% of the capacity with an unlimited number $m$ of transmissions and about 92\% of the capacity with $m=10$.
We also extend these results to a higher-SNR (8 dB) channel and use a larger 16 quadrature amplitude modulation (QAM) constellation. The capacity of the 8 dB 16-QAM AWGN channel is 2.68 bits per symbol. The VLF-with-CRC system with an unlimited number of transmissions achieves a throughput of 2.37 bits per symbol with a frame error probability of less than $10^{-3}$. This throughput corresponds to 88\% of the capacity in the blocklength regime of about 40 16-QAM symbols. Furthermore, we extend the results to a SNR-5dB BI-AWGN fading channel with the channel state information (CSI) available at the receiver. The capacity of this channel is 0.67 bits. The VLF-with-CRC system with an unlimited number of transmissions achieves a throughput of 0.60 with a frame error probability of less than $10^{-3}$. This throughput corresponds to 90\% of the capacity in the blocklength regime of about 140 bits.
The rest of the paper proceeds as follows: Section~\ref{sec:VLFTNBLDPC} provides an overview of the VLFT system with NB-LDPC codes and the reciprocal-Gaussian approximation for the probability mass function of the cumulative blocklengths.
Section~\ref{sec:VLFT with limited number of transmissions}
presents the sequential differential optimization algorithm (SDO) for optimizing the size of each incremental transmission in VLFT. Section~\ref{sec:VLFCRC} presents a VLF system with CRC and analyzes this system with an unlimited number of transmissions. Section~\ref{sec:VLFCRC_limited} extends the results of Section~\ref{sec:VLFCRC} to the system with a limited number of transmissions. Section \ref{sec:VLFoptimization} gives an overview of the two-phase VLF scheme and uses SDO to optimize the cumulative blocklength at each decoding attempt. Section \ref{sec:results} compares the throughput and the expected latency of NB-LDPC and convolutional codes in VLFT and VLF settings. Section \ref{sec:conclusion} concludes the paper.
\section{VLFT with Non-Binary LDPC Codes}
\label{sec:VLFTNBLDPC}
Feedback can facilitate capacity-approaching performance at significantly
shorter average blocklengths than systems without feedback. This improvement is made possible by capitalizing on favorable
noise realizations to decode early. In case of a bad channel realization, the communication rate is lowered by transmitting additional information until the attempted rate matches the instantaneous rate the channel supports.
In this paper, building on our precursor conference papers \cite{Kasraisit2014,Kasraitw2014}, we use high-rate protograph-based NB-LDPC codes for the initial transmission. See \cite{6404676} for a discussion of protograph-based LDPC
design. These short-blocklength codes are irregular, having mostly degree-2 and a few degree-1 variable nodes. Refer to \cite{Kasraisit2014} for more discussion on the specification of the codes.
For most of the analysis, the operating SNR in this paper is 2 dB, similar to the work of \cite{Chen_Feedback_Journal_2013,Kasraisit2014,Kasraitw2014}. However, to emphasize the generality of the approach in this paper, Section \ref{sec:Generality} shows results for higher-SNR AWGN and fading channels.
It is necessary that the initial transmission
has a rate higher than the capacity to take advantage of good channel realizations.
The coding rate is lowered until decoding is successful. For
example for SNR-2dB BI-AWGN channel, the initial code can have
a rate of 0.75 to 0.8 while the capacity of the channel is 0.685.
We will consider feedback systems that transmit incremental redundancy one bit at a time and also systems that transmit incremental redundancy in multiple-bit increments. For systems that use multiple-bit increments, a practical system may limit the maximum number $m$ of increments. In the context of a specified $m$, this paper optimizes the lengths of the $m$ possible increments to maximize throughput.
Section \ref{sec:Creating_a_bit} provides a detailed description of how we generate each bit of incremental redundancy for the NB-LDPC codes that we use. Then, Section \ref{sec:incremental} shows that in the context of this incremental redundancy, the coding rate that first produces successful decoding is closely approximated by a normal distribution.
Knowing a distribution that describes the coding rate of the first successful decoding facilitates optimization of the lengths of multiple-bit increments, as described in Section \ref{sec:VLFT with limited number of transmissions}.
\subsection{Creating a bit for incremental transmission}
\label{sec:Creating_a_bit}
In \cite{Kasraisit2014}, Vakilinia et al. use NB-LDPC codes in a VLFT system with 1-bit increments.
After the initial transmission, the transmitter sends one bit at a time until the decoder decodes correctly.
Traditionally, rate-compatible codes are designed by starting
with a low-rate mother code and increasing the rate by
puncturing the code. The proposed NB-LDPC coding scheme in \cite{Kasraisit2014}
does not explicitly involve puncturing. Rather, the design starts with
a short, high-rate NB-LDPC code for which all symbols
are transmitted in the initial transmission. Each subsequent
transmission is a single bit carefully selected to help the
decoder as much as possible given its current decoding state.
The rate is gradually lowered by sending these additional bits,
each of which is a function of selected bits in the binary
representation of the non-binary symbols.
A rate-$\frac{K}{N}$ NB-LDPC code over $GF(2^m)$ used in a binary communication link encodes an information sequence of size $Km$ bits into a sequence of size $Nm$ bits.
\begin{comment}
can be represented by a full-rank parity-check matrix $H$ of size $(N-K,N)$. This parity-check matrix is the null space of a $K \times N$ generator matrix $G$, which The elements of the matrix are from $GF(2^m)$.
\end{comment}
In order to use an NB-LDPC code with the primitive element $\alpha$ over binary-input channels, each $GF(2^m)=\{0,\alpha^0,\alpha^1,...,\alpha^{(2^m-2)}\}$ symbol is converted to $m$ bits.
\begin{comment}
as follows: There is a primitive element $\alpha$ associated with each Galois field. Each non-zero element can be represented as a power of $\alpha$ so that $GF(2^m)=\{0,\alpha^0,\alpha^1,...,\alpha^{(2^m-2)}\}$.
There is also at least one degree-$m$ primitive polynomial $\rho(x)$ associated with $GF(2^m)$, satisfying
$\rho(\alpha)=0$.
The primitive polynomial associated with $GF(2^m)$ allows the larger powers of $\alpha$ to be derived as polynomials of $\alpha$ with degrees of at most $m-1$, using the property $\alpha^{(2^m-1)}=1$.
These limited-degree polynomials yield an m-bit binary representation for each $GF(2^m)$ element.A primitive polynomial for $GF(2^3)$ is $\rho(x)=x^3+x+1$ so that $\alpha^3+\alpha+1=0$, which implies that $\alpha^3=\alpha+1 $.
\end{comment}
For example, consider $GF(2^3)$ with the primitive element of $\alpha$. Table \ref{tbl:binary_rep} shows how each element of $GF(2^3)$ can be uniquely represented in 3 bits $(g_3,g_2,g_1)$.
\begin{table}[h]
\begin{center}
\caption{Binary representation of $GF(8)$ elements }
\begin{tabular}{ |@{{~}~}c@{~}|@{~}c@{~}|@{~}c@{~}|@{~}c@{~}|@{~}c@{~}|@{~}c@{~}|@{~}c@{~}|@{~}c@{~}|@{~}c@{~}|}
\hline $\alpha^i$&0&1&$\alpha$&$\alpha^2$&$\alpha^3$&$\alpha^4$&$\alpha^5$&$\alpha^6$\\
\hline Poly.&0&1&$\alpha$&$\alpha^2$&$\alpha$$+$$1$&$\alpha^2$$+$$\alpha$&$\alpha^2$$+$$\alpha$$+$$1$&$\alpha^2$$+$$1$\\
\hline $g_3$$ g_2$$ g_1$&$000$&001&010&100&011&110&111&101\\
\hline
\end{tabular}
\label{tbl:binary_rep}
\end{center}
\end{table}
\begin{comment}
A valid codeword result in a zero vector when it is multiplied by this parity check matrix. \\
\begin{equation}
Codeword = \{x \in GF(2^m)^N | Hx = 0 \} \\
\end{equation}
\end{comment}
The rate-$\frac{K}{N}$ non-binary LDPC codes proposed here initially encode a sequence of $Km$ bits ($K$ $GF(2^m)$ symbols) into a codeword of length $Nm$ bits. Through incremental redundancy, the rate is lowered from $\frac{Km}{Nm}$ to $\frac{Km}{Nm+b}$ where $b$ is number of additional incremental bits. Each additional bit is created by an XOR $(\oplus)$ combination (summation in $GF(2)$) of bits in the binary representation of one $GF(2^m)$ symbol. For each variable node, the receiver computes the reliability of each of the $2^m$$-$$1$ possible combinations of the bits in the binary representation is computed. For example, in $GF(2^3)$ the reliabilities of the seven possible combinations $g_1,g_2,g_3,g_1 \oplus g_2,g_2 \oplus g_3,g_1 \oplus g_3, \text{and~} g_1 \oplus g_2\oplus g_3$ are computed for each variable node. Finally, the single combination bit that has the least reliability (e.g. considering all seven combinations for all variable nodes and choosing the least-reliable combination for a single variable node) is requested from the transmitter.
This is a form of active feedback in which relatively extensive feedback tells the transmitter \emph{what} to transmit in contrast to non-active feedback in which a single bit of feedback indicates {\emph{whether}} to transmit. This is a generalization of the ideas of active hypothesis testing \cite{JavidiPaper}. In \cite{Kasraisit2014} Vakilinia et al. compared the performance of a non-active feedback system and the active feedback system discussed earlier for NB-LDPC codes and showed significantly better performance with the active feedback system. The active feedback used in \cite{Kasraisit2014} tells the transmitter which bit combination to be transmitted next. This active feedback scheme does not require the receiver to transmit back the entire message, contrary to the analysis of \cite{Polyanskiy_IT_2011_NonAsym}. In the non-active feedback scheme of \cite{Kasraisit2014} the additional bits are selected at random.
This paper considers both active and non-active feedback. The non-active feedback in this paper corresponds to sending the XOR of all bits representing one of the variable nodes of the original rate-$k/N_0$ NB-LDPC code. This predetermined non-active feedback system performs close to the system with active feedback since the active feedback of \cite{Kasraisit2014} usually asks for the XOR of all bits for the subsequent transmissions. The figures and results in this paper indicate whether active or non-active feedback scheme was used to generate them.
The input frame consisting of $K$ $GF(2^m)$ information symbols is initially encoded by the rate-$\frac{K}{N}$ NB-LDPC encoder into a sequence of length $N$ $GF(2^m)$ symbols. These $GF(2^m)$ symbols are converted using their binary representations to bits. The $Nm$ bits are modulated using binary phase shift keying (BPSK) and transmitted over an AWGN channel. The additive noise is modeled as an independent, zero-mean Gaussian random sequence with variance $\sigma^2$.
As in~\cite{Chen_Feedback_Journal_2013}, SNR is calculated as $\frac{1}{\sigma^2}$, the ratio of the transmission power to the noise variance.
\subsection{Gaussian and reciprocal-Gaussian Approximations}
\label{sec:incremental}
\begin{figure}[t]
{\includegraphics[width=0.49\textwidth]{ArXiv_Plots/NCBL102.pdf}
\caption{Empirical probability mass function (p.m.f.) corresponding to the blocklength required for successful decoding for the first time in VLFT using $GF(256)$ NB-LDPC code over SNR-2dB AWGN channel. Also shown is the reciprocal-Gaussian approximation of \eqref{eqn:PDFNS} with $\mu_S=0.6374$ and $\sigma_S = 0.0579$. Smallest blocklength is $N_0=120$ bits with $k=96$ information bits so that the initial rate is $R_0=\frac{k}{N_0}=0.8$. }
\label{fig:NCBL}}
\vspace{0.1in}
{\includegraphics[width=0.49\textwidth]{ArXiv_Plots/NCRR2.pdf}
\caption{Empirical p.m.f. corresponding to $R_S=\frac{k}{N_S}$ computed from Fig.~\ref{fig:NCBL} and Gaussian approximation of \eqref{eqn:distribution_RS} with $\mu_S=0.6374$ and $\sigma_S = 0.0579$. }\label{fig:NCR}}
{\includegraphics[width=0.49\textwidth]{ArXiv_Plots/RS.pdf}
\caption{Empirical c.c.d.f. and the approximation on the tail of a normal distribution (Q-function) corresponding to the shaded area of Fig.~\ref{fig:NCR}.}\label{fig:NCR_ccdf}}
\end{figure}
Consider a stream of incremental redundancy as described in Section \ref{sec:Creating_a_bit} arriving one bit at a time at the receiver (after an initial transmission of a high-rate NB-LDPC code). We are interested in the statistical behavior of the random variable describing the blocklength of the first successful decoding and the corresponding random variable describing the coding rate of that first successful decoding.
For the system of \cite{Kasraisit2014}, the ``VLFT simulation active" plot in Fig.~\ref{fig:NCBL} shows the empirical p.m.f. of the blocklength of first successful decoding. The total blocklength $N_S$ includes the initial block and all incremental transmissions, (with active feedback) required for receiver to decode the NB-LDPC codeword correctly for the first time. The ``VLFT simulation active" plot in Fig.~\ref{fig:NCR} shows the empirical p.m.f. of the instantaneous rate $\left(\! R_S=\frac{k}{N_S} \!\right)$ at which decoding is successful for the first time. Fig.~\ref{fig:NCR} shows that $R_S$ is well-approximated by a normal distribution
\begin{IEEEeqnarray}{CCC}
f_{R_S}(r)=\frac{1}{\sqrt{2\pi\sigma_S^2}}e^{-\frac{\left(\!r-\mu_S \!\right)^2}{2\sigma^2_S}}
\label{eqn:distribution_RS}
\end{IEEEeqnarray}
with mean $\mu_S=\text{E}(R_S)$ and variance $\sigma^2_S=\text{Var}(R_S)$. The intuition behind these approximations is consistent with the ``normal approximation'' of the accumulated information density due to the law of large numbers (LLN) in \cite{Polyanskiy_CCR_2010}.
To maximize throughput, the initial code-rate of the NB-LDPC code is chosen so that almost no codeword is successfully decoded in the initial transmission. Thus, the empirical probability mass function (p.m.f.) of the number of additional increments required to decode correctly does not have a spike at zero.
Fig.~\ref{fig:NCR_ccdf} shows the complementary cumulative distribution function (c.c.d.f.) for the distribution of $R_S$ and the Gaussian approximation of Fig.~\ref{fig:NCR}. Fig.~\ref{fig:NCR_ccdf} confirms that the distribution of $R_S$ is well approximated by a Gaussian distribution. As discussed later, the empirical c.c.d.f is used to show that the Gaussian approximation is valid for a variety of AWGN channels including the high SNR ones using larger constellations and also for fading channels. The ``VLFT simulation active" plot in Fig.~\ref{fig:NCR_ccdf} shows the empirical c.c.d.f. of the instantaneous rate $\left(\! R_S=\frac{k}{N_S} \!\right)$ at which decoding is successful for SNR-2dB BI-AWGN of \cite{Kasraitw2014}. This c.c.d.f. plot shows the cumulative probability that the channel supports a rate higher than the rate on the $x$ axis. This higher rate means that the decoding has been successful with a lower number of transmitted bits. The c.c.d.f. plot corresponds to the shaded area of Fig.~\ref{fig:NCR}. The ``Gaussian Approximation" plot of Fig.~\ref{fig:NCR_ccdf} corresponds to the tail probability of the standard normal distribution of Fig.~\ref{fig:NCR}.
The parameters $\mu_S$ and $\sigma^2_S$ in \eqref{eqn:distribution_RS} for a particular code need to be determined through simulation and curve fitting. Having the p.m.f. of the $N_S$, the curve fitting process involves calculating the p.m.f. and c.c.d.f. of $R_S$ and solving a linear regression problem to obtain $\mu_S$ and $\sigma_S$.
Note that $\mu_S$ is \textit{not} the expected throughput but rather the average of the instantaneous rates supported by the channel.
The cumulative distribution function (c.d.f.) of $N_S$ is $F_{N_S}(n)=P(N_S \leq n)$, and we have
\begin{equation}
\small
F_{N_S}(n)=P\left(\!\frac{k}{R_S}\leq n\right)\! = P\left(\!R_S \geq \frac{k}{n}\right)\! = 1-F_{R_S}(\frac{k}{n}).
\end{equation}
Taking the derivative of $F_{N_S}$ using the Gaussian approximation of $F_{R_S}$ produces the following ``reciprocal-Gaussian" approximation for p.d.f. of $N_S$:
\begin{equation}
f_{N_S}(n) = \frac{k}{n^2\sqrt{2\pi\sigma_S^2}}e^{\frac{-\left(\!\frac{k}{n}-\mu_S\right)\!^2}{2\sigma_S^2}} \,. \label{eqn:PDFNS}
\end{equation}
This p.d.f as shown in Fig. \ref{fig:NCBL} closely approximates the empirical distribution of $N_S$. For $N_1<N_2$, the probability of the decoding attempt being successful at blocklength $N_2$ but not at $N_1$ using this approximation is
\begin{IEEEeqnarray}{lCl}
\int_{N_1}^{N_2}f_{N_S}(n)dn&=&\int_{N_1}^{N_2}\frac{k}{n^2\sqrt{2\pi\sigma_S^2}}e^{\frac{-\left(\!\frac{k}{n}-\mu_S \!\right)^2}{2\sigma_S^2}}dn
\label{eqn:n1n2int}
\\
\label{eqn:CDFNS}
&=&
Q\left(\!\frac{\frac{k}{N_2}-\mu_S}{\sigma_S}\right)\!-Q\left(\!\frac{\frac{k}{N_1}-\mu_S}{\sigma_S}\right)\! \, .
\end{IEEEeqnarray}
The increase in blocklength from $N_1$ to $N_2$ reduces the rate from $\frac{k}{N_1}$ to $\frac{k}{N_2}$. Note that (\ref{eqn:CDFNS}) gives the probability that the channel supports rate $\frac{k}{N_2}$ while not supporting the higher rate $\frac{k}{N_1}$.
The \textit{Q} functions in (\ref{eqn:CDFNS}) are due to the normally-distributed highest-rate-of-successful-decoding ($R_S$) at $\frac{k}{N_1}$ and $\frac{k}{N_2}$.
\subsection{General Applicability of the Normal Approximation}
\label{sec:Generality}
A similar Gaussian analysis is obtainable for other channels and different SNR values. Fig.~\ref{fig:QAM16} shows a similar complementary cumulative Gaussian approximation for the same $GF(256)$ NB-LDPC code of Fig.~\ref{fig:NCR} with an initial binary rate of 0.8 on SNR-8dB 16-QAM AWGN channel. Each non-binary element of the NB-LDPC code is mapped onto two 16-QAM symbols. Once again, the distribution of the instantaneous rate that the channel supports is well approximated by a normal distribution.
\begin{figure}[t]
{\includegraphics[width=0.49\textwidth]{ArXiv_Plots/QAM.pdf}
\caption{Empirical c.c.d.f. and the approximation on the tail of a normal distribution with $\mu_S=2.63$ and $\sigma_S = 0.19$ of the instantaneous rate $\left(\! R_S=\frac{k}{N_S} \!\right)$ at which decoding is successful for SNR-8dB 16-QAM AWGN channel.}\label{fig:QAM16}}
\end{figure}
Furthermore, Fig.~\ref{fig:WCSI} shows the complementary cumulative Gaussian approximation for the same $GF(256)$ NB-LDPC code of Fig.~\ref{fig:NCR} with an initial binary rate of 0.8 on SNR-5dB BI-AWGN fading channel with CSI knowledge at the receiver. The output of the channel, $Y=\beta X+N$ where the input $X$ is a binary phase-shift keying (BPSK) modulated signal and $N$ is the Gaussian noise with $\text{Var}(N)=\sigma^2$. The average SNR of this channel is $\frac{1}{\sigma^2}$. The coefficient $\beta$ is a Rayleigh distributed random variable satisfying $E[\beta^2]=1$. The value of $\beta$ is known at the receiver. The distribution of the instantaneous rate that the channel supports is again well approximated by a normal distribution. Since the normal distribution approximation is valid for various channels, most of the analyses in the subsequent sections of this paper are also valid for various channels with different SNR values.
\begin{figure}[t]
{\includegraphics[width=0.49\textwidth]{ArXiv_Plots/WCSI.pdf}
\caption{Empirical c.c.d.f. and the approximation on the tail of a normal distribution with $\mu_S=0.66$ and $\sigma_S = 0.05$ of the instantaneous rate $\left(\! R_S=\frac{k}{N_S} \!\right)$ at which decoding is successful for SNR-5dB AWGN fading channel.}\label{fig:WCSI}}
\end{figure}
To further discuss the generality of the Gaussian approximation on the rate that the channel supports in our feedback system, consider the accumulated information density $i(X,Y)$ at the receiver at the time of successful decoding. The expected value of $i(X,Y)$ is the capacity of the channel. For BI-AWGN channel, the $i(X,Y)$ is derived as follows:
{ \small \begin{IEEEeqnarray}{lll}
i(X,Y)&=&\log_2\frac{f_{Y|X}(y|x)}{f_{Y}(y)}
\\
&=&\log_2\frac{e^{-(y-x)^2/(2\sigma^2)}}{\frac{1}{2}(e^{-(y-1)^2/(2\sigma^2)}+e^{-(y+1)^2/(2\sigma^2)})}
\\
&=&\log_2\frac{e^{-z^2/(2\sigma^2)}}{\frac{1}{2}(e^{-z^2/(2\sigma^2)}+e^{-(z+2)^2/(2\sigma^2)})}
\\
&=&1-\log_2(1+e^{-2(z+1)/\sigma^2}).
\label{eqn:VLFEN2m2}
\end{IEEEeqnarray}}
For BI-AWGN channel, $i(X,Y)$ is a function only of the noise realization $z=y-x$ for $x=\pm 1$, and hence $i(X,Y)=i(z)$. For each transmitted bit from the NB-LDPC code over the channel, there is some amount of information density accumulated. The total amount of information density accumulation ($I$) at the receiver until the receiver decodes the message correctly
\begin{equation}
\small \label{eq:ami}
I=\sum^{N_s} _{k=1}i(z_k).
\end{equation}
The corresponding rate associated with the accumulated information density is $R_I=\frac{I}{N_s}$. As pointed out by \cite{Polyanskiy_CCR_2010}, \eqref{eq:ami} is a sum of independent random variables for which the central limit theorem will converge quickly to a normal distribution. An important consideration for our approach is whether the rate at which a practical decoder succeeds also follows a normal distribution. This hinges on the ability of a rate-compatible code family as in \cite{PBRLTCOM} to operate with a small gap from capacity over the rate range of interest.
For the previously discussed SNR-2dB BI-AWGN channel, Fig.~\ref{fig:MI} shows the c.c.d.f. of $R_I$ and the corresponding Gaussian approximation. The rate corresponding to the accumulated information density at the receiver until the decoding is successful also follows the Gaussian approximation.
\begin{figure}[t]
{\includegraphics[width=0.49\textwidth]{ArXiv_Plots/MI.pdf}
\caption{Empirical c.c.d.f. and the approximation on the tail of a normal distribution with $\mu_S=0.64$ and $\sigma_S = 0.06$ of the average accumulated information density $\left(\! R_I=\frac{I}{N_S} \!\right)$ at which decoding is successful for SNR-2dB AWGN channel.}\label{fig:MI}}
\end{figure}
Fig.~\ref{fig:MI2} shows the average accumulated information density for decoding correctly at a particular code rate for the NB-LDPC code. This figure shows on average, how much more information in number of bits the NB-LDPC code requires to decode the message correctly compared to the operating rate. The ``ideal decoder" plot in Fig.~\ref{fig:MI2} corresponds to the average accumulated information density being equal to the rate (the line of equality).
\begin{figure}[t]
{\includegraphics[width=0.49\textwidth]{ArXiv_Plots/MI2.pdf}
\caption{Average amount of the accumulated information density for decoding correctly at a particular code rate for the GF(256) NB-LDPC of Fig.~\ref{fig:NCBL} code over SNR-2dB AWGN channel.}\label{fig:MI2}}
\end{figure}
\section{Optimizing Transmission Lengths}
\label{sec:VLFT with limited number of transmissions}
Consider the scenario in which the number of increments (packets of incremental redundancy) associated with a codeword that can be accumulated at the receiver is limited to $m$. Using the p.d.f. of $N_S$ from (\ref{eqn:PDFNS}) we find the optimal blocklengths $\{N_1,N_2,\hdots,N_m\}$ to maximize the throughput. The initial blocklength $N_1$ satisfies $N_1 \ge N_0$ where $N_0$ is the smallest possible blocklength (of the original NB-LDPC code). Each of the additional bits beyond $N_0$ transmitted in the first transmission is the exclusive-or of all eight bits representing one of the variable nodes of the original rate-$k/N_0$ GF(256) NB-LDPC code.
The other transmissions use the scheme in Section~\ref{sec:Creating_a_bit} to generate the subsequent bits.
\subsection{Throughput optimization through exhaustive search}
An accumulation cycle (AC) is a set of $m$ or fewer transmissions and decoding attempts ending when decoding is successful or when the $m^{th}$ decoding attempt fails. If decoding is not successful after the $m^{th}$ decoding attempt, the accumulated transmissions are forgotten and the process starts over with a new transmission of the first block of $N_1$ symbols. From a strict optimality perspective, neglecting the symbols from the previous failed AC is sub-optimal. However, the probability of an AC failure is sufficiently small that the performance degradation is negligible. Neglecting these symbols greatly simplifies analysis.
Define the throughput as $R_T=\frac{E[K]}{E[N]}$, where $E[N]$ represents the expected number of channel uses in one AC and $E[K]$ is the effective number of \textit{information} bits transferred correctly over the channel in one AC.
The expression for $E[N]$ is
{ \small \begin{IEEEeqnarray}{lCl}
E[N]&=&N_1Q\left(\!\frac{\frac{k}{N_1}-\mu_S}{\sigma_S}\!\right)
\label{eqn:EN1}
\\%-Q\left(\!\frac{r_{0}-\mu}{\sigma}\!\right)]+\\
&&+\sum\limits_{i=2}^m N_{i}\left[Q\left(\!\frac{\frac{k}{N_{i}}-\mu_S}{\sigma_S}\!\right)-Q\left(\!\frac{\frac{k}{N_{i-1}}-\mu_S}{\sigma_S}\!\right)\right]
\label{eqn:EN2toM}
\\
&&+N_{m}\left[1-Q\left(\!\frac{\frac{k}{N_{m}}-\mu_S}{\sigma_S}\!\right)\right].
\label{eqn:ENM}
\end{IEEEeqnarray}}
The right hand side of (\ref{eqn:EN1}) shows the contribution to expected blocklength from successful decoding on the first attempt in the AC. $Q \left(\!\frac{\frac{k}{N_1}-\mu_S}{\sigma_S}\!\right)$ is the probability of decoding successfully with the initial block of $N_1$. Similarly, the terms in (\ref{eqn:EN2toM}) are the contributions to expected blocklength from decoding that is first successful at total blocklength $N_i$ (at the $i^{\text{th}}$ decoding attempt). Finally, the contribution to expected blocklength from not being able to decode even at $N_m$ is $1-Q \left(\!\frac{\frac{k}{N_{m}}-\mu_S}{\sigma_S}\!\right)$ which is shown in (\ref{eqn:ENM}). Even when the decoding has not been successful at $N_m$, the channel has been used for $N_m$ channel symbols.
The expected number of successfully transferred information bits $E[K]$ is
\begin{equation}
\small
E[K]=kQ \left(\! \frac{\frac{k}{N_{m}}-\mu_S}{\sigma_S} \right) \, ,
\label{eqn:EK}
\end{equation}
where $Q\left(\frac{\frac{k}{N_{m}}-\mu_S}{\sigma_S}\right)$ is the probability of successful decoding at some point in the AC. Note that $E[K]$ depends only upon $N_m$. In fact, for large values of $N_m$, $E[K]\approx k$ and thus not sensitive to the choice of $N_m$
Exhaustive search (ES) can be used to optimize \{$N_1,N_2,\hdots,N_m$\} to maximize $R_T=\frac{E[K]}{E[N]}$. The order of complexity for ES is $O\big({N_{max}-N_0+1 \choose m}\big)$, where $N_{max}$ is the maximum allowable overall blocklength for an AC. Since $E[K]\approx k$, maximization of $R_T$ is equivalent to minimization of $E[N]$.
\subsection{Sequential differential optimization}
Sequential differential optimization (SDO) is an extremely effective alternative to ES. Over a range of possible $N_1$ values, SDO optimizes \{$N_2,\hdots,N_m$\} to minimize $E[N]$ for each fixed value of $N_1$ by setting derivatives to zero as follows:
\begin{equation}
\small
\left \{N_2,\hdots,N_{m} : \frac{\partial E[N]}{\partial N_i}=0, ~\forall i=1,\hdots,m\!-\!1 \right\} \, . \label{derivative0}
\end{equation}
For each $i \in \{2, \ldots, m\}$, the optimal value of $N_i$ is found by setting $\frac{\partial E[N]}{\partial N_{i-1}}=0$, yielding a sequence of relatively simple computations. In other words, we select the $N_{i}$ that makes our previous choice of $N_{i-1}$ optimal in retrospect. For example to find $N_2$ we compute the derivative
\begin{equation}
\small
\dfrac{\partial E[N]}{\partial N_1}=Q\left(\!\frac{\frac{k}{N_1}-\mu_S}{\sigma_S}\!\right)+(N_1-N_2)Q^\prime\left(\!\frac{\frac{k}{N_1}-\mu_S}{\sigma_S}\!\right)=0
\end{equation}
and solve for $N_2$
as
\begin{equation}
\small
N_2=\frac{Q \left(\! \frac{\frac{k}{N_1}-\mu_S}{\sigma_S} \!\right) +N_1Q^\prime \left(\! \frac{\frac{k}{N_1}-\mu_S}{\sigma_S}\!\right)}{Q^\prime \left(\! \frac{\frac{k}{N_1}-\mu_S}{\sigma_S}\!\right)}\, ,
\label{eqn:N2D}
\end{equation}
where
\begin{equation}
Q^\prime\left(\!\frac{\frac{k}{N_i}-\mu_S}{\sigma_S}\!\right)=\!\frac{k}{N_i^2\sigma_S}\frac{1}{\sqrt{2\pi}}e^\frac{\left(\!\frac{k}{N_i}-\mu_S\!\right)^2}{2\sigma_S^2}.
\end{equation}
For $i>2$, $\frac{\partial E[N]}{\partial N_{i-1}}=0$ depends only on \{$N_{i-2},N_{i-1},N_{i}$\} as follows:
{\small \begin{equation}
\small
\frac{\partial E[N]}{\partial N_{i\!-\!1}}\!=\!Q\!\left(\!\!\frac{\frac{k}{N_{i\!-\!1}}\!-\!\mu}{\sigma}\!\right)\!+\left(\!N_{i\!-\!1}\!-\!N_{i}\!\right)\!Q^\prime\!\left(\!\!\frac{\frac{k}{N_{i\!-\!1}}\!-\!\mu}{\sigma}\!\right)\!- Q\!\left(\!\!\frac{\frac{k}{N_{i\!-\!2}}\!-\!\! \mu}{\sigma}\!\right)\!.\nonumber
\end{equation}}
Thus we can solve for $N_i$ as
\begin{equation}
\small
N_{i}=\frac{Q\!\left(\!\frac{\frac{k}{N_{i-1}}-\mu}{\sigma}\!\right)\!+\!N_{i-1} Q^\prime\!\left(\!\frac{\frac{k}{N_{i-1}}-\mu}{\sigma}\!\right)\!- \! Q\!\left(\!\frac{\frac{k}{N_{i-2}}-\mu}{\sigma}\!\right)}
{Q^\prime\!\left(\!\frac{\frac{k}{N_{i-1}}-\mu}{\sigma}\!\right)}\, .
\label{eqn:Ni1D}
\end{equation}
Actually, for each possible value of $N_1$, SDO can be used to produce an infinite sequence of $N_i$ values that solve \eqref{derivative0}. Each such sequence is an optimal sequence of increments for a given density of retransmission points on the transmission axis. As $N_1$ increases, the density decreases. Using SDO to compute the optimal $m$ points is equivalent to selecting the most dense SDO-optimal sequence that when truncated to $m$ points results in the highest throughput.
\subsection{Application to VLFT with $m$ transmissions}
Table~\ref{tbl:BFvsH} shows the optimized $\{N_1, N_2, \hdots, N_m\}$, resulting throughput $R_T$, and expected blocklength $\lambda=k/R_T$ for various $m$. The values obtained by SDO are very close to the values obtained by ES.
For $m=2,5,6, \text{and } 7$, the optimized blocklengths for both approaches are the same. For $m=3 \text{ and } 4$ the blocklengths differ only in the value of $N_m$ (shown in bold) and only by one bit. This small difference in $N_m$ causes a negligible difference in the maximum throughput $R_T$ and minimum expected blocklength $\lambda\!=\!\frac{k}{R_T}$. Since the complexity of ES is exponential in $m$, it is infeasible to obtain a globally optimal solution for $m>7$; whereas SDO, with complexity $O(N_{max}-N_0)$, can find a solution within seconds even for large $m$.
\begin{table}[t]
\renewcommand*{\arraystretch}{1.2}
\begin{center}
\caption{Optimized $\{N_1, N_2, \hdots, N_m\}$, $R_T$, and $\lambda$ from ES and SDO for $k $ = 96 bits for VLFT on a 2 dB SNR binary-input AWGN channel using $\mu_S=0.6374$ and $\sigma_S = 0.0579$.}
\begin{tabular}{@{}l|c|l|l|l@{}}
Alg. & $m$ & $\{N_1, N_2, \hdots, N_m\}$ & $R_T$ & $\lambda$\\
\hline
\hline
$\text{ES, SDO}$& 2& 158 , 188 &0.566&169.6 \\
\hline
$\text{ES}$& 3& 150, 167, \bf{194} &0.58638 &163.71\\
$\text{SDO}$& 3& 150, 167, \bf{195} &0.58635& 163.72 \\
\hline
$\text{ES}$& 4& 146, 158, 172, \bf{198} &0.59709&160.77 \\
$\text{SDO}$& 4& 146, 158, 172, \bf{197} &0.59707& 160.78 \\
\hline
$\text{ES, SDO}$& 5& 143, 153, 163, 176, 201 &0.603&159.2 \\
\hline
$\text{ES, SDO}$& 6& 140, 149, 157, 166, 179, 204 &0.608 &157.9\\
\hline
$\text{ES, SDO}$& 7& 139, 147, 154, 161, 170, 182, 206 & 0.611 &157.1\\
\end{tabular}
\label{tbl:BFvsH}
\end{center}
\end{table}
\begin{comment}
\begin{table}[h]
\begin{center}
\caption{Throughput percentage of RCSP-ML throughput achieved by non-binary LDPC codes in the VLFT setting}
\begin{tabular}{c|c|c|c|c|c|c}
$m$ & 2* & 3 & 4& 5* & 6* & 7* \\
\hline
$\text{brute force}$ &0.566 & 0.58638& 0.59709& 0.603& 0.608& 0.611 \\
\hline
$\text{\%heuristic}$ &0.566& 0.58635& 0.59707& 0.603 &0.608& 0.611 \\
\end{tabular}
\label{tbl:LDPC_CC_table}
\end{center}
\end{table}
\end{comment}
\begin{figure}[t]
{\includegraphics[width=0.49\textwidth]{ArXiv_Plots/RT_m2.pdf}
\label{SDArt}
}
{\includegraphics[width=0.49\textwidth]{ArXiv_Plots/lambda_m2.pdf}
\label{SDAlambda}}
\caption{Throughput ($R_T$) and the expected blocklength ($\lambda$) as a function of the number of transmissions $m$ achieved by non-binary LDPC codes in the VLFT setting for $k=96$.}
\label{fig:sdartlambda}
\end{figure}
Fig.~\ref{fig:sdartlambda} shows the optimum $R_T$ and $\lambda$ for various $m$ using SDO. The dashed lines show the maximum achievable $R_T$ and the corresponding minimum achievable $\lambda$ with an unlimited $m$ as in \cite{Kasraisit2014}. As a function of $m$, $R_T$ quickly converges to the $m=\infty$ asymptote and even for $m \approx 10$ the throughput is close to the value achievable with an unlimited number of increments. Correspondingly, the expected latency also converges quickly and for $m \approx 10$ the expected blocklength is close to the minimum $\lambda$ achievable by unlimited transmissions of one bit at a time.
\section{VLF with CRC}
\label{sec:VLFCRC}
In this section, instead of using NTC as a genie, cyclic redundancy check (CRC) codes are used as error-detecting codes to detect whether there is an error in the decoded message. In systems incorporating CRCs, a certain number of check bits, $L_{\text{crc}}$, are computed and added to the information message of length $k_{\text{inf}}$.
\begin{comment}
The checksum bits are computed by initially appending the information bits with an all-zero block of length $L_{\text{crc}}$ and then computing the remainder of the division by a divisor polynomial of degree $L_{\text{crc}}+1$. The total block containing the original information block and the additional CRC check bits are finally encoded by the NB-LDPC encoder and the encoded message is transmitted over the communication channel.
\end{comment}
At the receiver, the NB-LDPC decoder initially attempts to decode the received block. If decoding results in a codeword, the CRC check determines whether the check bits agree with the data by computing the checksum from the first $k_{\text{inf}}$ bits of the received sequence and comparing this checksum with the last $L_{\text{crc}}$ received bits. In order to achieve an undetected error probability of $\epsilon$, the CRC code length $L_{\text{crc}}$ is chosen so that the overall probability of error resulting from the NB-LDPC and CRC codes combined is smaller than $\epsilon$.
The transmitter terminates transmission when the receiver sends feedback indicating that the decoded message passes the CRC check. If the message is correctly decoded, it passes the CRC and the transmitter moves on to the next message. If the message is decoded incorrectly and the decoded message fails to pass CRC, the transmitter sends more bits to increase reliability of the bits already transmitted. If the receiver decodes the message incorrectly and the erroneously decoded message passes the CRC check, the transmitter moves on to the next message and the packet is decoded in error. This error is undetected by the receiver. In the case of unlimited transmissions ($m=\infty$), the transmitter transmits one bit at a time until the decoder either decodes the message correctly or until it decodes to a message that passes the CRC check.
\begin{figure}[t]
{\includegraphics[width=0.49\textwidth]{ArXiv_Plots/WBL11.pdf}
\caption{Empirical p.m.f. and reciprocal-Gaussian fit for the shortest cumulative blocklength ($N_E$) after which decoding never again converges to an incorrect codeword. The smallest blocklength for the GF(256) NB LDPC code is $N_0=120$ bits with $k=96$ information bits. Thus, the initial rate is $R_0=\frac{k}{N_0}=0.8$. }
\label{fig:EBL}}
\vspace{0.1in}
{\includegraphics[width=0.49\textwidth]{ArXiv_Plots/WR11.pdf}
\caption{Empirical p.m.f. and Gaussian approximation with $\mu_E=0.626$ and $\sigma_E^2=0.056 $ of $R_E$ in VLFT setting.
}\label{fig:ER}}
\end{figure}
With a limited number of transmissions, the blocklength corresponding to each transmission and the length of CRC are chosen to guarantee a probability of undetected error of at most $\epsilon$. If the message is not decoded correctly even after $m$ transmissions (and the NACKs are correctly received), the receiver deletes all received symbols and a new transmission cycle begins with the transmitter sending the original block of $N_1$ symbols.
Since the CRC as an error detection tool is used only when the decoder converges to a codeword, it is crucial to differentiate between \textit{erroneous} decoding and failure to converge to a codeword. Fig.~\ref{fig:EBL} shows the empirical p.m.f. of the required cumulative number of symbols ($N_E$) until the receiver will never again converge to an incorrect codeword. Note that Fig.~\ref{fig:EBL} is conditioned on the decoder initially decoding to a wrong codeword at $N_0=120$. The probability that the decoder decodes incorrectly at $N_0$ is $\gamma$. (For the experiment that produced the p.m.f. in Fig.~\ref{fig:EBL} $\gamma = 0.165$.)
For blocklengths larger than $N_E$, the decoder either decodes correctly or fails to converge to any codeword. This is a different condition than correct decoding, which was modeled in Figs. \ref{fig:NCBL} and \ref{fig:NCR}. Fig.~ \ref{fig:ER} shows the empirical p.m.f. of $R_E=\frac{k}{N_E}$, the instantaneous rate at which the decoder stops decoding to the wrong codeword, and the corresponding Gaussian approximation.
Fig.~\ref{fig:statediagram} shows the state diagram representing all the scenarios that can happen based on our simulations.
According to our simulations, if the decoder converges to a wrong codeword, it continues to decode to the same wrong codeword even with additional incremental transmissions. The increased reliability from incremental transmissions never moves the decoder from one wrong codeword to another wrong codeword. It only helps the decoder either to converge to the correct codeword or not to converge to any codeword at all. Figs.~\ref{fig:NCBL}, \ref{fig:NCR} correspond to the blocklength and rate of entry to state 3. Figs.~\ref{fig:EBL}, \ref{fig:ER} correspond to the blocklength and rate of leaving state 1.
\begin{figure}[t]
\includegraphics[width=0.49\textwidth]{ArXiv_Plots/Presentation15_cropped.pdf}
\caption{The state diagram corresponding to LDPC coding with incremental transmissions.}
\label{fig:statediagram}
\end{figure}
In this section, similar to the case of $m=\infty$ VLFT, the transmitter sends one bit of incremental redundancy at a time until the decoder converges to the correct codeword or converges to an incorrect codeword that passes the CRC check. We require an undetected error probability of smaller than $\epsilon$. If the transmission starts with a blocklength of length $N_0$, the total probability of error is $\gamma \times 2^{-L_{\text{crc}}}$, where $2^{-L_{\text{crc}}}$ is approximately the probability of error that the CRC checks for a wrong codeword.
This paper uses standard CRC codes. However, for the best error detection, the CRC codes can be designed specifically for a particular code as shown in \cite{Chung-Yu_tcom}.
For the error probability constraint of $\epsilon$, we choose the length of the CRC code so that $\gamma \times 2^{-L_{\text{crc}}} < \epsilon$. For example, if $\epsilon$ is set to be $10^{-3}$ and $\gamma = 0.165$, the length of the CRC code $L_{\text{crc}}=8$ is required to guarantee the overall probability of error, $\gamma \times 2^{-L_{\text{crc}}} = 6.25 \times 10^{-4} < \epsilon = 10^{-3}$.
As will be illustrated in the results section (Section \ref{sec:results}), the throughput of this scheme can be well predicted by the results obtained from VLFT with unlimited transmissions (Section \ref{sec:VLFT with limited number of transmissions}) modified by a factor of $\frac{k-L_{\text{crc}}}{k}$ that captures the back-off in rate due to the CRC overhead. For example, in our previous analysis from Table \ref{tbl:BFvsH} for $m=\infty$, the rate is 0.632 while with a CRC of length 8, for $k_{\text{inf}}=96-8 = 88$ the rate is predicted to be $\frac{96-8}{96} \times 0.632 = 0.579$. As the simulation results of Section~\ref{sec:results} show, the actual achieved rate is $0.575$ with an undetected error probability of $8.04 \times 10^{-4}$. We will discuss these results in more detail in Section \ref{sec:results}.
\section{VLF with CRC and Limited Transmissions}
\label{sec:VLFCRC_limited}
In VLF with a limited number of transmissions, the length of each incremental transmission should be selected to maximize
$R_T=\frac{E[K| L_{\text{crc}}]}{E[N]}$, where $E[N]$ is given by \eqref{eqn:EN1} and $E[K|L_{\text{crc}}]$ is the effective number of transmitted information bits, computed as
\begin{equation}
\small
E[K|L_{\text{crc}}]=(K-L_{\text{crc}})\left[Q \left(\! \frac{\frac{K}{N_{m}}-\mu_S}{\sigma_S} \right)-P_{N_1} 2^{-L_{\text{crc}}}~ \right],
\label{eqn:EK}
\end{equation}
under the constraint that the probability of undetected error $P_{N_1} ~ 2^{-L_{\text{crc}}}<\epsilon$. $P_{N_1}$ is the probability of converging to an incorrect codeword at blocklength $N_1$.
An approximation technique similar to the one used in optimizing the length of each incremental redundancy block in VLFT is used here: $ \left[Q \left(\! \frac{\frac{k}{N_{m}}-\mu_S}{\sigma_S} \right)-P_{N_1} ~ 2^{-L_{\text{crc}}}~ \right] \approx 1$. The optimization problem of maximizing $R_T=\frac{E[K| L_{\text{crc}}]}{E[N]}$ reduces to minimizing $E[N]$ for each $L_{\text{crc}}$. The SDO technique used in Section \ref{sec:VLFT with limited number of transmissions} can be used here under the additional constraint that $P_{N_1} ~ 2^{-L_{\text{crc}}}<\epsilon$.
\begin{table}[t]
\renewcommand*{\arraystretch}{1.13}
\begin{center}
\caption{Optimized $\{N_1,\hdots,N_m\}$ for $m$$=$$5$ in VLF-with-CRC using SDO for different values of $L_{\text{crc}}$. The exact same values were obtained by ES.}
\begin{tabular}{c|c|c|c|c}
$L_{\text{crc}}$ & $\{N_1, N_2, \hdots, N_5\}$ & $\lambda$& $R_T$& $\epsilon$ \\
\hline
1 & 193, 198, 205, 216, 241 & 193.27 & 0.49 & $8.95 \times 10^{-4}$\\
2 & 187, 192, 199, 210, 235 & 187.38 & 0.50 & $9.02 \times 10^{-4} $
\\
3 & 180, 185, 192, 203, 228 & 180.67 & 0.51 & 9.82 $ \times 10^{-4} $\\
4 & 174, 180, 187, 198, 222 & 175.14 & 0.52 & $9.14 \times 10^{-4}$ \\
5 & 166, 172, 180, 192, 216 & 168.48 & 0.54 & $9.62 \times 10^{-4}$ \\
6 & 157, 164, 172, 184, 209 & 162.68 & 0.55 & $9.58 \times 10^{-4}$ \\
7 & 143, 153, 163, 176, 201 & 159.14 & 0.56 & $9.44 \times 10^{-4}$ \\
8 & 143, 153, 163, 176, 201 & 159.07 & 0.55 &$ 4.72 \times 10^{-4}$ \\
9 & 143, 153, 163, 176, 201 & 159.04 & 0.54 &$ 2.36 \times 10^{-4}$ \\
10 & 143, 153, 163, 176, 201 & 159.02 & 0.54 & $1.18 \times 10^{-4}$
\end{tabular}
\label{tbl:VLFCRCN1N5}
\end{center}
\vspace{-0.1in}
\end{table}
For each $L_{\text{crc}}$, the optimized $\{N_1,\hdots,N_m\}$ values for this case are identical for SDO and ES and the values are given in Table \ref{tbl:VLFCRCN1N5}. For small values of $L_{\text{crc}}$ we need to use a large value of $N_1$ to make sure $P_{N_1} ~ 2^{-L_{\text{crc}}}<\epsilon$. As a larger value of $L_{\text{crc}}$ is selected, $N_1$ and consequently $\{N_2,\hdots,N_5\}$ decrease while the error probability constraint is still satisfied. For $L_{\text{crc}}=7$ the set of $\{N_1,\hdots,N_5\}=\{143, 153, 163, 176, 201\}$ minimizes the expected latency $\lambda$ and maximizes $R_T$. For larger values of $L_{\text{crc}}>7$, the set of optimum blocklengths does not change and only the overall probability of error decreases as the CRC length is increased.
The optimal set of blocklengths for $L_{\text{crc}} \ge 7$ and $m=5$ is the same as the set for VLFT and $m=5$ from Table \ref{tbl:BFvsH}. The intuition for this is that once $L_{\text{crc}}$ is large enough that decoding decisions are extremely reliable, the optimal blocklengths for VLF-with-CRC should match those of VLFT. Because the blocklengths are identical, the throughput $R_T$ for $m=5$ with $L_{\text{crc}}=7$ can be computed by reducing the $R_T$ in Table \ref{tbl:BFvsH} to account for the overhead of the CRC. The reduction from the $m=5$ VLFT rate $R_T =0.603$ is $\frac{96-7}{96}$ where $\frac{96-7}{96} \times 0.603 = 0.559$ which corresponds to the $R_T$ from Table~\ref{tbl:VLFCRCN1N5} for $L_{\text{crc}}=7$.
While both SDO and ES give the same values for different $L_{\text{crc}}$ values, the order of complexity for SDO is $O(L_{\text{crc}}(N_{max}-N_0))$ while with ES algorithm the complexity has the much larger order of $O\big(L_{\text{crc}}{N_{max}-N_0 \choose m}\big)$. As the simulation results of Section~\ref{sec:results} show, the actual achieved rate is $0.541$ with an undetected error probability of $5.75 \times 10^{-4}$.
For the simulations in Section~\ref{sec:results} the CRC code used for $L_{\text{crc}}=7$ has a polynomial representation of 0x09 ($x^7 + x^3 +1$). This CRC code has been used by Telecommunication Standardization Sector of the International Telecommunications (CCITT) which sets international communications standards. The CRC code used for $L_{\text{crc}}=8$ has a polynomial representation of 0x07 ($x^8 + x^2 + x + 1$) and is used in MultiMedia Cards (MMC) and Secure Digital (SD) cards.
\section{Two-phase VLF}
\label{sec:VLFoptimization}
Now we consider the two-phase VLF model in which the transmitter (source) uses the primary communication channel to confirm whether the receiver (destination) has decoded to the correct codeword. As in \cite{firing_genie}, the two-phase incremental redundancy scheme has a \textit{communication} phase followed by a \textit{confirmation} phase.
\begin{figure}[t]
\includegraphics[width=0.49\textwidth]{ArXiv_Plots/tree11.pdf}
\caption{Two-phase VLF block diagram and the forward transmission stages in two-phase VLF systems.}
\label{fig:tree}
\end{figure}
Fig.~\ref{fig:tree} shows a block diagram for the two-phase communication scheme. Starting at the left, a message block of size $N_1$ is transmitted (communication phase).
If the destination decodes correctly, the source sends a coded forward ``ACK" on the same forward noisy channel to confirm the successful decoding (confirmation phase). If the destination decodes incorrectly, the source sends a coded forward NACK. The ACKs and NACKs are repetition codes of length $A_1$ symbols and are transmitted over the same forward noisy channel from the transmitter (source) to the receiver (destination).
If the decoder does not converge to any codeword with $N_1$ symbols, the transmitter skips the unnecessary confirmation phase and immediately transmits the second increment of $N_2-N_1$ bits.
In the two-phase VLF setting, we use the probability distributions of $N_S$, $R_S$, $N_E$ and $R_E$ from Figs.~\ref{fig:NCBL}, \ref{fig:NCR}, \ref{fig:EBL}, and \ref{fig:ER}. The optimization problem is to maximize $R_T=\frac{E[K]}{E[N]}$ where
\begin{equation}
\small{
E[K]=k\left(\sum\limits_{i=1}^{m} P^{SS}_i\right)\approx k\left(\!Q\left(\!\frac{\frac{k}{N_m}\!-\!\mu_S}{\sigma_S}\!\right)-\!\sum\limits_{i=1}^{m-1} P^{EE}_i\!\right)\, ,
\label{eqn:VLFEK}}
\end{equation}
with $P_i^{EE}$ representing the probability the receiver decodes both the message and the NACK erroneously and $P_i^{SS}$ is the probability the receiver decodes both message and ACK successfully. Note that \eqref{eqn:VLFEK} assumes (consistent with our simulation results) that once the decoder is in state 3 of Fig. \ref{fig:statediagram}, it does not return to state 1 even if a forward ACK is incorrectly received as a forward NACK. In any case, as in Section \ref{sec:VLFTNBLDPC} we assume $E[K]\approx k$.
The expected number of symbols transmitted in an AC is
\begin{align}{\small
E[N] =& \sum\limits_{i=1}^m(N_i\!+\!A_i)\left[P^{SS}_i\!+\!P^{EE}_i\right]\!+\!A_i\left[P^{SE}_i\!+\!P^{ES}_i\right] \label{eqn:VLFEN2m21}\\
&\!{+N_m \left(\! 1-\left(\sum\limits_{i=1}^{m} P^{SS}_i+\sum\limits_{i=1}^{m} P^{EE}_i\right)\right)} \label{eqn:VLFENm} \, ,}\end{align}
where $P_i^{SE}$ is the probability of decoding the message successfully but decoding the ACK as a NACK. Conversely, $P_i^{ES}$ is the probability of decoding the message erroneously but decoding the NACK successfully. The term multiplying $N_m$ in (\ref{eqn:VLFENm}) is the probability that an AC ends without satisfying either of the stopping conditions. (\ref{eqn:VLFENm}) is also approximated to $N_m \left(\! 1- Q\left(\!\frac{\frac{k}{N_m}\!-\!\mu_S}{\sigma_S}\!\right)+ P^{SE}_m\!\right)$. The probabilities $P_i^{SS}$, $P_i^{EE}$, $P_i^{SE}$, and $P_i^{ES}$ are computed as follows:
{ \small \begin{IEEEeqnarray}{lll}
P^{SS}_i&=&\!\left[Q\!\left(\!\! \frac{\frac{k}{N_i}\!-\!\mu_S}{\sigma_S}\!\right)\!-\!Q\!\left(\!\frac{\frac{k}{N_{i\!-\!1}}\!-\!\mu_S\!}{\sigma_S}\!\right)\!\right]
\left[1\!-\!Q\!\left(\!\frac{\sqrt{A_i}}{\sigma_c}\!\right)\!\right]
\label{eqn:VLFEN2m1}
\\
P^{EE}_i&=&\,\left[\gamma\left(\! 1-Q\! \left(\! \frac{\frac{k}{N_i}\!-\!\mu_E}{\sigma_E}\!\right)\!\right)\right]\left[Q\!\left(\!\frac{\sqrt{A_i}}{\sigma_c}\!\right)\right] \label{eq:PEE}
\\
P^{SE}_i&=&\!\left[Q\left(\!\frac{\frac{k}{N_i}\!-\!\mu_S}{\sigma_S}\!\!\right)\!-\!Q\left(\!\frac{\frac{k}{N_{i-1}}\!-\!\mu_S}{\sigma_S}\!\right)\right]
\left[Q\left(\!\frac{\sqrt{A_i}}{\sigma_c}\!\right)\right]
\\
P^{ES}&=&\left[\gamma\left(\!1-Q\left(\!\frac{\frac{k}{N_i}\!-\!\mu_E}{\sigma_E}\!\right)\!\right)\right]\left[\left(\!1\!-\!Q\!\left(\!\frac{\sqrt{A_i}}{\sigma_c}\!\right)\right)\right] \, .
\label{eqn:VLFEN2m2}
\end{IEEEeqnarray}}
In \eqref{eqn:VLFEN2m1} the probability of decoding correctly at $N_i$ and not at blocklengths smaller than or equal to $N_{i-1}$ is $Q\left(\!\frac{\frac{k}{N_i}-\mu_S}{\sigma_S}\!\right)-Q\left(\!\frac{\frac{k}{N_{i-1}}-\mu_S}{\sigma_S}\!\right)$ and $Q\left(\!\frac{\sqrt{A_1}}{\sigma_c}\!\right)$ is the probability that the ACK is decoded as a NACK, where $\sigma_c$ is the standard deviation of the channel noise. In \eqref{eq:PEE}, $\gamma\left[1-Q\left(\!\frac{\frac{k}{N_i}\!-\!\mu_E}{\sigma_E}\!\right)\right]$ is the probability of decoding erroneously at $N_i$.
We optimize the blocklengths for two-phase VLF to maximize $R_T$ under the constraint that
$
\sum\limits_{i=1}^mP^{EE}_i<\epsilon \, ,
$
using both ES and SDO approaches from Section \ref{sec:VLFT with limited number of transmissions} for fixed values of $\{A_1,\hdots,A_m\}$. For ES we considered values of $N_1\leq N_2\leq \dots \leq N_m$ and constrained $N_m$ to be no larger than the blocklength corresponding to a rate-0.1 code ($N_m \leq 10k$). For SDO we considered $N_1$ values ranging from the initial coding length $N_0$ to $3k$, which was the range that gave useful values of $\epsilon$.
Table \ref{tbl:SDAESVLF} shows two sets of $\{N_1,\hdots,N_m\}$ with $m=5$ obtained for different $N_1$ in SDO with $\epsilon \!\approx\! 10^{-3}$. The optimized $\{N_1,\hdots,N_m\}$ with $\epsilon\!\leq\!10^{-3}$ from ES is close to the SDO optimized blocklengths. The optimized blocklengths from SDO can also be used as optimization limits for ES algorithm and significantly reduce the ES optimization space.
\begin{table}[b]
\renewcommand*{\arraystretch}{1.13}
\begin{center}
\caption{Optimized $\{N_1,\hdots,N_m\}$ for $m$$=$$5$ two-phase VLF using SDO and ES with $\{A_1,\hdots,A_5\}=\{5,4,3,3,3\}$.}
\begin{tabular}{c|c|c|c|c|c}
Alg. & $k$ & $\{N_1, N_2, \hdots, N_5\}$ & $\lambda$ & $R_T$ & $\epsilon$ \\
\hline
SDO & 96& 145, 156, 167, 180, 202 & 166.1 & 0.5779 & 1.2E-3\\
SDO & 96& 146, 158, 171, 188, 230 & 166.6 & 0.5762 & 9.4E-4\\
\hline
ES & 96 & 146, 158, 170, 184, 211 & 166.4 & 0.5771 & 9.9E-4 \\
\end{tabular}
\label{tbl:SDAESVLF}
\end{center}
\vspace{-0.1in}
\end{table}
\section{Results}
\label{sec:results}
\begin{figure}[t]
{\includegraphics[width=0.49\textwidth]{ArXiv_Plots/LDPC_CC5_7_VLFT3_m10.pdf}
\caption{$R_T$ vs. $\lambda$ for NB-LDPC and 1024-state convolutional codes for VLFT with $m=\infty$, $m=10$, and $m=5$.
}\label{fig:VLFT_rtlambda}}
\end{figure}
Fig.~\ref{fig:VLFT_rtlambda} shows $R_T$ versus $\lambda$ for NB-LDPC and convolutional codes using VLFT. In VLFT with an unlimited number of transmissions (1-bit increments), convolutional codes with ML decoders perform very well at short average blocklengths of up to 100 bits. VLFT schemes have throughputs greater than capacity at short blocklengths because of the NTC. Convolutional codes follow the marginal RCSP-ML (with unconstrained input) plot closely at short-blocklength with a small gap that is due to the binary input for convolutional codes. At longer blocklengths of about 200 bits, marginal RCSP-ML rate approaches the capacity. NB-LDPC codes outperform convolutional codes at longer blocklengths because the codeword error rate of convolutional codes increases once the blocklength exceeds twice the traceback depth \cite{Anderson_Traceback_TransIT_1989} whereas the NB-LDPC code performance continues to improve with blocklength. The gaps between the throughputs for $m=\infty$, $m=5$, and $m=10$ NB-LDPC codes are similar to the gap observed in Fig.~\ref{fig:sdartlambda} . For $m=10$ the performance of NB-LDPC codes in VLFT is much closer to the case of $m=\infty$. The NB-LDPC codes of Fig.~\ref{fig:VLFT_rtlambda} are over $GF(256)$. The shortest code for $k=96$ bits has an initial blocklength of 15 $GF(256)$ symbols (120 bits), corresponding to an initial rate of 0.8. The NB-LDPC codes for $k=192$ and $k=288$ have initial blocklengths of 256 and 384 bits, respectively. Some results for the finite-$m$ systems follow the non-active feedback scheme described in Section~\ref{sec:Creating_a_bit}.
\begin{figure}[t]
{\includegraphics[width=0.49\textwidth]
{ArXiv_Plots/LDPC_CC5_7_VLFTT3_4_555.pdf}
\caption{Percentage of VLFT $R_T$ that NB-LDPC achieves with $m=\infty$, $m=10$, and $m=5$.
}\label{fig:VLFT_rtpercentage}}
\end{figure}
Fig.~\ref{fig:VLFT_rtpercentage} shows the percentage of RCSP-ML rate for VLFT achieved by NB-LDPC and convolutional codes in VLFT. In the expected-blocklength range of 150-600 bits, NB-LDPC codes achieve a throughput of about $90\%$ of RCSP-ML throughput (and about 91\% and 96\% of unconstrained and binary-input capacity, respectively) with an unlimited number of transmissions. When the number of the transmissions is limited to $10$ and $5$, the throughput percentage decreases to about 90\% and 85\%, respectively. RCSP-ML analysis is applied to the unconstrained-input AWGN channel at SNR 2-dB, for which the capacity is 0.684. The capacity of BI-AWGN channel at 2-dB SNR is 0.642 which is about 6\% lower than the unconstrained-input AWGN capacity.
\begin{table}[t]
\renewcommand*{\arraystretch}{1.12}
\begin{center}
\caption{Optimized $\{N_1,\hdots,N_5\}$ for two-phase VLF and VLF-with-CRC with $m$$=$$5$ at SNR 2 dB, and corresponding $R_T$ and $\lambda$ values achieved in simulations. $\{A_1,\hdots,A_5\}=\{5,4,3,3,3\}$ for two-phase VLF using NB-LDPC codes. For the convolutional codes, $A_i =6$, $8$, and $9$ $\forall i$ for $k = 96$, $192$, and $288$ bits, respectively. }
\begin{tabular}{c|c|c|c|c|c}
Code & $k$ & $\{N_1, N_2, \hdots, N_5\}$ & $\lambda$ & $R_T$ & \%\\
\hline
CRC NB & 89 & 143,\! 153,\! 163,\! 176,\! 201 & 164.5 &0.541 &84.2\\
2-Phase NB& 96& 146,\! 158,\! 170,\! 184,\! 211 &170.4&0.563&87.7 \\
2-Phase CC& 96& 138,\! 153,\! 166,\! 180,\! 204 & 168.6 &0.569 & 88.6 \\
\hline
CRC NB & 185 & 293,\! 309,\! 325,\! 346,\! 386 & 323.4 &0.572 &89.1\\
2-Phase NB& 192& 301,\! 322,\! 344,\! 369,\! 408 &330.5&0.581&90.5 \\
2-Phase CC& 192& 287,\! 309,\! 331,\! 352,\! 384 &349.4&0.549&85.4\\
\hline
CRC NB &281& 459,\! 487,\! 518,\! 550,\! 597 & 491.3 &0.572 & 89.1\\
2-Phase NB& 288& 459,\! 487,\! 518,\! 550,\! 597 &495.7&0.581&90.5\\
2-Phase CC& 288& 416,\! 441,\! 463,\! 488,\! 532 & 599.6&0.480 &74.8 \\
\end{tabular}
\label{tbl:VLFLDPC_CC}
\end{center}
\end{table}
Table \ref{tbl:VLFLDPC_CC} summarizes the blocklengths that maximize the throughput in the two-phase VLF and VLF-with-CRC settings with $\epsilon$$=$$10^{-3}$, for both NB-LDPC codes and (for comparison) tail-biting convolutional codes. Blocklengths for the NB-LDPC codes are obtained from (\ref{eqn:VLFEK}-\ref{eqn:VLFENm}) using ES on an optimization space limited by initial SDO results. Blocklengths for the convolutional codes are based on the coordinate-descent algorithm in \cite{Williamson2014dissertation} using the assumption of rate-compatible sphere-packing. Table \ref{tbl:VLFLDPC_CC} also shows the percentage of BI-AWGN capacity obtained in the two-phase VLF setting with $m=5$ transmissions.
For $k=192 \text{ and } 288$, the NB-LDPC code obtains throughputs greater than $90\%$ of BI-AWGN capacity with an average blocklengths $\lambda$ of less than $500$ bits in the 2-phase setting. NB-LDPC codes in the VLF-with-CRC setting with $m=5$ achieve throughputs slightly lower than the ones in the 2-phase setting with $m=5$. However, similar to Fig.~\ref{fig:sdartlambda} if $m$ is increased to $10$, VLF-with-CRC results in higher throughputs. Large values of $m$ lead to a degradation in throughput performance for two-phase VLF due to the overhead associated with the more frequent forward ACK and NACK messages in the confirmation phase.
The rate-$1/3$ convolutional codes in Table \ref{tbl:VLFLDPC_CC}
have octal generator polynomials $(117, 127, 155)$ for the 64-state code and $(2325, 2731, 3747)$ for the 1024-state code \cite{Kasraisit2014}. The NB-LDPC codes are described completely online\footnote{UCLA Communication Systems Laboratory (CSL) website at http://www.seas.ucla.edu/csl/resources/index.htm}.
\begin{figure}[t]
{\includegraphics[width=0.49\textwidth]{ArXiv_Plots/newplot12.pdf}
\caption{$R_T$ vs. $\lambda$ for NB-LDPC with $m=\infty$ in VLF-with-CRC and 64 and 1024-state convolutional codes and NB-LDPC codes with $m=5$ in VLF.
}\label{fig:VLF_rtlambda}}
\end{figure}
Fig.~\ref{fig:VLF_rtlambda} shows the throughput obtained in the VLF setting for NB-LDPC codes, 64-state and 1024-state tail-biting convolutional codes with $m=5$, $m=10$, $m=\infty$ for $\epsilon=10^{-3}$. As the blocklength increases, as mentioned in \cite{Chen_Feedback_Journal_2013}, the performance of the codes in VLF gets closer to the performance in VLFT. The plots for $m=5$ are from Table \ref{tbl:VLFLDPC_CC}. With $m=\infty$, the $k=89$ the NB-LDPC code achieves a throughput greater than the random coding lower bound obtained from the analysis in \cite{Polyanskiy_IT_2011_NonAsym}.
\begin{figure}[t]
{\includegraphics[width=0.49\textwidth]
{ArXiv_Plots/VLF_percent_10.pdf}
\caption{Percentage of BI-AWGN capacity that NB-LDPC and convolutional codes achieve in VLF.
}\label{fig:VLF_rtpercentage}}
\end{figure}
Fig.~\ref{fig:VLF_rtpercentage} shows the percentage of the capacity of the BI-AWGN channel at 2-dB SNR achieved by NB-LDPC and convolutional codes using VLF. In the expected blocklength range of 300-500 bits, NB-LDPC codes with CRC achieve a throughput of about $94\%$ of capacity with an unlimited number of transmissions. When the number of the transmissions is limited to $10$, the throughput percentage decreases to about 93\%. For $m=5$, NB-LDPC codes perform slightly better in two-phase VLF setting than in VLF-with-CRC. Note that for $m=\infty$ or even $m=10$ two-phase VLF will not perform well because of the overhead associated with the confirmation messages.
As discussed in Section~\ref{sec:incremental} similar Gaussian approximation analysis can be done for higher-SNR AWGN channels. for instance, for SNR-8dB AWGN channel which uses a larger 16-QAM constellation, the VLF-with-CRC system with an unlimited number of transmissions achieves a throughput of 2.37 bits per symbol with a frame error probability of less than $10^{-3}$. This throughput corresponds to 88\% of capacity in the blocklength regime of 40 16-QAM (quadrature amplitude modulation) symbols. Furthermore, the VLF-with-CRC system on 5-dB BI-AWGN fading channel with an unlimited number of transmissions achieves a throughput corresponding to 90\% of capacity in the blocklength regime of about 140 bits.
\section{conclusion}
\label{sec:conclusion}
This paper uses the reciprocal-Gaussian approximation for the blocklength of first successful decoding to optimize the size of each incremental transmission to maximize throughput in VLFT and VLF settings. For feedback with a limitation on the number of transmissions, the sequential differential optimization (SDO) algorithm can be used quickly and accurately to find the optimal transmission lengths for a wide range of channels and codes. In this paper we applied SDO to non-binary LDPC codes for a variety of feedback systems. We focused on the binary-input AWGN channel but verified the effectiveness of the Gaussian approximation and SDO on the standard AWGN channel with a 16-QAM input and on a fading channel. In the 300-500 bit average blocklength regime, this paper reports the best VLFT and VLF throughputs yet. VLFT throughputs are higher than VLF, but VLF is more practical because it does not assume a noiseless transmitter confirmation symbol. For VLF-with-CRC with $m=\infty$, NB-LDPC codes with optimized blocklengths achieve about $94\%$ of the capacity of 2-dB BI-AWGN channel for an average blocklength of 300-500 bits. In the same blocklength regime, for VLF-with-CRC with $m=10$, NB-LDPC codes with optimized blocklengths achieve about $93\%$ of the capacity.
The performance results can also be considered in terms of SNR gap. In Fig.~\ref{fig:VLF_rtpercentage}, the random-coding lower bound for a system with feedback is 0.27 dB from the Shannon limit for $k=280$ with a blocklength of less than 500 bits. Looking at the VLF-CRC NB-LDPC codes for $k=280$ in Fig.~\ref{fig:VLF_rtpercentage}, the $m=\infty$ NB-LDPC code is 0.53 dB from Shannon limit. The NB-LDPC non-active feedback system in Fig.~\ref{fig:VLF_rtpercentage} uses ten rounds of single-bit feedback to operate within 0.65 dB of the Shannon limit with an average blocklength of less than 500 bits.
Similar analysis can also be done for higher-SNR AWGN and fading channels.
\bibliographystyle{IEEEtranTCOM}
{
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{"url":"https:\/\/serc.carleton.edu\/msu_nanotech\/nano_intro.html","text":"# Introduction to Nanoscience: Some Basics\n\nMost of what we think we know about the properties of materials and chemical or mechanical processes on the macroscale is either wrong, or at least incomplete, at the nanoscale. The \"rules\" are completely different. Materials of a given composition exhibit different chemical and mechanical properties on the nanaoscale compared to the macroscale. Surface energies are very large with respect to volumes, and therefore, the energetics of reactions (and rates) are different because of the increased influence of surface energies (e.g., interatomic coulombic interactions). As an introduction to this brave new world of nanoscience, start with these resources from the National Nanotechnology Initative (Nano.gov) Nanotechnology 101.\n\nSee the online presentations on:\n\n## Nanoscale--How Big?\n\nA nanometer is one-billionth of a meter. The National Nanotechnology Initiative provides some excellent comparisons of materials on different scales at Size of Nanoscale. The following are some graphics that demonstrate the relative size of numerous types of materials ranging from the nano- to the macro-scale. Nanomaterials have at least two dimensions that are between 1 and 100 nanometers in size. On this scale, interatomic (coulombic) forces become large, and must be considered when undertaking studies to characterize, experiment, and model the behaviors of nanomaterials.\n\n## Types of Nanoparticles\n\nNanomaterials (NMs) are functional materials consisting of particulates with at least one dimension below 100 nanometers (nm) (Grimsdale, A. C., and M\u00fcllen, K., 2005, The chemistry of organic nanomaterials: Angewandte Chemie International Edition, v. 44, no. 35, p. 5592-5629).\n\nHochella et al. (2019) Natural, incidental, and engineered nanomaterials and their impacts on the Earth system, Science, March 29, 2019, provide the following definitions:\n\n\u2022 Natural nanomaterials - A nanomaterial made by nature through (bio)geochemical or mechanical processes, without direct or indirect connection to a human activity or anthropogenic process.\n\u2022 Incidental nanomaterials -A nanomaterial unintentionally produced as a result of any form of direct or indirect human influence or anthropogenic process.\n\u2022 Engineered nanomaterials -A nanomaterial conceived, designed, and intentionally produced by humans. (Nanoparticles are being produced in any number of processes, and they are at large in the environment. We just haven't had the means to look for them\u2014but here they are, and they are doing work!)\n\u2022 Anthropogenic nanomaterials -Both incidental and engineered nanomaterials.\n\nHochella M., Aruguete, D., Kim, B., and Elwood Madden, A., 2012, Naturally occurring inorganic nanoparticles: general assessment and a global budget for one of earth's last unexplored major geochemical components, Pan Stanford Publishing Pte. Ltd., Nature's Nanostructures, 1-42 p. define the following:\n\n\u2022 Nanominerals are defined as minerals that only exist in the size range of approximately one to a few tens of nanometers inat least one dimension. Well-known examples include most clays and metal (oxyhydr)oxides (with ferrihydrite, an iron oxyhydroxide, as a type example).\n\u2022 Mineral nanoparticles are defined as minerals that have nano-dimensions, but these are minerals that can also exist in larger sizes.\n\u2022 Amorphous nanoparticles are the same, except without atomic structural order.\n\u2022 Thin Films and Coatings also occur on the nanoscale. These may be on the order from a few nanometers to a hundred nanometers in thickness, and may either be natural or engineered. Surface films and coatings commonly have compositions and structures that are not directly related to their underlying substrates. Thin films can be characterized by AFM imaging, and by a variety of analytical methods such as Auger Electron Spectroscopy, X-ray photoelectron spectroscopy, and Time-of-Flight SIMS.\n\u2022 Natural nanoparticles are ubiquitous in the Earth system. They have many sources; volcanic activity, cosmic dust, aeolian particles derived from weathering and wind erosion, chemical precipitation in many environments, biomineralization, and biomass combustion. Nanoparticles are important constituents of the atmosphere, oceans, soils, and in biota (produce or ingested by organisms). Detailed examples of Nanoparticles in the Earth system can be found on the related webpage on Nanoscience Topics in the Earth and Environmental Sciences.\n\n### Shapes of Nanoparticles\n\nHochella et al., (2008) Nanominerals, mineral nanoparticles, and earth systems: Science, v. 319, no. 5870, p. 1631-1635, provide some useful definitions for the shapes of nanomaterials: Nanoparticles \"are as small as ~1 nM and may range up to several tens of nanometers in at least one dimension\".\n\n\u2022 Nanosheets or nanofilms have one dimension in this size range; (e.g., clay minerals)\n\u2022 Nanorods have two dimensions in this size range\n\u2022 Nanoparticles have three dimensions in this size range\n\u2022 Nanotubes are nanoscale materials that have a tube-like structure; e.g., carbon nanotubes in the accompanying figure.\n\n## New Understanding of the Nano-World: Contexts and Concepts\n\nThe nanoscience revolution requires a fundamentally new understanding of the properties of matter at the nanoscale. These concepts, their significance, and applications are explored in more detail throughout this website. Here are some contexts and concepts that require special consideration when dealing with the nanoscale compared with micro- or mesoscale:\n\n\u2022 Quantum effects are dominant on the nanoscale. An example of this can be seen in the accompanying figure where the \"band gap\" between S and P orbitals changes significantly as a function of the number of atoms involved, thus changing the electronic properties from a semi-conductor (<18 atoms) to metallic. See: Thomas, O. C., Zheng, W., Xu, S., & Bowen Jr, K. H. (2002). Onset of metallic behavior in magnesium clusters. Physical review letters, 89(21), 213403. Interatomic coulombic forces also become important in understanding the properties of matter. See for example: Zhang, H., and Banfield, J. F., 2014, Interatomic Coulombic interactions as the driving force for oriented attachment: CrystEngComm, v. 16, no. 8, p. 1568-1578.\n\u2022 Properties of matter change as a function of particle size on the nanoscale. \"When particle size is made to be nanoscale, properties such as melting point, fluorescence, electrical conductivity, magnetic permeability, and chemical reactivity change as a function of the size of the particle\", (From Nano.gov).\n\u2022 \"Mineral nanoparticles also behave differently than larger micro- and macroscopic crystals of the same mineral. This observation violates aspects of the long-standing, formal definition of a mineral. Although definitions vary somewhat, depending on the source, the general consensus is that minerals are naturally occurring, crystalline substances having a characteristic and defined chemical composition (or compositional range in the case of solid solutions). For any particular composition, each mineral expresses a set of specific physical and chemical properties. Nanominerals and mineral nanoparticles satisfy these criteria, except that even with a fixed composition, they express a range of physical and chemical properties depending on their size and shape\". Hochella et al., (2008) Nanominerals, mineral nanoparticles, and earth systems: Science, v. 319, no. 5870, p. 1631-1635\n\u2022 Surface areas become quite large with respect to volume at the nanoscale.\n\u2022 Consequently, surface energies become quite large with respect to Gibbs Free Energy. Surface energies are mostly ignored in conventional thermodynamic studies of phase relations. See the pioneering work of Alex Navrotsky and her colleagues on the thermodynamics of nanoparticles. Nanoparticles contribute substantially to the overall energetics of a system, but this contribution is often overlooked.\n\u2022 Classical theory of nucleation and growth of crystals (one atom at a time into an ordered crystal structure) is not entirely accurate. Modern understanding of crystal growth indicates that crystal growth most commonly occurs by oriented aggregation of nanoparticles. See the recent work by DeYoreo, Carabello, Penn and Banfield among many others.\n\u2022 Control of Grain Size on Solubility on the Nanoscale: One of the most important properties of nanominerals is solubility. The size effect on dissolution has long been described by this modified version of Kelvin equation. S0 is the solubility of the bulk material. It is typically measured in conventional dissolution studies. S is the solubility of exceedingly fine particles. This equation indicates that, as the particle size decreases, the solubility is expected to exponentially increase. This plot shows the ratio of S to S0 vs particle radius for quartz grains. It was calculated according to this equation using the parameters of quartz grains. You can see that when the size is larger than 10-7 m that is 100nm, this ratio equals 1. Nothing changes with the size. Only when the size goes down to this nanoscale, the ratio substantially deviates from 1 and the size effect on solubility can be observed. However, this equation was proposed based on theoretic calculation. Very few experimental studies have been reported to support it. Image and explanation from Michael Hochella, Virginia Tech.\n\u2022 The contribution of nanoparticles to global bio\/geochemical cycling is rarely considered. The global budget for naturally occurring inorganic nanoparticles. All numbers are in units of terragrams (Tg = 1012 g). All italicized numbers are fluxes (Tg yr-1), and the numbers in rectangular boxes are reservoir sizes, if known. Some of the nanomaterial fluxes are listed as two components, explained as follows: for the volcanic input to the atmosphere, 20 Tg is due to SO2 aerosol formation, and 2 Tg is due to mineral ash. For the three aeolian inputs to the continents, continental shelves, and the open oceans, the first number is due to the 320 Tg continental mineral dust output, and the second number is due to the 22 Tg volcanic output. Image and explanation from Michael Hochella, Virginia Tech.\n\n## Processes that Occur on the Nanoscale\n\nThe interfaces between material surfaces and their environment at the atomic scale is where all the (chemical) action takes place! Some of the surface-mediated reactions that affect the rates and pathways of Earth processes and global biogeochemical cycling include:\n\n\u2022 Dissolution\/Precipitation Reactions\n\u2022 Catalytic Reactions\n\u2022 Sorption Reactions\n\u2022 Redox Reactions\n\n## Size Dependent Properties\n\n### Size Dependent Optical Properties\n\n\" Semiconductor nanocrystallites (quantum dots, QDs) whose radii are smaller than the bulk exciton Bohr radius constitute a class of materials intermediate between molecular and bulk forms of matter. Quantum confinement of both the electron and hole in all three dimensions leads to an increase in the effective band gap of the material with decreasing crystallite size (Figure 1). Consequently, both the optical absorption and emission of quantum dots shift to the blue (higher energies) as the size of the dots gets smaller.\" See the detailed description of this phenomenon from the Bawendi Research Group at MIT.\n\nSee the online presentation on Quantum Dots by Gerhard Klimeck, posted on nanoHUB, and Amiri et al., 2013, Preparation and Optical Properties Assessment of CdSe Quantum Dots. Materials Sciences and Applications, vol 4, p. 134-137.\n\nGold and silver nanoparticles also show size-dependence in their optical properties. An example of changing optical properties (color) as related to nanoparticle shape (prisms v. spheres) and size can be found at Dr. Shengli Zou's (Chemistry, University of Central Florida) website on Optical properties of nanoparticles and their applications. An explanation of this phenomenon, from the nanoComposix website on Nanoparticles: Optical Properties: \"Gold nanoparticles absorb and scatter light with extraordinary efficiency. Their strong interaction with light occurs because the conduction electrons on the metal surface undergo a collective oscillation when they are excited by light at specific wavelengths. This oscillation is known as a surface plasmon resonance (SPR), and it causes the absorption and scattering intensities of gold nanoparticles to be much higher than identically sized non-plasmonic nanoparticles. Gold nanoparticle absorption and scattering properties can be tuned by controlling the particle size, shape, and the local refractive index near the particle surface\". An example of the application of Au nanoparticles to biomedicine can be found in the article Gold and Silver Nanoparticles: Synthesis Methods, Characterization Routes and Applications towards Drugs byDhalid Alaquad and Tawfik Saleh, 2016, Journal of Environmental and Analytical Toxicology, 6:384. doi:10.4172\/2161-0525.1000384\n\n### Size Dependent Crystallization\n\nClassic theory of nucleation and growth of crystals assumes that crystals grow by ordering atoms (monomers) one at at a time in prescribed positions in the crystal structure. However, modern studies of growth mechanisms of crystals shows that crystals more typically grow by aggregation of nanoparticles. Crystallization pathways may involve formation of multi-ion complexes from dissociated ions, to organization of these complexes into isolated nanoparticles with very short range order, to oriented aggregates of nanoparticles, and ultimate formation of macroscopic crystals. Examples of this type of crystallization pathway of crystals forming from aggregates of nanoparticles can be found in calcium carbonates, calcium phosphates, ferric hydroxides and hydroxyl-sulfates, and aluminosilicate nanoparticles. For a more detailed description of crystallization by particle attachment (CPA), see DeYoreo et al., 2015, Crystallization by particle attachment in synthetic, biogenic, and geologic environments, Science, vol 349, issue 6247, aaa6760-1.\n\n### Size Dependent Dissolution\n\nOne of the most important properties of nanominerals is solubility. The size effect on dissolution has long been described by this modified version of Kelvin equation.The Kelvin equation may be written in the form:\nln(p\/p_0) = (2gammaV_m)\/(rRT)\nwhere p is the actual vapour pressure, p0 is the saturated vapour pressure, \u03b3 is the surface tension, Vm is the molar volume of the liquid, R is the universal gas constant, r is the radius of the droplet, and T is temperature. In the diagram, So is the solubility of the bulk (macro) material and can be substituted for Po. It is typically measured in conventional dissolution studies. S is the solubility of fine (nano-scale) particles and can be substituted for P. This equation indicates that, as the particle size decreases, the solubility is expected to exponentially increase. This plot shows the ratio of S to So vs particle radius for quartz grains. It was calculated according to this equation using the parameters of quartz grains. You can see that when the size is larger than 10-7 m that is 100nm, this ratio equals ~1. Nothing changes with the size. Only when the size goes down to the nanoscale, the ratio substantially deviates from 1 and the size effect on solubility can be observed. However, this equation was proposed based on theoretic calculation. Very few experimental studies have been reported to support it. The image and explanation was contributed by Michael Hochella, Virginia Tech, NanoEarth NNCI project.\n\nBut wait! There's more! In addition, the range of validity of the modified kelvin equation has been questioned. This equation is a correction to the kelvin equation. This added term shows the effect of surface charge. In this case, the increase in solubility is not infinite as the size is reduced. There is a critical size, which is always at the nanoscale. When the size is smaller than the critical value, the solubility will decrease with the shrinking of particle size. Moreover, a recent study found that dissolution slows down or even stop at undersaturation when the size of dissolution etch pit is smaller than a critical size. However, this study was conducted on bulk mineral surface, not on nanoparticles. Therefore, the size effect on dissolution is not totally understood. Our work is the first study to really see the dissolution of nanoparticles under microscope. (Tang et al., 2001, J. Am. Chem. Soc., 123).\n\nHere are some examples of size-dependent dissolution reactions at the nanoscale:\n\n\u2022 Schmidt, J., and Vogelsberger, W. (2006). Dissolution kinetics of titanium dioxide nanoparticles: The observation of an unusual kinetic size effect, J. Phys. Chem. B, 110, pp. 3955\u20133963.\n\u2022 Roelofs, F., and Vogelsberger, W. (2004). Dissolution kinetics of synthetic amorphous silica in biological-like media and its theoretical description, J. Phys. Chem. B, 108, pp. 11308\u201311316.\n\u2022 Yang, Z. H., and Xie, C. S. (2006). Zn2+ release from zinc and zinc oxide particles in simulated uterine solution, Colloids Surf. B, 47, pp. 140\u2013145.\n\u2022 Liu, J., Aruguete, D. A., Jinschek, J. R., Rimstidt, J. D., and Hochella, M. F. (2008). The non-oxidative dissolution of galena nanocrystals: Insights into mineral dissolution rates as a function of grain size, shape, and aggregation state, Geochim. Cosmochim. Ac., 72, pp. 5984\u20135996.\n\u2022 Erbs, J. J., Gilbert, B., and Penn, R. L. (2008). Influence of size on reductive dissolution of six-line ferrihydrite, J. Phys. Chem. C, 112, pp. 12127\u201312133.\n\u2022 Cwiertny, D. M., Hunter, G. J., Pettibone, J. M., Scherer, M. M., and Grassian, V. H. (2009). Surface chemistry and dissolution of alpha-FeOOH nanorods and microrods: environmental implications of size-dependent interactions with oxalate, J. Phys. Chem. C, 113, pp. 2175\u20132186.\n\u2022 Rubasinghege, G., Lentz, R. W., Park, H., Scherer, M. M., and Grassian, V. H. (2010). Nanorod dissolution quenched in the aggregated state, Langmuir, 26, pp. 1524\u20131527.\n\nWe have a lot to learn about the solubility of particles at the nanoscale!","date":"2023-04-01 20:52:52","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6027002334594727, \"perplexity\": 2707.925404756202}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2023-14\/segments\/1679296950247.65\/warc\/CC-MAIN-20230401191131-20230401221131-00489.warc.gz\"}"}
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Jerome Scott, 28, 130 Shady Oak Court, Houma, stop lamps and turn signals required on new motor vehicles, tail lamps required, possession with intent to distribute marijuana, possession or distribution of drug paraphernalia.
Latoya Brown, 29, 125 Park Ave., Thibodaux, contempt of court, tail lamps required, failure to display license plate, two counts child passenger restraint system.
Darrel Ledet, 48, 124 W. 142nd St., Cut Off, resisting an officer, domestic-abuse battery, domestic-abuse aggravated assault.
Monique Jones, 28, 275 Sanders St., Thibodaux, domestic-abuse battery with child endangerment.
Jesus Guiterrez, 19, 929 St. Charles St., Thibodaux, possession of marijuana.
Donald Lefleur, 59, 804 Manchester Manor Road, Thibodaux, operating a vehicle while under suspension for certain prior offenses, third-offense DWI, possession of alcoholic beverages in motor vehicles.
Rontrelle Square, 27, 900 Forty Appent Road, Thibodaux, contempt of court.
Humberto Ortiz, 32, 518 Sycamore St., Thibodaux, second-offense DWI, driving on roadway laned for traffic, no driver's license.
Brent Gervais, 46, 401 Louise Drive, Raceland, domestic-abuse battery with child endangerment, cruelty to juveniles with force.
Rod Blanchard, 59, 206 E. 85th St., Apt. 102, Cut Off, second-offense DWI, stop/yield signs.
|
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"redpajama_set_name": "RedPajamaC4"
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\section{Introduction}
\label{sec::introduction}
Knowledge Graphs (KGs) are large-scale collections of facts to represent real-world \textit{entities} and their \textit{interconnections}. KGs have gained widespread use in different domains including computer science, medicine, bioinformatics, education, and biology, among others~\cite{Refinement}. Several KGs such as Wikidata, YAGO, and Bio2RDF, to name a few, are openly available. As well, many private organizations such as Amazon have created KGs for different purposes such as \textit{similarity analysis}. The widespread adoption of KGs has highlighted the need for employing efficient Data Management Systems (DMSs). However, the rich diversity in \textit{variety} and \textit{size} of KG content pose challenges to DMSs to \textit{store} and \textit{query} KGs~\cite{Growing3,Growing2}. The individual \textit{queries} and applications executed on these systems have also become \textit{highly diverse}. These have been addressed in part through the development of a range of DMSs such as \textit{columnar-} and \textit{graph-based} stores. The consensus appears to be that a single, \textit{one-size-fits-all} DMS is unlikely to emerge for efficient KG query processing~\cite{watdiv}.
We develop a solution to this problem that is inspired by Ashby's First Law of Cybernetics~\cite{ashby} and Stonebraker et. al.~\cite{onesize2} which can be paraphrased in this context to state that the variety in the solution architecture should be greater than or at least equal to that of the variety displayed by the data and the queries. The \textit{requisite variety} is to be achieved through an architecture based on \textit{polyglot} model of query processing and access languages supported by a design that can analyze individual queries and match each to the likely best-performing database engine. From a conceptual standpoint, genuine polygloty will imply the requirement of employing multiple DMSs and friction-free translation across the different query and access languages at the polyglot access layer in order to optimize query execution performance. However, a review of the extant research and practitioner literature such as~\cite{fowler} reveals that polygloty has typically been interpreted as using different physical data stores depending on the type of application. For instance, traditional relational DMSs for financial data, document-stores for product catalog data, key-value stores for user activity logs, etc. However, it is not difficult to see that these interpretations typically suffer from lack of integration across the entire data which may lead to \textit{balkanized} data islands.
Supporting multiple data models against a single, integrated backend can potentially address the growing requirements for performance~\cite{lookliu}. Based on this, over the last few years, there has been growing interest in employing multiple DMSs for query processing. This interest led to the development of some open-source platforms such as Apache Drill\footnote{\url{https://drill.apache.org}} as well as some academic prototypes such as~\cite{bigdawg}. However, these solutions mainly focused on applications such as \textit{ETL}, \textit{machine learning}, \textit{stream processing}, \textit{OLAP}, data integration, etc., and somewhat less attention has been paid to efficient KG query processing which is the focus of this paper.
We present \textit{SymphonyDB}, a prototype that provides polyglot support for KG query processing. It is a multi-database approach supported by an access management layer to provide a unified query interface for accessing the underlying DMSs. This layer receives the incoming workloads in the form of SPARQL queries and routes each of them to one (or more than one) of the more likely to be efficiently matched DMSs. A suitable Just-In-Time (JIT) query translation may be needed if the underlying DMSs accept different query languages. Handling such a JIT translation process is the essential responsibility of the access layer. Currently, \textit{SymphonyDB} has included \textit{Virtuoso}, \textit{Blazegraph}, \textit{\mbox{RDF-3X}}, and \textit{\mbox{MongoDB}} as representative DMSs. SymphonyDB has the potential to achieve \textit{requisite variety}.
\newpage
Our contributions include:
\begin{itemize}
\item
Presenting a prototype, \textit{SymphonyDB} that can match KG query requirements with the best combination of DMS and storage representation to achieve consistent query execution performance.
\item
Experimental evaluation and comparative performance analysis of \textit{SymphonyDB} against representative single DMSs in supporting different archetypal KG query types. The source code, data, and other artifacts have been made available at \url{https://github.com/m-salehpour/SymphonyDB}.
\end{itemize}
\begin{figure}[ht]
\centering\includegraphics[width=0.4\textwidth]{Figs/kgnew.png}
\caption{An example of a simple Knowledge Graph (based on the LikedSPL KG).}
\label{fig::kg}
\end{figure}
\section{Background}
\label{sec::background}
In this section, we present some preliminary information about the archetypal KG query types using a \textit{human-readable} example depicted in \reffig{fig::kg}. This shows a small extract from the LinkedSPL KG which includes all sections of FDA-approved prescriptions and over-the-counter drug package inserts from DailyMed. The content of this KG subset can be represented by the following RDF triples\footnote{The content of a KG usually represents a large \textit{set of triples} of the form $<$subject predicate object$>$ creating a graph~\cite{CharacteristicsRDF}}:
\begin{figure}[ht]
\centering
\begin{lstlisting}[language=SQL,basicstyle=\scriptsize\ttfamily,numbers=left,tabsize=2,showstringspaces=false,numberstyle=\scriptsize\tt\color{gray}]
CISPLATIN UNII "Q20Q"
CISPLATIN adverse_reaction "Nausea"
CISPLATIN organization "Bedford Lab"
CISPLATIN FDA_Code "Table_1326"
Table_1326 marker "TPMT"
Table_1326 ID 305
Table_1326 CUI 2555
Table_1326 Xref "gene_PA356"
\end{lstlisting}
\label{fig:triples-KG-fda}
\end{figure}
An example of a query\footnote{Queries are formulated in the form of SPARQL. In this paper, we assume that the reader is familiar with the basic concepts of querying KG, e.g., the SELECT clauses.} is given below. It asks for ``UNII'' of a given drug (i.e., ``CISPLATIN''). In this query, ``?unii'' is a variable to return the result which is ``Q20Q''.
\begin{figure}[ht]
\centering
\begin{lstlisting}[language=SQL,basicstyle=\ttfamily,numbers=left,tabsize=2,showstringspaces=false,numberstyle=\scriptsize\tt\color{gray}]
SELECT ?unii
WHERE {
CISPLATIN UNII ?unii . }
\end{lstlisting}
\end{figure}
KG queries may contain a set of triple patterns such as \textit{\mbox{``CISPLATIN UNII ?unii''}} in which the subject, predicate, and/or object can be a variable. Each triple pattern typically returns a subgraph. This resultant subgraph can be further \textit{joined} with the results of other triple patterns to return the final resultset. In practice, there are three major types of join queries: (i) subject-subject joins (aka, star-like), (ii) subject-object joins (aka, chain-like or path), and (iii) tree-like (i.e., a combination of subject-subject and subject-object joins). These query types are explained below.
\textbf{Subject-subject joins.} A subject-subject join is performed by a DMS when a KG query has at least two triple patterns such that the predicate and object of each triple pattern is a given value (or a variable), but the subjects of both triple patterns are replaced by the \textit{same} variable. For example, the following query looks for all drugs for which their UNII and adverse reactions are equal to the given values (the result will be ``CISPLATIN'').
\begin{figure}[ht]
\centering
\begin{lstlisting}[language=SQL,basicstyle=\ttfamily,numbers=left,tabsize=2,showstringspaces=false,numberstyle=\scriptsize\tt\color{gray}]
SELECT ?x
WHERE {
?x UNII "Q20Q" .
?x adverse_reaction "Nausea" . }
\end{lstlisting}
\end{figure}
\textbf{Subject-object joins.} A subject-object join is performed by a DMS when a KG query has at least two triple patterns such that the subject of one of the triple patterns and the object of the other triple pattern are replaced by the same variable. For example, the following query looks for all drugs that their CUI\footnote{CUI (aka, RxCUI) is a unique drug identifier} is equal to 2555 (``CISPLATIN'' is the result).
\begin{figure}[ht]
\centering
\begin{lstlisting}[language=SQL,basicstyle=\ttfamily,numbers=left,tabsize=2,showstringspaces=false,numberstyle=\scriptsize\tt\color{gray}]
SELECT ?y
WHERE {
?x CUI 2555 .
?y FDA_Code ?x . }
\end{lstlisting}
\end{figure}
\textbf{Tree-like joins.} A tree-like join consists of a \textit{combination} of subject-subject and subject-object joins. For example, the following query looks for the ``Xref'' of all drugs with ``UNII'' of ``Q20Q'' that their ``CUI'' is equal to 2555 (the result will be ``gene\_PA356'').
\begin{figure}[ht]
\centering
\begin{lstlisting}[language=SQL,basicstyle=\ttfamily,numbers=left,tabsize=2,showstringspaces=false,numberstyle=\scriptsize\tt\color{gray}]
SELECT ?y
WHERE {
?x UNII "Q20Q" .
?x FDA_Code ?z .
?z CUI 2555 .
?z Xref ?y . }
\end{lstlisting}
\end{figure}
In addition to the query types, we provide a brief explanation of query \textit{selectivity} and \textit{optional} patterns. The selectivity of a query is typically represented by the fraction of triples matching the query pattern. Based on this, each query type can typically be either \textit{high-selective} or \textit{low-selective}. A query can be considered as low-selective when a large number of triples (as compared to the total number of stored triples) needs to be scanned before returning the resultset~\cite{selectivity}. Moreover, as explained previously, queries return resultsets only when the entire query pattern matches the content of the KG. However, some queries may contain optional patterns to allow KG queries to return a resultset even if the optional part of the query is not matched since completeness and adherence of KG content to their formal ontology specification is not always enforced.
\section{SymphonyDB}
\label{sec::SymphonyDB}
In this section, we present \textit{SymphonyDB}, a prototype that provides polyglot support for KG query processing. SymphonyDB is a multi-database solution supported by an access management layer. Currently, \textit{SymphonyDB} includes \textit{Virtuoso}, \textit{Blazegraph}, \textit{\mbox{RDF-3X}}, and \textit{\mbox{MongoDB}} as representative DMS types.
An abstract overview of the interactions across DMSs and the access layer is expressed using the pseudocode in~\refalgo{algo::polyglot}.
The polyglot access management layer is the entry point for query processing over stored data, providing a unified query interface for accessing the underlying KGs.
This layer receives the incoming SPARQL queries and labels each query based on its characteristics into three target taxonomies, namely: subject-subject, subject-object, and tree-like (line 1 in \refalgo{algo::polyglot}).
The labels are used for routing each query to one (or more than one) of the employed DMSs to increase optimal execution.
We defer details of the labelling approach to~\refsec{sec:label}.
After determining which DMS should be used for each query (line 2 in \refalgo{algo::polyglot}), a suitable JIT query translation may be needed for further query execution (line 3-5 in \refalgo{algo::polyglot}).
For example, the incoming query is written in SPARQL, but it is labeled to be run using MongoDB which cannot support SPARQL directly as an input query language.
In this case, the incoming query needs to be translated JIT into an equivalent JavaScript-like query, MQL, to be executed over MongoDB.
Handling this JIT translation process is one of the responsibilities of the access layer.
The following sections describe the detail of each step.
\begin{algorithm}[htbp]
\DontPrintSemicolon
\SetAlgoLined
\KwInput{SPARQL queries (applications' workload)}
\Initialization{Let q be an incoming SPARQL query}
$qLabel \longleftarrow$ \textbf{Query\_labeling}($q$)\;
$DMS \longleftarrow$ \textbf{DMS\_select}($qLabel$)\;
\If{ $DMS == MongoDB$}{
$q \longleftarrow$ \textbf{Translate\_MQL}($q$)\;
}
$qResult \longleftarrow$ \textbf{Route\_execute}($q$)\;
\textbf{Return\_result}($qResult$)\;
\caption{SymphonyDB KG query processing}
\label{algo::polyglot}
\end{algorithm}
\subsection{Database Management System Layer}
\label{sec:DMS}
Selection of adequate DMSs is vital in maximizing the power of a multi-database system. In our prototype, RDF-3X was selected as a candidate as it is an open-source system widely used in a range of studies such as~\cite{watdiv}.
One vital property is that RDF-3X creates exhaustive indexes on all permutations of triples along with their binary and unary projections.
Its query processor is designed to aggressively leverage cache-aware hash and merge joins.
Virtuoso was selected since it is already employed as the DMS of choice for a broad range of KGs, for example the Linked Data for the Life Sciences project\footnote{Available online: \url{https://bio2rdf.org/sparql}}.
Virtuoso's physical design is based on a relational table with three columns\footnote{In case of loading named graphs, another column is added, called C.} for S, P, and O (S: Subject, P: Predicate, and O: Object) and carries multiple bitmap indexes over that table to provide a number of different access paths.
Most recently, Virtuoso added columnar projections to minimize the on-disk footprint associated with RDF data storage. Blazegraph\footnote{It is alleged that the Amazon Neptune is based on Blazegraph.} was selected since it is the DMS behind Wikidata, i.e., a KG constructed from the content of Wikimedia sister projects including Wikipedia, Wikivoyage, Wiktionary, and Wikisource.
Blazegraph's physical design is based on \mbox{B+trees} to store KGs in the form of ordered data.
Blazegraph typically uses the following three indexes: SPO, POS, and OSP. MongoDB was selected as a representative document-store.
Its efficacy for executing queries over KGs has not been researched extensively but some academic prototypes such as~\cite{Jignesh,frankPhdThesis,tomaszuk2010document} have already shown document-stores efficacy in similar contexts.
\subsection{Query Labeling and Execution}
\label{sec:label}
Each SPARQL query can be viewed as a directed graph where nodes are formed by the subjects and objects of the query's triple patterns and edges are the properties of these patterns. Based on this, each SPARQL query can be classified into shape-specific categories.
At this stage, we confine our focus to the following query types: subject-subject, subject-object, and tree-like queries (more details can be found in~\refsec{sec::background}). Examples of these types are shown in~\reffig{fig::qlabel} where each node represents a variable connected through predicates as edges forming a graph-like structure.
\begin{figure}[hbt]
\centering\includegraphics[width=0.45\textwidth]{Figs/qlabel.png}
\caption{(a) Subject-subject query pattern (b) Subject-object query pattern (c) Tree-like query pattern}
\label{fig::qlabel}
\end{figure}
In general, each query has at least one source variable as shown in \reffig{fig::qlabel} using nodes represented with solid black colors. Any node with both incoming and outgoing edges represents a shared variable of a query. Query patterns are typically recognizable by the position of shared variables in a query.
Inspired by~\cite{peter-heuristic}, we utilize a heuristic-based approach to exploit the syntactic and the structural variations of patterns in a given SPARQL query in order to label it.
On this foundation, SymphonyDB begins by finding the source variable of a given query, looking for all immediate neighbor nodes, or shared variables, with one edge distance away.
From there, it then iteratively visits nodes further away until all nodes are visited, using a queue data structure to store visited nodes at each stage.
Upon finishing the traversal, the query will then be labeled according to the characteristics of its variables and graph.
A subject-subject label is applied if all nodes are only immediate neighbors of the source variable with one edge distance away (\mbox{\reffig{fig::qlabel} (a)}).
A subject-object label, however, follows a pattern where there is just one outgoing edge from a node at each stage (starting from the source variable) where there is a final node with no outgoing edge as depicted in \mbox{\reffig{fig::qlabel} (b)}.
Finally, queries that contain a combination of both patterns are labeled as tree-like (\mbox{\reffig{fig::qlabel} (c)}).
In addition to the query patterns, SymphonyDB checks the existence of modifiers as well, where modifiers are keywords such as ``\texttt{LIMIT}'' that are recognizable by the query parser and lexical analyzer implemented in SymphonyDB.
The following heuristics are then used to select one or more of the integrated DMSs.
Queries are routed to both MongoDB and RDF-3X if the following characteristics are present: (1)~it contains only a single triple pattern, (2)~it is a query with subject-subject joins, (3)~it contains no modifiers in the query, and (4)~it contains no optional patterns.
Alternatively, if a query contains subject-object joins, it is routed to Blazegraph, and finally all other queries (tree-like queries with or without optional patterns) are routed to Virtuoso.
As an intermediate step, any queries being routed to MongoDB must run through the JIT query translation to be translated from SPARQL to MQL. This is explained in the following section.
\subsection{Polyglot Access Management: Query Translation}
\label{sec:translate}
Query translation provides the extensibility for SymphonyDB to access various DMSs with differing query languages.
Currently, the only integrated DMS that requires this functionality is MongoDB as it provides its own query language MQL, which is a Javascript-like, object-oriented imperative language.
This contrasts with SPARQL, a domain-specific declarative language~\cite{tomaszuk2010document}, which is not supported natively by MongoDB, however, it is feasible to map SPARQL to MQL in most cases.
\reffig{fig::translation} depicts the logic flow of SymphonyDB's JIT query translation, showing that each query is analyzed lexically and tokenized based on the SPARQL query syntax.
The lexical analyzer dissects the SPARQL query into logical units of one or more characters that have a shared meaning, often referred to as \textit{tokens}.
For instance, ``\texttt{WHERE}'' is a token representing a keywords, whereas ``\texttt{.}'' is an identifier and ``\texttt{=}'' is a sign.
In parallel, it parses each query with regard to the grammatical description of the SPARQL language to generate the corresponding syntax tree.
The semantic analyzer then produces an \textit{operator graph} containing information about projection variables, join patterns, conditions and modifiers.
Finally, once the semantic analysis has completed, heuristic techniques are used to map the operator graph to MQL and translate the query.
\begin{figure}[h]
\centering\includegraphics[width=0.45\textwidth]{Figs/mql.png}
\caption{The query translation logic flow}
\label{fig::translation}
\end{figure}
Similar to~\cite{tomaszuk2010document,frankPhdThesis,frankMappping}, SymphonyDB maps each SPARQL to MQL using a collection of rules.
\reftab{table:rules1} shows SPARQL expressions and their equivalent MQL query string, along with additional set of rules to map SPARQL query patterns to MQL illustrated in~\reftab{table:rules2}.
For example, to translate subject-subject join queries, SymphonyDB uses the ``\texttt{\$match}'' aggregation pipeline operator of MongoDB to filters documents and pass a subset of the documents that match the specified condition(s) to the next pipeline stage.
It also uses ``\texttt{\$lookup}'' aggregation pipeline operator of MongoDB to translate joins.
\begin{table}[h]
\caption{SPARQL expressions representation and their equivalent MQL expressions}
\centering
\begin{tabular}{l | l }
\toprule
SPARQL & MQL \\ [0.7ex]
\hline\hline
Exists ($<$e1$>$) & $<$e1$>$:\{\$exists:true\} \\
Not Exists ($<$e1$>$) & $<$e1$>$:\{\$exists:false\} \\
($<$e1$>$ \&\& $<$e2$>$ ) & \{\$and:[\{$<$e1$>$\},\{$<$e2$>$\}] \\
($<$e1$>$ $||$ $<$e2$>$ ) & \{\$or:[\{$<$e1$>$\},\{$<$e2$>$\}] \\
!($<$e1$>$) & \{\$not:\{$<$e1$>$\} \\
($<$e1$>$ = $<$e2$>$ ) & \{\$eq:[\{$<$e1$>$\},\{$<$e2$>$\}] \\
($<$e1$>$ != $<$e2$>$ ) & \{\$ne:[\{$<$e1$>$\},\{$<$e2$>$\}] \\
($<$e1$>$ $>$ $<$e2$>$ ) & \{\$gt:[\{$<$e1$>$\},\{$<$e2$>$\}] \\
($<$e1$>$ $>=$ $<$e2$>$ ) & \{\$gte:[\{$<$e1$>$\},\{$<$e2$>$\}] \\
($<$e1$>$ $<$ $<$e2$>$ ) & \{\$lt:[\{$<$e1$>$\},\{$<$e2$>$\}] \\
($<$e1$>$ $<=$ $<$e2$>$ ) & \{\$lte:[\{$<$e1$>$\},\{$<$e2$>$\}] \\
\end{tabular}
\label{table:rules1}
\end{table}
\begin{table*}
\caption{Sample rules used to translate different join patterns of SPARQL queries to their equivalent MQL query string}
\resizebox{0.9\textwidth}{!}{
\centering
\begin{tabularx}{\textwidth}{l|l|l}
\cmidrule(lr){1-3}
\backslashbox{Pattern}{Query}& SPARQL (triple patterns) & MQL (aggregate pipeline)\\
\cmidrule(lr){1-3} \cmidrule(lr){1-3}
\multicolumn{1}{ c| }{\multirow{3}{*}{Single Triple Pattern} } &
\multicolumn{1}{ l| }{``subject'' ``predicate'' ``object''} & \{\$match:\{ subject\_id:``subject'', ``predicate'':``object''\}\}\\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{?subject ``predicate'' ``object''} & \{\$match:\{ subject\_id:\{\$exists:true\}, ``predicate'':``object''\}\}\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{``subject'' ``predicate'' ?object} & \{\$match:\{ subject\_id:``subject'', ``predicate'':\{\$exists:true\}\}\}\\
\cmidrule{1-3} \cmidrule{1-3}
\multicolumn{1}{ c| }{\multirow{9}{*}{Subject-subject join} } &
\multicolumn{1}{ l| }{\{?subject ``predicate1'' ``object1'' .} & \{\$match:\{ subject\_id:\{\$exists:true\}, ``predicate1'':``object1'',``predicate2'':``object2''\}\} \\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{~?subject ``predicate2'' ``object2'' .\}} & \\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{\{?subject ``predicate1'' ?object1~ .} & \{\$match:\{ subject\_id:\{\$exists:true\}, ``predicate1'':\{\$exists:true\},``predicate2'':``object2''\}\} \\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{~?subject ``predicate2'' ``object2'' .\}} & \\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{\{``subject'' ``predicate1'' ``object1'' .} & \{\$match:\{ subject\_id:``subject'', ``predicate1'':\{\$exists:true\},``predicate2'':``object2''\}\} \\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{~``subject'' ``predicate2'' ``object2'' .\}} & \\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \\
\cmidrule{1-3} \cmidrule{1-3}
\multicolumn{1}{ c| }{\multirow{12}{*}{Subject-object join} } &
\multicolumn{1}{ l| }{\{?subject ``predicate1'' ?object1~ .} & \{\{\$match:\{ subject\_id:\{\$exists:true\}\}\},\\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{~?object1 ``predicate2'' ``object2'' .\}} & \{\$lookup:\{ from: ``colc\_name'', localField: ``predicate1'', foreignField: ``subject\_id'', as: ``join\_field''\}\},\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \{\$match:\{ ``join\_field.predicate2'':``object2''\}\}\}\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{\{?subject ``predicate1'' ?object1~ .} & \{\{\$match:\{ subject\_id:\{\$exists:true\}\}\},\\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{~?object1 ``predicate2'' ?object2 .\}} & \{\$lookup:\{ from: ``colc\_name'', localField: ``predicate1'', foreignField: ``subject\_id'', as: ``join\_field''\}\},\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \{\$match:\{ ``join\_field.predicate2'':\{\$exists:true\}\}\}\}\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{\{``subject'' ``predicate1'' ?object1~ .} & \{\{\$match:\{ subject\_id:``subject''\}\},\\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{~?object1 ``predicate2'' ?object2 .\}} & \{\$lookup:\{ from: ``colc\_name'', localField: ``predicate1'', foreignField: ``subject\_id'', as: ``join\_field''\}\},\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \{\$match:\{ ``join\_field.predicate2'':\{\$exists:true\}\}\}\}\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \\
\cmidrule{1-3} \cmidrule{1-3}
\multicolumn{1}{ c| }{\multirow{12}{*}{Tree-like join} } &
\multicolumn{1}{ l| }{\{?subject ``predicate1'' ?object1~ .} & \{\{\$match:\{ subject\_id:\{\$exists:true\}, ``predicate1'':\{\$exists:true\}, ``predicate2'':\{\$exists:true\} \}\},\\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{~?subject ``predicate2'' ?object2~ .} & \{\$lookup:\{ from: ``colc\_name'', localField: ``predicate2'', foreignField: ``subject\_id'', as: ``join\_field''\}\},\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{~?object2 ``predicate3'' ``object3'' .\}} & \{\$match:\{ ``join\_field.predicate3'':``object3''\}\}\}\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{\{``subject'' ``predicate1'' ``object1''~ .} & \{\{\$match:\{ subject\_id:``subject'', ``predicate1'':``object1'', ``predicate2'':\{\$exists:true\} \}\},\\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{~``subject'' ``predicate2'' ?object2~ .} & \{\$lookup:\{ from: ``colc\_name'', localField: ``predicate2'', foreignField: ``subject\_id'', as: ``join\_field''\}\},\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{~?object2 ``predicate3'' ?object3 .\}} & \{\$match:\{ ``join\_field.predicate3'':\{\$exists:true\}\}\}\}\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{\{``subject'' ``predicate1'' ?object1~ .} & \{\{\$match:\{ subject\_id:``subject'', ``predicate1'':\{\$exists:true\} \}\},\\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{~?object1 ``predicate2'' ?object2~ .} & \{\$lookup:\{ from: ``colc\_name'', localField: ``predicate1'', foreignField: ``subject\_id'', as: ``join\_field''\}\},\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{~?object1 ``predicate3'' ``object3'' .\}} & \{\$match:\{ ``join\_field.predicate2'':\{\$exists:true\}, ``join\_field.predicate3'':``object3''\}\}\}\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \\
\cmidrule{1-3} \cmidrule{1-3}
\end{tabularx}
}
\label{table:rules2}
\end{table*}
\section{Experimental Evaluation}
\label{sec::Experimental-Setting}
In this section, we report the experimental setup and details of the KG benchmark datasets that are used in the experimental evaluation.
This includes detailed information about DMSs' configuration, indexing, data loading process as well as our computational platform.
The query performance of SymphonyDB and a range of DMSs are evaluated and presented below.
\subsection{Evaluation Datasets and Queries}
We select four well-known KG datasets with a collection of relevant queries that are publicly available, where a number have been used in previous studies~\cite{cellcycle,biobench}.
The datasets are as follows.
\textbf{Allie}\footnote{\url{http://allie.dbcls.jp/}} is a KG surrounding life sciences, containing abbreviations and long forms utilized within the field.
\textbf{Cellcycle}\footnote{\url{ftp://ftp.dbcls.jp/togordf/bmtoyama/cellcycle/}} contains orthology relations for proteins consiting of ten sub-graphs constituting the cell cycle.
In our experiments, however, we integrated all ten sub-graphs into a single KG dataset without modifying any content.
\textbf{DrugBank}\footnote{\url{https://download.bio2rdf.org/files/current/drugbank/drugbank.html}} contains bioinformatics and chemoinformatics resources which include detailed drug (chemical, pharmacological, pharmaceutical, etc.) and comprehensive drug targets (sequence, structure and pathway information) in the dataset.
\textbf{LinkedSPL\footnote{\url{https://download.bio2rdf.org/files/current/linkedspl/linkedspl.html}}} includes all sections of FDA-approved prescriptions and over-the-counter drug package inserts from DailyMed. \reftab{table::c5:kgs} depicts the statistical information related to the above KGs.
\begin{table}[tbp]
\centering
\caption{KGs that were used to run the experiments}
\begin{tabularx}{\linewidth}{l r r r r}
\toprule
\multirow{2}{*}{KG}& \multicolumn{4}{c}{Statistics} \\
\cmidrule{2-5}
& Sub. (\#) & Pre. (\#)& Obj. (\#) & Triples (\#)\\
\midrule
\midrule
Allie & 19,227,252 & 26 & 20,280,252 & 94,404,806 \\
Cellcycle & 21,745 & 18 & 142,812 & 322,751\\
DrugBank & 19,693 & 119 & 276,142 & 517,023 \\
LinkedSPL & 59,776 & 104 & 719,446 & 2,174,579 \\
\bottomrule
\end{tabularx}
\label{table::c5:kgs}
\end{table}
\begin{table*}
\caption{Types of the queries. $SS^{a*}$: Subject-subject join, $SO^{b*}$: Subject-object join, $Co^{c*}$: combination of $SS$ and $SO$, $OPT^{d*}$: Optional pattern, $Fil^{e*}$: Filter, $ORD^{f*}$: Order by, $Lim^{g*}$: Limit, $OFF^{h*}$: Offset, $STP^{i*}$: Single triple pattern (no join)}
\centering
\begin{tabular}{cccccccccccc}
\toprule
\multirow{3}{*}{Benchmark}& \multicolumn{11}{c}{Types} \\
\cmidrule{3-12}
& Query & $SS^{a*}$ & $SO^{b*}$ & $Co^{c*}$ & $OPT^{d*}$ & Selective & $Fil^{e*}$ & $ORD^{f*}$ & $Lim^{g*}$ & $OFF^{h*}$ & $STP^{i*}$\\
\hline\hline
\multicolumn{1}{ c }{\multirow{5}{*}{Allie} } &
\multicolumn{1}{ c }{Q1} & & & & & & & & & & \cmark\\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q2} & & & & & & \cmark & & & & \cmark\\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q3} & \cmark & & & & & & & & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q4} & & \cmark & & & & & & \cmark & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q5} & \cmark & & & & & & \cmark & \cmark & & \\ \cline{1-12}
\hline
\multicolumn{1}{ c }{\multirow{6}{*}{Cellcycle} } &
\multicolumn{1}{ c }{Q1} & & & \cmark & & & & & & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q2} & & & \cmark & \cmark & & & & & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q3} & & & \cmark & & \cmark & & & & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q4} & & & \cmark & & & & & & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q5} & \cmark & & & \cmark & & & & & & \\ \cline{1-12}
\hline
\multicolumn{1}{ c }{\multirow{5}{*}{DrugBank} } &
\multicolumn{1}{ c }{Q1} & \cmark & & & \cmark & & & & \cmark & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q2} & \cmark & & & \cmark & & & \cmark &\cmark &\cmark & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q3} & & & \cmark & & & & & & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q4} & & & \cmark & & & & & & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q5} & & & \cmark & & & & & \cmark & & \\ \cline{1-12}
\hline
\multicolumn{1}{ c }{\multirow{2}{*}{LinkedSPL} } &
\multicolumn{1}{ c }{Q1} & \cmark & & & & & & &\cmark & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q2} & & & \cmark & & & &\cmark & \cmark &\cmark & \\ \cline{1-12}
\end{tabular}
\label{table::queries}
\end{table*}
We selected 17 representative queries\footnote{Available through \url{https://github.com/m-salehpour/SymphonyDB}}. \reftab{table::queries} shows the classification of the 17 queries. A range of these queries have also been used in previous studies such as~\cite{biobench,cellcycle,saleem}.
\subsection{Evaluation Platform}
\noindent \textbf{Computational Environment.} Our benchmark system was a physical machine with a 3.4GHz Core i7-3770 Intel processor, running Ubuntu Linux (kernel version: 4.15.0-91-generic), with 16GB of main memory, 8 cores, 256K L2 cache, 1TB instance storage capacity.
\noindent \textbf{Data Management Systems (DMSs).} Our DMSs: (1) Virtuoso (version 07.20.3230), (2) Blazegraph (version 2.1.6), RDF-3X (version 0.3.8), and MongoDB (version 4.2.3). All or some of these DMSs have also been used in previous studies such as~\cite{watdiv,ISWC2013,Medha2,bsbm,saleem,biobench}.
We configured these DMSs based on their vendors' official recommendations.
We did not change the default indexing scheme of the DMSs since they usually create exhaustive indexes over all permutations of RDF triples. Note that creating alternate indexing schemes is feasible but will not be generally needed.\footnote{\url{http://docs.openlinksw.com/virtuoso/rdfperfrdfscheme}} For MongoDB, we created indexes on those name/value pairs of the JSON representations that were representatives of subjects and predicates. We loaded the RDF/N-Triples format of KGs into DMSs by using their native bulk loader functions. We converted the KG datasets from RDF/N-Triples syntax to JSON-LD using a parser designed and developed as part of this research\footnote{Available through \url{https://github.com/m-salehpour/SymphonyDB}} to load them into MongoDB.
\noindent \textbf{Measurement.}
The query times for cold-cache are reported below.
We dropped the cache using the following commands: \texttt{echo 3 > /proc/sys/vm/drop\_caches} and \mbox{\texttt{swapoff -a}}. The output of each query was verified to ensure that output results were correct and consistent across the different DMSs.
\begin{figure*}
\centering
\subfigure[]{\label{result:c5:a}\includegraphics[width=.375\textwidth]{Figs/allie.png}}
\subfigure[]{\label{result:c5:b}\includegraphics[width=.375\textwidth]{Figs/cellcycle.png}}
\subfigure[]{\label{result:c5:c}\includegraphics[width=.375\textwidth]{Figs/drug.png}}
\subfigure[]{\label{result:c5:d}\includegraphics[width=.375\textwidth]{Figs/linkedspl.png}}
\caption{ The execution times of different queries against each KG. The $X$ axis shows different queries. The $Y$ axis shows the execution time of each query in milliseconds (\textbf{log scale}). No value is shown when a query is not supported by a DMS (e.g., RDF-3X does not support queries of Cellcycle KG) or the returned result is different with others (e.g., Blazegraph's result for Allie-Q5).\label{fig:c5:result}}
\end{figure*}
\section{Results}
\label{sec::c5:results}
The query execution times over the KGs are presented in \reffig{fig:c5:result} where the $X$ axis shows the different queries and the $Y$ axis shows the execution times in milliseconds (\textbf{log scale}). These results suggest that RDF-3X offers several orders of magnitude performance advantages over others for queries with a single triple pattern (i.e., no join) and less complex triple patterns (e.g., no optional or complex filtering patterns) such as Allie-Q1 and Allie-Q3-Allie-Q5. However, this DMS could not execute Allie-Q2 as fast as others since this query contains a filtering pattern. Note that RDF-3X could not execute queries with complex triple patterns or offset modifiers (e.g.,~\reffig{result:c5:b}). In these cases, no value is shown.
Virtuoso exhibits around one order of magnitude better performance to run complex queries containing a combination of subject-subject and subject-object joins.
As compared to Virtuoso, Blazegraph showed relatively better performance to execute subject-object join queries like Allie-Q4.
MongoDB as a document-store could execute all the queries. For subject-subject join queries like DrugBank-Q1, DrugBank-Q2, and LinkedSPL-Q1, its performance is comparable with others.
Our results indicate that SymphonyDB performs \textit{consistently} across different datasets. Its performance is almost equal to the fastest DMSs in all cases. More specifically, SymphonyDB is consistently the second-best DMS (with a negligible difference as compared to the best DMS) for executing different queries.
\subsection{Discussion}
In this section, we provide a discussion on the experimental results, detailing further insight into the results, and providing key takeaways for SymphonyDB.
\subsubsection{Analysis}
\label{sec::analysis}
Although the results indicate SymphonyDB's consistency across ranging query types, various factors contribute to the performance differences of SymphonyDB with other DMSs.
SymphonyDB labels queries based on strict characteristics and heuristics, such as the number of triple patterns, modifiers, optional patterns, and a number of query join patterns.
This classification forms the basis for which underlying DMS is selected to route the given query to.
For instance, Allie KG queries were routed to RDF-3X and MongoDB (after translation) as two of these queries contained the \textit{single triple pattern}, shown in Allie-Q1 and Allie-Q2, and others contained subject-subject patterns with no modifiers or optional patterns.
This analysis is essential in routing queries to the most optimal DMS, even though it imposes minor overheads.
The overhead of query labeling increases with the need for query translation if routing to a DMS that requires this functionality.
It is plausible that this overhead influenced the performance of SymphonyDB, justifying that the performance difference between SymphonyDB and the best execution time for each query.
However, the significance of these overheads can be overlooked due to the consistency in performance across a range of query types exhibited by SymphonyDB.
Thus, the overheads observed as a result of labeling and translation are viewed as a small trade-off for added consistency in query execution performance.
\subsubsection{Limitations}
\label{sec::Discussion}
Although performance improvements were observed during experimental conditions, there are a number of hindrances experienced by multi-database environments that still pose challenges to increased performance.
Limitations experienced by SymphonyDB include:
\begin{enumerate}[itemsep=0pt,label=\roman*)]
\item \textbf{Replication of KG datasets} --- As multi-database systems employ multiple DMSs, it requires the datasets replicated on each system. The number of replications is determined by the number of DMSs utilized in the underlying layers, e.g. this number is equal to four for SymphonyDB.
For write-heavy applications, this replication can lead to increased latency during write operations and therefore decreased write performance.
However, most KG applications tend to be \textit{read-mostly} if not \textit{read-only}~\cite{Hexastore,RDF3x}. Thus, write latency is not a concern in most use cases.
\item \textbf{Efficiency of translations} --- All SPARQL queries may not be translated to (efficient) MQL queries due to the dissimilarity between the expressiveness of SPARQL and MQL~\cite{frankPhdThesis}. For instance, triple patterns whose predicates are replaced by variables could not be translated into an efficient query for being executed over MongoDB (in most cases). In this research, we did not have such queries and we carefully checked to ensure that our JIT query translation can produce correct and efficient MQL queries for the benchmark SPARQL queries.
However, future work entails further optimization and improvements on the translation to ensure optimality of the query.
\end{enumerate}
\section{Related Work}
\label{sec::related_work}
Over the last few years, there has been growing interest in multi-database solutions resulting in the research and development of open-source platforms, such as Apache Beam and Drill, as well as academic prototypes~\cite{bigdawg}.
In general, these proposals utilize a model consisting of multiple DMSs but require input from expert users to decide which specific DMS meets the requirement for a given application or query set.
For example, \cite{bigdawg} presents two commands, namely \texttt{scope} and \texttt{cast} which provides a user with information to select the most appropriate DMS for the query being analysed.
Recent works~\cite{rheemshort,Musketeershort,f1} present parallel cross-platform data processing systems to decouple application interaction from underlying platforms.
These systems follow a process that splits each given query into sub-queries, executing them on multiple platforms simultaneously to minimize the overall runtime.
Although providing a speedup, it is unclear, however, how much of the performance gain comes from minimizing inter-platform communication overheads by taking advantage of data locality for sub-query processing.
Various proposals take alternative approaches, presenting cross-platform \textit{stream processing}~\cite{lim2013} or building dynamic workload management through adaptable architecture design.
These various proposals show that current multi-database solutions primarily focus on applications such as data integration, ETL, machine learning, stream processing, etc. and pay little attention to employ multiple DMSs for high-performance KG query processing.
\section{Conclusion}
\label{sec::conclusion}
The increases in the heterogeneity of KG datasets have triggered the development of a range of DMSs broadly classified as key-value, document, columnar, and graph stores in addition to the relational. There exists no single DMS that meets the diverse performance requirements of KG query processing efficiently. In this paper, we have addressed some of the critical performance challenges associated with current DMSs in the context of KGs by proposing an architecture that can achieve polygloty for KG query processing supported by a unified access management layer. This approach potentially can be extendable to the non-monolithic conception of database processing in which the different components such as file systems, index structures compression, query processing engines, concurrency, consistency modules, etc. are made available in the cloud and communicate through high-performance networks~\cite{kossmannshort,Aurorashort}. Further steps will also include efforts to minimize the amount of data replications without negatively affecting the robustness and performance.
\balance
\small
\bibliographystyle{abbrv}
\section{Introduction}
\label{sec::introduction}
Knowledge Graphs (KGs) are large-scale collections of facts to represent real-world \textit{entities} and their \textit{interconnections}. KGs have gained widespread use in different domains including computer science, medicine, bioinformatics, education, and biology, among others~\cite{Refinement}. Several KGs such as Wikidata, YAGO, and Bio2RDF, to name a few, are openly available. As well, many private organizations such as Amazon have created KGs for different purposes such as \textit{similarity analysis}. The widespread adoption of KGs has highlighted the need for employing efficient Data Management Systems (DMSs). However, the rich diversity in \textit{variety} and \textit{size} of KG content pose challenges to DMSs to \textit{store} and \textit{query} KGs~\cite{Growing3,Growing2}. The individual \textit{queries} and applications executed on these systems have also become \textit{highly diverse}. These have been addressed in part through the development of a range of DMSs such as \textit{columnar-} and \textit{graph-based} stores. The consensus appears to be that a single, \textit{one-size-fits-all} DMS is unlikely to emerge for efficient KG query processing~\cite{watdiv}.
We develop a solution to this problem that is inspired by Ashby's First Law of Cybernetics~\cite{ashby} and Stonebraker et. al.~\cite{onesize2} which can be paraphrased in this context to state that the variety in the solution architecture should be greater than or at least equal to that of the variety displayed by the data and the queries. The \textit{requisite variety} is to be achieved through an architecture based on \textit{polyglot} model of query processing and access languages supported by a design that can analyze individual queries and match each to the likely best-performing database engine. From a conceptual standpoint, genuine polygloty will imply the requirement of employing multiple DMSs and friction-free translation across the different query and access languages at the polyglot access layer in order to optimize query execution performance. However, a review of the extant research and practitioner literature such as~\cite{fowler} reveals that polygloty has typically been interpreted as using different physical data stores depending on the type of application. For instance, traditional relational DMSs for financial data, document-stores for product catalog data, key-value stores for user activity logs, etc. However, it is not difficult to see that these interpretations typically suffer from lack of integration across the entire data which may lead to \textit{balkanized} data islands.
Supporting multiple data models against a single, integrated backend can potentially address the growing requirements for performance~\cite{lookliu}. Based on this, over the last few years, there has been growing interest in employing multiple DMSs for query processing. This interest led to the development of some open-source platforms such as Apache Drill\footnote{\url{https://drill.apache.org}} as well as some academic prototypes such as~\cite{bigdawg}. However, these solutions mainly focused on applications such as \textit{ETL}, \textit{machine learning}, \textit{stream processing}, \textit{OLAP}, data integration, etc., and somewhat less attention has been paid to efficient KG query processing which is the focus of this paper.
We present \textit{SymphonyDB}, a prototype that provides polyglot support for KG query processing. It is a multi-database approach supported by an access management layer to provide a unified query interface for accessing the underlying DMSs. This layer receives the incoming workloads in the form of SPARQL queries and routes each of them to one (or more than one) of the more likely to be efficiently matched DMSs. A suitable Just-In-Time (JIT) query translation may be needed if the underlying DMSs accept different query languages. Handling such a JIT translation process is the essential responsibility of the access layer. Currently, \textit{SymphonyDB} has included \textit{Virtuoso}, \textit{Blazegraph}, \textit{\mbox{RDF-3X}}, and \textit{\mbox{MongoDB}} as representative DMSs. SymphonyDB has the potential to achieve \textit{requisite variety}.
\newpage
Our contributions include:
\begin{itemize}
\item
Presenting a prototype, \textit{SymphonyDB} that can match KG query requirements with the best combination of DMS and storage representation to achieve consistent query execution performance.
\item
Experimental evaluation and comparative performance analysis of \textit{SymphonyDB} against representative single DMSs in supporting different archetypal KG query types. The source code, data, and other artifacts have been made available at \url{https://github.com/m-salehpour/SymphonyDB}.
\end{itemize}
\begin{figure}[ht]
\centering\includegraphics[width=0.4\textwidth]{Figs/kgnew.png}
\caption{An example of a simple Knowledge Graph (based on the LikedSPL KG).}
\label{fig::kg}
\end{figure}
\section{Background}
\label{sec::background}
In this section, we present some preliminary information about the archetypal KG query types using a \textit{human-readable} example depicted in \reffig{fig::kg}. This shows a small extract from the LinkedSPL KG which includes all sections of FDA-approved prescriptions and over-the-counter drug package inserts from DailyMed. The content of this KG subset can be represented by the following RDF triples\footnote{The content of a KG usually represents a large \textit{set of triples} of the form $<$subject predicate object$>$ creating a graph~\cite{CharacteristicsRDF}}:
\begin{figure}[ht]
\centering
\begin{lstlisting}[language=SQL,basicstyle=\scriptsize\ttfamily,numbers=left,tabsize=2,showstringspaces=false,numberstyle=\scriptsize\tt\color{gray}]
CISPLATIN UNII "Q20Q"
CISPLATIN adverse_reaction "Nausea"
CISPLATIN organization "Bedford Lab"
CISPLATIN FDA_Code "Table_1326"
Table_1326 marker "TPMT"
Table_1326 ID 305
Table_1326 CUI 2555
Table_1326 Xref "gene_PA356"
\end{lstlisting}
\label{fig:triples-KG-fda}
\end{figure}
An example of a query\footnote{Queries are formulated in the form of SPARQL. In this paper, we assume that the reader is familiar with the basic concepts of querying KG, e.g., the SELECT clauses.} is given below. It asks for ``UNII'' of a given drug (i.e., ``CISPLATIN''). In this query, ``?unii'' is a variable to return the result which is ``Q20Q''.
\begin{figure}[ht]
\centering
\begin{lstlisting}[language=SQL,basicstyle=\ttfamily,numbers=left,tabsize=2,showstringspaces=false,numberstyle=\scriptsize\tt\color{gray}]
SELECT ?unii
WHERE {
CISPLATIN UNII ?unii . }
\end{lstlisting}
\end{figure}
KG queries may contain a set of triple patterns such as \textit{\mbox{``CISPLATIN UNII ?unii''}} in which the subject, predicate, and/or object can be a variable. Each triple pattern typically returns a subgraph. This resultant subgraph can be further \textit{joined} with the results of other triple patterns to return the final resultset. In practice, there are three major types of join queries: (i) subject-subject joins (aka, star-like), (ii) subject-object joins (aka, chain-like or path), and (iii) tree-like (i.e., a combination of subject-subject and subject-object joins). These query types are explained below.
\textbf{Subject-subject joins.} A subject-subject join is performed by a DMS when a KG query has at least two triple patterns such that the predicate and object of each triple pattern is a given value (or a variable), but the subjects of both triple patterns are replaced by the \textit{same} variable. For example, the following query looks for all drugs for which their UNII and adverse reactions are equal to the given values (the result will be ``CISPLATIN'').
\begin{figure}[ht]
\centering
\begin{lstlisting}[language=SQL,basicstyle=\ttfamily,numbers=left,tabsize=2,showstringspaces=false,numberstyle=\scriptsize\tt\color{gray}]
SELECT ?x
WHERE {
?x UNII "Q20Q" .
?x adverse_reaction "Nausea" . }
\end{lstlisting}
\end{figure}
\textbf{Subject-object joins.} A subject-object join is performed by a DMS when a KG query has at least two triple patterns such that the subject of one of the triple patterns and the object of the other triple pattern are replaced by the same variable. For example, the following query looks for all drugs that their CUI\footnote{CUI (aka, RxCUI) is a unique drug identifier} is equal to 2555 (``CISPLATIN'' is the result).
\begin{figure}[ht]
\centering
\begin{lstlisting}[language=SQL,basicstyle=\ttfamily,numbers=left,tabsize=2,showstringspaces=false,numberstyle=\scriptsize\tt\color{gray}]
SELECT ?y
WHERE {
?x CUI 2555 .
?y FDA_Code ?x . }
\end{lstlisting}
\end{figure}
\textbf{Tree-like joins.} A tree-like join consists of a \textit{combination} of subject-subject and subject-object joins. For example, the following query looks for the ``Xref'' of all drugs with ``UNII'' of ``Q20Q'' that their ``CUI'' is equal to 2555 (the result will be ``gene\_PA356'').
\begin{figure}[ht]
\centering
\begin{lstlisting}[language=SQL,basicstyle=\ttfamily,numbers=left,tabsize=2,showstringspaces=false,numberstyle=\scriptsize\tt\color{gray}]
SELECT ?y
WHERE {
?x UNII "Q20Q" .
?x FDA_Code ?z .
?z CUI 2555 .
?z Xref ?y . }
\end{lstlisting}
\end{figure}
In addition to the query types, we provide a brief explanation of query \textit{selectivity} and \textit{optional} patterns. The selectivity of a query is typically represented by the fraction of triples matching the query pattern. Based on this, each query type can typically be either \textit{high-selective} or \textit{low-selective}. A query can be considered as low-selective when a large number of triples (as compared to the total number of stored triples) needs to be scanned before returning the resultset~\cite{selectivity}. Moreover, as explained previously, queries return resultsets only when the entire query pattern matches the content of the KG. However, some queries may contain optional patterns to allow KG queries to return a resultset even if the optional part of the query is not matched since completeness and adherence of KG content to their formal ontology specification is not always enforced.
\section{SymphonyDB}
\label{sec::SymphonyDB}
In this section, we present \textit{SymphonyDB}, a prototype that provides polyglot support for KG query processing. SymphonyDB is a multi-database solution supported by an access management layer. Currently, \textit{SymphonyDB} includes \textit{Virtuoso}, \textit{Blazegraph}, \textit{\mbox{RDF-3X}}, and \textit{\mbox{MongoDB}} as representative DMS types.
An abstract overview of the interactions across DMSs and the access layer is expressed using the pseudocode in~\refalgo{algo::polyglot}.
The polyglot access management layer is the entry point for query processing over stored data, providing a unified query interface for accessing the underlying KGs.
This layer receives the incoming SPARQL queries and labels each query based on its characteristics into three target taxonomies, namely: subject-subject, subject-object, and tree-like (line 1 in \refalgo{algo::polyglot}).
The labels are used for routing each query to one (or more than one) of the employed DMSs to increase optimal execution.
We defer details of the labelling approach to~\refsec{sec:label}.
After determining which DMS should be used for each query (line 2 in \refalgo{algo::polyglot}), a suitable JIT query translation may be needed for further query execution (line 3-5 in \refalgo{algo::polyglot}).
For example, the incoming query is written in SPARQL, but it is labeled to be run using MongoDB which cannot support SPARQL directly as an input query language.
In this case, the incoming query needs to be translated JIT into an equivalent JavaScript-like query, MQL, to be executed over MongoDB.
Handling this JIT translation process is one of the responsibilities of the access layer.
The following sections describe the detail of each step.
\begin{algorithm}[htbp]
\DontPrintSemicolon
\SetAlgoLined
\KwInput{SPARQL queries (applications' workload)}
\Initialization{Let q be an incoming SPARQL query}
$qLabel \longleftarrow$ \textbf{Query\_labeling}($q$)\;
$DMS \longleftarrow$ \textbf{DMS\_select}($qLabel$)\;
\If{ $DMS == MongoDB$}{
$q \longleftarrow$ \textbf{Translate\_MQL}($q$)\;
}
$qResult \longleftarrow$ \textbf{Route\_execute}($q$)\;
\textbf{Return\_result}($qResult$)\;
\caption{SymphonyDB KG query processing}
\label{algo::polyglot}
\end{algorithm}
\subsection{Database Management System Layer}
\label{sec:DMS}
Selection of adequate DMSs is vital in maximizing the power of a multi-database system. In our prototype, RDF-3X was selected as a candidate as it is an open-source system widely used in a range of studies such as~\cite{watdiv}.
One vital property is that RDF-3X creates exhaustive indexes on all permutations of triples along with their binary and unary projections.
Its query processor is designed to aggressively leverage cache-aware hash and merge joins.
Virtuoso was selected since it is already employed as the DMS of choice for a broad range of KGs, for example the Linked Data for the Life Sciences project\footnote{Available online: \url{https://bio2rdf.org/sparql}}.
Virtuoso's physical design is based on a relational table with three columns\footnote{In case of loading named graphs, another column is added, called C.} for S, P, and O (S: Subject, P: Predicate, and O: Object) and carries multiple bitmap indexes over that table to provide a number of different access paths.
Most recently, Virtuoso added columnar projections to minimize the on-disk footprint associated with RDF data storage. Blazegraph\footnote{It is alleged that the Amazon Neptune is based on Blazegraph.} was selected since it is the DMS behind Wikidata, i.e., a KG constructed from the content of Wikimedia sister projects including Wikipedia, Wikivoyage, Wiktionary, and Wikisource.
Blazegraph's physical design is based on \mbox{B+trees} to store KGs in the form of ordered data.
Blazegraph typically uses the following three indexes: SPO, POS, and OSP. MongoDB was selected as a representative document-store.
Its efficacy for executing queries over KGs has not been researched extensively but some academic prototypes such as~\cite{Jignesh,frankPhdThesis,tomaszuk2010document} have already shown document-stores efficacy in similar contexts.
\subsection{Query Labeling and Execution}
\label{sec:label}
Each SPARQL query can be viewed as a directed graph where nodes are formed by the subjects and objects of the query's triple patterns and edges are the properties of these patterns. Based on this, each SPARQL query can be classified into shape-specific categories.
At this stage, we confine our focus to the following query types: subject-subject, subject-object, and tree-like queries (more details can be found in~\refsec{sec::background}). Examples of these types are shown in~\reffig{fig::qlabel} where each node represents a variable connected through predicates as edges forming a graph-like structure.
\begin{figure}[hbt]
\centering\includegraphics[width=0.45\textwidth]{Figs/qlabel.png}
\caption{(a) Subject-subject query pattern (b) Subject-object query pattern (c) Tree-like query pattern}
\label{fig::qlabel}
\end{figure}
In general, each query has at least one source variable as shown in \reffig{fig::qlabel} using nodes represented with solid black colors. Any node with both incoming and outgoing edges represents a shared variable of a query. Query patterns are typically recognizable by the position of shared variables in a query.
Inspired by~\cite{peter-heuristic}, we utilize a heuristic-based approach to exploit the syntactic and the structural variations of patterns in a given SPARQL query in order to label it.
On this foundation, SymphonyDB begins by finding the source variable of a given query, looking for all immediate neighbor nodes, or shared variables, with one edge distance away.
From there, it then iteratively visits nodes further away until all nodes are visited, using a queue data structure to store visited nodes at each stage.
Upon finishing the traversal, the query will then be labeled according to the characteristics of its variables and graph.
A subject-subject label is applied if all nodes are only immediate neighbors of the source variable with one edge distance away (\mbox{\reffig{fig::qlabel} (a)}).
A subject-object label, however, follows a pattern where there is just one outgoing edge from a node at each stage (starting from the source variable) where there is a final node with no outgoing edge as depicted in \mbox{\reffig{fig::qlabel} (b)}.
Finally, queries that contain a combination of both patterns are labeled as tree-like (\mbox{\reffig{fig::qlabel} (c)}).
In addition to the query patterns, SymphonyDB checks the existence of modifiers as well, where modifiers are keywords such as ``\texttt{LIMIT}'' that are recognizable by the query parser and lexical analyzer implemented in SymphonyDB.
The following heuristics are then used to select one or more of the integrated DMSs.
Queries are routed to both MongoDB and RDF-3X if the following characteristics are present: (1)~it contains only a single triple pattern, (2)~it is a query with subject-subject joins, (3)~it contains no modifiers in the query, and (4)~it contains no optional patterns.
Alternatively, if a query contains subject-object joins, it is routed to Blazegraph, and finally all other queries (tree-like queries with or without optional patterns) are routed to Virtuoso.
As an intermediate step, any queries being routed to MongoDB must run through the JIT query translation to be translated from SPARQL to MQL. This is explained in the following section.
\subsection{Polyglot Access Management: Query Translation}
\label{sec:translate}
Query translation provides the extensibility for SymphonyDB to access various DMSs with differing query languages.
Currently, the only integrated DMS that requires this functionality is MongoDB as it provides its own query language MQL, which is a Javascript-like, object-oriented imperative language.
This contrasts with SPARQL, a domain-specific declarative language~\cite{tomaszuk2010document}, which is not supported natively by MongoDB, however, it is feasible to map SPARQL to MQL in most cases.
\reffig{fig::translation} depicts the logic flow of SymphonyDB's JIT query translation, showing that each query is analyzed lexically and tokenized based on the SPARQL query syntax.
The lexical analyzer dissects the SPARQL query into logical units of one or more characters that have a shared meaning, often referred to as \textit{tokens}.
For instance, ``\texttt{WHERE}'' is a token representing a keywords, whereas ``\texttt{.}'' is an identifier and ``\texttt{=}'' is a sign.
In parallel, it parses each query with regard to the grammatical description of the SPARQL language to generate the corresponding syntax tree.
The semantic analyzer then produces an \textit{operator graph} containing information about projection variables, join patterns, conditions and modifiers.
Finally, once the semantic analysis has completed, heuristic techniques are used to map the operator graph to MQL and translate the query.
\begin{figure}[h]
\centering\includegraphics[width=0.45\textwidth]{Figs/mql.png}
\caption{The query translation logic flow}
\label{fig::translation}
\end{figure}
Similar to~\cite{tomaszuk2010document,frankPhdThesis,frankMappping}, SymphonyDB maps each SPARQL to MQL using a collection of rules.
\reftab{table:rules1} shows SPARQL expressions and their equivalent MQL query string, along with additional set of rules to map SPARQL query patterns to MQL illustrated in~\reftab{table:rules2}.
For example, to translate subject-subject join queries, SymphonyDB uses the ``\texttt{\$match}'' aggregation pipeline operator of MongoDB to filters documents and pass a subset of the documents that match the specified condition(s) to the next pipeline stage.
It also uses ``\texttt{\$lookup}'' aggregation pipeline operator of MongoDB to translate joins.
\begin{table}[h]
\caption{SPARQL expressions representation and their equivalent MQL expressions}
\centering
\begin{tabular}{l | l }
\toprule
SPARQL & MQL \\ [0.7ex]
\hline\hline
Exists ($<$e1$>$) & $<$e1$>$:\{\$exists:true\} \\
Not Exists ($<$e1$>$) & $<$e1$>$:\{\$exists:false\} \\
($<$e1$>$ \&\& $<$e2$>$ ) & \{\$and:[\{$<$e1$>$\},\{$<$e2$>$\}] \\
($<$e1$>$ $||$ $<$e2$>$ ) & \{\$or:[\{$<$e1$>$\},\{$<$e2$>$\}] \\
!($<$e1$>$) & \{\$not:\{$<$e1$>$\} \\
($<$e1$>$ = $<$e2$>$ ) & \{\$eq:[\{$<$e1$>$\},\{$<$e2$>$\}] \\
($<$e1$>$ != $<$e2$>$ ) & \{\$ne:[\{$<$e1$>$\},\{$<$e2$>$\}] \\
($<$e1$>$ $>$ $<$e2$>$ ) & \{\$gt:[\{$<$e1$>$\},\{$<$e2$>$\}] \\
($<$e1$>$ $>=$ $<$e2$>$ ) & \{\$gte:[\{$<$e1$>$\},\{$<$e2$>$\}] \\
($<$e1$>$ $<$ $<$e2$>$ ) & \{\$lt:[\{$<$e1$>$\},\{$<$e2$>$\}] \\
($<$e1$>$ $<=$ $<$e2$>$ ) & \{\$lte:[\{$<$e1$>$\},\{$<$e2$>$\}] \\
\end{tabular}
\label{table:rules1}
\end{table}
\begin{table*}
\caption{Sample rules used to translate different join patterns of SPARQL queries to their equivalent MQL query string}
\resizebox{0.9\textwidth}{!}{
\centering
\begin{tabularx}{\textwidth}{l|l|l}
\cmidrule(lr){1-3}
\backslashbox{Pattern}{Query}& SPARQL (triple patterns) & MQL (aggregate pipeline)\\
\cmidrule(lr){1-3} \cmidrule(lr){1-3}
\multicolumn{1}{ c| }{\multirow{3}{*}{Single Triple Pattern} } &
\multicolumn{1}{ l| }{``subject'' ``predicate'' ``object''} & \{\$match:\{ subject\_id:``subject'', ``predicate'':``object''\}\}\\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{?subject ``predicate'' ``object''} & \{\$match:\{ subject\_id:\{\$exists:true\}, ``predicate'':``object''\}\}\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{``subject'' ``predicate'' ?object} & \{\$match:\{ subject\_id:``subject'', ``predicate'':\{\$exists:true\}\}\}\\
\cmidrule{1-3} \cmidrule{1-3}
\multicolumn{1}{ c| }{\multirow{9}{*}{Subject-subject join} } &
\multicolumn{1}{ l| }{\{?subject ``predicate1'' ``object1'' .} & \{\$match:\{ subject\_id:\{\$exists:true\}, ``predicate1'':``object1'',``predicate2'':``object2''\}\} \\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{~?subject ``predicate2'' ``object2'' .\}} & \\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{\{?subject ``predicate1'' ?object1~ .} & \{\$match:\{ subject\_id:\{\$exists:true\}, ``predicate1'':\{\$exists:true\},``predicate2'':``object2''\}\} \\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{~?subject ``predicate2'' ``object2'' .\}} & \\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{\{``subject'' ``predicate1'' ``object1'' .} & \{\$match:\{ subject\_id:``subject'', ``predicate1'':\{\$exists:true\},``predicate2'':``object2''\}\} \\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{~``subject'' ``predicate2'' ``object2'' .\}} & \\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \\
\cmidrule{1-3} \cmidrule{1-3}
\multicolumn{1}{ c| }{\multirow{12}{*}{Subject-object join} } &
\multicolumn{1}{ l| }{\{?subject ``predicate1'' ?object1~ .} & \{\{\$match:\{ subject\_id:\{\$exists:true\}\}\},\\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{~?object1 ``predicate2'' ``object2'' .\}} & \{\$lookup:\{ from: ``colc\_name'', localField: ``predicate1'', foreignField: ``subject\_id'', as: ``join\_field''\}\},\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \{\$match:\{ ``join\_field.predicate2'':``object2''\}\}\}\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{\{?subject ``predicate1'' ?object1~ .} & \{\{\$match:\{ subject\_id:\{\$exists:true\}\}\},\\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{~?object1 ``predicate2'' ?object2 .\}} & \{\$lookup:\{ from: ``colc\_name'', localField: ``predicate1'', foreignField: ``subject\_id'', as: ``join\_field''\}\},\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \{\$match:\{ ``join\_field.predicate2'':\{\$exists:true\}\}\}\}\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{\{``subject'' ``predicate1'' ?object1~ .} & \{\{\$match:\{ subject\_id:``subject''\}\},\\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{~?object1 ``predicate2'' ?object2 .\}} & \{\$lookup:\{ from: ``colc\_name'', localField: ``predicate1'', foreignField: ``subject\_id'', as: ``join\_field''\}\},\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \{\$match:\{ ``join\_field.predicate2'':\{\$exists:true\}\}\}\}\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \\
\cmidrule{1-3} \cmidrule{1-3}
\multicolumn{1}{ c| }{\multirow{12}{*}{Tree-like join} } &
\multicolumn{1}{ l| }{\{?subject ``predicate1'' ?object1~ .} & \{\{\$match:\{ subject\_id:\{\$exists:true\}, ``predicate1'':\{\$exists:true\}, ``predicate2'':\{\$exists:true\} \}\},\\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{~?subject ``predicate2'' ?object2~ .} & \{\$lookup:\{ from: ``colc\_name'', localField: ``predicate2'', foreignField: ``subject\_id'', as: ``join\_field''\}\},\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{~?object2 ``predicate3'' ``object3'' .\}} & \{\$match:\{ ``join\_field.predicate3'':``object3''\}\}\}\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{\{``subject'' ``predicate1'' ``object1''~ .} & \{\{\$match:\{ subject\_id:``subject'', ``predicate1'':``object1'', ``predicate2'':\{\$exists:true\} \}\},\\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{~``subject'' ``predicate2'' ?object2~ .} & \{\$lookup:\{ from: ``colc\_name'', localField: ``predicate2'', foreignField: ``subject\_id'', as: ``join\_field''\}\},\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{~?object2 ``predicate3'' ?object3 .\}} & \{\$match:\{ ``join\_field.predicate3'':\{\$exists:true\}\}\}\}\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{\{``subject'' ``predicate1'' ?object1~ .} & \{\{\$match:\{ subject\_id:``subject'', ``predicate1'':\{\$exists:true\} \}\},\\
\multicolumn{1}{ c| }{} &
\multicolumn{1}{ l| }{~?object1 ``predicate2'' ?object2~ .} & \{\$lookup:\{ from: ``colc\_name'', localField: ``predicate1'', foreignField: ``subject\_id'', as: ``join\_field''\}\},\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{~?object1 ``predicate3'' ``object3'' .\}} & \{\$match:\{ ``join\_field.predicate2'':\{\$exists:true\}, ``join\_field.predicate3'':``object3''\}\}\}\\
\multicolumn{1}{ c | }{} &
\multicolumn{1}{ l| }{} & \\
\cmidrule{1-3} \cmidrule{1-3}
\end{tabularx}
}
\label{table:rules2}
\end{table*}
\section{Experimental Evaluation}
\label{sec::Experimental-Setting}
In this section, we report the experimental setup and details of the KG benchmark datasets that are used in the experimental evaluation.
This includes detailed information about DMSs' configuration, indexing, data loading process as well as our computational platform.
The query performance of SymphonyDB and a range of DMSs are evaluated and presented below.
\subsection{Evaluation Datasets and Queries}
We select four well-known KG datasets with a collection of relevant queries that are publicly available, where a number have been used in previous studies~\cite{cellcycle,biobench}.
The datasets are as follows.
\textbf{Allie}\footnote{\url{http://allie.dbcls.jp/}} is a KG surrounding life sciences, containing abbreviations and long forms utilized within the field.
\textbf{Cellcycle}\footnote{\url{ftp://ftp.dbcls.jp/togordf/bmtoyama/cellcycle/}} contains orthology relations for proteins consiting of ten sub-graphs constituting the cell cycle.
In our experiments, however, we integrated all ten sub-graphs into a single KG dataset without modifying any content.
\textbf{DrugBank}\footnote{\url{https://download.bio2rdf.org/files/current/drugbank/drugbank.html}} contains bioinformatics and chemoinformatics resources which include detailed drug (chemical, pharmacological, pharmaceutical, etc.) and comprehensive drug targets (sequence, structure and pathway information) in the dataset.
\textbf{LinkedSPL\footnote{\url{https://download.bio2rdf.org/files/current/linkedspl/linkedspl.html}}} includes all sections of FDA-approved prescriptions and over-the-counter drug package inserts from DailyMed. \reftab{table::c5:kgs} depicts the statistical information related to the above KGs.
\begin{table}[tbp]
\centering
\caption{KGs that were used to run the experiments}
\begin{tabularx}{\linewidth}{l r r r r}
\toprule
\multirow{2}{*}{KG}& \multicolumn{4}{c}{Statistics} \\
\cmidrule{2-5}
& Sub. (\#) & Pre. (\#)& Obj. (\#) & Triples (\#)\\
\midrule
\midrule
Allie & 19,227,252 & 26 & 20,280,252 & 94,404,806 \\
Cellcycle & 21,745 & 18 & 142,812 & 322,751\\
DrugBank & 19,693 & 119 & 276,142 & 517,023 \\
LinkedSPL & 59,776 & 104 & 719,446 & 2,174,579 \\
\bottomrule
\end{tabularx}
\label{table::c5:kgs}
\end{table}
\begin{table*}
\caption{Types of the queries. $SS^{a*}$: Subject-subject join, $SO^{b*}$: Subject-object join, $Co^{c*}$: combination of $SS$ and $SO$, $OPT^{d*}$: Optional pattern, $Fil^{e*}$: Filter, $ORD^{f*}$: Order by, $Lim^{g*}$: Limit, $OFF^{h*}$: Offset, $STP^{i*}$: Single triple pattern (no join)}
\centering
\begin{tabular}{cccccccccccc}
\toprule
\multirow{3}{*}{Benchmark}& \multicolumn{11}{c}{Types} \\
\cmidrule{3-12}
& Query & $SS^{a*}$ & $SO^{b*}$ & $Co^{c*}$ & $OPT^{d*}$ & Selective & $Fil^{e*}$ & $ORD^{f*}$ & $Lim^{g*}$ & $OFF^{h*}$ & $STP^{i*}$\\
\hline\hline
\multicolumn{1}{ c }{\multirow{5}{*}{Allie} } &
\multicolumn{1}{ c }{Q1} & & & & & & & & & & \cmark\\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q2} & & & & & & \cmark & & & & \cmark\\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q3} & \cmark & & & & & & & & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q4} & & \cmark & & & & & & \cmark & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q5} & \cmark & & & & & & \cmark & \cmark & & \\ \cline{1-12}
\hline
\multicolumn{1}{ c }{\multirow{6}{*}{Cellcycle} } &
\multicolumn{1}{ c }{Q1} & & & \cmark & & & & & & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q2} & & & \cmark & \cmark & & & & & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q3} & & & \cmark & & \cmark & & & & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q4} & & & \cmark & & & & & & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q5} & \cmark & & & \cmark & & & & & & \\ \cline{1-12}
\hline
\multicolumn{1}{ c }{\multirow{5}{*}{DrugBank} } &
\multicolumn{1}{ c }{Q1} & \cmark & & & \cmark & & & & \cmark & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q2} & \cmark & & & \cmark & & & \cmark &\cmark &\cmark & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q3} & & & \cmark & & & & & & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q4} & & & \cmark & & & & & & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q5} & & & \cmark & & & & & \cmark & & \\ \cline{1-12}
\hline
\multicolumn{1}{ c }{\multirow{2}{*}{LinkedSPL} } &
\multicolumn{1}{ c }{Q1} & \cmark & & & & & & &\cmark & & \\
\multicolumn{1}{ c }{} &
\multicolumn{1}{ c }{Q2} & & & \cmark & & & &\cmark & \cmark &\cmark & \\ \cline{1-12}
\end{tabular}
\label{table::queries}
\end{table*}
We selected 17 representative queries\footnote{Available through \url{https://github.com/m-salehpour/SymphonyDB}}. \reftab{table::queries} shows the classification of the 17 queries. A range of these queries have also been used in previous studies such as~\cite{biobench,cellcycle,saleem}.
\subsection{Evaluation Platform}
\noindent \textbf{Computational Environment.} Our benchmark system was a physical machine with a 3.4GHz Core i7-3770 Intel processor, running Ubuntu Linux (kernel version: 4.15.0-91-generic), with 16GB of main memory, 8 cores, 256K L2 cache, 1TB instance storage capacity.
\noindent \textbf{Data Management Systems (DMSs).} Our DMSs: (1) Virtuoso (version 07.20.3230), (2) Blazegraph (version 2.1.6), RDF-3X (version 0.3.8), and MongoDB (version 4.2.3). All or some of these DMSs have also been used in previous studies such as~\cite{watdiv,ISWC2013,Medha2,bsbm,saleem,biobench}.
We configured these DMSs based on their vendors' official recommendations.
We did not change the default indexing scheme of the DMSs since they usually create exhaustive indexes over all permutations of RDF triples. Note that creating alternate indexing schemes is feasible but will not be generally needed.\footnote{\url{http://docs.openlinksw.com/virtuoso/rdfperfrdfscheme}} For MongoDB, we created indexes on those name/value pairs of the JSON representations that were representatives of subjects and predicates. We loaded the RDF/N-Triples format of KGs into DMSs by using their native bulk loader functions. We converted the KG datasets from RDF/N-Triples syntax to JSON-LD using a parser designed and developed as part of this research\footnote{Available through \url{https://github.com/m-salehpour/SymphonyDB}} to load them into MongoDB.
\noindent \textbf{Measurement.}
The query times for cold-cache are reported below.
We dropped the cache using the following commands: \texttt{echo 3 > /proc/sys/vm/drop\_caches} and \mbox{\texttt{swapoff -a}}. The output of each query was verified to ensure that output results were correct and consistent across the different DMSs.
\begin{figure*}
\centering
\subfigure[]{\label{result:c5:a}\includegraphics[width=.375\textwidth]{Figs/allie.png}}
\subfigure[]{\label{result:c5:b}\includegraphics[width=.375\textwidth]{Figs/cellcycle.png}}
\subfigure[]{\label{result:c5:c}\includegraphics[width=.375\textwidth]{Figs/drug.png}}
\subfigure[]{\label{result:c5:d}\includegraphics[width=.375\textwidth]{Figs/linkedspl.png}}
\caption{ The execution times of different queries against each KG. The $X$ axis shows different queries. The $Y$ axis shows the execution time of each query in milliseconds (\textbf{log scale}). No value is shown when a query is not supported by a DMS (e.g., RDF-3X does not support queries of Cellcycle KG) or the returned result is different with others (e.g., Blazegraph's result for Allie-Q5).\label{fig:c5:result}}
\end{figure*}
\section{Results}
\label{sec::c5:results}
The query execution times over the KGs are presented in \reffig{fig:c5:result} where the $X$ axis shows the different queries and the $Y$ axis shows the execution times in milliseconds (\textbf{log scale}). These results suggest that RDF-3X offers several orders of magnitude performance advantages over others for queries with a single triple pattern (i.e., no join) and less complex triple patterns (e.g., no optional or complex filtering patterns) such as Allie-Q1 and Allie-Q3-Allie-Q5. However, this DMS could not execute Allie-Q2 as fast as others since this query contains a filtering pattern. Note that RDF-3X could not execute queries with complex triple patterns or offset modifiers (e.g.,~\reffig{result:c5:b}). In these cases, no value is shown.
Virtuoso exhibits around one order of magnitude better performance to run complex queries containing a combination of subject-subject and subject-object joins.
As compared to Virtuoso, Blazegraph showed relatively better performance to execute subject-object join queries like Allie-Q4.
MongoDB as a document-store could execute all the queries. For subject-subject join queries like DrugBank-Q1, DrugBank-Q2, and LinkedSPL-Q1, its performance is comparable with others.
Our results indicate that SymphonyDB performs \textit{consistently} across different datasets. Its performance is almost equal to the fastest DMSs in all cases. More specifically, SymphonyDB is consistently the second-best DMS (with a negligible difference as compared to the best DMS) for executing different queries.
\subsection{Discussion}
In this section, we provide a discussion on the experimental results, detailing further insight into the results, and providing key takeaways for SymphonyDB.
\subsubsection{Analysis}
\label{sec::analysis}
Although the results indicate SymphonyDB's consistency across ranging query types, various factors contribute to the performance differences of SymphonyDB with other DMSs.
SymphonyDB labels queries based on strict characteristics and heuristics, such as the number of triple patterns, modifiers, optional patterns, and a number of query join patterns.
This classification forms the basis for which underlying DMS is selected to route the given query to.
For instance, Allie KG queries were routed to RDF-3X and MongoDB (after translation) as two of these queries contained the \textit{single triple pattern}, shown in Allie-Q1 and Allie-Q2, and others contained subject-subject patterns with no modifiers or optional patterns.
This analysis is essential in routing queries to the most optimal DMS, even though it imposes minor overheads.
The overhead of query labeling increases with the need for query translation if routing to a DMS that requires this functionality.
It is plausible that this overhead influenced the performance of SymphonyDB, justifying that the performance difference between SymphonyDB and the best execution time for each query.
However, the significance of these overheads can be overlooked due to the consistency in performance across a range of query types exhibited by SymphonyDB.
Thus, the overheads observed as a result of labeling and translation are viewed as a small trade-off for added consistency in query execution performance.
\subsubsection{Limitations}
\label{sec::Discussion}
Although performance improvements were observed during experimental conditions, there are a number of hindrances experienced by multi-database environments that still pose challenges to increased performance.
Limitations experienced by SymphonyDB include:
\begin{enumerate}[itemsep=0pt,label=\roman*)]
\item \textbf{Replication of KG datasets} --- As multi-database systems employ multiple DMSs, it requires the datasets replicated on each system. The number of replications is determined by the number of DMSs utilized in the underlying layers, e.g. this number is equal to four for SymphonyDB.
For write-heavy applications, this replication can lead to increased latency during write operations and therefore decreased write performance.
However, most KG applications tend to be \textit{read-mostly} if not \textit{read-only}~\cite{Hexastore,RDF3x}. Thus, write latency is not a concern in most use cases.
\item \textbf{Efficiency of translations} --- All SPARQL queries may not be translated to (efficient) MQL queries due to the dissimilarity between the expressiveness of SPARQL and MQL~\cite{frankPhdThesis}. For instance, triple patterns whose predicates are replaced by variables could not be translated into an efficient query for being executed over MongoDB (in most cases). In this research, we did not have such queries and we carefully checked to ensure that our JIT query translation can produce correct and efficient MQL queries for the benchmark SPARQL queries.
However, future work entails further optimization and improvements on the translation to ensure optimality of the query.
\end{enumerate}
\section{Related Work}
\label{sec::related_work}
Over the last few years, there has been growing interest in multi-database solutions resulting in the research and development of open-source platforms, such as Apache Beam and Drill, as well as academic prototypes~\cite{bigdawg}.
In general, these proposals utilize a model consisting of multiple DMSs but require input from expert users to decide which specific DMS meets the requirement for a given application or query set.
For example, \cite{bigdawg} presents two commands, namely \texttt{scope} and \texttt{cast} which provides a user with information to select the most appropriate DMS for the query being analysed.
Recent works~\cite{rheemshort,Musketeershort,f1} present parallel cross-platform data processing systems to decouple application interaction from underlying platforms.
These systems follow a process that splits each given query into sub-queries, executing them on multiple platforms simultaneously to minimize the overall runtime.
Although providing a speedup, it is unclear, however, how much of the performance gain comes from minimizing inter-platform communication overheads by taking advantage of data locality for sub-query processing.
Various proposals take alternative approaches, presenting cross-platform \textit{stream processing}~\cite{lim2013} or building dynamic workload management through adaptable architecture design.
These various proposals show that current multi-database solutions primarily focus on applications such as data integration, ETL, machine learning, stream processing, etc. and pay little attention to employ multiple DMSs for high-performance KG query processing.
\section{Conclusion}
\label{sec::conclusion}
The increases in the heterogeneity of KG datasets have triggered the development of a range of DMSs broadly classified as key-value, document, columnar, and graph stores in addition to the relational. There exists no single DMS that meets the diverse performance requirements of KG query processing efficiently. In this paper, we have addressed some of the critical performance challenges associated with current DMSs in the context of KGs by proposing an architecture that can achieve polygloty for KG query processing supported by a unified access management layer. This approach potentially can be extendable to the non-monolithic conception of database processing in which the different components such as file systems, index structures compression, query processing engines, concurrency, consistency modules, etc. are made available in the cloud and communicate through high-performance networks~\cite{kossmannshort,Aurorashort}. Further steps will also include efforts to minimize the amount of data replications without negatively affecting the robustness and performance.
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\section{Augmentations for Combinatorial Problems}
\label{sec: lpa}
\subsection{Definition and Motivation}
Intuitively, contrastive learning leads to representations that are invariant across augmentations. Thus, to guarantee downstream predictive performance, augmentations should (mostly) preserve the labels of downstream tasks \citep{arora2019theoretical, tosh2021contrastive, haochen2021provable, dubois2021lossy}.
The augmentations found in computer vision, e.g., cropping and color jittering of images, typically preserve the labels of classification tasks.
This is not the case for most previous graph contrastive frameworks \citep{you2020graph, hassani2020contrastive}, where simple graph augmentations like node dropping and link perturbation are used without consideration of the downstream task. For combinatorial problems, these \emph{label-agnostic augmentations} (LAAs\xspace) are likely to produce many false positive pairs. For example, in the SR dataset in NeuroSAT, each SAT can be turned into UNSAT by flipping one literal in one clause.
Thus, we aim to design \emph{label-preserving augmentations} (LPAs\xspace), which are transformations that preserve the instance label. Formally, given a problem family $\Phi$ and a labelling function $f$, an augmentation distribution $A$ is an LPA\xspace iff:
\begin{align*}f(\hat{\phi}) = f(\phi), \quad \forall \hat{\phi} \in \texttt{supp}\xspace\left(A(\cdot | \phi)\right), \forall \phi \in \Phi,\end{align*}
where $\texttt{supp}\xspace$ denotes the support of a distribution.
Importantly, LPAs\xspace are well-studied for many combinatorial problems and are much cheaper to obtain than labels. For SAT prediction, common preprocessing techniques from SAT solvers such as variable elimination can be used as LPAs\xspace, as discussed in Section \ref{subsec: augmentation_sat}. There are also LPAs\xspace for other combinatorial problems, such as, adding cuts \cite{achterberg2020presolve} for MILPs and deleting dominant vertices \citep{akiba2016branch} for Minimum Vertex Cover.
The type of LPAs\xspace is also crucial. Intuitively, LPAs\xspace that make more significant changes to an instance create harder positives from which better representations can be learned. Indeed, LPAs\xspace that lead to larger augmentation support $\texttt{supp}\xspace\left(A(\cdot | \phi)\right)$ will split $\Phi$ into coarser equivalences classes, and thus (roughly speaking) a classifier can be learned with fewer labelled instances.
\subsection{Label-preserving Augmentations for SAT}
\label{subsec: augmentation_sat}
The LPAs\xspace for {SAT} preserve satisfiability of any Boolean formula. In other words,
a \textit{(un)}satisfiable instance remains \textit{(un)}satisfiable after the applications of LPAs\xspace. We review some common LPAs for SAT below, with examples and time complexity results provided in Appendix \ref{appendix: example_lpas}.
\textbf{Unit Propagation (UP).} A clause is a \emph{unit clause} if it contains only one literal. If an instance $\phi$ contains a unit clause $\ell$, we can 1) remove all clauses in $\phi$ containing the literal $\ell$ and 2) delete $\neg \ell$ from all other clauses.
\textbf{Add Unit Literal (AU).} The inverse of UP: 1) construct a unit clause from a new literal $\ell$, 2) add its negation $\neg \ell$ to some other clauses and 3) create new clauses containing $\ell$.
\textbf{Pure Literal Elimination (PL).} A variable $v$ is called pure if it occurs with only one polarity in $\phi$. We can delete all clauses in $\phi$ containing $v$.
\textbf{Subsumed Clause Elimination (SC).} If a clause $c_1$ is a subset of $c_2$, i.e., all literals in $c_1$ are also in $c_2$, then deleting $c_2$ does not change satisfiability of $\phi$.
\textbf{Clause Resolution (CR).} Resolution produces a new clause implied by two clauses containing complementary literals:
\begin{equation}
\frac{\ell \lor a_1 \lor \cdots \lor a_n, \neg \ell \lor b_1 \lor \cdots b_m }{a_1 \lor \cdots \lor a_n \lor b_1 \lor \cdots b_m }
\label{equ: resolution}
\end{equation}
The new clause $c$ is called the \textit{resolvent} of $c_1$, $c_2$: $c = c_1 \otimes c_2$. Adding $c$ to $\phi$ does not change satisfiability.
\textbf{Variable Elimination (VE).}
Let $S_\ell$ be a set of clauses containing the literal $\ell$, and $S_{\neg \ell}$ be a set of clauses containing its negation $\neg \ell$. Then a new set $S$ is obtained by pairwise resolving on clauses of $S_\ell$ and $S_{\neg \ell}$: $S = \{c_1 \otimes c_2 | c_1 \in S_\ell, c_2 \in S_{\neg \ell}\}$. Replacing $S_\ell \cup S_{\neg \ell}$ with $S$ does not change satisfiability \cite{een2005effective}.
Note that some augmentations, such as VE, have the worst-case exponential complexity if run until convergence. However, in our paper, we only eliminate a small and fixed number of variables. Therefore, all LPAs listed here are cheap and have polynomial-time complexity.
\section{LPAs\xspace for SAT}
\label{appendix: example_lpas}
\subsection{Examples}
\begin{table}[h]
\centering
\label{tab: sat_aug}
\vskip 0.15in
\begin{tabular}{|l|l|}
\hline
Original & UP \\
\hline
$c_1: x_1$ & \st{$x_1$} \\
$c_2: x_2 \lor x_3 $ & $x_2 \lor x_3 $ \\
$c_3: x_1 \lor \neg x_3 \lor x_4 $ & \st{$x_1 \lor \neg x_3 \lor x_4 $} \\
$c_4: \neg x_1 \lor x_2 \lor x_3 \lor \neg x_4$ & \st{$\neg x_1 \lor$}$ x_2 \lor x_3 \lor \neg x_4$ \\
\hline
AU & SC \\
\hline
$\color{our_red} \neg x_5$ & \\
$\color{our_red} x_5 \lor$ $x_1$ & $x_1$ \\
$x_2 \lor x_3 $ & $x_2 \lor x_3 $ \\
$x_1 \lor \neg x_3 \lor x_4 $ & $x_1 \lor \neg x_3 \lor x_4 $ \\
$\neg x_1 \lor x_2 \lor x_3 \lor \neg x_4$ & \st{$\neg x_1 \lor x_2 \lor x_3 \lor \neg x_4$} \\
$\color{our_red} \neg x_5 \lor x_1 \lor \neg x_2 \lor x_3$& \\
\hline
CR & VE \\
\hline
$x_1$ & $x_1$ \\
$x_2 \lor x_3 $ & \st{$x_2 \lor x_3 $} \\
$x_1 \lor \neg x_3 \lor x_4 $ & \st{$x_1 \lor \neg x_3 \lor x_4 $} \\
$\neg x_1 \lor x_2 \lor x_3 \lor \neg x_4$ & \st{$\neg x_1 \lor x_2 \lor x_3 \lor \neg x_4$} \\
$\color{our_red} x_1 \lor x_2 \lor x_4 $ & $\color{our_red} x_1 \lor x_2 \lor x_4 $ \\
\hline
\end{tabular}
\caption{ Examples of LPAs\xspace \ for SAT. UP: remove $c_1, c_3$ and $\neg x$ from $c_4$. AU: add unit literal $\neg x_5$, add $x_5$ to $c_1$ and create a new random clauses containing $\neg x_5$ . SC: remove $c_4$ because $c_2 \subset c_4$. CR: add $c_2 \otimes c_3$. VE: eliminate $x_3$ by adding $c_2 \otimes c_3$ and $c_4 \otimes c_3$. }
\end{table}
\subsubsection{Time Complexity} \label{subsec: complexity}
Let $n$ denote the number of clauses in the instance.
\begin{table}[h]
\centering
\vskip 0.15in
\begin{tabular}{clc}
\toprule
Rule & Explanation & Time Complexity \\ \midrule
UP & Eliminate one Unit Clause & $O(n)$ \\
AU & Add one Unit Clause & $O(n)$ \\
PL & Eliminate one pure literal & $O(n)$ \\
SC & Eliminate all subsumed clauses & $O(n^2)$ \\
CR & Resolve one pair of clauses & $O(n)$ \\
VE & Eliminate one variable & $O(n^2)$ \\
\bottomrule
\end{tabular}
\caption{Complexity results of LPAs\xspace}
\end{table}
\section{NeuroSAT}\label{appendix: neurosat}
\paragraph{Encoder} We use $L^n$ and $C^n$ to denote the embeddings for literals and clauses at the message passing round $n$. In addition, we have hidden states for literals and clauses, denoted by $L_h^n, C_h^n$. Let $M$ be the bipartite adjacency matrix of the $\text{LIG}^{+}$. NeuroSAT encoder is parameterized by two MLPs ($L_{msg}, C_{msg}$) and two layer-norm LSTMs ($L_u, C_u$), which are shared for all rounds. Then for round $n$, embeddings are updated as:
\begin{align*}
(C^{n+1}, C_h^{n+1}) & \leftarrow C_u([C_h^n, M^TL_{msg}(L^n)])\\
(L^{n+1}, L_h^{n+1}) & \leftarrow L_u([L_h^n, FLIP(L^n), MC_{msg}(C^{n+1})]),
\end{align*}
where $FLIP$ swaps the literal and its negation in the embedding.
In this work, we always set the number of message passing rounds to $26$. We use $L^{26}$ as the final literal embedding. The graph-level representation can be computed as an average of the embeddings for all literals.
\paragraph{Aggregator} Given the embeddings $L^{N}$ from the encoder, we project each literal representation to a single scalar using an MLP $L_{vote}$: $V \leftarrow L_{vote} (L^N)$. Then we minimize the cross-entropy loss between the true label and $sigmoid(mean(L_{vote}))$
\section{Augmentation Rates for LAAs\xspace ans LPAs\xspace}
We also investigated the effect of augmentation rates for LAAs\xspace and LPAs\xspace, with results presented in Figure \ref{fig: gcl_hyper} and \ref{fig: our_hyper}.
\label{appendix: hyperparameter}
\begin{figure}[h]
\centering
\begin{tikzpicture}
\node at (0.65,0.1) {\includegraphics[width =.62\textwidth]{body/fig/percent_plot_gcl.pdf}};
\footnotesize
\draw (-3.4, 4.4) node {SR(10)};
\draw (-3.4 + 2.4, 4.4) node {PR(10)};
\draw (-3.4 + 2 * 2.4, 4.4) node {DP(20)};
\draw (-3.4 + 3 * 2.4, 4.4) node {PS(20)};
\draw (0, -4.3) node {Augmentation Rate (\%)};
\node[rotate=90] at (-5, 0) {Test Accuracy (\%)};
\end{tikzpicture}
\caption{The effect of rate of LAAs on the accuracy across datasets. The diamond markers indicate the maximum achieved accuracy which was the rate that was subsequently chosen to produce the heatmap of Figure \ref{fig: heatmap}.}
\label{fig: gcl_hyper}
\end{figure}
\begin{figure}[h]
\centering
\begin{tikzpicture}
\node at (0.65,0.1) {\includegraphics[width =.62\textwidth]{body/fig/percent_plot_ours.pdf}};
\footnotesize
\draw (-3.4, 3.4) node {SR(10)};
\draw (-3.4 + 2.3, 3.4) node {PR(10)};
\draw (-3.4 + 2 * 2.3, 3.4) node {DP(20)};
\draw (-3.4 + 3 * 2.3, 3.4) node {PS(20)};
\draw (0, -3.3) node {Augmentation Rate (\%)};
\node[rotate=90] at (-5, 0) {Test Accuracy (\%)};
\end{tikzpicture}
\caption{The effect of rate of our LPA augmentations on the accuracy across datasets. The diamond markers indicate the maximum achieved accuracy which was the rate that was subsequently chosen to produce the heatmap of Figure \ref{fig: heatmap}.}
\label{fig: our_hyper}
\end{figure}
\section{Datasets}\label{appendix: dataset}
\subsection{Brief Description}
\begin{itemize}
\item \textbf{SR \citep{selsam2018learning}} A random SAT generator proposed as a challenge for neural networks to learn intrinsic properties about satisfiability without cheating on some miscellaneous statistics about the dataset.
\item\textbf{PR \citep{ansotegui2009towards}} A random k-SAT generator where the frequency of each variable is sampled from a power-law distribution.
\item\textbf{DP \citep{ansotegui2009towards}} A pseudo-industrial generator based on PR with varying clause length
\item\textbf{PS \citep{giraldez2017locality}} A pseudo-industrial generator based on the notion of locality.
\item\textbf{UR} A random k-SAT generator where each variable is sampled uniformly.
\item\textbf{CA \citep{giraldez2015modularity}} A pseudo-industrial generator based on the notion of modularity.
\end{itemize}
\subsection{Parameters}
For the purpose of reproducibility, we show the parameters of each generator used in the paper below:
\begin{itemize}
\item \textbf{SR(10) / SR(40)} All parameters follow NeuroSAT.
\item \textbf{PR(10)} Number of variables: 10. Number of clauses: 41. Variable per clause: 3. Power-law exponents of variables: 1.7.
\item \textbf{PR(40)} Number of variables: 40. Number of clauses: 147. Variable per clause: 3. Power-law exponents of variables: 2.5.
\item \textbf{DP(20)} Number of variables: 20. Number of clauses: 34. Average variables per clause: 4. Power-law exponents of variables: 1.7.
\item \textbf{DP(40)} Number of variables: 40. Number of clauses: 75. Average variables per clause: 5. Power-law exponents of variables: 1.7.
\item \textbf{PS(20)} Number of variables: 20. Number of clauses: 58. Min variable per clause: 2. Average variables per clause: 4.
\item \textbf{PS(40)} Number of variables: 40. Number of clauses: 73. Min variable per clause: 2. Average variables per clause: 5.
\end{itemize}
\section{Does our method work with other GNN architectures?}
We also studied if our framework could be used with more common architectures, such as, graph convolutional networks (GCNs). We chose the GCN architecture in NeuralDiver \citep{nair2020solving} without the residual connection. The number of layers is set to be $10$. Table \ref{table: linear_gcn} shows that SSL + LPA still dominates in the low-label regime.
\label{appendix: other_gnn}
\begin{table}[h]
\centering
\label{table: linear_gcn}
\vskip 0.15in
\begin{tabular}{lllllll}
\toprule
& 2 & 10 & 100 & 1000 & 5000 & 10000 \\
\cmidrule{2-7}
Supervised & 49.82 & 50.08 & 50.68 & 50.23 & 51.25 & 78.82 \\
SSL + LPA & 60.35 & 70.26 & 73.17 & 75.29 & 75.27 & 77.54 \\
SSL + LAA & 51.32 & 51.29 & 48.93 & 50.72 & 50.41 & 63.24 \\
\bottomrule
\end{tabular}
\caption{Linear evaluation performance of different methods with GCNs on SR(10).}
\end{table}
\section{Does our method work with other contrastive loss functions?}\label{appendix: vicreg}
We replaced the SimCLR's loss function in Equation \ref{equ: contrastive} with the VICReg loss \citep{bardes2021vicreg}. We set $\lambda = 15, \mu = 1, \nu = 1$ for the hyperparameters of VICReg. As shown in Table \ref{table: vicreg}, linear evaluation performance for both objective functions are quite close.
\begin{table}[h]
\centering
\label{table: vicreg}
\vskip 0.15in
\begin{tabular}{lllllll}
\toprule
& 2 & 10 & 100 & 1000 & 5000 & 10000 \\
\cmidrule{2-7}
Ours + SimCLR & 79.53 & 88.32 & 92.23 & 93.32 & 93.01 & 95.12 \\
Ours + VICReg & 76.38 & 89.15 & 91.27 & 93.98 & 94.01 & 94.77 \\
\bottomrule
\end{tabular}
\caption{Linear evaluation performance of our methods with SimCLR and VICReg loss on SR(10)}
\end{table}
\section{More Transfer Learning Results}
See Figure \ref{fig: experiment_transfer}.
\label{appendix: transfer}
\begin{figure}[h]
\centering
\begin{tikzpicture}
\node at (0,0) {\includegraphics[scale=0.5]{body/fig/extrapolation.pdf}};
\end{tikzpicture}
\caption{ Each bar represents the SSL model trained on a different source dataset, and each subplot is the target dataset $4$ SSL models are evaluated on. The number on top of each bar is the linear evaluation accuracy with 20 labels from the target dataset. The blue line is the fully-supervised baseline trained on $20$ labels on the target baseline, and the red line is the supervised baseline trained on $10000$ labels. The $\blacklozenge$ symbol emphasizes that the train and test datasets are from the same distribution. \emph{Left}: Transfer to unseen problems of similar size. \emph{right}: Transfer to unseen problems of larger size.}
\label{fig: experiment_transfer}
\end{figure}
\section{Evaluation on Smaller Datasets}
We also performed linear evaluation and fine-tuning experiments for the smaller datasets in Section \ref{sec: augmentation}. The results are shown in Figure \ref{fig: small_result}.
\label{appendix: eval_small}
\begin{figure*}[h]
\centering
\begin{tikzpicture}
\node at (0,0) {\includegraphics[scale=0.41]{body/fig/small_LE.pdf}};
\node at (0,-4) {\includegraphics[scale=0.41]{body/fig/small_FT.pdf}};
\node[bag, rotate=90] at (-8.4, -0.2) {Linear Evaluation};
\node[bag, rotate=90] at (-8.4, -3.5) {Fine-tuning};
\scriptsize
\node[bag, rotate=90] at (-8, -0.2) {Accuracy (\%)};
\node[bag, rotate=90] at (-8, -3.5) {Accuracy (\%)};
\draw (-5.75, -5.8) node {No. of Training Labels};
\draw (-1.75, -5.8) node {No. of Training Labels};
\draw (2.5, -5.8) node {No. of Training Labels};
\draw (6.5, -5.8) node {No. of Training Labels};
\end{tikzpicture}
\caption{Our method (SSL + LPA) achieves significantly higher accuracy after linear evaluation and fine-tuning than baselines in low-label regime and is comparable to supervised models that have been given more labels. We vary the number of training labelled instances from $2$ to $10^4$ and report the average accuracy and standard error over 3 trials for all methods.}
\label{fig: small_result}
\end{figure*}
\section{Decision step} \label{appendix: decision}
We measured the decision steps of CryptoMiniSat \cite{soos2009extending} solvers before and after our augmentations. The result is shown in Table \ref{table: decision_step}.
\begin{table}[h]
\centering
\label{table: decision_step}
\vskip 0.15in
\begin{tabular}{lcccccccc}
\toprule
& \multicolumn{2}{c}{SR(40)} & \multicolumn{2}{c}{PR(40)} & \multicolumn{2}{c}{DP(40)} & \multicolumn{2}{c}{PS(40)} \\
\cmidrule{2-9}
& SAT & UNSAT & SAT & UNSAT & SAT & UNSAT & SAT & UNSAT \\
\cmidrule{2-9}
Before & $17.54$ & $13.25 $ & $18.21$ & $12.14$ & $32.78$ & $0.27$ & $33.26$ & $7.37$ \\
After & $16.36$ & $16.40$ & $21.61$ & $12.64$ & $28.47$ & $0.36$ & $34.83$ & $6.92$ \\
\bottomrule
\end{tabular}
\caption{Decision steps of CryptoMiniSat \cite{soos2009extending} solvers before and after augmentations used in Figure \ref{fig: main_result}. We use the same augmentation as the corresponding SSL model in Figure \ref{fig: main_result}. The statistics is computed over $1000$ instances for each dataset. }
\end{table}
\section{Empirical Study of Augmentations for SAT}
\label{sec: augmentation}
\begin{figure*}[htbp]
\centering
\begin{tikzpicture}
\node at (0,0) {\includegraphics[width=.75\textwidth]{body/fig/heat_ours.pdf}};
\node at (0,-3.2) {\includegraphics[width=.75\textwidth]{body/fig/heat_gcl.pdf}};
\footnotesize
\draw (-4.5, -5) node {SR(10)};
\draw (-1.7, -5) node {PR(10)};
\draw (1.3, -5) node {DP(20)};
\draw (4.2, -5) node {PS(20)};
\node[rotate=90] at (-6.6, -0.2) {LPA};
\node[rotate=90] at (-6.6, -3.5) {LAA};
\end{tikzpicture}
\caption{Contrastive learning with LPAs\xspace learns much better representations for SAT predictions than with LAAs\xspace.
We evaluated the linear classification performance of SSL models trained with different single augmentations (diagonal) or paired combinations (off-diagonal).
The four heatmaps on the first row show results using LPAs\xspace, and the ones on the second row using LAAs\xspace.
Each column denotes different datasets.
For off-diagonal entries on each heatmap, the row corresponds to the first augmentation applied.
}
\label{fig: heatmap}
\end{figure*}
Our first experiments study the role of augmentations in contrastive learning for SAT prediction. We followed the standard linear evaluation protocol to evaluate representations \citep{chen2020simple}, i.e., we report test accuracy of a linear classifier trained on top of frozen representations.
\textbf{Architecture.} We primarily used the encoder of NeuroSAT \citep{selsam2018learning} as the GNN architecture. We also re-ran a small number of experiments with another type of GNN in Appendix \ref{appendix: other_gnn}. Using the NeuroSAT architecture, we obtained the graph-level representations by average-pooling over all literal representations. We followed the design details of the original NeuroSAT paper.
\textbf{Experimental Setting.}
We used the contrastive loss in Equation \ref{equ: contrastive} with the temperature $0.5$. Appendix \ref{appendix: vicreg} also shows some experiments with other contrastive objectives. We used Adam optimizer with learning rate $2 \times 10^{-4}$ and weight decay $10^{-5}$. The batch size was $128$ and the maximum training epoch was $5000$. The generator produced a set of new unlabelled instances for each batch. We used 100 labelled instances to train our linear evaluators, and another $500$ as the validation set to pick the hyperparameters (ranging from $10^{-3}$ to $10^3$) of $L_2$ regularization. The test set consisted of $10^4$ instances.
\textbf{Datasets.} We experimented using four generators: SR \citep{selsam2018learning}, Power Random 3SAT (PR) \citep{ansotegui2009towards}, Double Power (DP) and Popularity Similarity (PS) \citep{giraldez2017locality}. SR and PR are the synthetic generators. DP and PS are pseudo-industrial generators producing instances that mimic real-world problems. We generated instances of 10 variables for SR and PR, and 20 for DP and PS. The number inside the parenthesis denotes the number of variables per instance. We also tweaked the parameters of the generators so that they produce roughly balanced SAT and UNSAT. More details are in Appendix \ref{appendix: dataset}.
\textbf{Augmentations.} We used four of the LPAs from Section \ref{subsec: augmentation_sat}, namely: AU\xspace, SC\xspace, CR\xspace and VE\xspace. The augmentations UP and PL were not studied because most of our instances originally do not contain unit clauses or pure literals.
For our baseline we adopted four LAAs\xspace from GraphCL \citep{you2020graph} with some adjustments. The issue with the original augmentations was that they operate directly on the graph. However, NeuroSAT requires the input to be in the $\text{LIG}^{+}$ format, and blindly applying these augmentations might break that structure. Thus, we adapted GraphCL augmentations to maintain the $\text{LIG}^{+}$ structure, resulting in the following augmentations: \textit{drop clauses} (DC\xspace), \textit{drop variables} (DV\xspace), \textit{link perturbation} (LP\xspace) and \textit{subgraph} (SG\xspace). DC\xspace and DV\xspace correspond to \textit{node dropping}. LP corresponds to \textit{edge perturbation}, where we randomly add/remove links between literals and clauses on a SAT instance. Lastly, SG is similar to the one in GraphCL where we do a random walk on an instance and keep the resulting subgraph.
All $8$ augmentations except SC\xspace are parameterized by $p$ which controls the intensity of perturbations. For SC\xspace, we eliminate all subsumed clauses. In this section, we chose the $p$ that achieved the best linear evaluation performance
when the corresponding augmentation was applied alone. The results for tuning $p$ are shown in Appendix \ref{appendix: hyperparameter}.
\subsection{Label-Preserving Augmentations are Necessary}
\begin{table}[tb]
\vskip 0.15in
\label{tab: random_resolution}
\small
\centering
\begin{tabular}{l|cccc}
\toprule
\diagbox{Type}{Dataset} & SR & PR & DP & PS \\
\midrule
Random & 55.6 & 52.8 & 64.7 & 68.9 \\
Resolution (ours) & \textbf{86.4} & \textbf{92.6} & \textbf{84.8} & \textbf{96.9} \\
\bottomrule
\end{tabular}
\caption{Adding SAT-preserving clauses (Resolution) leads to much higher accuracy than adding random ones (Random). The number of clauses added is controlled to be the same for both.}
\label{table:random_resolution}
\end{table}
\textbf{Do LPAs\xspace \ learn better representations than LAAs\xspace?}
To investigate the effect of using different augmentations, we evaluated the linear classification performance of SSL models trained with different single augmentations or paired combinations.
As Figure \ref{fig: heatmap} shows,
the accuracy ($\%$) for the best LPA pair is: $95.1$ for SR, $93.3$ for PR, $86.0$ for DP and $96.9 $ for PS, while the corresponding number ($\%$) for LAAs\xspace is much lower: $54.4, 59.1, 70.0$ and $65.5$. These results show that SSL models trained with LPAs\xspace learn significantly better representations than those with LAAs\xspace.
\textbf{Do our gains simply come from adding clauses?} The four LAAs from GraphCL do not add new clauses to the original instance. To investigate whether this explains the performance gap, we compared CR\xspace against adding the same number of randomly generated clauses (\textit{adding random clauses}). We eliminated subsumed clauses (SC) after both augmentations. As shown in Table \ref{table:random_resolution}, adding SAT-preserving resolvents achieved at least $20 \%$ higher accuracy than adding randomly generated clauses.
\textbf{Does combining LPAs\xspace \ and LAAs\xspace \ hurt performance?} In Table \ref{tab: nspp_after_spp}, we trained SSL models with VE\xspace \ followed by $7$ different augmentations on SR(10). We compared the accuracy change with the model trained with VE\xspace \ only. From Table \ref{tab: nspp_after_spp}, adding LAAs resulted in $30 \%$ drop in accuracy, while LPAs decreased the accuracy by at most $3 \%$.
In summary, the performance gap between LPAs and LAAs meets our intuition. LAAs do not guarantee preserving satisfiability, which could result in false positive pairs that hurt the SAT prediction performance.
\begin{table}[tb]
\vskip 0.15in
\label{tab: nspp_after_spp}
\centering
\small
\begin{tabular}{ccccccc}
\toprule
\multicolumn{3}{c}{LPAs\xspace} & \multicolumn{4}{c}{LAAs\xspace} \\
\cmidrule(lr){1-3} \cmidrule(lr){4-7}
AU\xspace & CR\xspace & SC\xspace & DC\xspace & DV\xspace & LP\xspace & SG\xspace \\
\midrule
\ccar{-1.5} & \ccay{0.7} & \ccay{1.2} & \textbf{\ccar{-34.4}} & \textbf{\ccar{-32.4}} & \textbf{\ccar{-33.7}} & \textbf{\ccar{-35.9}} \\
\bottomrule
\end{tabular}
\caption{Adding LAAs\xspace to LPAs\xspace significantly hurts the performance. Cells represent the difference of linear evaluation accuracy between training with VE followed by another aug and with VE alone on SR(10). Yellow indicates improved accuracy.}
\end{table}
\subsection{Type, Order, and Strengths of LPAs are Crucial}
Previous studies of contrastive learning for image \citep{chen2020simple} and graph \citep{you2020graph} have shown that the quality of learned representations relies heavily on finding the right type and composition of augmentations. We observe a similar pattern in SAT. In Figure \ref{fig: heatmap}, the accuracy gap between the best LPA combination and the worst is between $20 - 40 \%$ for all datasets.
The conjecture in SimCLR \citep{chen2020simple} is that stronger augmentations lead to better representations. Intuitively, weaker augmentations often create very correlated positive examples, providing shortcuts for neural networks to cheat in the contrastive task without learning meaningful representations.
Based on this conjecture, we study what type, order, and strengths of LPAs induce harder positive pairs and thus better representation quality.
\textbf{Resolution-based augmentations are the most powerful.} In Figure \ref{fig: heatmap}, the best pair for each dataset includes either CR\xspace \ or VE\xspace . Both of them are based on the resolution rule in Equation \ref{equ: resolution}.
Resolution is a powerful inference rule in propositional logic. In fact, we can build a sound and complete propositional theorem prover with only resolutions \citep{genesereth2013introduction}. The Davis–Putnam algorithm \citep{davis1960computing}, the basis of practical SAT solvers, iteratively applies resolution until reaching satisfiability certificates. Resolution-based augmentations in contrastive learning may help the neural work learn the essence of resolution, which leads to better satisfiability prediction.
\textbf{Composing different augmentations is beneficial across datasets.} The highest accuracy in Figure \ref{fig: heatmap} always comes from off-diagonal entries, i.e., composition of different augmentations. In PR, every singular augmentation failed to obtain accuracy higher than $60 \%$ by itself.
While singular augmentations achieved decent accuracy on SR and DP, combinations further improved the accuracy. Similar to image and graph, composing different augmentations resulted in harder positives and better representations.
\textbf{Eliminating subsumed clauses after adding resolvents is particularly helpful.} CR\xspace followed by SC\xspace \ consistently produced high-quality representations across all datasets. Using CR\xspace \ alone tended to perform substantially worse. The accuracy drop without SC\xspace is: $28.1 \%$ for SR, $33.4 \%$ for PR, $7.0 \%$ for DP, $25.3 \%$ for PS. Without SC\xspace, all original clauses are preserved by the augmentation, leading to trivial positive pairs. In other words, without SC\xspace the GNN may have solved the contrastive task by finding a common subgraph, not by learning anything meaningful about resolution rules.
Interestingly, swapping the order of this pair also decreased accuracy by $20-30 \%$ for all datasets. We conjecture that this may be an artefact of our specific generators. The percent of subsumed clauses in the original instances (over $1000$ samples) was $43, 0, 76, 69$ for SR, PR, DP and PS. Applying SC\xspace first had no effect on PR due to no subsumed clauses present. For the other three, SC\xspace may have eliminated too many clauses, which hurt the diversity of resolvents created in the next phase.
\begin{figure}[bt]
\centering
\begin{tikzpicture}
\node at (0,0) {\includegraphics[width =.2\textwidth]{body/fig/percent_plot_single.pdf}};
\footnotesize
\node[bag, rotate=90] at (-2.2, 0) {Linear Evaluation \\ Accuracy (\%)};
\draw (0.4, -2) node {Rate of Clause Resolution (\%)};
\end{tikzpicture}
\caption{Adding more resolvents improves the performance across all datasets. Each point represents linear evaluation accuracy of an SSL model trained with CR\xspace \ of the specified rate (i.e., \# of added resolvents divided by \# of original clauses) followed by SC\xspace.}
\label{fig: more_clause}
\end{figure}
\textbf{More resolvents lead to better representations.} Figure \ref{fig: more_clause} studies the effect of the number of resolvents added in CR\xspace. In general, adding more resolvents improved the linear evaluation performance across all datasets. More resolvents not only add more new clauses, but also may help SC\xspace \ delete more original clauses, which create harder positives.
\subsection{Do Our Augmentations Reveal the True Label?}
As mentioned before, CR\xspace and VE\xspace are powerful enough to build a sound and complete solver. Because our instances are relatively small, it is important to ask if our augmentations ``accidentally'' reveal the true label(SAT/UNSAT) to the model. We do not believe this is the case for the following reasons.
Our augmentations (polytime complexity) cannot in general determine whether a formula is satisfiable (exponential in the worst case). When applied to our datasets, VE\xspace only eliminated between 1-4 variables, and CR\xspace added between 10-20\% of the total number of original clauses. This amount of work alone is not enough to solve SAT.
To assess whether our augmentations are effectively solving our \emph{specific} instances, we measured the number of decision steps of CryptoMiniSat solvers \cite{soos2009extending} on different datasets before and after our LPAs\xspace. If our LPAs were close to solving the instance, the decision steps should have decreased dramatically after LPAs were applied. However, Table \ref{table: decision_step} in Appendix \ref{appendix: decision} shows the decision steps remain relatively stable after augmentations.
\section{Background}
\label{sec: background}
\textbf{SAT.} A Boolean formula in propositional logic consists of
Boolean variables composed by logical operators ``and'' ($\land$), ``or'' ($\lor$) and ``not'' ($\lnot$). A \emph{literal} is a variable $v$ or its negation $\lnot v$. A \emph{clause} is a disjunction of literals $\bigvee_{i=1}^n l_i$. A Boolean formula is in \emph{Conjunctive Normal Form} (CNF) if it is a conjunction of clauses. We assume that all formulas are in CNF.
A variable assignment satisfies a clause when at least one of its
literals is satisfied. A formula $\phi$ is satisfied under $\pi$ when all of its clauses are satisfied, and $\pi$ is a
\emph{satisfying assignment} for $\phi$. \emph{SAT} is the problem of deciding, for a given formula $\phi$, if there exists a satisfying assignment (SAT) or not (UNSAT).
We can represent a SAT formula $\phi$ by a bipartite graph
called \emph{literal-clause incidence graph} (LIG) (Figure \ref{fig:sat_representation}). The graph contains a node for every clause and literal of $\phi$. An edge connects a clause node and a literal node iff the clause contains that literal. We define $\text{LIG}^{+}$ when the literal nodes of the same variable in LIG are connected.
\begin{figure}[tb]
\centering
\includegraphics[width=0.6\columnwidth]{body/fig/bipartite.pdf}
\caption{A bipartite graph representation ($\text{LIG}^{+}$) of a SAT formula $\phi:=(x \vee \neg y \vee \neg z) \wedge (\neg x \vee y \vee z)$. The subgraph without \protect\tikz[baseline]{\protect\draw[line width=0.5mm,densely dashed] (0,.8ex)--++(0.5,0);} edges is the LIG of $\phi$.}
\label{fig:sat_representation}
\end{figure}
\textbf{NeuroSAT.} \citet{selsam2018learning} proposed NeuroSAT to classify satisfiability of Boolean formulas. NeuroSAT consists of two parts: \textbf{1) \emph{Encoder:}} a special \emph{Graph Neural Network} (GNN) that takes in the $\text{LIG}^{+}$ representation of a formula and produces an embedding for its literals;
\textbf{2) \emph{Aggregator}:} A function that maps each literal representation to a vote and then aggregates them into a prediction about satisfiability.
More details can be found in Appendix \ref{appendix: neurosat}.
\section{Conclusions and Outlook}
We studied the effect of data augmentations on contrastive learning for the Boolean satisfiability problem. We designed label-preserving augmentations using well-studied transformations, e.g., clause resolution or variable elimination, and confirmed the hypothesis that data augmentations should be label-preserving to help in downstream prediction. The design of our augmentations was critical; we found that resolution-based augmentations, which produced more distinct augmentations, were necessary for strong results. Our contrastive method was able to learn strong SAT predictors (at least as strong as our best supervised baselines) with $100\times$ fewer labelled training instances.
Although our results are restricted to Boolean satisfiability, they hold lessons for solver design more broadly. For example, the convex hull of MILP feasible sets is invariant to cuts, which suggests that contrastive pre-training could be used to improve Neural Diving \citep{nair2020solving}. In general, our results strongly suggest that studying and exploiting invariances can dramatically improve the sample complexity of heuristics learned via imitation learning.
The study of combinatorial problems is fruitful for the broader machine learning community, because these problems are non-trivial and so much is known about their invariances. In particular, experiments can be designed that exactly satisfy the assumptions of invariant learning theory, making combinatorial problems fantastic test beds for the emerging field of contrastive and self-supervised learning.
\section{Comparison with Other Methods}
\label{sec:experiment}
The second set of experiments compared our proposed framework with other baselines in the setups of linear evaluation, fine-tuning and few-shot transfer learning. In addition to our model (SSL with LPA), we studied 4 baselines: SSL with LAA, supervised models trained without augmentations, supervised models trained with LPA or LAA. The details of training SSL models follow Section \ref{sec: augmentation}. For supervised models, we used the same architecture as NeuroSAT. The hyperparameters for supervised were the same as SSL in Section \ref{sec: augmentation}, except the learning rate was chosen to be $2 \times 10^{-5}$, following \citet{selsam2018learning}. For each dataset, we chose the best augmentation combination for LPA and LAA according to Figure \ref{fig: heatmap}. The degree parameter associated with each augmentation was tuned separately for SSL and supervised models. We used $200$ instances as validation sets for early stopping of all methods.
Unless otherwise specified, all datasets used in this section have 40 variables per instance; in the Appendix \ref{appendix: eval_small} we report results for the same experiments on smaller instances. For SR datasets, following NeuroSAT, we trained on SR(U(10, 40)) and tested on SR(40). Training procedures for all models were the same as Section \ref{subsec: experiment_linear}, except the batch size was $80$ for SR(40) to avoid memory issues. When fine-tuning NeuroSAT, we used different learning rates for the encoder and aggregator, which were separately tuned for different models on each dataset.
\subsection{Linear Evaluation}
\label{subsec: experiment_linear}
We first evaluated the linear evaluation accuracy of all methods following the procedure in Section \ref{sec: augmentation}.
We varied the number of labelled instances from $2$ to $10^4$.
As shown in the first row of Figure \ref{fig: main_result}, our method achieved substantially higher accuracy than others across all datasets in the low-label regime. For example, with $10$ training labelled instances, the improvement ($\%$) of ours compared from the second best method was: $9.42$ for SR, $14.07$ for PR, $30.81$ for DP, $22.18$ for PS. Our model's accuracy was also on par with supervised models that had access to significantly more labels: for all datasets, the accuracy of our models trained with $100$ labels matched or exceeded the accuracy of our best supervised baselines trained with $10^4$ labels, a $100\times$ reduction in the number of labels needed.
On the other hand, SSL models trained with LAAs were not much better than random-initialized ones under linear evaluation. We also found that using LAAs for supervised models even hurt performance, possibly because LAAs add label noise.
For example, with $10000$ labels on DP(40), adding LAAs gave $20.21 \%$ lower accuracy than supervised without augmentations.
\subsection{Fine-tuning}
We evaluated the fine-tuning accuracy of different SSL models compared to the supervised baselines. Specifically, we took the pre-trained SSL models and optimized them end-to-end on labelled data for a few epochs.
As shown in the second row of Figure \ref{fig: main_result}, the fine-tuning results resemble those of linear evaluation. In the low-label regime, our model (SSL + LPA) dominated across all datasets. In contrast, SSL with LAAs did not improve much from supervised models that were randomly initialized. As with linear evaluation, we observed a reduction in sample complexity of at least $100\times$.
\subsection{Few-shot Transfer Learning}
\begin{figure}[bt]
\centering
\begin{tikzpicture}
\node at (0,0) {\includegraphics[width =.3\textwidth]{body/fig/extrapolation_main.pdf}};
\node[rotate=90] at (-3.2, -0.4) {Target Dataset};
\draw (0, -3.2) node {Source Dataset};
\footnotesize
\node[rotate=90] at (-2.6, 0.8) {UR(40)};
\node[rotate=90] at (-2.6, -1.5) {CA(40)};
\draw (-1.7, -2.7) node {SR(10)};
\draw (-1.7 + 1.1, -2.7) node {PR(10)};
\draw (-1.7 + 1.1 * 2 + 0.1, -2.7) node {DP(20)};
\draw (-1.7 + 1.1 * 3 + 0.1, -2.7) node {PS(20)};
\end{tikzpicture}
\caption{The representations of our model (solid bars) transfer better from source to unseen target dataset than those of supervised (hatched bars) trained on the same source. 10-shot transfer accuracy (\%) was evaluated for each setup. The blue line is the accuracy of fully-supervised models trained on each target dataset with $10k$ labels.}
\label{fig: transfer_supervised}
\end{figure}
We investigated how well the SSL models trained in Section \ref{subsec: experiment_linear} perform in 10-shot transfer to unseen and larger datasets, Uniform Random 3SAT (UR) and Community Attachment (CA) \citep{giraldez2015modularity}. UR is a synthetic generator, and CA is pseudo-industrial. We trained a logistic regression classifier on the fixed representations with $10$ SAT and $10$ UNSAT instances from the target dataset.
We compared transferability of representations from SSL with those from supervised representations trained on the same source dataset. In particular, the supervised baseline was trained using $100,000$ labelled source instances with the same augmentations as SSL. Only the encoder of the supervised models was used for extracting representations on target datasets.
\textbf{Do our representations transfer better than supervised?} As shown in Figure \ref{fig: transfer_supervised}, we found that our SSL models generally transferred better than the supervised baseline. This improvement also tended to be larger when the source and target domain were more distinct, such as from pseudo-industrial instances (DP, PS) to random 3 SAT (UR), and from random 3SAT (PR) to pseudo-industrial instances (CA). This meets our intuition, because SSL does not leverage labels in the source dataset, and thus reduces overfitting on source labels \cite{yang2020transfer}.
\textbf{Does transferability vary with the source dataset?} We also found that transferability of representations depended heavily on the source datasets. Generally speaking, models trained on pseudo-industrial generators, DP and PS, transferred worse than the synthetic generators, SR and PR. We conjecture that small industrial instances are not challenging enough for models to learn generalizable representation of unseen problem domains.
In Appendix \ref{appendix: transfer}, we also performed an experiment to show how the model trained on SR, PR, DP and PS transfer to each other, which supports similar conclusions: models trained on SR and PR transferred better to DP and PS than vice verse.
\section{Introduction}
\label{sec:introduction}
Combinatorial problems, e.g., Boolean satisfiability (SAT) or mixed-integer linear programming (MILP), have many applications in the industry and can encode many fundamental computational tasks. These problems are NP-complete, so solvers that perform efficiently in the worst case are not within reach. However, learning can be used to improve the average complexity of solvers on the population of combinatorial problems found in the wild \citep{nair2020solving}.
Supervised learning is a promising approach to combinatorial solver design \citep[e.g.,][]{selsam2019guiding,nair2020solving}. Unfortunately, the need for labels is a severe limitation. Many read-world problems, such as cryptography SAT instances, are extremely hard or, in some cases, even impossible to solve \citep{nejati2019cdcl}. Computing expert branching labels for large-scale MILPs requires sophisticated parallel solvers \citep{nair2020solving}.
In order to scale learning for combinatorial problems, we ask: how much can we learn from \emph{unlabelled} combinatorial instances?
In this work, we consider a contrastive learning approach, which begins by creating multiple ``views'' of every unlabelled instance, a process called augmentation. An encoder is trained to maximize the similarity between the representations of augmentations that come from the same instance, while minimizing the similarity between those of distinct ones \citep{chen2020simple}. This has been successful in computer vision: contrastive representations can be used with linear predictors to achieve competitive accuracies on ImageNet using a fraction of the labelled instances \citep{chen2020simple, he2020momentum, chen2020improved}.
Our key insight is that combinatorial problems have well-studied invariances that can be used to design extremely effective augmentations for contrastive learning. Contrastive learning theory indicates that augmentations should be (roughly) label-preserving in order to confer guarantees on downstream prediction \citep{arora2019theoretical, tosh2021contrastive, haochen2021provable}. This is in contrast to the majority of label-agnostic graph contrastive frameworks \citep[e.g.,][]{you2020graph,hassani2020contrastive}. For some combinatorial problems, label-preserving transformations are available from subroutines of existing solvers, e.g., variable elimination modifies SAT formulas while preserving their satisfiability. Our augmentations produce new formulas by randomly applying such satisfiability-preserving transformations. Crucially, our augmentations do not require full solves and are much cheaper to compute than the labels.
We study data augmentations for contrastive learning for Boolean satisfiability. We demonstrate that:
\begin{itemize}
\itemsep0em
\item Augmentation design is critical for contrastive pre-training on combinatorial problems. In particular, our augmentations are sufficient for strong performance, while existing graph augmentations are not.
\item Our contrastive method matches the best supervised baseline while using $100\times$ fewer training labels.
\item Our contrastive representations transfer better to new problem types than supervised representations.
\end{itemize}
\section{Framework Overview}
\begin{figure*}[th]
\centering
\begin{tikzpicture}
\node at (0,0) {\includegraphics[width=2\columnwidth]{body/fig/framework.pdf}};
\scriptsize
\draw (-7.7, 0.4) node[right] {$c_1:x_1$};
\draw (-7.7, 0.4 - 0.35) node[right] {$c_2: x_2 \lor x_3 $};
\draw (-7.7, 0.4 - 2 * 0.35) node[right] {$c_3: x_1 \lor \neg x_3 \lor x_4 $};
\draw (-7.7, 0.4 - 3 * 0.35) node[right] {$c_4: \neg x_1 \lor x_2 \lor x_3 \lor \neg x_4$};
\draw (-2.8, 1.45) node[right] {\st{$c_1: x_1$}};
\draw (-2.8, 1.45 - 0.35) node[right] {$c_2: x_2 \lor x_3$};
\draw (-2.8, 1.45 - 2 * 0.35) node[right] {\st{$c_3: x_1 \lor \neg x_3 \lor x_4 $}};
\draw (-2.8, 1.45 - 3 * 0.35) node[right] {$c_4\!\!: $\st{$\neg x_1 \lor$} $ x_2 \lor x_3 \lor \neg x_4$};
\draw (-2.8, -0.55) node[right] {$c_1: x_1$};
\draw (-2.8, -0.55 - 0.35) node[right] {\st{$c_2: x_2 \lor x_3$}};
\draw (-2.8, -0.55 - 2 * 0.35) node[right] {\st{$c_3: x_1 \lor \neg x_3 \lor x_4 $}};
\draw (-2.8, -0.55 - 3 * 0.35) node[right] {\st{$c_4: \neg x_1 \lor x_2 \lor x_3 \lor \neg x_4$}};
\draw (-2.8, -0.55 - 4 * 0.35) node[right] {\color{our_red} $c_5: x_1 \lor x_2 \lor x_4$};
\footnotesize
\draw (-5.3, 2.) node[left] {Label};
\draw (-5.3, 1.65) node[left] {Preserving};
\draw (-5.3, 1.35) node[left] {Augmentations};
\normalsize
\draw (5.8, 0.55) node {Encoder};
\draw (5.8, -1.4) node {Encoder};
\draw (2.5, 0.55) node {$G_1$};
\draw (2.5, -1.7) node {$G_2$};
\draw (1.2, 0.65) node {$\mathcal{E}$};
\draw (1.2, -1.45) node {$\mathcal{E}$};
\draw (6.8, 0) node {$\mathbf{z_1}$};
\draw (6.8, -1.95) node {$\mathbf{z_2}$};
\large
\draw (-6.0, -1.1) node {$\phi$};
\draw (-1.2, -0.1) node {$\hat{\phi}_1$};
\draw (-1.2, -2.4) node {$\hat{\phi}_2$};
\Large
\draw (7.8, -0.25) node {$\mathcal{L(.)}$};
\end{tikzpicture}
\caption{Our contrastive learning framework for combinatorial problems. Given an instance $\phi$, a pair of augmented samples $(\hat{\phi}_1$, $\hat{\phi}_2)$ are formed using label-preserving augmentations. $(\hat{\phi}_1$, $\hat{\phi}_2)$ are then transformed to graph formats $(G_1, G_2)$. An encoder is used to extract graph representations $(\mathbf{z}_1, \mathbf{z}_2)$. Lastly, a standard contrastive loss is applied over a mini-batch of instances.}
\label{fig:cocl_model}
\end{figure*}
Similar to contrastive algorithms in other domains (e.g., images), we learn representations by contrasting augmented views of the same instance against negative samples.
Our framework (Figure \ref{fig:cocl_model}) consists of four major components:
\textbf{Augmentations.}
Given a combinatorial instance $\phi$, a stochastic augmentation is applied to form a pair of positive samples, denoted by $(\hat{\phi}_1$, $\hat{\phi}_2)$. Our key insight is the tailored design of augmentations for combinatorial problems should preserve the label of the problem, e.g., satisfiability for SAT, as detailed in Section \ref{sec: lpa}.
\textbf{Format Transformation ($\mathcal{E}$).}
Formulas $(\hat{\phi}_1$, $\hat{\phi}_2)$ are transformed into (e.g., $\text{LIG}^{+}$) graphs $(G_1, G_2)$.
\textbf{Encoder.}
A neural encoder is trained to extract graph-level representations $\mathbf{z}_1, \mathbf{z}_2 \in \mathbb{R}^d$ from the augmented graphs $G_1$ and $G_2$. This encoder can be any GNN such as GCN \citep{kipf2016semi} and the encoder proposed by NeuroSAT.
\textbf{Contrastive Loss.}
In the end, SimCLR's \citep{chen2020simple} contrastive loss $\mathcal{L}$ is applied to the graph representations $\{ \mathbf{z}_i \}_{i=1}^{2n}$, obtained from a mini-batch of $n$ instances.
We follow \citet{chen2020simple} and use an MLP projection head to project each $\mathbf{z}_i$ to $\mathbf{m}_i$. For a positive pair of projected representations $(\mathbf{m}_i, \mathbf{m}_j)$, the loss $\mathcal{L}_{i, j}$ is a cross-entropy loss of differentiating the positive pair from the other $2(n-1)$ negative samples
(i.e., augmented samples of other instances):
\begin{align}
\mathcal{L}_{i, j} = -\log \frac{\exp(sim(\mathbf{m}_i, \mathbf{m}_j)\mathbin{/} \tau)}{\sum_{k=1}^{2N} \mathbf{1}_{k \neq i} \exp(sim(\mathbf{m}_i, \mathbf{m}_k)\mathbin{/} \tau)},
\label{equ: contrastive}
\end{align}
where $\tau$ is the temperature parameter, $\mathbf{1}$ is the indicator function, and $sim$ measures the similarity between two representations: $sim(\mathbf{m}_i, \mathbf{m}_j) := \mathbf{m}_i^T \mathbf{m}_j / \left\Vert \mathbf{m}_i \right\Vert \left\Vert \mathbf{m}_j \right\Vert$. The final loss $\mathcal{L}$ is the average of $\mathcal{L}_{i, j}$ over all positive pairs.
After training is completed, we only keep the encoder to extract the representation $\mathbf{z}_i$ for downstream tasks.
\iffalse
In the end, a contrastive loss function $\mathcal{L}$ is applied to the graph representations from a mini-batch of samples $\{ ( \mathbf{z}_i) \}_{i=1}^{2n}$. Unless otherwise specified, we use SimCLR's NT-Xent loss \citep{chen2020simple}, but our framework is flexible for any other contrastive loss. A multi-layer perceptron first projects each graph representation $\mathbf{z}_i$ to $\mathbf{m}_i$. Then the loss $\mathcal{L}_{i, j}$ is applied to a positive pair of the projected representations $(\mathbf{m}_i, \mathbf{m}_j)$:
\begin{align}
\mathcal{L}_{i, j} = -\log \frac{\exp(sim(\mathbf{m}_i, \mathbf{m}_j)\mathbin{/} \tau)}{\sum_{k=1}^{2N} \mathbf{1}_{k \neq i} \exp(sim(\mathbf{m}_i, \mathbf{m}_k)\mathbin{/} \tau)},
\label{equ: contrastive}
\end{align}
where $\tau$ is the temperature parameter, $\mathbf{1}$ is the indicator function, and $sim$ measures the similarity between two representations: $sim(\mathbf{m}_i, \mathbf{m}_j) := \mathbf{m}_i^T \mathbf{m}_j / \left\Vert \mathbf{m}_i \right\Vert \left\Vert \mathbf{m}_j \right\Vert$. Loss $\mathcal{L}_{i, j}$ quantifies how well the encoder can differentiate the positive pairs from the $2(n-1)$ negative samples.
The training objective is to minimize the total loss $\mathcal{L}~=~\frac{1}{n(n-1)} \sum_{i, j (i\neq j) }\mathcal{L}_{i, j}$.
After training is completed, we only keep the encoder to extract the representation for downstream tasks.
\fi
\section{Framework Overview}
\begin{figure*}[th]
\centering
\begin{tikzpicture}
\node at (0,0) {\includegraphics[width=2\columnwidth]{body/fig/framework.pdf}};
\scriptsize
\draw (-7.4, 0.4) node[right] {$c_1:x_1$};
\draw (-7.4, 0.4 - 0.35) node[right] {$c_2: x_2 \lor x_3 $};
\draw (-7.4, 0.4 - 2 * 0.35) node[right] {$c_3: x_1 \lor \neg x_3 \lor x_4 $};
\draw (-7.4, 0.4 - 3 * 0.35) node[right] {$c_4: \neg x_1 \lor x_2 \lor x_3 \lor \neg x_4$};
\draw (-2.8, 1.45) node[right] {\st{$c_1: x_1$}};
\draw (-2.8, 1.45 - 0.35) node[right] {$c_2: x_2 \lor x_3$};
\draw (-2.8, 1.45 - 2 * 0.35) node[right] {\st{$c_3: x_1 \lor \neg x_3 \lor x_4 $}};
\draw (-2.8, 1.45 - 3 * 0.35) node[right] {$c_4\!\!: $\st{$\neg x_1 \lor$} $ x_2 \lor x_3 \lor \neg x_4$};
\draw (-2.8, -0.55) node[right] {$c_1: x_1$};
\draw (-2.8, -0.55 - 0.35) node[right] {\st{$c_2: x_2 \lor x_3$}};
\draw (-2.8, -0.55 - 2 * 0.35) node[right] {\st{$c_3: x_1 \lor \neg x_3 \lor x_4 $}};
\draw (-2.8, -0.55 - 3 * 0.35) node[right] {\st{$c_4: \neg x_1 \lor x_2 \lor x_3 \lor \neg x_4$}};
\draw (-2.8, -0.55 - 4 * 0.35) node[right] {\color{our_red} $c_5: x_1 \lor x_2 \lor x_4$};
\footnotesize
\draw (-5.3, 2.) node[left] {Label};
\draw (-5.3, 1.65) node[left] {Preserving};
\draw (-5.3, 1.35) node[left] {Augmentations};
\normalsize
\draw (5.8, 0.55) node {Encoder};
\draw (5.8, -1.4) node {Encoder};
\draw (2.5, 0.55) node {$G_1$};
\draw (2.5, -1.7) node {$G_2$};
\draw (1.2, 0.65) node {$\mathcal{E}$};
\draw (1.2, -1.45) node {$\mathcal{E}$};
\draw (6.8, 0) node {$\mathbf{z_1}$};
\draw (6.8, -1.95) node {$\mathbf{z_2}$};
\large
\draw (-6.0, -1.1) node {$\phi$};
\draw (-1.2, -0.1) node {$\hat{\phi}_1$};
\draw (-1.2, -2.4) node {$\hat{\phi}_2$};
\Large
\draw (7.8, -0.25) node {$\mathcal{L(.)}$};
\end{tikzpicture}
\caption{Our contrastive learning framework for combinatorial problems. Given an instance $\phi$, a pair of augmented samples $(\hat{\phi}_1$, $\hat{\phi}_2)$ are formed using label-preserving augmentations. $(\hat{\phi}_1$, $\hat{\phi}_2)$ are then transformed to graph formats $(G_1, G_2)$. An encoder is used to extract graph representations $(\mathbf{z}_1, \mathbf{z}_2)$. Lastly, a standard contrastive loss is applied over a mini-batch of instances.}
\label{fig:cocl_model}
\end{figure*}
Similar to contrastive algorithms in other domains (e.g., images), we learn representations by contrasting augmented views of the same instance against negative samples. Our framework (Figure \ref{fig:cocl_model}) consists of four major components:
\textbf{Augmentations.}
Given a combinatorial instance $\phi$, a stochastic augmentation is applied to form a pair of positive samples, denoted by $(\hat{\phi}_1$, $\hat{\phi}_2)$. Our key insight is the tailored design of augmentations for combinatorial problems should preserve the label of the problem, e.g., satisfiability for SAT, as detailed in Section \ref{sec: lpa}.
\textbf{Format Transformation ($\mathcal{E}$).}
Formulas $(\hat{\phi}_1$, $\hat{\phi}_2)$ are transformed into (e.g., $\text{LIG}^{+}$) graphs $(G_1, G_2)$.
\textbf{Encoder.}
A neural encoder is trained to extract graph-level representations $\mathbf{z}_1, \mathbf{z}_2 \in \mathbb{R}^d$ from the augmented graphs $G_1$ and $G_2$. This encoder can be any GNN such as GCN \citep{kipf2016semi} and the encoder proposed by NeuroSAT.
\textbf{Contrastive Loss.}
In the end, SimCLR's \citep{chen2020simple} contrastive loss $\mathcal{L}$ is applied to the graph representations $\{ \mathbf{z}_i \}_{i=1}^{2n}$, obtained from a mini-batch of $n$ instances. We follow \citet{chen2020simple} and use an MLP projection head to project each $\mathbf{z}_i$ to $\mathbf{m}_i$. For a positive pair of projected representations $(\mathbf{m}_i, \mathbf{m}_j)$, the loss $\mathcal{L}_{i, j}$ is a cross-entropy loss of differentiating the positive pair from the other $2(n-1)$ negative samples
(i.e., augmented samples of other instances):
\begin{align}
\mathcal{L}_{i, j} = -\log \frac{\exp(sim(\mathbf{m}_i, \mathbf{m}_j)\mathbin{/} \tau)}{\sum_{k=1}^{2N} \mathbf{1}_{k \neq i} \exp(sim(\mathbf{m}_i, \mathbf{m}_k)\mathbin{/} \tau)},
\label{equ: contrastive}
\end{align}
where $\tau$ is the temperature parameter, $\mathbf{1}$ is the indicator function, and $sim$ measures the similarity between two representations: $sim(\mathbf{m}_i, \mathbf{m}_j) := \mathbf{m}_i^T \mathbf{m}_j / \left\Vert \mathbf{m}_i \right\Vert \left\Vert \mathbf{m}_j \right\Vert$. The final loss $\mathcal{L}$ is the average of $\mathcal{L}_{i, j}$ over all positive pairs.
After training is completed, we only keep the encoder to extract the representation $\mathbf{z}_i$ for downstream tasks.
\section{Related Work}
\label{sec:related_work}
\textbf{Graph Contrastive Learning.} Existing graph contrastive frameworks can, in principle, be used to learn representations for combinatorial problems. Common augmentations in graph contrastive frameworks include: 1) perturbing structures, such as, node dropping, subgraph sampling or graph diffusion, and 2) perturbing features, such as, masking or adding noise to the node features \citep{hassani2020contrastive}. These augmentations have achieved success in multiple graph-level tasks \citep{you2020graph, hassani2020contrastive}, as well as node-level tasks \citep{zhu2020deep,wan2021contrastive,tong2021directed}.
\textbf{Supervised Learning for Combinatorial Optimization (CO).} Almost all modern approaches to machine learning for CO, use variations of a GNN architecture. On the supervised front, the work of \citet{nowak2018revised} on
\emph{Quadratic Assignment Problem}, \citet{joshi2019efficient} and \citet{prates2019learning} on \emph{Travelling Salesman Problem} (TSP) have shown encouraging results.
\citet{gasse2019exact} trained a branching heuristic for MILPs by imitating an expert policy. Later, \citet{nair2020solving} extended those ideas
to make them scalable to substantially larger instances. In SAT, \citet{selsam2018learning} proposed a way to train GNNs to solve SAT problems in an end-to-end
fashion. The same architecture was used in \citet{selsam2019guiding} to guide variable branching across conventional solvers.
\textbf{Reinforcement Learning for CO.}
Supervised learning is bottlenecked by the need for labels. Consequently, many have explored the use of \emph{Reinforcement Learning} (RL). Node selection policy of \citet{dai2017learning} for TSP and \citet{kool2018attention} for \emph{Vehicle Routing Problem} are examples of that effort. \citet{LedermanRSL20} and \citet{YolcuP19} used REINFORCE to train variable branching heuristic for \emph{quantified Boolean formulas} and local search algorithm WalkSAT, respectively. \citet{kurin2019improving} used DQN for SAT and \citet{Vaezipoor_Lederman_Wu_Maddison_Grosse_Seshia_Bacchus_2021} improved a SOTA \#SAT solver via \emph{Evolution Strategy}. We refer to \citep{cappart2021combinatorial, bengio2021machine} for more comprehensive record of efforts in this area.
\textbf{Unsupervised Learning for CO.}
\citet{toenshoff2021graph} proposed an unsupervised approach to solve constrained optimization problems on graphs by minimizing a problem dependent loss function. \citet{Amizadeh2019LearningTS, amizadeh2019pdp} solved SAT and CircuitSAT problems by minimizing an energy function. Lastly, inspired by probabilistic method, \citet{karalias2020erdos} trained a GNN in an unsupervised way to act as a distribution over possible solutions of a given problem, by minimizing a probabilistic penalty loss. Our method is distinct from other unsupervised techniques in that we do not minimize a problem-dependent loss function and rather learn problem representations through contrastive learning. To the best of our knowledge this is the first attempt at applying contrastive learning in the domain of combinatorial optimization.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 7,847
|
Blake Williams
This Week in Film: Flight, How to Survive a Plague, Shana Moulton, Masao Adachi, Diaspora, Breast Fest, Toronto Reel Asian Film Festival, DVD & BluRay
This Week in Film rounds up noteworthy new releases in theatres, as well as key DVD / Blu-Ray releases, festivals, and other cinema-related events happening in Toronto.
Flight (Varsity, Rainbow Market Sq.)
Robert Zemeckis made some of the most critically-acclaimed films of the '80s, '90s, and early 2000's (Back to the Future I, II, & III, Who Framed Roger Rabbit, Forrest Gump, Cast Away, and so on). For some reason, the last decade of his output has been consumed by ill-advised digital animation projects like The Polar Express, Beowulf, and A Christmas Carol. So it's an exciting event in itself to see that Zemeckis has returned to live action drama, picking up where he left off in 2000 with a film about the repercussions of a plane crash. But where Tom Hanks was stranded on an island with a volleyball for four years after his plane went down in Cast Away, Flight creates a different kind of island for Denzel Washington after his plane miraculously stays up - albeit one no less isolating and life-altering. A solid and compelling drama, this.
How to Survive a Plague (TIFF Bell Lightbox)
There have been plenty of AIDS docs that have come around since shit hit the fan in the '80s. Even last year we got what has been widely considered to be one of the best in We Were Here, a talking heads doc in which surviving Castro residents recounted the tragic rapture of their friends and loved ones that occurred three decades ago. How to Survive a Plague is a largely different style of documentary than that one, and might be the first one ever to detail, in hindsight, the ACT UP (AIDS Coalition to Unleash Power) movement. Using archival video tape and voiceover narration from the group's participants, first-time director David France creates a near-encyclopedic account of the strenuous political swing that pulled the LGBT community out of catastrophe and toward something resembling resolution.
Also in theatres this week:
Luv Shuv Tey Chicken Khurana (Cineplex Yonge & Dundas)
Manborg (The Royal)
Midnight's Children (Varsity)
The Man With the Iron Fists (Carlton, Rainbow Market Sq., Cineplex Yonge & Dundas)
Wreck-It Ralph (Rainbow Market Sq., Cineplex Yonge & Dundas)
Pleasure Dome presents Shana Moulton in Person - Performance & Recent Videos (Friday, November 2 at 7PM; Tallulah's Cabaret, Buddies in Bad Times Theatre)
"Shana Moulton creates evocatively oblique narratives in her video and performance works. Combining an unsettling, wry humor with a low-tech, Pop sensibility, Moulton plays a character whose interactions with the everyday world are both mundane and surreal, in a domestic sphere just slightly askew. As her protagonist navigates the enigmatic and possibly magical properties of her home decor, Moulton initiates relationships with objects and consumer products that are at once banal and uncanny." If you like what you see, be sure to come back the next day, Saturday at 2PM, for Moulton's artist talk at CineCycle.
The Free Screen - It May Be That Beauty Has Strengthened Our Resolve: Masao Adachi (Wednesday, November 7 at 7PM; TIFF Bell Lightbox)
Philippe Grandrieux (La vie nouvelle, Un lac) made a name for himself with his first three films - all fiction - as someone who could translate cinema into a series of pure - usually terrifying - sensations. The subject of this, his first feature-length documentary intended for a theatrical environment, is Masao Adachi, who, like Grandrieux, is a radical filmmaker with avant-garde tendencies (albeit very different ones than Grandrieux). "Grandrieux's portrait follows Adachi through a twilit Tokyo, with gorgeously underlit, nearly narcotic images reinforcing the mystery of Adachi's steely inner resolve. Adachi's whispered words serve as a dialogue on the notion of art and revolution -- for him never a contradiction -- and a reflection on the possibilities of film to create both beauty and change." This screening is free for all.
International Diaspora Film Festival (November 1-6 at Innis Town Hall and The Carlton)
Founded eleven years ago in 2001, the International Diaspora Film Festival (IDFF) aims to provide Toronto audiences an experience of the present world through cinema. "Diaspora" refers to the dispersion and migration of a people from their homeland and the communities they form in new lands. The IDFF pays special attention to independent Canadian filmmakers from ethnic minorities. One of the highlights of this year's festival is Aliyah (Sunday, 4 November 2012 at 6PM), a prize-winner from this year's Director's Fortnight in Cannes. Tickets to all screenings are $10 and can be purchased online here.
Breast Fest (November 2-4 at Bloor Hot Docs Cinema)
Breast Fest is not only the world's first film & arts festival dedicated to breast cancer awareness--it's also, still, the only one. Using film and art to inspire dialogue and foster a community surrounding issues related to breast cancer, Breast Fest presents Toronto with three full days of films, comedy, and a speaker series. The film programme is comprised of seven titles, from docs (Pink Ribbons Inc.) to repertory classics (ClĂŠo from 5 to 7). Tickets to the films are $10 each and can be purchased online here.
Toronto Reel Asian International Film Festival (November 6-11 in Toronto, November 16-17 in Richmond Hill)
One of Toronto's largest and most exciting festivals, Reel Asian kicks off early next week for six days of high quality cinema from all over Asia, before taking a short break and then heading over to Richmond Hill for four more films. This year's line-up is just as promising as usual, including the feature-film debut by groundbreaking relational aesthetics artist Rirkrit Tiravanija (Lung Neaw Visits His Neighbours), Eric Khoo's animated documentary on gekiga pioneer Yoshihiro Tatsumi (Tatsumi), and a massive omnibus project with a new short by Hou Hsiao-hsien (10+10). Ticket prices range from $12-20 ($10-15 for students and seniors) and can be purchased online here.
American Nightmare [DVD]
Americano [DVD]
Arthur Christmas [BLU-RAY]
Beaches [BLU-RAY]
Ben Lee: Catch My Disease [DVD]
The Bonfire of the Vanities [BLU-RAY]
Booker's Place: A Mississippi Story [DVD]
The Client [BLU-RAY]
Corpo Celeste [DVD]
Cut to the Chase: The Charley Chase Collection [DVD]
The Day He Arrives [DVD]
Deathship [BLU-RAY]
Die Nibelungen [BLU-RAY]
Dinotasia [BLU-RAY]
Eleven Samurai [DVD]
Even the Rain [DVD]
Fritz Lang: The Early Works [DVD]
Guys and Dolls [BLU-RAY]
Hypothermia [DVD]
I Wish [DVD]
Jonsi: Go Quiet [DVD]
Little Nemo: Adventures in Slumberla [BLU-RAY]
The Muppet Christmas Carol [BLU-RAY]
The Nutcracker: The Untold Story [DVD]
The Paradise Lost Trilogy [DVD]
Planes, Trains & Automobiles [BLU-RAY]
Pray for Japan [BLU-RAY]
Rashomon [BLU-RAY]
Sunset Blvd. [BLU-RAY]
They Live [BLU-RAY]
360 [BLU-RAY]
Tomboy [DVD]
Trishna [BLU-RAY]
Von Ryan's Express [BLU-RAY]
What Happened to Kerouac? [DVD]
Your Sister's Sister [DVD]
Film still from Flight
Latest in Film
TIFF just unveiled its opening night movie for this year's festival
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People across Canada are now boycotting Cineplex for showing Unplanned movie
Chris Rock and Samuel L. Jackson are filming a movie in Toronto
An anti-abortion movie is screening in Toronto and people aren't happy
The cast from Saved by the Bell is coming to Toronto
These are all the free outdoor movies in Toronto this summer
Here are all the movies you can watch at Open Roof Festival in Toronto this summer
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 6,237
|
Root and shoot competition between established ryegrass and invading grass seedlings.
Authors: Snaydon, R. W. & Howe, C. D.
Seeds of Poa annua, P. trivialis and Festuca rubra were sown into gaps within established Lolium perenne swards, planted in systematic pattern to give...
Early seral communities in a limestone quarry: an experimental study of treatment effects on cover and richness of vegetation.
Authors: Davis, B. N. K. & Lakhani, K. H. & Brown, M. C. & Park, D. G.
Oolitic limestone workings on the Leicestershire/Lincolnshire border were sown in 1981 with Brachypodium sylvaticum and Lotus corniculatus; in a facto...
Physiological differences among populations of Anthoxanthum ordoratum L. collected from the Park Grass experiment, Rothamsted. IV. Response to potassium and magnesium.
Authors: Davies, M. S.
In pot experiments, populations of Anthoxanthum odoratum collected from acid soils (both natural and Park Grass), which had not been fertilized with K...
Physiological differences among populations of Anthoxanthum odoratum L. collected from the Park Grass Experiment, Rothamsted. 3. Response to phosphate.
Authors: Davies, M. S. & Snaydon, R. W.
A. odoratum populations from 8 plots of a long-term experiment given N, P, NP or no fertilizer, with or without lime, were grown in sand culture at 5 ...
Seed shed in the making of hay from mesotrophic grassland in a field in Northern England: effects of hay cut date, grazing and fertilizer in a split-split-plot experiment.
Authors: Smith, R. S. & Pullan, S. & Shiel, R. S.
A hay crop was cut at three different dates in 1990 from a mesotrophic (meadow) grassland in Upper Teesdale, under various fertilizer, grazing and cut...
Factors affecting the performance of seedlings and ramets of invading grasses in an established ryegrass sward.
Authors: Howe, C. D. & Snaydon, R. W.
Seedlings and 2-tiller ramets of Poa trivialis, P. annua, Agrostis stolonifera and Festuca rubra were planted into an established sward of Lolium pere...
Establishment of Ammophila arenaria (marram grass) from culms, seeds and rhizomes.
Authors: Putten, W. H. van der
In field trials on the coastal foredune ridge of Voorne Island, Netherlands in 1984, A. arenaria was sown with a mixture of nurse crops (Secale cereal...
Influence of irrigation and fertilization on the nematode community in a Swedish pine forest soil.
Authors: Sohlenius, B. & Wasilewska, L.
A study of a 20 to 25 year old stand of Scots pine in Central Sweden showed that the nematode fauna was influenced by both irrigation and fertilizatio...
The interactive effects of management on the productivity and plant community structure of an upland meadow: an 8-year field trial.
Authors: Smith, R. S. & Shiel, R. S. & Millward, D. & Corkhill, P.
Agricultural policy in the Pennine Dales Environmentally Sensitive Area in northern England aims to enhance plant species diversity in agriculturally ...
Agricultural intensification and the functional capacity of soil microbes on smallholder African farms.
Authors: Wood, S. A. & Bradford, M. A. & Gilbert, J. A. & McGuire, K. L. & Palm, C. A. & Tully, K. L. & Zhou, J. Z. & Naeem, S.
Fertilization may impact ecosystem processes that sustain agriculture, such as nutrient cycling, by altering the composition of soil microbial communi...
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 6,730
|
Search Results: 'charlottesville'
68 results for "charlottesville"
Legend for Messenger By Christopher Walker Hlad
This is a poem conceived in a rain storm in the middle of the night. It concerns the last night that Edgar Allan Poe was at the University of Virginia--in his room burning his furniture for firewood... More > in his fireplace and in the company of the University librarian--before fleeing into the woods the next morning to escape his creditors. It is a tribute to the author E.A. Poe and his tremendous, lasting voice. Christopher Walker Hlad.< Less
The Noble Salvage: A Song of the South By Matthew S. Farrell
Charlottesville novel, farce and fable. Naive and silly romp through downtown bars and country-side, country-parties, stuffed with literary mockery and local parodies. One of the very first... More > Charlottesville-set novels (Downtown Mall in the early 90s) after Darconville's Cat (Theroux); shares original publication date with Sidney Blair's wonderful Charlottesville novel, Buffalo. Matthew S. Farrell< Less
Concerning Big Fun By Gus Mueller
Nonfiction. Sociological, cultural, historical document. In the mid-90s three young women moved to Charlottesville and established a thriving youth alternative counter-culture around themselves,... More > centered on a large farmhouse in the countryside. "The Gus", Gus Mueller, was there at the outset and throughout, and lovingly, hilariously, and critically documented the growth and eventual dissolution of the community in what was to become one of the first-ever and closely-followed web blogs. Originally released in book-form by Hypocrite Press in the early phases of the blog's development, the entire text is provided in this new critical edition, expanded by lavish illustrations and photographs. Gus Mueller< Less
Ballads of the Sunlit Years By by James Lindsay Gordon Introduced by Michael Nace
Verse. This is a re-issue of the original edition of 900 copies printed in 1904. The author attended William and Mary and the University of Virginia. These poems, Southern-Fugitive in character,... More > powerfully evoke a distant time, and through them runs a deep sentimentality, nostalgia, and love for the Commonwealth in which the author lived and that the author served. James Lindsay Gordon, with an introduction by Michael Nace.< Less
Enactments by the Rector and Visitors of the University of Virginia for Constituting, Governing & Conducting that Institution For the Use of the University. By University of Virginia
Photographic facsimile reprint of original. All pages accounted for and legible. 12 pp. Multiple copies available.
Manufactured Housing: Sunrise Trailer Court By Andrew Thurlow
This publication, a two week effort, was created by Roger Williams University School of Architecture students, as an output of their initial analysis & research, during the Manufactured Housing... More > advanced architecture design studio [AADS] in the Spring of 2005 taught by Andrew Thurlow.< Less
Forgive Us Our Skins By Ginger Moran
An essay on the life and death of Chic Moran written by his daughter Ginger, writer and professor. Ginger and Chic lived together at the end of his life and shared a house with her sons and menagerie.
Paperback: $104.10
Arch 202 By Jargalmaa Enkhbold
Studio project about high density housing in Charlottesville.
<< < 1 2 3 4 5 6 7 > >>
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 1,877
|
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