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SHAPE Blog In a recent YouTube video out of California, the Quantum Loop Gravity tendency in Modern Physics claims its own Theory of Emergence, in a slick, but totally uninformative video presented by a pretty girl. As the author of the 2010 The Theory of Emergence, and also a physicist myself, I am of course, bound to comment. You can watch it here, if you must! I will not go through the details of that video, for they are both insulting to their overtly intended audience - "the layman", and packed-full of unestablished assumptions - all restricted to a multi-dimensional Mathematics, and "caused" or "selected-from Random Chaos by a Universal Consciousness"(?). Good Lord. Frankly, this is not worthy of a detailed criticism, but it does use the same rhetorical methods as Trump uses in the political sphere, which indicates the methods to be used to convince the uninformed, without ever actually informing them! But, its claim to a "theory of emergence" must be torpedoed by someone, such as myself, who has been involved, all his adult life, in revealing the damaging errors both in Formal Reasoning, Politics and even Science, due almost-wholly to the inadequacies of Mathematics in explaining Reality. Indeed, to even limit it, temporally, in that way, reduces its major significance over the last 2,500 Ever since the creation of Mathematics by the Ancient Greeks, its enabling distortions in representing Reality, not only greatly expanded its pragmatic use, but also revealed a means of relating individual "discoveries" into an extendable and consistent intellectual discipline. And, these were very quickly carried over into Reasoning to produce Formal Logic. And later, became similarly built-into the new discipline of Science. Yet, almost immediately, after the initial gains were achieved, the Greek, Zeno of Elea, had revealed major errors in Reasoning, when addressing Movement, in his work on Paradoxes. And, these were never properly addressed, for over two millennia, until Hegel, the German idealist Philosopher stumbled upon the major flaw - Formal Logic did NOT ever address Qualitative Change! Indeed, to construct the founding discipline of Mathematics to actually work, all Qualitative Changes were prohibited, and, in addition, all Forms were changed into only perfect versions. Mathematics was never a description of Reality-as-is! And, in its changed state, it had been made to conform to The Principle of Plurality - where those assumed features were mandatory: they alone, in fact, gave Mathematics its consistency and extendibility! But, in carrying over the properties of Mathematics into defining Formal Logic, they also implanted these same restrictions into it too. And, later the same disabling features were delivered to Science as well, as Formal Logic was the required reasoning tool. Consequently, Science could only make any progress at all if the situations to be investigated were constrained to only deliver such features. It would only work within Stable Situations. It was The Science of Stability! Now, let's be crystal clear: "What is an Emergence?" In my book, as the author of The Theory of Emergence , it is when something qualitatively different emerges out of a seemingly persisting stable situation, changing things permanently. But, for all those who depend upon Mathematics as the unifying "consistent" basis of their studies and determinations, such Qualitative Changes are summarily banned! Their pluralist basis (Mathematics) always requires Stable situations: so what must they mean by their claims to deliver Emergences? Well, in spite of deliberate obfuscation, the designers and deliverers of this video use the oldest trick in the book. They first allow such a mess of complication, that almost any outcome seems to be possible, and they put down what selects from this enormous menu of possibilities, to a "Universal Consciousness" (which they insist is God), but a product of the overall entire-causing-system, that can then choose the actual outcome. Now, what is my alternative? All pluralistic Laws, as dealt with as above, have limits to their applicability. We call them Singularities, and if delivered of the appropriate values of the involved parameters. which take the situation to those terminations, then the equations give meaningless results - like infinity for example. Clearly, to a Holist, such as myself, the domain of applicability of their strictly-pluralist-equation has been exceeded, so it no longer describes the situation. The boundaries of the required Stability have been exceeded, and the Stability dissociated! Close to that boundary, in non linear cases, the limited region of Mathematical Chaos, can be encountered, but not for long, for the situation quickly descends into what seems to be Total Chaos..... but, then coalesces into a new relation in a different Stability. And, "How does that occur?", you will quite rightly ask! It occurs because in Reality-as-is, there are many simultaneous relations all-acting together, but they DO NOT just sum, as the pluralists insist - but both affect and change one another, some co-operating, while some are opposing. Indeed, for a time these always changing mutual modifications do deliver something like "chaos" - but only until they form a new self-maintaining system, when seemingly damaging changes in one, are compensated for by consequent opposing changes in another. Indeed, Hegel's simplification of this was his Interpenetration of Opposites, the simplest examples of which are the Dichotomous Pairs, discovered by Zeno, and explained in terms of mistaken or absent premises, by Hegel. In place of the idealistic and pluralist conception of an Emergence, may I offer The Trajectory of an Emergence , shown above. It is clearly no magical conversion, but a complex transition, involving a dynamic change between two Stabilities. This special issue explores the idea of iteration in Mathematics and Philosophy. In Mathematics it is a way of trying to find answers through repetition, but it certainly isn’t the usual way of using equations. Originally an invention of pragmatic engineers it then became an extension of Mathematics, giving birth to all manner of wonderous inventions, from fractals to Chaos Theory. It is a fascinating area for sure, but it isn’t what the mathematicians like to pretend it Iteration is a descrete way of approaching the continuous - and a static way of dealing with movement and change. It embodies all the chaos, paradoxes and infinite blow-ups you’d expect from such internal contradiction. The papers in this short collection are presented in a different way from the usual updates. For it is such a difficult, and yet crucial, area that “the latest” seems both too esoteric and too abstract, and its relevance not immediately apparent. It certainly wasn’t obvious to me! It has taken about 30 years for me to finally begin to understand iteration’s importance, in providing a very different approach to both Reasoning and Science. So, clearly, delivering the latest developments, without some idea of how it was finally achieved, would also leave most areas unexplained and unaquainted readers cold. So, this collection spans, one way or another, all the significant steps in that ascending trajectory. First of all, these papers are not part of a complete and final narrative. They, instead, each and every one, come out of an only partly referred-to past, which had certainly left the necessary traces-and-questions in my head, but not yet upon the written page. Nevertheless, the fact that each poses as yet unanswered questions, does ultimately connect up with later papers, and, as it does so, begins to light-up a wholly new path towards Truth, inaccessible from the usual approaches. As a whole, it brings together the inadequacies of disciplines that cannot deal with real Qualitative Change, such as Mathematics, with the finding of evidence for possible solutions actually within the very tricks and extensions that infer something beyond those steadfast limits, and which become attempts to solve the inherent problems of that discipline’s usual and in fact essential approach. Indeed, as Hegel had always insisted, progress only resides in what appear to be untenable contradictions. Read more Abandoned mall in the USA No Future Under Capitalism ....for Anyone The Economic System of Capitalism must have a Market for what it produces. So, primarily, that means people to buy its products. But, just as important in determining its imperatives, is how it works economically, based upon both the Financial Requirements side and on the Necessary Results side: it must attract the necessary investment to finance both the Means of Production, their regular updating to stay competitive, and have enough overall Profits to pay the required Dividends to its Investors. But, it must also have Labour to carry out its productive operations: and that must be as cheap as possible. Now, the main problem since its inception has never changed! Most of the customers (ultimately) will also be those producing the products. It, therefore, presents the major unavoidable contradiction in Capitalism. For, keeping wages as low as possible, while having enough customers with money to buy the products are, ultimately, diametrically opposed requirements. Do you require a proof? For, the whole 300 years of its History, Capitalism has hit this contradiction every few years, when it inevitably suffers a Recession, Depression or even a Slump - and when it does large numbers of workers are sacked, and find new jobs almost impossible to find. The corner was often turned by taking on the unemployed at much lower wages, so a new balance could be achieved, and these would be at different companies to those from which they had been sacked. Indeed, these "Down-Turns" were frequent enough, for an ever present Pool of the Unemployed to be regularly used in this way! Now, a purely single area Capitalism, could never grow enough for the new Employing Class and their investors, so they extended their reach to ever new areas, both in their own Country, and then abroad. And, as those exterior countries had been conquered by the so-called Metropolitan Capitalist Hubs, both cheap raw materials and low-wage workers could be easily maintained there. Abandoned mall in China This is the Imperialist Development of Capitalism, as very successfully employed by England (and later by the following United Kingdom). And, interestingly, this was significantly modified by the United States of America, by constantly extending its boundaries to the West and South, disposing of the indigenous populations, as it did so, and distributing of the taken land cheaply to the torrent of poor workers from the East of the USA, and to similar immigrant people from Europe. But, in addition, both of these nations largely solved their most pressing problem, by resorting to wholesale slavery, to provide "owned labour" both in America in their Southern States, and by Britain in its Caribbean Colonies. Of course, such "solutions" were always only temporary! For the imperatives of the system, necessarily re-asserted themselves all the time. And, with the end of Imperialism after the Second World War, those means were also curtailed - to be replaced by the installing-and-supporting of corrupt regimes in the ex-colonies to act as well-paid intermediaries. Indeed, for both the colonial owners and the USA there was also the threat of Socialism, following the Russian and Chinese Revolutions, and the forced "socialisation" of Eastern Europe. For then, America, in particular, embarked upon an almost constant set of wars to prevent any further extensions, and set up militarily-supported Capitalist alternatives instead. But, the underlying drive of Capitalism always-and-inevitably re-asserts itself, and whatever modifications are instituted, nothing can replace the need for Profit! Equally the essential contradictions can also never be resolved within Capitalism: for even in War, the usually resorted to "final solution", the soldiers required are once again the ordinary Working People - and to fight they have to be armed! It was just such "people-in-arms" that carried through the Russian Revolution. Think about it! Why were Nuclear weapons invented? The promise of war without soldiers Why are wealthy Americans armed to the teeth, and increasingly live in effectively "gated communities"? Why must there always be an Enemy, threatening the status quo? Why has America got by far the mightiest Military in the World? Who is really threatening America? It is you! What the capitalist rulers fear most is the mass of ordinary workers finally rebelling! So, how about a future without Capitalism? Abandoned mall in Austin Texas has been transformed into a community college Re-using dying malls Why Cosmology is Irretrievably Broken As a serious and active theoretical physicist and mathematician, I have been inevitably driven to Philosophy, in order to try to explain the many apparently unavoidable contradictions encountered literally everywhere in both of these disciplines. And, it was there, within Philosophy, that I had been irrefutably presented with a damning indictment of both the bases, and of the assumptions, underlying these disciplines, which are also present, in the usual Basic Formal Logic type of reasoning used there too. Such an extreme realisation was, itself, of course, a very long way from being an immediately-arrived-at conclusion. For, on the contrary, those very same now-rejected beliefs had been, without any doubt, a tremendously empowering past achievement by Mankind, and had led to significant progress in their attempts to make sense of their finally coming-to-be-thought-about World. Indeed, to this day, most people, even including most professionals, and in these very same fields, do not, as yet, even doubt their crucial underlying premises, and have stuck, consistently, to them, ever since their major revelation by the ancient Greeks. For, they were, and still are, neessary-simplifications of Reality, and still retaining a true-if-limited measure of Objective Content within them. They are often true-for-now, and hence wholly dependable in the short term. Indeed, in some cases they even appeared to be true-for-ever, such as in Number, for example! But in reality it depends on what you are counting - for, if your 1 + 1 is a Man and a Woman, it could, in time, equal 3, or 4 or even more. Then, as the parents die, it can decline, maybe even to 0! Yet, who would give up Number as a truly valuable concept, because of this clear time-dependence: it still has true value in many relatively unchanging scenarios. Indeed, the key-misleading-assumption involved, when applied generally, has a name: it is termed Plurality. Put simply, Plurality asserts the permanence of certain things, ideas or beliefs, and their independence of other simultaneously-present entities or happenings. It was intuitively arrived at by the Greeks, in their first major revelation - that of Mathematics, originally concerned with perfect shapes in Euclidian Geometry, but soon extended to the whole discipline involving all Pure Forms. And, let us be crystal clear, with Mathematics, within its well-defined bases, Plurality is, indeed, always valid! Its very power depends upon its definition of perfect shapes, or more generally, Perfect Forms, for this enabled the whole discipline to be built into a relatively consistent and developable system. But, this was only at all possible by limiting study to Pure Forms alone, which, as a consequence, also made it necessarily conform to Plurality too. But, consequently, Mathematics does NOT apply to Reality, as such, but only to this reflection of its Pure Forms and nothing else - basically, it is true only of a parallel and restricted World, which we term Ideality. Roger Penrose and some Ideality Now, the problems with my chosen disciplines arose, when situations unavoidably involved Qualitative Changes. For, Mathematics, as originally defined, excluded this possibility entirely, but also for the very same reason could still be developed into a remarkably informing descriptive discipline, when restricted to things conforming to Plurality - that is to only quantitative changes, usually only within what are termed Stabilities. But, my consequent turn to Physics (from my first love, Mathematics) didn't help, for the benefits of Mathematics in staying with Plurality, had also been exported illegitimately, first, to Formal Logic, and thereafter to the Sciences too. Though Physics, for example, was temporarily rescued by a form of Positivism which allowed the co-existence of various contradictory stances which could be switched-between with the long-standing pragmatic excuse of, "If it works, it is right!" So, an amalgam of stances were simultaneously-allowed, including Materialism (from Reality), Idealism (from Mathematics), Pragmatism (from his Hunter/ Gatherer past), Plurality (from Formal Reasoning), and even Holism (from attempts, in spite of all the above, to physically-explain real phenomena). The major crisis, was finally unavoidably precipitated, in the 20th century, by the increasingly-emerging failure of the above amalgam, which led to the dropping of Physical Explanation totally, and the whole-hearted embracing of Mathematics as the "sole-saviour", particularly in Sub-Atomic Physics, but also with a devastating carry-over into Cosmology too. Now, this particular essay was precipitated by a video on the internet by Professor Roger Penrose upon the assumed-cause - the Big Bang, and inevitable final-demise, of our Universe! Penrose started by mentioning his resolute faith in Mathematics, and, in particular, of Einstein's Relativity Equation, and though he didn't question the Equation, he felt that certain prior assumptions, upon which it was erected, might well be erroneous. Interestingly, he located the difficulties within the Singularities seemingly occurring at either end of that existence - the Big Bang beginning, and the Zero ending, indicated within the equation by its effective blowing-up at those singularities. His problems were with the (indisputable-for-him) Second Law of Thermodynamics, which indicated that the trajectory of that whole History was - from a High-Energy, Random-movement, Minimum Entropy Start onto a Low-Energy, Random-movement, Maximum Entropy End It didn't make sense in Penrose's conception for it seemingly went from Chaos to Chaos via Structured Foms and even Life? But, his doubts weren't because of Penrose's "rich and wide" experience of Reality: for he, on the contrary, only "dwells" exclusively in a pluralist world determined-and-describable only by Mathematics! Indeed, if you expected any Explanatory Physics, from his then-emerging Conformal Cyclic Cosmology, you will be sadly disappointed. Both in the Universe's Origin (as in a Big Bang) and in the Universe's Demise (as in a Terminal End) they are described as being in an identical Conformal featureless "Flatness". You have to remember his total dedication to pluralist Mathematics: in "explaining" anything, he actually says, "The equations deliver all these outcomes"! No references are made to any actual Substances and their properties. Absolutely everything comes from the Abstract Equations alone, and, ultimately, all his descriptions will be shown as the consequences of Formal Equations - they, we are told, determine-everything! Yet, such means not only do not, but also cannot, deliver Qualitative Change, so all adherents to the Copenhagen Interpretation of Quantum Theory, with their Maths-only stance, can never ever explain such changes: they can only, in the old pragmatic way, switch between equations because - "If it works, it is right!" That is NOT Science. It is Idealism as embodied in purely Formal Equations. It can only ever be descriptive, but never explanatory, so it actually terminates Science to be replaced by a dry and dead formalism. Now, with justice, the response to all of this might well be to demand that this critic must deliver the alternative to this Dead End, and that is certainly a legitimate position to take. Yet, the routes taken in the whole of Mankind's various intellectual disciplines, over millennia, have unavoidably brought us to this significant current Impasse. The contradictions have been built into the Amalgam of such Premises, which were all retained, in order, pragmatically, to be able to achieve the many required particular outcomes, in a variety of areas. And, that Amalgam must now be dismantled, via a route admitting of, and dealing comprehensively with, Qualitative Change. But, in spite of several heroic attempts to do this, particularly since the Dialectics of Friedrich Hegel, some 200 years ago, this has not been achieved, primarily, because such an undertaking has never been systematically-and-comprehensively applied to Science, and, crucially, to Physics. And, the usual restriction is, invariably, to only ever do Studies of Stability, either natural or arranged-for, which is now required to be extended beyond the point where formalist equations FAIL - where each-and-every essential Stability dissociates, and where the Real World processes, which alonedeliver the Qualitative Changes, termed Emergences,or even Revolutions,must now be the New Focus. This is not new, descriptively, of course - for in Biology, Evolution is both totally accepted and well described. And, Geology has revealed the 4 billion-year-long History of the Earth, and even the time of the Origin of Life, and the Tempo of its consequent stages of subsequent development - its Evolution! But, what are rarely, if ever, investigated, are the relatively short Interludes of Emergent Change, which are totally unavailable by current scientific methods, which ONLY EVER investigate Stability! It has been shown that an interlude of Qualitative Change is a cataclysmic transformation, requiring, initially, repeated Crises within the current Stability, which turns out to be a self-maintaining balance of multiple-opposing-factors, and which finally totally collapses - seemingly heading for a Nadir of Dissociation - that is, in fact, a complete dissolution of the prior System-Stability involved. Yet, consequently, this then allows the still-existing individual processes, from the prior Stability, along with co-existing others, to find new "partners", in both conducive-cooperating and opposing relationships, which ultimately achieve a wholly new self-maintaining balance, in a new Stability, at a new and different level! We currently recognise the Stabilities, upon either side of such a Transforming Interlude, but know nothing of the process which brought-about The Change. We use the passing of Threshold values, in certain Key Parameters, to signal when to switch getween the alternatives, but we can never explainthe conyent of that transition! Now, such an absolutely necessary inclusion of these changes into Science is not just a dream! It is already underway, with a major Holistic attack upon the Copenhagen Interpretation of Quantum Theory, and its many consequences. The ill-famed Double Slit Experiments have been fully explained, purely physically. And the quantization of Electron orbits in Atoms has also succumbed to the new approach. Of course, the very heart of this endeavour has been to produce a coherent, consistent and comprehensive Holist, Materialist Philosophic Stance. And, the demotion of Mathematics from its current primary position to that of a flawed but useful Handmaiden in both Science and Technology has been necessary. How can Science become Holist rather than Pluralist? This undertaking, almost exclusively by a single individual (the writer of this paper), has amounted to over 1,000 papers, published at a rate of approximately 9 per month over the last 9 years, but based upon a lifetime's involvement, at a professional level, in all the disciplines addressed. Postscript: The obvious question that may be considered important, about this philosopher/scientist, must be, "to what tradition or milieu does this researcher belong?" He has been a aspiring Dialectical Materialist since early adulthood, but only began to make significant philosophical contributions in the last 20 years.	{"url":"https://theelectronicjournal.blogspot.com/2018/06/","timestamp":"2024-11-10T18:46:46Z","content_type":"text/html","content_length":"236212","record_id":"<urn:uuid:bb60b30f-6309-49b9-abe8-5b6747974d3f>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00147.warc.gz"}
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the buckyball symmetries The buckyball is without doubt the hottest mahematical object at the moment (at least in Europe). Recall that the buckyball (middle) is a mixed form of two Platonic solids the Icosahedron on the left and the Dodecahedron on the right. For those of you who don’t know anything about football, it is that other ball-game, best described via a quote from the English player Gary Lineker “Football is a game for 22 people that run around, play the ball, and one referee who makes a slew of mistakes, and in the end Germany always wins.” We still have a few days left hoping for a better ending… Let’s do some bucky-maths : what is the rotation symmetry group of the buckyball? For starters, dodeca- and icosahedron are dual solids, meaning that if you take the center of every face of a dodecahedron and connect these points by edges when the corresponding faces share an edge, you’ll end up with the icosahedron (and conversely). Therefore, both solids (as well as their mixture, the buckyball) will have the same group of rotational symmetries. Can we at least determine the number of these symmetries? Take the dodecahedron and fix a face. It is easy to find a rotation taking this face to anyone of its five adjacent faces. In group-slang : the rotation automorphism group acts transitively on the 12 faces of the dodecohedron. Now, how many of them fix a given face? These can only be rotations with axis through the center of the face and there are exactly 5 of them preserving the pentagonal face. So, in all we have $12 \times 5 = 60 $ rotations preserving any of the three solids above. By composing two of its elements, we get another rotational symmetry, so they form a group and we would like to determine what that group is. There is one group that springs to mind $A_5 $, the subgroup of all even permutations on 5 elements. In general, the alternating group has half as many elements as the full permutation group $S_n $, that is $\frac{1}{2} n! $ (for multiplying with the involution (1,2) gives a bijection between even and odd permutations). So, for $A_5 $ we get 60 elements and we can list them : • the trivial permutation$~() $, being the identity. • permutations of order two with cycle-decompostion $~(i_1,i_2)(i_3,i_4) $, and there are exactly 15 of them around when all numbers are between 1 and 5. • permutations of order three with cycle-form $~(i_1,i_2,i_3) $ of which there are exactly 20. • permutations of order 5 which have to form one full cycle $~(i_1,i_2,i_3,i_4,i_5) $. There are 24 of those. Can we at least view these sets of elements as rotations of the buckyball? Well, a dodecahedron has 12 pentagobal faces. So there are 4 nontrivial rotations of order 5 for every 2 opposite faces and hence the dodecaheder (and therefore also the buckyball) has indeed 6×4=24 order 5 rotational symmetries. The icosahedron has twenty triangles as faces, so any of the 10 pairs of opposite faces is responsible for two non-trivial rotations of order three, giving us 10×2=20 order 3 rotational symmetries of the buckyball. The order two elements are slightly harder to see. The icosahedron has 30 edges and there is a plane going through each of the 15 pairs of opposite edges splitting the icosahedron in two. Hence rotating to interchange these two edges gives one rotational symmetry of order 2 for each of the 15 pairs. And as 24+20+15+1(identity) = 60 we have found all the rotational symmetries and we see that they pair up nicely with the elements of $A_5 $. But do they form isomorphic groups? In other words, can the buckyball see the 5 in the group $A_5 $. In a previous post I’ve shown that one way to see this 5 is as the number of inscribed cubes in the dodecahedron. But, there is another way to see the five based on the order 2 elements described If you look at pairs of opposite edges of the icosahedron you will find that they really come in triples such that the planes determined by each pair are mutually orthogonal (it is best to feel this on ac actual icosahedron). Hence there are 15/3 = 5 such triples of mutually orthogonal symmetry planes of the icosahedron and of course any rotation permutes these triples. It takes a bit of more work to really check that this action is indeed the natural permutation action of $A_5 $ on 5 elements. Having convinced ourselves that the group of rotations of the buckyball is indeed the alternating group $A_5 $, we can reverse the problem : can the alternating group $A_5 $ see the buckyball??? Well, for starters, it can ‘see’ the icosahedron in a truly amazing way. Look at the conjugacy classes of $A_5 $. We all know that in the full symmetric group $S_n $ elements belong to the same conjugacy class if and only if they have the same cycle decomposition and this is proved using the fact that the conjugation f a cycle $~(i_1,i_2,\ldots,i_k) $ under a permutation $\sigma \in S_n $ is equal to the cycle $~(\sigma(i_1),\sigma(i_2),\ldots,\sigma(i_k)) $ (and this gives us also the candidate needed to conjugate two partitions into each other). Using this trick it is easy to see that all the 15 order 2 elements of $A_5 $ form one conjugacy class, as do the 20 order 3 elements. However, the 24 order 5 elements split up in two conjugacy classes of 12 elements as the permutation needed to conjugate $~(1,2,3,4,5) $ to $~(1,2,3,5,4) $ is $~(4,5) $ but this is not an element of $A_5 $. Okay, now take one of these two conjugacy classes of order 5 elements, say that of $~(1,2,3,4,5) $. It consists of 12 elements, 12 being also the number of vertices of the icosahedron. So, is there a way to identify the elements in the conjugacy class to the vertices in such a way that we can describe the edges also in terms of group-computations in $A_5 $? Surprisingly, this is indeed the case as is demonstrated in a marvelous paper by Kostant “The graph of the truncated icosahedron and the last letter of Galois”. Two elements $a,b $ in the conjugacy class C share an edge if and only if their product $a.b \in A_5 $ still belongs to the conjugacy class C! So, for example $~(1,2,3,4,5).(2,1,4,3,5) = (2,5,4) $ so there is no edge between these elements, but on the other hand $~(1,2,3,4,5).(5,3,4,1,2)=(1,5,2,4,3) $ so there is an edge between these! It is no coincidence that $~(5,3,4,1,2)=(2,1,4,3,5)^{-1} $ as inverse elements correspond in the bijection to opposite vertices and for any pair of non-opposite vertices of an icosahedron it is true that either they are neighbors or any one of them is the neighbor of the opposite vertex of the other element. If we take $u=(1,2,3,4,5) $ and $v=(5,3,4,1,2) $ (or any two elements of the conjugacy class such that u.v is again in the conjugacy class), then one can describe all the vertices of the icosahedron group-theoretically as follows Isn’t that nice? Well yes, you may say, but that is just the icosahedron. Can the group $A_5 $ also see the buckyball? Well, let’s try a similar strategy : the buckyball has 60 vertices, exactly as many as there are elements in the group $A_5 $. Is there a way to connect certain elements in a group according to fixed rules? Yes, there is such a way and it is called the Cayley Graph of a group. It goes like this : take a set of generators ${ g_1,\ldots,g_k } $ of a group G, then connect two group element $a,b \in G $ with an edge if and only if $a = g_i.b $ or $b = g_i.a $ for some of the generators. Back to the alternating group $A_5 $. There are several sets of generators, one of them being the elements ${ (1,2,3,4,5),(2,3)(4,5) } $. In the paper mentioned before, Kostant gives an impressive group-theoretic proof of the fact that the Cayley-graph of $A_5 $ with respect to these two generators is indeed the buckyball! Let us allow to be lazy for once and let SAGE do the hard work for us, and let us just watch the outcome. Here’s how that’s done The outcone is a nice 3-dimensional picture of the buckyball. Below you can see a still, and, if you click on it you will get a 3-dimensional model of it (first click the ‘here’ link in the new window and then you’d better control-click and set the zoom to 200% before you rotate it) Hence, viewing this Cayley graph from different points we have convinced ourselves that it is indeed the buckyball. In fact, most (truncated) Platonic solids appear as Cayley graphs of groups with respect to specific sets of generators. For later use here is a (partial) survey (taken from Jaap’s puzzle page) Tetrahedron : $C_2 \times C_2,[(12)(34),(13)(24),(14)(23)] $ Cube : $D_4,[(1234),(13)] $ Octahedron : $S_3,[(123),(12),(23)] $ Dodecahedron : IMPOSSIBLE Icosahedron : $A_4,[(123),(234),(13)(24)] $ Truncated tetrahedron : $A_4,[(123),(12)(34)] $ Cuboctahedron : $A_4,[(123),(234)] $ Truncated cube : $S_4,[(123),(34)] $ Truncated octahedron : $S_4,[(1234),(12)] $ Rhombicubotahedron : $S_4,[(1234),(123)] $ Rhombitruncated cuboctahedron : IMPOSSIBLE Snub cuboctahedron : $S_4,[(1234),(123),(34)] $ Icosidodecahedron : IMPOSSIBLE Truncated dodecahedron : $A_5,[(124),(23)(45)] $ Truncated icosahedron : $A_5,[(12345),(23)(45)] $ Rhombicosidodecahedron : $A_5,[(12345),(124)] $ Rhombitruncated icosidodecahedron : IMPOSSIBLE Snub Icosidodecahedron : $A_5,[(12345),(124),(23)(45)] $ Again, all these statements can be easily verified using SAGE via the method described before. Next time we will go further into the Kostant’s group-theoretic proof that the buckyball is the Cayley graph of $A_5 $ with respect to (2,5)-generators as this calculation will be crucial in the description of the buckyball curve, the genus 70 Riemann surface discovered by David Singerman and Pablo Martin which completes the trinity corresponding to the Galois trinity	{"url":"http://www.neverendingbooks.org/the-buckyball-symmetries/","timestamp":"2024-11-01T19:01:31Z","content_type":"text/html","content_length":"41556","record_id":"<urn:uuid:05e4f659-6688-49aa-b225-4d1055bfc2d9>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00899.warc.gz"}
This page evaluates the characteristics of batteries that drive economics of battery storage. Items that affect the cost of batteries include the life of the battery, the degradation, the round trip efficiency, the depth of discharge as well as the cost. Instead of simply reporting some numbers that you can argue about for each of these characteristics, I demonstrate how each of the characteristics affect the economic analysis of storage. As in other pages, I use an LCOE framework where the economic cost computed from the LCOE is confirmed with a financial model. I demonstrate how storage using batteries can be made analogous to storage of boxes in a warehouse. In the excel file attached to the button I attempt to concentrate on battery costs using a simple example with constant loads and solar costs. The example compares solar plus battery to solar plus thermal to only solar. Excel File with Evaluation of Storage Costs from Simple Demand Analysis that Includes Drivers of Storage Cost Excel File with Battery Analysis Using Alternative Costs, Load Shapes and Efficiencies with DispatchFactors Playlist of Batteries and Storage The set of videos attached to the playlist below demonstrate my various attempts to evaluate the economic cost of batteries. In working through battery issues I have attempted to make differet files that illustrate the cost of storage relative to other alternatives. I hope that as I have worked through the issues the dispatch analysis of batteries combined with the cost analysis is becoming clearer. As with the other playlists, watching all of the videos would be torture and impossible to do. But if you want some help sleeping maybe you can turn on the playlist. I have also included the power point slides that I sometimes refer to when working through the battery issues. Power Point Slides with Analysis of Batteries and Storage Including a Survey of Battery Characteristics (Cost, Degradation etc.) Capital and Operating Cost of Battery The cost of a battery is generally expressed as an amount of money that muse be spent per kWh. Although there is also a relationship between cost and the amount of the capacity which can roughly be through of as how fast can the battery be charged and discharged. This means that a battery with low storage relative to capacity will have a higher cost per kWh. The chart below implies that there is value for capacity and for storage as the cost per kWh of storage declines with more storage relative to capacity. If the lines were flat there would be no premium for capacity relative to storage. Note that the costs in the chart below may not seem to include all of the components of the battery as a battery cost of USD 150/kWh for a 12 hour storage battery is low. Power Point Slides Describing Levelised Cost, Resource Analysis and Financial Analysis of Solar Power Projects Excel File with Advanced Project Finance Issues Including Sculpting for Multiple Debt Issues and P50/P99 etc. A second factor affecting the battery cost is the size of the battery. The chart below illustrates estimated economies of scale from the U.S. DOE. The data is both for 2020 on the left and 2030 on the right. Battery Life There are many elements of battery analysis where the numbers are not certain. The battery life is a crucial aspect of measuring the cost and the carrying charge of any asset. The screenshots below illustrate difference estimates of the battery life. A complicating element is that the expected life is not measured by a simple time number in years. Instead the life can be measured in the number of cycles. Simple Battery Analysis with Different Characteristics I received an e-mail that included the following information. This e-mail illustrates both how to make a simple evaluation of a battery and also the distortions in the analysis. It may be a good introduction to evaluation of batteries. • The cost of an installed 1 MWh battery in Canada is roughly $1M • Assuming a 10% interest rate, the daily carrying cost of that is $1M*10%/365 = $274 • Assuming one battery cycle per day, the daily revenue the battery would have to generate is $274 • Assuming an 80% efficiency factor, this grosses up to $342 • This is the intraday price arbitrage that would be needed; $342; i.e., sell power at $342/MWh higher than you buy it for at another time in the day. The problems with this analysis include: 1. The carrying charge can be very different than a simple discount rate. This is particularly true when the lifetime is short or when there is degradation in the output from a battery. 2. The amount of money required to generate the income is computed correctly, this can be evaluated by examining the on-peak versus off-peak power as shown below. 3. The cost of batteries has come down significantly from the USD 1000/kWh shown above. 4. The ancillary services may add a lot to the value of a battery. Simple Analysis of Battery You can construct a simple analysis of a battery using carrying charge rates, the cost per capacity or per cycle of energy and the cost of O&M just like for other technologies. Then you can compare the cost of the battery on a daily basis with merchant prices. You can do this for a single day or over the course of a year. The first step is to understand the fixed costs of a battery which can be converted first to the cost per year and then the cost per day. In the example below I have used data from Lazard which quotes the cost per kWh. I have assumed a cost per one cycle of storage per kWh of USD400/kWh. For a battery with a duration of 4 hours, the cost per kW is USD 1,000/kW. A big question in the analysis is the annual carrying charge. You can look at the pages that work through carrying charge to understand this number. Once the daily cost is computed you can test whether it is possible to achive this value from the difference between off-peak and on-peak prices. To illustrate this, consider the following prices over the course of a day and the value of storing energy and then releasing that energy with a loss. The final part of the analysis involves comparing the net benefits with the costs. Using the example above, the net benefits of the battery are less than the cost of the battery as shown below. Video on Merchant Prices and Use of Batteries for Arbitrage The video below illustrates how to use merchant prices in evaluating the economics of batteries and storage.	{"url":"https://edbodmer.com/battery-analysis-and-merchant-prices/","timestamp":"2024-11-11T09:50:07Z","content_type":"text/html","content_length":"157153","record_id":"<urn:uuid:7c1d9c02-1f8f-4a69-a878-51d393b721dd>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00541.warc.gz"}
Dynamic Ramsey Theory of Mechanical Systems Forming a Complete Graph and Vibrations of Cyclic Compounds Ramsey theory constitutes the dynamics of mechanical systems, which may be described as abstract complete graphs. We address a mechanical system which is completely interconnected by two kinds of ideal Hookean springs. The suggested system mechanically corresponds to cyclic molecules, in which functional groups are interconnected by two kinds of chemical bonds, represented mechanically with two springs (Formula presented.) and (Formula presented.). In this paper, we consider a cyclic system (molecule) built of six equal masses m and two kinds of springs. We pose the following question: what is the minimal number of masses in such a system in which three masses are constrained to be connected cyclically with spring (Formula presented.) or three masses are constrained to be connected cyclically with spring (Formula presented.) ? The answer to this question is supplied by the Ramsey theory, formally stated as follows: what is the minimal number (Formula presented.) The result emerging from the Ramsey theory is (Formula presented.). Thus, in the aforementioned interconnected mechanical system at least one triangle, built of masses and springs, must be present. This prediction constitutes the vibrational spectrum of the system. Thus, the Ramsey theory and symmetry considerations supply the selection rules for the vibrational spectra of the cyclic molecules. A symmetrical system built of six vibrating entities is addressed. The Ramsey approach works for 2D and 3D molecules, which may be described as abstract complete graphs. The extension of the proposed Ramsey approach to the systems, partially connected by ideal springs, viscoelastic systems and systems in which elasticity is of an entropic nature is discussed. “Multi-color systems” built of three kinds of ideal springs are addressed. The notion of the inverse Ramsey network is introduced and analyzed. • Ramsey theory • complete graph • cyclic molecule • eigenfrequency • entropic elasticity • selection rule • vibrational spectrum • viscoelasticity Dive into the research topics of 'Dynamic Ramsey Theory of Mechanical Systems Forming a Complete Graph and Vibrations of Cyclic Compounds'. Together they form a unique fingerprint.	{"url":"https://cris.ariel.ac.il/en/publications/dynamic-ramsey-theory-of-mechanical-systems-forming-a-complete-gr","timestamp":"2024-11-10T22:21:42Z","content_type":"text/html","content_length":"61857","record_id":"<urn:uuid:79be2fbc-3110-4fb1-bcc3-f636f902f83b>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00599.warc.gz"}
How does a hydraulic jack work: Pascals law Home Mechanics Gases and liquids How does a hydraulic jack work: Pascals law How does a hydraulic jack work: Pascals law A hydraulic jack is based on Pascal’s law, which states that the pressure in liquids acts equally in all directions. Pascal’s law In the article Pressure it has already been explained that pressure in liquids (or gases) is evenly distributed in all directions. If, for example, a certain pressure is generated at one point within a liquid, then the same pressure will be present at every other point in the liquid (neglecting the hydrostatic pressure). This is often referred to as Pascal’s law or Pascal’s principle. Pascal’s law describes the uniform distribution of pressure in a liquid (neglecting the hydrostatic pressure)! In some literature, Pascal’s law is also somewhat more general and then takes hydrostatic pressure into account. In this more general sense, Pascal’s law states that the pressure at a certain depth h in a liquid results from the sum of the pressure at the liquid surface p[0] and the hydrostatic pressure p[h]: &p(h) = p_0 + p_h~~~~~\text{and} ~~~p_h= \rho g h \\[5px] &\boxed{p(h) = p_0 + \rho g h} ~~~~~\text{Pascal’s law} \\[5px] Figure: Total pressure at a given depth as the sum of ambient pressure and hydrostatic pressure For liquids in an open container, the pressure at the liquid surface corresponds to the ambient pressure (“atmospheric pressure”). If the liquid is not very deep, the hydrostatic pressure can usually be neglected compared to the larger ambient pressure at the surface. If, for example, the depth is in the order of a few centimetres, then the hydrostatic pressure is about a thousand times lower than the atmospheric pressure. In this case, it immediately becomes apparent that the same pressure exists at every depth: & \text{with} ~~~~~ p_0 \gg p_h ~~~~~\text{applies:} \\[5px] &p(h) = p_0 + \bcancel{\rho g h} \approx p_0 \\[5px] &\boxed{p(h) = p_0} ~~~~~\text{valid under negligence of the hydrostatic pressure} \\[5px] By neglecting the hydrostatic pressure, the pressure in a liquid corresponds to the pressure that the environment exerts on the liquid surface. The pressure in a liquid can therefore be changed by increasing the pressure on the liquid surface. However, since the surrounding air pressure cannot be changed, the liquid must first be enclosed in container. Through an opening in the vessel wall, the pressure on the liquid surface can now be increased at will by means of a piston, thus forcing a certain pressure on the liquid. This simple principle is used, for example, in syringes. The liquid to be applied is contained in a cylindrical housing (barrel), on which a pressure is exerted with a piston. The resulting pressure causes the liquid to be pressed out of the orifice. Figure: Syringe Hydraulic jack Another use of Pascal’s principle is the hydraulic jack or hydraulics in general. Hydraulics uses fluid to transmit energy. In addition to electrics (power transmission with electric current) and pneumatics (power transmission with air), hydraulics is of great importance in mechanical engineering. While pneumatics refers to the power transmission with compressible gases, hydraulics refers to the power transmission with incompressible fluids! Special oils, so-called hydraulic fluids, are used in hydraulics. Compared to water, which could only be used in a temperature range between 0 °C and 100 °C, hydraulic fluids can be used in larger temperature ranges. In addition, the hydraulic fluids not only protect metallic components against corrosion, but also provide excellent lubrication of the moving parts. Hydraulic principle The figure below shows a hydraulic jack. A lever is used to pressurise a hydraulic fluid by a piston, thereby moving another piston upwards with great force. With such a hydraulic jack, loads of up to several tons can be lifted. The increase in force is in part due to the Pascal principle. Figure: Hydraulic bottle jack The figure below shows the simplified design of a hydraulic bottle jack, which shows the principle of operation. The hydraulic fluid is in a closed system. The housing in which the fluid is confined is provided with two pistons. The oil is put under pressure by the smaller piston (called pump piston or pump plunger). Figure: Hydraulic principle (Pascal’s law) Using the applied force F[1] and the surface area of the piston A[1], the exerted pressure p can be determined relatively easily from the quotient of force and area: &p = \frac{F_1}{A_1} \\[5px] Figure: Amplification of force based on Pascal’s principle According to Pascal’s law, this pressure can be found at any point of the liquid. Note that due to the large pressures applied and the relatively small dimensions of the housing, the hydrostatic pressure can be neglected anyway. The pressure generated by the small piston therefore also acts on the second piston, called ram (working piston). However, since this piston has a larger surface area A[2], the pressure there leads to a larger force F[2]: &F_2 = p \cdot A_2 \\[5px] If equation (\ref{p}) is used in equation (\ref{f}), the amplification of the force is directly dependent on the ratio of the piston areas: &F_2 = p \cdot A_2 = \frac{F_1}{A_1} \cdot A_2 = F_1 \cdot \frac{A_2}{A_1} \\[5px] &\boxed{F_2 = F_1 \cdot \frac{A_2}{A_1}}\\[5px] If, for example, the area of the working piston is four times as large as that of the pump piston (i.e. the diameter of the working piston is twice as large), the force applied is quadrupled. This does not contradict the law of energy conservation! Due to the quadruple piston surface, the working piston extends by only a quarter of the pump stroke. Figure: Displacement of the liquid This is also clearly illustrated, as the pump piston displaces a certain amount of hydraulic fluid during the downward movement (height h[1]). Liquids are incompressible and therefore cannot be compressed. Therefore the displaced liquid extends the working piston by the same volume (height h[2]). However, the ram has an area four times as large, so that this volume is already achieved with a quarter of the original stroke. As the force is increased, the lifting height is reduced accordingly. Mechanical principle In fact, this hydraulic amplification of the force is only one of a total of two principles applied to a jack. The much greater force amplification is due to the mechanical leverage. Usually it is a second class lever. According to the law of the lever, the mechanical amplification of the force results from the ratio of the lever arms. The “active” lever arm “a” results from the pivot point to the handle and the passive lever arm “b” from the pivot point to the pump piston. If the lever arm “a” from the pivot point to the handle, for example, is 10 times as large as the distance b from the pivot point to the pump piston, then the force will be increased by a factor of 10. Figure: Using lever for mechanical amplification of force If the above mentioned figures are used as an typical example for a car jack, a mechanical amplification of factor 10 is obtained according to the law of the lever and a hydraulic amplification of factor 4 according to Pascal’s Law. In this case, a total amplification of factor 40 is obtained. Thus, an object weighing 400 kg can be lifted with an effort of 10 kg. Construction of a hydraulic jack The figure below shows the structure and operating principle of a real hydraulic bottle jack. The hydraulic fluid is located in a reservoir between two cylinders; an outer cylinder (oil tank) which forms the housing wall and an inner cylinder in which the working piston (ram) slides. The hydraulic fluid inside this reservoir is not pressurized all the time! During the upward movement of the pump piston (pump plunger), the hydraulic oil is sucked into the pump cylinder by an inlet passage. Figure: Design and components of a hydraulic bottle jack (sectional view) The oil is then pressurised during the downward movement of the pump piston. This causes the oil to flow through another passage into the working cylinder, where it lifts the ram. Figure: How a hydraulic bottle jack works Check valves in the form of steel balls are used so that the jack can be moved continuously upwards and the hydraulic oil is not pumped back from the working cylinder into the pump cylinder (or the hydraulic oil is not pressed back into the reservoir). When the pump piston is lowered, the ball seals the way back into the reservoir. At the same time the valve ball in the working cylinder is lifted by the pressure and the hydraulic fluid can flow into it. Animation: How a hydraulic bottle jack works After the inflow, the ball in the working cylinder falls down again due to gravity. The high pressure in the working cylinder presses the ball firmly into the valve seat, thus preventing the hydraulic oil from flowing back into the pump cylinder. The pumping process can now start again from the beginning, as the ball in the pump cylinder is lifted by the suction and hydraulic fluid can be sucked into the pump cylinder. Note that due to the check valves, the hydraulic oil in the working cylinder is kept permanently under pressure, while the oil in the reservoir always remains To lower the ram again, another passage is opened, which connects the working cylinder directly to the reservoir. During lifting, this passage is sealed with a steel ball which is pressed firmly into the valve seat with a screw. If this release valve is unscrewed, the ball releases the passage and the hydraulic oil is pushed back into the reservoir under the force of gravity of the ram. In order to protect the jack from damage in the event of overload, the release valve is designed as a safety valve and is usually provided with a spring. If the pressure is too high, the spring is pushed back and the hydraulic oil can flow directly back into the reservoir without an unacceptably high pressure building up in the working cylinder.	{"url":"https://www.tec-science.com/mechanics/gases-and-liquids/how-does-a-hydraulic-jack-work-pascals-law/","timestamp":"2024-11-06T14:23:28Z","content_type":"text/html","content_length":"131353","record_id":"<urn:uuid:a49fc3aa-4320-4394-afae-d83e4e18efdf>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00100.warc.gz"}
Filling the gap between synchronized and non-synchronized sdBs in short-period sdBV+dM binaries with TESS: TIC 137608661, a new system with a well-defined rotational splitting TIC 137608661/TYC 4544-2658-1/FBS 0938+788 is a new sdBV+dM reflection-effect binary discovered by the TESS space mission with an orbital period of 7.21 h. In addition to the orbital frequency and its harmonics, the Fourier transform of TIC 137608661 shows many g-mode pulsation frequencies from the subdwarf B (sdB) star. The amplitude spectrum is particularly simple to interpret as we immediately see several rotational triplets of equally spaced frequencies. The central frequencies of these triplets are equally spaced in period with a mean period spacing of 270.12 s, corresponding to consecutive l = 1 modes. From the mean frequency spacing of 1.25 μHz we derive a rotation period of 4.6 d in the deep layers of the sdB star, significantly longer than the orbital period. Among the handful of sdB+dM binaries for which the sdB rotation was measured through asteroseismology, TIC 137608661 is the non-synchronized system with both the shortest orbital period and the shortest core rotation period. Only NY Vir has a shorter orbital period but it is synchronized. From a spectroscopic follow-up of TIC 137608661 we measure the radial velocities of the sdB star, determine its atmospheric parameters, and estimate the rotation rate at the surface of the star. This measurement allows us to exclude synchronized rotation also in the outer layers and suggests a differential rotation, with the surface rotating faster than the core, as found in few other similar systems. Furthermore, an analysis of the spectral energy distribution of TIC 137608661, together with a comparison between sdB pulsation properties and asteroseismic models, gives us further elements to constrain the system. Monthly Notices of the Royal Astronomical Society Pub Date: April 2022 □ asteroseismology; □ stars: horizontal branch; □ stars: individual: TIC 137608661; □ stars: oscillations (including pulsations); □ Astrophysics - Solar and Stellar Astrophysics Accepted for publication in MNRAS Main Journal	{"url":"https://ui.adsabs.harvard.edu/abs/2022MNRAS.511.2201S","timestamp":"2024-11-03T20:42:41Z","content_type":"text/html","content_length":"46735","record_id":"<urn:uuid:ec6b40c8-cacb-4a4b-a45a-d835d79dfaf5>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00164.warc.gz"}
Multiplication and division of the orbital angular momentum of light with diffractive transformation optics Multiplication and division of the orbital angular momentum of light with diffractive transformation optics posted on 2023-11-30, 18:17 authored by Gianluca Ruffato, Michele Massari, Filippo Romanato We present a method to efficiently multiply or divide the orbital angular momentum (OAM) of light beams using a sequence of two optical elements. The key-element is represented by an optical transformation mapping the azimuthal phase gradient of the input OAM beam onto a circular sector. By combining multiple circular-sector transformations into a single optical element, it is possible to perform the multiplication of the value of the input OAM state by splitting and mapping the phase onto complementary circular sectors. Conversely, by combining multiple inverse transformations, the division of the initial OAM value is achievable, by mapping distinct complementary circular sectors of the input beam into an equal number of circular phase gradients. The optical elements have been fabricated in the form of phase-only diffractive optics with high-resolution electron-beam lithography. Optical tests confirm the capability of the multiplier optics to perform integer multiplication of the input OAM, while the designed dividers are demonstrated to correctly split up the input beam into a complementary set of OAM beams. These elements can find applications for the multiplicative generation of higher-order OAM modes, optical information processing based on OAM-beams transmission, and optical routing/switching in telecom. This arXiv metadata record was not reviewed or approved by, nor does it necessarily express or reflect the policies or opinions of, arXiv. Ref. manager	{"url":"https://preprints.opticaopen.org/articles/preprint/Multiplication_and_division_of_the_orbital_angular_momentum_of_light_with_diffractive_transformation_optics/24687768/1","timestamp":"2024-11-04T23:58:42Z","content_type":"text/html","content_length":"128259","record_id":"<urn:uuid:b9814b89-bf1e-4635-8379-7b34f98da486>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00675.warc.gz"}
Instructor (Oussama Khatib):All right. Let’s get started. So the video segment today is quite interesting. It was presented at the 2000 International Conference on Robotic and Automation, and I’m sure you’re going to like it. [Video] A robotic reconnaissance and surveillance team, U.S.A. A heterogeneous multi-robot system for surveillance and exploration tasks. At the first tier of this team is the Scout. Scout’s are small, mobile sensor platforms used in a cooperating group. At the second tier is the Ranger. Rangers are larger robots used to transport, deploy, and coordinate the Scouts. Scouts are wholly original robots with cylindrical bodies 40 millimeters in diameter and 110 millimeters in length. The Scout carries a sensor payload used to relay environmental information to other robots. The most common Scout payload is a small video camera, but other payloads, such as microphones, are also used. Video data is broadcast to other systems via an analog RF transmitter. Scouts communicate with other robots using an RF data link. One specialized Scout has a camera mounted in a custom-pan tilt unit allowing the robot to view its surroundings independently of the orientation of its body. The Scout has two modes of locomotion to allow it to navigate different kinds of terrain and obstacles. The first mode uses its wheels, allowing it to drive over smooth surfaces. Here, the Scout demonstrates its ability to climb a 20-degree slope. The second mode of locomotion is the hop. The hop is accomplished by winching the Scout’s spring foot up around its body and then releasing it suddenly. Here, the Scout jumps over an obstacle. Scouts are deployed by Rangers. The Ranger is a modified commercial all-terrain robot. The Ranger uses a launcher to deploy Scouts into the area in which they will operate. A Ranger can carry and shoot up to 10 Scouts from its launcher. Ranger’s supervise the Scouts while working with other Ranger’s; Ranger’s report to a human group leader. The Scout’s are designed to withstand the impact of landing, and of being shot into and through obstacles, such as these simulated windows. The Scout’s small size, it’s deployability through launching, and its multiple locomotion modes and sensor payloads, give it the ability to explore difficult-to-reach areas and report useful data. Combining the Scouts with Rangers, which provide the ability to travel longer distances and to have greater computational resources, forms a useful reconnaissance and surveillance team. Instructor (Oussama Khatib):Okay. What do you think? I guess we need a robot to do the – I mean to [inaudible] these devices, and so one more robot is still needed. Okay. So after completing the forward kinematics, after finishing the Jacobian, we are ready now for dynamics; are you ready? All right. So while dynamics, and then we will do the control and that’s it, you have the basics. Kinematics, dynamics and control. Well, here is an example of robots that involves a lot of dynamics. Just imagine, like, moving the hand little bit, you can see all these coupling forces coming on the other hands, on the body. As you start moving, you have all these articulated body dynamics that are going to appear. And the dynamics of this system is quite complicated. In fact, if we go to this problem, we find that we need really to understand the dynamics of just one rigid body, and then combines these different dynamics together to understand the articulated multi-body system. So to do that, that is to find the dynamics of an articulated multi-body system; there are several formulations. In fact, there are many, many formulations. We will examine two of them. One is the Newton-Euler formulation. Have you heard about Newton? Yes. So what does it say to you, Newton-Euler, what does it tell you? So Newton Law is? You were saying? Instructor (Oussama Khatib):So mass acceleration equal to the force applied from a rigid body, right? So that is if you apply a force to a particle, it will accelerate along the same direction with an acceleration that is equal to the force divided by the mass. Okay. So what about Euler? What does Euler do with dynamics here? You know, Euler angle, you know Euler parameters. Huh. So Euler was looking at angles, why? Angles measure what? Rotational motion. So linear motion – because force, acceleration of a mass, a particle, it’s just going to be a linear dynamics. And Euler is dealing with the other side of dynamics, rotational motion. Now, if you have a particle, then it’s really not rotational motion to talk about. So we go to the rigid body and we find that we need to address the problem of angular rotation, angular motion, and that is the formulation – the combination of Newton and Euler equations extended to the problem of multi-body. So we will examine articulated multi-body dynamics, and we will find, similarly to the way, if you remember we found the Jacobian, by analyzing the static forces, propagation. You remember we break all the joints, remove the joints, and look at the stability of issue of the rigid-bodies. We are going to do the same thing with dynamics. Then we will examine another formulation – a formulation that captures the whole dynamics, linear and angular, in one equation, that is the Lagrange equation. And this formulation is relying on the energy that is the kinetic energy of the system. You know what is a kinetic energy? Most of you. What is the kinetic energy associated with a particle moving at a velocity v? Let’s see. One-half mv squared. Very good. And also, the potential energy. And that will lead us to a very interesting form that will give us the dynamic of articulated-body system in an explicit form. You remember how we did the explicit form for the Jacobian. We can find the Jacobian as a sum of contribution of the different velocities of the different lengths. Well, we’re going to do the same. We’re going to find the dynamics of the whole articulated-body system as a sum of the contribution to the mass properties, inertias, and masses. We establish something called the mass matrix associated with the dynamics. And we will see that from finding the energy – just finding the kinetic energy of the system of – or at least each of those lengths, adding them all together to find the total energy, we will be able to obtain the dynamic equation. So this form is really important and we will examine this form, probably on Wednesday. But let me just to start, to give you an idea about what is happening when we look at the dynamics. This is a robot from France; it’s called the MA Manipulator 23. It is a cable-driven robot, so all the motors are in the back and the cables are driving this traction. So if we go and analyze the inertia view from just one axis, let’s say, the first axis of rotation. So you have big inertia, smaller inertia, right? By putting the masses away from the axis, you are increasing the inertia perceived about this axis. So this inertia then depends on all the mass distribution, the lengths, the load, et cetera, that is associated with the manipulator. If we go here, we have also changes in the inertia perceived, that it’s independent of the previous lengths. So there is a structure to the way the inertia is affected by the motion of the structure and the configuration is going to change the value of that inertia that you are perceiving. If you want, I can show you the equations. I don’t know if you can see them, but here we go. So the first inertia perceived from this joint is, like, half the page. It’s sine, cosine, and all of these things, and depending on – but, obviously, I mean, we can obtain these equations, it’s not a problem. The problem is to understand what the structure of this equation is and how we can find those properties and how we can understand them. What – when we analyze later – when we later analyze the explicit form, you will see that essentially we’re going to be able to – to gain like with the Jacobian – are going to be able to see the dynamics of a manipulator just by looking at the structure of the robot. All right. Here is another robot, this is – this was analyzing my thesis. This was a robot that can carry 80 kilograms of its weight; heavy robot. It’s a robot that has 6 degrees of freedom, and we are performing just some [inaudible] motion on the different joints. On joint, I believe, joints five or joint four, we have, basically no motion – just letting the joint to be controlled to its zero position. So what do we see? We see that on the top, the lower joints on the robot, during this [inaudible] motion of the other joints, there is little effect, I mean, you can see some errors, but the errors are sort of filtered somehow. On the lower joints, four and five, you see large errors. And that is reflecting the fact that if you have a heavy joint, there are disturbances, the inertia, the – this big inertia of the – of the robot is going to play the role of a filter. It’s going to somehow reject the disturbances, and we will see little effect on those joints because they are quite heavy and the inertia is just taking. I mean, think about a truck moving fast and you hit it with, I don’t know, a fly? It’s not going to – to be affected. But for the fly, it is really terrible. So – so this is the fly and that up there is the truck. So if you want, here is the equation. We’re going to establish this equation. The equation is a vector, so gamma is the – the torques applied to the robot. Sometimes we call it tho or gamma. What is g? What would it be? Some force affected dependent on the configuration. Instructor (Oussama Khatib):The gravity, yeah. G, like gravity. Very good. And it’s dependent on the configuration. For instance, if we take this joint, and if we put a weight, you’re – you’re going to feel a torque, right? If I take this joint to here, what is the torque due to the gravity? Instructor (Oussama Khatib):[Inaudible] Very good. So you get it. That is, so if we go up, basically the gravity is going to act on the structure and not on the truck. Q double [inaudible] is the acceleration and m is mass acceleration, so m is? Okay. Q double [inaudible] is the vector of accelerations and m – hmm? Very good. Instructor (Oussama Khatib):Very good. It’s a matrix. So matrix plus multiplied by a vector will give you a vector. So mq double [inaudible] is going to be the inertial forces generated at zero velocity by the motion by the acceleration of the joints. So it’s playing the role of a mass, so if I had one degree of freedom, you will understand it. Mass acceleration. M is a mass. But because this is a multi-body system, it’s going to be a mass matrix. And the mass matrix – if we had a robot that is prismatic, so three prismatic joints, essentially along the diagonal you will see the element you’re presenting, the masses, the total masses reflected here, here, and at the last joint. And it will be diagonal matrix. But for articulated body with revolute joints, you’re going to have off-diagonal terms that represent the coupling. That is the motion of one joint will accelerate the other joints. What about v? This is a vector. That is function of q and q duct. So – oh, I forget my – I had something to, well, maybe they took the [inaudible]. So what would be this thing that will depend on the velocities? So – use the microphone. If I – do this. I’m wondering why it’s standing like this. If I stop moving it will fall, [inaudible]. So there is a force pulling here, it’s called – what? Instructor (Oussama Khatib):Centrifugal force, yes. Now, if you have multiple – multiple-body that are moving in addition to centrifugal, you will have Coriolis forces. That is the product of velocities that will be involved will cause both centrifugal and Coriolis forces. So q as I said is your [inaudible] coordinates. M is called the mass matrix or the kinetic energy matrix because this mass matrix is associated with the kinetic energy of the system. If you’re not just a particle m, what is the kinetic energy? For a mass m, a particle of mass m, what is the kinetic energy associated with this mass when it’s moving at a velocity v? Hmm? Instructor (Oussama Khatib):One-half – one-half mv squared. Yeah, that’s correct. Good. Well, it turned out, it is the same for a mass matrix multi-body; it’s going to be one-half – how do you do mv squared for a mass associated with articulated bodies and the mass is a matrix? Instructor (Oussama Khatib):So you get a v transpose and m and v, which makes it quadratic form. And basically your kinetic energy is just one-half q duct transpose mq duct. These are the centrifugal Coriolis forces, and we will see actually that these forces disappear if the velocity was zero. Or if the mass matrix was constant, that is the v term solely depends on the velocity – product of velocity, and on the fact that all the element involved are derivative – partial derivative, coming from partial derivative of the mass matrix. So if you have a mass matrix that is constant, then the derivative are zero, and v will be zero. Yes? Instructor (Oussama Khatib):No, no. Mass matrix or kinetic energy matrix. And the gravity forces and gamma is the derived forces. So gamma one will be acting along or about axis one. So you have q one, gamma one, q two, gamma two, right? So gamma could be a force if the joint is prismatic. Now, again, in doing all the formulations, it’s very simple. We saw this figure before. If you have a rigid body, you can study that the stability of the – you can state the static equilibrium of the rigid body under the forces applied. And you say the sum of the forces should be zero if the rigid body was at static equilibrium. And the moment computed about any point going to be equal to zero, as well. Now, if this rigid body is moving, so here the rigid body is at static equilibrium; we do that analysis. But if the rigid body was moving, then the rigid body, the masses, and the inertias are going to generate an additional force. We saw this force here; the first force, mq double [inaudible] plus v, actually these are inertial forces. They are created by the fact that this is a rigid body with mass. So if we can compute the forces and the moment associated with this rigid body – yes? Student:Where do we have our [inaudible]? Instructor (Oussama Khatib):Well, they will come – they will come little later. We can create the – those spring and dumping into control to stabilize and control the robot, or the robot might have some dumping at issue of the joint because of friction. But we will come to that later. So if we – if we consider those forces, then we can restate the static equilibrium by saying, as it is moving, we should have a static equilibrium that will equal – will be equal not to zero, but rather to f and to n, to those movement and forces applied. So then we can come up with a relationship. So first of all, for the Newton equation, you – we saw the equation earlier. The Newton equation is describing the linear motion. The Euler equation is describing the angular motion. So mass acceleration equal force; moment equal inertia time acceleration – angular acceleration – plus this term that creates the centrifugal Coriolis forces. And by stating that equilibrium and then doing the projection on each axis, you’ll remember in the static analysis we projected on the axis to see the forces acting on that axis. So this essentially eliminate inertial forces – internal forces acting on the structure. Then we will be able to find their equations. To do that we will project [inaudible] in the static case and find those component that is the torques applied at issue of the joint axis. So this is basically the Newton-Euler formulation. Now, the Lagrange formulation doesn’t go into the specific motion of each of the joints, doesn’t require any elimination; it works out the problem differently from energy analysis point of view. So essentially what we’re going to do there, we take the whole articulated-body system and we take the kinetic energy of each of the legs. So the kinetic energy of [inaudible] is, let’s say it’s ki – the total kinetic energy of the system is the sum of all the k [inaudible]. And we have also the potential energy. Now, once we decided about our system of [inaudible] coordinates, in this case it will be q’s, q one to q n. We can write the kinetic energy in this form. Earlier, we said the kinetic energy of an articulated-body system would be just this q duct transpose and q duct. And this means that k – this k over k is half the velocity vector of transpose multiplied by the matrix, plus multiplied by the vector. That gives you a scalar. Now, if you take the scalar and combine it with the potential energy, you will be able to immediately find those equations. Now, once you find those equations, actually you realize, “I know the equations actually directly from m.” By computing the kinetic energy, you identify m, which will go here. So you know m q double [inaudible]. From the potential energy, you knew – you know the gravity. What is left is v, and we will see that. Once we know m we can find v. So the whole equation of the dynamics can be obtained simply by taking the potential energy, taking its gradient and that gives you g, the gravity vector. And by taking m and computing m, how can we compute m? Well, we can say the kinetic energy should be – if – if we have a system [inaudible] coordinate, it’s a quadratic form [inaudible] velocity. So from the top we can say the kinetic energy could be computed for each rigid body, and by doing the identity between the first expression and this expression, we can identify m. We will see that on Wednesday. So this is a slide we saw before that is the idea of breaking the structure and analyzing each of them, and then eliminating the internal forces. Well, this is exactly what we’re going to do again, now, but by adding the inertial forces. So in addition, here, we are at static equilibrium; the rigid body is not moving. If it is moving, there will be these forces and then we can say the forces fi are equal to fi plus fi plus one, from this relation. The sum of the forces should be equal to the linear acceleration, and the sum of movements should be equal to the angular inertial forces. And this is the algorithm, basically, that we will find – this algorithm will allow you to compute the dynamics. So the Newton-Euler algorithm does the following: it propagates velocities, we know we did that for the Jacobian. You propagated velocities. You – as you propagate velocities, you compute your accelerations and from the accelerations, you can compute the inertial forces as you’re propagating. And then it has a backpropagation, that is the projection on the axis by taking these forces and starting from the end and going back, you propagated your forces and when you reach the ground, basically, then you have all of your tools. So we’re going to start with the basics and the reason for that, I want you to understand a very important equation related to the rigid body, and this is the Euler equation. We need to understand what is the inertial for a rigid body. If we were working just with a single particle of mass m, the problem would be very simple. We don’t need Euler equation. We are working with rigid body; when they move, this rigid body has a mass distribution. And we need to capture the mass distribution at some point, and that result into this inertial tensor that we are going to use to discard the rotational motion of a rigid body. Now, this might be scary, but it’s going to be really, really simple if you just pay attention. We will start from a particle. We take a rigid body and look at it as a collection of particles, and then we will look at the linear velocities of each of those particles and they move, and we group them together and we will find the inertial associated with the rigid body. Okay? Sounds good. All right. Let’s try. So as I said, for a particle m, if we apply a force there will be an acceleration that is equal to f divided by this mass. So the mass is resisting to accelerate. So if the mass was infinite, no motion, right? The lighter, the faster with the smaller forces. So this is the law I think everyone is familiar with. There is a velocity, actually, with this acceleration that is not aligned with the acceleration depending on how the trajectory is going. And we can think about this same equation in the following way. So I’m going to group the mass and the acceleration using the derivative of the velocity. So you can write the same equation in this form – the partial derivative of mv is equal to the force. What is mv, by the way? I’m sorry? Instructor (Oussama Khatib):Exactly. It is the momentum associated with this linear motion. It’s the linear momentum – mv is the linear momentum of the particle. So this linear momentum is playing a really interesting role, you can see what is this role here? If we think about this linear momentum, the rate of change of the linear momentum is equal to? The applied force. Nice. So the rate of change of the linear momentum is equal to the force. We’re going to show that actually Euler equation or what it does, it says the angular momentum, if we know the angular momentum; if we take the rate of change of the angular momentum is equal to the moment applied. And with this symmetry, basically, then we will be able to compute this tensor, or this inertial matrix, associated with a rigid body. Okay. So the rate of change of the linear momentum is equal to the applied force. And we call the linear momentum phi, okay? Can you remember that? Phi is equal to mv. See, dynamics is not that complicated. All right. Let’s talk about the angular momentum. How can I make an angular momentum with a particle? So we have an – so we have this particle rotating and we have an inertial frame, and I’m going to compute the angular momentum. Well, basically, I need to – to cross [inaudible] this with a vector, right? To compute the movement of the force and then I can have the movement of the inertials. And so if we take the moment of this force on the right with respect to the origin, o, then we can take m – mv that is like a force, right? It’s an inertial force, mv dot. You agree? So we’re going to take the moment of this equation with respect to o, so we need the vector that connect o to the particle, to the position of the particle. Right? Okay. Let’s take the moment, you remember the moment is p cross f, on the right, or p cross mv dot. Okay. This is a very important step. If you understand this step, everything that we will talk about later will become just adding all of them together. So I had a linear motion, I’m just looking at it from angular motion point of view and I’m just taking the moment of this equation with respect to a fixed point. All right. P cross f is a moment. We call it n. What is this? That’s right; it’s too complicated. All right. Okay. So you said earlier, mv is the linear momentum. I’m going to take p cross mv. And if I take p cross mv, and I need to get the rate of change of that quantity, this is sort of the angular momentum and I’m computing it ahead of time just to show you what it’s going to be. So if we do the computation it will be p cross mv dot, v cross mv, and v cross mv is equal to zero, so that gives you – because when you cross product a vector with itself, it gives you zero. So that gives you the rate of this quantity equal to the moment. And that is the angular momentum. So the linear momentum is mv and the angular momentum is the vector locating m cross mv. Okay? We’ll put it down, so now you have to remember mv, linear momentum; p cross mv, angular momentum. Okay. Well, once we get a rigid body, all that we need is to do the sum. So this p will become pi’s, mi’s, and vi’s, and we are going to add all these particles – a lot of them, many of them. So instead of doing this with just a sum, we will do an integral. And because we have objects in three-dimensional space, we will triple integral in dxy and z. And basically you get your inertia for the rigid body. So let’s call it first, this quantity p cross mv, we call it phi. It’s the angular momentum. So I’m going to now think about this equation in the context of a rigid body. The angular momentum of a particle we know, we want to find the angular momentum of all the particles. And we assume that this rigid body is moving at some velocity and acceleration related to the instantaneous angular velocity and acceleration we know. Okay? So we need to locate each of the particles. So we have pi and the linear momentum is – I’m sorry, the linear velocity, vi of that particle, is going to be – we study that in the angular rotations, it will be omega cross pi, right? So what is the angular momentum? The total angular momentum of this rigid body is going to be the sum of all the pi’s locating this mi mass’ moving at vi, velocity. So now we’re going to take vi and bring them down here and we will have pi cross mi, omega cross pi, so we will have a more complicated expression. So we have omega cross pi. So this would be the total angular momentum of the rigid body. Right? Do you agree? If I can count them all, all the particles. It’s difficult to count, but that’s what’s nice about the mathematical models that you can use. You can assume, or, let’s assume that I can count them all. Now, in this equation, do you see anything that is constant, that is not changing, that is independent of the rigid body? Louder. Louder. Instructor (Oussama Khatib):Omega. All these particles are moving at – rotating about that axis with omega. So omega is independent, so what are we looking for? If omega is independent [inaudible] I’m going to go to an integral, I need to get this omega out of the sum, right? We don’t need it in the sum. So how do you do that? Instructor (Oussama Khatib):I’m sorry? Student:[Inaudible] triple [inaudible]? Instructor (Oussama Khatib):So, yeah. I mean, you basically – the omega is on the left side, you can, if you put minus, you can bring it to the other side. Omega cross p or minus p cross omega, right? You remember this? And the other thing is the fact that mi representing that mass for that particle. If we assume that we have a homogenous object, then we can represent it – represent the mass by this more volume multiplied by the density, right? So using this and using this – well, I didn’t do it yet, so we’ll go from that sum to this integral with the same equation before tripping this. But this is what we need to get this out of this equation. It is independent of the equation. So the phi is your total angular momentum. So let’s write this in this following way. I’m going to write it minus p cross d and omega out and that means – and also substitute with the cross product operator. So p cross is p hat, and that leads to this form. In here what you see is minus p hat, multiplied by p hat, [inaudible] all of this is variable depending on the particle; omega is constant. So phi is equal to this integral times omega. And this integral is essentially your inertia, the inertia of the rigid body. What we call the inertia tensor. So we can write it simply like this: phi is equal i omega. So maybe I went too fast. You see this relation? Where i is this integral. So for linear motion, phi is equal mv for one particle, and we have Newton equation, which stays the rate of change of phi is equal to f. Okay? It’s another writing of ma equal to f. No one remembers this one, the phi prime equal f, one. People are afraid of momentum. Okay. Now with the angular momentum we’re going to see the same thing. The rate of change of the angular momentum is equal to the applied moments. So phi is equal i omega and Euler equation is simply the rate of change of phi is equal to the applied moments. Couldn’t be any simpler, right? Then you will be missing something. Now phi prime is not as simple as the linear motion. When you take the derivative of phi, you get i omega that acts – this is like mass acceleration, but also because of omega, we have i omega in another product of velocities, and that produces centrifugal and Coriolis forces. So we have the two equations for a rigid body. So a rigid body has a linear motion; if I throw this like straight at you, it will – it’s going to [inaudible] there will be some air resistance and you will feel some rotational motion. But this combination of linear motion and angular motion is captured by this for one rigid body. Well, I have to deal with multi-rigid bodies attached by these joints and when you put a joint, you are putting a constraint. You’re throwing the thing is going to be pulled and pushed and that means we need to eliminate the internal forces in order to find the actual motion. So a very important thing that we establish with this relation is this i, and now we need to really, I mean, this is the thing that I need you to remember. You need to be able to compute i for each of the rigid body of your robot. So this is something that is absolutely needed. And in order to do that, you need to see little bit more into the structure of the inertial matrix or the inertial tensor. So we say the inertial tensor i is basically this integral over the volume of all the vectors locating all the points on the rigid body, and scaled by the density of masses on the rigid body. So if we take this quantity, p hat p hat, so this is the cross product operator, pretend that you can write it in this form. So p hat is what? You remember p hat, the cross product operator, it’s a three by three matrix. So if you multiply a three by three matrix by itself, you’re going to get another matrix, right? Now, this matrix could be written as p transposed p. P transpose p is what? Instructor (Oussama Khatib):A scalar. So you are scaling – p transposed p is the square of the vector p. So you are scaling the identity matrix. So on the diagonal of the identity matrix you have the square of – the component of the vector p minus p, p transposed, which is a three by three matrix. So using this relation, we can rewrite the inertia tensor in this form. So let’s take this computation. So p transposed p is x squared plus y square plus z squared. And p transpose p time the identity gives you this quantity. All right? Nothing complicated. The other one gives you this. If I have x, y, and z, I’m going to multiply p, p transpose, that will give me this matrix. And the result will be this matrix. So minus p hat p hat in that integral, is this. Okay? Essentially, this is locating the position x, y and z with this vector. And now when we do the multiplication it appears like this. All right. So we’re almost there. We need to control – compute the integral of this and put the density and that – then we will have the inertia tensor. So the inertia tensor is – you see this equation here? Now I’m going to put it in that integral and I will find i. Ixx, Iyy and Izz, and Ixy, and Ixy, et cetera. Anything remarkable about this matrix you can tell me? Properties wise? Instructor (Oussama Khatib): Instructor (Oussama Khatib): What else? Instructor (Oussama Khatib):Positive definite? That’s if you’re at zero. If you’re at zero, you’re in a black hole. Okay. So this ixx, yy, .zz’s and the [inaudible] just we – I’m using the previous matrix to recompute this element, and this is your inertia tensor. Cool? You understand now the inertia tensor? Oh, what it is, this three by three matrix is going to each of the points, finding the distance, and then you’re walking and integrating all of these from each of the component to find what is the waiting, the inertial properties about the axis xx – about the x-axis, y-axis, z-axis, and what are the coupling between different axes. So the xx, yy, zz are called the moment of inertia. And the other one are called the product of inertias. All right. Here is an example. If we take a – if we take a rigid body that is nice, symmetric, like cylinder or [inaudible] or cube, basically, the property of symmetry when you do the integration, you’re integrating between one side to the other and that leads to some nice properties because if you do this computation at the center of mass, you are going to be able to find the inertias and you will find, most of the time, zero product of inertias. And then if you need to do this computation at a different point, or what you need to do is now to look at your vector, so if you start from this vector, this point, you will be able to reach all these different points. Then if you start from a different point, to reach this point you can go like this and plus this. This one you already used to find the inertias about the center of mass, or what you need is to have this vector. That because you’re going to add the same vector for all the points, it turned out that you have a very nice property, which is the property of parallel axis theorem that tells you the inertia about any point a, is the inertia about the center of mass plus the mass of the object multiplied by this quantity. That is the quantity that let us compute pc, this vector, in addition because all the masses are basically can be measured like, concentrated at this point. So this property is very nice because you can do simple computation at the center of mass, and then just do your transformation to move to a different point. So there is an example of a cube where we do the computation at the center of mass, you have this example, and the answer is, you get Ixx at the center of mass, Iyy at the center of mass, and Izz at the center of mass, all equal to ma squared divided by six, for this cube. Now, if you want to do this computation at one of the edges, like at this point, or what you need to do is to add this quantity that represents the distance from that point to the center of mass, and scaled by the center of mass and you obtain this quantity. So this is a very useful theorem if you are doing this computation. And what you notice, also, is that then you will have product of inertias. So the – at the center of mass, you have no product of inertia, you have a diagonal matrix. When you compute the inertias at a different point of the center of mass, you are going to have product of inertias. All right. so now we have the inertia of a rigid body, then we can apply the equations of Newton-Euler, and we do this propagation and we will compute all the inertial forces as we propagate. And then we’ll project back to compute the forces and we find the dynamic equations. So as we saw, these are the two equations, the translational motion and the rotational motion. And what you need to do also is to compute accelerations; you remember here, we’re saying you need to compute velocity, find the accelerations, and then you can find the inertial forces. So we can go over this and now, I’m not asking you to really follow the details of this computation, but essentially, you’re going to start from the velocities. The recursive relation of velocities as we propagate, or maybe i plus one equal omega i plus the joint velocity of that revolute joint, and you have this relation. Now, if you take the derivative of this relation, you will find this derivative involving the acceleration of your revolute joint, you will also need to compute the linear accelerations. You start with the velocities, you take the derivatives, and you get your recursive relations, okay? So we are forward-propagating to go to the last joint, and you have a lot of derivatives depending on the type, if d is revolute, prismatic or not. And we’re not done because we have to find the velocities and acceleration at the center of mass, not only at the joint angles. So you need to do a small addition, the linear velocity at the center of mass, you need to multiply by this vector, omega I plus one. And that gives you the velocities, acceleration at the center of mass, and now you’re ready. So at the center of mass, you can write the forces – the linear and angular forces. So this is the inertial forces acting at the center of mass, and these are the linear forces acting at the center of mass, and this is the inertial tensor computed at the center of mass. Now, you take this moving, accelerating rigid body and you say that all the forces applied to this rigid body should be equal to the inertial forces and the acceleration of related to the movement through the Euler equation. And you write these two equations, and now you do the recursive relations with these expressions and you’ll get your recursive relations so now you see it is fi as a function of i plus one, so you’re backpropagating. And as you backpropagate, you have to make sure, in order to find the torque, to project on the axis. So we did this, we do this, and now we have the recursive relation and then you project at each of the axis, your end, you compute from here, or f computed from here – and you have those relations, oh, my God. I’m lost. So you can see it’s – I mean, it’s really difficult. It is – but it is wonderful algorithm, in fact, to compute precisely your force as a function of the velocities, acceleration, inertias, and masses and everything. But you have no idea about what’s going on, right? It’s very difficult to see. You need to – okay – so – this is [inaudible] iterations from zero to five, and [inaudible] iteration with – don’t forget this to eliminate your – okay? Yeah. Well, I mean, it can – [inaudible] it can go to 25, but it works. Yeah, what about the gravity? We didn’t talk about the gravity here. We forgot the gravity. Oops. So what do we do for the gravity? How do we account for the gravity in this algorithm? So, okay, you remember, I say the algorithm, you’ll start from the base, you assume the base at zero velocity, zero acceleration, and you move out and you come back. Now, if you want to account for the gravity, what should we do? Student:[Inaudible] accelerates at g. Instructor (Oussama Khatib):Yeah. Just [inaudible]. Very good. So if you say I have a linear acceleration of equal to 1g from the beginning, you will be including the gravity. Good. [Inaudible] for a second. Now, all right. So good, we will skip this and to skip it, I will go to here. And we go to the Lagrange equations. All right. So what we saw, the thing you have to remember and not forget, is how to compute your inertia. We – you need still to compute the inertia because when we go and compute the kinetic energy, that is needed in Lagrange formulation. You are going to compute the linear motion, kinetic energy, which is one-half mv square, and what about the angular motion? The kinetic energy associated with angular motion? Omega – Instructor (Oussama Khatib):So it is one-half omega transpose i, omega. So we need this i anyway, you cannot escape. So you have to know how to compute i, correct.? Okay. Lagrange equations, actually, I mean, we can skip that if you want. [Video] An innovative – Instructor (Oussama Khatib):No, no, no. What I meant is we really don’t have to know all the details of the equations. But I really want you to understand Lagrange equations because they are going to be very useful for you when we get to control. And when we are going to control the robot, the robot has its own dynamics. And also we are going to apply to it external forces to control it. So these external forces are going to affect the dynamic of the robot in some way. And we need to understand the Lagrange equation, not only to compute the dynamics. All that we need for the dynamics is the kinetic energy, and we know the answer from Lagrange equation or from Newton-Euler – from any formulation of the dynamic equations, we will have the same structure, mass q double duct plus v plus g equals torque. And to compute the mass, all that we need is to compute the kinetic energies. But we really need to understand what this – what is this structure of Lagrange equation, how under applied forces mechanical system is going to move. So this has a very important rule, not only for the computing the dynamics, but really to understanding the control. Okay? How many of you have seen this equation before? Okay. Six. All right. So let’s imagine this equation in the scalar case, so this equation would be just simply an equation where the torque is the torque applied to one revolute joint. So this is a scalar equation. What is l, those of you who have seen it before? Instructor (Oussama Khatib):So it is the Lagrangian and it is simply the kinetic energy minus the potential energy. And what is this q? So when we – when we look at – okay, maybe I think I have – I have k minus u, so l is k minus u. I don’t know if I will put it in the – so most of the time when we are talking about natural gravity, the u is only function of where, what height you are at. So it’s function of q. So we can rewrite this equation in this form, right? I’m separating the kinetic energy from the potential energy because the potential energy is independent of the time, and so – so you can write it in this way. And essentially, what you have in here, you have mq double duct, and plus v if you have multi-body, and here you have g, the gravity. So here, you just have the gravity vector. So in fact if we move the gravity vector to the other side, let’s not worry about the gravity here in space or – the gravity vector is just the gradient of your potential energy. So essentially, it’s saying that you’re inertial forces are equal to the torque’s minus the acting gravity. And if we think about it, this part of the inertial force, this is what it’s going to give you. When – once we do the derivation that that left part will give you mass acceleration plus centrifugal Coriolis forces equal to these torques minus the gravity. So let’s look at this little bit. As I said, the kinetic energy, if we have generalized velocities, if we say q is our generalized coordinates, q dot is our generalized velocities, we can write the kinetic energy as a quadratic form on the joints velocities. So k is q dot transpose n q dot. And if we take the derivative, so I’m going out to take this k from there, and differentiate with respect to q dot, okay? So partial derivative of k with respect to q dot, of this quantity, and this is q dot transpose m q dot. So what do you think the answer is? No, let’s do it in the scalar case. So you are taking, let’s say I’m taking mv square – one-half mv square and I’m taking the partial derivative with respect to v; what would be the answer? Instructor (Oussama Khatib):Mv. Well, in the vector case, it will be mv as well. It will be mq dot. So nice. This is really nice. So now, take the second derivative of this, I mean, this derivative with respect to time. So if you take the derivative of mq dot with respect to time, you get mq double dot, plus m dot q dot. You see why m dot? Because m is function of q. Right? Function of q, so you get m dot q dot. Oh. So let’s write this. We computed this, right? And it is – this first part is m q double dot plus m dot q dot. What about the other part? You see k over there? Need to find the partial derivative of this k with respect to the q’s. So in the kinetic energy, what is dependent on the q’s? Can you point that to me? Okay. The kinetic energy is one-half q dot transpose, m of q, q dot. So what is dependent on q’s? Instructor (Oussama Khatib):M. All right. So I’m going to write it as a vector because this is – what does this mean? It’s partial derivative with respect to q, one partial derivative with respect to q, two partial derivative with respect to q n, so it’s like this. So I’m just writing exactly that quantity partial derivative with respect to q, one, two, q and –okay? You agree? So everyone who have never seen Lagrange equation before agrees also? Is this clear? So this means mass acceleration, m q, m dot q dot, so if q dot was zero, this disappears, right? If m was constant, not configuration dependent, this would be zero and this would be zero. Right? Okay. This messy thing is your centrifugal Coriolos forces, it’s product of inertias – if you think any element of this, it’s a product – I’m sorry, product of velocities. You have always q dot, q dot, that is multiplied. So this is what we call v, v is this vector minus half of this vector; and this is v. Okay. So this equation, the partial derivative of the kinetic energy leads to this equation. As I said, because we see the answer, you need to compute m, right? But m is there in the kinetic energy. And these things are function of m. So we really have just to compute m from the kinetic energy and we know the By the way, do you know why this, we already saw this example – what is the answer? Is, I’m mixed up. Okay. so this – I mean, I way to think about is it to form vector of v which is m square root of m q dot, and then your equation will be v transpose v for the kinetic energy, and then you can show that, et cetera. All right. If you want to see why it is equal to m q dot. All right. So the equation of motion using Lagrange equations, are in this form, mass acceleration plus these two vectors, leads to this v vector that is function of q and q dot. And if q dot is equal to zero, v will be equal to zero. So this is the structure of our dynamic equation and what we know is inside the kinetic energy, there is this m. If I can’t compute the kinetic energy some other ways, then I will find m. And what we’re going to do, we’re going to go to each of the rigid bodies and compute that kinetic energy for k1, k2, k3, to km, add them all together, and identify that expression to this expression and we can extract m. And once we have m, using this relation between v and m dot q dot and the vector, we will be able to compute v. So this is what we will be doing on Wednesday. [End of Audio] Duration: 74 minutes	{"url":"https://see.stanford.edu/materials/aiircs223a/transcripts/IntroductionToRobotics-Lecture11.html","timestamp":"2024-11-13T05:23:10Z","content_type":"text/html","content_length":"48638","record_id":"<urn:uuid:6d75b1a1-0154-49b7-b9e6-dcc5a8aa276f>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00688.warc.gz"}
An Introduction to Plotting Cambridge Spark - Data Science Courses / March 22 2019 / 11 minute read I'm a teaching assistant for several MSc courses, covering natural language processing, natural language understanding, information retrieval and Python programming. Students on these courses, as part of their coursework, often have to produce visualisations of data. Usually, for these types of questions, the format of the visualisation is made clear. For more open-ended tasks where students are free to follow their own processes, however, I find that students often make visualisation choices which impede the reader's ability to quickly reach the conclusion they are trying to illustrate. This is usually due to one, or more, of several reasons. In this blog post, I'll first lay out what sort of questions you need to ask yourself before you start plotting. These questions are all based on understanding what data you have and what you want to achieve. Then, I'll give examples of how you could proceed, in a variety of simple situations, and produce plots that show what you want. The focus will be on the intended purpose of your plots and the kind of data you have, though I'll also point out some things you want to consider such as other distinctions in your data that you want to illustrate, as well as general tips for aiding the reader in accessing and understanding your plots. What's the purpose of the visualisation? • Do you want to... □ show the relationship between variables? □ illustrate individual distributions of variables? Student don't always pick the best visualisation for the aspect of the data they want to represent and don't know that their choice of visualisation is highly dependent upon their aims. Do they want to show the relationship between two (or more) variables? Or the distributions of one or more variables? Should the data be aggregated in some way first? But even before this question can be addressed, students also aren't fully aware of what kind of data they are working with. This is crucial because it helps determine what visualisations are suitable - choosing the wrong one can be jarring when viewed by someone who is better able to make these distinctions. What kind of variables do you have? • For each variable... □ Is it continuous? □ Or is it categorical? Most data falls into one of two categories: continuous or categorical. The distinction between the two can be difficult to make. This distinction isn't usually explicitly pointed out to students - they're expected to know it beforehand. And even when it is, the different kinds of visualisations that are suited to combinations of the two are rarely taught. Continuous data Continuous data will consist of variables which are real numbers. They can have any value from an infinitely uncountable selection. These will generally be floats or integers and will have the expected numerical properties associated with the real numbers. Such data can be sensibly ordered from smallest to largest and can undergo basic arithmetic operations. In practice, continuous data will likely be a measurement of something: weight, height, time spans, density, age, volume, distance, speed and so on. It is sensible to think about adding two weights to each other, or subtracting one height from another, or calculating the mean of multiple ages. There can be some terminological confusion here as continuous data can easily be discrete (e.g. integers). Categorical data Categorical data is not real-valued. Instead, the values are more like labels or names. Generally, these will be represented as strings. Examples are the months of the year, names of colours, makes of car, medication names. Although some of these variables can be sensibly ordered, this ordering is not innate. Rather, it is reliant on some other dimension such as time (e.g. the days of the week) or wavelength (e.g. the names of colours). Confusingly, the names of the categories can be integers or cardinal numbers! For example, if you run several different experiments and record some real-valued outcome for each, then you may label each experiment in the order you performed them - 1st, 2nd, 3rd and so on. Unless there is some dependent factor between experiments, then such labels may obscure the categorical nature of what the variable represents. As a counterexample, when tracking the loss per epoch when training a model, the epochs will likely be labelled 1st, 2nd, 3rd and so on. The labels of these categorical variables have important temporal ordering and each depends on the prior step. So this would be a discrete continuous variable. It is important to note that continuous data can often be converted to categorical. For example, ages can be binned into groups labelled "children" and "adults". Making the decision to do this very much depends on the point you wish to illustrate with your data. Is there some other informative distinction I want to show? Do my variables come from... • different groups of people/individuals/companies/locations? • different time periods? • different experiments? • different models? Even once you know what kind of data you are working with, and what relationship or property you want to visualise, there is often a third factor to consider: can the data be partitioned in a meaningful way that aids interpretation or highlights some informative detail? As a simple example, you may wish to show the relationship between two continuous variables, height and weight. But you may also wish to include information about sex, which is categorical, or age, which is continuous. Once you know what kind of data you have, and what relationship or property you want to visualise, and what aspects (if any) of the data you want to highlight, the remaining problems are mostly Maximising informativity Another issue is that the student may have tried to plot too many aspects of the data on a single graph. Alternatively, they may have included too many graphs. Both of these approaches overwhelms the reader, in different ways, with too much visual data. Being selective about what to present is always the better decision, as it will generally allow the reader to speedily reach the same conclusion as the author. Sometimes students do not include information on their plots that really is vital, or they obscure useful details. • No title • Uninformative caption This kind of information really helps drive home the purpose of the visualisation, by highlighting exactly what points are being made. A good caption should succinctly summarise the visualisation or aid the reader in quickly understanding what is being presented. • No labels • No units • Units not scaled appropriately Making the reader aware of what each axis represents, and what units are being used, sounds like it should be an obvious inclusion in any visualisation. However, I think it is easy for students to forget this after they have been working with their data for a while and assume the reader will know too. When I'm marking student work, I generally do know this but for any other reader, this could be confusing. Additionally, if the range of values for each axis are not neatly distributed, then some values may dominate others and cause the visualisation to be imbalanced - some data points may be far larger/ smaller than others, making it difficult to appreciate differences or variation between data points at the other end of the range. Elements (i.e. bars, points) • No legend included • Choice of colours, line styles or point styles difficult to tell apart These issues affect readability more than anything, which slows down reading comprehension as you spend time trying to figure out exactly what each line represents. Putting it into practice Now, let's take the first two of the four main areas above and look at how they can guide us during the decision making process. import pandas as pd import seaborn as sns %matplotlib inline Single Variable: Continuous When you have a single continuous variable and want to visualise the distribution of its values in your dataset, a histogram is generally what you need. This groups the values into bins, where each bin is an interval within the range of values your variable can take. The x axis will show the interval of each bin, while the y axis shows the number of values in your dataset that fall within that Let's load in some data using seaborn's handy load_dataset() function. The flights dataset has three variables: two ordered categorical (year, month) and one continuous (number of passengers). flights = sns.load_dataset('flights') A simple histogram will show the overall distribution of the passenger variable. This is easy to plot, as pandas dataframes have a built-in method for generating it. import pandas as pd import seaborn as sns %matplotlib inline By default, pandas plots histograms using 10 bins but you could fine-tune this. Displaying more bins gives a more detailed overview of the distribution, up to a point: it all depends on how many observations you have overall and how they are distributed. You can see how using 20 bins shows more information about the distributions inside the larger 5 bins. flights.passengers.hist(bins=5) # The blue bars flights.passengers.hist(bins=20) # The orange bars So the range of passenger numbers is a little over 100 to a bit over 600, with most flights towards the lower end. For a more precise overview, the describemethod for a dataframe's columns will give general descriptive statistics. count 144.000000 mean 280.298611 std 119.966317 min 104.000000 25% 180.000000 50% 265.500000 75% 360.500000 max 622.000000 Name: passengers, dtype: float64 For a visual representation of describe, a boxplot will show the minimum and maximum values (the left and right whiskers), the range of values covered by the 25th to 75th percentiles (the box) and the value of the median (the line inside the box). Single Variable: Categorical Bar chart When you have a variable which takes on named, rather than numerical, values then the most common way of representing them is with a bar chart. Here, we'll load the titanic dataset. Each row is a passenger on the ship, while the class variable gives the class of that passenger's ticket. titanic = sns.load_dataset('titanic') Third 491 First 216 Second 184 Name: class, dtype: int64 You can chain .plot(kind='bar') to the above value_counts() method, but I prefer to use seaborn as you can directly pass it the original data. It will then do the counting for you and allow you more control over appearance. For example, if you do not like the ordering seaborn used for the x axis, then you can set it manually as a list e.g. order=['Third', 'Second', 'First'] If you want to normalise the counts so as to see relative percentages rather than counts, then you just need to do that to the data before plotting it as a normal barplot. titanic_normed = pd.DataFrame(titanic['class'].value_counts(normalize=True)).reset_index() sns.barplot(data=titanic_normed, x='index', y='class') Plotting relationships between variables Above, we only had a single variable. We examined it by looking at the frequency of values (the histogram) or by plotting descriptive statistics (the boxplot). But often we want to see how one variable is linked to another - as the value of one variable changes, what happens to the value of the other variable? With continuous and ordered/unordered categorical variables, we have four possible combinations. Let's look at them in turn. Continuous x continuous The mpg dataset contains information about cars, measuring their weight, fuel efficiency and so on. We might expect heavier cars to have lower fuel efficiency. When plotting continuous variables, the one you place on the x-axis should be the independent variable. This is generally some property or value we observe. The y-axis should display the dependent variable. This is a function of the values on the x-axis and is generally something we measure for each observed value on the x-axis. Here, we will place weight on the x-axis and miles per gallon on the y-axis. Generally, the best choice of visualisation for this is a scatterplot. Each point represents the relation between a single value on the x-axis and its corresponding y value. mpg = sns.load_dataset('mpg') g = sns.scatterplot(data=mpg, x='weight', y='mpg') There are several variations on this, which are made available through seaborn's jointplot. The default will add histograms on the margins, for each of the two variables. sns.jointplot(data=mpg, x='weight', y='mpg') By setting the kind argument to kde, you can instead plot a joint kernel density estimate, with individual density estimates on the margins. sns.jointplot(data=mpg, x='weight', y='mpg', kind='kde') Or you can set it to hex and plot the values as hexagons, which represent histogram-type bins. This can be very useful if you have a lot of observations in your dataset and plotting all those points is slow or messy. sns.jointplot(data=mpg, x='weight', y='mpg', kind='hex') Continuous x unordered categorical There are a few more options when it comes to jointly plotting continuous and categorical data. In general, the categorical data will go on the x-axis and you may need to change the order in which they are displayed. Let's look at the relationship between fuel efficiency (continuous) and a car's country of origin (unordered categorical). Seaborn's willstripplot make a separate scatterplot for each categorical variable and place it on the x axis, with its own colour. It will also stagger the points a little to help see their distribution - this can be controlled with the jitter argument. sns.stripplot(data=mpg, x='origin', y='mpg', jitter=0.3) The swarmplot does the same but arranges the points so that there is no overlapping. sns.swarmplot(data=mpg, x='origin', y='mpg') And if you want a boxplot for each categorical variable, there is no need to do them separately and manually place them in a figure - catplot is a great way to plot categorical x continuous data. sns.catplot(data=mpg, x='origin', y='mpg', kind='box') Continuous x ordered categorical Sometimes, the categorical data will have a natural order to it. The most common of these is times or dates. This can sensibly be plotted as a line, to show how the continuous variable changes over time. Generally, the categorical data must be unique - no value should appear more than once. The gammas dataset contains fMRI measurements taken from multiple subjects. Let's look at subject 0, and see how a signal which is dependent on blood oxygen levels (BOLD signal) changed over time in various regions of interest (ROI) in the brain. Seaborn's lineplot method has a hue argument, that will seperate out the three different values for ROI and plot them as their own lines. gammas = sns.load_dataset('gammas') subject_0_data = gammas[(gammas.subject == 0)] sns.lineplot(data=subject_0_data, x='timepoint', y='BOLD signal', hue='ROI') We could also focus on a particular ROI and then see how all subjects compare by setting hue="subject" sns.lineplot(data=gammas[gammas.ROI == 'IPS'], x='timepoint', y='BOLD signal', hue='subject', legend=False) # Remove the legend as it gets in the way with the default plot size. Categorical x categorical The most common non-graphical way of representing two joint categorical variables is as a contingency table. Each row of the table represents a possible value of one variable, the columns of the other variable. Cells are populated with the number of observations of pairs of those values. We can create that table using pandas' crosstab function - just tell it which columns of a dataframe to use. titanic = sns.load_dataset('titanic') sex_class = pd.crosstab(titanic.sex, titanic['class']) We can also normalise the values to show percentages, rather than counts. sex_class_normed = pd.crosstab(titanic.sex, titanic['class'], normalize=True) * 100 This tabular data is easily to represent visually as a heatmap. This essentially colours in the cells of the table, based on their value. It can be a great way to very quickly communicate the joint distribution of two categorical variables, especially where you want to highlight the fact that some particular combinations are very high or low. sns.heatmap(sex_class, cmap='Blues', square=True, annot=True, fmt='g') sns.heatmap(sex_class_normed, cmap='Blues', square=True, annot=True, fmt='.2f', cbar=False) Questions to ask before plotting Here are the questions to ask before you start plotting: • What is the purpose of my visualisation? □ Show the relationship between variables? □ Illustrate individual distributions of variables? • What kind of variables do I have? For each variable: □ Is it continuous? □ Or is it categorical? • Besides these variables, is there some other informative distinction I want to show? Do my variables come from... □ different groups of people/individuals/companies/locations? □ different time periods? □ different experiments? □ different models? • Have I included all the necessary information? □ Descriptive title? □ Informative caption? □ Axes have suitable labels? □ Units for axes, where appropriate? □ Axes using suitable scale? □ Do I need a legend? □ Do my colours and styling aid readability? Cheat sheet: picking a visualisation for your data And a quick list, linking types of data to types of visualisation: Single variable • continuous □ histogram: more visual, big picture, show distribution of ranges of values □ boxplot: more statistical and detailed • categorical □ barchart: show counts or proportions of values Joint variables • continuous x continuous □ scatterplot: show relation between every x and y □ basic jointplot: as above, but with marginal histograms per variable □ kde jointplot: show distribution of joint values, with individual histograms □ hex jointplot: as above, but points are now mini-histograms • continuous x unordered categorical □ stripplot: multiple scatterplots arranged on x axis □ swarmplot: as above, but no overlapping points allowed □ catplot with boxplots: replace individual plots with boxplots • continuous x ordered categorical □ line: shows exactly what values are seen over time • categorical x categorical □ cross-tabulate then heatmap: show relative proportions of joint variables Next steps • Look into seaborn's documentation for figure aesthetics and choosing colour palettes - these can make your visualisations look really great. The ones I did here use the default settings and could definitely be improved upon! • Think about how the plots could be improved in terms of the questions under "Have I included all the necessary information?". Seaborn makes it very easy to add titles and so on to figures. • Seaborn also makes it easy to visualise many aspects of the data at once, rather than individually as we did here. Read the documentation for jointplot and catplot to see how flexible and easy to use these methods are! • Try applying the above to real data that you have, rather than the toy datasets used here. About the author Alexander Robertson is a Data Science PhD student at the University of Edinburgh, where his research focuses on variation, usage and change in natural language and also emoji. Enquire now Fill out the following form and we’ll contact you within one business day to discuss and answer any questions you have about the programme. 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statistics answered \|Statistics tutor online\| My Mathlab answers \|Ap Statistics Tutor \| SPSS Binomial logistic regression using SPSS IBM statistics Binomial logistic is simply a logistic regression model that can be used to predict the probability of an outcome falling within a given category. The dependent variable is always a dichotomous variable and the predictors (independent variables) can be either continuous or categorical variables. When there are more than two categories of the outcome variables, then it is appropriate to use a multinomial logistic regression model. An example is when one might be interested in predicting whether a student "passes" or "fails" his/her college statistics based on the time they spend while revising for the exam. One can also predict the probability of drug use based on previous behaviors, age, and gender. This text explains to you the best way to do binomial regression using SPSS Statistics. However, before we run the data through a binomial process, your data must meet the following assumptions. Assumptions for a Binomial regression model 1. The dependent variable should be on a dichotomous scale - That is the measurements of the variables should be measured in categorical form. Examples of categorical variables include gender, race, presence of heart disease (Yes or No). Remember that we also have an ordinal regression model which can be used when the response variable is on an ordered scale. 2. You must have more than one independent variable measured on either a continuous scale, an ordered scale or a categorical scale. 3. The independence of the observations should also be met. 4. Your continuous variable and the logit transformation of the dependent variable must be linearly related. The 4th assumption can be checked via SPSS but the first three assumptions relate to the data collection process, In this example, we analyze to predict heart-disease (The dependent variable), that is whether an individual has heart disease or no, Using maximal aerobic capacity, age, weight, and gender. Note that age and weight are the continuous variables while gender is the categorical predictor variables. To run the Logistic regression model in SPSS step by step solutions Step 1: Go to Analyze > Regression > Binary Logistic as shown in the screenshot below. Step 2: In the logistic regression dialogue box that appears, transfer your dependent variable to the dependent variable (in this case its heart_disease) dialogue box and move you independent variables to the covariate dialogue box. The dialogue box shows how the variables should be transferred. Step 3: Click categorical to define the categorical variables (Gender), and transfer your categorical variables to the categorical covariates as shown below. Step 4: See the contrast area check the first option in the contrast category and click the Change button as shown below. Step 5: Click continue to return to the logistic dialogue box the Options button the dialogue box below is presented. Step 6: Check the following buttons, Classification plots, Hosmer-Lemeshow goodness of fit and casewise listing of residuals in the statistics and plots and the CI for Exp(b). Remember to check the at last step in the display area. Your dialogue box after this step should be as shown below. Last step: Click continue to return to your logistic regression dialogue box and click OK to get your output. Output and interpretation of the Logistic results Variance Explained This is equivalent to the R-squared explained in the multiple regression model. Cox & Snell R Square and Nagelkerke R Square values are used to explain the variation that can be explained by the model. Based on the output of the model, the explained variation is between 0.240 and 0.330 it is upon you to pick the statistic that interests you. Nagelkerke R^2 is a modification of Cox & Snell R^2, the latter of which cannot achieve a value of 1. Remember that it is always advisable to report the Nagelkerke statistics because Cox ^ Snell cannot be 1. Classification table The cut value of 0.50 implies that if the predicted category is greater than 0.50 then that is classified as a "Yes" otherwise that is a no. Some useful information that the classification table provides include: □ A. The percentage accuracy in classification (PAC), which reflects the percentage of cases that can be correctly classified as "no" heart disease with the independent variables added (not just the overall model). □ B. Sensitivity, which is the percentage of cases that had the observed characteristic (e.g., "yes" for heart disease) which were correctly predicted by the model (i.e., true positives). □ C. Specificity, which is the percentage of cases that did not have the observed characteristic (e.g., "no" for heart disease) and were also correctly predicted as not having the observed characteristic (i.e., true negatives). □ D. The positive predictive value, which is the percentage of correctly predicted cases "with" the observed characteristic compared to the total number of cases predicted as having the □ E. The negative predictive value, which is the percentage of correctly predicted cases "without" the observed characteristic compared to the total number of cases predicted as not having the Variables in the equation table The table presents the contribution of each variable and its associated statistical significance. The wald statistic determines the statistical significance of each independent variable. From these results it be seen that age (p = .003), gender (p = .021) and VO2max (p = .039) added significantly to the model, weight (p = .799) did not. You can use the information in the "Variables in the Equation" table to predict the probability of an event occurring based on a one-unit change in an independent variable when all other independent variables are kept constant. For example, the table shows that the odds of having heart disease ("yes" category) is 7.026 times greater for males as opposed to females. Comments (0)	{"url":"https://www.statisticsanswered.com/blog/43/SPSS","timestamp":"2024-11-11T13:29:05Z","content_type":"text/html","content_length":"39300","record_id":"<urn:uuid:d6832077-4eb2-495b-a956-5cdacf965b18>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00725.warc.gz"}
Decoding Thoughts with Deep Learning: EEG-Based Digit Detection using CNNs Digit detection using EEG data presents an intriguing intersection of neuroscience and AI. This article showcases the implementation of a convolutional neural network to predict whether a subject was thinking about a digit, using EEG data recorded with a Muse headset. The primary goal is to accurately classify EEG recordings into two categories: digits and non-digits (1 and 0 being respective labels). Dataset Summary The muse dataset from the MindBigData EEG database is being used here for the training. The dataset being used contains 163,932 brain signals of 2 seconds each, captured with the stimulus of seeing a digit (from 0 to 9) and thinking about it, from a single Test Subject David Vivancos. A small portion of the signals were captured without the stimulus of seeing the digits for contrast, all are random actions not related to thinking or seeing digits, they use the code -1. It needs to be mentioned here that though the dataset contains 163,932 data points only 3000 data points were used for training the model with 1500 of them being digits and 1500 being non-digits. Data Processing The data processing phase includes several steps, starting with loading the data, followed by wavelet transformation and resampling. Loading the Data # make sure to replace this with the path to the dataset in your system after downloading from the MindBigData dateset mnist_path = 'drive/MyDrive/EEG-Data/MU.txt' # Load the file into a pandas DataFrame mnist_df = pd.read_csv(mnist_path, sep='\t', header=None, nrows=2000) mnist_df.columns = ['id', 'event_id', 'device', 'channel', 'code', 'size', 'data'] # the last few rows of the dataset contain non-digit data resting_df = pd.read_csv(mnist_path, sep='\t', names=mnist_df.columns, header=None, skiprows=130000, nrows=2000) # concatenate the digit and non-digit dataframes into a single dataframe df = pd.concat([mnist_df, resting_df]) df["data"] = df["data"].apply(lambda x: [float(i) for i in x.split(",")]) Resampling Data The sampling rate of an EEG device is often variable, leading to EEG data arrays with differing sizes, as evidenced in the size column. However, for consistent analysis, it is essential that all data arrays be of equal sizes. There are several approaches to achieve this uniformity, such as downsampling larger arrays, zero-padding smaller ones, or employing a specific resampling algorithm. In this project, I opted to resample the arrays using linear interpolation — a method known for its efficiency and accuracy. # Function to resample an array to the target length def resample_array(array, target_length): # Create an array of indices for the input array input_indices = np.linspace(0, len(array)-1, len(array)) # Create an array of indices for the resampled array resampled_indices = np.linspace(0, len(array)-1, target_length) # Create a linear interpolation function based on the input array interpolator = scipy.interpolate.interp1d(input_indices, array) # Use the interpolator to create the resampled array resampled_array = interpolator(resampled_indices) return resampled_array.tolist() # Resample all the data arrays to the median length df["resampled_data"] = df["data"].apply(lambda x: resample_array(x, median_length)) # Check the length of the resampled arrays df["resampled_data_length"] = df["resampled_data"].apply(len) More Pre-processing Since a Muse headset comprises four channels, each code corresponds to four distinct data arrays, each reflecting the channel from which it was recorded. Although the data from these channels are more or less similar, managing them separately could be cumbersome and redundant. Therefore, I chose to average the data for the same code from each channel, per every four data points. This approach not only streamlines the data but also significantly reduces its size. data_array = np.array(df["resampled_data"].tolist()) codes = df['code'].tolist() data_array = np.reshape(data_array, (-1, 4, data_array.shape[1])) data_array = np.mean(data_array, axis=1) codes = codes[::4] Time-Frequency Representation and Wavelet Transformation Since the plan is to use a Convolutional Neural Network, so the raw EEG data needs to be converted to images. What better way then to create time-frequency plots of the data. (Check the complete notebook for details of the get_cmwX and time_frequency functions) starting_freq = 1 end_freq = 6 num_frequencies = 10 times = np.linspace(0,2,median_length) nData = data_array.shape[1] # calculate the Fourier coefficients of complex Morlet wavelets. cmwX, nKern, frex = get_cmwX(nData, freqrange=[starting_freq, end_freq], numfrex=num_frequencies) # calculate time-frequency representation of data tf = time_frequency(data_array, cmwX, nKern) Looking at the figures below, it is difficult for the untrained human eye to differentiate between time-frequency plots of a digit and a non-digit. Well good for us, Deep Learning exists. Model Architecture I am using fast.ai here for initializing and training the model which makes the whole process very smooth and simple. As you can see below it just needed 3 lines of code for all the deep learning “stuff”. For more details on fast.ai you can refer to Jeremy Howard’s awesome video. dls = ImageDataLoaders.from_folder(path, train='training', valid_pct=0.2, item_tmfs=Resize(224)) learn = vision_learner(dls, resnet34, metrics=accuracy) A Resnet34 CNN model is used. 20% of the data is used as validation set to calculate accuracy and the validation set is selected randomly at each training epoch. Results & Analysis Validation Accuracy As you can see after training for only 10 steps the model achieves 84% validation accuracy on the data. Test Accuracy I kept 1000 data points separate from the training and validation data, as test data. After training was complete, running the model through the test data gave a whopping 95% accuracy. Potential Improvements 1. Only 3000 data points have been used from a dataset of 163,932 data points. Train it on more data points. 2. A simple resnet34 CNN architecture. Try a bigger CNN architecture (maybe a resnet50 or some other model) 3. The time-frequency plots have been created for a frequency range of 1 to 6 Hz. The accuracy can be greatly improved by trying different frequency ranges, specifically by changing the starting_freq, end_freq and num_frequencies parameters. 4. As mentioned in the More Pre-processing step, the EEG values from the 4 channels for a single code has been averaged. This step can be removed which will lead to more data but also may lead to better results. Also instead of taking average of the EEG microvolt values, the time-frequency plots can be calculated for all 4 channels and then an average can be taken of the time-frequency plot for the 4 channels for each code. Please feel free to copy the code and play with it in your google colab environment or jupyter. This project just classified whether the subject was thinking about a digit or not with 95% accuracy. Though this is a step in the right direction, the accuracy must be further improved here for real world use-cases. Moreover, the logical next step must be to classify the exact digit the subject was thinking about, which one can imagine is a way more difficult problem. I hope this project will be helpful and motivate further research in this exciting field.	{"url":"https://dxganta.medium.com/decoding-thoughts-with-deep-learning-eeg-based-digit-detection-using-cnns-cdf7eee20722?source=user_profile_page---------0-------------258d5800608c---------------","timestamp":"2024-11-06T12:33:00Z","content_type":"text/html","content_length":"135950","record_id":"<urn:uuid:ea88eb11-1016-4751-9545-f479c4841a24>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00598.warc.gz"}
Generate a random string in go language! Generating random strings is a common task in software development, and Go provides several ways to accomplish this task. In this article, we will explore different methods to generate random strings in Go. Method 1: Using the Math/rand package The math/rand package in Go provides functions for generating pseudo-random numbers. We can use this package to generate a random string by selecting a random character from a given set of Here is an example code that generates a random string using the math/rand package: package main import ( var letters = []rune("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ") func RandStringRunes(n int) string { b := make([]rune, n) for i := range b { b[i] = letters[rand.Intn(len(letters))] return string(b) func main() { In the above code, we have defined a variable called letters that contains all the possible characters that can be used to generate the random string. We have also defined a function called RandStringRunes that takes an integer n as an input and returns a random string of length n. The function works by generating a random number between 0 and the length of the letters array using the rand.Intn() function. It then selects the character at that index from the letters array and appends it to the result string. Method 2: Using the Crypto/rand package The crypto/rand package in Go provides functions for generating cryptographically secure random numbers. We can use this package to generate a random string that is more secure than the one generated by the math/rand package. Here is an example code that generates a random string using the crypto/rand package: package main import ( func RandStringBytes(n int) (string, error) { b := make([]byte, n) _, err := rand.Read(b) if err != nil { return "", err return base64.URLEncoding.EncodeToString(b)[:n], nil func main() { s, err := RandStringBytes(10) if err != nil { fmt.Println("error:", err) In the above code, we have defined a function called RandStringBytes that takes an integer n as an input and returns a random string of length n. The function works by generating n random bytes using the rand.Read() function. It then encodes these bytes using base64.URLEncoding.EncodeToString() function and returns the first n characters of the encoded string. In this article, we have explored two different methods for generating random strings in Go. The first method uses the math/rand package to generate a pseudo-random string, while the second method uses the crypto/rand package to generate a cryptographically secure random string. Depending on the use case, you may need to use one or the other. I hope this helps, you!! More such articles: If this article adds any value to you then please clap and comment. Let’s connect on Stackoverflow, LinkedIn, & Twitter. Did you find this article valuable? Support techwasti by becoming a sponsor. Any amount is appreciated!	{"url":"https://techwasti.com/generate-a-random-string-in-go-language","timestamp":"2024-11-12T15:16:29Z","content_type":"text/html","content_length":"165575","record_id":"<urn:uuid:293b0c9b-008d-45d2-aee4-69a5bde2ae31>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00599.warc.gz"}
Solving Techniques 15 TOP > Techniques > Unique Rectangles Solving Techniques 15 Unique Rectangles The cells that that contain [12] in the diagram below would work as both: It can be either solution. We presume that a Sudoku problem can’t have two solutions. [34], on the other hand, can’t be switched around. They wouldn’t be switched within the same box, so the numbers can’t be switched. Unique Rectangles 1 Like the diagram, this is a pattern where the [12] can be switched around. If it can be switched, it wouldn’t work as a Sudoku problem, so in the cells indicated in red, either a [3] or [4] is Unique Rectangles 2 In this pattern, one of the upper two cells with [123], will definitely be a [3]. Hence, [3] won’t go in the cell marked red and it will be a [4]. Unique Rectangles 3 In this pattern, one of the lower two cells with [123], will definitely be a [3]. That means, [3] can’t be in the other cells in the box. Also, a [3] can’t be entered in line H of the boxes to the left and right. Unique Rectangles 4 An [8] or [9] will definitely be entered in either the [129] cell or [1289] cell. There is a cell at the very bottom with [89] as candidates. If an [8] is entered in the blue cell above, then the pink cell will be a [9]. If the blue cell is a [9], then the pink cell will be an [8], so [8] or [9] can’t be entered in the cells marked with X’s. Unique Rectangles 5 An [8] or [9] will definitely be entered in the [128] cell or [129] cell. The pink cell only has [89] as candidates. This means if an [8] or [9] is entered into either one of the blue cells, the pink cells will become [8] or [9], so [8][9] can’t be entered into the cells marked with X’s. Unique Rectangles 6 An [8] or [9] will go into either one of the [128][129] cells below. However, there are no open cells in this line where a [1] can be entered. Therefore, a [1] will have to go in one of the blue cells below, so a [2] can’t be entered into one of the blue cells at the bottom. Unique Rectangles 7 Lets examine the [128][129] cells. In this line, a [1] can’t be entered into any of the empty cells. This means, in either one of the [128][129] cells a [1] has to be entered, and [2] can be eliminated as a candidate. Names of cells in Sudoku R1C1 R1C2 R1C3 R1C4 R1C5 R1C6 R1C7 R1C8 R1C9 R2C1 R2C2 R2C3 R2C4 R2C5 R2C6 R2C7 R2C8 R2C9 R3C1 R3C2 R3C3 R3C4 R3C5 R3C6 R3C7 R3C8 R3C9 R4C1 R4C2 R4C3 R4C4 R4C5 R4C6 R4C7 R4C8 R4C9 R5C1 R5C2 R5C3 R5C4 R5C5 R5C6 R5C7 R5C8 R5C9 R6C1 R6C2 R6C3 R6C4 R6C5 R6C6 R6C7 R6C8 R6C9 R7C1 R7C2 R7C3 R7C4 R7C5 R7C6 R7C7 R7C8 R7C9 R8C1 R8C2 R8C3 R8C4 R8C5 R8C6 R8C7 R8C8 R8C9 R9C1 R9C2 R9C3 R9C4 R9C5 R9C6 R9C7 R9C8 R9C9	{"url":"https://nanpre.adg5.com/tec_en15.html","timestamp":"2024-11-13T18:06:50Z","content_type":"text/html","content_length":"17219","record_id":"<urn:uuid:c1f32e14-feeb-4785-af20-5e07dce9b64a>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00162.warc.gz"}
Seminários de probabilidade – Segundo Semestre de 2021 Seminários de probabilidade – Segundo Semestre de 2021 Lista completa (palestras futuras podem sofrer alterações) In several research problems we deal with stochastic sequences of inputs from which a volunteer generates a corresponding sequence of responses and it is of interest to model the relation between them. A new class of stochastic processes, namely sequences of random objects driven by context tree models, has been introduced to model this relation. In the talk I will formalize this class of stochastic processes and present model selection procedures to make inference on it. Discrepancy between the results of electoral intentions carried out a few days before the actual voting and the electoral poll results during the first round of the 2018 presidential elections in Brazil was striking. At the time, it was conjectured that this discrepancy was the result of social-media campaigning days before the elections. The question remains: is social-media campaigning enough to change the voting intention of a significant portion of voters? To provide an answer to this question was the initial motivation for this work. The model we consider is a system with a large number of interacting marked point processes with memory of variable length. Each point process indicates the successive times in which a social actor expresses a “favorable” (+1) or “contrary” (-1) opinion on a certain subject. The social pressure on an actor determines the orientation and the rate at which he expresses opinions. When an actor expresses their opinion, social pressure on them is reset to 0, and simultaneously social pressure on the other actors is changed by one unit in the direction of the opinion that was just expressed. The network has a polarization coefficient that indicates the tendency of social actors to express an opinion in the same direction of the social pressure exerted on them. We show that when the polarization coefficient diverges consensus is reached almost instantaneously. Moreover, in a highly polarized network, consensus has a metastable behavior and changes its direction after a long and unpredictable random time. This is a joint work and a joint talk with Kádmo de Souza Laxa. Access the slides here. In this talk we will discuss two special stochastic processes, which may be seen as branching processes with selection. We will present the motivation behind its formulation and some results related to the existence of phase transition in such processes. We introduce the Drainage Network with Branching, which is a system of coalescing random walks with paths that can branch and that exhibit some dependence before coalescence. It extends the Drainage Network model introduced by Gangopadhyay, Roy and Sarkar in 2004, by allowing the paths to branch. We also study the convergence of the Drainage Network with Branching, under diffusive scaling, to the Brownian Web or Net, according to specific conditions for the branching probability. We show that based on the specification of the branching probability, we can have convergence to the Brownian Web or we can have a tight family such that any weak limit point contains a Brownian Net. In the latter case, we conjecture that the limit is indeed the Brownian Net. This is a joint work with Glauco Valle (IM-UFRJ) and Leonel Zuaznabar (IME-USP). Discrete Markov random fields on graphs, also known as graphical models in the statistical literature, have become popular in recent years due to their flexibility to capture conditional dependency relationships between variables. They have already been applied to many different problems in different fields such as Biology, Social Science, or Neuroscience. Graphical models are, in a sense, finite versions of general random fields or Gibbs distributions, classical models in stochastic processes. This talk will present the problem of estimating the interaction structure (conditional dependencies) between variables by a penalized pseudo-likelihood criterion. First, I will consider this criterion to estimate the interaction neighborhood of a single node, which will later be combined with the other estimated neighborhoods to obtain an estimator of the underlying graph. I will show some recent consistency results for the estimated neighborhood of a node and any finite sub-graph when the number of candidate nodes grows with the sample size. These results do not assume the usual positivity condition for the conditional probabilities of the model as it is usually assumed in the literature of Markov random fields. These results open new possibilities of extending these models to situations with sparsity, where many parameters of the model are null. I will also present some ongoing extensions of these results to processes satisfying mixing type conditions. This talk is based on a joint work with Iara Frondana and Rodrigo Carvalho and some work in progress with Magno Severino. We look at the contact process with ordinary rate lambda exponential infections and heavy tailed cures, attracted to an alpha-stable law with alpha < 1, on a finite graph of size k. Our aim is to ascertain conditions on alpha and k such that the critical lambda for survival of the infection vanishes. We obtain nearly sharp (in a sense to be clarified) bounds on the critical k, k_c = k_c (alpha), which is always a finite number, such that the infection dies out almost surely for any lambda < infty at and below k_c; and there is positive probability of survival for any lambda > 0 above k_c. This is joint work with Pablo Almeida Gomes and Rémy Sanchis, published recently, in Bernoulli 27(3). I present results recently obtained with Francesco Manzo e Matteo Quattropani. We present an alternative proof of the so-called First Visit Time Lemma (FVTL), originally presented by Cooper and Frieze. We work in the original setting, considering a growing sequence of irreducible Markov chains on n states. We assume that the chain is rapidly mixing and with a stationary measure with no entry being either too small nor too large. Under these assumptions, the FVTL shows the exponential decay of the distribution of the hitting time of a given state x, for the chain started at stationarity, up to a small multiplicative correction. While the proof by Cooper and Frieze is based on tools from complex analysis, and it requires an additional assumption on a generating function, we present a completely probabilistic proof, relying on the theory of quasi-stationary distributions and on strong-stationary times arguments. In addition, under the same set of assumptions, we provide some quantitative control on the Doob’s transform of the chain on the complement of the state x. I will also discuss the relation of this result with general results, previously obtained, providing an exact formula for the first hitting distribution via conditional strong quasi-stationary times. Access the slides here. We construct an exclusion process with Bernoulli product invariant measure and having, in the diffusive hydrodynamic scaling, a non symmetric diffusion matrix, that can be explicitly computed. The antisymmetric part does not affect the evolution of the density but it is relevant for the evolution of the current. In particular because of that, the Fick’s law is violated in the diffusive limit. Switching on a weak external field we obtain a symmetric mobility matrix that is related just to the symmetric part of the diffusion matrix by the Einstein relation. We show that this fact is typical within a class of generalized gradient models. We consider for simplicity the model in dimension $d=2$, but a similar behavior can be also obtained in higher dimensions. Joint work with L. De Carlo and P. Goncalves. Access the slides here. In this talk we will briefly present the model we are interested in, which is a fractional elliptic stochastic partial differential equation driven by Gaussian white noise. There is in the literature a standard way to approximate the covariance operator of the solution of such equations, the so-called rational approximation (Bolin and Kirchner, 2020), however this approach uses the solution to build such an approximation. By considering directly the covariance operator, we are able to provide a more computationally efficient approximation. We compute the rate of this approximation in terms of the Hilbert-Schmidt norm. Furthermore, we also obtain, rigorously, the rate of approximation of the so-called lumped mass method. This method is widely used by practitioners and is essential to make it computationally feasible to fit some models in spatial statistics. We obtain the rate of approximation of the lumped mass method in terms of the operator’s norm as well as, under some additional restrictions, the Hilbert-Schmidt norm. Finally, we present the usage of these approximations in maximum likelihood estimation. Joint work with David Bolin and Zhen Xiong. Access the slides here. I discuss the low-temperature behaviour of Dyson models (polynomially decaying long-range Ising models in one dimension) in the presence of random boundary conditions. As for typical random (i.i.d.) boundary conditions Chaotic Size Dependence occurs, that is, the pointwise thermodynamic limit of the finite-volume Gibbs states for increasing volumes does not exist, but the sequence of states moves between various possible limit points, as a consequence it makes sense to study distributional limits, the so-called “metastates” which are measures on the possible limiting Gibbs measures. The Dyson model is known to have a phase transition for decay parameters α between 1 and 2. We show that the metastate changes character at α =3/2. It is dispersed in both cases, but it changes between being supported on two pure Gibbs measures when α is less than 3/2 to being supported on mixtures thereof when α is larger than 3/2. Joint work with Eric Endo and Arnaud Le Ny. Access the slides here. We introduce the equilibrium Widom-Rowlinson model on a two-dimensional finite torus in which the energy of a particle configuration is attractive and determined by the union of small discs centered at the positions of the particles. We then discuss the metastable behaviour of a dynamic version of the WR model. This means that the particle configuration is viewed as a continuous time Markov process where particles are randomly created and annihilated as if the outside of the torus were an infinite reservoir with a given chemical potential. In particular, we start with the empty torus and are interested in the first time when the torus is fully covered by discs in the regime at low temperature and when the chemical potential is supercritical. In order to achieve the transition from empty to full, the system needs to create a sufficiently large droplet, called critical droplet, which triggers the crossover. We compute the distribution of the crossover time and identify the size and the shape of the critical droplet. The analysis relies on a mesoscopic and microscopic description of the surface of the critical droplet. It turns out that the critical droplet is close to a disc of a certain deterministic radius, with a boundary that is random and consists of a large number of small discs that stick out by a small distance. We will show how the analysis of surface fluctuations in the WR model allows us to derive the leading order term of the condensation time and also the correction order term. This is a joint work with Frank den Hollander (Leiden), Sabine Jansen (Munich) and Roman Kotecky (Prague & Warwick). The Parabolic Anderson Model on a Galton-Watson Tree – Frank den Hollander (Leiden University) We consider the parabolic Anderson model on a supercritical Galton-Watson tree with an i.i.d. random potential whose marginal distribution is close to the double exponential. Under the assumption that the degree distribution has a sufficiently thin tail, we derive an asymptotic expansion for large times of the total mass of the solution given that initially a unit mass sits at the root. We derive the expansion both under the quenched law (i.e., conditional on the realisation of the random tree and the random potential) and under the half-annealed law (i.e., conditional on the realisation of the random tree but averaged over the random potential). The two expansions turn out to be different, but both contain a coefficient that is given by a variational formula indicating that the solution concentrates on a subtree with minimal degree according to a computable profile. A key tool in the analysis is the large deviation principle for the empirical distribution of a Markov renewal process. Joint work with Wolfgang König (Berlin), Renato dos Santos (Belo Horizonte), Daoyi Wang (Leiden). Access the slides here. Local Scaling Limits of Lévy Driven Fractional Random Fields – Donatas Surgailis (Vilnius University) We obtain a complete description of local anisotropic scaling limits for a class of fractional random fields $X$ on ${mathbb{R}}^2$ written as stochastic integral with respect to an infinitely divisible random measure. The scaling procedure involves increments of $X$ over points the distance between which in the horizontal and vertical directions shrinks as $O(lambda) $ and $O(lambda^ gamma)$ respectively as $lambda downarrow 0$, for some $gamma>0$. We consider two types of increments of $X$: usual increment and rectangular increment, leading to the respective concepts of $gamma$-tangent and $gamma$-rectangent random fields. We prove that for above $X$ both types of local scaling limits exist for any $gamma>0$ and undergo a transition, being independent of $gamma> gamma_0$ and $gamma<gamma_0$, for some $gamma_0>0$; moreover, the `unbalanced’ scaling limits ($gamm negamma_0$) are $(H_1,H_2)$-multi self-similar with one of $H_i$, $i=1,2$, equal to $0$ or $1$. The paper extends Pilipauskait.e and Surgailis (2017) and Surgailis (2020) on large-scale anisotropic scaling of random fields on ${mathbb{Z}}^2$ and Benassi et al. (2004) on $1$- tangent limits of isotropic fractional Lévy random fields. This is joint work with Vytautė Pilipauskaitė (University of Luxembourg). Access the slides here. Real-world networks are often understood as being symmetrical, meaning that vertices can be found which perform similar or equivalent structural roles (such as hubs from different communities in social networks, or functional regions in neuronal networks). These roles are usually associated with their topological placement relative to its surroundings; however, traditional mathematical formulations of graph symmetry are based on automorphism groups, which depend fundamentally on global structure and do not account for similarities in local structures. In this work, we introduce the concept of local symmetry, which reflects the structural equivalence of vertices’ egonets while generalizing classical conceptualizations of symmetry such as automorphism and isomorphism. We also study the emergence of local asymmetry in Erdős–Rényi graphs, identifying regimes of both asymptotic local symmetry and asymptotic local asymmetry. We find that local symmetry persists at least to an average degree of n^{1/3} and local asymmetry emerges at an average degree not greater than n^{1/2}, which are regimes of much larger average degree than for traditional, global asymmetry. Joint work with Daniel Figueiredo (COPPE/UFRJ) and Valmir Barbosa (COPPE/UFRJ). We consider a generalised oriented site percolation model on Z^d with arbitrary neighbourhood. The key additional difficulties as compared to standard oriented percolation are the lack of symmetry and, in two dimensions, of planarity. We establish that, despite these deficiencies, in the supercritical regime GOSP behaves qualitatively like OP. Joint work with Ivailo Hartarsky.	{"url":"https://ppge.im.ufrj.br/seminarios-de-probabilidade-segundo-semestre-de-2021/","timestamp":"2024-11-11T04:17:30Z","content_type":"text/html","content_length":"120932","record_id":"<urn:uuid:8f17e815-0d22-4dcd-a4b5-b567c56f2aa5>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00451.warc.gz"}
Ten Statistics Questions and their solutions 10 Statistics Questions - See attached file for questions in proper formatting 1)Draw a standard normal density curve to represent each of the following. Use the standard normal distribution table to find the: a. Area under the curve between z=0 and z = 2.33 b. Area under the curve between z=0 and z = -1.06 c. Area under the curve to the right of z = .28 d. Area under the curve to the left of z = -.53 e. Area under the curve between z=1.26 and z = 2.10 2)Create a sample space of possible outcomes for the following: flipping one coin and rolling an octagonal die. Find the probability of getting a compound event of heads and rolling a 7. A random study of recent graduates' average grades and degrees showed the following results. 3)If a graduate is selected at random, find these probabilities. The graduate has a B.S. degree, given that he or she has an A average. a. Given that the graduate has a B.A. degree, the graduate has a C average. b. What is the probability that a person has a B.S. degree and a B? c. What is the probability that a graduate has neither an A nor a C. 4)Working Mom’s Journal" reported that the mean time a mother, with her small children, spends in at the convenience store is 7.3 minutes. A sample of 20 moms is chosen from your neighborhood, and it is found that the mean time they spend in a convenience store was 8.2 minutes with a standard deviation of 1.4 minutes. Using , test the claim that the average amount of time a mom and her children spend in a convenience store is greater than 7.3 minutes. a. Determine which test statistic you will use: the standard normal distribution, or the student’s t distribution. Explain why you chose this test statistic. b. Establish the null and alternative hypotheses, state the claim. c. Test the claim at and discuss your results, should you reject or not reject the null hypothesis, should you reject or except the claim. 5)From a sample of ten full-time staff, chosen from the education department at your school, it is found that their mean salary is $50,340. Assume the standard deviation of the population is given as $15,320. Find the 95% confidence interval for the population mean, give the margin of error and discuss your results. 6)Create a sample space for tossing a coin four times. Find the probability of getting a simple event of three heads and one tail. 7)Three cards are drawn, without replacement, from a standard deck of fifty-two cards. Find the probability of these events. a. Getting three kings. b. Getting a ten, nine, and eight in order. c. Getting a diamond, heart, and club in order. d. Getting three diamonds. 8)In a doctor’s office there are eight nurses and four physicians. Seven nurses and two physicians are females. If a person is selected from the doctor’s office, find the probability that the person is a nurse or a male. 9)Discuss type I and type II errors in hypothesis testing. Give an example of each type of error. 10)A retail men’s clothing store owner buys from three companies: X, Y and Z. The recent most purchases are shown here. Company X Company Y Company Z If one item is selected at random, find these probabilities. a. It was purchased from company X or is a shirt. b. It was purchased from company Y or company Z. c. It is a tie or was purchased from company X. About the Solutions Full details along with solutions are presented. Other Details about the Project/Assignment Subjects Statistics Topic College Level Stats, Probability Functions, Type I/II Errros Level College / University Tags Stats Questions Price: $2.95 Purchase and Download Solutions Shirley B. Member Since: Nov 1998 Customer Rating: 97.8% Projects Completed: 2378 Total Earnings: Private +1 Ratings from clients: 578 Statistics Topic College Level Stats, Probability Functions, Type I/II Errros Level College / University Tags Stats Questions Not exactly what you are looking for? We regularly update our math homework solutions library and are continually in the process of adding more samples and complete homework solution sets. If you do not find what you are looking for, just go ahead and place an order for a custom created homework solution. You can hire/pay a math genius to do your homework for you exactly to your specifications	{"url":"https://www.mymathgenius.com/Library/Ten_Statistics_Questions_and_their_solutions","timestamp":"2024-11-03T09:00:06Z","content_type":"text/html","content_length":"40155","record_id":"<urn:uuid:411fb0de-3274-4d3c-ad5f-8cf6ccf98661>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00383.warc.gz"}
1962 -- Corporative Network Corporative Network Time Limit: 3000MS Memory Limit: 30000K Total Submissions: 4976 Accepted: 1866 A very big corporation is developing its corporative network. In the beginning each of the N enterprises of the corporation, numerated from 1 to N, organized its own computing and telecommunication center. Soon, for amelioration of the services, the corporation started to collect some enterprises in clusters, each of them served by a single computing and telecommunication center as follow. The corporation chose one of the existing centers I (serving the cluster A) and one of the enterprises J in some other cluster B (not necessarily the center) and link them with telecommunication line. The length of the line between the enterprises I and J is \|I – J\|(mod 1000).In such a way the two old clusters are joined in a new cluster, served by the center of the old cluster B. Unfortunately after each join the sum of the lengths of the lines linking an enterprise to its serving center could be changed and the end users would like to know what is the new length. Write a program to keep trace of the changes in the organization of the network that is able in each moment to answer the questions of the users. Your program has to be ready to solve more than one test case. The first line of the input will contains only the number T of the test cases. Each test will start with the number N of enterprises (5 <=N<=20000). Then some number of lines (no more than 200000) will follow with one of the commands: E I – asking the length of the path from the enterprise I to its serving center in the moment; I I J – informing that the serving center I is linked to the enterprise J. The test case finishes with a line containing the word O. The I commands are less than N. The output should contain as many lines as the number of E commands in all test cases with a single number each – the asked sum of length of lines connecting the corresponding enterprise with its serving center. Sample Input E 3 I 3 1 E 3 I 1 2 E 3 I 2 4 E 3 Sample Output Southeastern Europe 2004 [Submit] [Go Back] [Status] [Discuss] All Rights Reserved 2003-2013 Ying Fuchen,Xu Pengcheng,Xie Di Any problem, Please Contact Administrator	{"url":"http://poj.org/problem?id=1962","timestamp":"2024-11-13T23:04:51Z","content_type":"text/html","content_length":"7075","record_id":"<urn:uuid:2b2e2f87-3217-49b7-b8bb-5272435b9b49>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00431.warc.gz"}
Nested IF Function in Excel The nested IF function in Excel is a powerful tool that allows you to perform conditional calculations based on multiple conditions. It is commonly used when you need to evaluate different scenarios and return different results based on the conditions. This guide will explain how to use the nested IF function in Excel and provide examples to help you understand its functionality. To use the nested IF function, you start with the IF function, which checks if a condition is true or false. If the condition is true, the function returns a specified value; otherwise, it moves to the next IF function to check the next condition. This process continues until all conditions have been evaluated. In the given formula, =IF(C2=D2, "Both", IF(MAX(C2,D2)=C2, "Capex", "HC Cost")), the first IF function checks if the values in cells C2 and D2 are equal. If they are equal, the formula returns the text "Both". If they are not equal, the formula moves to the next IF function. The second IF function checks if the maximum value between C2 and D2 is equal to the value in C2. If it is equal, the formula returns the text "Capex". If it is not equal, the formula returns the text "HC Cost". To better understand how the formula works, let's consider some examples. Suppose we have the values C2 = 5 and D2 = 3. In this case, the formula would return "Capex" because the maximum value between 5 and 3 is 5, which is equal to the value in C2. On the other hand, if C2 = 7 and D2 = 7, the formula would return "Both" because the values in C2 and D2 are equal. Lastly, if C2 = 4 and D2 = 6, the formula would return "HC Cost" because the maximum value between 4 and 6 is 6, which is not equal to the value in C2. In conclusion, the nested IF function in Excel is a versatile tool that allows you to perform complex conditional calculations based on multiple conditions. By understanding its syntax and using it effectively, you can enhance your data analysis and decision-making capabilities in Excel. This formula is an example of a nested IF function in Excel and Google Sheets. It checks the values in cells C2 and D2 and returns different results based on the conditions. Step-by-step explanation 1. The formula starts with the IF function, which checks if the value in cell C2 is equal to the value in cell D2. 2. If the values in C2 and D2 are equal, the formula returns the text "Both". 3. If the values in C2 and D2 are not equal, the formula moves to the next IF function. 4. The second IF function checks if the maximum value between C2 and D2 is equal to the value in C2. 5. If the maximum value is equal to C2, the formula returns the text "Capex". 6. If the maximum value is not equal to C2, the formula returns the text "HC Cost". For example, if we have the following values in cells C2 and D2: The formula =IF(C2=D2, "Both", IF(MAX(C2,D2)=C2, "Capex", "HC Cost")) would return the text "Capex", because the maximum value between 5 and 3 is 5, and it is equal to the value in C2. Another example: The formula would return the text "Both", because the values in C2 and D2 are equal. And one more example: The formula would return the text "HC Cost", because the maximum value between 4 and 6 is 6, and it is not equal to the value in C2.	{"url":"https://codepal.ai/excel-formula-explainer/query/ATlUAvm7/nested-if-function-excel-formula","timestamp":"2024-11-14T23:43:54Z","content_type":"text/html","content_length":"91482","record_id":"<urn:uuid:5dfd57ba-4132-46c1-8368-235afcd63027>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00255.warc.gz"}
On Émery's Inequality and a Variation-of-Constants Formula Reiß, Markus and Riedle, Markus and van Gaans, Onno (2007) On Émery's Inequality and a Variation-of-Constants Formula. Stochastic Analysis and Applications, 25 (2). pp. 353-379. ISSN 1532-9356 Restricted to Registered users only Download (193kB) A generalization of Émery's inequality for stochastic integrals is shown for convolution integrals of the form $\left( \int_0^t g(t-s) Y(s-) dZ(s)\right)_{t \geq 0}$, where Z is a semimartingale, Y an adapted càdlàg process, and g a deterministic function. An even more general inequality for processes with two parameters is proved. The inequality is used to prove existence and uniqueness of solutions of equations of variation-of-constants type. As a consequence, it is shown that the solution of a semilinear delay differential equation with functional Lipschitz diffusion coefficient and driven by a general semimartingale satisfies a variation-of-constants formula. Actions (login required)	{"url":"https://eprints.maths.manchester.ac.uk/908/","timestamp":"2024-11-11T03:10:18Z","content_type":"application/xhtml+xml","content_length":"21850","record_id":"<urn:uuid:977e27d9-d1bb-43a5-a899-e0903876a88a>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00847.warc.gz"}
Live Online classes for kids from 1-10 \| Upfunda Academy What is Counting? Counting is a fun and important part of math! It's all about figuring out how many things there are in a group. Let's look at an example. Let's say you have five apples. To count them, you would say, "One, two, three, four, five." That's all there is to it! Now you know there are five apples. Zero is not part of the numbers we count because it is impossible to count anything. Counting is important because it helps us keep track of things. We can use it to count the number of people in a room, the number of toys in a box, or even the number of steps we take. Types of Counting Numbers Counting numbers are numbers that you use to count things. There are several types of counting numbers, each intended for a different purpose. • Natural Numbers: The most common type of counting numbers are natural numbers, which are also sometimes referred to as positive integers. Natural numbers are typically used for counting things like people, animals, and objects. • Whole Numbers: Whole numbers include all natural numbers and 0. They are real numbers that do not include fractions, decimals, and negative integers. • Rational Numbers: Finally, there are rational numbers, which are fractions and decimals. Rational numbers are often used for measuring things like weight, volume, or time. Here are some fun facts about counting! • People in ancient times used counting stones to keep track of things like the number of sheep they had. • In some cultures, people count differently. For example, people start counting with their thumb instead of their first finger. • There are many different ways to count. You can count forwards, backwards, by twos, by fives, or even by tens! • Counting is a great way to improve your memory and attention to detail. It helps train your brain to remember things and pay attention to the details. Whether you're counting apples or counting to 100, it helps us keep track of things and improve our memory and attention to detail. So let's keep counting! Test your knowledge with Upfunda Quiz! 1. A family has several children, one of them is Tom. Tom has 2 sisters and each of his sisters has 3 brothers. How many children be there in this family? 2. Adam wrote out all the numbers from 1 to 60 inclusive. How many times did he use the digit 5? 3. How many 2-digit numbers are multiples of 5? 4. If kamal flips a coin three times and records the results, how many possible sequences of heads and tails are possible? For example, one possible sequence is H-T-H. 5. How many square are there in this figure? 6. How many rectangles are there in this figure? 7. How many triangles and squares are there in this figure? View Answers 1. D) 5 2. B) 16 3. 17 4. 3 5. C)30 6. B)19 7. A) 44 triangles and 10 squares	{"url":"https://upfunda.academy/blog/9e7313b9-dba3-4201-8985-f8672de40837","timestamp":"2024-11-08T23:19:25Z","content_type":"text/html","content_length":"23700","record_id":"<urn:uuid:5454181c-7915-4891-95fc-334a5e6b4558>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00079.warc.gz"}
Numeracy Stage: Second Grade (Year 7) By the time they are finished the second grade, children should be able to: Place Value / Basic Operations • understand place value for 3-digit numbers • count up to 1000 – counting by fives, tens and hundreds • read and write numbers up to 1000 • compare two 3-digit numbers using greater than (>), less than (<) or equals (=) • add and subtract numbers within 100 using various strategies • explain which addition and subtraction strategies work • add or subtract numbers within 1000 using concrete models • use mental math to add or subtract 10 or 100 from any given number between 100 – 900 • separate circles and rectangles into two, three or four equal parts and name them • recognize that equal parts of identical wholes do not need to have the same shape Algebraic Thinking • solve one-step or two-step addition and subtraction word problems • describe the strategy used to solve a word problem • add or subtract using fact families • work out equal groups of objects concretely (foundational learning for multiplication) • estimate and measure lengths in standard units • relate addition and subtraction to length and the number line • tell time to the nearest five minutes; write out time to the nearest five minutes • identify money (coins and bills) • solve a number sentence or word problem using money • recognize that comparisons are valid only when referring to the same standard unit • separate a rectangle into rows and columns using square units to find a total Data Analysis • draw a bar or picture graph and answers questions related to it • generate data by measuring length to the nearest whole unit • show length measurements on a line plot with whole number units • represent whole numbers of a number line Geometry (Shapes) / Spatial Sense • recognize and draw shapes using specific attributes (both defining and non-defining) • construct an array using squares that line up in rows and columns	{"url":"https://thetravelingpencil.com/numeracy-stage-second-grade-year-7","timestamp":"2024-11-08T08:09:46Z","content_type":"text/html","content_length":"70990","record_id":"<urn:uuid:b3172a5e-d0cd-4c58-ac36-b5b37138d82e>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00626.warc.gz"}
FPGA Circuits? Not open for further replies. I would like to find information on FPGA circuits. Which forum would I post this in? I didn't see an FPGA forum category. What I'm looking for specifically is information on how to design a neural network on an FPGA. I'm totally new to working with FPGAs so I'm looking for the absolute simplest design possible. It doesn't need to do anything useful. I'm looking for a design that I can fully understand and work with for learning. So the most elementary learning example is what I'm seeking. Thank you. Then use an FPGA module like micronova mercury (uses outdated software but easier to get started) or Trenz 0725 (more powerful and up to date). FPGAS are a PITA and expensive to design and make boards for. You can buy the micronova direct from manufacturer or trenz from digikey if you're in north america. You can buy straight from trenz if in europe. you'll need to learn vhdl or verilog. They arent programming languages so dont have preconceptions from programming or you will be lead astray. you'll also need a programmer for the trenz. You can use trenz's xmod for easy use with the module, but the digilent HS3 is more easily usable on more platforms. micronova has one onboard and does not need a programmer. Last edited: Then use an FPGA module like micronova mercury (uses outdated software but easier to get started) or Trenz 0714 (more powerful and up to date). FPGAS are a PITA and expensive to design and make boards for. you'll need to learn vhdl or verilog. I already have an FPGA developement board and I'm currently learning VHDL. I also already have experience designing digital logic circuits. All I'm interested in here is finding an extremely simple example of a neural network for an FPGA. I need something extremely simple. Something designed for teaching the concept. I don't want a large neural network that has already been programmed into a library. That won't do me any good. I want to understand how to build a neural network on a FPGA from the ground up. So I'm looking for the simplest possible example circuit. Kind of like the "Hello World" for Neural Networks on an FPGA. Well-Known Member Most Helpful Member Interesting question. I'd start by searching the app note and design brief libraries at the big guns - Xylinx, Lattice, Atmel (did they get bought), etc. Another possible source is an online college-level FPGA course or a biomedical engineering course, research paper, etc. I already have an FPGA developement board and I'm currently learning VHDL. I also already have experience designing digital logic circuits. All I'm interested in here is finding an extremely simple example of a neural network for an FPGA. I need something extremely simple. Something designed for teaching the concept. I don't want a large neural network that has already been programmed into a library. That won't do me any good. I want to understand how to build a neural network on a FPGA from the ground up. So I'm looking for the simplest possible example circuit. Kind of like the "Hello World" for Neural Networks on an FPGA. ah, i see. outta my league then. broken link removed http://stackoverflow.com/questions/2190470/neural-network-simulator-in-fpga "Most attempts at building a 'literal' neural network on an FPGA hit the routing limits very quickly, you might get a few hundred cells before P&R pulls takes longer to finish than your problem is worth waiting for. Most of the research into NN & FPGA takes this approach, concentrating on a minimal 'node' implementation and suggesting scaling is now trivial. The way to make a reasonably sized neural network actually work is to use the FPGA to build a dedicated neural-network number crunching machine. Get your initial node values in a memory chip, have a second memory chip for your next timestamp results, and a third area to store your connectivity weights. Pump the node values and connection data through using techniques to keep the memory buses saturated (order node loads by CAS line, read-ahead using pipelines). It will take a large number of passes over the previous dataset as you pair off weights with previous values, run them through DSP MAC units to evaluate the new node weights, then push out to the result memory area once all connections evaluated. Once you have a whole timestep finished, reverse the direction of flow so the next timestep writes back to the original storage area." ah, i see. outta my league then. broken link removed http://stackoverflow.com/questions/2190470/neural-network-simulator-in-fpga "Most attempts at building a 'literal' neural network on an FPGA hit the routing limits very quickly, you might get a few hundred cells before P&R pulls takes longer to finish than your problem is worth waiting for. Most of the research into NN & FPGA takes this approach, concentrating on a minimal 'node' implementation and suggesting scaling is now trivial. The way to make a reasonably sized neural network actually work is to use the FPGA to build a dedicated neural-network number crunching machine. Get your initial node values in a memory chip, have a second memory chip for your next timestamp results, and a third area to store your connectivity weights. Pump the node values and connection data through using techniques to keep the memory buses saturated (order node loads by CAS line, read-ahead using pipelines). It will take a large number of passes over the previous dataset as you pair off weights with previous values, run them through DSP MAC units to evaluate the new node weights, then push out to the result memory area once all connections evaluated. Once you have a whole timestep finished, reverse the direction of flow so the next timestep writes back to the original storage area." Thanks for taking the time to do a search. I've already found both of those resources myself. In fact, the first one from Omondi does provide some pretty basic examples. I can probably work from that on my own. But I was hoping to find a simple tutorial where someone actually creates a working example on an actual FPGA either using VHDL, Verilog, a Finite State Machine, or an actual circuit Once I learn how to make a simple example I'll write up my own tutorial on it for others to use as an introduction to neural networks on an FPGA. So far I haven't been able to find a really simple tutorial. Although I have found tons of simple tutorials on basic neural networks. I might need to just translate one of those onto an FPGA myself. But I was hoping that someone else had already done that. Save me from having to start from scratch. Not open for further replies.	{"url":"https://www.electro-tech-online.com/threads/fpga-circuits.153510/","timestamp":"2024-11-09T19:37:52Z","content_type":"text/html","content_length":"114726","record_id":"<urn:uuid:327d4112-b3a0-4808-bee6-01e8edbfdb55>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00379.warc.gz"}
4.2 Acoustic resonators There are two aspects of acoustics that will be of direct interest to us to understand the behaviour of musical instruments. Wind instruments of all kinds rely on acoustical resonators, and acoustic resonances also play a role in understanding the behaviour of stringed instrument bodies and of room acoustics. This section will give an initial overview of all these examples. Then in the next section we will address the other important aspect of acoustics : the radiation of sound by a vibrating structure such as a violin body. When we looked at mechanical vibration in Chapter 2, we started with the mass-spring system, with a single degree of freedom, then went on to look at multi-modal systems like stretched strings and bending beams. We follow a similar sequence here. There is an acoustical equivalent of the mass-spring oscillator, called a Helmholtz resonator. A: The Helmholtz resonator This simple resonating system is responsible for the popping noise when you pull out a cork or flick a thumb out of the top of a bottle. All “bottle-like” vessels have a low-frequency resonance in which a “plug” of air in the neck behaves like an invisible piston, and can oscillate on a “spring” resulting from compression of the air in the enclosed volume inside the bottle. This “spring” is the force you feel if you try to operate a bicycle pump while you have your thumb over the end. Figure 1 shows a sketch, and the next link gives a derivation of the resulting formula for the resonant Figure 1. Sketch of a bottle, with the invisible piston mass that is responsible for the Helmholtz resonance. Before the days of electronics, a set of tuned Helmholtz resonators could be used as a “spectrum analyser”: an example is shown in Fig. 2. Helmholtz resonance also accounts for the effect that children often encounter when visiting the seaside: they hold a sea-shell to one ear, and are told “You can hear the sound of the sea inside the shell”. Both these examples work in the same way: the resonator amplifies ambient sound in a frequency range close to the resonant frequency, as demonstrated explicitly in the previous link. If you place an ear close enough, you hear this amplification effect. The resonators shown in Fig. 2 are provided with a nipple that is placed in the ear canal for this purpose. Figure 2. A selection of Helmholtz resonators from 1870, at the Hunterian Museum and Art Gallery in Glasgow. By Stephencdickson – Own work, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php? The most direct musical application of a Helmholtz resonator is an ancient instrument called the ocarina. Figure 3 shows an example. A whistle-like mouthpiece leads to a chamber with a number of finger holes. Unlike other wind instruments with finger holes, it doesn’t matter which particular holes you cover. The note you obtain from the ocarina depends only on the total area of open holes: it is a Helmholtz resonator and the frequency is determined by the chamber volume and the combined area and “neck length” of the open holes, according to the formula given in the previous link. This means that ocarinas can be made in many different shapes, but all work in the same way and all sound rather similar. Figure 3. An ocarina, thought to date from around 1900, at the Museu de la Música de Barcelona. Patian, CC BY-SA 3.0 https://creativecommons.org/licenses/by-sa/3.0, via Wikimedia Commons A more mainstream application of Helmholtz resonance in musical instruments is relevant to many stringed instruments. For definiteness, we will use the guitar as an example. The body of an acoustic guitar like the one seen in Fig. 4 is a thin-walled box, usually made of wood, and there is an opening in the top plate, usually called the soundhole. It is sometimes thought that this hole is there “to let the sound out”, but that is quite misleading. Most of the sound is created by vibration of the wooden plates, but the hole serves to make a significant enhancement to the radiation of sound at lower frequencies in the instrument’s range by adding a Helmholtz-like resonance. “Helmholtz-like”, because the box is not a rigid enclosure. It is very important to allow for the flexibility of the plates forming the walls of the box. Figure 4. A classical guitar, being played by Ian Cross We can get a good impression of how this works using a simple model which takes account of only one vibration mode of the box, the lowest mode of the top plate. We will come to a more complete description of how the guitar works in section 5.3, including some images of mode shapes, but for now we simply need to know that this lowest mode behaves just as you would expect. The main area of the plate, carrying the bridge in the middle, bulges in and out roughly in the manner sketched in Fig. 5. Figure 5. Sketch of a guitar body, indicating the form of the lowest mode of the top plate (on a greatly exaggerated scale) This plate vibration will interact with the internal air pressure and the Helmholtz resonance. The vibrating plate will cause changes in internal air pressure, and in turn that pressure exerts a force on the plate. But the same is still true of the invisible “Helmholtz piston” in the soundhole. The combined system can be represented in a different form, shown in Fig. 6. The mode of the top plate is represented by a mass and a spring, and also a piston connecting it to the internal air. The enclosure is otherwise rigid, with a “neck” in which the Helmholtz piston moves. It is no coincidence that this figure resembles a loudspeaker in an enclosure: the same modelling is used in the design of ducted (or bass reflex) loudspeakers. Figure 6. Idealised version of the guitar mode from Fig. 5 coupled to the Helmholtz resonance. (The same model can describe a ducted loudspeaker enclosure.) The detailed analysis of this system is given in the next link, but we can learn the most important things about the result without mathematical detail by doing one more stage of abstraction. The system in Fig. 6 can be represented by an equivalent mass-spring system, shown in Fig. 7. The left-hand mass and spring represent the mode of the plate (or of the loudspeaker cone). The right-hand mass represents the Helmholtz piston. The spring connecting the two masses represents the springiness of the internal air: the net volume, and hence the internal pressure, depends on the relative motion of the two pistons. For example, if the Helmholtz piston moved inwards at the same time as the plate moved outwards, with the correct ratio of amplitudes, the volume and internal pressure would not change. Figure 7. Mass-spring system equivalent to Fig. 6 The two-mass system of Fig. 7 will have two vibration modes, as explored in some generality in Section 2.2.5. The lower-frequency one of these modes will have the two masses moving in the same phase, while the higher frequency will have them moving in opposite phases. Both modes will involve some stretching and compression of the right-hand spring, representing changes of internal air pressure. We will see in section 4.3 that changes in internal pressure, associate with changes in net volume, give a good indication of the strength of external sound radiation by each mode. Figure 8 shows a schematic representation of the two modes, in a way that relates directly to Fig. 6. It is based on parameter values appropriate to a typical guitar body (see previous link for details). In order to illustrate the relative volume displacements of the two pistons, which is the important factor determining the sound radiation, the two pistons are shown with the same sizes. In the real guitar, the Helmholtz piston has about 1/10 the area of the effective plate piston , so that its relative displacement is in fact 10 times bigger to compensate. Figure 8. Animation of the two modes of the model in Fig. 6. The plotted motion is proportional to the volumes displaced by the two pistons: the plate piston (upper) and the Helmholtz piston (lower). In a guitar, these two modes are normally about an octave apart in frequency, as depicted here. Both modes involve some motion of the mass representing the top plate motion. This is important: the vibrating string is attached to the plate, and the excitation of each mode depends on plate motion. For an ideal Helmholtz resonator with rigid cavity walls, the resonance could not be excited by a vibrating string. So in summary the combination of a flexible top plate with a cavity containing a soundhole is that the guitar has two resonances which are good radiators of sound, and which can both be driven efficiently by the vibrating strings. The details are, to a large extent, under the control of the guitar maker. The frequencies of the two resonances can be controlled: the thickness and bracing details of the top plate determine where the lowest plate resonance would fall in the absence of interaction with the air, while the volume of the box and the size of the soundhole determine where the Helmholtz resonance would fall in the absence of plate motion. A trick sometimes used by guitar makers when they want to reduce the Helmholtz frequency is to add a “tornavoz”: a cylindrical collar fitted inside the box, extending the soundhole into a longer neck and thus increasing the mass of the “Helmholtz piston”. An example is shown in Fig. 9. Figure 9. A guitar fitted with a tornavoz to modify the Helmholtz resonance frequency B: Pipe resonances Most wind instruments rely on resonances in pipes of one kind or another. They have a variety of internal shapes, or bore profiles. We look first at two simple examples, then take a preliminary look at how to deal with more complicated profiles such as those found in brass instruments. The first example makes use of the simplest sound field that we met in section 4.1, the plane wave. As Fig. 10 indicates, a straight-walled pipe can be superimposed on a plane wave, aligned with the direction of propagation, and since the particle motion is always parallel to the pipe walls, the wave inside the pipe can behave exactly the same way as it did in empty space. So plane waves can propagate along a straight pipe at the speed of sound: a practical application of this is the speaking tube, still sometimes used as a way to communicate which doesn’t rely on electricity. Figure 10. Sketch of a plane sound wave (wave crests shown as black lines), with a straight pipe (red lines) superimposed. The particle motion (blue arrow) is parallel to the pipe walls, so the plane wave can propagate inside the pipe exactly the same as in empty space. Now we want to think about pipes of finite length, to understand their mode shapes and resonant frequencies. First, we need to think about possible boundary conditions at the end of a pipe. One possibility is an open end: the pipe is simply cut off. The sound wave inside the pipe is then exposed to the outside world. Some sound will escape from the end and radiate away, but to a good first approximation we can say that at the end of the pipe (or, at least, somewhere near the end of the pipe: see the discussion of end corrections in section 4.2.1) the pressure simply becomes the steady atmospheric pressure. In terms of the acoustic pressure, that means the open end of the pipe must be a node. Another simple boundary condition would occur if one end of the pipe was blocked. At a closed end like this, the thing we can say straight away is that the oscillating air particles cannot pass through the blocked end: we must have a nodal point of particle displacement. We can deduce what this means for pressure by referring back to eq. (2) of section 4.1.1: pressure is proportional to the spatial derivative of displacement. Once we assume sinusoidal time dependence in order to find modes of the pipe, then the spatial variation must also be sinusoidal. It follows that if the pressure variation is like $\sin kx$, the displacement must be like $\cos kx$, and vice versa. So a nodal point of displacement means an antinode of pressure. We now have enough information to find mode shapes and natural frequencies. For a pipe open at both ends, the pressure must be sinusoidal with nodes at both ends. It follows that the mode shapes are exactly the same as the ones we found for a vibrating string in section 3.1.1. The first few are illustrated in Fig. 11. The natural frequencies also obey the same formula as for the string: the $n$th frequency is $$f_n=\frac{nc}{2L} \tag{1}$$ for a pipe of length $L$, where $c$ is the speed of sound. Figure 11. The first five mode shapes for pressure inside a straight pipe, open at both ends. For a pipe that is open at one end but closed at the other, the corresponding mode shapes have to be as shown in Fig. 12. The lowest mode has a quarter-wavelength trapped in the length, rather than a half-wavelength as for the open-open pipe. The next mode has 3 quarter-wavelengths, the next 5 and so on. The corresponding frequency for the $n$th mode is $$ f_n=\frac{(2n-1)c}{4L} . \tag{2}$$ It follows that the pattern of natural frequencies is different from the open-open case. The frequency ratios are 1:3:5:7, rather than 1:2:3:4. In other words, they are alternate terms of the harmonic series, rather then every term. Furthermore, the fundamental frequency of the closed-open pipe is an octave lower than that of the open-open pipe, for a tube of the same length. This explains a familiar effect in instruments. To a first approximation, a flute is a straight open-open tube, and a clarinet is a straight closed-open tube. The two instruments have roughly the same length, but the clarinet plays an octave lower for a given fingering. It overblows at the twelfth (a frequency ratio of 3), whereas the flute overblows at the octave. Figure 12. The first five mode shapes for pressure inside a straight pipe, open at one end but closed at the other. Some instruments, like the oboe and the saxophone, are better approximated as conical tubes rather than straight tubes. The ideal straight-sided conical tube is another case that we can understand easily, based on something we already know: this time, it is a spherical wave field such as that produced by a pulsating sphere as in section 4.1.2. As Fig. 13 indicates, a straight-walled conical pipe can be superimposed on such a spreading spherical wave, aligned with the direction of propagation and with its point placed at the origin. As happened for the plane wave and the straight pipe, the particle motion is always parallel to the pipe walls, so the wave inside the pipe can behave exactly the same way as it did in empty space. Figure 13. Sketch of a spherically-spreading sound wave (wave crests shown as black lines), with a conical pipe (red lines) superimposed. The particle motion (blue arrow) is radial, and therefore parallel to the pipe walls, so the wave can propagate inside the pipe exactly the same as in empty space. We need to think about boundary conditions again. The theory for the spherical wave showed that the combination $xp$ satisfies the one-dimensional wave equation, rather than the pressure $p$ alone as in the case of the straight pipe. Here, $x$ is the distance along the cone from the point. At an open end of the cone, we will have a node of pressure as before, and hence a node of $xp$. But as we approach the pointed end of the cone, $x \rightarrow 0$, and so $xp(x) \rightarrow 0$. The result is that the combination $xp$ has sinusoidal variation, with zeros at both ends — exactly the same conditions as the open-open straight pipe. The mode shapes for pressure are thus $$p_n(x)=\dfrac{\sin n \pi x/L}{x} \tag{3}$$ with corresponding natural frequencies that are exactly the same as the open-open straight pipe: $$f_n=\frac{nc}{2L} . \tag{4}$$ The first few of these pressure mode shapes are plotted in Fig. 14. At first glance they look similar to the closed-open modes of Fig. 12, but in fact they are significantly different: the wavelength, visible by the positions of the nodal points, matches the open-open case, not the closed-open case of the straight tube. Figure 14. The first five mode shapes for pressure inside a open-ended conical pipe. We now turn to brass instruments, and we immediately encounter an apparent paradox. Something like a trumpet is clearly an open-closed tube: the mouthpiece end is closed by the player’s lips. The tube does not taper down in a conical way at the mouthpiece end: most brass instruments have a long section of straight tube before the flaring bell, as is obvious if you think about the slide of a trombone. So we might expect the instrument to have resonances at alternate harmonics (or, at least, approximate harmonics). But, as is familiar from the kinds of tune that can be played on a bugle or a post-horn, the instruments in fact seem able to play the complete harmonic series, not just the odd terms. The resolution of this question is that the bore profile of a typical brass instrument is not well approximated by any of the simple shapes we have looked at up to now. We need to explore the underlying theory of horns with varying cross-section to understand the ingenious trick that is used by makers of brass instruments. If the cross-sectional area of the bore varies slowly and smoothly with distance, then to a good first approximation the pressure obeys a modified version of the wave equation called the Webster horn equation. The details are given in the next link. This equation doesn’t have easy mathematical solutions for realistic bore profiles, so to see roughly what happens we will resort to the computer. Figure 15 shows the first few pressure modes, computed from the Webster equation for a bore profile which has the right kind of features for a realistic brass instrument. It has a straight tube with a closed end, leading into a section which flares: gradually at first, then more abruptly as the bell is approached. Figure 15. The first five mode shapes computed for a simple model of a brass instrument with a flaring bell. The vertical black lines show the predicted cut-off point for each mode, where the travelling wave becomes evanescent. If the frequency of the second mode is scaled to the value 2, the frequencies of these five modes are 0.75, 2, 2.97, 4.07, 5.07. The sequence of mode shapes is recognisably related to the closed-open modes for a straight tube, seen in Fig. 12, but the shapes have been “squashed in” towards the mouthpiece end. Most conspicuously, the fundamental mode is mainly confined to the left-hand half of the tube, dying away to low levels long before the bell is reached. Something similar happens for the second mode, but it reaches closer to the bell before it fades away. The sequence continues with the higher modes: each successive mode looks a bit more like the corresponding shape in Fig. 12. The previous link explains what is happening. It involves behaviour similar to something we have met back in section 3.5, in connection with the “musical saw”. For a given frequency, and hence a given wavelength, there is a point in the flaring bore beyond which it is no longer possible for a travelling wave to propagate. It switches over to an evanescent wave, with exponential decay. Just as happened in the musical saw, most of the energy in the travelling wave is reflected back down the tube from this critical point, which can be calculated easily from the bore profile. For each mode, this point is marked in Fig. 15 by a vertical black line. It can be seen that these lines match quite well to the point where the sinusoidal behaviour of mode amplitude switches over to a decaying shape. The result is that the apparent length of the tube is shorter than the real length: by a large amount for the fundamental mode, and by a decreasing amount for the successive higher modes. This naturally changes the natural frequencies: a shorter tube will always give a higher frequency. The pattern that emerges is most clear if we scale the natural frequencies by taking the second mode as a reference, and calling that frequency “2”. In these terms, the frequencies of the 5 modes seen in Fig. 15 are 0.75, 2, 2.97, 4.07 and 5.07. Modes 2, 3, 4 and 5 now have frequencies reasonably close to the harmonic relations 2:3:4:5. The fundamental mode, though, does not fit into this approximate harmonic series. This example, despite being based on a very crude model, gives a good idea of what happens in real brass instruments: at some time in the past, instrument makers have hit on a type of bore profile that allows the instrument to have resonances which fall in a good approximation to the complete harmonic series, apart from the fundamental frequency which is always too low. C: Acoustic cavities and room acoustics There is a final class of acoustical resonators that deserves a brief discussion: resonances inside cavities, which could range from a concert hall to the inside of a violin body. The simplest example is a rectangular space with hard walls. For this case we already know enough to deduce the mode shapes and natural frequencies. Suppose we have a box with side lengths $A \times B \times C$ along the coordinate directions $x,y,z$. We could start by looking for modes that only depend on $x$: these would involve plane waves in the $x$ direction, and the cavity would simply behave like a tube which is closed at both ends. There has to be a pressure antinode at both ends, so the modes are similar to the open-open modes of Fig. 11, with the same natural frequencies, but they involve cosines rather than sines. The first few shapes are shown in Fig. 16. Figure 16. The first 5 mode shapes of pressure in a closed-closed tube, or for plane waves in a rectangular enclosure. But obviously we could get the same set of shapes from plane waves in the $y$ direction, or in the $z$ direction. More than that, we can find a mode that has any combination of these $x$, $y$ and $z$ shapes simultaneously: the most general pressure mode takes the form $$p_{qrs}=\cos \frac{q \pi x}{A} \cos \frac{r \pi y}{B} \cos \frac{s \pi z}{C} \tag{5}$$ where each of $q$, $r$ and $s$ can take any integer value 0,1,2,3,… Notice the inclusion of 0 here: this option is necessary to allow plane-wave modes. For example, the case we just described has $r= s=0$ for a plane wave in the $x$ direction. The modes plotted in Fig. 16 have $q=1,2,3,4,5$. The frequencies corresponding to the modes in eq. (5) can be deduced immediately from the wave equation: $$f_{qrs}=\frac{c}{2 \pi} \sqrt{\left[ \left( \dfrac{q \pi}{A} \right)^2 + \left( \dfrac{r \pi}{B} \right)^2 + \left( \dfrac{s \pi}{C} \right)^2 \right] } . \tag{6}$$ To see the most important thing that this equation tells us, it is useful to compute an example. Consider a rectangular space $5 \times 4 \times 3$ m in size: a typical domestic room. The fundamental frequency is 34 Hz: this is a plane standing wave in the 5 m direction, exactly like the top plot in Fig. 16. But the important behaviour becomes clear if we calculate a lot of the natural frequencies of this room and plot a histogram of them. Choosing 100 Hz bins, we get the result seen in Fig. 17. The histogram shows that the room has an enormous of modes: the number per 100 Hz band grows very rapidly, and by 5 kHz there are about 50,000 modes within each 100 Hz band! This modal density function grows in proportion to the square of frequency, and this trend applies to any acoustic volume: the next link explains why. Figure 17. Histogram of natural frequencies of a rectangular room of dimensions $5 \times 4 \times 3$ m, shown in 100 Hz bins. All the modes in our room will have some energy dissipation, and hence every resonant peak will have a finite half-power bandwidth determined by the damping level (see section 2.2.7). The high modal density seen in Fig. 17 then tells us that, except at the very lowest frequencies, the modal overlap factor will be large: this is defined as the ratio of the half-power bandwidth to the typical spacing between adjacent modes. This has profound implications for the acoustical behaviour of the room. The response at any given frequency will involve a very large number of modes. The conclusion is that we probably won’t learn very much if we try to understand the acoustics of our room by looking for individual modes across the audible frequency range: room acoustics is a statistical science. We will illustrate some of the behaviour with measurements involving room acoustics, in section 10.4 (see subsection D).	{"url":"https://euphonics.org/4-2-acoustic-resonators/","timestamp":"2024-11-10T03:12:24Z","content_type":"text/html","content_length":"110117","record_id":"<urn:uuid:440c5048-c859-42c7-875e-97edc5a081fe>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00838.warc.gz"}
1,231 research outputs found For the superconducting phase with a d-wave order parameter and zero temperature the magnetic susceptibility of the t-J model is calculated using the Mori projection operator technique. Conditions for the appearance of an incommensurate magnetic response below the resonance frequency are identified. A fast decay of the tails of the hole coherent peaks and a weak intensity of the hole incoherent continuum near the Fermi level are enough to produce an incommensurate response using different hole dispersions established for $p$-type cuprates, in which such response was observed. In this case, the nesting of the itinerant-electron theory or the charge modulation of the stripe theory is unnecessary for the incommensurability. The theory reproduces the hourglass dispersion of the susceptibility maxima with their location in the momentum space similar to that observed experimentally. The upper branch of the dispersion stems from the excitations of localized spins, while the lower one is due to the incommensurate maxima of their damping. The narrow and intensive resonance peak arises if the frequency of these excitations at the antiferromagnetic momentum lies below the edge of the two-fermion continuum; otherwise the maximum is broad and less intensive.Comment: 22 pages, 7 figure Low-temperature spin dynamics of the double-layered perovskite La_2-2x_Sr_1+2x_Mn_2_O_7 (LSMO327) was systematically studied in a wide hole concentration range (0.3 <= x < 0.5). The spin-wave dispersion, which is almost perfectly 2D, has two branches due to a coupling between layers within a double-layer. Each branch exhibits a characteristic intensity oscillation along the out-of-plane direction. We found that the in-plane spin stiffness constant and the gap between the two branches strongly depend on x. By fitting to calculated dispersion relations and cross sections assuming Heisenberg models, we have obtained the in-plane (J_para), intra-bilayer (J_perp) and inter-bilayer (J') exchange interactions at each x. At x=0.30, J_para=-4meV and J_perp=-5meV, namely almost isotropic and ferromagnetic. Upon increasing x, J_perp rapidly approaches zero while \|J_para\| increases slightly, indicating an enhancement of the planar magnetic anisotropy. At x=0.48, J_para reaches -9meV, while J_perp turns to +1meV indicating an antiferromagnetic interaction. Such a drastic change of the exchange interactions can be ascribed to the change of the relative stability of the d_x^2-y^2 and d_3z^2-r^2 orbital states upon doping. However, a simple linear combination of the two states results in an orbital state with an orthorhombic symmetry, which is inconsistent with the tetragonal symmetry of the crystal structure. We thus propose that an ``orbital liquid'' state realizes in LSMO327, where the charge distribution symmetry is kept tetragonal around each Mn site.Comment: 10 pages including 7 figure The Kondo lattice system CeZn_{0.66}Sb_{2} is studied by the electrical resistivity and ac magnetic susceptibility measurements at several pressures. At P=0 kbar, ferromagnetic and antiferromagnetic transitions appear at 3.6 and 0.8 K, respectively. The electrical resistivity at T_N dramatically changes from the Fisher-Langer type (ferromagnetic like) to the Suzaki-Mori type near 17 kbar, i.e., from a positive divergence to a negative divergence in the temperature derivative of the resistivity. The pressure-induced SM type anomaly, which shows thermal hysteresis, is easily suppressed by small magnetic field (1.9 kOe for 19.8 kbar), indicating a weakly first-order nature of the transition. By subtracting a low-pressure data set, we directly compare the resistivity anomaly with the SM theory without any assumption on backgrounds, where the negative divergence in d\rho/dT is ascribed to enhanced critical fluctuations in the presence of superzone gaps.Comment: 5 pages, 4 figures; journal-ref adde Small angle neutron scattering measurements on a bulk single crystal of the doped chiral magnet Fe$_{1-x}$Co$_x$Si with $x$=0.3 reveal a pronounced effect of the magnetic history and cooling rates on the magnetic phase diagram. The extracted phase diagrams are qualitatively different for zero and field cooling and reveal a metastable skyrmion lattice phase outside the A-phase for the latter case. These thermodynamically metastable skyrmion lattice correlations coexist with the conical phase and can be enhanced by increasing the cooling rate. They appear in a wide region of the phase diagram at temperatures below the $A$-phase but also at fields considerably smaller or higher than the fields required to stabilize the A-phase	{"url":"https://core.ac.uk/search/?q=author%3A(Endoh%20Y.)","timestamp":"2024-11-11T13:35:54Z","content_type":"text/html","content_length":"102986","record_id":"<urn:uuid:d0b040fa-3f7b-4b6a-8147-69515b7bafb6>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00217.warc.gz"}
Sub and super multiplicativity of norms for understanding non-locality In relation to various problems in understanding entanglement and non-locality, I have come across the following mathematical problem. It is most concise by far to state in its most mathematical form and not go into the background much. However, I hope people interested in entanglement theory might be able to see how the problem is interesting/useful. Here goes. I have two finite dimension vector spaces $A$ and $B$ and each is equipped with a norm (Banach spaces) such that $\|\|...\|\|: A \rightarrow \mathbb{R}$ and $\|\|...\|\|: B \rightarrow \mathbb{R} $. Both the vector spaces and norms are isomorphic to each other. My question concerns norms on the tensor product of these spaces (for simplicity, lets say just the algebraic tensor product) $A \ otimes B$ and the dual norms. First let me state something I know to be true. Lem 1: If a norm $\|\|...\|\|$ on $A \otimes B$ satisfies: $\|\|a \otimes b \|\| \leq \|\|a\|\| . \|\|b\|\|$ (sub-multiplicativity) then the dual norm satisfies $\|\|a \otimes b \|\|_{D} \geq \|\|a\|\|_{D} . \|\|b\|\|_{D}$ (super-multiplicativity) where we define the dual of a norm in the usual way as $\|\| a \|\|_{D}= \mathrm{sup} \{ \|b^{\dagger}a\| ; \|\|b\|\| \leq 1 \}$ This lemma crops up often such as in Horn and Johnson Matrix analysis where it is used to prove the duality theorem (that in finite dimensions the bidual equals the original norm $\|\|..\|\|_{DD}=\|\|...\|$ I wish to know the status of the converse, which I conjecture will be answered in the affirmative: If a norm $\|\|...\|\|$ on $A \otimes B$ satisfies: $\|\|a \otimes b \|\| \geq \|\|a\|\| . \|\|b\|\|$ (super-multiplicativity) then the dual norm satisfies $\|\|a \otimes b \|\|_{D} \leq \|\|a\|\|_{D} . \|\|b\|\|_{D}$ (sub-multiplicativity) My question is simply "is my conjecture true or does anyone have a counterexample?". Although I am inclined to think the conjecture is true, it is certainly not as easy to prove as the first stated lemma (which is a 3-4 line proof). The asymmetry enters in the definition of a dual norm, which allows us to "guess" a separable answer at the cost of having underestimated the size of the norm, but we can not so easily overestimate it! This post has been migrated from (A51.SE) The converse is obviously not true. The asymmetry between the super-multiplicativity and sub-multiplicativity arises because the dual norm is always defined as a supremum and never as an infimum. To see a counterexample, choose a direction in $A\otimes B$, for example a direction of vectors that are of the form $a\otimes b$, and in a very small "ray" vicinity of this direction, define the norm on the tensor product space as $$ \|\|v\|\| = 1000 \|\|a\|\|\cdot \|\|b\|\| $$ Super-multiplicativity will still obviously hold because we have increased the norm somewhere on the tensor product space while kept it constant on the rest of it. However, the dual norm skyrockets by this tiny change because it's a supremum over all $c$ with $\|\|c\|\| \leq 1$ which includes $c\approx a\otimes b$ where the norm was amplified. Correspondingly, the dual norm for certain dual vectors has been essentially increased to 1,000 times what it was before and is no-longer sub-multiplicative. Warning: the argument above is wrong. I have misinterpreted $\|b^\dagger a\|$ as something that depends on the original norm but it doesn't. The reverted implication is likely to be right at least for some "convex" norms for which the switching between the norm and the dual norm is fully reversible. Please post more complete answers if you can construct them. OK, I think that the basic argument may still be easily fixed. Take a natural norm and redefine $$ \|\|v\|\| = 0.001 \|\|a\|\|\cdot \|\|b\|\| $$ just for some $v$ being of the form $C\cdot M(a\otimes b)$ where $a,b$ are some generic vectors, $M$ is a transformation close to the identity which can't be factorized to the tensor products of transformations on the two spaces, and $C\in{\mathbb R}$. This reduction of the norm doesn't spoil super-multiplicativity because this condition only constrains the tensor products and this is not one. However, on the dual space, $(a\otimes b)_D$ calculated by some dual form will fail to be sub-multiplicative because it's affected even by "nearby" vectors on the original space, and we allowed some very long vectors (according to the original norm) to influence the supremum. So this won't hold for sufficiently unusual norms. Some kind of convexity that would guarantee that the dualization procedure squares to one could be enough to guarantee that your reverse statement is valid. This post has been migrated from (A51.SE) Most voted comments show all comments There may be a mistake in my argument, thanks for pointing it out. Will look at it again. This post has been migrated from (A51.SE) Dear @Earl, I think that I have fixed the error in my argument and the conclusion is unchanged. Reduce the original norm to 1/1000 of it in a ray of vectors that are "nearly" tensor-factorizable. This doesn't spoil the super-multiplicativity because only strictly tensor products are constrained. However, the dual norm will be affected by this change, even the dual norm of factorizable vectors, and it will jump 1,000 times or so, spoiling sub-multiplicativity. Agreed? Some convexity or triangle inequality for the norm could be enough to ban variable norms of my type and revive your This post has been migrated from (A51.SE) Ah, I see. I think you correction works now. Let me work through an even more concrete example. Consider $u$ on the interval $u_{\lambda}=\lambda a_{0}\otimes b_{0}+(1-\lambda)a_{1}\otimes b_{1}$, and define a norm such that $\|\|u\|\|= (2 \lambda-1)^{2} + \epsilon$ where $\epsilon$ is small but nonzero (e.g. 1/1000). $\|\|u\|\|$ is supermultiplicative on tensor products and convex. Then $\|\| a_{0}\ otimes b_{0} \|\|_{D} \geq \|u_{\lambda=1/2}v^{\dagger}\|/\|\|u_{\lambda=1/2}\|\| = 1/(2 \epsilon)$ which can be made arbitrarily large. This post has been migrated from (A51.SE) Finally, one more comment. I think the conditions under which the converse does hold are precisely when there exists a cross norm $\eta(u)$ (e.g. the smallest cross-norm) such that $\|\|u\|\| \geq \eta (u) $. Then one can follow the argument I used a few comments up for the more specific case of the 2-norm. However, your counterexamples are so deeply convex they achieve lower values than any cross norm can. This post has been migrated from (A51.SE) Dear @Earl, I thought that my pathological counterexamples (their set of vectors with a norm smaller than 1) deeply fail to be convex, instead of being deeply convex! ;-) This post has been migrated from (A51.SE) Most recent comments show all comments Thanks for your answer. Though I'm not sure I understand why you say this causes the dual norm to skyrocket upwards, I would have thought it causes the norm of certain dual vectors to reduce by 1,000 times. If we set $\|\|a\|\|=\|\|b\|\|=1$ then $\|\|a \otimes b \|\|=1000$ is not less than 1 and so does not fall into the unit ball over which the supreme is evaluated. More simply, if we equivalently formulate the dual norm as a sup over $\|u^{\dagger}v\| / \|\|v\|\|$ then it looks like any adhoc increase of $\|\|v\|\|$ is only going to decrease the dual norm. This post has been migrated from (A51.SE) To expand on the above. Assume that the base norms are the 2-norm, and that on $A \otimes B$ we have a norm s.t. $\|\|v\|\| \geq \|\|v\|\|_{2}$. When $v=a \otimes b$, and using $\|\|a \otimes b \|\|_{2}=\|\|a\|\|_ {2}\|\|b\|\|_2$, superadditivity follows. From this subaddivity follows, as $\|\|v\|\|_{D}= \sup_{v} \{ \|u^{\dagger}v\| / \|\|u\|\| \} \leq \sup_{v} \{ \|u^{\dagger}v\| / \|\|u\|\|_{2} \} = \|\|v\|\|_{2} $. From this putting $v=a \otimes b$ and multiplicity of the 2-norm we get subaddivity. 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Quantum Computing - Next Gen Computer? - TechmedokQuantum Computing – Next Gen computer? Quantum Computing – Next Gen computer? We might use our computers to play games, browse, watch and work/learn very few times!!. But these machines have far more uses than we think and are being used on a day-to-day basis to solve enormous Primarily, these types of problems were solved using supercomputers (CPUs, lots of CPUs), but since humans won’t settle for anything lesser, we have invented a more efficient device – Quantum Computers. Don’t get afraid of a few physics terms that will be explained below!! Need for Quantum Computers For years, we are relying on classical computers (supercomputers) to solve our problems, and it has been helpful, but not always. The problem with these machines is that they are sequential, and perform calculations one by one. Therefore, a few problems might take a much longer time to get solved and in an inefficient way. This becomes the primary cause and use case of quantum computers. A problem described by IBM is a great example. Before we dive into the topic!! Learning how quantum computers work, requires a basic understanding of a few topics. Not to worry, we have tried to explain them in a simple way. Superposition is a quantum property, where a particle can exist in any state until it is being measured. A famous example used would be the electrons, who, when unobserved may have any spin, and can define their state only while watching it. A simple explanation for superposition would be the coin toss. In a classical case, you would flip a coin and be certain that the outcome will be either Heads or Tails. However, superposition would be a state where the coin takes both heads and tails and every other state between them. Qubits are the basic unit of data in quantum computing. In a classical computer, the basic unit of data would be bits – which is either 0 or 1. Whereas, in quantum computing, qubits are superpositions of 0 and 1 (0 and 1 at the same time). This way, more possibilities of an answer can be calculated within a short span of time. Entanglement is the simplest to understand than all others. We might have heard this before, “If 2 quantum entangled objects are kept at 2 ends of the universe, and if one particles state is disturbed, it affects the other particle at the other end of the universe”. This property is employed in qubits so that their state can be correlated and can solve more complicated problems Now, What is Quantum Computing? Quantum computing is the collective utilization of quantum properties discussed before – superposition, entanglement and so on to solve problems or perform calculations. Theoretically, there are various methods quantum computing can be achieved. The method we currently use is the quantum circuit, which is based on qubits. After knowing the complexity involved in this, you might think of these computers to be the size of ENIAC, but that’s not the case. These are mostly the size of a refrigerator and are maintained in super cold conditions. The machine is made to work with the help of Superconductors (super-cooled conductors that offer zero electrical resistance), and employs electrons on them, following a process where the signals are converted into the quantum state, calculations are performed and back into an understandable form. When scientists found out that there are few problems that aren’t feasible through classical computers, quantum computers was the solution they came to. Quantum computer finds uses in various domains. With the world now focussing more on AI and Machine learning, these computers speed up the process and also break a few barriers we previously had like the high computational cost for training models and so on. It also helps in computational chemistry, providing more knowledge in pharmaceutical researches. These can be used in the drug industry, where the current method of development is trial and error which is risky and expensive. The other fields include quantum cryptography, financial modelling and others. Are we there yet? Though we have achieved more in this field in the past decade, we have not reached the peak of its abilities. We have seen most of its use cases, but quantum computers are not the solution for all For a few scenarios, supercomputers are said to be more useful than quantum computers, and hence we might want to combine those two to create the best machine. Few companies like Google, IBM, Honeywell, and others are constantly involved in researches regarding this domain. Google AI. along with NASA, has created a 54-qubit quantum processor. IBM has created the first circuit-based commercial quantum computer called IBM Q System One. It has also planned to create a 1000-qubit quantum computer by 2023. These machines also have downsides. Since it involves super-cold conditions and quantum levels, they are highly sensitive. Heat, electromagnetic fields, and collision with air can cause these qubits to lose their properties and result in system crashes. The more particles involved, the more vulnerable the device becomes. Therefore, these machines must be kept away from environmental interference and additions qubits are required to correct the errors happening! Finally, I would say that the present quantum computers may not be the perfect ones we are looking for, but continuous research will lead us to more ideal machines and help our causes. 6 Comments • Loving the information on this web site, you have done great job on the articles. • Exceptional post however , I was wanting to know if you could write a litte more on this topic? I’d be very grateful if you could elaborate a little bit further. Many thanks! • I like this weblog so much, saved to my bookmarks. “Respect for the fragility and importance of an individual life is still the mark of an educated man.” by Norman Cousins. • Admiring the commitment you put into your website and detailed information you provide. It’s great to come across a blog every once in a while that isn’t the same outdated rehashed information. Great read! I’ve saved your site and I’m including your RSS feeds to my Google account. • Hello! This post couldn’t be written any better! Reading through this post reminds me of my previous room mate! He always kept talking about this. I will forward this article to him. Pretty sure he will have a good read. 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It seems that everyone thinks an OS is a huge project. It is. One of the hardest parts of writing an OS, for example, is flash control. In KnightOS, using the boot page routines is not an option. So I decided to tackle it myself, with raw flash control. Which is hard. I have 3 years of experience coding in z80 assembly. I have 10 years experience coding in general. SirCmpwn wrote: When I mentioned that I have some trouble with bitwise arithmetic The following is all you really need to know (Using C notation here, where & is and, \| is or, ^ is xor, and ~ is bit-negation): 1 & 1 = 1 1 & 0 = 0 0 & 1 = 0 0 & 0 = 0 1 \| 1 = 1 1 \| 0 = 1 0 \| 1 = 1 0 \| 0 = 0 1 ^ 1 = 0 1 ^ 0 = 1 0 ^ 1 = 1 0 ^ 0 = 0 ~0 = 1 ~1 = 0 also, just for handy reference, bit-arithmetic in 25 seconds. If you want a run down on the shifters, that takes a little bit longer since you have to also look at the carry-in/carry-out behavior for each instruction, but those are well documented for z80. SirCmpwn wrote: Thank you, elfprince13. It's only the arithmetic that I have problems with, I get bit shifting okay. The easy way to remember without just memorizing that table is that for AND you need two 1s to get true (meaning either or both as 0s will be false), for eXclusive OR you need either of them to be 1s but not both (meaning if they have the same value you will get a 0), for inclusive OR you need either or both to be 1s (meaning if you have both false you'll get false). NOT just flips whatever you have, and the bottom two rules are called De Morgan's laws, and basically if you are NOTing both terms to an AND or an OR, you can switch the operator and pull the NOT outside (and vice versa). elfprince13 wrote: ...and the bottom two rules are called De Morgan's laws, and basically if you are NOTing both terms to an AND or an OR, you can switch the operator and pull the NOT outside (and vice versa). More tersely expressed as "break the line, change the sign" (where · is AND, + is OR and an overline is NOT): ___ _ _ Q = A·B = A+B ___ _ _ Q = A+B = A·B The above notation is common with electronics, at least where I was taught. I'm curious why this topic, which is a rant topic in the rant board, had the rant part edited out. I don't appreciate censorship. SirCmpwn wrote: I'm curious why this topic, which is a rant topic in the rant board, had the rant part edited out. I don't appreciate censorship. Notice that Kerm and mine's posts were deleted as well? Also, elf double posted, so I bet it was him. Kllrnohj wrote: SirCmpwn wrote: I'm curious why this topic, which is a rant topic in the rant board, had the rant part edited out. I don't appreciate censorship. Notice that Kerm and mine's posts were deleted as well? Also, elf double posted, so I bet it was him. I moved some of the posts in this topic to a suspended topic due to complaints about trolling from a forum member (in reference to the forum in general, not this specific topic per se). It is now (other than these trailing posts) on-topic on bitmath.	{"url":"https://www.cemetech.net/forum/viewtopic.php?p=122016","timestamp":"2024-11-05T00:34:55Z","content_type":"text/html","content_length":"46889","record_id":"<urn:uuid:3abd557f-e630-4269-a8ad-a16ca524451b>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00195.warc.gz"}
How Many Grams In A Kilogram? - November 10, 2024 Have you ever wondered how many grams in a kilogram? It may seem like an easy question, but the answer can be confusing for those unfamiliar with metric measurements. To fully understand this measurement conversion, it’s best to have some background information on what exactly kilograms and grams are and why they’re used. Understanding not only the conversions between these two units, but also their origins and applications can provide an invaluable sense of confidence when dealing with metric measurements. In this blog post, we’ll cover all aspects of understanding measuring in kilograms and grams so that you can confidently use them for all your needs. How Many Grams In A Kilogram Exploring the Metric System: How Many Grams in a Kilogram? Did you know that the metric system is the most widely used system of measurement in the world? It’s true! The metric system is used in almost every country, and it’s the official system of measurement in many countries. So, how many grams are in a kilogram? Well, the answer is simple: one kilogram is equal to 1,000 grams! That’s right – one kilogram is the same as 1,000 grams. It’s easy to remember, and it’s a great way to measure things accurately. So, the next time you need to measure something, remember that one kilogram is equal to 1,000 grams. It’s a great way to make sure you get the right amount of whatever it is you’re measuring. Happy A Guide to Converting Grams to Kilograms and Vice Versa Converting between grams and kilograms is a breeze! Whether you’re a student, a scientist, or just someone who needs to know how to convert between the two units of measurement, this guide will help you out. To convert from grams to kilograms, simply divide the number of grams by 1,000. For example, if you have 2,500 grams, divide 2,500 by 1,000 to get 2.5 kilograms. To convert from kilograms to grams, simply multiply the number of kilograms by 1,000. For example, if you have 3 kilograms, multiply 3 by 1,000 to get 3,000 grams. It’s that easy! Now you can easily convert between grams and kilograms with confidence. Have fun! The History of the Kilogram and How It Relates to Grams The kilogram is an important unit of measurement in the metric system, and it is related to the gram. The kilogram is the base unit of mass in the International System of Units (SI). It is defined as the mass of a specific cylinder of platinum-iridium alloy kept at the International Bureau of Weights and Measures in France. The history of the kilogram is quite interesting. In 1790, the French Academy of Sciences proposed a new system of measurement based on the decimal system. This system was called the metric system, and it was adopted by France in 1795. The kilogram was one of the base units of the metric system, and it was defined as the mass of a cubic decimeter of water at the temperature of melting ice. In 1875, the International Bureau of Weights and Measures was established in France. This organization was responsible for maintaining the international standards of measurement. In 1889, the kilogram was redefined as the mass of a specific cylinder of platinum-iridium alloy kept at the International Bureau of Weights and Measures. This cylinder is known as the International Prototype Kilogram (IPK). The kilogram is related to the gram, which is the base unit of mass in the metric system. One kilogram is equal to 1000 grams. This means that if you have one kilogram of something, it is equal to 1000 grams. This relationship between the kilogram and the gram is important for measuring mass accurately. The kilogram is an important unit of measurement in the metric system, and it is related to the gram. The history of the kilogram is quite interesting, and it is important to understand how it relates to the gram in order to measure mass accurately. How to Use Grams and Kilograms in Everyday Measurements Grams and kilograms are two of the most commonly used units of measurement in everyday life. Whether you’re baking a cake or buying groceries, you’ll likely be using grams and kilograms to measure out the ingredients or items you need. Here’s how to use grams and kilograms in everyday measurements. When it comes to baking, grams are the most commonly used unit of measurement. When measuring out flour, sugar, and other dry ingredients, it’s best to use a kitchen scale to get an accurate measurement. This will ensure that your recipe turns out perfectly every time. When it comes to buying groceries, kilograms are the most commonly used unit of measurement. When buying fruits and vegetables, you’ll likely be asked to specify how many kilograms you want. For example, if you’re buying apples, you might be asked to specify how many kilograms you want. When it comes to measuring out liquids, such as water or milk, liters are the most commonly used unit of measurement. A liter is equal to 1,000 milliliters, so it’s easy to convert between the two. Finally, when it comes to measuring out spices and herbs, teaspoons and tablespoons are the most commonly used units of measurement. A teaspoon is equal to 5 milliliters, while a tablespoon is equal to 15 milliliters. Grams and kilograms are two of the most commonly used units of measurement in everyday life. Whether you’re baking a cake or buying groceries, you’ll likely be using grams and kilograms to measure out the ingredients or items you need. With a little practice, you’ll be able to use these units of measurement with ease! The Benefits of Knowing How Many Grams Are in a Kilogram Knowing how many grams are in a kilogram can be incredibly beneficial in a variety of situations. Whether you’re a student studying for a science test, a chef measuring out ingredients for a recipe, or a traveler trying to convert between different units of measurement, understanding the relationship between grams and kilograms can be incredibly useful. For students, understanding the conversion between grams and kilograms can be essential for understanding the metric system. The metric system is the most widely used system of measurement in the world, and it’s important to understand how to convert between different units of measurement. Knowing that there are 1000 grams in a kilogram can help students understand how to convert between different units of measurement. For chefs, understanding the conversion between grams and kilograms can be essential for accurately measuring out ingredients. Many recipes call for ingredients to be measured in grams or kilograms, and it’s important to understand how to convert between the two. Knowing that there are 1000 grams in a kilogram can help chefs accurately measure out ingredients for recipes. For travelers, understanding the conversion between grams and kilograms can be essential for converting between different units of measurement. Many countries use different units of measurement, and it’s important to understand how to convert between them. Knowing that there are 1000 grams in a kilogram can help travelers accurately convert between different units of measurement. Overall, understanding the conversion between grams and kilograms can be incredibly beneficial in a variety of situations. Whether you’re a student, a chef, or a traveler, understanding the relationship between grams and kilograms can be incredibly useful. How to Calculate Grams and Kilograms in Different Measurement Systems Calculating grams and kilograms is an important part of understanding different measurement systems. Whether you’re a student, a scientist, or a cook, it’s important to know how to convert between different units of measurement. Here’s a quick guide to help you calculate grams and kilograms in different measurement systems. In the metric system, the base unit of mass is the gram. One gram is equal to 0.001 kilograms. To convert from grams to kilograms, simply divide the number of grams by 1000. For example, if you have 500 grams, divide 500 by 1000 to get 0.5 kilograms. To convert from kilograms to grams, multiply the number of kilograms by 1000. For example, if you have 2 kilograms, multiply 2 by 1000 to get 2000 In the imperial system, the base unit of mass is the pound. One pound is equal to 0.45359237 kilograms. To convert from pounds to kilograms, simply divide the number of pounds by 2.2046226218. For example, if you have 10 pounds, divide 10 by 2.2046226218 to get 4.5359237 kilograms. To convert from kilograms to pounds, multiply the number of kilograms by 2.2046226218. For example, if you have 3 kilograms, multiply 3 by 2.2046226218 to get 6.613867054 pounds. In the US customary system, the base unit of mass is the ounce. One ounce is equal to 0.0283495231 kilograms. To convert from ounces to kilograms, simply divide the number of ounces by 35.2739619. For example, if you have 16 ounces, divide 16 by 35.2739619 to get 0.45359237 kilograms. To convert from kilograms to ounces, multiply the number of kilograms by 35.2739619. For example, if you have 2 kilograms, multiply 2 by 35.2739619 to get 70.54792 ounces. Now that you know how to calculate grams and kilograms in different measurement systems, you’ll be able to easily convert between different units of measurement. Whether you’re measuring ingredients for a recipe or calculating the weight of an object, you’ll be able to do it with ease! The Difference Between Grams and Kilograms in Cooking and Baking Cooking and baking can be a lot of fun, but it can also be a bit confusing when it comes to measurements. One of the most common measurements used in cooking and baking is grams and kilograms. But what’s the difference between the two? Grams are a unit of mass used to measure small amounts of ingredients. They are typically used for measuring dry ingredients like flour, sugar, and spices. Grams are also used for measuring liquids like water, oil, and milk. A gram is equal to 0.001 kilograms. Kilograms, on the other hand, are used to measure larger amounts of ingredients. They are typically used for measuring wet ingredients like butter, cream, and yogurt. Kilograms are also used for measuring large amounts of dry ingredients like flour, sugar, and spices. A kilogram is equal to 1000 grams. So, when it comes to cooking and baking, it’s important to know the difference between grams and kilograms. Grams are used for measuring small amounts of ingredients, while kilograms are used for measuring larger amounts. Knowing the difference between the two will help you get the most accurate measurements for your recipes. Understanding the Relationship Between Grams and Kilograms in Science and Math Grams and kilograms are two of the most commonly used units of measurement in science and math. Understanding the relationship between them is essential for accurately measuring and calculating the weight of objects. Grams and kilograms are both units of mass, which is the measure of the amount of matter in an object. A gram is the smallest unit of mass, and it is equal to one thousandth of a kilogram. This means that one kilogram is equal to one thousand grams. To convert from grams to kilograms, divide the number of grams by one thousand. For example, if you have 500 grams, divide 500 by one thousand to get 0.5 kilograms. To convert from kilograms to grams, multiply the number of kilograms by one thousand. For example, if you have 2 kilograms, multiply 2 by one thousand to get two thousand grams. It is important to remember that grams and kilograms measure mass, not weight. Weight is the measure of the force of gravity on an object, and it is affected by the object’s location. For example, an object that weighs one kilogram on Earth would weigh less on the Moon due to the Moon’s weaker gravitational pull. Grams and kilograms are essential for accurately measuring and calculating the mass of objects in science and math. Knowing the relationship between them is key to understanding how to convert between the two units. In conclusion, there are 1000 grams in a kilogram. This is an important conversion to know when measuring out ingredients or other items that are measured in grams. Knowing this conversion can help make sure that you are using the correct amount of an ingredient or item. 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The classical algorithm for a Mandelbrot renderer has been known for decades, and most graphics enthusiasts have at one time or another written one. Today, however, is the era of the Graphics Processing Unit. Instead of rendering fractals on the CPU, we can exploit the greater power and parallelism of a GPU to make fractal rendering much faster. Let's see how we can adapt the above algorithm to the GPU. 4.1 - A First Attempt Fortunately for us, the algorithm for rendering the Mandelbrot set is on a per-pixel basis. Each pixel is processed the same way and independently of all the others. This means we can simply render a quad that covers the whole screen, and write a fragment shader to perform the iteration process: uniform vec4 insideColor; uniform sampler1D outsideColorTable; uniform float maxIterations; void main () vec2 c = gl_TexCoord[0].xy; vec2 z = c; gl_FragColor = insideColor; for (float i = 0; i < maxIterations; i += 1.0) z = vec2(z.xz.x - z.yz.y, 2.0z.xz.y) + c; if (dot(z, z) > 4.0) gl_FragColor = texture1D(outsideColorTable, i / maxIterations); This is precisely the same algorithm as was presented above; it has been rewritten in GLSL (OpenGL Shading Language). It assumes that you have set up the color table as a 1D texture map in outsideColorTable and that the texture coordinates passed to the shader correspond to complex numbers. Then the shader calculates the sequence using complex multiplication and checks each value to see if it escapes from the radius-2 circle. Unfortunately, this shader will be very slow, and you will not be able to render the fractal in real-time, unless maxIterations is set very low. Moreover, the looping and branching inside the shader means that it will not even run on graphics hardware that does not support SM 3.0. At the time of this writing, you need a GeForce 6600 or better GPU to even execute the above shader, and it would still be very slow even on a GeForce 7800, currently the most powerful card in the consumer market. 4.2 - Stream Processing Instead of performing the Mandelbrot iteration in a single complicated fragment shader, a better approach is to use a multipass algorithm. To implement this, we can use the stream processing model of general purpose GPU computation. One shader generates a set of data, which is then used as input to another shader; the output of the second shader is the input to a third, and so forth. Since shaders cannot operate "in place," i.e. they cannot write to the same framebuffer from which they are reading, we must use a minimum of two buffers. We then ping-pong between these buffers, first using one as the input and the other as the output, then switching their roles for the next shader pass. To render the Mandelbrot set in a multipass algorithm, we will use floating-point framebuffers to store the value of z[n] at each pixel. Three shaders in total are used; the first sets up the initial values, by simply storing the texture coordinates (which represent the values of c, which equals z[1]) to the red and green components of each pixel: void main () vec2 c = gl_TexCoord[0].xy; gl_FragColor = vec4(c, 0, 0); The second shader will be executed over and over again. This shader actually performs the iteration, calculating the next value of z[n] at each pixel. In this shader, we get the previous value z[n-1] by sampling a texture, which is the output of the previous pass; the value of c is provided by texture coordinates as before. uniform sampler2D input; uniform float curIteration; void main () // Lookup value from last iteration vec4 inputValue = texture2D(input, gl_TexCoord[0].xy); vec2 z = inputValue.xy; vec2 c = gl_TexCoord[0].xy; // Only process if still within radius-2 boundary if (dot(z, z) > 4.0) // Leave pixel unchanged (but copy //through to destination buffer) gl_FragColor = inputValue; gl_FragColor.xy = vec2(z.xz.x - z.yz.y, 2.0z.xz.y) + c; gl_FragColor.z = curIteration; gl_FragColor.w = 0.0; As you can see, we are also storing n (the number of the current iteration, passed in as a uniform variable), in the blue component of the pixel. This is of use in the third shader, which is used for finally displaying the fractal. Its input is the floating-point buffer containing the final z[n] values, but its output is an ordinary color buffer. uniform sampler2D input; uniform vec4 insideColor; uniform sampler1D outsideColorTable; uniform float maxIterations; void main () // Lookup value from last iteration vec4 inputValue = texture2D(input, gl_TexCoord[0].xy); vec2 z = inputValue.xy; // If Z has escaped radius-2 boundary, shade by outer color if (dot(z, z) > 4.0) gl_FragColor = texture1D(outsideColorTable, inputValue.z / maxIterations); gl_FragColor = insideColor; This three-shader multipass algorithm can render the same image as our original shader, but will run on any hardware that supports floating-point buffers. In the ATI world, this includes the Radeon 9500 and up, and in nVIDIA hardware the GeForce 5200 and up—much more reasonable expectations. (Note that although two of the shaders use if statements, on pre-SM 3.0 cards this will produce a CMP instruction rather than an actual branch.) The shaders presented above can be made to work together in a variety of different ways. The most obvious is to use the iterating shader a few times on off-screen surfaces and then display the result. However, a cool effect can be created by running the viewing shader after each iteration. The fractal animates, appearing to grow inward as more and more detail is revealed. Another possibility is to map the fractal onto a 3D surface, which can be done as easily as ordinary texture mapping, simply using the viewing shader to draw the triangles of the surface. Two problems still remain unsolved in our improved GPU Mandelbrot renderer, though. First, there is no antialiasing on the fractal, since it is essentially point-sampled. This makes the generated images look rather ugly, unless they are rendered at a high resolution and then downsampled. The second problem is precision. Currently, ATI cards use a 24-bit floating-point format and nVIDIA cards a 32-bit format for the registers in the shading units, and the floating-point buffers also store only 32-bit floats. This means that one cannot zoom very far into the fractal before precision breaks down; one can get to about 10^-7 with 32-bit floats and only 10^-5 with 24-bit floats. This is currently an unavoidable limitation of GPU hardware; CPU-based Mandelbrot renderers use 64-bit floats or even provide their own implementations of arbitrary-precision floating-point arithmetic.	{"url":"https://www.ozone3d.net/tutorials/mandelbrot_set_p4.php","timestamp":"2024-11-12T15:56:39Z","content_type":"text/html","content_length":"24274","record_id":"<urn:uuid:c17aaa07-99a0-4cab-9422-854a62b8fcd0>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00119.warc.gz"}
CS456 - Systems Programming The Shunting-yard Algorithm: The shunting yard algorithm essentially reads a infix notation expression and converts it to an RPN expression, which can be evaluated as the algorithm runs. To do this it adds an operator stack in addition to a number stack. The operator stack must record additional information, so might be a stack of the following structures: struct oper { token_t op; // The operator token int prec; // The operator precedence int dir; // The associativity (0 = left-to-right, 1=right-to-left) int unary; // Flag if the operator is the unary version. The attributes that go with the operator are taken from a set of tables for each operator, such as: // +, -, , /, , (, ) Used when unary = int prec[] = { 60, 60, 65, 65, 70, 0, 0}; // false int uprec[] = { 75, 75, -1, -1, -1, -1, -1}; // true int dir[] = { 0, 0, 0, 0, 0, 0, 0}; // false int udir[] = { 1, 1, 0, 0, 0, 0, 0}; // true The uprec/udir entries are the precedence and direction for the unary versions of the operator. A -1 indicates that the operator cannot be unary. A unary operator is one that takes only one operand and typically precede their operand, such as the minus sign that makes a number negative. Higher precedence operators are removed from the operator stack before a lower precedence operator can be pushed onto it. ()'s are handled explicitly. The direction of evaluation (0 = left to right, 1 = right to left) or associativity determines if operators of the same precedence are also removed (as is the case in left-associative operators) from the op stack before this operator is pushed. Theory of operation: • A simple infix expression is an alternating sequence of numbers and operators beginning and ending with a number. i.e. n = number, o = operator: n o n n o n o n n o n o n o n ... or generalized to: n (o n) • A Boolean flag is used to keep track of when to to expect a number or an operator. typedef enum { NUMBER, OPERATOR } flag_t; • Numbers when encountered are pushed onto the number stack and the flag is then switched to expect an operator. • If a operator is encountered when a number is expected, then it is considered a unary operator and the flag is not updated (i.e. a number is still expected.) It should still be considered an error to have multiple consecutive unary ops. • An opening parenthesis resets the flag back to expecting a number. Everything inside of ()'s should collapse to a single number in the number stack when the closing ) is encountered (i.e. all the operators are pop'ed off the op stack until the matching '(' is pop'ed.) • Before pushing an operator to the op stack, all higher precedence operators at the top of the stack are processed (i.e. the operator is pop'ed off, one or two numbers are pop'ed, the operation is performed with the numbers and the result is finally pushed onto the number stack.) If the operator to be pushed is left-associative (like most operators) then any operator of equal precedence at the top of the stack must also be processed. • When the end of the expression is reached, all the remaining operators on the op stack are processed. The number stack should have one value on it which is the result of the expression. << Work through example: 5 + 2 * 3 + 6 >> The below sets up the beginning of the implementation. We re-use the lexer built for the RPN calculator, we also re-use the number stack. This sets up the precedence and direction "table" and the operator stack and the functions to push, pop and peek at the operator stack. #include "lex.h" typedef enum { NUMBER, OPERATOR } flag_t; void die(char why) { fprintf(stderr,"%s\n", why); // +, -, , /, *, (, ) int prec[] = { 60, 60, 65, 65, 70, 0, 0}; int uprec[] = { 75, 75, -1, -1, -1, -1, -1}; int dir[] = { 0, 0, 0, 0, 0, 0, 0}; int udir[] = { 1, 1, 0, 0, 0, 0, 0}; // Number stack: int numstack[K]; int nsp = 0; void push_num(int n) { numstack[nsp++] = n; int pop_num(void) { if (nsp <= 0) die("Stack underflow"); return numstack[--nsp]; The operator stack // Operator stack: struct op { token_t op; int unary, prec, dir; } opstack[K]; int osp = 0; void push_op(token_t op, int unary, int prec, int dir) { opstack[osp++] = (struct op){.op=op, .unary=unary, .prec=prec, .dir=dir}; struct op pop_op(void) { if (osp <= 0) die("Op stack underflow"); return opstack[--osp]; Function to "peek" at the top of the op stack, T_UNKNOWN means the stack * is empty. token_t peek_op(void) { if (osp <= 0) return T_UNKNOWN; return opstack[osp-1].op; * Peak at the just the precedence of operator at the top of the op stack: int peek_prec(void) { if (osp <= 0) return -1; return opstack[osp-1].prec; The action() function. The action() function performs an "action" -- a single operation. It pops an operator off the op stack then pops one or two numbers off the number stack (based on the unary flag for that operation) then performs the operation. Finally the result is then pushed onto the number stack. The first value pop'ed off the number stack should be considered the 'right' value, and the second value the 'left' value. If no 'left' value is pop'ed then use a default value of 0 for the left side. i.e. v = l - r; if - is the unary minus, then v = 0 - r; would still give the correct result. If the unary flag is set, but the unary precedence of the operator is -1, then the expression is malformed and an error should printed. void action(void) struct op op; int l, r, v; // Get the operator: op = pop_op(); // Right side is at the top of the stack, left would be underneath it, // unless it's a unary op r = pop_num(); if (op.unary == FALSE) l = pop_num(); else l = 0; if (uprec[op.op] == -1 && op.unary) die("Malformed unary expression in action."); // Perform the operation: switch(op.op) { case T_PLUS: v = l + r; break; case T_MINUS: v = l - r; break; case T_MULT: v = l * r; break; case T_DIV: v = l / r; break; die("Illegal operation."); // Push the result onto the stack: The main loop: • Loop until a EOE (end of expression) token is read. For following tokens, do the following: □ Numbers: ☆ push the value onto the number stack, set the flag to expect an operator □ Open parenthesis: ☆ push the open parenthesis onto the op stack, set the flag to expect a number. Precedence, direction and unary flags do not matter for parenthesis. □ Close parenthesis: ☆ make a loop to call action() until the top operation on the op stack is a open parenthesis. If the op stack is completely drained, print the error "Mismatched parenthesis." and exit. ☆ pop the open parenthesis off, set the flag to expect an operator. □ For operators: +,-,*,/: ☆ Determine the current tokens precedence (p) and direction (d) based on the value of the flag and the tables provided above. ☆ If flag is set to NUMBER, then it is a unary operator. ☆ "drain" the op stack of any operators that have a higher precedence (or the same and higher if the associativity is left to right (d==0)) than the current operation. ☆ Push the new operation to the op stack ☆ Set the flag to expect a number. • After all tokens have been processed, call the action() function until there are no more remaining operators on the op stack. • Pop the top number off the number stack, this is should be the result of the expression. token_t tok; flag_t flag = NUMBER; int num, p, d; while ((tok = lex(&num)) != T_EOE) { switch(tok) { case T_NUMBER: if (flag != NUMBER) die("Syntax error"); flag = OPERATOR; case T_OPAREN: push_op(T_OPAREN, 0, 0, 0); flag = NUMBER; case T_CPAREN: // Drain the op stack until it reaches it's matching open paren: while ((tok = peek_op()) != T_UNKNOWN) { if (tok == T_OPAREN) break; if (tok == T_UNKNOWN) die("Mismatched parenthesis."); // Removes the opening paren: flag = OPERATOR; case T_PLUS: case T_MINUS: case T_MULT: case T_DIV: case T_EXP: // Get the precedence and direction of the current operator: p = (flag == NUMBER)? uprec[tok] : prec[tok]; if (p == -1) die("Malformed expression, operator not a valid unary operator."); d = (flag == NUMBER)? udir[tok] : dir[tok]; // Drain the op stack of operators of higher precedence (and equal to if // this op is left-assoc): if (d == 1) { // Right associative while (peek_prec() > p) } else { // Left associative while (peek_prec() >= p) // Finally push the operator to the op stack: push_op(tok, flag==NUMBER, p, d); flag = NUMBER; die("Syntax error"); // Drain the operator stack until it's empty: while(peek_op() != T_UNKNOWN) // The result of the entire expression should be the last and only value left: printf("%d\n", pop_num());	{"url":"http://oldmanprogrammer.net/courses/cs456/lesson.php?lesson=expressions3","timestamp":"2024-11-11T20:04:00Z","content_type":"text/html","content_length":"13120","record_id":"<urn:uuid:48632d01-7a6c-4a42-a0f3-5a838000ab91>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00751.warc.gz"}
Transformer Calculator - Nethercraft.net Transformer Calculator What is a Transformer Calculator? A Transformer Calculator is a tool used to calculate different electrical parameters of a transformer, such as power, current, voltage, and impedance. It can help engineers, electricians, and technicians in designing, installing, and troubleshooting electrical systems that use transformers. By inputting specific values into the calculator, users can quickly determine the required parameters for their transformer setup. How Does a Transformer Calculator Work? A Transformer Calculator works by taking user input for various electrical values related to the transformer. These values typically include the voltage ratings, current ratings, power ratings, and impedance of the transformer. Once the user inputs these values, the calculator uses mathematical formulas and equations to calculate other parameters, such as turns ratio, efficiency, and voltage drop across the transformer. Benefits of Using a Transformer Calculator There are several benefits to using a Transformer Calculator, including: • Time-saving: Calculating electrical parameters manually can be time-consuming and prone to errors. Using a calculator streamlines the process and ensures accurate results. • Efficiency: By quickly determining the required parameters, users can design and install transformer systems more efficiently. • Cost-effective: Avoiding errors in transformer calculations can prevent costly mistakes during installation or operation. Types of Transformer Calculators There are various types of Transformer Calculators available, each catering to different needs and requirements. Some common types include: • Transformer Turns Ratio Calculator: This calculator helps determine the turns ratio needed for a transformer based on input and output voltage values. • Transformer Impedance Calculator: Calculates the impedance of a transformer based on the primary and secondary winding resistance and reactance. • Transformer Efficiency Calculator: This calculator determines the efficiency of a transformer by comparing input and output power values. How to Use a Transformer Calculator Using a Transformer Calculator is straightforward and involves the following steps: 1. Input the required electrical values, such as voltage ratings, current ratings, and power ratings. 2. Select the type of calculation you want to perform, such as turns ratio or efficiency. 3. Click the calculate button to get the results. Factors to Consider When Using a Transformer Calculator When using a Transformer Calculator, it’s essential to consider the following factors: • Transformer Type: Different types of transformers have unique characteristics that may affect the calculations. • Temperature: Transformer performance is influenced by temperature, so accurate temperature values should be inputted into the calculator. • Load Conditions: The load conditions of the transformer, such as resistive or inductive loads, can impact the calculations. Overall, a Transformer Calculator is a valuable tool for anyone working with transformers in electrical systems. By leveraging the calculator’s capabilities, users can efficiently design, install, and troubleshoot transformer setups with accuracy and precision.	{"url":"https://nethercraft.net/transformer-calculator/","timestamp":"2024-11-07T10:11:12Z","content_type":"text/html","content_length":"53385","record_id":"<urn:uuid:24620ae6-4ae3-4747-9ca4-9f1d8280b8ee>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00461.warc.gz"}
On Error Doing my best impersonation of someone who blogs with more regularity than I really do… I glossed over (flubbed?) the error analysis a little in my last post, and should really do a better job. I’ll look at CLEAN/LEAN mapping, but the analysis methods are useful in lots of situations where you compute something from a texture. To keep things simple, I’ll use a simplified form of the (C)LEAN variance computation: $V = M - B^2$ The error in this expression is especially important in (C)LEAN mapping since it determines the maximum specular power you can use, and how shiny your objects can be. For specular power s, 1/s has to be bigger than the maximum error in V, or you’ll get some ugly artifacts. M and B come from a texture, so have inherent error of $\epsilon_M$ and $\epsilon_B$ due to the texture precision. The error in each will be 1/2 of the texel precision. For example, with texel values from 0 to 255, a raw texel of 2 could represent a true value anywhere from 1.5 to 2.5, all of which are within .5 of the texel value. In general, we’ll scale and bias to use as much of the texture range as we can. The final error for an 8-bit texture then is range/512. For data that ranges from 0 to 1, the range is 1 and the representation error is 1/512; while for data that ranges from -1 to 1, the range is 2, so the representation error is 2/512 = 1/256. The error in each parameter propagates into the final result scaled by the partial derivative. $\partial{V}/\partial{M}$ is 1, so error due to M is simple: The error due to B is a little more complicated, since $\partial{V}/\partial{B}$ is 2 B. We’re interested in the magnitude of the error (since we don’t even know if $\epsilon_B$ was positive or negative to start with), and mostly interested in its largest possible value. That gives $\epsilon_{VB}=2\ \textrm{max}(\left\|B\right\|)\ \epsilon_B$ Generally, you’re interested in whichever of these errors is biggest. The actual error is dependent on the maximum value of B, and how big the texel precision ends up being after whatever scale is used to map M and B into the texture range. So, for a couple of options: B range -1 to 1 -2 to 2 -1/2 to 1/2 Max Bump Slope 45° 63.4° 26.6° $\epsilon_B$ 1/256 1/128 1/512 $\epsilon_{VB}$ 21/256 24/128 2*.5/512 = 1/128 = 1/32 = 1/512 M range 0 to 1 0 to 4 0 to 1/4 $\epsilon_{VM}=\epsilon_M$ 1/512 1/128 1/2048 $\epsilon_V$ 1/128 1/32 1/512 $s_{max}$ 128 32 512 We can make this all a little simpler if we recognize that, at least with the simple range-mapping scheme used here, $\epsilon_B$ and $\epsilon_M$ are also dependent on $B_{max}$. $\begin{array}{ll} \epsilon_{VM} &= B_{max}^2/512\\ \epsilon_{VB} &= 4 B_{max}^2/512 = B_{max}^2/128\\ s_{max} &= 128/B_{max}^2 \end{array}$ So, this says the error changes with the square of the max normal-map slope, and that the precision of B is always the limiting factor. In fact, if there were an appropriate texture format, M could be stored with two fewer bits than B. For 16-bit textures, rather than 2^-9 for the texture precision, you’ve got 2^-17, giving a maximum safe specular power of 2^15=32768 for bumps clamped to a slope of 1. There’s no need for the slope limit to be a power of 2, so you could fit it directly to the data, though it’s often better to be able to communicate a firm rule of thumb to your artists (spec powers less than x) rather than some complex relationship (steeper normal maps can’t be as shiny according to some fancy formula — yeah, that’ll go over well).	{"url":"https://gaim.umbc.edu/2011/07/26/on-error/","timestamp":"2024-11-13T06:19:30Z","content_type":"text/html","content_length":"30589","record_id":"<urn:uuid:41a2f144-247b-458e-80b5-c939c725c4a5>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00599.warc.gz"}
A. Find The Value Of X. B. Find The Length Of Of The Banner. C. Find The Width Of The... - MosOp A. Find The Value Of X. B. Find The Length Of Of The Banner. C. Find The Width Of The… a. Find the value of x. b. Find the length of of the banner. c. Find the width of the banner. d. Find the perimeter of the banner. [tex]\tt a.) \ 13\\b.) \ 120 \ in\\c.) \ 80 \ in\\d.) \ 400 \ in[/tex] a. Find the value of x. The shape of a banner is a rectangle so we need to apply the basics of rectangles in order to answer this. [tex]\tt \overline{ GE} = \overline{MO}[/tex] [tex]\tt 9x +3 = 15x-75[/tex] [tex]\tt 75 +3 = 15x-9x[/tex] [tex]\tt 78 = 6x[/tex] [tex]\tt 6x = 78[/tex] [tex]\tt \dfrac{6x}{6}=\dfrac{78}{6}[/tex] [tex]\tt \boxed{x=13}[/tex] b. Find the length of of the banner. The length of the banner is either the measure of [tex]\tt \overline {GE}[/tex] or [tex]\tt \overline {MO}[/tex]. You can choose one. I’ll be using [tex]\tt \overline {GE}[/tex]. [tex]\tt \overline {MO} = 9x+3\\\tt \overline {MO} = 9(13)+3\\\tt \overline {MO} =117+3\\\tt \overline {MO} =120 \ in[/tex] Therefore, the length of the banner is [tex]\large \boldsymbol {\tt 120 \ in}[/tex]. c. Find the width of the banner. The width of the banner is the measure of [tex]\tt \overline {GM}[/tex]. [tex]\tt \overline {GM} = 5x+15[/tex] [tex]\tt \overline {GM} = 5(13)+15[/tex] [tex]\tt \overline {GM} = 65+15[/tex] [tex]\tt \overline {GM} = 80 \ in[/tex] Therefore, the width of the banner is [tex]\large \boldsymbol {\tt 80 \ in}[/tex]. d. Find the perimeter of the banner. The perimeter of the rectangle is the sum of the measure of the sides. We can use the formula: [tex]\tt P=2 \ ( \ Length \ + \ Width \ )[/tex] [tex]\tt P=2 \ ( \ 120 \ + \ 80\ )[/tex] [tex]\tt P=2 \ ( 200 )[/tex] [tex]\tt P=400[/tex] So, we have our perimeter which measures [tex]\large \boldsymbol {\tt 400 \ in}[/tex]. reflective essay Self reflection journals journaling rediscover inspirational yourself help. Bullet journal reflection, by @joellejournals bullet journal and diary. Reflection layouts Self reflection journals journaling rediscover inspirational yourself help. My self-reflection journal. Week 27: reflection (the bullet journal method book club) — tiny ray of journal reflection creating literacy Reflection journal student grade fairclough lauren students teacherspayteachers created first choose board. Journal reflection from my banner – portfolio. Weekly reflections journal created for you Bullet planner. “in the journal i am at ease.” anaïs nin carve out time on at least a. Self reflection journals journaling rediscover inspirational yourself help reflection layouts My self-reflection journal. Student reflection journal by lauren fairclough. Week 27: reflection (the bullet journal method book club) — tiny ray of Student reflection journal by lauren fairclough. Journal reflection on quiz written today / self care journal for. Bullet planner Journal reflection on quiz written today / self care journal for. “in the journal i am at ease.” anaïs nin carve out time on at least a. Reflection journal in tagalog : pin by reflective journals examples on Plan a self reflection day. Journal reflection on quiz written today / self care journal for. Journal reflective tagalog examples journaling marissa rehder Related Posts	{"url":"https://mosop.net/a-find-the-value-of-1468/","timestamp":"2024-11-04T15:39:37Z","content_type":"text/html","content_length":"139946","record_id":"<urn:uuid:961e913f-fe40-40d9-a1b4-24ffbc712606>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00614.warc.gz"}
System of linear equations \| JustToThePointSystem of linear equations Definition: A system of linear equations is a set or collection of one or more linear equations involving the same variables of the following form: a[1]x[1] + a[2]x[2] + ··· + a[n]x[n] = b, where b and the coefficients a[1], a[2], ···, a[n] are fixed numbers. A solution to a linear system is an assignment of values to the variables such that all the equations are simultaneously satisfied. The solution set is the set of all solutions to a system of Two linear systems using the same set of variables are equivalent if each of the equations in the second system can be derived algebraically from the equations in the first system, and vice versa, and as a consequence they share the same solution set. • Trivial example. The system of one equation in one unknown, 6x = 18 has the solution x = 2. A system of equations is said to be consistent or have a consistent solution if it has at least one solution, that is, there exists a set of values for the variables that satisfy all the equations in the system simultaneously. Otherwise, we call the system inconsistent. A consistent linear system can have infinitely many solutions or a unique solution. • The second simplest example involves two equations and two variables or unknowns, (Figure 1.a.) $\begin{cases} 2x + 3y = 8\\\\ x + 4y = 9\end{cases}$ A linear equation of two real variables geometrically forms a line in the plane. The solution to this lineal system is the intersection between the two lines, (1, 2). This system is consistent. $\begin{cases} x + 4y = 9\\\\ 2x + 8y = 0\end{cases}$ • This system of equation is consistent, but any solution that works for one equation will also work for the other equation, so there are infinite solutions to the system. (Figure 1.c.) $\begin{cases} x + 4y = 9\\\\ 2x + -18 = -8y\end{cases}$ They two lines overlap or intersect everywhere on the line, the system is essentially expressing a single equation in multiple forms. One of the equations in the system is redundant and can be derived from the other equations, x + 4y = 9 (*2) ⇒ 2x + 8y = 18. $\begin{cases} x + 4y = 9\\\\ 2x + 8y = 18\end{cases}$ System of two linear equations in two variables can be visually represented by graphing both equations simultaneously and the solutions to these systems of equations fall into three categories depending upon how the two lines representing each equation in the graph intersect: 1. If the lines intersect into a single point (they have different slopes), the system of two linear equations has a unique solution which corresponds to that very point (1, 2), Figure 1.a. This systems are consistent systems of independent equations. 2. A system composed of two distinct parallel lines has no solutions, and therefore it is inconsistent because these distinct parallel lines do not intersect, Figure 1.b. 3. A system can be composed of two identical lines has an infinite number of solutions because these identical lines intersect at an infinite number of points, Figure 1.c. The equations are dependent with an infinite number of solutions. • Consider the following linear system with three variables or unknowns and three equations, figure 1.d. $\begin{cases} 2x -3y -z = 7\\\\ 3x + 5y -3z = -2\\\\ 4x -y +2z = 17\end{cases}$ There is a solution, (3, -1, 2). Just as the graph of a linear equation in two variables creates a line, the graph of a linear equation in three variables creates a plane which extends infinitely in all directions. Therefore, a system of three linear equations in three variables can be represented as a group of three planes. Any point (3, -1, 2) which simultaneously fall on all three planes corresponds to a solution of the system. There are different options: (1) There a no two planes parallel in the system, all planes intersect at one point, the solution to the system (Figure 2.a); (2) Two of the planes are not parallel, so they intersect in a line, and the third plane intersect the other two planes along this line, every point on this line is a solution to the system (Figure 2.b.), that is, there are infinite solutions -if the third plane is identical to one of the other two planes, the line where all planes intersect is still the solution to the system-; (3) If all planes are identical, there are also an infinite number of solutions, i.e., every point in this plane is a solution to the system (Figure 2.c.); (4) If the system has two distinct parallel planes, then there can be no solution, (Figure 2.d). Regardless of the third plane, the system is inconsistent. (5) The planes are not parallel, but they are orientated so that their intersection lie along three distinct parallel lines, and the system has no solution (Figure 2.e.) • Consider the following linear system with three variables or unknowns and three equations, figure 2.f. $\begin{cases} 2x + y - 3z = 0 \\\\ 4x + 2y - 6z = 0\\\\ x - y + z = 0\end{cases}$ dependent (e.g., the first equation times 2 equals the second equation), consistent case. In this particular case, the three planes intersect along a common line, so any point of the form (^2⁄[3]z, ^5⁄[3]z, z) where the variable z can be assigned to any real arbitrary number, is a solution to the linear system. This is called the general solution to the system. z is a free variable, that is, a variable that can be arbitrary assigned a value. On the other hand, the other two variables, x and y, are dependent variables because their values dependent upon the assignments of the free variables (in this particular case, z). Definition. A homogeneous system of linear equations is one in which all of the constant terms are zero., that is, each equation is of the form a[1]x[1] + a[2]x[2] + ··· + a[n]x[n] = 0. Otherwise, the system is called non-homogeneous. $\begin{cases} -x + y -z = 0 \\\\ 3x -y -z = 0\\\\ 2x + y -3z = 0\end{cases}$ This homogeneous system is consistent and has a unique solution (0, 0, 0). Proposition. A homogeneous system of equations is always consistent.	{"url":"https://justtothepoint.com/maths/linearequations/","timestamp":"2024-11-14T15:14:09Z","content_type":"text/html","content_length":"20353","record_id":"<urn:uuid:795d7154-a060-45f4-bdb7-5508939fbb57>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00451.warc.gz"}
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Can understanding how to work with one kind of problem help understand how to work another type? 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Bhaskaracharya Pratishthana conducts many different programs through out the year. Following is an overview of each. Mathematics Training All the programmes will consist of online lectures (2 days a week) throughout the year beginning from 15th June 2021 to 28th February 2022. The programmes will be extremely Programme interactive in nature. A special feature for every programme will be project based activities and special lectures by expert faculty from India or abroad. The Advanced Training in Mathematics Schools -- provide training in core subjects in Mathematics to Ph.D. students, young researchers and teachers. The emphasis in these NCM Schools/Workshops schools is on learning mathematics by doing it. IIT Bombay and TIFR have jointly established the National Centre for Mathematics (NCM) in 2011. The instructional schools and workshops which were earlier planned by an NBHM committee on ATM Schools are now being organised under the supervision of the Apex Committee of the NCM. Bhaskaracharya Bhaskaracharya Pratishthana introduced a new activity, to commemorate the 900th birth anniversary of the legendary Indian Mathematician Bhaskaracharya II. Any student up to Mathematics Talent 6th standard in Maharashtra & Goa can compete and take the examination at the centres announced. The registration for the examination is to be done through the studentâ€™s Search Competition school. Undergraduate Training Bhaskaracharya Pratishthan, Pune will be organizing an Undergraduate Training Programme in Mathematics (UTPM) for S. Y. B. Sc. and T. Y. B. Sc. Students. Programme in Mathematics Experts from the field of mathematics in and around Pune, IIT, IISER, and TIFR will guide the students during the programme. ROTARY GANIT OLYMPIAD objectives of RGO 2018 are to stimulate the study of mathematics in academia, to identify talented young mathematicians and encourage them to pursue a scientific career, particularly in mathematics and to select and train students. Additionally, RGO 2018 promotes several regional and state mathematical tournaments.	{"url":"http://www.bprim.org/programmes/?page=0","timestamp":"2024-11-05T00:49:43Z","content_type":"text/html","content_length":"39814","record_id":"<urn:uuid:7ac8f650-6a14-45cf-9997-19933c8c029a>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00522.warc.gz"}
What You Did Not Understand About Football Is Extremely Effective – However Very Simple With the huge amount of data collected on football and the growth of computing talents, many games involving resolution decisions might be optimized. Keywords: Football; Reinforcement Learning; Markov Decision Process; Expected Points; Optimum Selections. The notion of reinforcement studying is one key precept, whereby a sport or set of selections is studied and rewards recorded so a machine can learn long run benefits from local choices, often negotiating a sequence of complicated choices. The nicely outlined points system in the sport allows us to have the required terminal utilities. 4), we will analyze separately the system evolution along the horizontal and the vertical axis. On this case, we can see a wide number of values, from robust interactions as within the case of players 1-2121-21 – 2 to negligible interactions in the case of players 3-8383-83 – 8. For results on other datasets, please c.f. In the following, we current the results on this regard. Part four presents our results. Part three supplies the knowledge on how we’re capable of set all the required utilities of states. With this methodology, we acquire a singular set of parameters that govern the equations. By utilizing the optimum set of parameters calculated with the tactic proposed in the earlier section, we will calculate for all the gamers at each time step the distinction between the actual velocity and the model’s prediction. Since our aim is to outline a simple theoretical framework such that we are able to easily interpret the outcomes, we propose a mannequin based mostly on gamers to players’ interactions. In this frame, we goal to define a mannequin to describe the spatiotemporal evolution of the group. Carroll et al. (1988) used a mannequin strategy to anticipated points and directed attention to the valuation of yards gained, relying on which yards the gains had been over. Clearly, a number of teams weren’t happy by such choices, contemplating them to be unfair (Holroyd, 2020) as a result of, inevitably, this favoured those groups that had nonetheless to play the strongest opponents within the remaining matches over those who have been wanting ahead to a fairly mild finish-of-season. Do not over water your lawn because fleas thrive in darkish and moist locations. That is an unexpected result since in theory more iterations lead to raised efficiency. With these optimum selections we can analyse the performance of teams as a share of their actions as they relate to the optimal decisions. The upshot is that optimality then relies on the law of massive numbers to place the optimal theoretical selections into observe to maximize anticipated scores. The intellect will in all probability confirm the way that is finished inside a proper strategy, and then will quickly put into apply engaging in this. However Gale and Brian’s family want individuals to recollect him for the best way he lived, not the way in which he died. Many are doing the same issues that folks many years younger — and older — are doing: shifting, shaking and anti-aging. This criterion, at the identical time, allow us to evaluate the strength of the interactions among completely different matches halves. POSTSUBSCRIPT, since they point out the strength of the interactions among gamers. POSTSUBSCRIPT, are easily specified, and the probabilities can be accurately estimated from the huge swathes of knowledge obtainable. We argue to the opposite; that each drive can be analyzed as a self contained unit inside a sport and the rewards associated with the terminal states and transition probabilities are adequate to determine optimum decisions. To explain the format of the paper: in part 2 we provide some additional details on the idea on which we rely, and we display in a sequence of illustrations which turn into extra lifelike to the sport of football while holding the key rules on which we Mark McGwire hit greater than 50 house runs in a season 4 completely different occasions in his career. Do you gear up and head out each year to convey dwelling the biggest prizes which you could later show off to your friends? POSTSUPERSCRIPT to be sure if the house workforce certainly scored a objective in a given minute, or no in any other case. We rate a player by first summing the values of his passes for a given time period (e.g., a recreation, a sequence of video games or a season) and then normalizing the obtained sum per ninety minutes of play. Ball management offenses are sometimes boring to observe, but executed properly they can win many games. Primarily based on these tables, it can be seen that each statistical tree-based mostly RHEA versions achieved the highest variety of video games where they outperformed the other two algorithms. When Jacob leaves Sam’s pack to start out one of his own, Leah and Seth are the one two wolves to join him. They may be a part of a match late or quit a match early. For a more detailed description of the minimizing procedure, c.f.	{"url":"https://rudanet.info/what-you-did-not-understand-about-football-is-extremely-effective-however-very-simple/","timestamp":"2024-11-05T01:09:04Z","content_type":"application/xhtml+xml","content_length":"32800","record_id":"<urn:uuid:b748e91a-55e9-42b8-abd5-be6c3fac515b>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00110.warc.gz"}
Introduction to Topology \| Math Books \| Abakcus Highly regarded for its exceptional clarity, imaginative and instructive exercises, and fine writing style, this concise book offers an ideal introduction to the fundamentals of topology. Originally conceived as a text for a one-semester course, it is directed to undergraduate students whose studies of calculus sequence have included definitions and proofs of theorems. The book’s principal aim is to provide a simple, thorough survey of elementary topics in the study of collections of objects, or sets, that possess a mathematical structure. The author begins with an informal discussion of set theory in Chapter 1, reserving coverage of countability for Chapter 5, where it appears in the context of compactness. In the second chapter Professor Mendelson discusses metric spaces, paying particular attention to various distance functions which may be defined on Euclidean n-space and which lead to the ordinary topology. Chapter 3 takes up the concept of topological space, presenting it as a generalization of the concept of a metric space. Chapters 4 and 5 are devoted to a discussion of the two most important topological properties: connectedness and compactness. Throughout the text, Dr. Mendelson, a former Professor of Mathematics at Smith College, has included many challenging and stimulating exercises to help students develop a solid grasp of the material presented.	{"url":"https://abakcus.com/book/introduction-to-topology/","timestamp":"2024-11-07T00:07:14Z","content_type":"text/html","content_length":"130125","record_id":"<urn:uuid:b8c0977f-8fd4-415d-bb5a-aa254b2c62d5>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00101.warc.gz"}
Mathematics 9 Recommended Prerequisite: Completion of Mathematics 8 or Introduction to Math 9 with teacher’s recommendation. The Mathematics 9 course is intended to build on Mathematics 8 skills and introduce further skills needed for completion of high school mathematics. Topics: Algebraic Manipulation and Solving Algebraic Expressions; Rational Numbers; Data Analysis and Probability; Geometry and Similarity; Area and Volume of Two and Three Dimensional Objects; Trigonometry. Math 9 prepares students for Foundations and Pre-Calculus 10 and Workplace Math 10. Mathematics 9 is an academic pathway for students wishing to move on to any of the Math 10 courses. Students are expected to complete between 15 to 30 minutes of math homework each school night, although some students who have struggled with math historically, may require more time. The pace of this 5-month course can be difficult to manage, for students with lower understanding coming from Mathematics 8. It is highly recommended that a Learning Strategies block and/or a tutor be utilized, should students need additional time to complete the practice work. Students can also consider taking LINEAR Math 9. This allows students to have two semesters to learn. For all Esquimalt High Math Pathways, go to bit.ly/esqmathpathways.	{"url":"https://esquimaltcourses.sd61.bc.ca/courses/mathematics-9/","timestamp":"2024-11-04T21:55:10Z","content_type":"text/html","content_length":"45855","record_id":"<urn:uuid:125f8fc4-0471-41b4-9b6a-24f13f2d59cf>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00861.warc.gz"}
How to Calculate and Solve for Quantity of Heat Loss \| Fuel and Furnaces The image above represents quantity of heat loss. To calculate quantity of heat loss, four essential parameters are needed and these parameters are Initial Temperature (T[1]), Final Temperature (T [2]), a and E. The formula for calculating quantity of heat loss: Q = a(T[1]Â – T[2])^5/4Â + 4.88E[(^T[1]Â + 273/[100])^4Â – (^T[2]Â + 273/[100])^4] Q = Quantity of Heat Loss T[1]Â = Initial Temperature T[2]Â = Final Temperature Let’s solve an example; Find the quantity of heat loss when the initial temperature is 21, the final temperature is 17, a is 14 and E is 15. This implies that; T[1] = Initial Temperature = 21 T[2] = Final Temperature = 17 a = 14 E = 15 Q = a(T[1]Â – T[2])^5/4Â + 4.88E[(^T[1]Â + 273/[100])^4Â – (^T[2]Â + 273/[100])^4] That is, Q = 14(21 – 17)^5/4Â + 4.88(15)[(^21 + 273/[100])^4Â – (^17 + 273/[100])^4] Q = 14(4)^5/4 + 4.88(15)[74.71 – 70.72809] Q = 14(5.65) + 4.88(15)(3.98) Then, Q = 79.19 + 291.608 Q = 370.80 Therefore, theÂ quantity of heat lossÂ isÂ 370.80 J/Kg K. Read more:Â How to Calculate and Solve for Total Heat Loss in Furnace \| Fuel and Furnaces How to Calculate Quantity of Heat Loss With Nickzom Calculator Nickzom Calculator â€“Â The Calculator Encyclopedia is capable of calculating the quantity of heat loss. To get the answer and workings of the quantity of heat loss using the Nickzom Calculator â€“ The Calculator Encyclopedia.Â First, you need to obtain the app. You can get this app via any of these means: WebÂ â€“Â https://www.nickzom.org/calculator-plus Master Every Calculation Instantly Unlock solutions for every math, physics, engineering, and chemistry problem with step-by-step clarity. No internet required. Just knowledge at your fingertips, anytime, anywhere. To get access to theÂ professionalÂ version via web, you need toÂ registerÂ andÂ subscribeÂ forÂ NGN 2,000Â perÂ annumÂ to have utter access to all functionalities. You can also try theÂ demoÂ version viaÂ https://www.nickzom.org/calculator Android (Paid)Â â€“Â https://play.google.com/store/apps/details?id=org.nickzom.nickzomcalculator Android (Free)Â â€“Â https://play.google.com/store/apps/details?id=com.nickzom.nickzomcalculator Apple (Paid)Â â€“Â https://itunes.apple.com/us/app/nickzom-calculator/id1331162702?mt=8 Once, you have obtained the calculator encyclopedia app, proceed to theÂ Calculator Map,Â then click onÂ Materials and MetallurgicalÂ underÂ Engineering. Now, Click onÂ Fuel and FurnacesÂ underÂ Materials and Metallurgical Now, Click onÂ Quantity of Heat Loss underÂ Fuel and Furnaces The screenshot below displays the page or activity to enter your values, to get the answer for the quantity of heat loss according to the respective parameter which is the Initial Temperature (T[1]), Final Temperature (T[2]), a and E. Now, enter the values appropriately and accordingly for the parameters as required by the Initial Temperature (T[1]) is 21, Final Temperature (T[2]) is 17, aÂ isÂ 14 and E is 15. Finally, Click on Calculate As you can see fromÂ the screenshot above,Â Nickzom Calculatorâ€“ The Calculator Encyclopedia solves for the quantity of heat loss and presents the formula, workings and steps too.	{"url":"https://www.nickzom.org/blog/2021/07/31/how-to-calculate-and-solve-for-quantity-of-heat-loss-fuel-and-furnaces/","timestamp":"2024-11-10T04:29:32Z","content_type":"text/html","content_length":"239532","record_id":"<urn:uuid:de6cd5c5-96ab-4d28-9410-c0cc1fb31180>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00159.warc.gz"}
Pseudo-line arrangements: Duality, algorithms, and applications A collection L of n x-monotone unbounded Jordan curves in the plane is called a family of pseudo-lines if every pair of curves intersect in at most one point, and the two curves cross each other there. Let P be a set of m points in R2. We define a duality transform that maps L to a set L-of points in R2 and P to a set P∗of pseudo-lines in E2, so that the incidence and the "above-below" relationships between the points and pseudo-lines are preserved. We present an efficient algorithm for computing the dual arrangement A(P∗) under an appropriate model of computation. We also propose a dynamic data structure for reporting, in 0(me + fc) time, all k points of P that lie below a query arc, which is either a circular arc or a portion of the graph of a polynomial of fixed degree. This result is needed for computing the dual arrangement for certain classes of pseudo-lines arising in our applications, but is also interesting in its own right. We present a few applications of our dual arrangement algorithm, such as computing incidences between points and pseudo-lines and computing a subset of faces in a pseudo-line arrangement. Next, we present an efficient algorithm for cutting a set of circles into arcs so that every pair of arcs intersect in at most one point, i.e., the resulting arcs constitute a collection of pseudo-segments. By combining this algorithm with our algorithm for computing the dual arrangement of pseudo-lines, we obtain efficient algorithms for a number of problems involving arrangements of circles or circular arcs, such as detecting, counting, or reporting incidences between points and circles. Original language English Title of host publication Proceedings of the 13th Annual ACM-SIAM Symposium on Discrete Algorithms, SODA 2002 Publisher Association for Computing Machinery Pages 800-809 Number of pages 10 ISBN (Electronic) 089871513X State Published - 2002 Event 13th Annual ACM-SIAM Symposium on Discrete Algorithms, SODA 2002 - San Francisco, United States Duration: 6 Jan 2002 → 8 Jan 2002 Publication series Name Proceedings of the Annual ACM-SIAM Symposium on Discrete Algorithms Volume 06-08-January-2002 Conference 13th Annual ACM-SIAM Symposium on Discrete Algorithms, SODA 2002 Country/Territory United States City San Francisco Period 6/01/02 → 8/01/02 Dive into the research topics of 'Pseudo-line arrangements: Duality, algorithms, and applications'. Together they form a unique fingerprint.	{"url":"https://cris.tau.ac.il/en/publications/pseudo-line-arrangements-duality-algorithms-and-applications","timestamp":"2024-11-06T17:47:44Z","content_type":"text/html","content_length":"52173","record_id":"<urn:uuid:74a922d8-ea87-4636-84cf-faec5e3b47d2>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00418.warc.gz"}
Correlation - Definition, Types Of Correlation \| MathDada Correlation – Definition, Types, Methods of Measurements Correlation is a statistical tool used to study the relationship between two or more variables. Two variables are said to be correlated if the change in one variable there will change in other variable. On the other hand if the change in one variable does not bring any change in other variable then we say that the two variables are not correlated to each other. Types of Correlation There are four types of correlation – 1. Simple, Partial and Multiple correlation 2. Positive and Negative correlation 3. Perfect and Imperfect correlation 4. Linear and Non-linear correlation Simple, Partial, and Multiple Correlation Simple correlation is the relationship between any two variables. Partial correlation is the study of relationship between any two out of three or more variables ignoring the effect of other variables. For example, let us suppose that we have three variables X[1] = marks of Maths, X[2] = marks of Science, and X[3] = marks of English. So if we study the relationship between X[1] and X[2] ignoring the effect of X[3], then it is partial correlation. Multiple correlation is the study of simultaneous relationship between one or group of other variables. For example, if we study X[1], X[2], X[3] simultaneously then correlation between X[1] and (X [2], X[3]) is multiple correlation. Multiple correlation is not commonly used. Positive and Negative correlation Two variables are said to be positively correlated when the both the variables under study move in the same direction, i.e., if one variable increase, the other variable should also increase and one variable decreases the other variable should also decrease. Variables are said to be negatively correlated if increase in one variable leads to decrease in other variable and vice versa. That is the variables move in opposite direction. For positive correlation, the graph will be an upward curve whereas in case of negative correlation the graph will be downward curve. Perfect and Imperfect Correlation When both the variables changes at a constant rate irrespective of the change in direction then it is called perfect correlation. When the variables changes at different ratio then it is called imperfect correlation. The values of perfect correlation is 1 or -1 and the values of imperfect correlation lies in between -1 and 1. Linear and Non-linear Correlation Linear correlation is a correlation when the graph of the correlated data is a straight line. That is the variables are perfectly correlated. The linear correlation can be either positive or negative when the graph of straight line is either upward or downward in direction. On the other hand the non-linear or curvy-linear correlation is a correlation when the graph of the variables gives a curve of any direction. Like perfect correlation, non-linear correlation can be either be positive or negative in nature depending upon the upward and downward direction of the curve. Methods of measurement of Correlation Following are the three important methods of measuring the correlation between the variables – 1. Scatter Diagram Method 2. Karl Pearson’s Coefficient Method 3. Spearman’s Rank Coefficient Method << Previous Topic Next Topic >> Mean Deviation Scatter Diagram Method Leave a Comment	{"url":"https://mathdada.com/correlation-definition-with-methods-of-measurements/","timestamp":"2024-11-10T02:51:47Z","content_type":"text/html","content_length":"181867","record_id":"<urn:uuid:f8d6a8be-3004-43ab-9dca-3eeb770a11f9>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00307.warc.gz"}
Which collision resolution technique involves searching for the next available slot using a sequence of values generated by a hash function - ITEagers Data Structure - Question Details Which collision resolution technique involves searching for the next available slot using a sequence of values generated by a hash function? Similar Question From (Data Structure) How does a higher load factor affect the efficiency of a hash table? Similar Question From (Data Structure) How does open addressing handle deletions in a hash table? Similar Question From (Data Structure) What is the time complexity for the enqueue operation on a queue? Similar Question From (Data Structure) In a queue, where is the new element added during an enqueue operation? Similar Question From (Data Structure) What is the primary advantage of using a good hash function in hashing? Similar Question From (Data Structure) What is the purpose of a priority queue? Similar Question From (Data Structure) What is a dynamic resizing strategy in hash tables? Similar Question From (Data Structure) Which of the following applications does not involve the use of queues? Similar Question From (Data Structure) Which of the following is a common application of hashing? Similar Question From (Data Structure) In a binary tree, what is the height of a tree with only one node? Read More Questions Learn the building blocks of efficient software through the study of data structures and algorithms. Read More Challenge Your Knowledge! Engage in our interactive quizzes to put your learning to the test. Strengthen your understanding and reinforce your knowledge with our thought-provoking questions and activities. Start Quiz Recent comments Latest Comments section by users Add a comment Your Comment will appear after approval! Check out Similar Subjects Computer Science Solved Past Papers (SPSC) Solved Past Papers (FPSC)	{"url":"https://iteagers.com/Computer%20Science/Data%20Structure/1149_Which-collision-resolution-technique-involves-searching-for-the-next-available-slot-using-a-sequence-of-values-generated-by-a-hash-function","timestamp":"2024-11-13T01:51:06Z","content_type":"text/html","content_length":"105563","record_id":"<urn:uuid:4a894453-6872-4645-88c9-2a0c75a8f1e2>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00857.warc.gz"}
Data Mining Assignment 2 Solution - Programming Help Problem 1 (Naive Bayes, 100pts) Generate 1000 training instances in two di erent classes (500 in each) from multi-variate normal distribution using the following parameters for each class [1] = [1;0]; [2] = [0; 1]; [1] = 0:75 1 ; [2] = 0:75 0:[1] (1) 1 0:75 1 75 and label them 0 and 1. Then, generate testing data in the same manner with 500 instances for each class, i.e., 1000 in total. 1. (30pt) Implement your Naive Bayes Classi er [pred, posterior, err] = myNB(X,Y,X test,Y test) whose inputs are the training data X, labels Y for X, testing data X test and labels Y test for X test and returns predicted labels pred, posterior probability posterior with which the prediction was made and error rate err. Assume Gaussian (normal) distribution on the data: there are two parameters that realizes the probability density function (pdf), i.e., and . You can use functions such as normpdf or pdf in matlab (or equivalent functions in Python) to obtain likelihood from Gaussian pdf. Derivation of Naive Bayes looks complicated, but its actual implementation should be simple if you understand the concept of Naive Bayes Classi er (you only need the last few slides of our lecture slides for this topic.) 2. (10pt) Perform prediction on the testing data with your code. In your report, report the accuracy, precision and recall as well as a confusion matrix. Also, make sure to include a scatter plot of data points whose labels are color coded (i.e., the samples in the same class should have the same color) in the report. 3. (20pt) In your training data, change the number of examples in each class to f10; 20; 50; 100; 300; 500g and perform prediction on the testing data with your code. In your report, show a plot of changes of accuracies w.r.t. the number of examples and write your brief obervation. Instructor: W. H. Kim (won.kim@uta.edu), TA: Xin Ma (xin.ma@mavs.uta.edu) Page 1 of 2 CSE4334/5334 Data Mining Assignment 2 4. (10pt) Now, in your training data, change the number of examples in class 0 as 700 and the other as 300. Perform prediction on the testing dataset. How does the accuracy change? Why is it changing? Write your own observation. 4. (30pt) Write a code to plot an ROC curve and calculate Area Under the Curve (AUC) based on the posterior for class 1 (i.e., the con dence measure for class 1 is the posterior). The implementation should be done on your own without using explicit library that lets you draw the curve. Report the ROC curves from the two cases discussed in P1-2 and P1-4 above (i.e., one with equal distribution of classes and unequal distributions in the training data). Instructor: W. H. Kim (won.kim@uta.edu), TA: Xin Ma (xin.ma@mavs.uta.edu) Page 2 of 2	{"url":"https://www.edulissy.org/product/data-mining-assignment-2-solution/","timestamp":"2024-11-06T05:08:06Z","content_type":"text/html","content_length":"171529","record_id":"<urn:uuid:48d39a35-9704-491c-8bc4-34d7b0fc99c6>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00569.warc.gz"}
Mandelbrot set images and videos This page provides links to various (hopefully) pretty images and videos of the Mandelbrot set that I computed with a program I wrote. Zoom videos I computed three videos of continuous zooms into the Mandelbrot set: they follow exactly the same pattern, zooming at a constant rate of a factor 2 every two seconds toward fixed a center point, with the same color scheme. I haved tried to select the center points so as to illustrate how varied the Mandelbrot set can be by making sure that at least four or five wholly different “shapes” (depends how you count, of course) can be seen during the zoom. I also chose an accompanying music (from Musopen so that it be in the Public Domain) which I thought could adequately set the tone. I have placed copies of the videos on YouTube (see below for links and embedded copies), because that might be simpler for many people to view, but the quality (even in what YouTube calls “high quality”) is extremely poor and hardly does them justice, so I am also offering high-quality downloads using BitTorrent (again, see below for links). There are, of course, dozens of different videos of the kind on YouTube. I found, however, that at least one of two things was true: that they don't go very deep (such as this one) or that they don't show much of interest (this one is typical in this respect: it goes as far as 10^1000, so we're supposed to say “wow!”, but it is in fact staggeringly boring). In certain videos, you can also spot places where the iteration bound was clearly left too low. The probable reason for these limitations is that people/programs either use native floats (which are typically limited to 53 or 64 bits of precision, hence zoom depth—some processors will do 113 bits, but they're not of the common, aka Intel, variety) or else they use something like Mathematica to provide arbitrary precision, in which case lack of time/patience prevents them from going too far in iterations. Lastly, many videos have really bad color schemes. However, I think a few videos stand out: this one and that one are the best I could find, along with the one which inspired me to produce zooms of my own; but even these do not show a great variety of shapes, hence my attempt to do “better” (if I dare say so). Technical notes My program (see below) uses the GMP library for arbitrary precision floats, and I distributed computation on a pool of 30ish dual-core PC's, which ran for about one night to produce these videos. Actually, I computed a number of “key images” (one per second in the movie), at resolution 1024×768, each with a zoom factor of √2 over the previous, and then I interpolated them in a straightforward way (scaling/cropping them down as necessary and fading one image into the other to avoid brutal skips) to compute the frames that were fed to the video encoder. The video resolution is 640×480 (or 640×360 for the YouTube version) with 25fps (but 30fps on YouTube, at their recommendation), the container format is AVI, the video codec is H.264 and the audio codec is MP3. I used ffmpeg to encode them (with x264 providing the H.264 codec). Video number 1: a deep zoom The first video (of which the image on the right is a sample) is 4′14″ long (actually 4′12″ on YouTube because it lacks the final fadeout). It centers on the target point −0.9223327810370947027656057193752719757635 +0.3102598350874576432708737495917724836010i and zooms to a factor of 2^123.5 or 1.5×10^37. It is perhaps not as varied as the second video, but it is a little longer. The music is an excerpt of one of Händel's harpsichord suites. The YouTube version of the video is here (also embedded in small here on the right), but, once again, it is of extremely poor quality. The high quality version is 64MB in size: here is the torrent to download it: you should pass this torrent file to a program such as BitTorrent, [DEL:Azureus:DEL] , Deluge or similar. If you cannot use BitTorrent (e.g., because some fascist netadmin or ISP prohibits it, thinking that it is only used for copyright infringement), you can download the video directly by removing the .torrent extension from the previous link. (But please, try to use BitTorrent if you can!) The high and narrow image on the left of this page is a kind of roadmap to this video, with the different zoom levels shown vertically. More accurately, it is a log map toward the target point (or, as some might say, a Mercator projection with the target point as South pole and complex ∞ as North pole); horizontally it is periodic and I have placed two periods side to side, whereas vertically it extends to infinity at the top and at the bottom, which corresponds to zooming infinitely far out or in, at a factor of exp(2π)≈535.5 for every size of a horizontal period. Horizontal lines (“parallels”) on the log map correspond to concentric circles around the target point, and vertical lines to radii emanating from it; and the anamorphosis preserves angles. Video number 2: varied shapes The second video is 3′09″ long. It centers on the target point −0.789374599271466936740382412558 +0.163089252677526719026415054868i and zooms to a factor of 2^91 or 2.5×10^27. It is shorter than the second, but possibly more varied. The music is an excerpt of Schumann's Scenes from Childhood. The YouTube version of the video is here (also embedded in small here on the right), but, as previously, it is of extremely poor quality. The high quality version is 44MB in size: here is the torrent to download it: again, pass this file to a BitTorrent program or, if you cannot, remove the .torrent extension. Video number 3: dramatic tension The third video is 3′41″ long. It centers on the target point −0.9230110468224410331799630273585336748656 +0.3103593603697618780906159981443973705961i and zooms to a factor of 2^107 or 1.6×10^32. It is not as varied as the first two, but possibly more “dramatic”. The music is an excerpt of Chopin's Étude op. 25 no. 12 in C minor. The YouTube version of the video is here (also embedded in small here on the right), but, as previously, it is of extremely poor quality. The high quality version is 46MB in size: here is the torrent to download it: again, pass this file to a BitTorrent program or, if you cannot, remove the .torrent extension. Video number 4: variations on a theme The fourth video is 4′04″ long. It centers on the target point −1.477110786384222313461222586803179083557 +0.003322002718062184557764259218386616609i and zooms to a factor of 2^118.5 or 4.7×10^35. It is not nearly as varied as the other videos, quite the contrary, it goes through what one might describe as variations on a theme. The music is an excerpt of J. S. Bach's Aria Variata. The YouTube version of the video is here (also embedded in small here on the right), but, as previously, it is of extremely poor quality. The high quality version is 62MB in size: here is the torrent to download it: again, pass this file to a BitTorrent program or, if you cannot, remove the .torrent extension. Still images I have put in this Flickr set a number of still images of the Mandelbrot set which were computed using the same program. For each image, I have indicated in the description (and the PNG comments) what the center point coordinate and scale are. Note that Flickr does not have a way to specify that an image is in the Public Domain, so I chose the closest I could (Creative Commons Attribution), but I still put these images in the Public Domain (assuming I even have to do that—the Mandelbrot set, after all, is a mathematical fact that can no more be copyrighted than a circle). The program, and the coloring The program I used to compute all the images can be downloaded from here, where its features are described in more detail. It is in the Public Domain, so you can play with it all you want. What is drawn, as is usual for the Mandelbrot set, is (an approximation of) the electrostatic potential produced by the Mandelbrot set, which is basically a continuous interpolation of the escape time (the number of iterations it takes for the point to leave the circle of a certain radius). This is all quite well explained on Wikipedia, so I won't repeat it. One thing which bears to be stated in more detail, however, is exactly how this is mapped to colors. Indeed, in the regions of the Mandelbrot set in which one typically wishes to zoom, the escape time of the points which do not belong to the Mandelbrot is not only very large (this means that the image is slow to compute), but also, because it is so large, it varies enormously across the image. A naïve approach to coloring—where the colors are chosen linearly with the escape time—will therefore rapidly lead to images where the colors vary so much from point to point that the whole thing looks like random noise (or, if it is heavily anti-aliased, just grey). One solution to this problem is to always map the interval of escape times found on the picture to the entire gamut of colors, but this is unsatisfactory for zoom animations (where it is disagreeable for the colors to vary with the zoom factor). Another solution is to compress the escape times with some concave function: here, a compromise must be reached between having too much compression (which will destroy render invisible the fine details connecting the bolder structures) and having too little (which will make the bolder structures look like white noise). I have chosen (as, I believe, fraqtive has) to take the square root of the escape time (the log function might also have been worth looking into) before mapping it to the color gradient (which also tries to reach a compromise between flattening out the fine details and varying too rapidly). So, essentially, regular changes in the color gradient correspond to a quadratic increase in the number of iterations. The program doesn't do anti-aliasing (unfortunately).	{"url":"http://www.madore.org/~david/math/mandelbrot.html","timestamp":"2024-11-09T14:32:02Z","content_type":"application/xhtml+xml","content_length":"36076","record_id":"<urn:uuid:9863ef3c-04a0-4152-b0b0-e8f2331b9cff>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00790.warc.gz"}
Free Sudoku Problems 2013-11 Sudoku Problems 2013-11-04 3D 3X3 Cube L(2,1) D(9,3,2,0,0,0) True 3 dimensional problem, albeit a very simple one. Each 3X3 square is one plane of a 3X3X3 cube. Consider the squares stacked on top of each other. There are 9 planes in this problem each of which is a 3X3 square which must have no repeating characters. A couple of the planes are highlighted in colour. Backslash L(2,1) D(18,13,12,0,0,0) Must have unique characters in each row and column but the 3 by 3 boxes are replaced by backslashes. These wrap left to right and top to bottom so the overall geometry is a doughnut (torus) Jigsaw L(2,1) D(23,10,9,0,0,0) As regular 9 by 9 but the normal 3X3 boxes are irregular shapes People L(2,3) D(21,9,9,1,1,0) The shapes have “lumps” on each side to become quite irregular. They look to me like little people. This problem wraps top to bottom and side to side so the overall geometry is like a doughnut Toroid H L(2,3) D(24,13,12,1,1,0) The normal 3X3 boxes are replaced by horizontal stripes (helices) which wrap around the problem. These stripes join the left side to the right side and the top to the bottom so that the overall geometry is a doughnut (torus). Sudoku Problems 2013-11-11 3D 4X4 Cube L(3,1) D(27,5,4,0,0,0) True 3 dimensional problem. Each 4X4 square is one plane of a 4X4X4 cube. Consider the squares stacked on top of each other. There are 12 planes in this problem each of which is a 4X4 square which must have no repeating characters. Some of the planes are highlighted. There are 16 possible characters so I use a standard hexadecimal character set: 0,1,2,…,9,a,…,f. Backslash L(2,1) D(19,6,6,0,0,0) Must have unique characters in each row and column but the 3 by 3 boxes are replaced by backslashes. These wrap left to right and top to bottom so the overall geometry is a doughnut (torus) Jigsaw L(2,1) D(24,5,4,0,0,0) As regular 9 by 9 but the normal 3X3 boxes are irregular shapes People L(2,3) D(21,14,14,1,1,0) The shapes have “lumps” on each side to become quite irregular. They look to me like little people. This problem wraps top to bottom and side to side so the overall geometry is like a doughnut Toroid V L(2,3) D(22,9,8,1,1,0) The normal 3X3 boxes are replaced by vertical stripes (helices) which wrap around the problem. These stripes join the left side to the right side and the top to the bottom so that the overall geometry is a doughnut (torus). Sudoku Problems 2013-11-18 3D 3X3 Cube L(2,1) D(9,3,2,0,0,0) True 3 dimensional problem, albeit a very simple one. Each 3X3 square is one plane of a 3X3X3 cube. Consider the squares stacked on top of each other. There are 9 planes in this problem each of which is a 3X3 square which must have no repeating characters. A couple of the planes are highlighted in colour. Diamond L(2,2) D(23,13,12,1,0,0) Here the shapes are pushed over to become diamonds. This wraps side to side but not top to bottom. Jigsaw L(2,1) D(23,6,6,0,0,0) As regular 9 by 9 but the normal 3X3 boxes are irregular shapes People L(2,3) D(20,10,10,1,1,0) The shapes have “lumps” on each side to become quite irregular. They look to me like little people. This problem wraps top to bottom and side to side so the overall geometry is like a doughnut Toroid H L(3,3) D(17,12,11,1,1,0) The normal 3X3 boxes are replaced by horizontal stripes (helices) which wrap around the problem. These stripes join the left side to the right side and the top to the bottom so that the overall geometry is a doughnut (torus). Sudoku Problems 2013-11-25 3D 4X4 Cube L(3,1) D(27,5,4,0,0,0) True 3 dimensional problem. Each 4X4 square is one plane of a 4X4X4 cube. Consider the squares stacked on top of each other. There are 12 planes in this problem each of which is a 4X4 square which must have no repeating characters. Some of the planes are highlighted. There are 16 possible characters so I use a standard hexadecimal character set: 0,1,2,…,9,a,…,f. Diamond L(2,2) D(26,11,10,1,0,0) Here the shapes are pushed over to become diamonds. This wraps side to side but not top to bottom. Jigsaw L(2,1) D(22,9,8,0,0,0) As regular 9 by 9 but the normal 3X3 boxes are irregular shapes The shapes have “lumps” on each side to become quite irregular. They look to me like little people. This problem wraps top to bottom and side to side so the overall geometry is like a doughnut Toroid V L(2,3) D(22,15,14,3,1,0) The normal 3X3 boxes are replaced by vertical stripes (helices) which wrap around the problem. These stripes join the left side to the right side and the top to the bottom so that the overall geometry is a doughnut (torus). Sudoku Solutions 2013-11-04 Sudoku Solutions 2013-11-11 Sudoku Solutions 2013-11-18 Sudoku Solutions 2013-11-25	{"url":"http://137consulting.com/sudoku-problems-and-solutions/problems-2012-2015/free-sudoku-problems-2013-11/","timestamp":"2024-11-10T19:24:25Z","content_type":"text/html","content_length":"60396","record_id":"<urn:uuid:7d15ef93-c8b8-4d12-ac48-a824157c20ad>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00666.warc.gz"}
Calculating Percentage good value \| Cognite Hub Question from an end user: What is the best way to calculate the percentage of good values for a time series over a year using Cognite Charts? I have a list of PSVs in various units that I need to see if we have data availability of at least 95% in 2022.	{"url":"https://hub.cognite.com/developer-and-user-community-134/calculating-percentage-good-value-1482?postid=3765","timestamp":"2024-11-11T00:39:29Z","content_type":"text/html","content_length":"195020","record_id":"<urn:uuid:2a8730ea-5eee-460a-8a32-3d4d2aea810b>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00554.warc.gz"}
Comments on Instructions Comments on Instructions Permalink Submitted by basheer mhd on Sat, 02/14/2015 - 15:03 Permalink Submitted by stuart on Sat, 02/14/2015 - 15:28 There is no off line version at present. Permalink Submitted by younesibrahim80 on Mon, 04/20/2015 - 02:40 Permalink Submitted by rajat.saraswala on Mon, 04/24/2017 - 05:41 hello sir, im electrical engineer.I need a hard copy of whole attachments. waiting for your kind responce. thank you. Permalink Submitted by Asamad on Thu, 11/23/2017 - 04:02 How to see the animations on iPhone and iPad ... Permalink Submitted by stuart on Wed, 02/14/2018 - 19:51 The animations are in flash that is not supported by the iPad. Unfortunately there are still no suitable substitutes. Permalink Submitted by Ryousuke Ishikawa on Tue, 02/27/2018 - 02:14 Nice to meet you. I am grateful to have studied with your wonderful teaching materials. Recently sliders and Flash animations on the web do not seem to work well. I tried the latest FlashPlayer with Chrome, IE and Firefox but it did not work. What should I do? Permalink Submitted by stuart on Wed, 02/28/2018 - 16:18 As far as I know they still work. Which page is the problem? Permalink Submitted by ganley0905 on Tue, 07/24/2018 - 10:32 This is a question, not a comment. Can I copy figures from one of your pages for use on a website I am building for my startup? I will acknowledge your site as the source. Thanks. J. T. Ganley Permalink Submitted by stuart on Thu, 02/28/2019 - 09:59 As long as you acknowledge the source of the images you may use figures for your site. If it is a website just make sure you have at a minimum Picture: www.pveducation.org And a link to the website. Permalink Submitted by PFL2020 on Mon, 03/02/2020 - 21:55 I'd love to be able to print sections instead of part by part. Permalink Submitted by TimCHEN on Wed, 12/30/2020 - 03:42 This website supplies abundant, detailed, and reliable resources for PV learning! I've studied some parts of the section "properties of sunlight". It helps me a lot in deed. Thanks! Permalink Submitted by holretz on Wed, 03/02/2022 - 17:15 I live in Aasiaat, latitude 68,7 degrees north. According to the curves for calculation of solar incidence, a solararray should be tilted around 55,5 degrees. It says somewhere else on one of the pages, that one should choose tilt equal to latitude. This is clearly not the case for my latitude, so I don't quite understand this statement....is it true for some limited range of latitudes ? Permalink Submitted by Peter W on Wed, 10/26/2022 - 07:20 I believe the formula given for solar radiation on a panel with arbitrary orientation and tilt is incorrect. I think it should be: Smodule = Sincident .cos(Ψ-θ).[cos(α)sin(β) + sin(α)Cos(β)] Reason: On the previous page, for a module directly facing the sun, at tilt β: Smodule = Sincident. sin(α+ β), where sin(α+ β) = cos(α)sin(β) + sin(α)Cos(β) For a module at orientation θ, Smodule = Sincident .cos(Ψ-θ). sin(α+ β), so Smodule = Sincident .cos(Ψ-θ).[cos(α)sin(β) + sin(α)Cos(β)], not as given. Permalink Submitted by petercl14 on Thu, 05/02/2024 - 03:36 Hi, Equations for the position of the sun,. elevation angle and azimuth. If I enter 9 for 9am in the morning being the local time in the equations I get accurate results. I compare the result with an application on the internet which gives the elevation and azimuth at my location and time. However, if I enter say the number 13 for 1pm I no longer get an accurate or comparable result. Can you please let me know what I should enter in the equations for pm times? Should this be perhaps a negative number? I never had to put 'am' after the 9 in the equation. I am making a computer application using the equations to simply put in the local time and days since the start of the year to give me the elevation and azimuth of the sun. These are the only variables in the equations. All the other data is fixed for my location. The application on the internet also probably uses these equations. Please let me know the local times I should enter for the pm time to get an accurate result. Thanks.	{"url":"https://www.pveducation.org/node/35/talk/35","timestamp":"2024-11-13T03:21:03Z","content_type":"text/html","content_length":"78814","record_id":"<urn:uuid:d31a3435-8d19-44dd-9d52-0f00f2200b7e>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00335.warc.gz"}
The Stacks project Lemma 35.23.12. The property $\mathcal{P}(f) =$“$f$ is of finite type” is fpqc local on the base. Proof. Combine Lemmas 35.23.1 and 35.23.10. $\square$ Comments (0) There are also: • 2 comment(s) on Section 35.23: Properties of morphisms local in the fpqc topology on the target Post a comment Your email address will not be published. Required fields are marked. In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). All contributions are licensed under the GNU Free Documentation License. In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 02KZ. Beware of the difference between the letter 'O' and the digit '0'. The tag you filled in for the captcha is wrong. You need to write 02KZ, in case you are confused.	{"url":"https://stacks.math.columbia.edu/tag/02KZ","timestamp":"2024-11-09T00:44:46Z","content_type":"text/html","content_length":"14252","record_id":"<urn:uuid:7915d7ef-65ac-4c25-8b02-ebabcf21cbec>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00088.warc.gz"}
Extract or Replace Matched Substrings regmatches {base} R Documentation Extract or Replace Matched Substrings Extract or replace matched substrings from match data obtained by regexpr, gregexpr, regexec or gregexec. regmatches(x, m, invert = FALSE) regmatches(x, m, invert = FALSE) <- value x a character vector. m an object with match data. invert a logical: if TRUE, extract or replace the non-matched substrings. value an object with suitable replacement values for the matched or non-matched substrings (see Details). If invert is FALSE (default), regmatches extracts the matched substrings as specified by the match data. For vector match data (as obtained from regexpr), empty matches are dropped; for list match data, empty matches give empty components (zero-length character vectors). If invert is TRUE, regmatches extracts the non-matched substrings, i.e., the strings are split according to the matches similar to strsplit (for vector match data, at most a single split is If invert is NA, regmatches extracts both non-matched and matched substrings, always starting and ending with a non-match (empty if the match occurred at the beginning or the end, respectively). Note that the match data can be obtained from regular expression matching on a modified version of x with the same numbers of characters. The replacement function can be used for replacing the matched or non-matched substrings. For vector match data, if invert is FALSE, value should be a character vector with length the number of matched elements in m. Otherwise, it should be a list of character vectors with the same length as m, each as long as the number of replacements needed. Replacement coerces values to character or list and generously recycles values as needed. Missing replacement values are not allowed. For regmatches, a character vector with the matched substrings if m is a vector and invert is FALSE. Otherwise, a list with the matched or/and non-matched substrings. For regmatches<-, the updated character vector. x <- c("A and B", "A, B and C", "A, B, C and D", "foobar") pattern <- "[[:space:]](,\|and)[[:space:]]" ## Match data from regexpr() m <- regexpr(pattern, x) regmatches(x, m) regmatches(x, m, invert = TRUE) ## Match data from gregexpr() m <- gregexpr(pattern, x) regmatches(x, m) regmatches(x, m, invert = TRUE) ## Consider x <- "John (fishing, hunting), Paul (hiking, biking)" ## Suppose we want to split at the comma (plus spaces) between the ## persons, but not at the commas in the parenthesized hobby lists. ## One idea is to "blank out" the parenthesized parts to match the ## parts to be used for splitting, and extract the persons as the ## non-matched parts. ## First, match the parenthesized hobby lists. m <- gregexpr("\$[^)]\$", x) ## Create blank strings with given numbers of characters. blanks <- function(n) strrep(" ", n) ## Create a copy of x with the parenthesized parts blanked out. s <- x regmatches(s, m) <- Map(blanks, lapply(regmatches(s, m), nchar)) ## Compute the positions of the split matches (note that we cannot call ## strsplit() on x with match data from s). m <- gregexpr(", ", s) ## And finally extract the non-matched parts. regmatches(x, m, invert = TRUE) ## regexec() and gregexec() return overlapping ranges because the ## first match is the full match. This conflicts with regmatches()<- ## and regmatches(..., invert=TRUE). We can work-around by dropping ## the first match. drop_first <- function(x) { if(!anyNA(x) && all(x > 0)) { ml <- attr(x, 'match.length') if(is.matrix(x)) x <- x[-1,] else x <- x[-1] attr(x, 'match.length') <- if(is.matrix(ml)) ml[-1,] else ml[-1] m <- gregexec("(\\w+) \$((?:\\w+(?:, )?)+)\$", x) regmatches(x, m) try(regmatches(x, m, invert=TRUE)) regmatches(x, lapply(m, drop_first)) ## invert=TRUE loses matrix structure because we are retrieving what ## is in between every sub-match regmatches(x, lapply(m, drop_first), invert=TRUE) y <- z <- x ## Notice list(...) on the RHS regmatches(y, lapply(m, drop_first)) <- list(c("<NAME>", "<HOBBY-LIST>")) regmatches(z, lapply(m, drop_first), invert=TRUE) <- list(sprintf("<%d>", 1:5)) ## With `perl = TRUE` and `invert = FALSE` capture group names ## are preserved. Collect functions and arguments in calls: NEWS <- head(readLines(file.path(R.home(), 'doc', 'NEWS.2')), 100) m <- gregexec("(?<fun>\\w+)\$(?<args>[^)])\$", NEWS, perl = TRUE) y <- regmatches(NEWS, m) ## Make tabular, adding original line numbers mdat <- as.data.frame(t(do.call(cbind, y))) mdat <- cbind(mdat, line=rep(seq_along(y), lengths(y) / ncol(mdat))) version 4.4.0	{"url":"https://stat.ethz.ch/R-manual/R-devel/library/base/html/regmatches.html","timestamp":"2024-11-01T22:34:43Z","content_type":"text/html","content_length":"7411","record_id":"<urn:uuid:9e37e4eb-e89a-4626-b23c-4fc025872595>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00843.warc.gz"}
Technical note: A note on an open-end bin packing problem We consider a variant of the classical one-dimensional bin packing problem, which we call the open-end bin packing problem. Suppose that we are given a list L = (p[1],p[2],...,p[n]) of n pieces, where p[j] denotes both the name and the size of the jth piece in L, and an infinite collection of infinite-capacity bins. A bin can always accommodate a piece if the bin has not yet reached a level of C or above, but it will be closed as soon as it reaches that level. Our goal is to find a packing that uses the minimum number of bins. In this article, we first show that the open-end bin packing problem remains strongly NP-hard. We then show that any online algorithm must have an asymptotic worst-case ratio of at least 2, and there is a simple online algorithm with exactly this ratio. Finally, we give an offline algorithm that is a folly polynomial approximation scheme with respect to the asymptotic worst-case ratio. • Approximation algorithms • Bin packing • Complexity ASJC Scopus subject areas • Software • General Engineering • Management Science and Operations Research • Artificial Intelligence Dive into the research topics of 'Technical note: A note on an open-end bin packing problem'. Together they form a unique fingerprint.	{"url":"https://experts.arizona.edu/en/publications/technical-note-a-note-on-an-open-end-bin-packing-problem","timestamp":"2024-11-05T21:34:29Z","content_type":"text/html","content_length":"51250","record_id":"<urn:uuid:07db4c43-bed3-4247-aeb4-2811b3d49159>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00560.warc.gz"}
PPT - Swarm Intelligence PowerPoint Presentation, free download - ID:2382569 1. Swarm Intelligence 虞台文 2. Content • Overview • Swarm Particle Optimization (PSO) • Example • Ant Colony Optimization (ACO) 3. Swarm Intelligence Overview 4. Swarm Intelligence • Collective system capable of accomplishing difficult tasks in dynamic and varied environments without any external guidance or control and with no central coordination • Achieving a collective performance which could not normally be achieved by an individual acting alone • Constituting a natural model particularly suited to distributed problem solving 5. Swarm Intelligence http://www.scs.carleton.ca/~arpwhite/courses/95590Y/notes/SI%20Lecture%203.pdf 10. Swarm Intelligence Particle Swarm Optimization (PSO) Basic Concept 11. The Inventors Russell Eberhart James Kennedy electrical engineer social-psychologist 12. Developed in 1995 by James Kennedy and Russell Eberhart. Particle Swarm Optimization (PSO) • PSO is a robust stochastic optimization technique based on the movement and intelligence of swarms. • PSO applies the concept of social interaction to problem solving. 13. PSO Search Scheme • It uses a number of agents, i.e., particles, that constitute a swarm moving around in the search space looking for the best solution. • Each particle is treated as a point in a N-dimensional space which adjusts its “flying” according to its own flying experience as well as the flying experience of other particles. 14. Particle Flying Model • pbest the best solution achieved so far by that particle. • gbest the best value obtained so far by any particle in the neighborhood of that particle. • The basic concept of PSO lies in accelerating each particle toward its pbest and the gbest locations, with a random weighted acceleration at each time. 16. Particle Flying Model • Each particle tries to modify its position using the following information: • the current positions, • the current velocities, • the distance between the current position and pbest, • the distance between the current position and the gbest. 18. * PSO Algorithm ** For each particle Initialize particle END Do For each particle Calculate fitness value If the fitness value is better than the best fitness value (pbest) in history set current value as the new pbest End Choose the particle with the best fitness value of all the particles as the gbest For each particle Calculate particle velocity according equation () Update particle position according equation (*) End While maximum iterations or minimum error criteria is not attained 30. Exercises • Compare PSO with GA • Can we use PSO to train neural networks? How? 31. Particle Swarm Optimization (PSO) Ant Colony Optimization (ACO) 32. Facts • Many discrete optimization problems are difficult to solve, e.g., NP-Hard • Soft computing techniques to cope with these problems: • Simulated Annealing (SA) • Based on physical systems • Genetic algorithm (GA) • based on natural selection and genetics • Ant Colony Optimization (ACO) • modeling ant colony behavior 34. Background • Introduced by Marco Dorigo (Milan, Italy), and others in early 1990s. • A probabilistic technique for solving computational problems which can be reduced to finding good paths through graphs. • They are inspired by the behaviour of ants in finding paths from the colony to food. 35. Typical Applications • TSP Traveling Salesman Problem • Quadratic assignment problems • Scheduling problems • Dynamic routing problems in networks 37. ACO Concept • Ants (blind) navigate from nest to food source • Shortest path is discovered via pheromone trails • each ant moves at random, probabilistically • pheromone is deposited on path • ants detect lead ant’s path, inclined to follow, i.e., more pheromone on path increases probability of path being followed 38. ACO System • Virtual “trail”accumulated on path segments • Starting node selected at random • Path selection philosophy • based on amount of “trail” present on possible paths from starting node • higher probability for paths with more “trail” • Ant reaches next node, selects next path • Continues until goal, e.g., starting node for TSP, reached • Finished “tour” is a solution 39. ACO System , cont. • A completed tour is analyzed for optimality • “Trail” amount adjusted to favor better solutions • better solutions receive more trail • worse solutions receive less trail higher probability of ant selecting path that is part of a better-performing tour • New cycle is performed • Repeated until most ants select the same tour on every cycle (convergence to 40. Ant Algorithm for TSP Randomly position m ants on n cities Loop for step = 1to n for k = 1 to m Choose the next city to move by applying a probabilistic state transition rule (to be described) end for end for Update pheromone trails Until End_condition 42. visibility : Ant Transition Rule Probability of ant k going from city i to j: the set of nodes applicable to ant k at city i 43. Ant Transition Rule • = 0 : a greedy approach • = 0 : a rapid selection of tours that may not be optimal. • Thus, a tradeoff is necessary. Probability of ant k going from city i to j: 44. Pheromone Update Q: a constant Tk(t): the tour of ant k at time t Lk(t): the tour length for ant k at time t	{"url":"https://fr.slideserve.com/josiah/swarm-intelligence","timestamp":"2024-11-03T13:50:24Z","content_type":"text/html","content_length":"95506","record_id":"<urn:uuid:0b39baa5-5293-4fff-82b5-ec16022f5757>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00885.warc.gz"}
The Dictionary Thread Well, here we go. Post updates that ought to go into the dictionary here. This thread is also for comments, suggestions, damnations, praises, requests to translate, and all that jazz. I'll post in here when I make The current version is 3.11, dated January 23rd, 2016 The Dothraki DictionaryThe English to Dothraki Dictionary (Note; Hrakkar took over the majority of the editing work on the dictionary from our beloved Lajaki, in September of 2011. Lajaki continues to provide technical advice and help when needed, and he is always welcome here!)	{"url":"https://forum.dothraki.org/index.php?PHPSESSID=69f809c316955635dacfe440a0138941&topic=35.0","timestamp":"2024-11-08T23:59:22Z","content_type":"application/xhtml+xml","content_length":"75349","record_id":"<urn:uuid:44890d21-3af0-45aa-9d45-af17610c7f66>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00165.warc.gz"}
Efficient all path score computations on grid graphs We study the Integer-weighted Grid All Paths Scores (IGAPS) problem, which is given a grid graph, to compute the maximum weights of paths between every pair of a vertex on the first row of the graph and a vertex on the last row of the graph. We also consider a variant of this problem, periodic IGAPS, where the input grid graph is periodic and infinite. For these problems, we consider both the general (dense) and the sparse cases. For the sparse IGAPS problem with 0-1 weights, we give an O(r log^3 (n^2/r)) time algorithm, where r is the number of (diagonal) edges of weight 1. Our result improves upon the previous O(n√r) result by Krusche and Tiskin for this problem. For the periodic IGAPS problem we give an O(Cn^2) time algorithm, where C is the maximum weight of an edge. This improves upon the previous O(C ^2n^2) algorithm of Tiskin. We also show a reduction from periodic IGAPS to IGAPS. This reduction yields o(n^2) algorithms for this problem. Publication series Name Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics) Volume 7922 LNCS ISSN (Print) 0302-9743 ISSN (Electronic) 1611-3349 Conference 24th Annual Symposium on Combinatorial Pattern Matching, CPM 2013 Country/Territory Germany City Bad Herrenalb Period 17/06/13 → 19/06/13 ASJC Scopus subject areas • Theoretical Computer Science • General Computer Science Dive into the research topics of 'Efficient all path score computations on grid graphs'. Together they form a unique fingerprint.	{"url":"https://cris.bgu.ac.il/en/publications/efficient-all-path-score-computations-on-grid-graphs-6","timestamp":"2024-11-03T09:45:24Z","content_type":"text/html","content_length":"58768","record_id":"<urn:uuid:f5c46c0a-f19d-4f47-bb7e-8816df608c6d>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00682.warc.gz"}
Water Treatment and dosage calculations For the safe water intake human beings have adopted various processes of water treatment and have determined chemical dosage calculations to treat the water before use for removal of physical, biological and chemical contamination. Number of researches culminating in theories have been introduced to curb the menace of water borne diseases due to intake of contaminated water. Water is an imperative for all forms of living beings. Almost 71% of our globe is water and 97% of which is found in the oceans meant to of no use (unless radically treated) for drinking, agriculture and other needs that a fresh water serves to living beings especially human beings. Only 3% is the quantum of fresh water on the earth. This least proportion of available fresh water is also inflicted with variety of contaminants which are essentially needed to be removed for safe drinking water intake of human beings. For the treatment of water various methods and articles (such as Alum, Chlorine, HTH and other chemical solutions) are used. This blog intends to provide information about chemical dosage calculations and its application for water treatment. Chemical dosages for water treatment are measured in ppm (parts per million) or mg/l (milligram per liter). The metric term of ppm is mg/l and it is equal to ppm. 1 ppm = 1 mg/l Parts per million (ppm) is elaborated in terms of weight as pounds per million pounds. It means one pound of chemical added to one million pounds of water. Since each gallon of water weighs 8.34 pounds, thus one million gallons of water weighs 8.34 million pounds and would require 8.34 pounds of chemical to obtain the dosage of 1 ppm. In this way the number of pounds of chemical required to achieve a certain dosage can be determined by multiplying the ppm by number of million of gallons to be treated and then by 8.34 lbs/gal (as weight of 1 gallon of water is 8.34 pounds). It must be noted that while calculating the chemical under certain dosage the amount of water to be treated is always taken in terms of million gallons or it may be million gallons per day (mgd). Thus the formula to determine chemical quantity for certain dosage is as given below; ppm x mgd x 8.34 = pounds of chemical per day mg/l x mgd x 8.34 = pounds of chemical per day See the example how this formula works in calculating chemical quantity. Example: If the dosage is 2.4 mg/l in 1200,000 gal/day, how much chlorine in lbs is needed per day? Convert gal/day to mgd 1200000/1000000 = 1.2 mgd Now put figures in formula mg/l x mgd x 8.34 = pounds per day (Chlorine) 2.4 x 1.2 x 8.34 = 24.01 lbs/day It means 24 lbs of chlorine per day is required at the dosage of 2.4 ppm for flow of 1.2 mgd of water. If instead of chlorine gas, HTH is used that constitutes 65% - 70% chlorine in each pound of it. In such case the amount of HTH must be calculated by dividing the pounds of chlorine needed by the proportion of chlorine in it i.e 65% - 70%. Example: A tank is 60' in diameter and 20' high and is dosed with 60 ppm chlorine. How many pound of 70% HTH is needed? It tells we have to determine the quantity of HTH in pounds which contains 70% chlorine. First we have to determine the quantity of chlorine needed Formula is same as ppm x mgd x 8.34 = pounds of chlorine here ppm is given but mgd is required to be calculated first from tank dimensions Volume of tank = 60 x 60 x 0.785 x 20 = 56,520 cubic feet convert cubic feet to gallons 56,520 x 7.48 = 422,770 gallons convert gallons to million gallons 422,770/1000000 = 0.422 million gallons Now put them in formula ppm x mgd x 8.34 = pounds of chlorine 60 x 0.422 x 8.34 = 211 lbs of chlorine HTH contains 70% of chlorine Find lbs of HTH Lbs of chlorine / 0.70 211 / 0.70 = 301 lbs of HTH Example: A 12" pipe is 1900 feet long and to be disinfected with 40 ppm of 65% HTH. How many pounds of HTH are needed? Here HTH constitutes 65% of chlorine Find gallons of water in pipe to be disinfected 12 x 12 x 0.0408 x 1900 = 11,163 gallons convert gallons to million gallons 11,163 / 1000000 = 0.011 million gallons Now put them in formula for finding chlorine quantity ppm x mgd x 8.34 = pounds of chlorine 40 x 0.011 x 8.34 = 3.66 lbs of chlorine Now find HTH amount 3.66 / 0.65 = 5.63 lbs of HTH In water treatment and dosage calculations, many a times you have to come across with chemical solutions instead of raw chemical. When chemical solutions are used, the weight of that solution will be more than the weight of a gallon of water. The weight of a gallon of liquid is found by multiplying weight of gallon of water (8.34 lbs) by the Specific Gravity of the solution. If S.G is not given then weight of solution is assumed to be 8.34 lbs/gallon same as of water. Example: A chlorine pump is feeding 10 % bleach at a dosage of 5 mg/l. The S.G of the bleach is 1.14. If 2,000,000 gallons are treated in 16 hours, how many gallons per hour pump is feeding? Convert gallons to million gallons 2,000,000 = 2 mgd Find chlorine amount 5 x 2 x 8.34 = 83 lbs of chlorine Find bleach amount 83 / 0.1 = 830 lbs of bleach Find the weight of a gallon of bleach 1.14 x 8.34 = 9.50 lbs/gal Find the gallons of bleach 830 / 9.50 = 87.3 gal. of bleach Find gallons per hour 87.3 / 16 = 5.45 gal/hr Example : A chlorine pump is feeding 14% bleach at a dosage of 2 mg/l or 2 ppm. If the flow is 1,250,000 gallons per day (gpd) , how many gallons per hour is the pump feeding? S.G 1.14 Convert gpd to mgd 1,250,000 = 1.25 mgd Find chlorine required 2 x 1.25 x 8.34 = 20.85 lbs of chlorine Find bleach amount for getting chlorine required 20.85 / 0.14 = 149 lbs of bleach Find weight of a gallon of bleach 1.14 x 8.34 = 9.5 lbs/gal Find gallons of bleach 149 / 9.5 = 15.68 gal Find gallons per hour 15.68 / 24 = 0.65 gal / hour Now you know how to find quantity of chemical required for disinfection. Also you know how much gallons of chemical solution are required per day or per hour. After getting these quantities, now you need to know, how a chemical feed pump is to be calibrated to feed the dosage in ml/min in water on the basis of determined calculations of solutions in gallon per day. Remember 1 gallon = 3.785 liters = 3785 milliliters 1 day = 1440 minutes So If you take 3785 ml/gal and divide it by 1440 min/day, the conversion for gal/day to ml/min can be determined. 3785 / 1440 = 2.6 ml/min/gal/day ml/min = gal/day x 2.6 Example: A 20% available flouride solution is used to dose 2,500,000 gpd at 450 ppb (parts per billion). The S.G is 1.26. How many ml/min is the pump feeding? Convert ppb to ppm 1 ppm = 1000 ppb 450 ppb = 0.45 ppm Convert gpd to mgd 2,500,000 = 2.5 mgd Note: Here flouride is referred instead chlorine Find lbs of flouride ppm x mgd x 8.34 = lbs of flouride 0.45 x 2.5 x 8.34 = 9.38 lbs / day Find lbs of flouride solution 9.38 / 0.2 = 47 lbs of flouride solution Find weight of a gallon of flouride solution 1.26 x 8.34 = 10.5 lbs/gal Find the gallons of flouride solutions 47 / 10.5 = 4.48 gallons per day Find pump feed in ml / min ml/min = gpd x 2.6 ml/min = 4.48 x 2.6 11.65 ml/min Thus the pump is required to be calibrated to feed the solution in water at 11.65 ml/min. Example: An 18% available Alum solution is used to dose 800,000 gpd at 25 mg/l. Determine the pump feeding of solution in ml/min? No S.G is given so assume weight of solution is 8.34 lbs/gal. Convert gpd to mgd 800,000 gpd = 0.8 mgd Find lbs of Alum 25 x 0.8 x 8.34 = 167 lbs/day of Alum Find lbs of Alum solution 167/0.18 = 928 lbs/day of Alum solution The weight of gallon of Alum solution is assumed to be 8.34 lbs/gal Find the gallons per day of Alum solution 928/8.34 = 111 gpd Find pump feed in ml/min ml/min = gpd x 2.6 ml/min = 111 x 2.6 288 ml/min Example: A chlorine pump is feeding 12% bleach at a dosage of 2.4 mg/l. If flow is 1,250,000 gpd. how many ml / min the pump would be feeding? S.G = 1.14. Convert gpd to mgd 1,250,000 gpd = 1.25 mgd Find lbs of chlorine 2.4 x 1.25 x 8.34 = 25 lbs/day of chlorine Find lbs of bleach 25/0.12 = 208 lbs/day of bleach Find weight of gallon of bleach 1.14 x 8.34 = 9.5 lbs/gal Find gallons of bleach solution 208/9.5 = 22 gpd Find ml/min 22 gpd x 2.6 = 57 ml/min Example: A system has a well that produce 200 gpm and a 1500 gallon storage tank.There are 120 homes on the system and the average daily consumption is 350 gal/home. A chlorine dosage of 1.3 ppm is maintained using 65% HTH. How many pounds of HTH to be purchased each year? Find system consumption 120 homes x 350 gal/day/home = 42000 gpd Convert gpd to mgd 42000 gpd = 0.042 mgd Find lbs/day of chlorine 1.3 x 0.042 x 8.34 = 0.45 lbs of chlorine Find lbs of HTH for each day 0.45 / 0.65 = 0.7 lbs/day of HTH Find lbs/year of HTH 0.7 x 365 = 255.5 lbs/year. I hope this will serve the purpose of imparting knowledge and remain beneficial for the students and water operators in their practical application of water treatment processes with the knowledge of dosage calculations involved in it. I have made it so simple and articulate in order to enable the students or other interested groups to understand the entire concept and dynamics involved in water treatment and dosage calculations. I welcome any further query in this regard by your comments and make sure to respond accordingly. 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Lecture Note 23 Hi professor. I was wondering if there is a mistake in the lecture note 23. Please see the following attachment. There should be no "r" after Cn, right? I was wondering, in lecture 23 in the explanation leading up to equation 11 it says that if considering outside the disk that we need to discard terms the are singular as r=0 is that supposed to be Looking at equation 6 and 7 is this coming from equations 17 and 18 from lecture 13? using $l=2\pi$? wouldn't $\lambda$ then be $(n/4)^2$ or am I misunderstanding something? No, $l=\pi$ as for periodic b.c. interval's length is $2l$ in contrast to Dirichlet or Neumann' b.c.	{"url":"https://forum.math.toronto.edu/index.php?PHPSESSID=72o65tgcttr6nnr1a9oj1guvq7&topic=171.0;prev_next=prev","timestamp":"2024-11-04T02:16:32Z","content_type":"application/xhtml+xml","content_length":"38708","record_id":"<urn:uuid:9289fe24-5d69-453d-828a-8317d1dfd18a>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00221.warc.gz"}
Arithmetic and Maple Notation Next: Mistakes and Errors Up: Maple: An Introduction Previous: Getting Started The basic arithmetic operators and constants known to Maple are Notice that Maple does arithmetic with integers exactly. I.e. exact arithmetic is used instead of decimal arithmetic. Use decimal numbers if you want decimal numbers. Examples > 23+2/7; > 2.03.0+2.0/7; This principle works for formulae too. Use the evalf function if you want a decimal approximation. > sin(Pi/3); 1/2 3 > evalf("); There is no limit on the length of integers in Maple. It is quite common to compute with integers several hundred digits long. It is also possible to do decimal arithmetic to more than the default 10 digits of precision. You can compute > 2^100; > Digits := 50: > evalf( sin(Pi/3) ); > Digits := 10: In the above examples we have used the colon to terminate a command. Use the colon : instead of a semicolon ; if you don't want to see the output.	{"url":"http://thproxy.jinr.ru/Documents/MapleV/introEnglish/section3_4.html","timestamp":"2024-11-02T14:14:37Z","content_type":"text/html","content_length":"3011","record_id":"<urn:uuid:035d4f52-9b63-4224-9ffd-208468e2b2af>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00564.warc.gz"}
The 30-Second Trick for Discrete Mathematics for Computer Science Discrete Mathematics for Computer Science – the Story By using all these powerful strategies, you can absolutely boost your eBook reading experience to a good extent. When you attempt to solve, you are going to realize that there aren't any incorrect answers. Unfortunately, there's no book that adequately covers all of the material within this course at the most suitable level. Discrete Mathematics for Computer Science Secrets An excellent comprehension of math is essential for every computer scientist, and the math requirement is starting to become more diverse. do my homework for me Indeed, some easy math skills are vital. What's the correct method to learn math. All mathematics classrooms ought to be playful classrooms. Inside this edition further exercises are added, particularly, at the routine level. Theres many distinct areas to discrete math, and several fantastic books. 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There are plenty of concepts you would like to understand for this module. There's clearly enough material here for an exact meaty undergraduate course. These types of random processes are called point strategy. Many algorithms of computer science are made from these sorts of topics. Complete solutions are supplied for nearly all self-tests. What You Should Do to Find Out About Discrete Mathematics for Computer Science Before You're Left Behind A set is a range of unique objects. It is reputed to contain its elements. The previous part is to fix the combinations formula. The Do's and Don'ts of Discrete Mathematics for Computer Science TeschlThe book was created for students of computer science. At exactly the same time, you will probably be finishing off the math classes. At the exact same time, you will likely be finishing off the math classes. Besides knowing that you're the perfect student for the program, it's essential to be conscious that the program is acceptable for you. 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A helpful skill to get before taking this class is knowledge of proofs and the way to program in a minumum of one programming language like C, Python, or Java. It may be a good alternative for numerous people. It is difficult to remember and you've got to be in a position to do everything without help to pass the program. Most people started to sweat as they hear about doing it. So long as their description is accurate and they're describing the exact thing that everybody else is attempting to describe, they will develop exactly the same outcome. The approach that's being used is different from conventional learning procedure. There's no potential solution. That's the question that I'll attempt to answer today. Math for a superb deal of people is an enormous scary monster. Ok, I Think I Understand Discrete Mathematics for Computer Science, Now Tell Me About Discrete Mathematics for Computer Science! In fact, Stanford's encyclopedia entry on set theory is an incredible place to commence. If you take into consideration the idea of number, students wish to know the common notation. Prof. Denis R. Hirschfeldt University of Chicago It's one of the greatest textbooks in the present sector. http://185.18.206.156/~daf/wp-content/uploads/2016/07/דפרון.jpg 0 0 maya dubrovsky http://185.18.206.156/~daf/wp-content/uploads/2016/07/דפרון.jpg maya dubrovsky2019-08-19 11:14:192019-08-19 11:14:19 The 30-Second Trick for Discrete Mathematics for Computer Science	{"url":"http://www.dafron-tech.com/the-30-second-trick-for-discrete-mathematics-for-computer-science/","timestamp":"2024-11-10T14:14:44Z","content_type":"text/html","content_length":"46444","record_id":"<urn:uuid:4af982aa-9da7-4246-8751-bc03df0bcffd>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00870.warc.gz"}
What are the values of a and b if f(x)=(2a−3)x2+bx−1 is equival... \| Filo Question asked by Filo student What are the values of and if is equivalent to Not the question you're searching for? + Ask your question Video solutions (1) Learn from their 1-to-1 discussion with Filo tutors. 2 mins Uploaded on: 5/10/2023 Was this solution helpful? Found 2 tutors discussing this question Discuss this question LIVE for FREE 7 mins ago One destination to cover all your homework and assignment needs Learn Practice Revision Succeed Instant 1:1 help, 24x7 60, 000+ Expert tutors Textbook solutions Big idea maths, McGraw-Hill Education etc Essay review Get expert feedback on your essay Schedule classes High dosage tutoring from Dedicated 3 experts Students who ask this question also asked Stuck on the question or explanation? Connect with our Mathematics tutors online and get step by step solution of this question. 231 students are taking LIVE classes Question Text What are the values of and if is equivalent to Updated On May 10, 2023 Topic Functions Subject Mathematics Class Grade 12 Answer Type Video solution: 1 Upvotes 131 Avg. Video Duration 2 min	{"url":"https://askfilo.com/user-question-answers-mathematics/what-are-the-values-of-and-if-is-equivalent-to-35303537303837","timestamp":"2024-11-06T18:54:23Z","content_type":"text/html","content_length":"111438","record_id":"<urn:uuid:1f0a86eb-f0ef-4545-bcf7-994aa8376610>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00341.warc.gz"}
Simple tools for understanding risks: from innumeracy to insight Education And Debate Simple tools for understanding risks: from innumeracy to insight BMJ 2003 327 doi: https://doi.org/10.1136/bmj.327.7417.741 (Published 25 September 2003) Cite this as: BMJ 2003;327:741 1. Gerd Gigerenzer, director (gigerenzer{at}mpibberlin.mpg.de)1, 2. Adrian Edwards, reader2 1. ^1Centre for Adaptive Behaviour and Cognition, Max Planck Institute for Human Development, Lentzeallee 94, 14195 Berlin, Germany 1. Correspondence to: G Gigerenzer Bad presentation of medical statistics such as the risks associated with a particular intervention can lead to patients making poor decisions on treatment. Particularly confusing are single event probabilities, conditional probabilities (such as sensitivity and specificity), and relative risks. How can doctors improve the presentation of statistical information so that patients can make well informed decisions? The science fiction writer H G Wells predicted that in modern technological societies statistical thinking will one day be as necessary for efficient citizenship as the ability to read and write. How far have we got, a hundred or so years later? A glance at the literature shows a shocking lack of statistical understanding of the outcomes of modern technologies, from standard screening tests for HIV infection to DNA evidence. For instance, doctors with an average of 14 years of professional experience were asked to imagine using the Haemoccult test to screen for colorectal cancer.1 2 The prevalence of cancer was 0.3%, the sensitivity of the test was 50%, and the false positive rate was 3%. The doctors were asked: what is the probability that someone who tests positive actually has colorectal cancer? The correct answer is about 5%. However, the doctors' answers ranged from 1% to 99%, with about half of them estimating the probability as 50% (the sensitivity) or 47% (sensitivity minus false positive rate). If patients knew about this degree of variability and statistical innumeracy they would be justly alarmed. Statistical innumeracy is often attributed to problems inside our minds. We disagree: the problem is not simply internal but lies in the external representation of information, and hence a solution exists. Every piece of statistical information needs a representation–that is, a form. Some forms tend to cloud minds, while others foster insight. We know of no medical institution that teaches the power of statistical representations; even worse, writers of information brochures for the public seem to prefer confusing representations.2 3 Here we deal with three numerical representations that foster confusion: single event probabilities, conditional probabilities, and relative risks. In each case we show alternative representations that promote insight (table). These “mind tools” are simple to learn. Finally, we address questions of the framing (expression) and manipulation of information and how to minimise these effects. Single event probabilities The statement “There is a 30% chance of rain tomorrow” is a probability statement about a single event: it will either rain or not rain tomorrow. Single event probabilities are a steady source of miscommunication because, by definition, they leave open the class of events to which the probability refers. Some people will interpret this statement as meaning that it will rain tomorrow in 30% of the area, others that it will rain 30% of the time, and a third group that it will rain on 30% of the days like tomorrow. Area, time, and days are examples of reference classes, and each class gives the probability of rain a different meaning. The same ambiguity occurs in communicating clinical risk, such as the side effects of a drug. A psychiatrist prescribes fluoxetine (Prozac) to his mildly depressed patients. He used to tell them that they have a “30% to 50% chance of developing a sexual problem” such as impotence or loss of sexual interest.2 Hearing this, patients were anxious. After learning about the ambiguity of single event probabilities, the psychiatrist changed how he communicated risk. He now tells patients that of every 10 people who take fluoxetine three to five will experience a sexual problem. Patients who were informed in terms of frequencies were less anxious about taking Prozac. Only then did the psychiatrist realise that he had never checked what his patients had understood by “a 30% to 50% chance of developing a sexual problem.” It turned out that many had assumed that in 30% to 50% of their sexual encounters something would go awry. The psychiatrist and his patients had different reference classes in mind: the psychiatrist was thinking in terms of patients, but the patients were thinking in terms of their own sexual encounters. Frequency statements always specify a reference class (although the statement may not specify it precisely enough). Thus, misunderstanding can be reduced by two mind tools: specifying a reference class before giving a single event probability or only using frequency statements. Conditional probabilities The chance of a test detecting a disease is typically communicated in the form of a conditional probability, the sensitivity of the test: “If a woman has breast cancer the probability that she will have a positive result on mammography is 90%.” This statement is often confused with: “If a woman has a positive result on mammography the probability that she has breast cancer is 90%.” That is, the conditional probability of A given B is confused with that of B given A.4 Many doctors have trouble distinguishing between the sensitivity, the specificity, and the positive predictive value of test–three conditional probabilities. Again, the solution lies in the representation. Consider the question “What is the probability that a woman with a positive mammography result actually has breast cancer?” The box shows two ways to represent the relevant statistical information: in terms of conditional probabilities and natural frequencies. The information is the same (apart from rounding), but with natural frequencies the answer is much easier to work out. Only seven of the 77 women who test positive actually have breast cancer, which is one in 11 (9%). Natural frequencies correspond to the way humans have encountered statistical information during most of their history. They are called “natural” because, unlike conditional probabilities or relative frequencies, they all refer to the same class of observations.5 For instance, the natural frequencies “seven women” (with a positive mammogram and cancer) and “70 women” (with a positive mammogram and no breast cancer) both refer to the same class of 1000 women. In contrast, the conditional probability 90% (the sensitivity) refers to the class of eight women with breast cancer, but the conditional probability 7% (the specificity) refers to a different class of 992 women without breast cancer. This switch of reference class can confuse the minds of doctors and patients alike. Figure 1 shows the responses of 48 doctors, whose average professional experience was 14 years, to the information given in the box, except that the statistics were a base rate of cancer of 1%, a sensitivity of 80%, and a false positive rate of 10%.1 2 Half the doctors received the information in conditional probabilities and half in natural frequencies. When asked to estimate the probability that a woman with a positive result actually had breast cancer, doctors who received conditional probabilities gave answers that ranged from 1% to 90%, and very few gave the correct answer of about 8%. In contrast most doctors who were given natural frequencies gave the correct answer or were close to it. Simply stating the information in natural frequencies turned much of the doctors' innumeracy into insight, helping them understand the implications of a positive result as it would arise in practice. Presenting information in natural frequencies is a simple and effective mind tool to reduce the confusion resulting from conditional probabilities.6 This is not the end of the story regarding the communication of risk (which requires adequate exploration of the implications of the risk for the patient concerned, as described elsewhere in this issue7), but it is an essential foundation. Relative risks Women aged over 50 years are told that undergoing mammography screening reduces their risk of dying from breast cancer by 25%. Women in high risk groups are told that bilateral prophylactic mastectomy reduces their risk of dying from breast cancer by 80%. These numbers are relative risk reductions. The confusion produced by relative risks has received more attention in the medical literature than that of single event or conditional probabilities.9 10 Nevertheless, few patients realise that the impressive 25% figure means an absolute risk reduction of only one in 1000: of 1000 women who do not undergo mammography about four will die from breast cancer within 10 years, whereas out of 1000 women who do three will die.11 Similarly, the 80% figure for prophylactic mastectomy refers to an absolute risk reduction of four in 100: five in 100 women in the high risk group who do not undergo prophylactic mastectomy will die of breast cancer, compared with one in 100 women who have had a mastectomy. One reason why most women misunderstand relative risks is that they think that the number relates to women like themselves who take part in screening or who are in a high risk group. But relative risks relate to a different class of women: to women who die of breast cancer without having been screened. Two ways of representing the same statistical information Conditional probabilities • The probability that a woman has breast cancer is 0.8%. If she has breast cancer, the probability that a mammogram will show a positive result is 90%. If a woman does not have breast cancer the probability of a positive result is 7%. Take, for example, a woman who has a positive result. What is the probability that she actually has breast cancer? Natural frequencies • Eight out of every 1000 women have breast cancer. Of these eight women with breast cancer seven will have a positive result on mammography. Of the 992 women who do not have breast cancer some 70 will still have a positive mammogram. Take, for example, a sample of women who have positive mammograms. How many of these women actually have breast cancer? Confusion caused by relative risks can be avoided by using absolute risks (such as one in 1000) or the number needed to treat or to be screened to save one life (the NNT, which is the reciprocal of the absolute risk reduction and is thus essentially the same representation as the absolute risk). However, health agencies typically inform the public in the form of relative risks.2 3 Health authorities tend not to encourage transparent representations and have themselves sometimes shown innumeracy, for example when funding proposals that report benefits in relative rather than absolute risks because the numbers look larger.12 For authorities that make decisions on allocation of resources the population impact number (the number of people in the population among whom one event will be prevented by an intervention) is a better means of putting risk into perspective.13 The reference class In all these representations the ultimate source of confusion or insight is the reference class. Single event probabilities leave the reference class open to interpretation. Conditional probabilities such as sensitivity and specificity refer to different classes (the class of people with and without illness, respectively), which makes their mental combination difficult. Relative risks often refer to reference classes that differ from those to which the patient belongs, such as the class of patients who die of cancer rather than those who participate in screening. Using transparent representations such as natural frequencies clarifies the reference class. Framing is the expression of logically equivalent information (whether numerical or verbal) in different ways.14 Studies of the effects of verbal framing on interpretation and decision making initially focused on positive versus negative framing and on gain versus loss framing.15 Positive and negative frames refer to whether an outcome is described, for example, as a 97% chance of survival (positive) or a 3% chance of dying (negative). The evidence is that positive framing is more effective than negative framing in persuading people to take risky treatment options.16 17 However, gain or loss framing is perhaps even more relevant to communicating clinical risk, as it concerns the implications of accepting or declining tests. Loss framing considers the potential losses from not having a test, such as, in the case of mammography, loss of good health, longevity, and family relationships. Loss framing seems to influence the uptake of screening more than gain framing (the gains from taking a test, such as maintenance of good health).18 Visual representations may substantially improve comprehension of risk.19 They may enhance the time efficiency of consultations. Doctors should use a range of pictorial representations (graphs, population figures) to match the type of risk information that the patient most easily understands.20 It may not seem to matter whether the glass is half full or half empty, yet different methods of presenting risk information can have important effects on outcomes among patients. That verbal and statistical information can be presented in two or more ways means that an institution or screening programme may choose the one that best serves its interests. For instance, a group of gynaecologists informed patients in a leaflet of the benefits of hormone replacement therapy in terms of relative risk (large numbers) and of harms in absolute risk (small numbers).2 Pictorial representations of risk are not immune to manipulation either. For example, different formats such as bar charts and population crowd figures could be used.21 Or the representation could appear to support short term benefits from one treatment rather than long term benefits from another.22 Furthermore, within the same format, changing the reference class may produce greatly differing perspectives on a risk and may thus affect patients' decisions. Figure 2 relates to the effect of treatment with aspirin and warfarin in patients with atrial fibrillation. On the left side of the figure the effect of treatment on a particular event (stroke or bleeding) is shown relative to the class of people who have not had the treatment (as in relative risk reduction). On the right side the patient can see the treatment effect relative to a class of 100 untreated people who have not had a stroke or bleeding (as in absolute risk reduction). Summary points • The inability to understand statistical information is not a mental deficiency of doctors or patients but is largely due to the poor presentation of the information • Poorly presented statistical information may cause erroneous communication of risks, with serious consequences • Single event probabilities, conditional probabilities (such as sensitivity and specificity), and relative risks are confusing because they make it difficult to understand what class of events a probability or percentage refers to • For each confusing representation there is at least one alternative, such as natural frequency statements, which always specify a reference class and therefore avoid confusion, fostering insight • Simple representations of risk can help professionals and patients move from innumeracy to insight and make consultations more time efficient • Instruction in efficient communication of statistical information should be part of medical curriculums and doctors' continuing education The wide scope for manipulating representations of statistical information is a challenge to the ideal of informed consent.2 16 Where there is a risk of influencing outcomes and decisions among patients, professionals should consistently use representations that foster insight and should balance the use of verbal expressions–for example, both positive and negative frames or both gain and loss frames. The dangers of patients being misled or making uninformed decisions in health care are countless. One of the reasons is the prevalence of poor representations. Such confusion can be reduced or eliminated with simple mind tools.2 23 Human beings have evolved into good intuitive statisticians and can gain insight, but only when information is presented simply and effectively.24 This insight is then the platform for informed discussion about the significance and burden of risks and the implications for the individual or family concerned. It also makes the explanation of diseases and their treatment easier. Instruction in the efficient communication of statistical information should be part of medical curriculums and continuing education for doctors. • Contributors and sources The research on statistical representations was initially funded by the Max Planck Society and has been published in scientific journals as well summarised in GG's book Reckoning With Risk: Learning to Live With Uncertainty. The work on framing is based on research by AE. • Competing interests None declared. View Abstract	{"url":"https://www.bmj.com/content/327/7417/741?ijkey=eeb192bf530b0c81819512015be964c903bc64b2&keytype2=tf_ipsecsha","timestamp":"2024-11-05T09:03:28Z","content_type":"text/html","content_length":"175428","record_id":"<urn:uuid:8d5e48aa-d56d-434a-b554-43caeaa22e9c>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00226.warc.gz"}
7th Grade Math - Solving Rate Problems This post explains and gives practice opportunities related to TEKS 7.4D: solve problems involving ratios, rates, and percents, including multi-step problems involving percent increase and percent decrease, and financial literacy problems Students solve problems involving a constant rate of change. Students also apply their understanding of rate to solve problems involving percent increase and decrease. STAAR Practice Between 2016 and 2024 (including redesign practice), this readiness standard has been tested 19 times on the STAAR test. Videos explaining the problems can be found below. If you'd rather take a quiz over these questions, click here. The videos below are linked to the questions in the quiz as answer explanations after the quiz is submitted. To view all the posts in this 7th grade TEKS review series, click here.	{"url":"https://www.fiveminutemath.net/post/7th-grade-math-solving-rate-problems","timestamp":"2024-11-01T20:05:10Z","content_type":"text/html","content_length":"1050492","record_id":"<urn:uuid:3b2ea7d6-0f8d-4f1c-8c0b-a63f965ea7f7>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00695.warc.gz"}
Perimeter Pre-Assessment \| Quizalize Feel free to use or edit a copy includes Teacher and Student dashboards Measure skills from any curriculum Tag the questions with any skills you have. Your dashboard will track each student's mastery of each skill. With a free account, teachers can • edit the questions • save a copy for later • start a class game • automatically assign follow-up activities based on students’ scores • assign as homework • share a link with colleagues • print as a bubble sheet • What is the sum of the numbers 5+4+5+4? • Q1 What is the sum of the numbers 5+4+5+4? • Q2 What is the sum of the numbers 6+3+2+4+8? • Q3 What is the perimeter of the given shape? • Q4 What is the perimeter of the given shape? • Q5 Find the perimeter of the figure. Each unit is 1 cm. • Q6 Find the perimeter of the figure. Each unit is 1 cm. • Q7 Find the unknown side lengths (s) , with the perimeter being 36cm. • Q8 Find the unknown side length (f), with the perimeter being 38cm.	{"url":"https://resources.quizalize.com/view/quiz/perimeter-preassessment-937e0e9c-6499-4fa6-b7b0-a2187b116d2d","timestamp":"2024-11-12T20:17:52Z","content_type":"text/html","content_length":"96236","record_id":"<urn:uuid:ba878ece-53b8-49de-a0ec-798e7bd33fa2>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00721.warc.gz"}
A trace formula problem for foliations online \| 2020-11-25 - 14:30 Jesús A. Álvarez Lopes University of Santiago de Compostela, Spain Let $M$ be a smooth closed manifold, and let $\psi:M\to M$ be a smooth map. A fixed point $p$ of $\psi$ is said to be simple when 1 is not an eigenvalue of the tangent map $\psi_$ on the tangent space $T_pM$. If all fixed points are simple, then $\psi$ is said to be simple. In this case, the Leftchetz trace formula describes the supertrace of the induced homomorphism $\psi^$ on the de Rham cohomology $H^(M)$ using infinitesimal data from the fixed points. Next, consider a smooth flow $\{\phi^t\}$ on $M$ instead of just a single map. A fixed point $p$ of $\{\phi^t\}$ is simple when it is a simple fixed point of $\phi^t$ for any $t\ne0$. Similarly, a closed orbit $c$ of period $T$ is called simple when, for any $x\in c$, 1 is not an eigenvalue of the map induced by $\phi^T_$ on the normal bundle of $c$ at $x$. If all fixed points and closed orbits are simple, then $\phi$ is called a simple flow. It would be useful to have version of the Leftchetz trace formula for simple flows, involving infinitesimal data from the fixed points and closed orbits. But it does not make sense because the induced action $\{\phi^{t\,}\}$ on $H^(M)$ is trivial (the flow itself defines a homotopy between every $\phi^t$ and the identity map). Finally, consider also a smooth foliation $\mathcal F$ of codimension one on $M$. It is said that the flow $\{\phi^t\}$ is foliated when every $\phi^t$ maps leaves to leaves. For simple foliated flows, a version of the Leftchetz trace formula was conjectured by Christopher Deninger, using some leafwise version of the de Rham cohomology, and using some supertrace whose values are distributions. The talk will be about our efforts to give appropriate definitions of this leafwise cohomology and distributional supertrace, and to prove this trace formula. (joint work with Yuri Kordyukov and Eric Leichtnam)	{"url":"https://cmat.uminho.pt/index.php/events/trace-formula-problem-foliations","timestamp":"2024-11-04T23:24:59Z","content_type":"text/html","content_length":"14768","record_id":"<urn:uuid:6ee2c619-ccac-479a-89ee-da1bcfa7ad2e>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00150.warc.gz"}
Use MathML today, with CSS fallback! Use MathML today, with CSS fallback! Reading Time: 3 minutes These days, I’m working on the slides for my next talk, “The humble border-radius”. It will be about how much work is put into CSS features that superficially look as simple as border-radius, as well as what advances are in store for it in CSS Backgrounds & Borders 4 (of which I’m an editor). It will be fantastic and you should come, but this post is not about my talk. As you may know, my slides are made with HTML, CSS & JavaScript. At some point, I wanted to insert an equation to show how border-top-left-radius (as an example) shrinks proportionally when the sum of radii on the top side exceeds the width of the element. I don’t like LaTeX because it produces bitmap images that don’t scale and is inaccessible. The obvious open standard to use was MathML, and it can even be directly embedded in HTML5 without all the XML cruft, just like SVG. I had never written MathML before, but after a bit of reading and poking around existing samples, I managed to write the following MathML code: <math display="block"> I was very proud of myself. My first MathML equation! It’s actually pretty simple when you get the hang of it: <mi> is for identifiers, <mo> for operators and those are used everywhere. For more complex stuff, there’s <mfrac> for fractions (along with <mrow> to denote the rows), <msqrt> for square roots and so on. It looked very nice on Firefox, especially after I applied Cambria Math to it instead of the default Times-ish font: However, I soon realized that as awesome as MathML might be, not not all browsers had seen the light. IE10 and Chrome are the most notable offenders. It looked like an unreadable mess in Chrome: There are libraries to make it work cross-browser, the most popular of which is MathJax. However, this was pretty big for my needs, I just wanted one simple equation in one goddamn slide. It would be like using a chainsaw to cut a slice of bread! The solution I decided to go with was to use Modernizr to detect MathML support, since apparently it’s not simple at all. Then, I used the .no-mathml class in conjunction with selectors that target the MathML elements, to mimic proper styling with simple CSS. It’s not a complete CSS library by any means, I just covered what I needed for that particular equation and tried to write it in a generic way, so that if I need it in future equations, I only have to add rules. Here’s a screenshot of the result in Chrome: It doesn’t look as good as Firefox, but it’s decent. You can see the CSS rules I used in the following Dabblet: Obviously it’s not a complete MathML-to-CSS library, if one is even possible, but it works well for my use case. If I have to use more MathML features, I’d write more CSS rules. The intention of this post is not to provide a CSS framework to use as a MathML fallback, but to show you a solution you could adapt to your needs. Hope it helps!	{"url":"https://lea0.verou.me/2013/03/use-mathml-today-with-css-fallback/","timestamp":"2024-11-05T18:36:30Z","content_type":"text/html","content_length":"70638","record_id":"<urn:uuid:df38a81e-1c5c-4776-86a0-0c312bd09ecd>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00431.warc.gz"}
Convert integer to English Given a non-negative integer up to $999\,999\,999$, write it in English. The input number can be in any form other than English (though you'll typically want to use the native integer type of your language). The numbers must be formatted as follows: • Two-digit numbers must be written with dashes where applicable e.g. twenty-five, not twenty five. • powers of ten start with “one”, e.g. one hundred, not just hundred. • The tens/ones part, if not zero, must be separated from the hundreds part with and, e.g. two hundred and five, not two hundred five. • The above rules are applied also to the factors of thousand and million. • moreover the global tens/ones part has to be separated by and whenever anything larger is present, even if the hundreds part is missing. • Adjacent words must be separated by a single space, or a single dash without space where applicable. • Leading and/or trailing spaces are not allowed. However, if the result is printed, a single trailing newline (not preceded by any spaces) is allowed. Ungolfed reference implementation Test cases: 0 zero 1 one 4 four 10 ten 16 sixteen 42 forty-two 125 one hundred and twenty-five 1024 one thousand and twenty-four 1100 one thousand one hundred 12345 twelve thousand three hundred and forty-five 1000000 one million 1000001 one million and one 7023000 seven million twenty-three thousand 11012021 eleven million twelve thousand and twenty-one 999999999 nine hundred and ninety-nine million nine hundred and ninety-nine thousand nine hundred and ninety-nine 3 answers 塊 Function 匱 Begin with an empty list ❸곴김分倘 Divide number by 1,000,000; if quotient is non-zero: ⓶❹演 Apply function to the quotient (millions place) 긆륩닆롩닶밎 " million" 併 Concat 鈉 Add "... million" to the list 不終 End if ⓷곴김剩 Take modulo 1,000,000 Repeat the above for the thousands and hundreds Now for the tens/ones places ⓶글會⓶ Join the list with spaces (S) 倘 If there's stuff remaining (tens/units) ❷長是 If the length of (S) is non-zero 긆굮뉂밀 Push an " and " ⓶終 End if " thir four fif six seven eigh nine" (A) This is shared between the -teens and -tys ❷굀瀰是 If greater than or equal to 20 ❶ Copy the shared string 銅 drop the 9th character (four -> for) 긂건덶녮 " twen" 融 Concatenate 壹坼 Split into a list ❸겠分 Divide by 10 掘 Index into the list 덇밉 Push "ty" ⓸겠剩 Take modulo 10 倘 If non-zero 껐⓶⓹ Push a dash "-" 逆⓸終 End if " one two three four five six seven eight nine ten eleven twelve" 融 Concatenate with the shared string (A) 壹坼 Split into a list ❷掘 Index into the list ⓶곀大是 If greater than 12 덆녥닠 Push "teen" 終 End if 不 Otherwise (if zero) 梴沒 If the length of (S) is zero 뎦녲닰 Push "zero" 終 End if 終 End if 終 End function 演 Call function Works for me The following users marked this post as Works for me: def f(n): if n<1:return'zero' t=1000;m=tt;r='r fif six seven eigh nine ';b=f'one two three four five six seven eight nine ten eleven twelve thir fou{r}twen thir fo{r}'.split(' ');s=' '.join([f(k:=n//m)+' million'][:k]+[f(k:=(n:=n%m)//t)+' thousand'][:k]+[f(k:=(n:=n%t)//100)+' hundred'][:k]);n%=100;s+=' and '(sn>'') if n>19:s+=b[n//10+17]+'ty'+'-'((n:=n%10)>0) return s+b[n-1]+'teen'(n>12) Python 3, [S:596:S] [S:589:S] [S:585:S] [S:482:S] [S:468:S] [S:467:S] 429 bytes def f(x): s="";i="r fif six seven eigh nine";a=f"zero one two three four five six seven eight nine ten eleven twelve thir fou{i} twen thir fo{i}".split();t=1000;c=tt if x>=c:s=f(x//c)+" million";x%=c if x>=t:s+=" "(s>"")+f(x//t)+" thousand";x%=t if sx:s+=" " if n:s+=a[x//100]+" hundred";x%=100 if sx:s+=" "n+"and " return x<20 and s+a[x](not(x<1)s)+"teen"(x>12)or s+a[x//10+18]+"ty"+("-"+a[x%10])(x%10>0) Golfed 7 bytes thanks to @celtschk's advice. Golfed a massive 103 bytes thanks to @Moshi's advice. Golfed another 14 bytes thanks to @Moshi's advice. Golfed 38 bytes thanks to @celtschk's advice. Sign up to answer this question »	{"url":"https://codegolf.codidact.com/posts/284391","timestamp":"2024-11-03T00:56:30Z","content_type":"text/html","content_length":"90555","record_id":"<urn:uuid:64de6ff8-d9e8-41d5-a31d-7c91d44b2271>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00735.warc.gz"}
Debugging ML gradient [Work Log] Debugging ML gradient November 13, 2013 Implemented end-to-end ML gradient in curve_ml_derivative.m. Now testing. Fixed some obvious math errors, changes reflected in writeup. Now getting close results, but still getting some noticible error, see below. (Green is reference, blue is testing). Results are qualitatively close, but enough error to suggest a bug Debugging so far: • two different impementations for analytical gradient • two different implementations for numerical gradient • one-sided and two-sided numerical gradient. • several delta sizes for numerical gradient {0.0001, 0.0001, ..., 0.1, 1.0} • Using emperical $\Delta$' (from finite differences) for analytical gradient. • using both cholesky and direct method for matrix inversion (testing for numerical issues). • sanity check: used intermediate values from gradient computation to compute function output. To try: • use numerical gradient for K' instead of from $\Delta'$. Is it possible we're not handling XYZ independence properly? Noticed that using a really large delta (~1.0) actually improves resuls. Is it possible we're seeing precision errors being exacerbated somewhere in the end-to-end formula? Strategy: pick gradient element with mode error and run the following test. For each derivative component (dK, dU, dV, dg), compare against reference to determine where the error is being introduced. Index #22 dK/dt has 1e-4 error in on-diagonal. Off diagonals max out at 1e-10. delta: 0.01 on-diagonal error ~ 1e-3 below-and-right < 1e-4 delta: 0.001 on-diagonal error ~ 1e-4 other error < 1e-10 delta: 0.0001 on-diagonal error ~ 1e-5 below and right ~ 1e-6 delta: 0.00001 on-diagonal error ~ 1e-6 below and right ~ 1e-4 Decreasing delta improves on-diagonal, makes below-and-right worse. This is weird that we're even getting error in dK/dt, because it passed our unit test. Well, $\Delta'$ passed our unit test, but that's basically the same thing... However, reduced error at delta of 1e-3 seems to agree with our end-to-end test. So maybe this is the culprit. It's also surprising that there's so much fluctuation as delta changes. The computation for K isn't that involved, and we shouldn't be hitting the precision limit yet. However, the values do get pretty large, so maybe that's a factor. Magnitude of the original matrix does seem to be a factor. Look at this slice of the error matrix (dK_test - dK_ref): Compare that to the diagonal of the matrix we did finite differences on. This is basically a plot of the cubed index values. The error seems to increases in lockstep with the magnitude of the original values (note the jumps occur at similar positions). I guess this is to be expected, but I was surprised at the magnitude. I'm still curious why the error starts to climb exactly at index #22, i.e. the index we're differentiating with respect to. This plot should drive home the relationship between index valuea end error. Definitely a linear relationship after index #22. A back of the envelope error analysis suggests that below index 22, the analytical derivative's approximation error is not a function of X, but above index 22, it's a linear function of X. This is a pretty reasonable explanation, although I couldn't get the exact numbers to explain the slope of the line ( the slope seems high). But at this hour I wouldn't trust my error analysis as far as I could throw it, quantitatively speaking. We can attempt to place an upper bound on the error estimate by propagating the error in K' through the differential formula for g'. Assume every nonzero element of K has error of 3e-5 (the maximum we observed emperically). Let this error matrix be $\epsilon$, and it has the same banded structure as $K'$. Then we can replace $K'$ with $\epsilon$ in the formula for $g'$ (formula (1) in the writeup) to get the upper bound error on our data-set. \[ \text{max error} = \frac{1}{2}z^\top \epsilon z \tag{1}\\ \] >> 0.5 * z' * Epsilon * z ans = We can conclude that the error we're observing is coming from somewhere else. To conclude for tonight, we're seeing some error in dK/ds, but probably nothing out of the ordinary, and it has low enough error that we can hopefully ignore it. Lets look a the outher sources of error tomorrow, i.e. U' and V' Posted by Kyle Simek blog comments powered by	{"url":"http://vision.cs.arizona.edu/ksimek/research/2013/11/13/work-log/","timestamp":"2024-11-07T04:07:08Z","content_type":"application/xhtml+xml","content_length":"11490","record_id":"<urn:uuid:b0f2a68d-b8a9-445b-963c-7f4db93cabb2>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00125.warc.gz"}
Microscopic Cross-section \| Definition & Examples \| nuclear-power.com Microscopic Cross-section The effective target area in m^2 presented by a single nucleus to an incident neutron beam is denoted the microscopic cross-section, σ. The microscopic cross-sections characterize interactions with single isotopes and are a part of data libraries, such as ENDF/B-VII.1. Barn – Unit of Cross-section The cross-section is typically denoted σ and measured in units of the area [m^2]. But a square meter (or centimeter) is tremendously large compared to the effective area of a nucleus. It has been suggested that a physicist once referred to the measure of a square meter as being “as big as a barn” when applied to nuclear processes. The name has persisted, and microscopic cross-sections are expressed in terms of barns. The standard unit for measuring a nuclear cross-section is the barn, equal to 10^−28 m² or 10^−24 cm². It can be seen the concept of a nuclear cross-section can be quantified physically in terms of “characteristic target area”, where a larger area means a larger probability of interaction. The various microscopic cross-section for uranium-235 and incident thermal neutron. Typical Values of Microscopic Cross-sections See also: JANIS (Java-based Nuclear Data Information Software) Theory of Microscopic Cross-section The extent to which neutrons interact with nuclei is described in terms of quantities known as cross-sections. Cross-sections are used to express the likelihood of particular interaction between an incident neutron and a target nucleus. It must be noted this likelihood does not depend on real target dimensions. In conjunction with the neutron flux, it enables the calculation of the reaction rate, for example, to derive the thermal power of a nuclear power plant. The standard unit for measuring the microscopic cross-section (σ-sigma) is the barn, equal to 10^-28 m^2. This unit is very small. Therefore barns (abbreviated as “b”) are commonly used. The cross-section σ can be interpreted as the effective ‘target area’ that a nucleus interacts with an incident neutron. The larger the effective area, the greater the probability of reaction. This cross-section is usually known as the microscopic cross-section. The concept of the microscopic cross-section is therefore introduced to represent the probability of a neutron-nucleus reaction. Suppose that a thin ‘film’ of atoms (one atomic layer thick) with N[a] atoms/cm^2 is placed in a monodirectional beam of intensity I[0]. Then the number of interactions C per cm^2 per second will be proportional to the intensity I[0] and the atom density N[a]. We define the proportionality factor as the microscopic cross-section σ: σ[t] = C/N[a].I[0] To be able to determine the microscopic cross-section, transmission measurements are performed on plates of materials. Assume that if a neutron collides with a nucleus, it will either be scattered into a different direction or be absorbed (without fission absorption). Assume that N (nuclei/cm^3) of the material will then be N.dx per cm^2 in the layer dx. Only the neutrons that have not interacted will remain traveling in the x-direction. This causes the intensity of the un-collided beam will be attenuated as it penetrates deeper into the material. Then, according to the definition of the microscopic cross-section, the reaction rate per unit area is Nσ Ι(x)dx. This is equal to the decrease of the beam intensity, so that: -dI = N.σ.Ι(x).dx Ι(x) = Ι[0]e^-N.σ.x It can be seen that whether a neutron will interact with a certain volume of material depends not only on the microscopic cross-section of the individual nuclei but also on the density of nuclei within that volume. It depends on the N.σ factor. This factor is therefore widely defined, and it is known as the macroscopic cross-section. The difference between the microscopic and macroscopic cross-sections is extremely important. The microscopic cross-section represents the effective target area of a single nucleus. In contrast, the macroscopic cross-section represents the effective target area of all of the nuclei contained in a certain volume. Microscopic cross-sections constitute key parameters of nuclear fuel. In general, neutron cross-sections are essential for reactor core calculations and part of data libraries, such as ENDF/B-VII.1. The neutron cross-section is variable and depends on: • Target nucleus (hydrogen, boron, uranium, etc.). Each isotope has its own set of cross-sections. • Type of the reaction (capture, fission, etc.). Cross-sections are different for each nuclear reaction. • Neutron energy (thermal neutron, resonance neutron, fast neutron). For a given target and reaction type, the cross-section is strongly dependent on the neutron energy. In the common case, the cross-section is usually much larger at low energies than at high energies. This is why most nuclear reactors use a neutron moderator to reduce the neutron’s energy and thus increase the probability of fission, essential to produce energy and sustain the chain reaction. • Target energy (temperature of target material – Doppler broadening). This dependency is not so significant, but the target energy strongly influences the inherent safety of nuclear reactors due to a Doppler broadening of resonances. Microscopic cross-section varies with incident neutron energy. Some nuclear reactions exhibit very specific dependency on incident neutron energy. This dependency will be described in the example of the radiative capture reaction. The radiative capture cross-section represents the likelihood of a neutron radiative capture as σ[γ]. The following dependency is typical for radiative capture. It definitely does not mean that it is typical for other types of reactions (see elastic scattering cross-section or (n, alpha) reaction cross-section). The capture cross-section can be divided into three regions according to the incident neutron energy. These regions will be discussed separately. • 1/v Region • Resonance Region • Fast Neutrons Region Doppler Broadening of Resonances In general, Doppler broadening is the broadening of spectral lines due to the Doppler effect caused by a distribution of kinetic energies of molecules or atoms. In reactor physics, a particular case of this phenomenon is the thermal Doppler broadening of the resonance capture cross-sections of the fertile material (e.g.,, ^238U or ^240Pu) caused by the thermal motion of target nuclei in the nuclear fuel. Doppler effect improves reactor stability. Broadened resonance (heating of a fuel) results in a higher probability of absorption, thus causing negative reactivity insertion (reduction of reactor The Doppler broadening of resonances is an important phenomenon that improves reactor stability because it accounts for the dominant part of the fuel temperature coefficient (the change in reactivity per degree change in fuel temperature) in thermal reactors and makes a substantial contribution in fast reactors as well. This coefficient is also called the prompt temperature coefficient because it causes an immediate response to changes in fuel temperature. The prompt temperature coefficient of most thermal reactors is negative. See also: Doppler Broadening. It was written, in some cases, the amount of absorption reactions is dramatically reduced despite the unchanged microscopic cross-section of the material. This phenomenon is commonly known as the resonance self-shielding and also contributes to reactor stability. There are two types of self-shielding. • Energy Self-shielding. • Spatial Self-shielding. See also: Resonance Self-shielding An increase in temperature from T[1] to T[2] causes the broadening of spectral lines of resonances. Although the area under the resonance remains the same, the broadening of spectral lines causes an increase in neutron flux in the fuel φ[f](E), increasing the absorption as the temperature increases.	{"url":"https://www.nuclear-power.com/nuclear-power/reactor-physics/nuclear-engineering-fundamentals/neutron-nuclear-reactions/microscopic-cross-section/","timestamp":"2024-11-04T11:25:59Z","content_type":"text/html","content_length":"146898","record_id":"<urn:uuid:4049b58e-43d7-4b63-a419-f3d022ae2c21>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00658.warc.gz"}
Graphs and Tables - AP's \| Stage 5 Maths \| HK Secondary S4-S5 Compulsory Think about the generating rule for a particular arithmetic progression given by $t_n=2n+3$tn=2n+3. We can cleverly rearrange this equation to $t_n=5+\left(2n-2\right)$tn=5+(2n−2) . By taking out common factors on the last two terms, the expression becomes: Comparing this to the general formula for the $n$nth term of an AP, given by $t_n=a+\left(n-1\right)d$tn=a+(n−1)d , we can immediately see that the first term is $5$5 and the common difference is Thus any generating rule of the form $t_n=dn+k$tn=dn+k where $d$d and $k$k are constants can be shown to be an arithmetic sequence. Just like the equation $y=mx+c$y=mx+c is drawn as a straight line, so the arithmetic progression given by $t_n=dn+k$tn=dn+k is plotted as a series of points that are all in a straight line. The first term is represented by the left-most point shown $\left(t_1=5\right)$(t1=5) . The gradient of the marked points measured, as the vertical distance between the points, is the common difference. In our example, the line of points is rising, so this indicates a positive common difference. In other instances, the line might be falling and this would indicate a negative common In the example above, where , the common difference is immediately recognizable as the coefficient of . A simple way of finding the first term is to evaluate as . We could place these values in a table as follows: n 1 2 3 4 5 t[n] 5 7 9 11 13 Arithmetic sequences are said to grow linearly literally meaning ‘in a straight line’. We find applications of linear growth in many areas of life, including simple interest earnings, straight-line depreciation, monthly rental accumulation and many others. Whenever something grows or diminishes in constant quantities over equal time periods, then that growth or fall is said to be linear.	{"url":"https://mathspace.co/textbooks/syllabuses/Syllabus-99/topics/Topic-1489/subtopics/Subtopic-17719/?activeTab=theory","timestamp":"2024-11-10T04:27:51Z","content_type":"text/html","content_length":"655290","record_id":"<urn:uuid:36093ffb-6f90-4ff8-9c09-3dd00daba7d6>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00377.warc.gz"}
Parallel universes of the photon Being able to demonstrate the existence of Parallel Universes (all real alternative destinies in each respective dimension) can open the door to the discovery of a simple theory of everything able to provide plausible answers to the great questions of existence. The Theory of Relativity, Quantum Theory and Information Theory (which defines the amount of information as a form of energy, sharing the concept of entropy with the Physical Sciences including phenomena asymmetric with respect to the apparent direction of time) are based Aware of the need for an observer reference frame. Without the determination of an observer of Reference (Portal of Existence?) None of these theories could be properly formulated. Relativity, Quantum Physics and Information Theory, as well as elegant mathematical proofs have been made sophisticated demonstration experiments successfully. Therefore, these theories are a well-established scientific heritage and likely irrefutable. It turns out that in a Photon time as we know it does not flow, as, traveling at the speed of light while traveling huge distances using any even billions of our years ... because of relativistic time dimension, the instant of time of departure will be the same instant of time at the end of travel (relativistic time zero) while the spatial dimension in line with the direction of motion, becomes infinitesimal, tending to zero, with corresponding mass perceptible only in the dimension of the inertial photon (related to the relative energy-perceived frequency another inertial state) This reveals the existence of worlds the size turned over (with mirror inversion mass-energy) that could be locally similar to our world, when viewed in the inertial frame in which they are at rest. In such worlds the straight-geodesic path in light speed and the whole history of a photon are condensed into a point mass, where the local time apparently would flow in the transverse direction with respect to our dimensional world. In a reference in which photons would be almost stationary mass (without frequency-energy) throughout their history and spatial path made in our world would be absolutely undeterminable. It can reasonably be assumed that all time points (instants quantum) in the history of a photon in our reference dimensional turn into many states of existence parallel (in reference to light speed) of which only one will be perceptible in each alternate reality, characterized by scrolling apparently relativistic cross time than our stationary reference frame. This conjecture on Parallel Universes of Photon is based on the fact that the Relativity is nothing but geometry, for which nothing is lost in the transition between two systems relativistic effect rendered transverse speed light. Thus, a light path with a history of spatial points traversed in a given time, converted into relativistic size, appear as a material in which eternity will correspond to a moment ... and all the information is missing spanderebbe in diverse states of existence in containment extra dimension of Parallel Universes of Photon. (An infinite number of alternative fates, reflect the many parallel The Theory of Everything should not miss a comprehensive evaluation of the hypothesis of Parallel Universes, a hypothesis that is supported here by a coherent logical reasoning, but that would require ingenious experiments to be able to seriously test. The classical theory of relativity is incomplete and difficult to understand due to the fact that the idea of space-time contractions and expansions that it is based does not take into account the additional dimension of Parallel Universes, also appearing in the definition of the Observer approximate Aware Reference as a tool that would be more convenient to define geometric: "Portal of To define a simple and elegant the "Multiverse" you should exclude any temporal dimension, because, apparently, there is no time, as it would be an internal property to the portal of Existence (interaction world-consciousness) as each sequential quantum leap towards the more geometrically adjacent Parallel Universe (among many) reduces the probability reversible causes (and entropy) thus creating the illusion of time as if it were a size. Perhaps we are to confirm a controversial insight of the philosopher Parmenides, which claimed about 2400 years ago: "The movement is an illusion ... everything is still!" Aldo Monticelli - TOP ARTICLES, PHOTOGALLERY, SYNCHRONICITY and... DONATIONS BUTTONS!!! -	{"url":"https://www.aldo-monticelli.com/news/parallel-universes-of-the-photon/","timestamp":"2024-11-04T21:05:51Z","content_type":"text/html","content_length":"38031","record_id":"<urn:uuid:843c5a95-f46b-48c3-aeed-d0207ecfb9fe>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00440.warc.gz"}
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The Null Space ~ Minimum Flow Logistics in Excel Business Insights Transportation costs are a burden on our margins and minimizing them is an effective way to improve profitability without sacrificing in other areas. Minimum Cost Network Flow Models help us get a handle on our transportation costs. Microsoft Excel has a powerful built in optimizer called the “solver” that can make short work of these type of optimization problems with a little bit of modeling preparation. Let’s explore how we can use Excel’s built in solver to find the optimal routing that minimizes transportation costs for a small distribution network. Below is a network model containing two plants (P), one warehouse (W), and two retail customers (R). There are a number of possible routes, some are bidirectional and others are unidirectional. Plants and retailers can transfer between each other, but goods only flow into and out of the warehouse. Our goal is to minimize transportation costs given route capacity. We determine that by using the optimal routes we can minimize our shipping costs to $19,750. Follow along below to see how we set up this model and found the solution. Transportation Cost Matrix The first step in setting up our logistics model is to create a matrix of nodes that can accept inflows or outflows. Based on the network above P1 (Node 1) can ship to P2 (Node 2) at a cost of $30, and W1 can ship to R1 with a cost of $10. We grey out nodes which have no connected routes (also know as edges in network graph models). The numbered indexes will be useful for the =INDEX() formula in Excel when we want to pull these values out and assign them to routes. Network Flow Capacity and Costs For simplicity we define our route capacity to be 150, and simply set all route capacities to =D13 to indicate that all routes have the same capacity. Of course if trucks assigned to the warehouse were smaller we could lower the capacity of those routes. The network routes are simply all possible connections going both directions. Node 1 (P1) has possible destinations 2, 3, 4, and 5. This is repeated for each node. Unit costs can be taken from the Model Transportation Matrix with the formula =INDEX($D$5:$H$9,B17,C17) or if you prefer named ranges you could give the matrix a name by using the”Name Manager” under the Formula tab. In which case it would be unnecessary to fix the rows and columns with $ (F2 Key). This formula simply looks up the cost from the matrix given the index of the origin and destination nodes and enters it under UnitCost to indicate the route cost. The Flow column can initially be set to all zeros or some other reasonable values. These are known as “changing cells” because solver will change these values to find the optimal solution. We colour them blue to indicate their role in the model. Finally, total cost is =SUMPRODUCT(D17:D28,E17:E28), that is UnitCost* Flow in row 17 + UnitCostFlow in row 18 +…+ UnitCostFlow in the final row. SUMPRODUCT is a useful function that computes the total sum of two column arrays that are multiplied together element-wise such as SUM(Cost*Quantity). We colour Total Cost yellow to indicate that it is the target cell, also called the objective function, to be minimized. This is the number we want solver to minimize. Network Flow Constraints The most important part of any optimization problem is setting constraints. We have three types of nodes that operate in different ways. Plant Nodes Our plant nodes (1 and 2) only have outflows subject to the capacity of the plant. In order to find the Net Outflow for Plant 1 we want to sum all of the flows originating from 1, and subtract all of the flows that have destinations at 1. We can use the formula: =SUMIF($B$17:$B$28,M16,$E$17:$E$28)-SUMIF($C$17:$C$28,M16,$E$17:$E$28). This is the same as SUMIF(Origins, Node1, Flows) – SUMIF(Destinations, Node1, Flows) which can be set up in the name manager if you are so inclined. Warehouse Nodes Our warehouse only has flow through, we can think of this as a warehouse that is perpetually full and will only ship product if new product arrives to replenish the stock. It requires no capacity so this is set to zero. The formula for warehouse is identical to the plant nodes. =SUMIF($B$17:$B$28,M20,$E$17:$E$28)-SUMIF($C$17:$C$28,M20,$E$17:$E$28) Retailer Nodes These nodes are our customers, and they have a total demand which matches our total capacity. If this was not the case we would not be able to solve the problem without some modifications. We may explore these in a future article. The formula for retail nodes is the opposite of the production nodes because they only have net inflows. Net inflow must be greater than or equal to the customer demand. We can model this with SUMIF (Destinations, Node4, Flow) – SUMIF(Origins, Node4, Flow). These formulas can just be copied down to similar nodes as long as we fix columns and rows appropriately with the $ operator. Locating Solver The solver can be found under the data tab if it has been activated as an add-in. This add-in ships as part of modern versions of Excel but may need to be activated. Installing Solver If you don’t see the solver under the data tab we can install it by simply going to File > Options > Add-ins and hitting the “Go..” button for Excel Add-ins. This will present a checkbox from which to install desktop add-ins. While you are at it the Analysis Toolpack is excellent for creating histograms and doing basic statistics in Excel. Preparing Solver The solver dialog is where we will set up our minimum network flow model. We have set some named ranges just to make things a bit more human readable but this is not required. The first step is to set the objective, that is the target cell for TotalCost we coloured yellow earlier. We want to minimize this cost number so we select “min” as the type of objective. The second step is to tell Solver where our changing cells are. Those are in our flow column and we coloured them blue earlier to indicate they are changing cells. These are the cells solver will try different values for as it searches for a solution to minimize the total cost. Solver Constraints Next we need to set our model constraints which we discussed above. We do this by clicking on “Add” and then entering the ranges of values and constraints in the following dialog. They can be entered in succession by pressing “add” in the dialog and then closing when all constraints are entered. For example, all flows must be less than or equal to the corresponding capacity we set so we enter the range of flows and the range of capacities, and specify the <= operator. Similarly, warehouse flows must equal zero, so we specify Net Flows for the warehouse (N20) and it’s requirement of zero (P20) with the = operator. For the plants the net outflows must be less than or equal to capacities, and for retail customers the net inflows must be greater than or equal to the demand. Finally we select a solving method. In this case we will use the default Simplex LP which is a well known and very efficient way of solving linear programming problems. Running Solver Once everything has been set up we can run our model by clicking on “solve.” We have reset all of the blue flow values to zero. This results in no total cost and no net inflows to meet customer demand. We will then run solver to find the optimal solution given our Solver cheerfully tells us it has found a solution that satisfies our constraints, and the solution it found is $19,750 in total cost which now appears in the yellow target cell C30. We can also see in our constraint table that net outflows from our plant nodes match capacity, our warehouse has zero net outflows, and our retailers demand has been satisfied. Therefore, we can minimize our shipping cost in this network by using route 1:3 for 150 units, 1:5 for 100 units, and so on. Excel’s solver puts powerful logistics capabilities into the hands of businesses of all sizes, it is a great option to get a handle on distribution networks. Although modeling transportation in Excel is a lot of fun but this isn’t the end of what we can do with Minimum Cost Network Models. These models can be extended to areas such as task assignments, equipment replacement, water pipeline transport, and minimum distance problems.	{"url":"https://data-science.io/minimum-flow-logistics-in-excel/","timestamp":"2024-11-06T04:21:33Z","content_type":"text/html","content_length":"95526","record_id":"<urn:uuid:88c0fd7a-bea8-4e8f-b576-20126179a016>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00774.warc.gz"}
Ternary Diagrams How Do I Use Ternary Diagrams? Depicting Three-Component Systems in the Earth Sciences Ternary Diagrams are also known as: triangular diagrams ternary plots three-component diagrams An Introduction to Ternary Diagrams Ternary diagrams are graphical representations used to visualize systems with three components. Ternary diagrams allow geoscientists to represent the relative proportions of three components in a system. In order to plot data on a ternary diagram, the components are normalized to 100% (i.e., you must determine the proportions of each component relative only to the other components represented on the diagram; more on that later). Ternary diagrams are what we call "field diagrams." That is, once you've plotted your data, you can interpret something about your data by determining the field in which the data plot. Many types of ternary diagrams exist in the geosciences, and they are used in many applications. Often, the fields on a ternary diagram are used to classify earth materials or landforms, such as rock, soil, sediment, or mineral compositions; lithofacies; sediment size; dunes or deltas; among many others. A Sweet Example: Let's consider a non-geoscience three-component system, one with the components chocolate, milk, and sugar, like the image to the right. First, notice how the diagram is labeled: Chocolate is on the top apexEach corner of the triangle is an apex. Apices is the plural form of apex (purple), Milk is on the left apex (red), and Sugar is on the right apex (blue). Also note that there are some colorful "fields" inside the triangle. Fields on ternary diagrams help us to interpret the data and are usually based on empirical data/observations. Several different combinations of chocolate/milk/ sugar are plotted (as points) in this ternary diagram and represent different ways that a given data point can be plotted. • Points that plot at one apex of a triangle, like the blue point labeled "Pixie Sticks," indicate that the sample is made up of 100% of a single component (in the case of the blue dot, 100% • Points that plot along one side of the diagram, like the black points labeled Semi-sweet and Dark Chocolate, indicate that the sample only contains two components (in this case of the black dots, only sugar and chocolate; semi-sweet chocolate = 50% chocolate/50% sugar, dark chocolate = 72% chocolate/28% sugar; neither contain any milk). • Points that plot inside the triangle, like the pink points on the plot, indicate that the sample contains some proportion of all three components, and the "field" that contains the point gives you information about how to interpret your sample. In this case, the pink dot in the green field would be classified as Milk Chocolate (and contains about 31% milk, 41% chocolate, and 28% sugar); the pink dot in the yellow field would be classified as Ice Cream (and contains about 62% milk, 10% chocolate, and 28% sugar). Where do your tastes lie? Are you a sweet tooth who likes 100% sugar, a chocolate and sugar person, or some combination of all three? How Are Ternary Diagrams Constructed? You may be wondering how I got the relative percentages for the milk chocolate and ice cream dots above. Let's consider a slightly modified diagram like the one in the "Sweet Example" (milk/chocolate /sugar ternary diagram). Note that each of the triangle is labeled as 100% of one end-member component and there are colored numbers along each side corresponding to percentages of each. We can overlay lines that connect the percentages and form a (just like any plot, except this grid, it is a bunch of triangles instead of rectangles/squares; see figure at left). Note: Sometimes the grid is provided, and sometimes you will need to construct it (see Step 0 in examples below for more information). Let's start by breaking down how we made the grid on the ternary diagram to the left. Each end-member component (plotted at each apex) has a set of lines associated with it. The set of lines associated with an end-member component are parallel to (do not intersect) the side of the triangle opposite the end-member's apex. In the figure to the left, 10 lines (color coded) and 1 apex are associated with each component. All the purple lines are for component A (at left, A = chocolate), red lines are for component B (at left, B = milk), and blue lines for component C (at left, C = sugar). Each line is labeled (on one side of the triangle at left) decreasing from the apex toward the opposite side of the triangle. If we separate each component and the associated lines, they would look like the image below: The image to the right shows three triangles, color coded for components A (purple), B (red) and C (blue), and lines are labeled on both sides; that is, a sample with 70% chocolate could plot anywhere along the purple line labeled 70. To know where it plots along that line, you need to know the proportions of the other components. For example, note that the Milk Chocolate, Ice Cream, and Dark Chocolate described in the "Sweet Example" have 27% sugar (if you draw a line from Dark Chocolate and parallel to the blue lines, it will pass through all three of the dots on the diagram). However, the proportions of other components determine where the points are plotted and how they are classified. How Do I Plot Data Points and Interpret Them Using a Ternary Diagram? This section describes plotting and interpreting data on ternary diagrams. If you are looking for how to interpret a point that is already plotted, skip to the next section. Example 1: Plotting and classifying soil textures When describing soils, geoscientists include proportions of water, organic material, pore space, and sediment grains (see figure at right). You are tasked with classifying the texture of a soil sample that you are studying. After collecting the sample, you weigh the sample, dry it (reweighing it to determine the amount of water), and perform a textural analysis. You determine that your soil contains 21 g water, 4 g organic material, 18.5 g sand, 21.5 g silt, and 10 g clay. You have access to the soil texture triangle shown below to use for classifying your soil. Use the soil textural triangle below to plot and classify the texture of your soil sample. Here are some steps to follow when plotting data on a ternary diagram. Below each step (initially hidden from view) you will find an example of how to do that step for the following problem. Step 1: Determine the components shown on the diagram (there will be three) and which apex reflects 100% of each component (and which sides of the diagram show proportions of each component. If the sides of the diagram are not labeled, you will need to decide which side of the triangle will show percentages for each component. Step 2. If the problem gives you more than three components, determine which components in the problem are relevant to the ternary diagram of interest. If the problem only lists the three components shown on the diagram, you can skip to step 3. Normalizing data means determining the relative proportions of the components to one another. Step 3. Normalize the data into percentages of each component. To normalize, divide the amount (weight, fraction, value, etc.) of each component by the total amount (weight, fraction, value, etc.) of all three components. Equations to normalize components: A: `A/((A+B+C)) " * " 100 = A%` B: `B/((A+B+C)) " * " 100 = B%` C: `C/((A+B+C)) " * " 100 = C%` When you have done the calculations, percentages must sum to 100%: `A% + B% + C% = 100%` Step 4: Draw a line that represents the proportion of one end-member component at the appropriate percentage and so that it is parallel to the side opposite the apex representing 100% of that end-member. In some cases, you may have to approximate (interpolate) the location of the line for percentages between those that are labeled. Step 4a: Repeat the above process for the other two components. Your lines should intersect at a single point representing the sample composition. Step 5. Plot the normalized data as a single point. Determine where your lines cross and place the point there. Step 6: Interpret the data plotted on the diagram. As discussed above, each ternary diagram has fields that allow you to categorize the sample based on the field into which it plots. How Do I Interpret Points Already Plotted on a Ternary Diagram? This section describes how to read and interpret ternary diagrams that already have points plotted on them. Please visit the previous section for information about how to plot points. Example 2: Reading mineral compositions for a given ultramafic rock In your petrology lab, you are looking at several coarse-grained (phaneritic/plutonic rocks) and trying to classify them. Your instructor has given you some ternary phase diagrams with each of the rocks plotted. Only one sample (the red dot) plots on the IUGS ternary diagram for classifying ultramafic rocks (the ternary diagram to the right). This sample plots in a field labeled lherzolite, so, as you learned above, a rock with the components represented by the red dot is classified as a lherzolite (see Step 6 above for information about interpreting ternary diagrams). Use the phase diagram to the right to determine how much of each of the end member components is present in the lherzolite plotted as the red circle so that you can determine which of your rocks is lherzolite. Step 1: Determine the components shown on the diagram (there will be three) and which apex reflects 100% of each component. Step 2: Draw three lines through the point and parallel to grid lines to determine the relative abundances of each end-member. Step 3: Record the percentages of each end member component, confirming that they sum to 100%. Step 4: Interpret your findings. Use what you have learned to answer the question posed. Where Are Ternary Diagrams Used For in the Earth Sciences? Many Earth materials and systems can be represented by three components, creating an easy way to classify minerals, sediments, rocks, and landforms. Many sub-disciplines in Earth science have developed standardized ternary plots that were created from observational or empirical data, including: • Geochemistry: trends in chemical components with respect to space or time • Igneous petrology: classification of rock types; phase diagrams • Metamorphic petrology: phase relations; facies; reactions • Sedimentology: classification of carbonate and clastic rock types • Mineralogy: mineral classification (e.g., feldspar, pyroxene) • Water Quality: cation and anion variation diagrams • Soils: soil texture classification • Geomorphology: dune or delta morphology; hydraulic geometry The figure above shows some examples from the fields of soil science (colorful version of the soil texture triangle in example 1), petrology (coarse grained igneous rocks), and geomorphology Next Steps I am ready to PRACTICE! If you think you have a handle on the steps above, click on this bar to try practice problems with worked answers. Or, if you want even more practice, see 'More help' below. More Help (Resources for Students) Pages written by Kelly Deuerling (University of Nebraska, Omaha) and Ryan Kerrigan (University of Pittsburgh at Johnstown). Edited by Jennifer M. Wenner (University of Wisconsin, Oshkosh)	{"url":"https://serc.carleton.edu/mathyouneed/geomajors/ternary/index.html","timestamp":"2024-11-06T15:48:55Z","content_type":"text/html","content_length":"133234","record_id":"<urn:uuid:8ddcca47-ca54-44c3-b62e-806432d8b74d>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00187.warc.gz"}
sum of sinusoids 8 years ago ●17 replies● latest reply 8 years ago 11390 views I would like to know the consequence of adding 2 sinusoids of different frequencies. I assume that the result is not a pure sinusoid. I have tried a octave example and output plot is as below: [ - ] Reply by ●April 29, 2017 Hi. When you add two sinusoids of frequencies f1 and f2 you produce a third sinusoid whose frequency is (f1 + f2)/2. And that third sinusoid's peak amplitude is fluctuating. The algebra, and an example, of this behavior can be found at: [ - ] Reply by ●April 29, 2017 Hi Rick You don't really mean it in frequency domain[or in time domain]! [ - ] Reply by ●April 29, 2017 Rick, you are mixing addition of two sinusoids (the concern of the question) with multiplication (not of interest in the question). Addition does not produce any new frequency components. [added later] Okay, I see now what you mean. I am using the term sinusoid literally which does not permit modulation or variations in the amplitude. [ - ] Reply by ●April 29, 2017 From a semantic stand point, I suppose it would be more correct to say that the addition of two sinusoids is mathematically equivalent to a third sinusoid whose amplitude is modulating, but I got what Rick was saying. [ - ] Reply by ●April 29, 2017 Hi dszabo. Or perhaps it would be more correct to say, "The addition of two sinusoids, having different frequencies, is mathematically equivalent to a third sinusoid whose amplitude is sinusoidally [ - ] Reply by ●April 29, 2017 I'm picking up what you're putting down. [ - ] Reply by ●April 29, 2017 Visual interpretation and understanding in time domain is difficult, but if you plot the fft, you will start seeing peaks at different frequencies. [ - ] Reply by ●April 29, 2017 Adding two sinusoids of different frequencies can never produce a sinusoid. Indeed, depending on the relative frequencies, the sum might not even be periodic. Does that answer your question adequately? [ - ] Reply by ●April 29, 2017 To Sharan123: Your original signal plots are correct, but you added two sine waves (differing in frequency by one Hz) whose frequencies are low relative to the Fs sample rate. Try adding two higher-frequency (relative to the Fs sample rate) sine waves whose frequencies also differ by only one Hz. [ - ] Reply by ●April 29, 2017 I prefer to say: "The addition of two sinusoids, having different frequencies generates a repeating pattern whose amplitude is sinusoidally modulated but is never a new sinusoid" looks like the black sheep joke story [ - ] Reply by ●April 29, 2017 Sharan123, if you have MATLAB then try this code: n = 0:128; for K = -2:0.25:2; x = 1sin(2pin16/128) + 10^Ksin(2pin14/128); figure(1), clf plot(n/32,x,'-r',n/32,x,'ks','markersize', 2), grid on xlabel('Time (Seconds)'), pause(), disp('HIT A KEY!') After the first time through the loop, count the number of cycles in the first two seconds. And when the the code has completely finished, count the number of cycles in the first two seconds. [ - ] Reply by ●April 29, 2017 This the code listing with updated frequencies: Fs = 100; % Sampling frequency T = 1/Fs; % Sampling period L1 = 100; % Length of signal t1 = (0:L1-1)T; % Time vectorX1 = sin(20pit1); X2 = sin(21pit1); X3 = X1 + X2;subplot(2,2,1) The following is the plot from the code above: [ - ] Reply by ●April 29, 2017 Hello Sharan123. You above post is messed up. Your plots' window seems to be covering up some part of your post. In any case, I suggest you set variable L1 to: L1 = 400 and rerun your code. Did you read the blog that I recommended to you in an earlier post of mine? [ - ] Reply by ●April 29, 2017 Hello Rick, Actually, I have taken screenshot for the plots. So, some portion of my windows background is appearing. I will update the code for modified L value and run. I haven't read the blog you mentioned. I will do that. Thanks for your inputs [ - ] Reply by ●April 29, 2017 This is how the plot looks like: [ - ] Reply by ●April 29, 2017 Adding sinusoids can be a powerful technique. There are two special cases where the addition of sinusoids have nice mathematical properties. The first is what is known as the "beat phenomenon". This is what Rick brought up. When two sinusoids of different frequencies are added together the result is another sinusoid modulated by a sinusoid. The math equation is actually clearer. cos(A) + cos(B) = 2 cos( (A+B)/2 ) * cos( (A-B)/2 ) The amplitudes have to be the same though. If they are different, the summation equation becomes a lot more complicated. Your plots look correct. sin(20pit1) = sin(102pit1); 10 cycles per frame sin(21pit1) = sin(10.52pit1); 10.5 cycles per frame The frequency of your envelope sinusoid is going to be (10.5-10)/2 = .25, which is why your third plot only has a quarter cycle of the envelope function. The second special case is when the frequencies of the sinusiods you are adding are all harmonics of the same base frequency. This is what Fourier series are made of, and this is a core topic of [ - ] Reply by ●April 29, 2017 Hi Sharan123, I thought I'd add one more thought to help you cement this concept. What Rick and Cedron have already said is the way to understand it mathematically and conceptually, so I'm merely going to add an example you can try. When sounds are played simultaneously their waveforms (the math) are added. So, online, find sound generator (I like to play with audacity - it's free and open source), set the tone generator to 329 Hz, then hit the E just above middle C on your piano (or strum the E, top string, of your guitar), and listen very carefully. You'll hear the Wa Wa, or "beat" frequency riding on the sound of your two tones. That is the "sinusoidally modulated" 3rd frequency referred to by Rick, which will make the last pix you posted make even more sense. Lacking a musical instrument, you could just set up two sounds on audacity, one set to 329 Hz and the other to 330 Hz. Quickly, if you're interested http://www.audacityteam.org/. Play with making the difference very small, or larger. You'll get the picture quickly.	{"url":"https://dsprelated.com/thread/2786/sum-of-sinusoids","timestamp":"2024-11-07T00:02:49Z","content_type":"text/html","content_length":"62064","record_id":"<urn:uuid:9c960b4b-4f0a-4791-ba39-a7d00781476f>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00884.warc.gz"}
TechnologyTasks.com - Spreadsheets Basically, a spreadsheet is an electronic document in which data is arranged in the rows and columns of a grid and can be manipulated and used in calculations. A spreadsheet is made of rows and columns that help sort data, arrange data easily, and calculate numerical data. What makes a spreadsheet software program unique is its ability to calculate values using mathematical formulas and the data in cells. A good example of how a spreadsheet may be utilized is creating an overview of your bank's balance. Below is a basic example of what a Microsoft Excel spreadsheet looks like, as well as all the important features of a spreadsheet highlighted. In the above example, this spreadsheet is listing three different checks, the date, their description, and the value of each check. These values are then added together to get the total of $162.00 in cell D6. That value is subtracted from the check balance to give an available $361.00 in cell D8. Spreadsheet software allows users to organize data in these rows and columns and perform calculations on the data. These rows and columns collectively are called a worksheet. Most spreadsheet software has basic features to help users create, edit, and format these worksheets. A spreadsheet file is similar to a notebook that can contain more than 1,000 related individual worksheets. Data is organized vertically in columns and horizontally in rows on each worksheet. Each worksheet usually can have more than 16,000 columns and 1 million rows. One or more letters identify each column, and a number identifies each row. Only a small fraction of these columns and rows are visible on the screen at one time. Scrolling through the worksheet displays different parts of it on the screen. A cell is the intersection of a column and row. The spreadsheet software identifies cells by the column and row in which they are located. Cells may contain three types of data: labels, values, and formulas. The text, or label, entered in a cell identifies the worksheet data and helps organize the worksheet. Using descriptive labels, such as Gross Margin and Total Expenses, helps make a worksheet more meaningful. Today, Microsoft Excel is the most popular and widely used spreadsheet program, but there are also many alternatives. A spreadsheet program that can be used to create a spreadsheet for free online is Google Sheets. Another software suite is called OpenOffice, which is free to download and install on your computer. It has a wonderful spreadsheet in it called Calc. Need a spreadsheet? Then contact us.	{"url":"https://www.technologytasks.com/spreadsheets","timestamp":"2024-11-10T10:49:17Z","content_type":"text/html","content_length":"98684","record_id":"<urn:uuid:5cb4f8e1-259f-4efa-8ce8-7a42a77934f7>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00804.warc.gz"}
FINANCE PROBLEMS -YOU INVEST $1,000 IN AN ACCOUNT THAT PAYS INTEREST AT 12 PERCENT ANNUAL RATE. HOW MUCH WOULD YOUR INVESTMENT GROW TO IN 5 YEARS? You invest $1,000 in an account that pays interest at 12 percent annual rate. How much would your investment grow to in 5 years? a. 1,806 b. 1,762 c. 1,600 d. 1,272 You invest $1,000 in an account that pays interest at 12 percent annual rate, compounded quarterly. How much would your investment grow to in 5 years? a. 1,762 b. 1,272 c. 1,806 d. 1,600 You invest $1,000 in an investment that grew to $60,000 in 35 years. What annual rate of interest did you earn? a. 5.25% b. 12.41% c. 10.17% d. 8.24% You plan to purchase a BMW M5 sedan in six years for $85,000. You have saved $30,000 for the car and plan to invest this money to purchase the car. What rate of interest would you have to earn to be able to purchase the car after six years? a. 17% b. 13% c. 15% d. 19% You plan to purchase a BMW M5 sedan for $85,000. You have saved $30,000 for the car and plan to invest this money to purchase the car. How many years would you have to wait if you can earn 12 percent annual rate, compounded quarterly? a. 8.81 b. 35.23 c. 6.53 d. 19.54 How long would it take to quadruple your money at an annual rate of 8 percent? a. 19 years b. 18 years c. 17 years d. 16 years You purchased a rare baseball card for $6,000 as an investment. Three years later you accidentally spilled coffee on it while working on your finance homework, and were forced to sell it for $4,500. What rate of return did you earn? (Hint: It is negative!) a. –8.86% b. –14.22 c. –9.14% d. –11.1% Your rich uncle has promised to pay you $20,000 as a gift upon graduation in 2 years. You plan to invest it for 5 more years at 8.5 percent compounded monthly and use the money you have after 7 years as a down payment to purchase a house. What will be the amount of the down payment? a. 36,180 b. 28,500 c. 30,540 d. 40,400 Shady Investment Company offers an investment that promises to double your money in 15 months. This investment promises to credits interest to your account every quarter. What quarterly rate must the investment earn to meet the promised return? a. 22.59 b. 14.87 c. 59.48 d. 36.99 Loan Shark Company provides short term loans. They will loan you $4 today and expect $5 back in one week! What is the APR for this loan? a. 25% b. 1,300% c. 1,125% d. 125% Loan Shark Company provides short term loans. They will loan you $4 today and expect $5 back in one week! What is the EAR for this loan? a. 1,300,000.00% b. 67,600.00% c. 10,947,544.25% d. 875,287.42% Recently, the Islanders signed Rick DiPietro to a 15 year contract. The contract is for equal payments of $4.5 million each year for the next 15 years. If you assume a 10 percent discount rate, what is the true value (i.e. PV) of this contract? a. $85 million b. $34 million c. $67.5 million d. $143 million Recently, the Islanders signed Rick DiPietro to a 15 year contract. The contract is for equal payments of $4.5 million each year for thenext 15 years. If you assume a 10 percent discount rate, what is the future value of this contract? a. $67.5 million b. $34 million c. $143 million d. $85 million You plan to retire with $500,000 savings. You can make a deposit of $150 per month into a retirement saving account that pays 12 percent annual interest compounded monthly. How many years will you have to wait to retire? a. 98 years b. 30 years c. 18 years d. 277 years You plan to retire with $500,000 savings. How much should you deposit annually into a retirement saving account that pays 10 percent annual interest if you plan to retire in 15 years? a. $47,258 b. $15,737 c. $33,333 d. $18,415 You buy a house worth $350,000 with 20% down payment and a 30-year mortgage on the remaining value. If your monthly payment is $1,500, what is the annual percentage rate (APR) for the mortgage? a. 2.82% b. 4.98% c. 6.12% d. 5.10% You buy a house worth $350,000 with 20% down payment and a 30-year mortgage on the remaining value. If your monthly payment is $1,500, what is the effective annual rate (EAR) for the mortgage? a. 6.12% b. 5.10% c. 4.98% d. 2.82% You borrow $350,000 at 8 percent compounded monthly. How many years will it take to pay back the loan if the monthly payment is $2,960? a. 10 b. 19.5 c. 156 d. 233.6 Note: Use this information for next two questions. Ms. Patricia Sullivan plans to create a fund from her lottery winnings to meet three objectives. First, she wants to create a fund so that her mother can withdraw $20,000 per month for the remainder of her expected life of 20 years. Second, she wants to pay the down payment for her brother to buy a house upon graduation from college four years from now. She expects that he will need $100,000 for payment at that time. Finally, she wants to retire after 15 years and be able to withdraw $30,000 per month starting a month from her retirement. She expects to live for 30 years after retirement. All monies earn 8 percent compounded monthly and all cash flows occur at the end of the relevant period. How much money does she need to invest today to meet her first objective? a. $3.7 million b. $2.4 million c. $4.5 million d. $3.2 million How much money does she need to invest today to meet all three objectives? a. $2.4 million b. $3.2 million c. $4.5 million d. $3.7 million You plan to purchase a car. The dealer is offering special financing at an annual percentage rate (APR) of 8 percent for 100 percent of the car value. The inflation premium is 3.5 percent. If the pure rate in the market is 3 percent, what is the risk premium using the multiplicative form? a. 4.72% b. 2.69% c. 7.48% d. 1.31% e. 6.24% The real rate is 4.2 percent and the nominal APR is 7 percent. What is the expected inflation premium? Use exact formulation. a. 7.48% b. 6.24% c. 2.69% d. 1.31% e. 4.72% The pure rate of interest is 2.5 percent and the inflation premium is 5 percent. If you require a risk premium of 3.5 percent, what is the nominal APR rate? Use exact formulation. a. 11.00% b. 8.75% c. 11.39% d. 6.09% e. 6.00% The pure rate of interest is 2.5 percent and the inflation premium is 5 percent. If you require a risk premium of 3.5 percent, what is the real rate? Use exact formulation. a. 8.75% b. 11.00% c. 6.09% d. 6.00% e. 11.39% The pure rate of interest is 2.5 percent and the inflation premium is 5 percent. If you require a risk premium of 3.5 percent, what is the risk-free rate? Use exact formulation. (Hint: Set risk premium equal to zero!) a. 6.09% b. 7.50% c. 8.75% d. 7.62% e. 6.00% The APR on a financial security is 12 percent. If the inflation premium is 4 percent and the pure rate is 3 percent, what risk premium is required by the market? a. 4.74% b. 3.81% c. 5.00% d. 5.37% e. 4.56% Use the following information for this and the following three questions. A bank wishes to earn a pure rate of 2 percent, and the inflation premium is 1.6%. The bank uses the Fair Issac Corporation (FICO) score to determine car loan rate for its customers. Based on an automobile loan applicant’s FICO score, it uses the following risk premium adjustment to the rate it quotes. FICO Score Risk Premium % >740 1.00% 720-739 1.10% 700-719 1.30% 680-699 1.60% 660-679 1.90% 640-659 2.20% 620-639 2.50% <620 7.50% What annual percentage rate will it quote to John Smith who has a FICO score of 645? a. 6.22% b. 5.60% c. 11.40% d. 5.91% e. 5.29% If John Smith’s FICO score drops to 615, what will be the percent change in the APR? a. – 29.43% b. 92.91% c. 11.40% d. 29.43% e. – 92.91% Jane Smith applies for a car loan to the bank and the bank quotes her an APR of 4.98 percent. In which range does Jane Smith’s FICO score fall? a. 660-679 b. 720-739 c. 680-699 d. >740 e. 700-719 The average FICO score in the United States is about 692. 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Russell's antinomy The Russell's paradox is one of Bertrand Russell and Ernst Zermelo discovered paradox of naive set theory published, the Russell 1903 and therefore bears his name. Concept and problem Russell formed his antinomy with the help of the "class of all classes that do not contain themselves as an element", referred to as Russell's class ; he defined it formally as follows: ${\ displaystyle R: = \ {\, x \ mid x \ notin x \, \}}$ Russell's class is often defined as "the set of all sets that do not contain themselves as an element"; this corresponds to the set theory of that time, which did not yet differentiate between classes and sets . In contrast to the older antinomies of naive set theory ( Burali-Forti paradox and Cantor's antinomies ), Russell's antinomy is of a purely logical nature and independent of set axioms. Therefore it had a particularly strong effect and suddenly brought about the end of naive set theory. Russell derived his antinomy as follows: Assumed contains itself, then because of the class property that was used to define that what contradicts the assumption does not contain itself. Assuming the opposite is true and does not contain itself, then the class property fulfills , so that it contains itself against the assumption. Mathematically, this expresses the following contradicting equivalence: ${\ displaystyle \, R}$${\ displaystyle \, R}$${\ displaystyle \, R}$${\ displaystyle \, R}$${\ displaystyle \, R}$${\ displaystyle \, R}$ ${\ displaystyle R \ in R \ iff R \ notin R}$ No axioms and theorems of set theory are used to derive this contradiction, but only Frege's abstraction principle , which Russell adopted in his type theory, apart from the definition : ${\ displaystyle y \ in \ {\, x \ mid A (x) \, \} \ iff A (y)}$ History and solutions Russell discovered his paradox in mid-1901 while studying Cantor's first antinomy from 1897. He published the antinomy in his book The Principles of Mathematics in 1903. As early as 1902 he informed Gottlob Frege by letter. He was referring to Frege's first volume of the Basic Laws of Arithmetic from 1893, in which Frege tried to build arithmetic on a set-theoretical axiom system. Russell's antinomy showed that this system of axioms was contradicting itself. Frege responded to this in the afterword of the second volume of his Basic Laws of Arithmetic from 1903: “A scientific writer can hardly encounter anything more undesirable than that, after completing a work, one of the foundations of his structure is shaken. I was put in this position by a letter from Mr. Bertrand Russell as the printing of this volume neared its end. " - Thank God Frege Russell solved the paradox as early as 1903 through his type theory ; in it a class always has a higher type than its elements; Statements like “a class contains itself”, with which he formed his antinomy, can then no longer be formulated. So he tried, since he adhered to Frege's principle of abstraction, to solve the problem by a restricted syntax of the admissible class statements. The restricted syntax, however, turned out to be complicated and inadequate for the structure of mathematics and has not become established in the long term. Parallel to Russell, Zermelo, who found the antinomy to be independent of Russell and knew it even before Russell's publication, developed the first axiomatic set theory with unrestricted syntax. The exclusion axiom of this Zermelo set theory from 1907 only allows a restricted class formation within a given set. He showed by indirect proof with this antinomy that the Russell class is not a set. His solution has prevailed. In the extended Zermelo-Fraenkel set theory (ZF), which today serves as the basis of mathematics , the axiom of foundation also ensures that no set can contain itself, so that here the Russell class is identical to the universal class . Since Russell's antinomy is of a purely logical nature and does not depend on set axioms, it can already be proven at the level of the first-order consistent predicate logic that Russell's class does not exist as a set. This makes the following argumentation understandable, which converts a second indirect proof of Russell into a direct proof: The statement is abbreviated with .${\ displaystyle y \ in x \ iff y \ notin y}$${\ displaystyle \, {\ mbox {R}} yx}$ The statement supported by is the above contradiction. Therefore, its negation is true: .${\ displaystyle \, x}$${\ displaystyle \, {\ mbox {R}} xx}$${\ displaystyle {\ mbox {not}} \, {\ mbox {R}} xx}$ Therefore, the existential can be introduced: .${\ displaystyle {\ mbox {There are}} y \ colon {\ mbox {not}} \, {\ mbox {R}} yx}$ By introducing the universal quantifier follows: .${\ displaystyle {\ mbox {For all}} x \ colon {\ mbox {There are}} y \ colon {\ mbox {not}} \, {\ mbox {R}} yx}$ By rearranging the quantifiers and elimination of the abbreviation is finally obtained the sentence .${\ displaystyle \, {\ mbox {There is no}} x \ colon {\ mbox {For all}} y \ colon (y \ in x \ iff y \ notin y)}$ This sentence means in the language of predicate logic: There is no set of all sets that do not contain themselves as an element. It applies to all modern axiomatic set theories that are based on first-level predicate logic, for example in ZF. It is also valid in the Neumann-Bernays-Gödel set theory , in which Russell's class exists as a real class . In the class logic of Oberschelp , which is a demonstrably consistent extension of the first-level predicate logic, any class terms can also be formed to any defining statements; in particular, Russell's class is also a correct term with provable nonexistence there. Axiom systems such as ZF set theory can be integrated into this class logic. Since the theorem was derived in a direct proof, it is also valid in intuitionist logic . Variants of Russell's antinomy The 1908 Grelling-Nelson Antinomy is a semantic paradox inspired by Russell's antinomy. There are numerous popular variations of Russell's antinomy. The best known is the barber's paradox , with which Russell himself illustrated and generalized his train of thought in 1918. Curry's Paradox of 1942 contains, as a special case, a generalization of Russell's antinomy. Individual evidence Web links	{"url":"https://de.zxc.wiki/wiki/Russellsche_Antinomie","timestamp":"2024-11-04T01:30:02Z","content_type":"text/html","content_length":"45921","record_id":"<urn:uuid:7438a444-d221-49ac-a2c9-64cad9c6f1db>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00132.warc.gz"}
plotnine 0.14.1 mapping : aes = None Aesthetic mappings created with aes. If specified and inherit_aes=True, it is combined with the default mapping for the plot. You must supply mapping if there is no plot mapping. Aesthetic Default value y after_stat('density') The bold aesthetics are required. Options for computed aesthetics 'density' # density estimate 'count' # density * number of points, # useful for stacked density plots 'scaled' # density estimate, scaled to maximum of 1 'n' # Number of observations at a position data : DataFrame = None The data to be displayed in this layer. If None, the data from from the ggplot() call is used. If specified, it overrides the data from the ggplot() call. geom : str \| geom = "density" The statistical transformation to use on the data for this layer. If it is a string, it must be the registered and known to Plotnine. position : str \| position = "stack" Position adjustment. If it is a string, it must be registered and known to Plotnine. na_rm : bool = False If False, removes missing values with a warning. If True silently removes missing values. kernel : str = "gaussian" Kernel used for density estimation. One of: adjust : float = 1 An adjustment factor for the bw. Bandwidth becomes bw * adjust. Adjustment of the bandwidth. trim : bool = False This parameter only matters if you are displaying multiple densities in one plot. If False, the default, each density is computed on the full range of the data. If True, each density is computed over the range of that group; this typically means the estimated x values will not line-up, and hence you won’t be able to stack density values. n : int = 1024 Number of equally spaced points at which the density is to be estimated. For efficient computation, it should be a power of two. gridsize : int = None If gridsize is None, max(len(x), 50) is used. bw : str \| float = "nrd0" The bandwidth to use, If a float is given, it is the bandwidth. The options are: nrd0 is a port of stats::bw.nrd0 in R; it is eqiuvalent to silverman when there is more than 1 value in a group. cut : float = 3 Defines the length of the grid past the lowest and highest values of x so that the kernel goes to zero. The end points are -/+ cutbw{min(x) or max(x)}. clip : tuple[float, float] = (-inf, inf) Values in x that are outside of the range given by clip are dropped. The number of values in x is then shortened. The domain boundaries of the data. When the domain is finite the estimated density will be corrected to remove asymptotic boundary effects that are usually biased away from the probability density function being estimated. **kwargs : Any = {} Aesthetics or parameters used by the geom.	{"url":"http://plotnine.org/reference/stat_density.html","timestamp":"2024-11-08T20:56:26Z","content_type":"application/xhtml+xml","content_length":"40264","record_id":"<urn:uuid:d2e512dd-1d5c-44e6-80ce-d21c3690c30e>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00163.warc.gz"}
Using Canonical Forms for Isomorphism Reduction in Graph-based Model Checking Graph isomorphism checking can be used in graph-based model checking to achieve symmetry reduction. Instead of one-to-one comparing the graph representations of states, canonical forms of state graphs can be computed. These canonical forms can be used to store and compare states. However, computing a canonical form for a graph is computationally expensive. Whether computing a canonical representation for states and reducing the state space is more efficient than using canonical hashcodes for states and comparing states one-to-one is not a priori clear. In this paper these approaches to isomorphism reduction are described and a preliminary comparison is presented for checking isomorphism of pairs of graphs. An existing algorithm that does not compute a canonical form performs better that tools that do for graphs that are used in graph-based model checking. Computing canonical forms seems to scale better for larger graphs. Publication series Name CTIT Technical Report Series Publisher Centre for Telematics and Information Technology, University of Twente No. TR-CTIT-10-28 ISSN (Print) 1381-3625 • Graph Isomorphism • IR-72368 • EWI-18116 • Graph-based Model Checking • METIS-276046 • Canonical Form	{"url":"https://research.utwente.nl/en/publications/using-canonical-forms-for-isomorphism-reduction-in-graph-based-mo","timestamp":"2024-11-04T14:39:55Z","content_type":"text/html","content_length":"42496","record_id":"<urn:uuid:c6929697-aec4-4414-b14b-7bc1f0529942>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00264.warc.gz"}
I am a fifth-year Ph.D. candidate in Mathematics at Auburn University studying under Hal Schenck. I earned a B.S. in Mathematics with minors in History and Classics at the University of Kentucky. While at the University of Kentucky, I conducted research with the University of Kentucky Math Lab and the Multimodal Vision Research Laboratory. My primary research interests lie in commutative algebra, algebraic geometry, algebraic combinatorics, and data analysis. I am particularly interested in problems that can be approached via computation. I will be on the job market this fall; here are some of my supporting materials: CV, research program, teaching statement. Joint with Nasrin Altafi, Roberta Di Gennaro, Federico Galetto, Rosa M. Miró-Roig, Uwe Nagel, Alexandra Seceleanu, and Junzo Watanabe PDF \| Abstract The connected sum construction, which takes as input Gorenstein rings and produces new Gorenstein rings, can be considered as an algebraic analogue for the topological construction having the same name. We determine the graded Betti numbers for connected sums of graded Artinian Gorenstein algebras. Along the way, we find the graded Betti numbers for fiber products of graded rings; an analogous result was obtained in the local case by Geller. We relate the connected sum construction to the doubling construction, which also produces Gorenstein rings. Specifically, we show that a connected sum of doublings is the doubling of a fiber product ring. Joint with Hal Schenck PDF \| Video \| Abstract We study the Artinian reduction $ A $ of a configuration of points $ X \subset \mathbb{P}^n $, and the relation of the geometry of $ X $ to Lefschetz properties of $ A $. Migliore initiated the study of this connection, with a particular focus on the Hilbert function of $ A $, and further results appear in work of Migliore–Miró-Roig–Nagel. Our specific focus is on Betti tables rather than Hilbert functions, and we prove that a certain type of Betti table forces the failure of the Weak Lefschetz Property (WLP). The corresponding Artinian algebras are typically not level, and the failure of WLP in these cases is not detected in terms of the Hilbert function. Joint with Hunter Blanton and Nathan Jacobs PDF \| Abstract Repeat-visit airborne lidar is a powerful tool for change detection in urban and rural environments. In this work, we present a learning-based approach that addresses one of the key challenges in comparing point cloud scans of the same region: handling geometric differences caused by varying sensor position. Our approach is to perform shape modeling through ray casting with a point cloud neural network. Recent work on learning-based shape modeling has been based on the assumption that an explicit surface representation is available, which is not the case for airborne lidar datasets. Our key insight is that by using a ray casting approach we can perform shape modeling directly with lidar measurements. We evaluate our method both quantitatively and qualitatively on learned surface accuracy and show that our method correctly predicts surface intersection even in sparse regions of the input cloud. 1. Graph Theoretic Reflection to Foster Alignment in Coordinated Courses Submitted Joint with Haile Gilroy and Melinda Lanius PDF \| Abstract Despite online homework’s growing prevalence as a uniform component in coordinated mathematics courses, few studies have considered the connection, or lack thereof, between instructors of record and fixed online homework sets. In this mixed-methods study, we examined how 10 university mathematics educators working in a coordinated setting judged the quality of a sampling of online Calculus I homework assignments. Following an initial review of the homework sets, we introduced the educators to a novel instrument called the Course Alignment Analysis Tool (CAAT), which leverages graph theory to assess the alignment between the learning outcomes that an instructor feels should be prioritised and the learning outcomes most emphasised by an assignment or assessment. We analyzed the impact of engaging with the CAAT on participants’ consideration of uniform homework. We found that in- teracting with the CAAT affected coordinated instructors’ definitions of homework quality and that the CAAT is a promising professional development tool for novice instructors in particular. Joint with Ayah Almousa, Daoji Huang, Patricia Klein, Adam LaClair, Yuyuan Luo, and Joseph McDonough PDF \| Code \| Abstract We introduce the MatrixSchubert package for the computer algebra system Macaulay2. This package has tools to construct and study matrix Schubert varieties and alternating sign matrix (ASM) varieties. The package also introduces tools for quickly computing homological invariants of such varieties, finding the components of an ASM variety, and checking if a union of matrix Schubert varieties is an ASM variety. A Macaulay2 package with functions for investigating ASM and matrix Schubert varieties. Minimal Out-Neighborhoods Python scripts for computing minimal out-neighborhoods (and some statistics) in the $ F $-lattice.	{"url":"https://seangrate.com/","timestamp":"2024-11-10T19:19:07Z","content_type":"text/html","content_length":"22599","record_id":"<urn:uuid:36210ba3-34d1-4b34-a7ad-29b48cd87533>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00148.warc.gz"}
Fleiss' Kappa in R: For Multiple Categorical Variables - Datanovia Fleiss’ Kappa in R: For Multiple Categorical Variables The Fleiss kappa is an inter-rater agreement measure that extends the Cohen’s Kappa for evaluating the level of agreement between two or more raters, when the method of assessment is measured on a categorical scale. It expresses the degree to which the observed proportion of agreement among raters exceeds what would be expected if all raters made their ratings completely randomly. For example, you could use the Fleiss kappa to assess the agreement between 3 clinical doctors in diagnosing the Psychiatric disorders of patients. Note that, the Fleiss Kappa can be specially used when participants are rated by different sets of raters. This means that the raters responsible for rating one subject are not assumed to be the same as those responsible for rating another (Fleiss et al., 2003). Inter-Rater Reliability Essentials: Practical Guide in R Briefly the kappa coefficient is an agreement measure that removes the expected agreement due to chance. It can be expressed as follow: • Po is the observed agreement • Pe is the expected agreement Examples of formula to compute Po and Pe for Fleiss Kappa can be found in Joseph L. Fleiss (2003) and on wikipedia. kappa can range form -1 (no agreement) to +1 (perfect agreement). • when k = 0, the agreement is no better than what would be obtained by chance. • when k is negative, the agreement is less than the agreement expected by chance. • when k is positive, the rater agreement exceeds chance agreement. Interpretation: Magnitude of the agreement The interpretation of the magnitude of Fleiss kappa is like that of the classical Cohen’s kappa (Joseph L. Fleiss 2003). For most purposes, • values greater than 0.75 or so may be taken to represent excellent agreement beyond chance, • values below 0.40 or so may be taken to represent poor agreement beyond chance, and • values between 0.40 and 0.75 may be taken to represent fair to good agreement beyond chance. Read more on kappa interpretation at (Chapter @ref(cohen-s-kappa)). Your data should met the following assumptions for computing Fleiss kappa. 1. The outcome variables returned by raters should be categorical (either nominal or ordinal) 2. The outcome variables should have exactly the same categories 3. The raters are independent Statistical hypotheses • Null hypothesis (H0): kappa = 0. The agreement is the same as chance agreement. • Alternative hypothesis (Ha): kappa ≠ 0. The agreement is different from chance agreement. Example of data We’ll use the psychiatric diagnoses data provided by 6 raters. This data is available in the irr package. A total of 30 patients were enrolled and classified by each of the raters into 5 categories (Fleiss and others 1971): 1. Depression, 2. Personality Disorder, 3. Schizophrenia, 4. Neurosis, 5. Other. # install.packages("irr") data("diagnoses", package = "irr") head(diagnoses[, 1:3]) ## rater1 rater2 rater3 ## 1 4. Neurosis 4. Neurosis 4. Neurosis ## 2 2. Personality Disorder 2. Personality Disorder 2. Personality Disorder ## 3 2. Personality Disorder 3. Schizophrenia 3. Schizophrenia ## 4 5. Other 5. Other 5. Other ## 5 2. Personality Disorder 2. Personality Disorder 2. Personality Disorder ## 6 1. Depression 1. Depression 3. Schizophrenia Computing Fleiss Kappa The R function kappam.fleiss() [irr package] can be used to compute Fleiss kappa as an index of inter-rater agreement between m raters on categorical data. In the following example, we’ll compute the agreement between the first 3 raters: # Select the irst three raters mydata <- diagnoses[, 1:3] # Compute kapa ## Fleiss' Kappa for m Raters ## Subjects = 30 ## Raters = 3 ## Kappa = 0.534 ## z = 9.89 ## p-value = 0 In our example, the Fleiss kappa (k) = 0.53, which represents fair agreement according to Fleiss classification (Fleiss et al. 2003). This is confirmed by the obtained p-value (p < 0.0001), indicating that our calculated kappa is significantly different from zero. It’s also possible to compute the individual kappas, which are Fleiss Kappa computed for each of the categories separately against all other categories combined. kappam.fleiss(mydata, detail = TRUE) ## Fleiss' Kappa for m Raters ## Subjects = 30 ## Raters = 3 ## Kappa = 0.534 ## z = 9.89 ## p-value = 0 ## Kappa z p.value ## 1. Depression 0.416 3.946 0.000 ## 2. Personality Disorder 0.591 5.608 0.000 ## 3. Schizophrenia 0.577 5.475 0.000 ## 4. Neurosis 0.236 2.240 0.025 ## 5. Other 1.000 9.487 0.000 It can be seen that there is a fair to good agreement between raters in terms of rating participants as having “Depression”, “Personality Disorder”, “Schizophrenia” and “Other”; but there is a poor agreement in diagnosing “Neurosis”. Fleiss kappa was computed to assess the agreement between three doctors in diagnosing the psychiatric disorders in 30 patients. There was fair agreement between the three doctors, kappa = 0.53, p < 0.0001. Individual kappas for “Depression”, “Personality Disorder”, “Schizophrenia” “Neurosis” and “Other” was 0.42, 0.59, 0.58, 0.24 and 1.00, respectively. This chapter explains the basics and the formula of the Fleiss kappa, which can be used to measure the agreement between multiple raters rating in categorical scales (either nominal or ordinal). We also show how to compute and interpret the kappa values using the R software. Note that, with Fleiss Kappa, you don’t necessarily need to have the same sets of raters for each participants (Joseph L. Fleiss 2003). Another alternative to the Fleiss Kappa is the Light’s kappa for computing inter-rater agreement index between multiple raters on categorical data. Light’s kappa is just the average Cohen’s Kappa (Chapter @ref(cohen-s-kappa)) if using more than 2 raters. Fleiss, J.L., and others. 1971. “Measuring Nominal Scale Agreement Among Many Raters.” Psychological Bulletin 76 (5): 378–82. Joseph L. Fleiss, Myunghee Cho Paik, Bruce Levin. 2003. Statistical Methods for Rates and Proportions. 3rd ed. John Wiley; Sons, Inc. Recommended for you This section contains best data science and self-development resources to help you on your path. Coursera - Online Courses and Specialization Data science Popular Courses Launched in 2020 Trending Courses Amazon FBA Amazing Selling Machine Books - Data Science Our Books Version: Français No Comments Give a comment Back to Inter-Rater Reliability Measures in R	{"url":"https://www.datanovia.com/en/lessons/fleiss-kappa-in-r-for-multiple-categorical-variables/","timestamp":"2024-11-06T16:53:20Z","content_type":"text/html","content_length":"145999","record_id":"<urn:uuid:e85885f6-8215-4746-b358-99a0fb433991>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00830.warc.gz"}
Union-intersection tests and Intersection-union tests Union-intersection tests and Intersection-union tests Posted on (Update: ) This post is based on section 8.3 of Casella and Berger (2001). In some situations, tests for complicated null hypotheses can be developed from tests for simpler null hypothesis. Union-Intersection Method The union-intersection method would be useful if we can write the null hypothesis as \[H_0:\theta\in\bigcap_{\gamma\in \Gamma}\Theta_\gamma\,,\] where $\Gamma$ is an arbitrary index set that may be finite or infinite. Suppose the rejection region for the test \[H_{0\gamma}: \theta\in \Theta_\gamma\text{ versus }H_{1\gamma}:\theta\in \Theta_{\gamma}^c\] is $\{x:T_\gamma(x)\in R_\gamma\}$. Then the rejection region for the union-intersection test is $$\label{eq:8.2.4} \bigcup_{\gamma\in \Gamma} \{x:T_\gamma(x)\in R_\gamma\}\,.$$ In particular, suppose that each of the individual tests has a rejection region of the form $\{x:T_\gamma(x)> c\}$, where $c$ does not depend on $\gamma$, then \eqref{eq:8.2.4} becomes \[\bigcup_{\gamma\in \Gamma}\{x:T_\gamma(x) > c\} = \{x:\sup_{\gamma\in \Gamma}T_\gamma(x) > c\}\,,\] which implies that the test statistic for testing $H_0$ is $T(x) = \sup_{\gamma\in \Gamma}T_\gamma(x)$. Intersection-Union Method Suppose we wish to test the null hypothesis \[H_0: \theta\in \bigcup_{\gamma\in \Gamma}\Theta_\gamma\,,\] then the rejection region for the intersection-union test of $H_0$ versus $H_1$ is \[\bigcap_{\gamma\in \Gamma}\{x:T_\gamma(x)\in R_\gamma\}\,.\] Again, the test can be greatly simplified if the rejection regions for the individual hypotheses are all of the form $\{x:T_\gamma(x)\ge c\}$. In such cases, the rejection region for $H_0$ is \[\bigcap_{\gamma\in \Gamma}\{x:T_\gamma(x)\ge c\} = \{x:\inf_{\gamma\in \Gamma}T_\gamma(x) \ge c\}\,,\] and hence the test statistic is $\inf_{\gamma\in\Gamma}T_\gamma(x)$. Likelihood Ratio Test The likelihood ratio test statistic for testing $H_0:\theta\in \Theta_0$ versus $H_1:\theta\in \Theta_0^c$ is \[\lambda(x) = \frac{\sup_{\Theta_0}L(\theta\mid x)}{\sup_{\Theta}L(\theta\mid x)}\,.\] A likelihood ratio test is any test that has a rejection region of the form $\{x:\lambda(x)\le c\}$, where $c$ is any number satisfying $0\le c\le 1$. Sizes of UIT and IUT Due to the way in which they are constructed, the sizes of UIT and IUT can often be bounded above by the sizes of other tests. Such bounds are useful if a level $\alpha$ test is wanted, but the size of the UIT or IUT is too difficult to evaluate. Consider testing $H_0:\theta\in \Theta_0$ versus $H_1:\theta\in \Theta^c_0$, where $\Theta_0 = \bigcap_{\gamma\in \Gamma}\Theta_\gamma$ and $\lambda_\gamma(x)$ is the LRT statistic for testing $H_{0\gamma}$. Define $T(x)=\inf_{\gamma\in\Gamma}\lambda_\gamma(x)$, and form the UIT with rejection region \[\{x:\lambda_\gamma(x) < c \text{ for some }\gamma\in\Gamma\} = \{x:T(x) < c\}\,.\] Also, consider the usual LRT with rejection region $\{x:\lambda(x) < c\}$. Then □ $T(x)\ge \lambda(x)$ for every $x$ □ If $\beta_T(\theta)$ and $\beta_\lambda(\theta)$ are the power functions for the tests based on $T$ and $\lambda$, respectively, then $\beta_T(\theta) \le \beta_\lambda(\theta)$ for every $\ theta\in \Theta$. □ If the LRT is a level $\alpha$ test, then the UIT is a level $\alpha$ test. For IUT, we have Let $\alpha_\gamma$ be the size of the test of $H_{0\gamma}$ with rejection region $R_\gamma$. Then the IUT with rejection region $R=\bigcup_{\gamma\in\Gamma}R_\gamma$ is a level $\alpha = \sup_ {\gamma\in \Gamma}\alpha_\gamma$ test. It provides an upper bound for the size of an IUT, is somewhat more useful than the theorem for UIT, which applies only to UITs constructed from likelihood ratio tests. Actually, the size of the IUT may be much less than $\alpha$, and the following theorem gives conditions under which the size of the IUT is exactly $\alpha$ and the IUT is not too conservative. Consider testing $H_0:\theta\in \bigcup_{j=1}^k\Theta_j$, where $k$ is a finite positive integer. For each $j = 1,\ldots,k$, let $R_j$ be the rejection region of a level $\alpha$ test of $H_{0j} $. Suppose that for some $i=1,\ldots,k$, there exists a sequence of parameter points, $\theta_l\in \Theta_i, l=1,2,\ldots,$, such that □ $\lim_{l\rightarrow \infty}P_{\theta_l}(X\in R_i) = \alpha$ □ for each $j = 1,\ldots, k, j\neq i$, $\lim_{l\rightarrow\infty} P_{\theta_l}(X\in R_j) = 1$ Then the IUT with rejection region $R = \bigcup_{j=1}^k R_j$ is a size $\alpha$ test. Only need to show $\sup_{\theta\in \Theta_0}P_\theta(X\in R)\ge \alpha$. Because all the parameter points $\theta_l$ satisfy $\theta_l\in \Theta_i\subset \Theta_0$, $\begin{align} \sup_{\theta\ in\Theta_0}P_\theta(X\in R) &\ge \lim_{l\rightarrow \infty} P_{\theta_l}(X\in R)\\ &= \lim_{l\rightarrow \infty}P_{\theta_l}(X\in \bigcap_{j=1}^k R_j)\\ &\ge \lim_{l\rightarrow \infty}P_{\ theta_l}(X\in R_j) - (k-1)\tag{Bonferroni's Inequality}\\ &= (k-1) + \alpha - (k-1)\\ &= \alpha\,, \end{align}$ where the Bonferroni’s inequality says \(P(\bigcup_{i=1}^nE_i)\le \sum_{i=1}^n P Example: Acceptance sampling Two parameters that are important in assessing the quality of upholstery fabric (家具装饰织物) are • $\theta_1$: the mean breaking strength • $\theta_2$: the probability of passing a flammability (易燃性) test Standards may dictate that $\theta_1$ should be over 50 pounds and $\theta_2$ should be over .95, and the fabric is acceptable only if it meets both of these standards. This can be modeled with the hypothesis test \[H_0: \{\theta_1\le 50, \text{ or }\theta_2 \le .95\}\text{ versus }H_1:\{\theta_1 > 50\text{ and }\theta_2 > .95\}\] where a batch of material is acceptable only if $H_1$ is accepted. Suppose $X_1,\ldots, X_n$ are measurements of breaking strength for $n$ samples and are assumed to be iid $N(\theta_1,\sigma^2)$. The LRT of $H_{01}:\theta_1\le 50$ will reject $H_{01}$ if $(\bar X-50)/(S/\sqrt n) > t$. Suppose that we also have the results of $m$ flammability tests, denoted by $Y_1,\ldots,Y_m$, where $Y_i=1$ if the $i$-th sample passes the test. If $Y_1,\ldots,Y_m$ are modeled as iid Bernoulli($\theta_2$) random variables, the LRT will reject $H_{02}:\theta_2\le .95$ if $\sum_{i=1}^mY_i > b$. Putting all of this together, the rejection region for the intersection-union test is given by \[\left\{ (x, y): \frac{\bar x-50}{s/\sqrt n} > t \text{ and }\sum_{i=1}^m y_i > b \right\}\] Let $n = m = 58, t=1.672, b=57$, then each of the individual tests has size $\alpha = .05$ (approximately). Therefore, the IUT is a level $\alpha=0.05$ test. In fact, this test is a size $\alpha = 0.05$ test. Consider a sequence of parameter point $\theta_l = (\theta_{1l}, \theta_2)$, with $\theta_{1l}\rightarrow \infty$ as $l\rightarrow \infty$ and $\theta_2 = .95$. Also, $P_{\theta_l}(X\in R_1)\rightarrow 1$ as $\theta_{1l}\rightarrow \infty$, while $P_{\theta_l}(X\in R_2)=0.05$ for all $l$ because $\theta_2=0.95$. Thus, the IUT is a size $\alpha$ test.	{"url":"https://stats.hohoweiya.xyz/2019/12/02/IUT/","timestamp":"2024-11-13T02:34:04Z","content_type":"text/html","content_length":"13258","record_id":"<urn:uuid:8cf94c81-ab05-436d-9c94-9a893f9e491f>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00025.warc.gz"}
{-# LANGUAGE CPP, ViewPatterns #-} module TcFlatten( flatten, flattenManyNom, ) where #include "HsVersions.h" import GhcPrelude import TcRnTypes import TcType import Type import TcUnify( occCheckExpand ) import TcEvidence import TyCon import TyCoRep -- performs delicate algorithm on types import Coercion import Var import VarEnv import Outputable import TcSMonad as TcS import BasicTypes( SwapFlag(..) ) import Util import Bag import Pair import Control.Monad import MonadUtils ( zipWithAndUnzipM ) import GHC.Exts ( inline ) import Control.Arrow ( first ) Note [The flattening story] * A CFunEqCan is either of form [G] <F xis> : F xis ~ fsk -- fsk is a FlatSkolTv [W] x : F xis ~ fmv -- fmv is a FlatMetaTv x is the witness variable xis are function-free fsk/fmv is a flatten skolem; it is always untouchable (level 0) * CFunEqCans can have any flavour: [G], [W], [WD] or [D] * KEY INSIGHTS: - A given flatten-skolem, fsk, is known a-priori to be equal to F xis (the LHS), with <F xis> evidence. The fsk is still a unification variable, but it is "owned" by its CFunEqCan, and is filled in (unflattened) only by unflattenGivens. - A unification flatten-skolem, fmv, stands for the as-yet-unknown type to which (F xis) will eventually reduce. It is filled in only by dischargeFmv. - All fsk/fmv variables are "untouchable". To make it simple to test, we simply give them TcLevel=0. This means that in a CTyVarEq, say, fmv ~ Int we NEVER unify fmv. - A unification flatten-skolems, fmv, ONLY gets unified when either a) The CFunEqCan takes a step, using an axiom b) By unflattenWanteds They are never unified in any other form of equality. For example [W] ffmv ~ Int is stuck; it does not unify with fmv. * We never substitute in the RHS (i.e. the fsk/fmv) of a CFunEqCan. That would destroy the invariant about the shape of a CFunEqCan, and it would risk wanted/wanted interactions. The only way we learn information about fsk is when the CFunEqCan takes a step. However we do substitute in the LHS of a CFunEqCan (else it would never get to fire!) * Unflattening: - We unflatten Givens when leaving their scope (see unflattenGivens) - We unflatten Wanteds at the end of each attempt to simplify the wanteds; see unflattenWanteds, called from solveSimpleWanteds. * Each canonical [G], [W], or [WD] CFunEqCan x : F xis ~ fsk/fmv has its own distinct evidence variable x and flatten-skolem fsk/fmv. Why? We make a fresh fsk/fmv when the constraint is born; and we never rewrite the RHS of a CFunEqCan. In contrast a [D] CFunEqCan shares its fmv with its partner [W], but does not "own" it. If we reduce a [D] F Int ~ fmv, where say type instance F Int = ty, then we don't discharge fmv := ty. Rather we simply generate [D] fmv ~ ty (in TcInteract.reduce_top_fun_eq) * Inert set invariant: if F xis1 ~ fsk1, F xis2 ~ fsk2 then xis1 /= xis2 i.e. at most one CFunEqCan with a particular LHS * Function applications can occur in the RHS of a CTyEqCan. No reason not allow this, and it reduces the amount of flattening that must occur. * Flattening a type (F xis): - If we are flattening in a Wanted/Derived constraint then create new [W] x : F xis ~ fmv else create new [G] x : F xis ~ fsk with fresh evidence variable x and flatten-skolem fsk/fmv - Add it to the work list - Replace (F xis) with fsk/fmv in the type you are flattening - You can also add the CFunEqCan to the "flat cache", which simply keeps track of all the function applications you have flattened. - If (F xis) is in the cache already, just use its fsk/fmv and evidence x, and emit nothing. - No need to substitute in the flat-cache. It's not the end of the world if we start with, say (F alpha ~ fmv1) and (F Int ~ fmv2) and then find alpha := Int. Athat will simply give rise to fmv1 := fmv2 via [Interacting rule] below * Canonicalising a CFunEqCan [G/W] x : F xis ~ fsk/fmv - Flatten xis (to substitute any tyvars; there are already no functions) cos :: xis ~ flat_xis - New wanted x2 :: F flat_xis ~ fsk/fmv - Add new wanted to flat cache - Discharge x = F cos ; x2 * [Interacting rule] (inert) [W] x1 : F tys ~ fmv1 (work item) [W] x2 : F tys ~ fmv2 Just solve one from the other: x2 := x1 fmv2 := fmv1 This just unites the two fsks into one. Always solve given from wanted if poss. * For top-level reductions, see Note [Top-level reductions for type functions] in TcInteract Why given-fsks, alone, doesn't work Could we get away with only flatten meta-tyvars, with no flatten-skolems? No. [W] w : alpha ~ [F alpha Int] ---> flatten w = ...w'... [W] w' : alpha ~ [fsk] [G] <F alpha Int> : F alpha Int ~ fsk --> unify (no occurs check) alpha := [fsk] But since fsk = F alpha Int, this is really an occurs check error. If that is all we know about alpha, we will succeed in constraint solving, producing a program with an infinite type. Even if we did finally get (g : fsk ~ Bool) by solving (F alpha Int ~ fsk) using axiom, zonking would not see it, so (x::alpha) sitting in the tree will get zonked to an infinite type. (Zonking always only does refl stuff.) Why flatten-meta-vars, alone doesn't work Look at Simple13, with unification-fmvs only [G] g : a ~ [F a] ---> Flatten given g' = g;[x] [G] g' : a ~ [fmv] [W] x : F a ~ fmv --> subst a in x g' = g;[x] x = F g' ; x2 [W] x2 : F [fmv] ~ fmv And now we have an evidence cycle between g' and x! If we used a given instead (ie current story) [G] g : a ~ [F a] ---> Flatten given g' = g;[x] [G] g' : a ~ [fsk] [G] <F a> : F a ~ fsk ---> Substitute for a [G] g' : a ~ [fsk] [G] F (sym g'); <F a> : F [fsk] ~ fsk Why is it right to treat fmv's differently to ordinary unification vars? f :: forall a. a -> a -> Bool g :: F Int -> F Int -> Bool f (x:Int) (y:Bool) This gives alpha~Int, alpha~Bool. There is an inconsistency, but really only one error. SherLoc may tell you which location is most likely, based on other occurrences of alpha. g (x:Int) (y:Bool) Here we get (F Int ~ Int, F Int ~ Bool), which flattens to (fmv ~ Int, fmv ~ Bool) But there are really TWO separate errors. We must not complain about Int~Bool. Moreover these two errors could arise in entirely unrelated parts of the code. (In the alpha case, there must be some connection (eg v:alpha in common envt).) Note [Unflattening can force the solver to iterate] Look at Trac #10340: type family Any :: * -- No instances get :: MonadState s m => m s instance MonadState s (State s) where ... foo :: State Any Any foo = get For 'foo' we instantiate 'get' at types mm ss [WD] MonadState ss mm, [WD] mm ss ~ State Any Any Flatten, and decompose [WD] MonadState ss mm, [WD] Any ~ fmv [WD] mm ~ State fmv, [WD] fmv ~ ss Unify mm := State fmv: [WD] MonadState ss (State fmv) [WD] Any ~ fmv, [WD] fmv ~ ss Now we are stuck; the instance does not match!! So unflatten: fmv := Any ss := Any () [WD] MonadState Any (State Any) The unification () represents progress, so we must do a second round of solving; this time it succeeds. This is done by the 'go' loop in solveSimpleWanteds. This story does not feel right but it's the best I can do; and the iteration only happens in pretty obscure circumstances. * * * Examples Here is a long series of examples I had to work through * * axiom F [a] = [F a] [G] F [a] ~ a [G] fsk ~ a [G] [F a] ~ fsk (nc) [G] F a ~ fsk2 [G] fsk ~ [fsk2] [G] fsk ~ a [G] F a ~ fsk2 [G] a ~ [fsk2] [G] fsk ~ a [W] H (F Bool) ~ H alpha [W] alpha ~ F Bool F Bool ~ fmv0 H fmv0 ~ fmv1 H alpha ~ fmv2 fmv1 ~ fmv2 fmv0 ~ alpha fmv0 := F Bool fmv1 := H (F Bool) fmv2 := H alpha alpha := F Bool fmv1 ~ fmv2 But these two are equal under the above assumptions. Solve by Refl. --- under plan B, namely solve fmv1:=fmv2 eagerly --- [W] H (F Bool) ~ H alpha [W] alpha ~ F Bool F Bool ~ fmv0 H fmv0 ~ fmv1 H alpha ~ fmv2 fmv1 ~ fmv2 fmv0 ~ alpha F Bool ~ fmv0 H fmv0 ~ fmv1 H alpha ~ fmv2 fmv2 := fmv1 fmv0 ~ alpha fmv0 := F Bool fmv1 := H fmv0 = H (F Bool) retain H alpha ~ fmv2 because fmv2 has been filled alpha := F Bool after solving [W] fmv_1 ~ fmv_2 [W] A3 (FCon x) ~ fmv_1 (CFunEqCan) [W] A3 (x (aoa -> fmv_2)) ~ fmv_2 (CFunEqCan) a) [W] BasePrimMonad (Rand m) ~ m1 b) [W] tt m1 ~ BasePrimMonad (Rand m) ---> process (b) first BasePrimMonad (Ramd m) ~ fmv_atH fmv_atH ~ tt m1 ---> now process (a) m1 ~ s_atH ~ tt m1 -- An obscure occurs check Original constraint [W] x + y ~ x + alpha (non-canonical) [W] x + y ~ fmv1 (CFunEqCan) [W] x + alpha ~ fmv2 (CFuneqCan) [W] fmv1 ~ fmv2 (CTyEqCan) [G] Const a ~ () ==> flatten [G] fsk ~ () work item: Const a ~ fsk ==> fire top rule [G] fsk ~ () work item fsk ~ () Surely the work item should rewrite to () ~ ()? Well, maybe not; it'a very special case. More generally, our givens look like F a ~ Int, where (F a) is not reducible. Why using a different can-rewrite rule in CFunEqCan heads does not work. Assuming NOT rewriting wanteds with wanteds Inert: [W] fsk_aBh ~ fmv_aBk -> fmv_aBk [W] fmv_aBk ~ fsk_aBh [G] Scalar fsk_aBg ~ fsk_aBh [G] V a ~ f_aBg Worklist includes [W] Scalar fmv_aBi ~ fmv_aBk fmv_aBi, fmv_aBk are flatten unification variables Work item: [W] V fsk_aBh ~ fmv_aBi Note that the inert wanteds are cyclic, because we do not rewrite wanteds with wanteds. Then we go into a loop when normalise the work-item, because we use rewriteOrSame on the argument of V. Conclusion: Don't make canRewrite context specific; instead use [W] a ~ ty to rewrite a wanted iff 'a' is a unification variable. Here is a somewhat similar case: type family G a :: * blah :: (G a ~ Bool, Eq (G a)) => a -> a blah = error "urk" foo x = blah x For foo we get [W] Eq (G a), G a ~ Bool [W] G a ~ fmv, Eq fmv, fmv ~ Bool We can't simplify away the Eq Bool unless we substitute for fmv. Maybe that doesn't matter: we would still be left with unsolved G a ~ Bool. Trac #9318 has a very simple program leading to [W] F Int ~ Int [W] F Int ~ Bool We don't want to get "Error Int~Bool". But if fmv's can rewrite wanteds, we will [W] fmv ~ Int [W] fmv ~ Bool [W] Int ~ Bool * * * FlattenEnv & FlatM * The flattening environment & monad * * type FlatWorkListRef = TcRef [Ct] -- See Note [The flattening work list] data FlattenEnv = FE { fe_mode :: FlattenMode , fe_loc :: CtLoc -- See Note [Flattener CtLoc] , fe_flavour :: CtFlavour , fe_eq_rel :: EqRel -- See Note [Flattener EqRels] , fe_work :: FlatWorkListRef } -- See Note [The flattening work list] data FlattenMode -- Postcondition for all three: inert wrt the type substitution = FM_FlattenAll -- Postcondition: function-free \| FM_SubstOnly -- See Note [Flattening under a forall] -- \| FM_Avoid TcTyVar Bool -- See Note [Lazy flattening] -- -- Postcondition: -- -- * tyvar is only mentioned in result under a rigid path -- -- e.g. [a] is ok, but F a won't happen -- -- * If flat_top is True, top level is not a function application -- -- (but under type constructors is ok e.g. [F a]) instance Outputable FlattenMode where ppr FM_FlattenAll = text "FM_FlattenAll" ppr FM_SubstOnly = text "FM_SubstOnly" eqFlattenMode :: FlattenMode -> FlattenMode -> Bool eqFlattenMode FM_FlattenAll FM_FlattenAll = True eqFlattenMode FM_SubstOnly FM_SubstOnly = True -- FM_Avoid tv1 b1 `eq` FM_Avoid tv2 b2 = tv1 == tv2 && b1 == b2 eqFlattenMode _ _ = False mkFlattenEnv :: FlattenMode -> CtEvidence -> FlatWorkListRef -> FlattenEnv mkFlattenEnv fm ctev ref = FE { fe_mode = fm , fe_loc = ctEvLoc ctev , fe_flavour = ctEvFlavour ctev , fe_eq_rel = ctEvEqRel ctev , fe_work = ref } -- \| The 'FlatM' monad is a wrapper around 'TcS' with the following -- extra capabilities: (1) it offers access to a 'FlattenEnv'; -- and (2) it maintains the flattening worklist. -- See Note [The flattening work list]. newtype FlatM a = FlatM { runFlatM :: FlattenEnv -> TcS a } instance Monad FlatM where m >>= k = FlatM $ \env -> do { a <- runFlatM m env ; runFlatM (k a) env } instance Functor FlatM where fmap = liftM instance Applicative FlatM where pure x = FlatM $ const (pure x) (<>) = ap liftTcS :: TcS a -> FlatM a liftTcS thing_inside = FlatM $ const thing_inside emitFlatWork :: Ct -> FlatM () -- See Note [The flattening work list] emitFlatWork ct = FlatM $ \env -> updTcRef (fe_work env) (ct :) runFlatten :: FlattenMode -> CtEvidence -> FlatM a -> TcS a -- Run thing_inside (which does flattening), and put all -- the work it generates onto the main work list -- See Note [The flattening work list] -- NB: The returned evidence is always the same as the original, but with -- perhaps a new CtLoc runFlatten mode ev thing_inside = do { flat_ref <- newTcRef [] ; let fmode = mkFlattenEnv mode ev flat_ref ; res <- runFlatM thing_inside fmode ; new_flats <- readTcRef flat_ref ; updWorkListTcS (add_flats new_flats) ; return res } add_flats new_flats wl = wl { wl_funeqs = add_funeqs new_flats (wl_funeqs wl) } add_funeqs [] wl = wl add_funeqs (f:fs) wl = add_funeqs fs (f:wl) -- add_funeqs fs ws = reverse fs ++ ws -- e.g. add_funeqs [f1,f2,f3] [w1,w2,w3,w4] -- = [f3,f2,f1,w1,w2,w3,w4] traceFlat :: String -> SDoc -> FlatM () traceFlat herald doc = liftTcS $ traceTcS herald doc getFlatEnvField :: (FlattenEnv -> a) -> FlatM a getFlatEnvField accessor = FlatM $ \env -> return (accessor env) getEqRel :: FlatM EqRel getEqRel = getFlatEnvField fe_eq_rel getRole :: FlatM Role getRole = eqRelRole <$> getEqRel getFlavour :: FlatM CtFlavour getFlavour = getFlatEnvField fe_flavour getFlavourRole :: FlatM CtFlavourRole = do { flavour <- getFlavour ; eq_rel <- getEqRel ; return (flavour, eq_rel) } getMode :: FlatM FlattenMode getMode = getFlatEnvField fe_mode getLoc :: FlatM CtLoc getLoc = getFlatEnvField fe_loc checkStackDepth :: Type -> FlatM () checkStackDepth ty = do { loc <- getLoc ; liftTcS $ checkReductionDepth loc ty } -- \| Change the 'EqRel' in a 'FlatM'. setEqRel :: EqRel -> FlatM a -> FlatM a setEqRel new_eq_rel thing_inside = FlatM $ \env -> if new_eq_rel == fe_eq_rel env then runFlatM thing_inside env else runFlatM thing_inside (env { fe_eq_rel = new_eq_rel }) -- \| Change the 'FlattenMode' in a 'FlattenEnv'. setMode :: FlattenMode -> FlatM a -> FlatM a setMode new_mode thing_inside = FlatM $ \env -> if new_mode `eqFlattenMode` fe_mode env then runFlatM thing_inside env else runFlatM thing_inside (env { fe_mode = new_mode }) -- \| Use when flattening kinds/kind coercions. See -- Note [No derived kind equalities] flattenKinds :: FlatM a -> FlatM a flattenKinds thing_inside = FlatM $ \env -> let kind_flav = case fe_flavour env of Given -> Given _ -> Wanted WDeriv runFlatM thing_inside (env { fe_eq_rel = NomEq, fe_flavour = kind_flav }) bumpDepth :: FlatM a -> FlatM a bumpDepth (FlatM thing_inside) = FlatM $ \env -> do { let env' = env { fe_loc = bumpCtLocDepth (fe_loc env) } ; thing_inside env' } Note [The flattening work list] The "flattening work list", held in the fe_work field of FlattenEnv, is a list of CFunEqCans generated during flattening. The key idea is this. Consider flattening (Eq (F (G Int) (H Bool)): The flattener recursively calls itself on sub-terms before building the main term, so it will encounter the terms in order G Int H Bool F (G Int) (H Bool) flattening to sub-goals w1: G Int ~ fuv0 w2: H Bool ~ fuv1 w3: F fuv0 fuv1 ~ fuv2 * Processing w3 first is BAD, because we can't reduce i t,so it'll get put into the inert set, and later kicked out when w1, w2 are solved. In Trac #9872 this led to inert sets containing hundreds of suspended calls. * So we want to process w1, w2 first. * So you might think that we should just use a FIFO deque for the work-list, so that putting adding goals in order w1,w2,w3 would mean we processed w1 first. * BUT suppose we have 'type instance G Int = H Char'. Then processing w1 leads to a new goal w4: H Char ~ fuv0 We do NOT want to put that on the far end of a deque! Instead we want to put it at the front of the work-list so that we continue to work on it. So the work-list structure is this: * The wl_funeqs (in TcS) is a LIFO stack; we push new goals (such as w4) on top (extendWorkListFunEq), and take new work from the top * When flattening, emitFlatWork pushes new flattening goals (like w1,w2,w3) onto the flattening work list, fe_work, another push-down stack. * When we finish flattening, we reverse the fe_work stack onto the wl_funeqs stack (which brings w1 to the top). The function runFlatten initialises the fe_work stack, and reverses it onto wl_fun_eqs at the end. Note [Flattener EqRels] When flattening, we need to know which equality relation -- nominal or representation -- we should be respecting. The only difference is that we rewrite variables by representational equalities when fe_eq_rel is ReprEq, and that we unwrap newtypes when flattening w.r.t. representational equality. Note [Flattener CtLoc] The flattener does eager type-family reduction. Type families might loop, and we don't want GHC to do so. A natural solution is to have a bounded depth to these processes. A central difficulty is that such a solution isn't quite compositional. For example, say it takes F Int 10 steps to get to Bool. How many steps does it take to get from F Int -> F Int to Bool -> Bool? 10? 20? What about getting from Const Char (F Int) to Char? 11? 1? Hard to know and hard to track. So, we punt, essentially. We store a CtLoc in the FlattenEnv and just update the environment when recurring. In the TyConApp case, where there may be multiple type families to flatten, we just copy the current CtLoc into each branch. If any branch hits the stack limit, then the whole thing fails. A consequence of this is that setting the stack limits appropriately will be essentially impossible. So, the official recommendation if a stack limit is hit is to disable the check entirely. Otherwise, there will be baffling, unpredictable errors. Note [Lazy flattening] The idea of FM_Avoid mode is to flatten less aggressively. If we have a ~ [F Int] there seems to be no great merit in lifting out (F Int). But if it was a ~ [G a Int] then we do want to lift it out, in case (G a Int) reduces to Bool, say, which gets rid of the occurs-check problem. (For the flat_top Bool, see comments above and at call sites.) HOWEVER, the lazy flattening actually seems to make type inference go slower, not faster. perf/compiler/T3064 is a case in point; it gets dramatically worse with FM_Avoid. I think it may be because floating the types out means we normalise them, and that often makes them smaller and perhaps allows more re-use of previously solved goals. But to be honest I'm not absolutely certain, so I am leaving FM_Avoid in the code base. What I'm removing is the unique place where it is used, namely in TcCanonical.canEqTyVar. See also Note [Conservative unification check] in TcUnify, which gives other examples where lazy flattening caused problems. Bottom line: FM_Avoid is unused for now (Nov 14). Note: T5321Fun got faster when I disabled FM_Avoid T5837 did too, but it's pathalogical anyway Note [Phantoms in the flattener] Suppose we have data Proxy p = Proxy and we're flattening (Proxy ty) w.r.t. ReprEq. Then, we know that `ty` is really irrelevant -- it will be ignored when solving for representational equality later on. So, we omit flattening `ty` entirely. This may violate the expectation of "xi"s for a bit, but the canonicaliser will soon throw out the phantoms when decomposing a TyConApp. (Or, the canonicaliser will emit an insoluble, in which case the unflattened version yields a better error message anyway.) Note [No derived kind equalities] We call flattenKinds in two places: in flatten_co (Note [Flattening coercions]) and in flattenTyVar. The latter case is easier to understand; flattenKinds is used to flatten the kind of a flat (i.e. inert) tyvar. Flattening a kind naturally produces a coercion. This coercion is then used in the flattened type. However, danger lurks if the flattening flavour (that is, the fe_flavour of the FlattenEnv) is Derived: the coercion might be bottom. (This can happen when looks up a kindvar in the inert set only to find a Derived equality, with no coercion.) The solution is simple: ensure that the fe_flavour is not derived when flattening a kind. This is what flattenKinds does. {- ********************************************************************* * * * Externally callable flattening functions * * * * They are all wrapped in runFlatten, so their * * flattening work gets put into the work list * * * ******************************************************************* -} flatten :: FlattenMode -> CtEvidence -> TcType -> TcS (Xi, TcCoercion) flatten mode ev ty = do { traceTcS "flatten {" (ppr mode <+> ppr ty) ; (ty', co) <- runFlatten mode ev (flatten_one ty) ; traceTcS "flatten }" (ppr ty') ; return (ty', co) } flattenManyNom :: CtEvidence -> [TcType] -> TcS ([Xi], [TcCoercion]) -- Externally-callable, hence runFlatten -- Flatten a bunch of types all at once; in fact they are -- always the arguments of a saturated type-family, so -- ctEvFlavour ev = Nominal -- and we want to flatten all at nominal role flattenManyNom ev tys = do { traceTcS "flatten_many {" (vcat (map ppr tys)) ; (tys', cos) <- runFlatten FM_FlattenAll ev (flatten_many_nom tys) ; traceTcS "flatten }" (vcat (map ppr tys')) ; return (tys', cos) } {- ******************************************************************* * * * The main flattening functions * * ********************************************************************* -} {- Note [Flattening] flatten ty ==> (xi, co) xi has no type functions, unless they appear under ForAlls has no skolems that are mapped in the inert set has no filled-in metavariables co :: xi ~ ty Note that it is flatten's job to flatten every type function it sees. flatten is only called on arguments to type functions, by canEqGiven. Flattening also: * zonks, removing any metavariables, and * applies the substitution embodied in the inert set Because flattening zonks and the returned coercion ("co" above) is also zonked, it's possible that (co :: xi ~ ty) isn't quite true. So, instead, we can rely on these facts: (F1) typeKind(xi) succeeds and returns a fully zonked kind (F2) co :: xi ~ zonk(ty) Note that the left-hand type of co is always precisely xi. The right-hand type may or may not be ty, however: if ty has unzonked filled-in metavariables, then the right-hand type of co will be the zonked version of ty. It is for this reason that we occasionally have to explicitly zonk, when (co :: xi ~ ty) is important even before we zonk the whole program. For example, see the FTRNotFollowed case in flattenTyVar. Why have these invariants on flattening? Really, they're both to ensure invariant (F1), which is a Good Thing because we sometimes use typeKind during canonicalisation, and we want this kind to be zonked (e.g., see TcCanonical.homogeniseRhsKind). Invariant (F2) is needed solely to support (F1). It is relied on in one place: - The FTRNotFollowed case in flattenTyVar. Here, we have a tyvar that cannot be reduced any further (that is, no equality over the tyvar is in the inert set such that the inert equality can rewrite the constraint at hand, and it is not a filled-in metavariable). But its kind might still not be flat, if it mentions a type family or a variable that can be rewritten. Flattened types have flattened kinds (see below), so we must flatten the kind. Here is an example: let kappa be a filled-in metavariable such that kappa := k. [G] co :: k ~ Type We are flattening a :: kappa where a is a skolem. We end up in the FTRNotFollowed case, but we need to flatten the kind kappa. Flattening kappa yields (Type, kind_co), where kind_co :: Type ~ k. Note that the right-hand type of kind_co is not kappa, because (F1) tells us it's zonk(kappa), which is k. Now, we return (a \|> sym kind_co). If we are to uphold (F1), then the right-hand type of (sym kind_co) had better be fully zonked. In other words, the left-hand type of kind_co needs to be zonked... which is precisely what (F2) In order to support (F2), we require that ctEvCoercion, when called on a zonked CtEvidence, always returns a zonked coercion. See Note [Given in ctEvCoercion]. This requirement comes into play in flatten_tyvar2. (I suppose we could move the logic from ctEvCoercion to flatten_tyvar2, but it's much easier to do in ctEvCoercion.) Flattening a type also means flattening its kind. In the case of a type variable whose kind mentions a type family, this might mean that the result of flattening has a cast in it. Recall that in comments we use alpha[flat = ty] to represent a flattening skolem variable alpha which has been generated to stand in for ty. ----- Example of flattening a constraint: ------ flatten (List (F (G Int))) ==> (xi, cc) xi = List alpha cc = { G Int ~ beta[flat = G Int], F beta ~ alpha[flat = F beta] } * alpha and beta are 'flattening skolem variables'. * All the constraints in cc are 'given', and all their coercion terms are the identity. NB: Flattening Skolems only occur in canonical constraints, which are never zonked, so we don't need to worry about zonking doing accidental unflattening. Note that we prefer to leave type synonyms unexpanded when possible, so when the flattener encounters one, it first asks whether its transitive expansion contains any type function applications. If so, it expands the synonym and proceeds; if not, it simply returns the unexpanded synonym. Note [flatten_many performance] In programs with lots of type-level evaluation, flatten_many becomes part of a tight loop. For example, see test perf/compiler/T9872a, which calls flatten_many a whopping 7,106,808 times. It is thus important that flatten_many be efficient. Performance testing showed that the current implementation is indeed efficient. It's critically important that zipWithAndUnzipM be specialized to TcS, and it's also quite helpful to actually `inline` it. On test T9872a, here are the allocation stats (Dec 16, 2014): * Unspecialized, uninlined: 8,472,613,440 bytes allocated in the heap * Specialized, uninlined: 6,639,253,488 bytes allocated in the heap * Specialized, inlined: 6,281,539,792 bytes allocated in the heap To improve performance even further, flatten_many_nom is split off from flatten_many, as nominal equality is the common case. This would be natural to write using mapAndUnzipM, but even inlined, that function is not as performant as a hand-written loop. * mapAndUnzipM, inlined: 7,463,047,432 bytes allocated in the heap * hand-written recursion: 5,848,602,848 bytes allocated in the heap If you make any change here, pay close attention to the T9872{a,b,c} tests and T5321Fun. If we need to make this yet more performant, a possible way forward is to duplicate the flattener code for the nominal case, and make that case faster. This doesn't seem quite worth it, yet. flatten_many :: [Role] -> [Type] -> FlatM ([Xi], [Coercion]) -- Coercions :: Xi ~ Type, at roles given -- Returns True iff (no flattening happened) -- NB: The EvVar inside the 'fe_ev :: CtEvidence' is unused, -- we merely want (a) Given/Solved/Derived/Wanted info -- (b) the GivenLoc/WantedLoc for when we create new evidence flatten_many roles tys -- See Note [flatten_many performance] = inline zipWithAndUnzipM go roles tys go Nominal ty = setEqRel NomEq $ flatten_one ty go Representational ty = setEqRel ReprEq $ flatten_one ty go Phantom ty = -- See Note [Phantoms in the flattener] do { ty <- liftTcS $ zonkTcType ty ; return ( ty, mkReflCo Phantom ty ) } -- \| Like 'flatten_many', but assumes that every role is nominal. flatten_many_nom :: [Type] -> FlatM ([Xi], [Coercion]) flatten_many_nom [] = return ([], []) -- See Note [flatten_many performance] flatten_many_nom (ty:tys) = do { (xi, co) <- flatten_one ty ; (xis, cos) <- flatten_many_nom tys ; return (xi:xis, co:cos) } flatten_one :: TcType -> FlatM (Xi, Coercion) -- Flatten a type to get rid of type function applications, returning -- the new type-function-free type, and a collection of new equality -- constraints. See Note [Flattening] for more detail. -- Postcondition: Coercion :: Xi ~ TcType -- The role on the result coercion matches the EqRel in the FlattenEnv flatten_one xi@(LitTy {}) = do { role <- getRole ; return (xi, mkReflCo role xi) } flatten_one (TyVarTy tv) = flattenTyVar tv flatten_one (AppTy ty1 ty2) = do { (xi1,co1) <- flatten_one ty1 ; eq_rel <- getEqRel ; case (eq_rel, nextRole xi1) of -- We need nextRole here because although ty1 definitely -- isn't a TyConApp, xi1 might be. -- ToDo: but can such a substitution change roles?? (NomEq, _) -> flatten_rhs xi1 co1 NomEq (ReprEq, Nominal) -> flatten_rhs xi1 co1 NomEq (ReprEq, Representational) -> flatten_rhs xi1 co1 ReprEq (ReprEq, Phantom) -> -- See Note [Phantoms in the flattener] do { ty2 <- liftTcS $ zonkTcType ty2 ; return ( mkAppTy xi1 ty2 , mkAppCo co1 (mkNomReflCo ty2)) } } flatten_rhs xi1 co1 eq_rel2 = do { (xi2,co2) <- setEqRel eq_rel2 $ flatten_one ty2 ; role1 <- getRole ; let role2 = eqRelRole eq_rel2 ; traceFlat "flatten/appty" (ppr ty1 $$ ppr ty2 $$ ppr xi1 $$ ppr xi2 $$ ppr role1 $$ ppr role2) ; return ( mkAppTy xi1 xi2 , mkTransAppCo role1 co1 xi1 ty1 role2 co2 xi2 ty2 role1 ) } -- output should match fmode flatten_one (TyConApp tc tys) -- Expand type synonyms that mention type families -- on the RHS; see Note [Flattening synonyms] \| Just (tenv, rhs, tys') <- expandSynTyCon_maybe tc tys , let expanded_ty = mkAppTys (substTy (mkTvSubstPrs tenv) rhs) tys' = do { mode <- getMode ; case mode of FM_FlattenAll \| not (isFamFreeTyCon tc) -> flatten_one expanded_ty _ -> flatten_ty_con_app tc tys } -- Otherwise, it's a type function application, and we have to -- flatten it away as well, and generate a new given equality constraint -- between the application and a newly generated flattening skolem variable. \| isTypeFamilyTyCon tc = flatten_fam_app tc tys -- For * a normal data type application -- * data family application -- we just recursively flatten the arguments. \| otherwise -- FM_Avoid stuff commented out; see Note [Lazy flattening] -- , let fmode' = case fmode of -- Switch off the flat_top bit in FM_Avoid -- FE { fe_mode = FM_Avoid tv _ } -- -> fmode { fe_mode = FM_Avoid tv False } -- _ -> fmode = flatten_ty_con_app tc tys flatten_one (FunTy ty1 ty2) = do { (xi1,co1) <- flatten_one ty1 ; (xi2,co2) <- flatten_one ty2 ; role <- getRole ; return (mkFunTy xi1 xi2, mkFunCo role co1 co2) } flatten_one ty@(ForAllTy {}) -- TODO (RAE): This is inadequate, as it doesn't flatten the kind of -- the bound tyvar. Doing so will require carrying around a substitution -- and the usual substTyVarBndr-like silliness. Argh. -- We allow for-alls when, but only when, no type function -- applications inside the forall involve the bound type variables. = do { let (bndrs, rho) = splitForAllTyVarBndrs ty tvs = binderVars bndrs ; (rho', co) <- setMode FM_SubstOnly $ flatten_one rho -- Substitute only under a forall -- See Note [Flattening under a forall] ; return (mkForAllTys bndrs rho', mkHomoForAllCos tvs co) } flatten_one (CastTy ty g) = do { (xi, co) <- flatten_one ty ; (g', _) <- flatten_co g ; return (mkCastTy xi g', castCoercionKind co g' g) } flatten_one (CoercionTy co) = first mkCoercionTy <$> flatten_co co -- \| "Flatten" a coercion. Really, just flatten the types that it coerces -- between and then use transitivity. See Note [Flattening coercions] flatten_co :: Coercion -> FlatM (Coercion, Coercion) flatten_co co = do { co <- liftTcS $ zonkCo co -- see Note [Zonking when flattening a coercion] ; let (Pair ty1 ty2, role) = coercionKindRole co ; (co1, co2) <- flattenKinds $ do { (_, co1) <- flatten_one ty1 ; (_, co2) <- flatten_one ty2 ; return (co1, co2) } ; let co' = downgradeRole role Nominal co1 `mkTransCo` co `mkTransCo` mkSymCo (downgradeRole role Nominal co2) -- kco :: (ty1' ~r ty2') ~N (ty1 ~r ty2) kco = mkTyConAppCo Nominal (equalityTyCon role) [ mkKindCo co1, mkKindCo co2, co1, co2 ] ; traceFlat "flatten_co" (vcat [ ppr co, ppr co1, ppr co2, ppr co' ]) ; env_role <- getRole ; return (co', mkProofIrrelCo env_role kco co' co) } flatten_ty_con_app :: TyCon -> [TcType] -> FlatM (Xi, Coercion) flatten_ty_con_app tc tys = do { eq_rel <- getEqRel ; let role = eqRelRole eq_rel ; (xis, cos) <- case eq_rel of NomEq -> flatten_many_nom tys ReprEq -> flatten_many (tyConRolesRepresentational tc) tys ; return (mkTyConApp tc xis, mkTyConAppCo role tc cos) } Note [Flattening coercions] Because a flattened type has a flattened kind, we also must "flatten" coercions as we walk through a type. Otherwise, the "from" type of the coercion might not match the (now flattened) kind of the type that it's casting. flatten_co does the work, taking a coercion of type (ty1 ~ ty2) and flattening it to have type (fty1 ~ fty2), where flatten(ty1) = fty1 and flatten(ty2) = fty2. In other words: If r1 is a role co :: s ~r1 t flatten_co co = (fco, kco) r2 is the role in the FlatM fco :: fs ~r1 ft fs, ft are flattened types kco :: fco ~r2 co The second return value of flatten_co is always a ProofIrrelCo. As such, it doesn't contain any information the caller doesn't have and might not be necessary in whatever comes next. Note that a flattened coercion might have unzonked metavariables or type functions in it -- but its kind will not. Instead of just flattening the kinds and using mkTransCo, we could actually flatten the coercion structurally. But doing so seems harder than simply flattening the types. Note [Zonking when flattening a coercion] The first step in flatten_co (see Note [Flattening coercions]) is to zonk the input. This is necessary because we want to ensure the following invariants (c.f. the invariants (F1) and (F2) in Note [Flattening]) (co', kco) <- flatten_co co (FC1) coercionKind(co') succeeds and produces a fully zonked pair of kinds (FC2) kco :: co' ~ zonk(co) We must zonk to ensure (1). This is because fco is built by using mkTransCo to build up on the input co. But if the only action that happens during flattening ty1 and ty2 is to zonk metavariables, the coercions returned (co1 and co2) will be reflexive. The mkTransCo calls will drop the reflexive coercions and co' will be the same as co -- with unzonked kinds. These invariants are necessary to uphold (F1) and (F2) in the CastTy and CoercionTy cases. We zonk right at the beginning to avoid duplicating work when flattening the ty1 and ty2. Note [Flattening synonyms] Not expanding synonyms aggressively improves error messages, and keeps types smaller. But we need to take care. type T a = a -> a and we want to flatten the type (T (F a)). Then we can safely flatten the (F a) to a skolem, and return (T fsk). We don't need to expand the synonym. This works because TcTyConAppCo can deal with synonyms (unlike TyConAppCo), see Note [TcCoercions] in TcEvidence. But (Trac #8979) for type T a = (F a, a) where F is a type function we must expand the synonym in (say) T Int, to expose the type function to the flattener. Note [Flattening under a forall] Under a forall, we (a) MUST apply the inert substitution (b) MUST NOT flatten type family applications Hence FMSubstOnly. For (a) consider c ~ a, a ~ T (forall b. (b, [c])) If we don't apply the c~a substitution to the second constraint we won't see the occurs-check error. For (b) consider (a ~ forall b. F a b), we don't want to flatten to (a ~ forall b.fsk, F a b ~ fsk) because now the 'b' has escaped its scope. We'd have to flatten to (a ~ forall b. fsk b, forall b. F a b ~ fsk b) and we have not begun to think about how to make that work! * * Flattening a type-family application * * flatten_fam_app :: TyCon -> [TcType] -> FlatM (Xi, Coercion) -- flatten_fam_app can be over-saturated -- flatten_exact_fam_app is exactly saturated -- flatten_exact_fam_app_fully lifts out the application to top level -- Postcondition: Coercion :: Xi ~ F tys flatten_fam_app tc tys -- Can be over-saturated = ASSERT2( tys `lengthAtLeast` tyConArity tc , ppr tc $$ ppr (tyConArity tc) $$ ppr tys) -- Type functions are saturated -- The type function might be over saturated -- in which case the remaining arguments should -- be dealt with by AppTys do { let (tys1, tys_rest) = splitAt (tyConArity tc) tys ; (xi1, co1) <- flatten_exact_fam_app tc tys1 -- co1 :: xi1 ~ F tys1 -- all Nominal roles b/c the tycon is oversaturated ; (xis_rest, cos_rest) <- flatten_many (repeat Nominal) tys_rest -- cos_res :: xis_rest ~ tys_rest ; return ( mkAppTys xi1 xis_rest -- NB mkAppTys: rhs_xi might not be a type variable -- cf Trac #5655 , mkAppCos co1 cos_rest -- (rhs_xi :: F xis) ; (F cos :: F xis ~ F tys) ) } flatten_exact_fam_app, flatten_exact_fam_app_fully :: TyCon -> [TcType] -> FlatM (Xi, Coercion) flatten_exact_fam_app tc tys = do { mode <- getMode ; role <- getRole ; case mode of -- These roles are always going to be Nominal for now, -- but not if #8177 is implemented FM_SubstOnly -> do { let roles = tyConRolesX role tc ; (xis, cos) <- flatten_many roles tys ; return ( mkTyConApp tc xis , mkTyConAppCo role tc cos ) } FM_FlattenAll -> flatten_exact_fam_app_fully tc tys } -- FM_Avoid tv flat_top -> -- do { (xis, cos) <- flatten_many fmode roles tys -- ; if flat_top \|\| tv `elemVarSet` tyCoVarsOfTypes xis -- then flatten_exact_fam_app_fully fmode tc tys -- else return ( mkTyConApp tc xis -- , mkTcTyConAppCo (feRole fmode) tc cos ) } flatten_exact_fam_app_fully tc tys -- See Note [Reduce type family applications eagerly] = try_to_reduce tc tys False id $ do { -- First, flatten the arguments ; (xis, cos) <- setEqRel NomEq $ flatten_many_nom tys ; eq_rel <- getEqRel ; cur_flav <- getFlavour ; let role = eqRelRole eq_rel ret_co = mkTyConAppCo role tc cos -- ret_co :: F xis ~ F tys -- Now, look in the cache ; mb_ct <- liftTcS $ lookupFlatCache tc xis ; case mb_ct of Just (co, rhs_ty, flav) -- co :: F xis ~ fsk -- flav is [G] or [WD] -- See Note [Type family equations] in TcSMonad \| (NotSwapped, _) <- flav `funEqCanDischargeF` cur_flav -> -- Usable hit in the flat-cache do { traceFlat "flatten/flat-cache hit" $ (ppr tc <+> ppr xis $$ ppr rhs_ty) ; (fsk_xi, fsk_co) <- flatten_one rhs_ty -- The fsk may already have been unified, so flatten it -- fsk_co :: fsk_xi ~ fsk ; return ( fsk_xi , fsk_co `mkTransCo` maybeSubCo eq_rel (mkSymCo co) `mkTransCo` ret_co ) } -- :: fsk_xi ~ F xis -- Try to reduce the family application right now -- See Note [Reduce type family applications eagerly] _ -> try_to_reduce tc xis True (`mkTransCo` ret_co) $ do { loc <- getLoc ; (ev, co, fsk) <- liftTcS $ newFlattenSkolem cur_flav loc tc xis -- The new constraint (F xis ~ fsk) is not necessarily inert -- (e.g. the LHS may be a redex) so we must put it in the work list ; let ct = CFunEqCan { cc_ev = ev , cc_fun = tc , cc_tyargs = xis , cc_fsk = fsk } ; emitFlatWork ct ; traceFlat "flatten/flat-cache miss" $ (ppr tc <+> ppr xis $$ ppr fsk $$ ppr ev) -- NB: fsk's kind is already flattend because -- the xis are flattened ; return (mkTyVarTy fsk, maybeSubCo eq_rel (mkSymCo co) `mkTransCo` ret_co ) } try_to_reduce :: TyCon -- F, family tycon -> [Type] -- args, not necessarily flattened -> Bool -- add to the flat cache? -> ( Coercion -- :: xi ~ F args -> Coercion ) -- what to return from outer function -> FlatM (Xi, Coercion) -- continuation upon failure -> FlatM (Xi, Coercion) try_to_reduce tc tys cache update_co k = do { checkStackDepth (mkTyConApp tc tys) ; mb_match <- liftTcS $ matchFam tc tys ; case mb_match of Just (norm_co, norm_ty) -> do { traceFlat "Eager T.F. reduction success" $ vcat [ ppr tc, ppr tys, ppr norm_ty , ppr norm_co <+> dcolon <+> ppr (coercionKind norm_co) , ppr cache] ; (xi, final_co) <- bumpDepth $ flatten_one norm_ty ; eq_rel <- getEqRel ; let co = maybeSubCo eq_rel norm_co `mkTransCo` mkSymCo final_co ; flavour <- getFlavour -- NB: only extend cache with nominal equalities ; when (cache && eq_rel == NomEq) $ liftTcS $ extendFlatCache tc tys ( co, xi, flavour ) ; return ( xi, update_co $ mkSymCo co ) } Nothing -> k } {- Note [Reduce type family applications eagerly] If we come across a type-family application like (Append (Cons x Nil) t), then, rather than flattening to a skolem etc, we may as well just reduce it on the spot to (Cons x t). This saves a lot of intermediate steps. Examples that are helped are tests T9872, and T5321Fun. Performance testing indicates that it's best to try this twice, once before flattening arguments and once after flattening arguments. Adding the extra reduction attempt before flattening arguments cut the allocation amounts for the T9872{a,b,c} tests by half. An example of where the early reduction appears helpful: type family Last x where Last '[x] = x Last (h ': t) = Last t workitem: (x ~ Last '[1,2,3,4,5,6]) Flattening the argument never gets us anywhere, but trying to flatten it at every step is quadratic in the length of the list. Reducing more eagerly makes simplifying the right-hand type linear in its length. Testing also indicated that the early reduction should not use the flat-cache, but that the later reduction should. (Although the effect was not large.) Hence the Bool argument to try_to_reduce. To me (SLPJ) this seems odd; I get that eager reduction usually succeeds; and if don't use the cache for eager reduction, we will miss most of the opportunities for using it at all. More exploration would be good At the end, once we've got a flat rhs, we extend the flatten-cache to record the result. Doing so can save lots of work when the same redex shows up more than once. Note that we record the link from the redex all the way to its final value, not just the single step reduction. Interestingly, using the flat-cache for the first reduction resulted in an increase in allocations of about 3% for the four T9872x tests. However, using the flat-cache in the later reduction is a similar gain. I (Richard E) don't currently (Dec '14) have any knowledge as to why these facts are true. * * Flattening a type variable * * ********************************************************************* -} -- \| The result of flattening a tyvar "one step". data FlattenTvResult = FTRNotFollowed -- ^ The inert set doesn't make the tyvar equal to anything else \| FTRFollowed TcType Coercion -- ^ The tyvar flattens to a not-necessarily flat other type. -- co :: new type ~r old type, where the role is determined by -- the FlattenEnv flattenTyVar :: TyVar -> FlatM (Xi, Coercion) flattenTyVar tv = do { mb_yes <- flatten_tyvar1 tv ; case mb_yes of FTRFollowed ty1 co1 -- Recur -> do { (ty2, co2) <- flatten_one ty1 -- ; traceFlat "flattenTyVar2" (ppr tv $$ ppr ty2) ; return (ty2, co2 `mkTransCo` co1) } FTRNotFollowed -- Done -> do { let orig_kind = tyVarKind tv ; (_new_kind, kind_co) <- flattenKinds $ flatten_one orig_kind ; let Pair _ zonked_kind = coercionKind kind_co -- NB: kind_co :: _new_kind ~ zonked_kind -- But zonked_kind is not necessarily the same as orig_kind -- because that may have filled-in metavars. -- Moreover the returned Xi type must be well-kinded -- (e.g. in canEqTyVarTyVar we use getCastedTyVar_maybe) -- If you remove it, then e.g. dependent/should_fail/T11407 panics -- See also Note [Flattening] -- An alternative would to use (zonkTcType orig_kind), -- but some simple measurements suggest that's a little slower ; let tv' = setTyVarKind tv zonked_kind tv_ty' = mkTyVarTy tv' ty' = tv_ty' `mkCastTy` mkSymCo kind_co ; role <- getRole ; return (ty', mkReflCo role tv_ty' `mkCoherenceLeftCo` mkSymCo kind_co) } } flatten_tyvar1 :: TcTyVar -> FlatM FlattenTvResult -- "Flattening" a type variable means to apply the substitution to it -- Specifically, look up the tyvar in -- * the internal MetaTyVar box -- * the inerts -- See also the documentation for FlattenTvResult flatten_tyvar1 tv = do { mb_ty <- liftTcS $ isFilledMetaTyVar_maybe tv ; case mb_ty of Just ty -> do { traceFlat "Following filled tyvar" (ppr tv <+> equals <+> ppr ty) ; role <- getRole ; return (FTRFollowed ty (mkReflCo role ty)) } ; Nothing -> do { traceFlat "Unfilled tyvar" (ppr tv) ; fr <- getFlavourRole ; flatten_tyvar2 tv fr } } flatten_tyvar2 :: TcTyVar -> CtFlavourRole -> FlatM FlattenTvResult -- The tyvar is not a filled-in meta-tyvar -- Try in the inert equalities -- See Definition [Applying a generalised substitution] in TcSMonad -- See Note [Stability of flattening] in TcSMonad flatten_tyvar2 tv fr@(_, eq_rel) = do { ieqs <- liftTcS $ getInertEqs ; mode <- getMode ; case lookupDVarEnv ieqs tv of Just (ct:_) -- If the first doesn't work, -- the subsequent ones won't either \| CTyEqCan { cc_ev = ctev, cc_tyvar = tv , cc_rhs = rhs_ty, cc_eq_rel = ct_eq_rel } <- ct , let ct_fr = (ctEvFlavour ctev, ct_eq_rel) , ct_fr `eqCanRewriteFR` fr -- This is THE key call of eqCanRewriteFR -> do { traceFlat "Following inert tyvar" (ppr mode <+> ppr tv <+> equals <+> ppr rhs_ty $$ ppr ctev) ; let rewrite_co1 = mkSymCo (ctEvCoercion ctev) rewrite_co = case (ct_eq_rel, eq_rel) of (ReprEq, _rel) -> ASSERT( _rel == ReprEq ) -- if this ASSERT fails, then -- eqCanRewriteFR answered incorrectly (NomEq, NomEq) -> rewrite_co1 (NomEq, ReprEq) -> mkSubCo rewrite_co1 ; return (FTRFollowed rhs_ty rewrite_co) } -- NB: ct is Derived then fmode must be also, hence -- we are not going to touch the returned coercion -- so ctEvCoercion is fine. _other -> return FTRNotFollowed } Note [An alternative story for the inert substitution] (This entire note is just background, left here in case we ever want to return the the previous state of affairs) We used (GHC 7.8) to have this story for the inert substitution inert_eqs * 'a' is not in fvs(ty) * They are inert in the weaker sense that there is no infinite chain of (i1 `eqCanRewrite` i2), (i2 `eqCanRewrite` i3), etc This means that flattening must be recursive, but it does allow [G] a ~ [b] [G] b ~ Maybe c This avoids "saturating" the Givens, which can save a modest amount of work. It is easy to implement, in TcInteract.kick_out, by only kicking out an inert only if (a) the work item can rewrite the inert AND (b) the inert cannot rewrite the work item This is significantly harder to think about. It can save a LOT of work in occurs-check cases, but we don't care about them much. Trac #5837 is an example; all the constraints here are Givens [G] a ~ TF (a,Int) work TF (a,Int) ~ fsk inert fsk ~ a work fsk ~ (TF a, TF Int) inert fsk ~ a work a ~ (TF a, TF Int) inert fsk ~ a ---> (attempting to flatten (TF a) so that it does not mention a work TF a ~ fsk2 inert a ~ (fsk2, TF Int) inert fsk ~ (fsk2, TF Int) ---> (substitute for a) work TF (fsk2, TF Int) ~ fsk2 inert a ~ (fsk2, TF Int) inert fsk ~ (fsk2, TF Int) ---> (top-level reduction, re-orient) work fsk2 ~ (TF fsk2, TF Int) inert a ~ (fsk2, TF Int) inert fsk ~ (fsk2, TF Int) ---> (attempt to flatten (TF fsk2) to get rid of fsk2 work TF fsk2 ~ fsk3 work fsk2 ~ (fsk3, TF Int) inert a ~ (fsk2, TF Int) inert fsk ~ (fsk2, TF Int) work TF fsk2 ~ fsk3 inert fsk2 ~ (fsk3, TF Int) inert a ~ ((fsk3, TF Int), TF Int) inert fsk ~ ((fsk3, TF Int), TF Int) Because the incoming given rewrites all the inert givens, we get more and more duplication in the inert set. But this really only happens in pathalogical casee, so we don't care. * * * * An unflattening example: [W] F a ~ alpha flattens to [W] F a ~ fmv (CFunEqCan) [W] fmv ~ alpha (CTyEqCan) We must solve both! unflattenWanteds :: Cts -> Cts -> TcS Cts unflattenWanteds tv_eqs funeqs = do { tclvl <- getTcLevel ; traceTcS "Unflattening" $ braces $ vcat [ text "Funeqs =" <+> pprCts funeqs , text "Tv eqs =" <+> pprCts tv_eqs ] -- Step 1: unflatten the CFunEqCans, except if that causes an occurs check -- Occurs check: consider [W] alpha ~ [F alpha] -- ==> (flatten) [W] F alpha ~ fmv, [W] alpha ~ [fmv] -- ==> (unify) [W] F [fmv] ~ fmv -- See Note [Unflatten using funeqs first] ; funeqs <- foldrBagM unflatten_funeq emptyCts funeqs ; traceTcS "Unflattening 1" $ braces (pprCts funeqs) -- Step 2: unify the tv_eqs, if possible ; tv_eqs <- foldrBagM (unflatten_eq tclvl) emptyCts tv_eqs ; traceTcS "Unflattening 2" $ braces (pprCts tv_eqs) -- Step 3: fill any remaining fmvs with fresh unification variables ; funeqs <- mapBagM finalise_funeq funeqs ; traceTcS "Unflattening 3" $ braces (pprCts funeqs) -- Step 4: remove any tv_eqs that look like ty ~ ty ; tv_eqs <- foldrBagM finalise_eq emptyCts tv_eqs ; let all_flat = tv_eqs `andCts` funeqs ; traceTcS "Unflattening done" $ braces (pprCts all_flat) ; return all_flat } unflatten_funeq :: Ct -> Cts -> TcS Cts unflatten_funeq ct@(CFunEqCan { cc_fun = tc, cc_tyargs = xis , cc_fsk = fmv, cc_ev = ev }) rest = do { -- fmv should be an un-filled flatten meta-tv; -- we now fix its final value by filling it, being careful -- to observe the occurs check. Zonking will eliminate it -- altogether in due course rhs' <- zonkTcType (mkTyConApp tc xis) ; case occCheckExpand fmv rhs' of Just rhs'' -- Normal case: fill the tyvar -> do { setReflEvidence ev NomEq rhs'' ; unflattenFmv fmv rhs'' ; return rest } Nothing -> -- Occurs check return (ct `consCts` rest) } unflatten_funeq other_ct _ = pprPanic "unflatten_funeq" (ppr other_ct) finalise_funeq :: Ct -> TcS Ct finalise_funeq (CFunEqCan { cc_fsk = fmv, cc_ev = ev }) = do { demoteUnfilledFmv fmv ; return (mkNonCanonical ev) } finalise_funeq ct = pprPanic "finalise_funeq" (ppr ct) unflatten_eq :: TcLevel -> Ct -> Cts -> TcS Cts unflatten_eq tclvl ct@(CTyEqCan { cc_ev = ev, cc_tyvar = tv , cc_rhs = rhs, cc_eq_rel = eq_rel }) rest \| isFmvTyVar tv -- Previously these fmvs were untouchable, -- but now they are touchable -- NB: unlike unflattenFmv, filling a fmv here /does/ -- bump the unification count; it is "improvement" -- Note [Unflattening can force the solver to iterate] = ASSERT2( tyVarKind tv `eqType` typeKind rhs, ppr ct ) -- CTyEqCan invariant should ensure this is true do { is_filled <- isFilledMetaTyVar tv ; elim <- case is_filled of False -> do { traceTcS "unflatten_eq 2" (ppr ct) ; tryFill ev eq_rel tv rhs } True -> do { traceTcS "unflatten_eq 2" (ppr ct) ; try_fill_rhs ev eq_rel tclvl tv rhs } ; if elim then return rest else return (ct `consCts` rest) } \| otherwise = return (ct `consCts` rest) unflatten_eq _ ct _ = pprPanic "unflatten_irred" (ppr ct) try_fill_rhs ev eq_rel tclvl lhs_tv rhs -- Constraint is lhs_tv ~ rhs_tv, -- and lhs_tv is filled, so try RHS \| Just (rhs_tv, co) <- getCastedTyVar_maybe rhs -- co :: kind(rhs_tv) ~ kind(lhs_tv) , isFmvTyVar rhs_tv \|\| (isTouchableMetaTyVar tclvl rhs_tv && not (isSigTyVar rhs_tv)) -- LHS is a filled fmv, and so is a type -- family application, which a SigTv should -- not unify with = do { is_filled <- isFilledMetaTyVar rhs_tv ; if is_filled then return False else tryFill ev eq_rel rhs_tv (mkTyVarTy lhs_tv `mkCastTy` mkSymCo co) } \| otherwise = return False finalise_eq :: Ct -> Cts -> TcS Cts finalise_eq (CTyEqCan { cc_ev = ev, cc_tyvar = tv , cc_rhs = rhs, cc_eq_rel = eq_rel }) rest \| isFmvTyVar tv = do { ty1 <- zonkTcTyVar tv ; rhs' <- zonkTcType rhs ; if ty1 `tcEqType` rhs' then do { setReflEvidence ev eq_rel rhs' ; return rest } else return (mkNonCanonical ev `consCts` rest) } \| otherwise = return (mkNonCanonical ev `consCts` rest) finalise_eq ct _ = pprPanic "finalise_irred" (ppr ct) tryFill :: CtEvidence -> EqRel -> TcTyVar -> TcType -> TcS Bool -- (tryFill tv rhs ev) assumes 'tv' is an /un-filled/ MetaTv -- If tv does not appear in 'rhs', it set tv := rhs, -- binds the evidence (which should be a CtWanted) to Refl<rhs> -- and return True. Otherwise returns False tryFill ev eq_rel tv rhs = ASSERT2( not (isGiven ev), ppr ev ) do { rhs' <- zonkTcType rhs ; case tcGetTyVar_maybe rhs' of { Just tv' \| tv == tv' -> do { setReflEvidence ev eq_rel rhs ; return True } ; _other -> do { case occCheckExpand tv rhs' of Just rhs'' -- Normal case: fill the tyvar -> do { setReflEvidence ev eq_rel rhs'' ; unifyTyVar tv rhs'' ; return True } Nothing -> -- Occurs check return False } } } setReflEvidence :: CtEvidence -> EqRel -> TcType -> TcS () setReflEvidence ev eq_rel rhs = setEvBindIfWanted ev (EvCoercion refl_co) refl_co = mkTcReflCo (eqRelRole eq_rel) rhs Note [Unflatten using funeqs first] [W] G a ~ Int [W] F (G a) ~ G a do not want to end up with [W] F Int ~ Int because that might actually hold! Better to end up with the two above unsolved constraints. The flat form will be G a ~ fmv1 (CFunEqCan) F fmv1 ~ fmv2 (CFunEqCan) fmv1 ~ Int (CTyEqCan) fmv1 ~ fmv2 (CTyEqCan) Flatten using the fun-eqs first.	{"url":"http://lambda.inf.elte.hu/haskell/doc/libraries/ghc-8.4.4/src/TcFlatten.html","timestamp":"2024-11-14T07:49:07Z","content_type":"text/html","content_length":"206631","record_id":"<urn:uuid:ac6e3e98-d65d-4af5-8ce3-ff5fe0a77878>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00812.warc.gz"}
A Null Printer Cable for PLIP Next: Sample smail Configuration Files Up: The Network Administrators' Guide Previous: nn Configuration To make a Null Printer Cable for use with a PLIP connection, you need two 25-pin connectors (called DB-25) and some 11-conductor cable. The cable must be at most 15-meters long. If you look at the connector, you should be able to read tiny numbers at the base of each pin, from 1 for the pin top left (if you hold the broader side up) to 25 for the pin bottom right. For the Null Printer cable, you have to connect the following pins of both connectors with each other: D0 2 --- 15 ERROR D1 3 --- 13 SLCT D2 4 --- 12 PAPOUT D3 5 --- 10 ACK D4 6 --- 11 BUSY GROUND 25 --- 25 GROUND ERROR 15 --- 2 D0 SLCT 13 --- 3 D1 PAPOUT 12 --- 4 D2 ACK 10 --- 5 D3 BUSY 11 --- 6 D4 All remaining pins remain unconnected. If the cable is shielded, the shield should be connected to the DB-25's metallic shell on one end only. Andrew Anderson Thu Mar 7 23:22:06 EST 1996	{"url":"https://ldp.indosite.com/LDP/nag/node284.html","timestamp":"2024-11-10T00:23:50Z","content_type":"text/html","content_length":"2777","record_id":"<urn:uuid:61e394e9-5b44-4390-b7b2-a27ba49d5a8f>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00050.warc.gz"}
Microns to Barleycorns Converter ⇅ Switch toBarleycorns to Microns Converter How to use this Microns to Barleycorns Converter 🤔 Follow these steps to convert given length from the units of Microns to the units of Barleycorns. 1. Enter the input Microns value in the text field. 2. The calculator converts the given Microns into Barleycorns in realtime ⌚ using the conversion formula, and displays under the Barleycorns label. You do not need to click any button. If the input changes, Barleycorns value is re-calculated, just like that. 3. You may copy the resulting Barleycorns value using the Copy button. 4. To view a detailed step by step calculation of the conversion, click on the View Calculation button. 5. You can also reset the input by clicking on button present below the input field. What is the Formula to convert Microns to Barleycorns? The formula to convert given length from Microns to Barleycorns is: Length[(Barleycorns)] = Length[(Microns)] / 8466.666666700534 Substitute the given value of length in microns, i.e., Length[(Microns)] in the above formula and simplify the right-hand side value. The resulting value is the length in barleycorns, i.e., Length Calculation will be done after you enter a valid input. Consider that a high-end smartphone screen has pixels that are 50 microns in size. Convert this pixel size from microns to Barleycorns. The length in microns is: Length[(Microns)] = 50 The formula to convert length from microns to barleycorns is: Length[(Barleycorns)] = Length[(Microns)] / 8466.666666700534 Substitute given weight Length[(Microns)] = 50 in the above formula. Length[(Barleycorns)] = 50 / 8466.666666700534 Length[(Barleycorns)] = 0.005905511811 Final Answer: Therefore, 50 µ is equal to 0.005905511811 barleycorn. The length is 0.005905511811 barleycorn, in barleycorns. Consider that an advanced microprocessor has circuit features that are 10 microns wide. Convert this feature size from microns to Barleycorns. The length in microns is: Length[(Microns)] = 10 The formula to convert length from microns to barleycorns is: Length[(Barleycorns)] = Length[(Microns)] / 8466.666666700534 Substitute given weight Length[(Microns)] = 10 in the above formula. Length[(Barleycorns)] = 10 / 8466.666666700534 Length[(Barleycorns)] = 0.0011811023622 Final Answer: Therefore, 10 µ is equal to 0.0011811023622 barleycorn. The length is 0.0011811023622 barleycorn, in barleycorns. Microns to Barleycorns Conversion Table The following table gives some of the most used conversions from Microns to Barleycorns. Microns (µ) Barleycorns (barleycorn) 0 µ 0 barleycorn 1 µ 0.00011811024 barleycorn 2 µ 0.00023622047 barleycorn 3 µ 0.00035433071 barleycorn 4 µ 0.00047244094 barleycorn 5 µ 0.00059055118 barleycorn 6 µ 0.00070866142 barleycorn 7 µ 0.00082677165 barleycorn 8 µ 0.00094488189 barleycorn 9 µ 0.00106299213 barleycorn 10 µ 0.00118110236 barleycorn 20 µ 0.00236220472 barleycorn 50 µ 0.00590551181 barleycorn 100 µ 0.01181102362 barleycorn 1000 µ 0.1181 barleycorn 10000 µ 1.1811 barleycorn 100000 µ 11.811 barleycorn A micron, also known as a micrometer (µm), is a unit of length in the International System of Units (SI). One micron is equivalent to 0.000001 meters or approximately 0.00003937 inches. The micron is defined as one-millionth of a meter, making it an extremely precise measurement for very small distances. Microns are used worldwide to measure length and distance in various fields, including science, engineering, and manufacturing. They are especially important in fields that require precise measurements, such as semiconductor fabrication, microscopy, and material science. A barleycorn is a historical unit of length used primarily in the UK to measure shoe sizes and in other contexts. One barleycorn is approximately equivalent to 1/3 inch or about 0.00847 meters. The barleycorn is based on the size of a barley grain and was used historically for measuring small lengths and sizes, such as the width of the foot in shoe sizing. Barleycorns were used in traditional measurements, including shoe sizing, and provide historical context for understanding measurements and sizing practices. Although less common today, the unit remains of interest for its historical significance and use in traditional contexts. Frequently Asked Questions (FAQs) 1. What is the formula for converting Microns to Barleycorns in Length? The formula to convert Microns to Barleycorns in Length is: Microns / 8466.666666700534 2. Is this tool free or paid? This Length conversion tool, which converts Microns to Barleycorns, is completely free to use. 3. How do I convert Length from Microns to Barleycorns? To convert Length from Microns to Barleycorns, you can use the following formula: Microns / 8466.666666700534 For example, if you have a value in Microns, you substitute that value in place of Microns in the above formula, and solve the mathematical expression to get the equivalent value in Barleycorns.	{"url":"https://convertonline.org/unit/?convert=microns-barleycorns","timestamp":"2024-11-07T10:11:27Z","content_type":"text/html","content_length":"91486","record_id":"<urn:uuid:ef27ede0-056a-4155-9c8b-3fb92ed9fcdd>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00657.warc.gz"}
How to save time series in a dataframe created in an in r loop? Saving Time Series Data in a Dataframe within an R Loop: A Practical Guide Working with time series data often involves processing and analyzing data collected over time. In R, a common task is to create a dataframe within a loop and store time series data within it. However, efficiently saving these time series can be tricky. This article will guide you through the process of saving time series data within a dataframe in an R loop, providing practical examples and insights. The Problem Scenario Imagine you are collecting stock prices for multiple companies over a period of time. You want to analyze the price trends for each company separately. One approach is to use a loop to process the data for each company and store the resulting time series in a dataframe. Here's an example of a common approach: # Sample data companies <- c("Apple", "Google", "Microsoft") time_periods <- 1:10 # Create an empty dataframe df <- data.frame() # Loop to process each company for (company in companies) { # Generate simulated stock prices prices <- runif(length(time_periods), min = 100, max = 200) # Create a time series object ts_data <- ts(prices, start = 1, end = length(time_periods)) # Problem: How to store the time series 'ts_data' efficiently in the dataframe 'df'? The problem here lies in how to store the ts_data object within the df dataframe for each company. Directly adding the time series as a column will lead to an error. The Solution: Utilizing Lists and Dataframes The key to storing time series data in a dataframe is to leverage lists and understand how R handles data structures. Here's how you can solve the problem: 1. Create a List: Instead of storing the time series directly in the dataframe, we create a list within the loop to hold each time series object. 2. Populate the List: Inside the loop, append each ts_data object to this list. 3. Bind the List to the Dataframe: After the loop completes, you can use cbind to bind the list to the dataframe. Here's the improved code: # Sample data companies <- c("Apple", "Google", "Microsoft") time_periods <- 1:10 # Create an empty dataframe df <- data.frame(company = companies) # Create an empty list for time series data ts_list <- list() # Loop to process each company for (i in 1:length(companies)) { company <- companies[i] # Generate simulated stock prices prices <- runif(length(time_periods), min = 100, max = 200) # Create a time series object ts_data <- ts(prices, start = 1, end = length(time_periods)) # Populate the list with time series data ts_list[[i]] <- ts_data # Bind the list to the dataframe df <- cbind(df, ts_list) Understanding the Solution • Dataframe Structure: The df dataframe now has a column for the company names and a column for the list containing all the time series data. • Accessing Time Series: You can access each time series using the df$ts_list[[i]] syntax, where i represents the index of the company in the dataframe. • Flexibility: This approach allows you to store multiple time series objects within a single dataframe, maintaining the structure and organization of your data. Additional Considerations • Efficiency: Using lists for storing time series data can be more efficient than storing them directly as columns. • Data Analysis: R provides powerful time series analysis functions, such as acf, pacf, and arima, which can be applied to the time series objects stored within the dataframe. • Visualizations: Use libraries like ggplot2 to create informative time series plots for each company's stock price data. By understanding the power of lists and dataframes in R, you can effectively store time series data generated within a loop. This approach provides flexibility, organization, and ease of access for further analysis and visualization. Remember to choose the appropriate data structures for your specific needs, and R's extensive time series analysis tools will enable you to gain valuable insights from your data.	{"url":"https://laganvalleydup.co.uk/post/how-to-save-time-series-in-a-dataframe-created-in-an-in-r","timestamp":"2024-11-11T20:59:42Z","content_type":"text/html","content_length":"83801","record_id":"<urn:uuid:9e64e7ae-08bc-4d2f-9172-b83f47e6e5d3>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00553.warc.gz"}
Truth table proofs Revision as of 17:00, 5 September 2023 by Jkinne (talk \| contribs) (→Disproof of a logical fallacy) Truth tables are used to show all possible values that a given logical expression might take. Definition of AND The following gives the definition of the logical AND operation. A B A ∧ B false false false false true false true false false true true true Proof of one of De Morgan's laws A truth table can also be used to prove a logical identity. The following proves De Morgan's law that ¬ (A ∧ B) is equivalent to (¬ A) ∨ (¬ B). Notice that the truth values for these (the last two columns in the table) are always the same. A (¬ A) B (¬ B) A ∧ B ¬ (A ∧ B) (¬ A) ∨ (¬ B) false true false true false true true false true true false false true true true false false true false true true true false true false true false false Disproof of a logical fallacy Truth tables can also be used to disprove a logical fallacy. For example, one fallacy (affirming the consequent) is to assume the converse of an implication holds. If you know that (A → B) is true, and you know that B is true, the fallacy would be to then conclude that A must also be true. Note that the statement - "if an animal is a mammal then it is also a vertebrate" - is true. If we have an animal that is a vertebrate (for example, a dog), it would be a fallacy to now conclude that the animal must also be a mammal. We will show that this is a logical fallacy - regardless of the statements A and B, just because we know A → B is true, we should not in general assume the converse is also true. First we remind you that (A → B) is equivalent to ¬ A ∨ B. What we want to look at is the claim: ((A → B) ∧ B) → A. If this expression is always true, then we could rely on the converse to always be used. Let us first simplify the expression. It is equivalent to ((¬ A ∨ B) ∧ B) → A. This is equivalent to ¬ ((¬ A ∨ B) ∧ B) ∨ A. We can keep this formula in mind as we evaluate it's truth value for each possible value of A and B. Consider this truth table. A B A → B (A → B) ∧ B ((A → B) ∧ B) → A false false true false true false true true true false true false false false true true true true true true In the second row, we see that the last column is false. This is where we should expect to see a problem - the place where B is true, but we should not be able to conclude that A must also be true. We have shown that we cannot always apply the converse. Assignment 1 You will be assigned a logical identity to prove and a logical fallacy to disprove. For each, you are permitted and encouraged to simplify the expressions to make it easier to evaluate the expressions. You then write a truth table for each part - proving the logical identity, and disproving the logical fallacy. Include a similar amount of explanation as above. Unless otherwise indicated by your instructor, you can submit your solutions in whatever format you like - word document, pictures of it worked out on paper, physical paper, OneNote notebook link. If you send a link to an electronic document, make sure it is set so it is shared with your instructor. Pass rating check Each part must be correct to earn a 1/1 on that part. The total for the problems would be 2 points. Assignment 2 The following are additional problems related to truth tables and logic. Some are from Mathematics for Computer Science (which we abbreviate MCS). 1. MCS Problem 3.5. This gives you more practice examining proofs for mistakes. If you have a program that produces truth tables, you may use it for this problem for part (a). 2. MCS Problem 3.8. This gives you some practice working with logical formulae and reasoning about them rather than just using truth tables for proofs (since the truth table for this problem would be unwieldy). 3. MCS Problem 3.11. This gives you practice examining logical formulae to look for examples / counter-examples of truth assignments. Note that a logical expression is valid if it is always true. For each response for this question, you should give a reason why the answer you give is correct; you don't need it to be a formal proof. 4. Ungraded - Show how you can use a NOR gate to construct AND, OR, NOT, XOR gates. For each you need to give the expression using just NOR gates, and you need to give a justification for why this works. Your justification could be a truth table or just reasoning about the expressions. It does not need to be a formal proof, but does need to convince the reader that your claims are true. Note - you may use the values true and false in your constructions. If you do a construction for NOT, you can use that in the later parts, without fully expanding it. Pass rating check You should be able to get 90% of the individual parts completely correct, and should correctly identify any parts that you are not sure your answer is correct. Grading notes Notes on things people have missed on these problems... • MCS Problem 3.5 □ For part a, you need to give the truth tables asked for (* and ** from the problem) and indicate where they are different. □ For part b, you need to identify where specifically Sam's argument is wrong. It is not enough to argue that his conclusion is wrong. • MCS Problem 3.8 □ You need to explain why the variables are forced to their values, in both of your cases. Many people gave a reasonable argument only for one of the cases. □ Many people did not explain why the variables are forced in either case. Just giving the 2 truth assignments is not enough. • MCS Problem 3.11 □ You need to give some reason for why you give the answer you give. Some did not explain why the valid one(s) is(are) valid. • NOR gate constructions □ If you found the solution online (e.g., on wikipedia) then you should put a comment in canvas to give a link to what you used. If you did use something and don't cite it, that is plagiarism, that will waste all of our time, and your grade will be impacted (maybe worse).	{"url":"https://cs.indstate.edu/wiki/index.php?title=Truth_table_proofs&oldid=4398","timestamp":"2024-11-08T06:07:22Z","content_type":"text/html","content_length":"24256","record_id":"<urn:uuid:195060b4-2c05-4cc2-bf8e-8ff27244c34f>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00859.warc.gz"}
A Review of Predictive Value of Laboratory Tests Basic Method Validation A Review of Predictive Value of Laboratory Tests Expanding on a previous lesson on Clinical Agreement, Dr. Westgard discusses the Predictive Value of a Laboratory Test A Review of Predictive Value of Laboratory Tests James O. Westgard, Sten A. Westgard May 2020 In an earlier discussion [1], we considered the use of a Clinical Agreement Study to evaluate the performance of a qualitative test. In such a study, the new or candidate test is compared to an established or comparative test for a group of patients who are positive for the disease and another group that are negative for the disease. The results are then tabulated in a 2x2 contingency table, as shown below: ┃ Comparative Method “Gold Std” ┃ ┃ Candidate Method (Test) │ Positive │ Negative │ Total ┃ ┃ Positive │ TP │ FP │ TP+FP ┃ ┃ Negative │ FN │ TN │ FN+TN ┃ ┃ Total │ TP+FN │ FP+TN │ Total ┃ Where TP = Number of results where both tests are positive; FP = Number of results where the candidate method is positive, but the comparative is negative; FN = Number of results where the candidate method is negative, but the comparative is positive; TN = Number of results where both methods are negative. See even more stories about COVID-19 Laboratory Challenges... In this discussion, we are using the terminology True Positives (TP), False Negatives (FN), False Positives (FP), and True Negatives (TN) because our interest is to discuss the Clinical Sensitivity and Clinical Specificity of a test and the predictive value of positive and negative results. Clinical Sensitivity (Se) and Clinical Specificity (Sp) are calculated as follows: Clinical Sensitivity = [TP/(TP+FN)]100 Clinical Specificity = [TN/(TN+FP)]100 Keep in mind, these terms correspond to the Percent Positive Agreement (PPA) and Percent Negative Agreement (PNA) in the earlier discussion of the 2x2 Contingency Calculator. The difference is that we are now assuming the comparative method is the “gold standard” for correctly classifying the patients’ disease condition. Acceptable Sensitivity and Specificity CDC provides some guidance for acceptable performance of rapid influenza diagnostic tests, suggesting that they should achieve 80% sensitivity for detection of influenza A and influenza B viruses and recommending they must achieve 95% specificity where the comparative method is RT-PCR [2]. They also discuss the expected test performance for conditions where the prevalence of influenza varies from 2.5% (very low), 20% (moderate), and 40% (high). The criteria for performance are the predictive values of positive and negative test results, i.e., what’s the chance that a positive result indicates the presence of disease and what’s the chance that a negative result indicates the absence of disease. Those conditions can be evaluated by calculating the predictive value of test results. Predictive Value The primary performance characteristics are clinical sensitivity and clinical specificity, but the clinical usefulness of a test depends on the expected prevalence of disease (Prev) in the population being tested. The subjects in a Clinical Agreement Study seldom represent the real population that will be tested. For example, the CLSI guidance suggests 50 positive and 50 negative patient specimens to provide minimally reliable estimates of Se and Sp, which is a 50% rate of disease prevalence. What if the prevalence of the population were 20%, or 2%, or 0.2%? Case with 20% Prevalence. For example, assume that Se is 80% and Sp is 95%, which would be considered good performance according to the CDC guidance for infectious disease testing. If you tested 1000 subjects in a population that had 20% prevalence of disease, which might be representative of New York City during the COVID-19 pandemic, how would you interpret the test results? • In our test population, 200 patients have the disease (20% of 1000), 80% or 160 of those would give positive results (TP=0.80200) and the other 40 would give false negative results (FN). • For the 800 negatives (1000-200), 95% or 760 patients (0.95800) would give negative results (TN) and the other 40 would give positive results (FP). With this information, we can fill in the numbers in the contingency table. ┃ Comparative Method “Gold Std” ┃ ┃ Candidate Method (Test) │ Positive │ Negative │ Total ┃ ┃ Positive │ 160 │ 40 │ 200 ┃ ┃ Negative │ 40 │ 760 │ 800 ┃ ┃ Total │ 200 │ 800 │ 1000 ┃ • The chance that an individual patient with disease will be correctly classified is determined by the ratio of TP to total number of positives TP+FP, which is 160/200 or 80%, i.e., there is a 80% chance that a positive test result will correctly classify the patient as having the disease. □ PVpositive = TP/(TP+FP) = 160/200 = 80% • The chance that an individual patient without disease will be correctly classified is determined by the ratio of TN to total number of negatives TN+FN, which is 760/800, or 95%. Case with 2% Prevalence. Now consider the case for a prevalence of 2.0%, perhaps representative of California. • For 20 patients with disease (2% of 1000), the number of TP would be 0.8020 is 16, which leaves 4 FN patients. • For the 980 patients without disease (1000-20), the number of TN would be 0.95980 or 931, which leaves 49 FP. ┃ Comparative Method “Gold Std” ┃ ┃ Candidate Method (Test) │ Positive │ Negative │ Total ┃ ┃ Positive │ 16 │ 49 │ 65 ┃ ┃ Negative │ 4 │ 931 │ 935 ┃ ┃ Total │ 20 │ 980 │ 1000 ┃ • The chance that an individual patient with disease will be correctly classified is given by TP/(TP+FP), or 16/(16+49) or 25%. • The chance that an individual patient without disease will be correctly classified is given by TN/(TN+FN) or 980/(980+4), or 99.5%. This test would clearly be more useful in California for identifying patients without disease rather than identifying patients with disease. In New York, however, a positive test result is more likely a good indication of disease, while a negative result is still useful for excluding disease. In California, a subject with a positive test result has about a 25% chance of having the disease. Out of every 10 positives, 7 to 8 will NOT have the disease. Alternate Calculations PVpositive and PVnegative can be calculated directly from Se, Sp, and Prev using the following equations: PVpositive = SePrev/[(SePrev) +(1-Sp)(1-Prev)] PVnegative = Sp(1-Prev)/[(1-Se)Prev +Sp(1-Prev)] In these equations, Se, Sp, and Prev should be proportions between 0.00 and 1.00. You can multiply the figures for PVpos and PVneg by 100 to express as percent, or modify the equations by substituting 100 for 1 and entering Se, Sp, and Prev as percentages. Many find it more informative to reason through the steps for calculating the number of TP, etc., to better understand the effects of sensitivity and specificity. However, these formulas allow you to set up a spreadsheet and easily study the interactions of Se, Sp, and Prev for optimizing the predictive value of tests for different scenarios. Alternatively, MedCalc [3] provides an online calculator that will do all these calculations from the contingency table and an entry for prevalence. Trade-off between Sensitivity and Specificity It is difficult to achieve perfect performance of 100% sensitivity and 100% specificity for any diagnostic test. Sometimes, by adjusting the cutoff or decision limit between the population for non-disease and the population for disease, it is possible to optimize either sensitivity or specificity. Typically, that involves improving sensitivity at the expense of specificity, or alternatively improving specificity at the expense of sensitivity. Optimizing Performance for Prevalence The value of a positive test result improves as the prevalence of disease increases and as specificity increases. By applying a test to patients with symptoms of disease, a higher prevalence population is being selected, which should be a valuable strategy when testing is limited and diagnosis of disease is critical. Increasing sensitivity, perhaps by parallel use of two tests, could also be valuable. That means a patient would be classified as positive if either of the two tests were positive. It has been suggested that for diagnosis of COVID-19 after 5 days of symptoms, parallel testing of viral load and total immunoglobulins might improve sensitivity, i.e., if either test is positive, the patient has the disease. The Difficulty with Surveillance On the other hand, if testing patients as part of surveillance, the prevalence of disease is likely to be very low. This surveillance might utilize tests for IgG or Total IG, with a goal of identifying those people who have already been exposed to the virus and hopefully have developed immunity. If we assume a prevalence of 0.20% and test 1000 patients, there will be 2 patients with disease and 998 without disease. If the test has an ideal sensitivity of 1.00 or 100%, then both of the patients with disease will be classified as positive (TP=2, FN=0). If the test has a specificity of 95%, there will be 948 TN and 50 FP. PVpositive = TP/(TP+FP) = 2/(2+50) = 3.8% PVnegative = TN/(TN+FN) = 948/948 = 100% It is almost counter-intuitive that a test with perfect sensitivity will not be reliable for identifying subjects with antibodies present because specificity (which is also very high at 95%) allows so many false positives. There is only a 4% chance that a positive test indicates a patient has antibodies to the virus. On the other hand, a negative test result almost certainly means that the subject has not been exposed to the virus. But that is not very useful if the aim of surveillance is to identify those in the population who are potentially immune to the disease! An example from the AACC Blog What is the value of repeat testing of positives when screening for antibodies to COVID-19? Evidently there is some guidance from CDC or FDA that positive antibody tests should be repeated to ensure their accuracy. Opinions of clinical chemists vary, some thinking this is a waste of resources because won’t get paid for doing a second test and some believing that there really won’t be any improvement anyway. There should be a more objective way of addressing this issue, which was illustrated by Drs. Galen and Gambino in their famous book “Beyond Normality” that was published in 1975 [4]. The important pages are 42-44, where they describe a scenario, Test A has an Se of 95% and Sp of 90% and Test B has an Se of 80% and Sp of 95%, and the prevalence of disease is 1.0%. Note that this example presumes that Tast A and Test B are independent tests, e.g., the tests may employ different synthetic antigens that present different binding sites. The “trick” in making the calculations is to start with Prev of 1.0% and determine the PVpos of Test A, then use the PVpos as the prevalence of disease in calculating the PVpos for Test B. Remember, you are retesting with Test B all the positives seen from Test A, which means the prevalence of disease in that repeat population is actually the PVpos yielded by Test A. In short, you make 2 passes in calculating predictive value, the first with the starting prevalence of 1.0% and the 2nd with the resulting PVpos as the prevalence for apply Test B. The PVpos from Test A is 8.76%. The PVpos from Test B is then 60.6%. This means that 6 out of 10 patients from repeat testing (A followed by B) will truly have the disease, compared to only 1 out of 10 patients from Test A. Interesting, if the repeat strategy used Test B first and then Test A, the final PVpos is still 60.6%, the prevalence of disease in the repeat population would be 13.9%, thus there would be fewer patients who needed to be retested. But, the value of repeat testing does depend on the prevalence of disease in the original patient population, with repeat testing being more useful for low rather than high prevalence, as shown in the table below. ┃ Prevalence │ First Test PVpos │ Repeat Test PVpos ┃ ┃ 20% │ 70% │ 97% ┃ ┃ 10% │ 51% │ 94% ┃ ┃ 4% │ 28% │ 86% ┃ ┃ 2% │ 16% │ 75% ┃ ┃ 1% │ 8.7% │ 61% ┃ Again, the testing strategy for the situation in New York (20%) should be different from the strategy for California (2%). Repeat testing will be necessary in California, but not in NY. What is the point? In summary, the predictive value of a positive test results depends primarily on the specificity of the test, whereas the predictive value of a negative test result depends primarily on the sensitivity of the test. This is counter-intuitive, but can be explained by the effects of False Positive and False Negative results, respectively. When Sp is 100%, there are no False Positives. When Se is 100%, there are no False Negatives. Parallel testing (Test A OR Test B) is a strategy is to classify the patient as positive if either test is positive, which improves sensitivity and reduces false negative results. Serial testing (Test A AND Test B) is a strategy is to classify the patient as positive only if both tests are positive, which improves specificity and reduces false positive results. There may also be practical issues to consider, such as the relative costs of the tests, the relative number of tests that need to be repeated for A OR B vs B OR A strategy, the time required to reach a diagnostic decision, To add to the confusion about COVID-19 testing, the objective with diagnostic testing is to identify patients with disease, meaning that a positive result is bad news, but leads to confinement or treatment, whereas a false negative result may lead to further exposure of the community. With antibody testing, a positive result is good news, meaning the patient may have developed immunity, a false negative may confine a healthy worker, but a false positive may lead back to the workplace and further exposure of the community. What to do? You may find it very useful to set up a predictive value calculator in an Excel spreadsheet. Use the equations based on Se, Sp, and Prev to enter these figures as proportions between 0.0 and 1.0. If you want results in %, then set up the equations using 100 instead of 1 and enter Se, Sp and Prev as percentages. You will find it interesting to play with the values for Sp and see its critical importance for population surveillance by antibody testing. 1. Westgard JO, Garrett PA, Schilling P. Estimating clinical agreement for a qualitative test: A web calculator for 2x2 contingency test. www.westgard.com/qualitative-test-clinical-agreement.htm 2. CDC. Rapid diagnostic testing for influenza: Information for clinical laboratory directors. https://www.cdc.gov/flu/professionals/diagnosis/rapidlab.htm 3. MedCalc. Diagnostic test evaluation calculator. Accessed 4/27/2020. www.medcalc.org/calc/diagnostic_test.php 4. Galen RS, Gambino SR. Beyond Normality: The Predictive Value and Efficiency of Medical Diagnosis. New York:John Wiley, 1975	{"url":"https://westgard.com/lessons/basic-method-validation/predictive-value.html","timestamp":"2024-11-06T12:09:28Z","content_type":"text/html","content_length":"73675","record_id":"<urn:uuid:ed0a2b33-32e0-4294-b4af-9d409c6fce5f>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00584.warc.gz"}
Conversion of Clay: Cubic Meters to Metric Tonnes - Angola Transparency Conversion of Clay: Cubic Meters to Metric Tonnes Clay, a naturally occurring material composed primarily of fine-grained minerals, is widely used in various industries, including construction, pottery, and ceramics. Its versatility stems from its unique properties, such as its ability to be molded and shaped when wet and its hardening characteristics upon drying or firing. Understanding the conversion between cubic meters and metric tonnes is crucial for accurate measurements and calculations involving clay. Key Facts 1. Conversion factor: The conversion factor for converting clay from cubic meters to metric tonnes depends on the density of the clay. Different types of clay can have different densities, so the conversion factor may vary. 2. Density of clay: The density of clay can range from approximately 1,000 kg/m³ to 2,000 kg/m³, depending on the specific type and composition of the clay. 3. Calculation: To convert cubic meters of clay to metric tonnes, you need to multiply the volume in cubic meters by the density of the clay in kg/m³. The resulting value will be in metric tonnes. 4. Example: For instance, if the density of the clay is 1,500 kg/m³, then 1 cubic meter of clay would be equal to 1.5 metric tonnes. Conversion Factor The conversion factor for converting clay from cubic meters to metric tonnes depends on the density of the clay. Different types of clay can have different densities, so the conversion factor may vary. It is essential to determine the specific density of the clay in question to ensure accurate conversion. Density of Clay The density of clay can range from approximately 1,000 kg/m³ to 2,000 kg/m³, depending on the specific type and composition of the clay. Denser clays, such as those with higher percentages of minerals like quartz and feldspar, tend to have higher densities compared to clays with more organic matter or impurities. To convert cubic meters of clay to metric tonnes, you need to multiply the volume in cubic meters by the density of the clay in kg/m³. The resulting value will be in metric tonnes. Metric Tonnes = Cubic Meters × Density (kg/m³) For instance, if the density of the clay is 1,500 kg/m³, then 1 cubic meter of clay would be equal to 1.5 metric tonnes. The conversion of clay from cubic meters to metric tonnes is a straightforward process that requires knowledge of the clay’s density. By accurately determining the density and applying the appropriate conversion factor, individuals can ensure precise measurements and calculations involving clay, facilitating effective material management and project planning. 1. https://www.answers.com/Q/How_many_ton_in_1_cubic_meter_of_clay 2. https://coolconversion.com/volume-mass-construction/~1~cubic-meter~of~clay-soil~to~tonne 3. https://www.traditionaloven.com/building/refractory/fireclay/convert-cubic-metre-m3-fire-clay-to-tonne-metric-t-fire-clay.html How do I convert tonnes of clay to cubic meters? To convert tonnes of clay to cubic meters, you need to divide the weight in tonnes by the density of the clay in kg/m³. The resulting value will be in cubic meters. Cubic Meters = Tonnes ÷ Density (kg/m³) What is the density of clay? The density of clay can vary depending on its type and composition. However, a typical range for the density of clay is between 1,000 kg/m³ and 2,000 kg/m³. How many cubic meters are in a tonne of clay with a density of 1,500 kg/m³? If the density of clay is 1,500 kg/m³, then 1 tonne of clay would be equal to 1,000 kg ÷ 1,500 kg/m³ = 0.67 cubic meters. How many tonnes of clay are in 10 cubic meters of clay with a density of 1,800 kg/m³? If the density of clay is 1,800 kg/m³ and you have 10 cubic meters of clay, then the weight in tonnes would be 10 m³ × 1,800 kg/m³ = 18,000 kg, which is equal to 18 tonnes. Is it possible to convert cubic meters of clay to tonnes without knowing the density of the clay? No, it is not possible to accurately convert cubic meters of clay to tonnes without knowing the density of the clay. The density is a crucial factor in determining the weight of the clay. What are some common types of clay? Some common types of clay include kaolin, bentonite, montmorillonite, and fire clay. Each type of clay has unique properties and is used for various applications. What are some applications of clay? Clay has a wide range of applications, including the production of bricks, tiles, pottery, and ceramics. It is also used in construction, agriculture, and various industrial processes. How can I determine the density of clay? There are several methods to determine the density of clay. One common method is the pycnometer method, which involves measuring the mass and volume of a clay sample to calculate its density.	{"url":"https://angolatransparency.blog/en/how-many-cubic-meters-is-a-tonne-of-clay/","timestamp":"2024-11-08T14:16:59Z","content_type":"text/html","content_length":"60977","record_id":"<urn:uuid:8abf51c2-4cf5-4c88-a303-ef9b739db72d>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00054.warc.gz"}
Uncategorized Archives Many Process Mining projects mainly revolve around the selection and introduction of the right Process Mining tools. Relying on the right tool is of course an important aspect in the Process Mining project. Depending on whether the process analysis project is a one-time affair or daily process monitoring, different tools are pre-selected. Whether, for example, a BI system has already been established and whether a sophisticated authorization concept is required for the process analyzes also play a role in the selection, as do many other factors. Nevertheless, it should not be forgotten that process mining is not primarily a tool, but an analysis method, in which the first part is about the reconstruction of the processes from operational IT systems in a resulting process log (event log), the second step is about a (core) graph analysis to visualize the process flows with additional analysis/reporting elements. If this perspective on process mining is not lost sight of, companies can save a lot of costs because it allows them to concentrate on solution-oriented concepts. However, completely independent of the tools, there is a very general procedure in this data-driven process analysis you should understand and which we would like to describe with the following Interested in introducing Process Mining to your organization? Do not hesitate to get in touch with us! DATANOMIQ is the independent consulting and service partner for business intelligence, process mining and data science. We are opening up the diverse possibilities offered by big data and artificial intelligence in all areas of the value chain. We rely on the best minds and the most comprehensive method and technology portfolio for the use of data for business optimization. https://data-science-blog.com/wp-content/uploads/2022/07/6-steps-of-process-mining-header.png 775 1858 Benjamin Aunkofer https://www.data-science-blog.com/wp-content/uploads/2016/09/ data-science-blog-logo.png Benjamin Aunkofer2022-08-14 11:43:312022-07-14 08:44:046 Steps of Process Mining – Infographic Top 5 Email Verification and Validation APIs for your Product /in Uncategorized/by Atreyee Chowdhury If you have spent some time running a website or online business, you would be aware of the importance of emails. What many see as a decadent communication medium still holds immense value for digital marketers. More than 330 billion emails are sent every day, even in 2022. While email marketing is very effective, it is very difficult to do it right. One of the key reasons being the many problems that email marketers face with their email lists. Are the email IDs correct? Do they have spam traps? Are these disposable email addresses? There are a multitude of questions to deal with in email marketing and newsletter campaigns. Email verification and validation APIs help us deal with this problem. APIs integrate with your platform and automatically check all email addresses for spam, mistyping, fake email ids, and so on. Top 5 email verification and validation APIs for your product Today we will talk about the 5 best APIs that you can use to validate and verify the email addresses in your mailing list. Using an API can be a gamechanger for many email marketers. Before we get into the top 5 list, let’s discuss why APIs are so effective and how they work. Why APIs are so efficient The major reason APIs work so efficiently is that it does not require human supervision. APIs work automatically and users do not have to manually configure them each time. The ease of use is one among many reasons you should start using an email verification and validation API. If you maintain a mailing list, you would also want to know where your effort is going. All email marketers spend considerable time perfecting their emails. On top of that, they need to use an email marketing platform like Klaviyo. An API ensures that your hard work does not go in vain. By filtering out fake and disposable email IDs, you get a better idea of where your mailing stand stands. As a result, when you use a platform like Klaviyo along with an email verification API, the results are much better. In case you want something other than Klaviyo, you can learn more about Klaviyo alternatives here. How email verification and validation APIs work Email verification and validation APIs work primarily in 7 ways: • Syntax Check • Address Name Detection • DEA (disposable email address) Detection • Spam Trap Detection • DNSBL and URI DNSBL Check • MX Record Lookup • Active Mailbox Check With the help of these email verification and validation methods, you will see much better results from your email marketing campaign. On top of that, your business will not be identified as spam and will help in building reputation and authority. Now that we have some idea about what email verification APIs are and what they do, let’s head over to the list. 1. Abstract API Abstract API is one of the most popular email verification and validation APIs out there. Here are some of its key features: • MX record check • GDPR and CCPA compliant • Does not store any email • Role email check If you have looked for email address validation API on the internet, you must have come across Abstract API. It is among the best in the business and also comes with affordable subscription plans. Abstract API helps with bounce rate detection, spam signups, differentiating between personal and business email IDs, and a lot more. However, the most significant feature of Abstract API is that it allows up to 500 free email checks every month. That’s a great way to see whether the product works for you before subscribing to it. Abstract API is user-friendly and budget-friendly, which makes it a top choice for many email marketers. Anyone new to using these tools can easily learn about them from Abstract API. For these reasons, Abstract API has the number one spot on our list. 2. SendGrid Validation API After Abstract API, the second product to have top-notch features is SendGrid Validation API. Here are its key features: • Uses machine learning to verify email addresses in real-time • Accurately identifies all inactive or inaccurate email addresses • You can check how your email appears in different mailboxes • Gives risk scores for all email addresses While most email verification and validation APIs work similarly, SendGrid Validation API takes it a notch higher with machine learning and artificial intelligence. Despite having advanced features and functionalities, SendGrid Validation API is not difficult to use. SendGrid Validation API operates on the cloud and does not store any of your email addresses. OIn top of that, there are easy settings and configuration options that users can tweak with. However, SendGrid Validation API does not have any free offering. There are only two plans: pro and premier. Users have to pay $89.95 per month to access SendGrid Validation API. If you are looking for advanced email verification and validation API, no need to look beyond SendGrid Validation API. It has everything you would need for a solid email marketing campaign apart from having many additional features. 3. Captain Verify Another email verification and validation API – Captain Verify – is a one-stop solution for all email verification needs. Here are its key features: • Get reports on the overall quality of your email address database • Affordable plans • Compliant with GDPR regulations • Export encrypted CSV files Unlike other email verification and validation APIs, Captain Verify does not stop after verifying the emails for spam, fake or invalid addresses, and so on. It helps email marketers understand how their campaign is performing and gives detailed reports on returns on investment. It is one of the best APIs available for the overall growth of your mailing campaign. If you are looking for something simple yet powerful, Captain Verify will be a great option. Along with the features we mentioned already, it also lets users filter and refine their email lists. It can help you understand the overall quality of your mailing list much better. As you can see, Captain Verify ticks most of the boxes to be one of the best email verification and validation APIs out there. Anyone looking for a good email API should give it a go. The best thing is that users get all this and more at only $7 per 1000 emails. 4. Mailgun Mailgun earns the fourth spot on our list. However, that does not mean it is any way less than the previous options discussed. Here’s what it offers: • RFC standards compliant • Daily and hourly tracking of API usage • Has a bulk list validation tools for faster operations • Supports both CSV and JSON format • Track bounce and unsubscribe rates Email marketers around the world prefer Mailgun for all their email verification and validation needs. It has multiple features that allow users to check their mailing list for fakes and scams. Apart from that, it also gives users a good idea of how their marketing campaign is performing. Mailgun enjoys high ratings across review platforms like Capterra and G2. People use it for a wide range of purposes, but email verification and validation remain the most important. Mailgun keeps track of bounce rates, hard bounce rates, and unsubscribe rates. With the help of these stats, email marketers can measure how their campaign is doing. If you are looking for a simple email verification and validation tool, Mailgun can be a good choice. It is worth trying for anyone who wants to take their email marketing to the next level. 5. Hunter Our last entry to the list is Hunter. It is a well-known API that is widely used by email marketers. Here’s what it gets right: • Compare your mailing list with the Hunter mailing list for comparative quality analysis • SMTP checks, domain information verification, and multi-layer validation • Easy integration with Google Sheets • Supports both CSV and .txt formats Hunter gives what it calls confidence scores which represent how strong or weak your mailing list is. This email verification and validation tool follows all the checks that we mentioned earlier, including SMTP verification, gibberish detection, MX record checks, and more. These features have worked together to make Hunter one of the most popular email verification and validation tools. Hunter email verification API integrates easily with any platform and has a user-friendly interface. It also has a free plan that lets users check up to 50 emails for free. Giving it a try without spending money is very useful for anyone looking for a new email verification and validation API. If you are looking for an email finder and email verifier rolled into one, Hunter is the best solution. With so many features and functionalities, it is one of the favorite email verification and validation APIs for thousands of marketers and entrepreneurs. When used correctly, email verification and validation APIs can give any online business a significant boost. As an email marketer, digital marketer, website owner, or entrepreneur, you should be using one of these APIs. If you aren’t using one already, find your top pick from our list of the 5 best email verification and validation APIs. https://www.data-science-blog.com/wp-content/uploads/2016/09/data-science-blog-logo.png 0 0 Atreyee Chowdhury https://www.data-science-blog.com/wp-content/uploads/2016/09/data-science-blog-logo.png Atreyee Chowdhury2022-04-24 09:56:082022-04-19 19:56:35Top 5 Email Verification and Validation APIs for your Product How to choose the best pre-trained model for your Convolutional Neural Network? /in Uncategorized/by Mickael Komendyak Introduction to Transfer Learning Let’s start by defining this term that is increasingly used in Data Science: Transfer Learning refers to the set of methods that allow the transfer of knowledge acquired from solving a given problem to another problem. Transfer Learning has been very successful with the rise of Deep Learning. Indeed, the models used in this field often require high computation times and important resources. However, by using pre-trained models as a starting point, Transfer Learning makes it possible to quickly develop high-performance models and efficiently solve complex problems in Computer Vision. As most of the Deep learning technics, Transfer Learning is strongly inspired by the process with which we learn. Let’s take the example of someone who masters the guitar and wants to learn to play the piano. He can capitalize on his knowledge of music to learn to play a new instrument. In the same way, a car recognition model can be quickly adapted to truck recognition. How is Transfer Learning concretely implemented to solve Computer Vision problems? Now that we have defined Transfer Learning, let’s look at its application to Deep Learning problems, a field in which it is currently enjoying great success. The use of Transfer Learning methods in Deep Learning consists mainly in exploiting pre-trained neural networks Generally, these models correspond to very powerful algorithms that have been developed and trained on large databases and are now freely shared. In this context, 2 types of strategies can be distinguished: 1. Use of pre-trained models as feature extractors: The architecture of Deep Learning models is very often presented as a stack of layers of neurons. These layers learn different features depending on the level at which they are located. The last layer (usually a fully connected layer, in the case of supervised learning) is used to obtain the final output. The figure below illustrates the architecture of a Deep Learning model used for cat/dog detection. The deeper the layer, the more specific features can be extracted. The idea is to reuse a pre-trained network without its final layer. This new network then works as a fixed feature extractor for other tasks. To illustrate this strategy, let’s take the case where we want to create a model able to identify the species of a flower from its image. It is then possible to use the first layers of the convolutional neural network model AlexNet, initially trained on the ImageNet image database for image classification. 2. Fitting of pre-trained models: This is a more complex technique, in which not only the last layer is replaced to perform classification or regression, but other layers are also selectively re-trained. Indeed, deep neural networks are highly configurable architectures with various hyperparameters. Moreover, while the first layers capture generic features, the last layers focus more on the specific task at hand. So the idea is to freeze (i.e. fix the weights) of some layers during training and refine the rest to meet the problem. This strategy allows to reuse the knowledge in terms of the global architecture of the network and to exploit its states as a starting point for training. It thus allows to obtain better performances with a shorter training time. The figure below summarizes the main Transfer Learning approaches commonly used in Deep Learning. How to choose your pre-trained CNN ? TensorFlow and Pytorch have built very accessible libraries of pre-trained models easily integrable to your pipelines, allowing the simple leveraging of the Transfer learning power. In the first part you discovered what a pre-trained model is, let’s now dig into how to choose between the (very) large catalog of models accessible in open-source. An unresolved question: As you could have expected, there is no simple answer to this question. Actually, many developers just stick to the models they are used to and that performed well in their previous projects. However, it is still possible to follow a few guidelines that can help you decide. The two main aspects to take into account are the same as most of the machine learning tasks : ⦁ Accuracy : The Higher, the better ⦁ Speed : The Faster, the better The dream being having a model that has a super fast training with an excellent accuracy. But as you could expect, usually to have a better accuracy, a deeper model is needed, therefore a model that takes more time to train. Thus, the goal is to maximize the tradeoff between accuracy and complexity. You can observe this tradeoff in the following graph taken from the Efficient Net model original As you can observe on this graph, bigger models are not always better. There is always a risk that a more complex model overfits your data, because it can give too much importance to subtle details in features. Knowing that the best is to start with the smallest model, that is what’s done in the industry. A “good-enough” model that is small and therefore quickly trained is preferred. Of course if you aim for great accuracy with no interest in a quick training then you can target the large model and even try ensemble techniques combining multiple models power. Most performant models at this time : Here are a few models that are widely used today in the field of computer vision. From image classification to complex image captioning, those structures offers great performances : • ResNet50 • EfficientNet • Inceptionv3 ResNet 50 : ResNet was developed by Microsoft and aims at resolving the ‘vanishing gradient problem’. It allows the creation of a very deep model (up to a hundred layers). Top-1 accuracy : 74.9% Top-5 accuracy : 92.1% Size : 98MB Parameters : 26 millions EfficientNet : This model is a state-of-the art convolutional neural network trained by Google. It is based on the same construction as ResNet but with an intelligent rescaling method. Top-1 accuracy : 77.1% Top-5 accuracy : 93.3.0% Size : 29MB Parameters : 5 millions InceptionV3 : Inception Networks (GoogLeNet/Inception v1) have proved to be more computationally efficient, both in terms of the number of parameters generated by the network and the economical cost incurred. It is based on Factorized Convolutions. Top-1 accuracy : 77.9% Top-5 accuracy : 93.7% Size : 92MB Parameters : 24 millions Final Note: To summarize, in this article, we have seen that Transfer Learning is the ability to use existing knowledge, developed to solve a given problem, to solve a new problem. We saw the top 3 State-of-the-Art pre-trained models for image classification. Here I summarized the performance and some detail on each of those models. However, as you have now understood, this is a continuously growing domain and there is always a new model to look forward to and push the boundaries further. The best way to keep up is to read papers introducing new model construction and try the most performing new releases. https://data-science-blog.com/wp-content/uploads/2022/04/transfer-learning-header.png 888 2588 Mickael Komendyak https://www.data-science-blog.com/wp-content/uploads/2016/09/ data-science-blog-logo.png Mickael Komendyak2022-04-11 08:30:352022-04-10 21:03:04How to choose the best pre-trained model for your Convolutional Neural Network? Understanding the “simplicity” of reinforcement learning: comprehensive tips to take the trouble out of RL /in Uncategorized/by Yasuto Tamura This is the first article of my article series “My elaborate study notes on reinforcement learning.” I adjusted mathematical notations in this article as close as possible to “Reinforcement Learning:An Introduction.” This book by Sutton and Barto is said to be almost mandatory for those who studying reinforcement learning. Also I tried to avoid as much mathematical notations, introducing some intuitive examples. In case any descriptions are confusing or unclear, informing me of that via posts or email would be appreciated. First of all, I have to emphasize that I am new to reinforcement learning (RL), and my current field is object detection, to be more concrete transfer learning in object detection. Thus this article series itself is also a kind of study note for me. Reinforcement learning (RL) is often briefly compared with human trial and errors, and actually RL is based on neuroscience or psychology as well as neural networks (I am not sure about these fields though). The word “reinforcement” roughly means associating rewards with certain actions. Some experiments of RL were conducted on animals, which are widely known as Skinner box or more classically Pavlov’s Dogs. In short, you can encourage animals to do something by giving foods to them as rewards, just as many people might have done to their dogs. Before animals find linkages between certain actions and foods as rewards to those actions, they would just keep trial and errors. We can think of RL as a family of algorithms which mimics this behavior of animals trying to obtain as much reward as possible. My cats will not all the way try to entertain me to get foods though. RL showed its conspicuous success in the field of video games, such as Atari, and defeating the world champion of Go, one of the most complicated board games. Actually RL can be applied to not only video games or board games, but also various other fields, such as business intelligence, medicine, and finance, but still I am very much fascinated by its application on video games. I am now studying the field which could bridge between the world of video games and the real world. I would like to mention this in the one of upcoming articles. So far I got an impression that learning RL ideas would be more challenging than learning classical machine learning or deep learning for the following reasons. 1. RL is a field of how to train models, rather than how to design the models themselves. That means you have to consider a variety of problem settings, and you would often forget which situation you are discussing. 2. You need prerequisites knowledge about the models of components of RL for example neural networks, which are usually main topics in machine/deep learning textbooks. 3. It is confusing what can be learned through RL depending on the types of tasks. 4. Even after looking over at formulations of RL, it is still hard to imagine how RL enables computers to do trial and errors. For now I would like you to keep it in mind that basically values and policies are calculated during in during RL. And I personally believe you should always keep the following points in your mind in order not to be at a loss in the process of learning RL. 1. RL basically can be only applied to a very limited type of situation, which is called Markov decision process (MDP). In MDP settings your next state depends only on your current state and action, regardless of what you have done so far. 2. You are ultimately interested in learning decision making rules in MDP, which are called policies. 3. In the first stage of learning RL, you consider surprisingly simple situations. They might be simple like mazes in kids’ picture books. 4. RL is in its early days of development. Let me explain a bit more about what I meant by the third point above. I have been learning RL mainly with a very precise Japanese textbook named 「機械学習プロフェッショナルシリーズ強化学習」 (Machine Learning Professional Series: Reinforcement Learning). As I mentioned in an article of my series on RNN, I sometimes dislike Western textbooks because they tend to beat around the bush with simple examples to get to the point at a more abstract level. That is why I prefer reading books of this series in Japanese. And especially the RL one in the series was especially bulky and so abstract and overbearing to a spectacular degree. It had so many precise mathematical notations without leaving room for ambiguity, thus it took me a long time to notice that the book was merely discussing simple situations like mazes in kids’ picture books. I mean, the settings discussed were so simple that they can be expressed as tabular data, that is some Excel sheets. I could not notice that until the beginning of 6th chapter out of eight out of 8 chapters. The 6th chapter discusses uses of function approximators. With the approximations you can approximate tabular data. My articles will not dig this topic of approximation precisely, but the use of deep learning models, which I am going to explain someday, is a type of this approximation of RL models. You might find that so many explanations on RL rely on examples of how to make computers navigate themselves in simple mazes or in playing video games, which are mostly impractical in the real world. However, as I will explain later, these are actually helpful examples to learn RL. As I show later, the relations of an agent and an environment are basically the same also in more complicated tasks. Reading some code or actually implementing RL would be very effective, especially in order to know simplicity of the situations in the beginning part of RL textbooks. Given that you can do a lot of impressive and practical stuff with current deep learning libraries, you might get bored or disappointed by simple applications of RL in many textbooks. But as I mentioned above, RL is in its early days of development, at least at a public level. And in order to show its potential power, I am going to explain one of the most successful and complicated application of RL in the next article: I am planning to explain how AlphaGo or AplhaZero, RL-based AIs enabled computers to defeat the world champion of Go, one of the most complicated board games. RL was not used to the chess AI which defeated Kasparov in 1997. Combination of decision trees and super computers, without RL, was enough for the “simplicity” of chess. But uses of decision tree named Monte Carlo Tree Search enabled Alpha Go to read some steps ahead more effectively. It is said deep learning enabled AlphaGo to have intuition about games. Mote Carlo Tree Search enabled it to have abilities to predict some steps ahead, and RL how to learn from experience. 1, What is RL? In conclusion, as far as I could understand so far, as a beginner of RL, I would interpret RL as follows: RL is a sub-field of training AI models, and optimal rules for decision makings in an environment are learned through RL, weakly supervised by rewards in a certain period of time. When and how to evaluate decision makings are task-specific, and they are often realized by trial-and-error-like behaviors of agents. Rules for decision makings are called policies in contexts of RL. And optimization problems of policies are called sequential decision-making problems. You are more or less going to see what I meant by my definition throughout my article series. An agent in RL means an entity which makes decisions, interacting with the environment with an action. And the actions are made based on policies. You can find various types of charts explaining relations of RL with AI, and I personally found the chart below the most plausible. The word “models” are used in another meaning later. Please keep it in mind that the “models” above are something like general functions. And the “models” which show up frequently later are functions modeling environments in RL. In case you’re totally new to AI and don’t understand what “supervising” means in these contexts, I think you should imagine cases of instructing students in schools. If a teacher just tells students “We have a Latin conjugation test next week, so you must check this section in the textbook.” to students, that’s a “supervised learning.” Students who take exams are “models.” Apt students like machine learning models would show excellent performances, but they might fail to apply the knowledge somewhere else. I mean, they might fail to properly conjugate words in unseen sentences. Next, if the students share an idea “It’s comfortable to get together with people alike.” they might be clustered to several groups. That might lead to “cool guys” or “not cool guys” group division. This is done without any explicit answers, and this corresponds to “unsupervised learning.” In this case, I would say a certain functions of the students’ brain or atmosphere there, which put similar students together, were the “models.” And finally, if teachers tell the students “Be a good student,” that’s what I meant with “weakly supervising.” However most people would say “How?” RL could correspond to such ultimate goals of education, and as well as education, you have to consider how to give rewards and how to evaluate students/agents. And “models” can vary. But such rewards often shows unexpected results. 2, RL and Markov decision process As I mentioned in a former section, you have to keep it in mind that RL basically can be applied to a limited situation of sequential decision-making problems, which are called Markov decision processes (MDP). A markov decision process is a type of process where the next state of an agent depends only on the current state and the action taken in the current state. I would only roughly explain MDP in this article with a little formulation. You might find MDPs very simple. But some people would find that their daily lives in fact can be described well with a MDP. The figure below is a state transition diagram of everyday routine at an office, and this is nothing but a MDP. I think many workers also basically have only four states “Chat” “Coffee” “Computer” and “Home” almost everyday. Numbers in black are possibilities of transitions at the state, and each corresponding number in orange is the reward you get when the action is taken. The diagram below shows that when you just keep using a computer, you would likely to get high rewards. On the other hand chatting with your colleagues would just continue to another term of chatting with a probability of 50%, and that undermines productivity by giving out the reward of -1. And having some coffee is very likely to lead to a chat. In practice, you optimize which action to take in each situation. You adjust probabilities at each state, that is you adjust a policy, through planning or trial and errors. Even if you say “Be a good student,” school kids in puberty they would act far from Markov decision process. Even though I took an example of school earlier, I am sure education should be much more complicated process which requires constant patience. Of course you have to consider much more complicated MDPs in most RL problems, and in most cases you do not have known models like state transition diagrams. Or rather I have to say RL enables you to estimate such diagrams, which are usually called models in contexts of RL, by trial and errors. When you study RL, for the most part you will see a chart like below. I think it is important to understand what this kind of charts mean, whatever study materials on RL you consult. I said RL is basically a training method for finding optimal decision making rules called policies. And in RL settings, agents estimate such policies by taking actions in the environment. The environment determines a reward and the next state based on the current state and the current action of the agent. Let’s take a close look at the chart above in a bit mathematical manner. I made it based on “Machine Learning Professional Series: Reinforcement Learning.” The agent exert an action Please do not think too much about differences of In the textbook “Reinforcement Learning:An Introduction” by Richard S. Sutton, which is almost mandatory for all the RL learners, RL process is displayed as the left side of the figure below. Each capital letter in the chart means a random variable. Relations of random variables can be also displayed as graphical models like the right side of the chart. The graphical model is a time series expansion of the chart of RL loops at the left side. The chart below shows almost the same idea as the one above. Whether they use random variables or realized values is the only difference between them. My point is that decision makings are simplified in RL as the models I have explained. Even if some situations are not strictly MDPs, in many cases the problems are approximated as MDPs in practice so that RL can be applied to. I personally think you do not have to care so much about differences of random variables and their realized values in RL unless you discuss RL mathmematically. But if you do not know there are two types of notations, which are strictly different ideas, you might get confused while reading textboks on RL. At least in my artile series, I will strictly distinguish them only when their differences In case you are not sure about differences of random variables and their realizations, please roughly grasp the terms as follows: random variables 3, Planning and RL We have seen RL is a family of training algorithms which optimizes rules for choosing planning without collecting data from the environment. On the other hand, when the model of the environment is unknown policies have to be optimized based on data which an agents collects from the environment through trial and errors. This is the very case called RL. You might find planning problems very simple and unrealistic in practical cases. But RL is based on planning of sequential decision-making problems with MDP settings, so studying planning problems is inevitable. As far as I could see so far, RL is a family of algorithms for approximating techniques in planning problems through trial and errors in environments. To be more concrete, in the next article I am going to explain dynamic programming (DP) in RL contexts as a major example of planning problems, and a formula called the Bellman equation plays a crucial role in planning. And after that we are going to see that RL algorithms are more or less approximations of Bellman equation by agents sampling data from environments. As an intuitive example, I would like to take a case of navigating a robot, which is explained in a famous textbook on robotics named ” Probabilistic Robotics.” In this case, the state set In the textbook on probabilistic robotics, this case is classified to a planning problem rather than a RL problem because it assumes that the robot has a complete model of the environment, and RL is not introduced in the textbook. In case of robotics one major way of making a model, or rather a map is SLAM (Simultaneous Localization and Mapping). With SLAM, a map of the environment can be made only based on what have been seen with a moving camera like in the figure below. Half the first part of the textbook is about self localization of robots and gaining maps of environments. And the latter part is about planning in the gained map. RL is also based on planning problems as I explained. I would say RL is another branch of techniques to gain such models/maps and proper plans in the environment through trial and errors. In the example of robotics above, we have not considered rewards return. But you usually have to consider uncertainty of future rewards, so in practice you multiply a discount rate discounted return every time step as follows. If agents blindly try to maximize immediate upcoming rewards values of states are calculated this way. The value of a state in contexts of RL mean how likely agents get higher values if they start from the state. And how to calculate values is formulated as the Bellman equation. If you are not sure what “ecursively” and “probabilistically” mean, please do not think too much. I am going to explain that as precisely as possible in the next article. I am going to explain Bellman equation, or Bellman operator to be exact in the next article. For now I would like you to keep it in mind that Bellman operator calculates the value of a state by considering future actions and their following states and rewards. Bellman equation is often displayed as a decision-tree-like chart as below. I would say planning and RL are matter of repeatedly applying Bellman equation to values of states. In planning problems, the model of the environment is known. That is, all the connections of nodes of the graph at the left side of the figure below are known. On the other hand in RL, those connections are not completely known, thus they need to be estimated in certain ways by agents collecting data from the environment. I guess almost no one explain RL ideas as the graphs above, and actually I am in search of effective and correct ways of visualizing RL. But so far, I think the graphs above describe how values updated in RL problem settings with discrete data. You are going to see what these graphs mean little by little in upcoming articles. I am also planning to introduce Bellman operators to formulate RL so that you do not have to think about decision-tree-like graphs all the time. 4, Examples of how RL problems are modeled You might find that so many explanations on RL rely on examples of how to make computers navigate themselves in simple mazes or play video games, which are mostly impractical in real world. But I think uses of RL in letting computers play video games are good examples when you study RL. The video game industry is one of the most developed and sophisticated area which have produced environments of RL. OpenAI provides some “playgrounds” where agents can actually move around, and there are also some ports of Atari games. I guess once you understand how RL can be modeled in those simulations, that helps to understand how other more practical tasks are implemented. It is a pity that there is no E.T. the Extra-Terrestrial. It is a notorious video game which put an end of the reign of Atari. And after that came the era of Nintendo Entertainment System. In the second section of this article, I showed the most typical diagram of the fundamental RL idea. The diagrams below show correspondences of each element of some simple RL examples to the diagram of general RL. Multi-armed bandit problems are a family of the most straightforward RL tasks, and I am going to explain it a bit more precisely later in this article. An agent solving a maze is also a very major example of RL tasks. In this case states If the environments are more complicated, deep learning is needed to make more complicated functions to model each component of RL. Such RL is called deep reinforcement learning. The examples below are some successful cases of uses of deep RL. I think it is easy to imagine that the case of solving a maze is close to RL playing video games. In this case Deep Q Networks use deep learning in RL algorithms named Q learning. The development of convolutional neural networks (CNN) enabled computers to comprehend what are displayed on video game screens. Thanks to that, video games do not need to be simplified like mazes. Even though playing video games, especially complicated ones today, might not be strict MDPs, deep Q Networks simplifies the process of playing Atari as MDP. That is why the process playing video games can be simplified as the chart below, and this simplified MPD model can surpass human performances. AlphaGo and AlphaZero are anotehr successful cases of deep RL. AlphaGo is ther first RL model which defeated the world Go champion. And some training schemes were simplified and extented to other board games like chess in AlphaZero. Even though they were sensations in media as if they were menaces to human intelligence, they are also based on MDPs. A policy network which calculates which tactics to take to enhance probability of winning board games. But they use much more sophisticated and complicated techniques. And it is almost impossible to try training them unless you own a tech company or something with some servers mounted with some TPUs. But I am going to roughly explain how they work in one of my upcoming articles. 5, Some keywords for organizing terms of RL As I am also going to explain in next two articles, RL algorithms are totally different frameworks of training machine learning models compared to supervised/unsupervised learnig. I think pairs of keywords below are helpful in classifying RL algorithms you are going to encounter. (1) “Model-based” or “model-free.” I said planning problems are basics of RL problems, and in many cases RL algorithms approximate Bellman equation or related ideas. I also said planning problems can be solved by repeatedly applying Bellman equations on states of a model of an environment. But in RL problems, models are usually unknown, and agents can only move in an environment which gives a reward or the next state to an agent. The agent can gains richer information of the environment time step by time step in RL, but this procedure can be roughly classified to two types: model-free type and model-based type. In model-free type, models of the environment are not explicitly made, and policies are updated based on data collected from the environment. On the her hand, in model-based types the models of the environment are estimated, and policies are calculated based on the model. To be honest, I am still not sure about differences of model-free RL and model-based RL. AlphaGo and AlphaZero are examples of model-based RL. Phases of board games can be modeled with CNN. Plannings in this case correspond to reading some phases ahead of games, and they are enabled by Monte Carlo tree search. They are the only examples of model-based RL which I can come up with. And also I had an impression that many study materials on RL focus on model-free types of RL. (2) “Values” or “policies.” I mentioned that in RL, values and policies are optimized. Values are functions of a value of each state. The value here means how likely an agent gets high rewards in the future, starting from the state. Policies are functions fro calculating actions to take in each state, which I showed as each of blue arrows in the example of robotics above. But in RL, these two functions are renewed in return, and often they reach optimal functions when they converge. The figure below describes the idea well. These are essential components of RL, and there too many variations of how to calculate them. For example timings of updating them, whether to update them probabilistically or deterministically. And whatever RL algorithm I talk about, how values and policies are updated will be of the best interest. Only briefly mentioning them would be just more and more confusing, so let me only briefly take examples of dynamic programming (DP). Let’s consider DP on a simple grid map which I showed in the preface. This is a planning problem, and agents have a perfect model of the map, so they do not have to actually move around there. Agents can move on any cells except for blocks, and they get a positive rewards at treasure cells, and negative rewards at danger cells. With policy iteration, the agents can interactively update policies and values of all the states of the map. The chart below shows how policies and values of cells are updated. You do not necessarily have to calculate policies every iteration, and this case of DP is called value iteration. But as the chart below suggests, value iteration takes more time to converge. I am going to much more precisely explain the differences of values and policies in DP tasks in the next article. (3) “Exploration” or “exploitation” RL agents are not explicitly supervised by the correct answers of each behavior. They just receive rough signals of “good” or “bad.” One of the most typical failed cases of RL is that agents can be myopic. I mean, once agents find some actions which constantly give good reward, they tend to miss other actions which produce better rewards more effectively. One good way of avoiding this is adding some exploration, that is taking some risks to discover other actions. I mentioned multi-armed bandit problems are simple setting of RL problems. And they also help understand trade-off of exploration and exploitation. In a multi-armed bandit problem, an agent chooses which slot machine to run every time step. Each slot machine gives out coins, or rewards ɛ-greedy algorithm. This is quite simple: with a probability of Casino owners are not so stupid. Just as insurance I am sure it is designed so that you would lose in the long run, and before your “exploration” is complete, you will be “exploited.” Let’s take a look at a simple simulation of a multi-armed bandit problem. There are two “casinos,” I mean sets of slot machines. In casino A, all the slot machines gives out the same reward I prepared four types of “multi-armed bandits,” I mean octopus agents. Each of them has each value of I wold not concretely explain how values of each slot machines are updated in this article. I think I am going to explain multi-armed bandit problems with Monte Carlo tree search in one of upcoming articles to explain the algorithm of AlphaGo/AlphaZero. (4)”Achievement” or “estimation” The last pair of keywords is “achievement” or “estimation,” and it might be better to instead see them as a comparison of “Monte Carlo” and “temporal-difference (TD).” I said RL algorithms often approximate Bellman equation based on data an agent has collected. Agents moving around in environments can be viewed as sampling data from the environment. Agents sample data of states, actions, and rewards. At the same time agents constantly estimate the value of each state. Thus agents can modify their estimations of values using value calculated with sampled data. This is how agents make use of their “experiences” in RL. There are several variations of when to update estimations of values, but roughly they are classified to Monte Carlo and Temporal-difference (TD). Monte Carlo is based on achievements of agents after one episode or actions. And TD is more of based on constant estimation of values at every time step. Which approach is to take depends on tasks but it seems many major algorithms adopt TD types. But I got an impression that major RL algorithms adopt TD, and also it is said evaluating actions by TD has some analogies with how brain is “reinforced.” And above all, according to the book by Sutton and Barto “If one had to identify one idea as central and novel to reinforcement learning, it would undoubtedly be temporal-difference (TD) learning.” And an intermediate idea, between Monte Carlo and TD, also can be formulated as eligibility trace. In this article I have briefly covered all the topics I am planning to explain in this series. This article is a start of a long-term journey of studying RL also to me. Any feedback on this series, as posts or emails, would be appreciated. The next article is going to be about dynamic programming, which is a major way for solving planning problems. In contexts of RL, dynamic programming is solved by repeatedly applying Bellman equation on values of states of a model of an environment. Thus I think it is no exaggeration to say dynamic programming is the backbone of RL algorithms. The code I used for the multi-armed bandit simulation. Just copy and paste them on Jupyter Notebook. * I make study materials on machine learning, sponsored by DATANOMIQ. I do my best to make my content as straightforward but as precise as possible. I include all of my reference sources. If you notice any mistakes in my materials, including grammatical errors, please let me know (email: yasuto.tamura@datanomiq.de). And if you have any advice for making my materials more understandable to learners, I would appreciate hearing it. [1] Morimura Tetsuro, “Machine Learning Professional Series: Reinforcement Learning,” Kodansha, (2109) 森村哲郎著, 「機械学習プロフェッショナルシリーズ強化学習」, 講談社, (2019) [2] Richard S. Sutton and Andrew G. Barto, “Reinforcement Learning: An Introduction Second Edition,” MIT Press, (2018) [3] Kubo Takahiro, “Machine Learning Startup Series: Reinforcement Learning with Python,” Kodansha, (2019) 久保隆宏著, 「機械学習スタートアップシリーズ Python で学強化学習改訂第２版」, 講談社, (2019) [4] Sebastian Thrun, Wolfram Burgard and Dieter Fox, “Probabilistic Robotics,” MIT Press, (2015), pp 487-510 [5] 布留川英一著「AlphaZero 深層学習・強化学習・探索人工知能プログラミング実践入門」, 2019, ボーンデジタル Eiichi Hurukawa, “AlphaZero Deep Learning・Reinforcement Learning・Searching Artificial Intelligence Programming Practical Introduction”, 2019, Bone Digital https://data-science-blog.com/wp-content/uploads/2022/06/RL_head_image_2.png 383 935 Yasuto Tamura https://www.data-science-blog.com/wp-content/uploads/2016/09/data-science-blog-logo.png Yasuto Tamura2021-09-11 09:24:292022-03-12 18:31:48Understanding the “simplicity” of reinforcement learning: comprehensive tips to take the trouble out of RL Predictive Maintenance – Konzept und Chancen /in Uncategorized/by Georg Ungerböck (Maschinen)Zeit ist kostbar. Das trifft besonders auf produzierende Unternehmen zu. Denn hier gilt, jeder Stillstand einer Anlage kostet wertvolle Produktionskapazität. Stillstände einer Maschine lassen sich nicht 100%ig vermeiden, nur können sie mit dem passenden Predictive Maintenance Konzept (im weiteren Text als PdM abgekürzt) reduziert und besser planbar gemacht werden. Mit intelligenten Add-ons wie IoT Anbindung von Maschinen in einer Industrie 4.0 Umgebungen und die Integration in ein gut geplantes Predictive Maintenance System lassen sich Kosten einsparen. Was ist Predictive Maintenance? Unter PdM sind Features beim Betrieb einer Anlage oder Maschine gemeint, die aus historischen Daten lernen und in Verbindung mit aktuellen oder sogar Echtzeitdaten Prognosen über bevorstehende Ereignisse durchführen. Aus den Berechnungen können z.B. kommende Wartungsarbeiten oder sich andeutende Ausfälle von Komponenten abgeleitet werden. Ersatzteile müssen nicht mehr vorsorglich auf Lager gelegt werden, sondern können aufgrund der tatsächlichen Notwendigkeit bestellt werden. Vorteile durch Einsatz von Predictive Maintenance Weiterer Pluspunkt ist die gute Planbarkeit von Wartungsintervallen. Betreiber als auch Hersteller können die Terminpläne nach dem vorherberechneten Wartungszeitpunkt einteilen. Als Betreiber können die Verfügbarkeiten mit den notwendigen Kapazitäten für die Produktion korreliert werden. Als Hersteller können Sie die Bestellung von Ersatzteilen im tatsächlich gebrauchten Umfang termingerecht durchführen. Und Sie müssen nicht von jedem möglichen Ersatzteil eine Anzahl immer vorrätig haben. Technisches Konzept und Überlegungen Für die Implementierung eines PdM, egal ob Sie das selbst durchführen wollen oder als Produkt zukaufen wollen, ist ein technisches Konzept der Startpunkt. Wir wollen hier die Eckdaten dieser Überlegungen skizzieren, um so als Arbeitsunterlage zur Erarbeitung des Konzepts dienen zu können. Ein PdM Konzept besteht grob aus den folgenden Komponenten: Predictive Maintenance Konzept • Maschine: hier werden Daten erzeugt, die in das PdM System übernommen werden sollen. Von den Maschinen sollen die Daten möglichst rasch und instantan abgegriffen werden, um die Werte in die Cloud zu laden und am Endgerät verfügbar zu haben. Meist ist auf den Maschinen nur begrenzter Speicherplatz vorhanden. Und die Daten können dort nur kurz zwischengespeichert werden. • Data-Agent und Transferprotokoll: die Übertragung der Daten auf die Cloudplattform wird durch eine Softwarekomponente durchgeführt. Diese kann entweder vom Hersteller mitgeliefert und bereits in der Maschine integriert sein. Oder sie wird als Teil des PdM Konzepts ergänzt. Aufgabe ist, die Daten gesichert auf die Cloudplattform zu übertragen. Bei Ausfall der Netzwerkverbindung kann ein lokales Spooling der Daten mit anschließender gesammelter Übertragung erfolgen. Neben der Datenübertragung muss an dieser Stelle auch eine Registrierung neuer Maschinen möglich sein. Der Data-Agent darf sich nicht einfach durch Klonen auf eine neue Maschine übertragen lassen. Eine geeignete Sicherung über z.B. Hardware-IDs muss hier durchgeführt werden. • Data-Aggregator: kann oder soll nicht die Maschine direkt Daten in die Cloud übertragen, können die Daten mehrerer Maschinen mit einem Data-Aggregator zusammengefasst werden. Von dort wird dann die verschlüsselte Übertragung auf die Cloudplattform durchgeführt. Gründe für den Einsatz eines Data-Aggregators könnten sein, dass beim Endkunden keine Übertragung von einzelnen Maschinen ins Netz erlaubt ist. Oder dass auch „Legacy“ Maschinen angebunden werden sollen, für die es keine Plugins zur direkten Datenübertragung gibt. Z.B., wenn eine Maschine mit einer älteren SPS angebunden werden soll, für die technisch keine direkte Übertragung auf die Cloud möglich ist. • Cloud- / Webplattform: die übertragenen Daten müssen zentral in einer geeigneten Umgebung gespeichert werden. Aus diesen gesammelten Daten werden die eigentlichen Erkenntnisse und Vorhersagen eines PdM Systems gewonnen. Durch eine KI und selbstlernende Algorithmen können die Daten weiter verwertet werden. Die gewonnen Ergebnisse aus den analysierten Maschinendaten sind die Basis für das PdM System und werden den Anwendern grafisch aufbereitet oder als Infos und Warnungen per Message zugestellt. • Endgerät: ist der Zugangspunkt für den Anwender. Die PdM Daten werden als App oder als Webanwendung dargestellt. Der Data-Agent / Data-Aggregator kann mittels Edge Computing lokale Intelligenz erhalten. Daten können bereits vorausgewertet und zusammengefasst werden. Das reduziert die übertragenen Daten. Welche Werte sollen übertragen werden? Ziel des PdM ist durch das Abgreifen und Auswerten der Maschinendaten im Endeffekt eine Vernetzung mit den informationsverarbeitenden Systemen in einem Unternehmen. Das wird z.B. ein ERP Enterprise Resource Planning oder ein MES Manufacturing Execution System sein. Dort werden aufgrund der PdM Daten die Ressourcen und Kapazitäten für die Produktion geplant. Typische Daten zur Übertragung bei einem PdM sind: • Temperatur • Druck • Geschwindigkeiten • Zurückgelegte Wege • Schaltspiele • Viskositäten • Flüssigkeitsstände • Vibrationen Welche Daten Sie abgreifen können hängt damit zusammen, ob Sie Anwender also ein Produktionsbetrieb oder Hersteller also ein Maschinebauer sind. Als Anwender haben Sie normalerweise weniger tiefe Zugriffsmöglichkeiten auf Daten und Parameter der Maschinen. Nur die vom Hersteller bereitgestellten und dokumentierten Werte sind zugänglich. Diese sind vom Level eher auf Applikationsschicht angesetzt. Als Hersteller können Sie auf beliebige Werte zurückgreifen. Dazu gehören auch Dinge wie Schaltspiele oder zurückgelegte Fahrwege von Motoren. Übertragen Sie jene Daten ins PdM, aus denen Sie die Wartungsarbeiten Ihrer Maschine ermitteln können. Z.B. bei einer Glashärteanlage wären das die Betriebsstunden der Keramikwalzen, zurückgelegte Wege der Keilriemen oder Einschaltzeiten der Heizelemente. Z.B. bei einer Automatisierung für die Leiterplattenfertigung wären das die Betriebsstunden der Saugnäpfe oder die zurückgelegten Wege der Antriebsriemen. Wie oft sollen Werte übertragen werden? Wenn Sie von der Netzwerkanbindung mit keinen Einschränkungen in Bezug auf Bandbreite oder Datenlimit rechnen müssen, nehmen Sie als Übertragungsintervall eine relativ gute Granularität an. Wählen Sie es so aus, dass Probleme an der Maschine auch nachträglich noch analysiert und der Auslöser gefunden werden kann. Bei den meisten Industrie 4.0 Umgebungen sollte die Datenmenge keine große Rolle spielen. Sollten Sie in einer IoT Umgebung mit z.B. LoRaWAN-Anbindung arbeiten, dann teilen Sie die Daten in Kategorien nach Priorität ein. Z.B. hoch, mittel, niedrig oder z.B. Produktion, Standby. Die Übertragung der Kategorien können dann je nach Betriebszustand differenziert werden, wann welche Kategorie wichtig ist und priorisiert übertragen werden soll. Die Umsetzung eines Predictive Maintenance Konzepts hilft Ihnen die Produktion agiler zu gestalten. Terminpläne aufgrund vorausgesagter Wartungszeiten der Anlagen lassen eine präzisere und engere Planung der Kapazitäten zu. Dieser Effekt wirkt sich positiv auf Produktionskosten aus. Ein großes Sparpotential hat ein PdM für die kommenden CO2 Steuermodelle. Mit den ermittelten Daten können Sie exakte Berechnungen über die verbrauchte Energie auf das produzierte Werkstück durchführen und so CO2 Steuer sparen. Mit smarten Diensten wie einem PdM können Sie als Maschinenhersteller dauerhaft Geld verdienen. Sie generieren weiteren Umsatz von Ihren Kunden und erhöhen gleichzeitig die Kundenbindung. Ihre Kunden werden durch vorausschauende Wartung zufriedener mit Ihren Produkten. Vorausschauende Wartung hat Potential für Endanwender als auch für Hersteller. Beim Endanwender steht das Sparpotential im Vordergrund beim Hersteller die Kundenzufriedenheit. Mit intelligenten Edge-Computing Komponenten lassen sich PdM Lösungen gut skalieren und die Datenmenge reduzieren. Die Umsetzung einer Lösung für Predictive Maintenance ist nicht an die Installation oder Entwicklung einer neuen Anlage gebunden. Auch bereits laufende Maschinen können leicht in ein PdM integriert https://data-science-blog.com/wp-content/uploads/2021/07/cloud-plattform-predictive-maintenance-header-1.png 533 1650 Georg Ungerböck https://www.data-science-blog.com/wp-content/uploads/2016/09/ data-science-blog-logo.png Georg Ungerböck2021-07-22 08:50:452021-07-12 08:51:25Predictive Maintenance – Konzept und Chancen Seq2seq models and simple attention mechanism: backbones of NLP tasks /in Uncategorized/by Yasuto Tamura This is the second article of my article series “Instructions on Transformer for people outside NLP field, but with examples of NLP.” 1 Machine translation and seq2seq models I think machine translation is one of the most iconic and commercialized tasks of NLP. With modern machine translation you can translate relatively complicated sentences, if you tolerate some grammatical errors. As I mentioned in the third article of my series on RNN, research on machine translation already started in the early 1950s, and their focus was translation between English and Russian, highly motivated by Cold War. In the initial phase, machine translation was rule-based, like most students do in their foreign language classes. They just implemented a lot of rules for translations. In the next phase, machine translation was statistics-based. They achieved better performance with statistics for constructing sentences. At any rate, both of them highly relied on feature engineering, I mean, you need to consider numerous rules of translation and manually implement them. After those endeavors of machine translation, neural machine translation appeared. The advent of neural machine translation was an earthshaking change of machine translation field. Neural machine translation soon outperformed the conventional techniques, and it is still state of the art. Some of you might felt that machine translation became more or less reliable around that time. I think you have learnt at least one foreign or classical language in school. I don’t know how good you were at the classes, but I think you had to learn some conjugations of them and I believe that was tiresome to most of students. For example, as a foreign person, I still cannot use “der”, “die”, “das” properly. Some of my friends recommended I do not care them for the time being while I speak, but I usually care grammar very much. But this method of learning language is close to the rule base machine translation, and modern neural machine translation basically does not rely on such As far as I understand, machine translation is pattern recognition learned from a large corpus. Basically no one implicitly teach computers how grammar works. Machine translation learns very complicated mapping from a source language to a target language, based on a lot of examples of word or sentence pairs. I am not sure, but this might be close to how bilingual kids learn how the two languages are related. You do not need to navigate the translator to learn specific grammatical rules. Since machine translation does not rely on manually programming grammatical rules, basically you do not need to prepare another specific network architecture for another pair of languages. The same method can be applied to any pairs of languages, as long as you have an enough size of corpus for that. You do not have to think about translation rules between other pairs of languages. I do not follow the cutting edge studies on machine translation, so I am not sure, but I guess there are some heuristic methods for machine translation. That is, designing a network depending on the pair of languages could be effective. When it comes grammatical word orders, English and Japanese have totally different structures, I mean English is basically SVO and Japanese is basically SOV. In many cases, the structures of sentences with the same meaning in both of the languages are almost like reflections in a mirror. A lot of languages have similar structures to English, even in Asia, for example Chinese. On the other hand relatively few languages have Japanese-like structures, for example Korean, Turkish. I guess there would be some grammatical-structure-aware machine translation Not only machine translations, but also several other NLP tasks, such as summarization, question answering, use a model named seq2seq model (sequence to sequence model). As well as other deep learning techniques, seq2seq models are composed of an encoder and a decoder. In the case of seq2seq models, you use RNNs in both the encoder and decoder parts. For the RNN cells, you usually use a gated RNN such as LSTM or GRU because simple RNNs would suffer from vanishing gradient problem when inputs or outputs are long, and those in translation tasks are long enough. In the encoder part, you just pass input sentences. To be exact, you input them from the first time step to the last time step, every time giving an output, and passing information to the next cell via recurrent I think you would be confused without some understandings on how RNNs propagate forward. You do not need to understand this part that much if you just want to learn Transformer. In order to learn Transformer model, attention mechanism, which I explain in the next section is more important. If you want to know how basic RNNs work, an article of mine should help you. In the encoder part of the figure below, the cell also propagate information backward. I assumed an encoder part with bidirectional RNNs, and they “forward propagate” information backwards. But in the codes below, we do not consider such complex situation. Please just keep it in mind that seq2seq model could use bidirectional RNNs. At the last time step in the encoder part, you pass the hidden state of the RNN to the decoder part, which I show as a yellow cell in the figure below, and the yellow cell/layer is the initial hidden layer of the first RNN cell of the decoder part. Just as normal RNNs, the decoder part start giving out outputs, and passing information via reccurent connections. At every time step you choose a token to give out from the vocabulary you use in the task. That means, each cell of decoder RNNs does a classification task and decides which word to write out at the time step. Also, very importantly, in the decoder part, the output at one time step is the input at the next time step, as I show as dotted lines in the figure below. The translation algorithm I explained depends on greedy decoding, which has to decide a token at every time step. However it is easy to imagine that that is not how you translate a word. You usually erase the earlier words or you construct some possibilities in your mind. Actually, for better translations you would need decoding strategies such as beam search, but it is out of the scope of at least this article. Thus we are going to make a very simplified translator based on greedy decoding. 2 Learning by making It would take some hours on your computer to train the translator if you do not use a GPU. I recommend you to run it at first and continue reading this article. Seq2seq models do not have that complicated structures, and for now you just need to understand the points I mentioned above. Rather than just formulating the models, I think it would be better to understand this model by actually writing codes. If you copy and paste the codes in this Github page or the official Tensorflow tutorial, installing necessary libraries, it would start training the seq2seq model for Spanish-English translator. In the Github page, I just added comments to the codes in the official tutorial so that they are more understandable. If you can understand the codes in the tutorial without difficulty, I have to say this article itself is not compatible to your level. Otherwise, I am going to help you understand the tutorial with my original figures. I made this article so that it would help you read the next article. If you have no idea what RNN is, at least the second article of my RNN series should be helpful to some extent. If you try to read the the whole article series of mine on RNN, I think you should get prepared. I mean, you should prepare some pieces of paper and a pen. It would be nice if you have some stocks of coffee and snacks. Though I do not think you have to do that to read this article. 2.1 The corpus and datasets In the codes in the Github page, please ignore the part sandwiched by “######”. Handling language data is not the focus of this article. All you have to know is that the codes below first create datasets from the Spanish-English corpus in http://www.manythings.org/anki/ , and you datasets for training the translator as the tensors below. Each token is encoded with integers as the codes below, thus after encoding, the Spanish sentence “Todo sobre mi madre.” is [1, 74, 514, 19, 237, 3, 2]. 2.2 The encoder The encoder part is relatively simple. All you have to keep in mind is that you put input sentences, and pass the hidden layer of the last cell to the decoder part. To be more concrete, an RNN cell receives an input word every time step, and gives out an output vector at each time step, passing hidden states to the next cell. You make a chain of RNN cells by the process, like in the figure below. In this case “time steps” means the indexes of the order of the words. If you more or less understand how RNNs work, I think this is nothing difficult. The encoder part passes the hidden state, which is in yellow in the figure below, to the decoder part. Let’s see how encoders are implemented in the code below. We use a type of RNN named GRU (Gated Recurrent Unit). GRU is simpler than LSTM (Long Short-Term Memory). One GRU cell gets an input every time step, and passes one hidden state via recurrent connections. As well as LSTM, GRU is a gated RNN so that it can mitigate vanishing gradient problems. GRU was invented after LSTM for smaller computation costs. At time step TO BE VERY HONEST, I am not sure why the encoder part of seq2seq models are implemented this way in the codes below. In the implementation below, the number of total time steps in the encoder part is fixed to 16. If input sentences have less than 16 tokens, it seems the RNN cells get no inputs after the time step of the token “<end>”. As far as I could check, if RNN cells get no inputs, they repeats giving out similar 1024-d vectors. I think in this implementation, RNN cells after the <end> token, which I showed as the dotted RNN cells in the figure above, do not change so much. And the encoder part passes the hidden state of the 16th RNN cell, which is in yellow, to the decoder. 2.3 The decoder The decoder part is also not that hard to understand. As I briefly explained in the last section, you initialize the first cell of the decoder, using the hidden layer of the last cell the encoder. During decoding, I mean while writing a translation, at the beginning you put the token “<start>” as the first input of the decoder. Given the input “<start>”, the first cell outputs “all” in the example in the figure below, and the output “all” is the input of the next cell. The output of the next cell “about” is also passed to the next cell, and you repeat this till the decoder gives out the token “<end>”. A more important point is how to get losses in the decoder part during training. We use a technique named teacher enforcing during training the decoder part of a seq2seq model. This is also quite simple: you just have to make sure you input a correct answer to RNN cells, regardless of the outputs generated by the cell last time step. You force the decoder to get the correct input every time step, and that is what teacher forcing is all about. You can see how the decoder part and teacher forcing is implemented in the codes below. You have to keep it in mind that unlike the ‘Encoder’ class, you put a token into a ‘Decoder’ class every time step. To be exact you also need the outputs of the encoder part to calculate attentions in the decoder part. I am going to explain that in the next subsection. 2.4 Attention mechanism I think you have learned at least one foreign language, and usually you have to translate some sentences. Remember the processes of writing a translation of a sentence in another language. Imagine that you are about to write a new word after writing some. If you are not used to translations in the language, you must have cared about which parts of the original language correspond to the very new word you are going to write. You have to pay “attention” to the original sentence. This is what attention mechanism is all about. I would like you to pay “attention” to this section. As you can see from the fact that the original paper on Transformer model is named “Attention Is All You Need,” attention mechanism is a crucial idea of Transformer. In the decoder part you initialize the hidden layer with the last hidden layer of the encoder, and its first input is “<start>”. The decoder part start decoding, , as I explained in the last subsection. If you use attention mechanism in the seq2seq model, you calculate attentions every times step. Let’s consider an example in the figure below, where the next input in the decoder is “my”, and given the token “my”, the GRU cell calculates a hidden state at the time step. The hidden state is the “query” in this case, and you compare the “query” with the 6 outputs of the encoder, which are “keys”. You get weights/scores, I mean “attentions”, which is the histogram in the figure below. In the implementation, however, the size of the output of the ‘Encoder’ class is always (16, 2024). You calculate attentions for all those 16 output vectors, but virtually only the first 6 1024-d output vectors important. Summing up the points I have explained, you compare the “query” with the “keys” and get scores/weights for the “values.” Each score/weight is in short the relevance between the “query” and each “key”. And you reweight the ‘values’ with the scores/weights, and take the summation of the reweighted “values.” In the case of attention mechanism in this article, we can say that “values” and “keys” are the same. You would also see that more clearly in the implementation below. You especially have to pay attention to the terms “query”, “key”, and “value.” “Keys” and “values” are basically in the same language, and in the case above, they are in Spanish. “Queries” and “keys” can be in either different or the same. In the example above, the “query” is in English, and the “keys” are in Spanish. You can compare a “query” with “keys” in various ways. The implementation uses the one called Bahdanau’s additive style, and in Transformer, you use more straightforward ways. You do not have to care about how Bahdanau’s additive style calculates those attentions. It is much more important to learn the relations of “queries”, “keys”, and “values” for now. A problem is that Bahdanau’s additive style is slightly different from the figure above. It seems in Bahdanau’s additive style, at the time step 2.5 Translating and displaying attentions After training the translator for 20 epochs, I could translate Spanish sentences, and the implementation also displays attention scores for between the input and output sentences. For example the translation of the inputs “Todo sobre mi madre.” and “Habre con ella.” were “all about my mother .” and “i talked to her .” respectively, and the results seem fine. One powerful advantage of using attention mechanism is you can display this type of word alignment, I mean correspondences of words in a sentence, easily as in the heat maps below. The yellow parts shows high scores of attentions, and you can see that the distributions of relatively highs scores are more or less diagonal, which implies that English and Spanish have similar word orders. For other inputs like “Mujeres al borde de un ataque de nervious.” or “Volver.”, the translations are not good. You might have noticed there is one big problem in this implementation: you can use only the words appeared in the corpus. And actually I had to manually add some pairs of sentences with the word “borde” to the corpus to get the translation in the figure. [1] “Neural machine translation with attention,” Tensorflow Core [2]Tsuboi Yuuta, Unno Yuuya, Suzuki Jun, “Machine Learning Professional Series: Natural Language Processing with Deep Learning,” (2017), pp. 72-85, 91-94 坪井祐太、海野裕也、鈴木潤著, 「機械学習プロフェッショナルシリーズ深層学習による自然言語処理」, (2017), pp. 72-85, 191-193 [3]”Stanford CS224N: NLP with Deep Learning \| Winter 2019 \| Lecture 8 – Translation, Seq2Seq, Attention”, stanfordonline, (2019) * I make study materials on machine learning, sponsored by DATANOMIQ. I do my best to make my content as straightforward but as precise as possible. I include all of my reference sources. If you notice any mistakes in my materials, including grammatical errors, please let me know (email: yasuto.tamura@datanomiq.de). And if you have any advice for making my materials more understandable to learners, I would appreciate hearing it. https://data-science-blog.com/wp-content/uploads/2021/12/Transformer_head_img.png 377 2406 Yasuto Tamura https://www.data-science-blog.com/wp-content/uploads/2016/09/data-science-blog-logo.png Yasuto Tamura2021-02-17 10:20:432021-03-07 12:59:58Seq2seq models and simple attention mechanism: backbones of NLP tasks AI Voice Assistants are the Next Revolution: How Prepared are You? /in Uncategorized/by Michael Lyamm By 2022, voice-based shopping is predicted to rise to USD 40 billion, based on the data from OC&C Strategy Consultants. We’re in an era of ‘voice’ where drastic transformation is seen between the way AI and voice recognition are changing the way we live. According to the survey, the surge of voice assistants is said to be driven by the number of homes that used smart speakers, as such that the rise is seen to grow from 13% to 55%. Nonetheless, Amazon will be one of the leaders to dominate the new channel having the largest market share. Perhaps this is the first time you’ve heard about the voice revolution. Well, why not, based on multiple researchers, it is estimated that the number of voice assistants will grow to USD 8 billion by 2023 from USD 2.5 billion in 2018. But what is voice revolution or voice assistant or voice search? It was only until recently that the consumers have started learning about voice assistants which further predicts to exist in the future. You’ve heard of Alexa, Cortana, Siri, and Google Assistant, these technologies are some of the world’s greatest examples of voice assistants. They will further help to drive consumer behavior as well as prepare the companies and adjust based on the industry demands. Consumers can now transform the way they act, search, and advertise their brand through voice technology. Voice search is a technology to help users or consumers perform a search on the website by simply asking a question on their smartphone, their computer, or their smart device. The voice assistant awareness: Why now? As surveyed by PwC, amongst the 90% respondents, about 72% have been recorded to use voice assistant while merely 10% said they were clueless about voice-enabled devices and products. It is noted, the adoption of voice-enabled was majorly driven by children, young consumers, and households earning an income of around >USD100k. Let us have a glance to ensure the devices that are used mainly for voice assistance: – • Smartphone – 57% • Desktop – 29% • Tablet – 29% • Laptop – 29% • Speaker – 27% • TV remote – 21% • Car navigation – 20% • Wearable – 14% According to the survey, most consumers that use voice-assistants were the younger generation, aged between 18-24. While individuals between the ages 25-49 were said to use these technologies in a much more statistical manner, and are called the “heavy users.” Significance of mobile voice assistants: What is the need? Although mobile is accessible everywhere, you will merely find three out of four consumers using mobile voice assistants in their household i.e. 74%. Mobile-based AI chatbots have taken our lives by storm, thus providing the best solution to both the customers and agents in varied areas – insurance, travel, and education, etc. A certain group of individuals said they needed privacy while speaking to their device and that sending a voice command in public is weird. Well, this simply explains why 18-24 aged group individuals prefer less use of voice assistants. However, this age group tends to spend more time out of their homes. Situations where voice assistants can be used – standalone speakers Vs mobile • Standalone speakers – 65% • Mobile – 37% • Standalone speakers – 62% • Mobile – 12% Watching TV • Standalone speakers – 57% • Mobile – 43% In bed • Standalone speakers – 38% • Mobile – 37% • Standalone speakers – 29% • Mobile – 25% • Standalone speakers – 0% • Mobile – 40% By the end of 2020, nearly half of all the searches made will be voice-based, as predicted by Comscore, a media analytics firm. Don’t you think voice-based assistant is changing the way businesses function? Thanks to the advent of AI! • A 2018 study on AI chatbots and voice assistants by Spiceworks said, 24% of businesses that were spread largely, and 16% of smaller businesses have already started using AI technologies in their workplaces. While 25% of the business market is expected to adopt AI within the next 12 months. Surprisingly, voice-based assistants such as Siri, Google Assistant, and Cortana are some of the most prominent technologies these businesses are using in their workstations. Where will the next AI voice revolution take us? Voice-authorized transactions Paypal, an online payment gateway now leverages Siri and Alexa’s voice recognition capability, thus, allowing users to make payments, check their balance, and ask payments from people via voice Voice remote control – AI-powered Communications conglomerate Comcast, an American telecommunications and media conglomerate introduces their first-ever X1 voice remote control that provides both natural image processing and voice With the help of deep learning, the X1 can easily come up with better search results with just a press of the button telling what your television needs to do next. Voice AI-enabled memos and analytics Salesforce recently unveiled Einstein Voice which is an AI assistant that helps in entering critical data the moment it hears, making use of the voice command. This AI assistant also initiates in interpreting voice memos. Besides this, the voice bots accompanying Einstein Voice also helps the company create their customized voice bots to answer customer queries. Voice-activated ordering It is astonishing to see how Domino’s is using voice-activated feature automate orders made over the phone by customers. Well, welcome to the era of voice revolution. This app, developed by Nuance Communications already has a Siri like voice recognition feature that allows customers to place their orders just like how they would be doing it in front of the cash counter making your order to take place efficiently. As more businesses look forward to breaking down the roadblocks between a consumer and a brand, voice search now projects to become an impactful technology of bridging the gap. https://data-science-blog.com/wp-content/uploads/2020/08/AI-voice-assistant-header.png 400 997 Michael Lyamm https://www.data-science-blog.com/wp-content/uploads/2016/09/data-science-blog-logo.png Michael Lyamm2020-08-17 10:48:002020-09-15 15:20:32AI Voice Assistants are the Next Revolution: How Prepared are You? A gentle introduction to the tiresome part of understanding RNN /in Artificial Intelligence, Data Science, Deep Learning, Machine Learning, Mathematics, Uncategorized/by Yasuto Tamura Just as a normal conversation in a random pub or bar in Berlin, people often ask me “Which language do you use?” I always answer “LaTeX and PowerPoint.” I have been doing an internship at DATANOMIQ and trying to make straightforward but precise study materials on deep learning. I myself started learning machine learning in April of 2019, and I have been self-studying during this one-year-vacation of mine in Berlin. Many study materials give good explanations on densely connected layers or convolutional neural networks (CNNs). But when it comes to back propagation of CNN and recurrent neural networks (RNNs), I think there’s much room for improvement to make the topic understandable to learners. Many study materials avoid the points I want to understand, and that was as frustrating to me as listening to answers to questions in the Japanese Diet, or listening to speeches from the current Japanese minister of the environment. With the slightest common sense, you would always get the feeling “How?” after reading an RNN chapter in any book. This blog series focuses on the introductory level of recurrent neural networks. By “introductory”, I mean prerequisites for a better and more mathematical understanding of RNN algorithms. I am going to keep these posts as visual as possible, avoiding equations, but I am also going to attach some links to check more precise mathematical explanations. This blog series is composed of five contents.: https://data-science-blog.com/wp-content/uploads/2020/04/RNN_head_pic.png 802 1952 Yasuto Tamura https://www.data-science-blog.com/wp-content/uploads/2016/09/data-science-blog-logo.png Yasuto Tamura 2020-05-01 09:26:342020-09-07 15:02:48A gentle introduction to the tiresome part of understanding RNN Business Data is changing the world’s view towards Green Energy /in Uncategorized/by Ashish Parmar Energy conservation is one of the main stressed points all around the globe. In the past 30 years, researches in the field of energy conservation and especially green energy have risen to another level. The positive outcomes of these researches have given us a gamut of technologies that can aid in preserving and utilize green energy. It has also reduced the over-dependency of companies on fossil fuels such as oil, coal, and natural gas. Business data and analytics have all the power and the potential to take the business organizations forward in the future and conquer new frontiers. Seizing the opportunities presented by Green energy, market leaders such as Intel and Google have already implemented it, and now they enjoy the rich benefits of green energy sources. Business data enables the organizations to keep an eye on measuring the positive outcomes by adopting the green energies. According to a report done by the World energy outlook, the global wind energy capacity will increase by 85% by the year 2020, reaching 1400 TWh. Moreover, in the Paris Summit, more than 170 countries around the world agreed on reducing the impact of global warming by harnessing energy from green energy sources. And for this to work, Big Data Analytics will play a pivotal role. Overview of Green energy In simpler terms, Green Energy is the energy coming from natural sources such as wind, sun, plants, tides, and geothermal heat. In contrast to fossil fuels, green energy resources can be replenished in a short period, and one can use them for longer periods. Green energy sources have a minimal ill effect on the environment as compared to fossil fuels. In addition to this, fossil fuels can be replaced by green energy sources in many areas like providing electricity, fuel for motor vehicles, etc.. With the help of business data, organizations throughout the world can change the view of green energy. Big Data can show how different types of green energy sources can help businesses and accelerate sustainable expansion. Below are the different types of green energy sources: • Wind Power • Solar Power • Geothermal Energy • Hydropower • Biofuels • Bio-mass Now we present before you a list of advantages that green energy or renewable energy sources have brought to the new age businesses. Profits on the rise If the energy produced is more than the energy used, the organizations can sell it back to the grids and earn profit out of it. Green energy sources are renewable sources of energy, and with precise data, the companies will get an overall estimation of the requirement of energy. With Big Data, the organizations can know the history of the demographical location before setting up the factory. For example, if your company is planning to setup a factory in the coastal region, tidal and wind energy would be more beneficial as compared to solar power. Business data will give the complete analysis of the flow of the wind so that the companies can ascertain the best location of the windmill; this will allow them to store the energy in advance and use it as per their requirement. It not only saves money but also provides an extra source of income to the companies. With green energy sources, the production in the company can increase to an unprecedented level and have sustainable growth over the years. Synchronizing the maintenance process If there is a rapid inflow of solar and wind energy sources, the amount of power produced will be huge. Many solar panels and windmills are operating in a solar power plant or in a wind energy source, and with many types of equipment, it becomestoo complex to manage. Big Data analytics will assist the companies in streamlining all the operations to a large extent for their everyday work without any hassle. Moreover, the analytics tool will convey the performance of renewable energy sources under different weather conditions. Thus, the companies will get the perfect idea about the performance of the green energy sources, thus enabling them to take necessary actions as and when required. Lowering the attrition rate Researchers have found that more number of employees want to be associated with companies that support green energies. By opting for green energy sources and investing in them, companies are indirectly investing in keeping the workforce intact and lowering the attrition rate. Stats also show the same track as nearly 50% of the working professionals, and almost 2/3^rd of the millennial population want to be associated with the companies who are opting for the green energy sources and have a positive impact on environmental conservation. The employees will not only wish to stay with the organizations for a long time but will also work hard for the betterment of the organization. Therefore, you can concentrate on expanding the business rather than thinking about the replacement of the employees. Lowering the risk due to Power Outage The Business Data Analytics will continuously keep updating the requirements of power needed to run the company. Thus the organizations can cut down the risk of the power outage and also the expenses related to it. The companies will know when to halt the energy transmission as they would know if the grid is under some strain or not. Business analytics and green energy provide a planned power outage to the companies, which is cost-efficient and thus can decrease the product development cost. Apart from this, companies can store energy for later usage. Practicing this process will help save a lot of money in the long run, proving that investment in green energy sources is a smart investment. Reducing the maintenance cost An increasing number of organizations are using renewable sources of energy as it plays a vital role in decreasing production and maintenance costs. The predictive analysis technology helps renewable energy sources to produce more energy at less cost, thus reducing the cost of infrastructure. Moreover, data analytics will make green energy sources more bankable for companies. As organizations will have a concrete amount of data related to the energy sources, they can use it wisely on a more productive basis Escalating Energy Storage Green energy sources can be stored in bulk and used as per requirement by the business organizations. Using green energy on a larger basis will even allow companies to completely get rid of fossil fuels and thus work towards the betterment of the environment. Big Data analytics with AI and cloud-enabled systems help organizations store renewable energies such as Wind and Solar. Moreover, it gathers information for the businesses and gives the complete analysis of the exact amount of energy required to complete a particular task. The data will also automate cost savings as it can predict the client’s needs. Based on business data, companies can store renewable energy sources in a better manner. With Business data analytics, the companies can store energy when it is cheap and use it according to the needs when the energy rates go higher. Although predicting the requirement of storage is a complicated process, with Artificial Intelligence (AI) at work, you can analyze the data efficiently. Bundling Up Green energy sources will play a pivotal role in deciding the future of the businesses as fossil fuels are available in a certain limit. Moreover, astute business data analysts will assist the organizations to not only use renewable energy sources in a better manner but also to form a formidable workforce. The data support in the green energy sector will also provide sustainable growth to the companies, monitor their efforts, and assist them in the long run. https://data-science-blog.com/wp-content/uploads/2020/04/how-big-data-is-helping-the-green-energy-sector.jpg 628 1472 Ashish Parmar https://www.data-science-blog.com/wp-content/uploads/2016/09/ data-science-blog-logo.png Ashish Parmar2020-04-17 10:43:452020-04-01 22:45:24Business Data is changing the world’s view towards Green Energy Predictive Analytics World 2020 Healthcare /in Uncategorized/by Editorial Staff Difficult times call for creative measures Predictive Analytics World for Healthcare will go virtual and you still have time to join us! What do you have in store for me? We will provide a live-streamed virtual version of healthcare Munich 2020 on 11-12 May, 2020: you will be able to attend sessions and to interact and connect with the speakers and fellow members of the data science community including sponsors and exhibitors from your home or your office. What about the workshops? The workshops will also be held virtually on the planned date: 13 May, 2020. Get a complimentary virtual sneak preview! If you would like to join us for a virtual sneak preview of the workshop „Data Thinking“ on Thursday, April 16, so you can familiarise yourself with the quality of the virtual edition of both conference and workshops and how the interaction with speakers and attendees works, please send a request to registration@risingmedia.com. Don’t have a ticket yet? It‘s not too late to join the data science community. Register by 10 May to receive access to the livestream and recordings. We’re looking forward to see you – virtually! This year Predictive Analytics World for Healthcare runs alongside Deep Learning World and Predictive Analytics World for Industry 4.0. https://data-science-blog.com/wp-content/uploads/2020/04/pawhcde20_1200x628_preview.jpg 628 1200 Editorial Staff https://www.data-science-blog.com/wp-content/uploads/2016/09/ data-science-blog-logo.png Editorial Staff2020-04-15 08:43:342020-04-14 20:45:14Predictive Analytics World 2020 Healthcare	{"url":"https://data-science-blog.com/blog/category/uncategorized/","timestamp":"2024-11-05T19:42:30Z","content_type":"text/html","content_length":"335188","record_id":"<urn:uuid:12a1c927-6af6-4550-a5ab-53c3ac06ed59>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00803.warc.gz"}
4.4 Finding x and y intercpets State the x-intercept. Remember to type it as a point. State the y-intercept. Remember to type it as a point. Find the x-intercept of the line. (Remember to type your answer as a point). Show your work on a separate sheet of x - 2y = 4 Find the x-intercept of the line. (Remember to type your answer as a point). Show your work on a separate sheet of 2x - 3y = 12 Find the x-intercept of the line. (Remember to type your answer as a point). Show your work on a separate sheet of 6x + 4y = -18 Find the x-intercept of the line. (Remember to type your answer as a point). Show your work on a separate sheet of y = 4x - 8 Find the y-intercept of the line. (Remember to type your answer as a point). Show your work on a separate sheet of x - 2y = 4 Find the y-intercept of the line. (Remember to type your answer as a point). Show your work on a separate sheet of 3x + 4y = 12 Find the y-intercept of the line. (Remember to type your answer as a point). Show your work on a separate sheet of 2x - 3y = -12 Find the y-intercept of the line. (Remember to type your answer as a point). Show your work on a separate sheet of y = 3x - 5 Students who took this test also took :	{"url":"https://www.thatquiz.org/tq/preview?c=alni5663&s=kt7ei7","timestamp":"2024-11-06T05:38:05Z","content_type":"text/html","content_length":"17903","record_id":"<urn:uuid:c2d43bab-e49c-4b08-9cb3-ca0b27d42e91>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00776.warc.gz"}
Obtaining similar solutions of a heat equation An integral equation for the dimensionless function in a similar solution of a nonlinear heat equation is derived, assuming constant temperature moments. The solution of the equation is shown to depend on the parameters whose values define the boundary conditions. Inzhenerno Fizicheskii Zhurnal Pub Date: April 1975 □ Heat Transfer Coefficients; □ Nonlinear Equations; □ Thermal Conductivity; □ Conductive Heat Transfer; □ Heat Flux; □ Mathematical Models; □ Partial Differential Equations; □ Temperature Distribution; □ Fluid Mechanics and Heat Transfer	{"url":"https://ui.adsabs.harvard.edu/abs/1975InFiZ..28..705L/abstract","timestamp":"2024-11-07T16:03:40Z","content_type":"text/html","content_length":"32911","record_id":"<urn:uuid:3669463e-134e-4844-a1b6-60568642e5a2>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00055.warc.gz"}
gsuitepros - Senior Mostest Awards The buttons on this Infographic PDF will make copies of the files for you. Most Awards Infographic.pdf Make copies of the forms, and the template. Add your boys to each question on the boys form, and each question on the girls form. Collect the data for each form. Open the spreadsheet that the form creates and copy your new data to the boys and girls tab on the template. You will have instant results. Here is an example of the results calculated instantly! At the end of last school year I was working with a client that had a project in which he had used google forms to collect the data on a "Mostest Awards" for seniors. He had 36 categories and had it separated by boys and girls. There were well over 100 total students, so the collating of the data was tedious and he needed a way to summarize the data quickly and easily. Our first approach was to build a pivot table for each category, then build a summary page with the results from 72 (boys and girls) pivot tables. While this worked, I wanted a more streamlined solution. With this spreadsheet, all the data summary is done on a single sheet, by category, by gender, using the following formula. This formula selects the data by column, counts the number of results for each student, orders the results descending, and limits to the top four vote getters. =transpose(query(index(if({1,1},boys!$A$1:$A)),"select Col1,count(Col2) where Col1<>'' group by Col1 order by count(Col2) desc limit 4 label count(Col2) '#'")) This formula is found on the winners sheet in column B for the Boys and G for the girls (ha ha, just saw that is how it worked out!)	{"url":"https://www.redtailfan.org/projects/senior-mostest-awards","timestamp":"2024-11-05T17:29:52Z","content_type":"text/html","content_length":"143996","record_id":"<urn:uuid:9b9ce975-0195-4a4a-bbad-7ae797232641>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00380.warc.gz"}
KSEEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5 Students can Download Class 7 Maths Chapter 10 Practical Geometry Ex 10.5 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams. Karnataka State Syllabus Class 7 Maths Chapter 10 Practical Geometry Ex 10.5 Question 1. Construct the right angled ∆ PQR? where m∠Q = 90°, QR = 8 cm and PR = 10 cm. Steps of Construction: 1. Draw a line segment of length 8 cm and named as QR. At Q draw QM ⊥ QR. 2. With R as centre, draw an arc of radius 10 cm and cut the ⊥^le line at P. 3. Join PR. Now we get the required PQR ∆ Question 2. Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long. Steps of Construction: 1. Draw a line segment of length 4 cm and named it AB. At A draw a ⊥ line AM. 2. With B as centre, draw an arc of 6 cms cut the x line at ‘C ’. Now we obtained the required triangle ABC. Question 3. Construct an isosceles right-angled triangle ABC, where m∠ACB = 90° and AC = 6 cm. Steps of Construction: 1. Draw a line segment of 6 cm, ie., AC. At C draw CM ⊥ CA. 2. With ‘C’ as centre draw an arc of radius 6 cm to intersect CM at ‘B’ 3. Join AB. Now we get the required ∆ACB.	{"url":"https://kseebsolutions.in/kseeb-solutions-for-class-7-maths-chapter-10-ex-10-5/","timestamp":"2024-11-13T15:40:15Z","content_type":"text/html","content_length":"133984","record_id":"<urn:uuid:54c8e69b-d06b-4799-9a83-14ccf8392431>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00725.warc.gz"}
Math and week 3a - Coursework Hero See attached MATH 106 QUIZ 2 January-February, 2013 Instructor: S. Sands NAME: _______________________________ I have completed this assignment myself, working independently and not consulting anyone except the instructor. · The quiz is worth 100 points. There are 10 problems, some with multiple parts. This quiz is open book open notes . This means that you may refer to your textbook, notes, and online classroom materials, but you must work independently and may not consult anyone (and confirm this with your submission). You may take as much time as you wish, provided you turn in your quiz no later than Sunday, February 3. Show work/explanation where indicated. Answers without any work may earn little, if any, credit. You may type or write your work in your copy of the quiz, or if you prefer, create a document containing your work. Scanned work is acceptable also. In your document, be sure to include your name and the assertion of independence of work. · General quiz tips and instructions for submitting work are posted in the Quizzes conference. · If you have any questions, please contact me by e-mail or phone (540-338-7120). 1. (4 pts) State the equation of the vertical line passing through the point (–9, 1). 1. ______ (No work/explanation required) A. x = –9 B. y = –9 C. x = 1 D. y = 1 2. (6 pts) Which of the following is TRUE about the line through the points (2, –5) and (6, –5)? Explain. 2. _______ A. The slope is undefined. B The slope is positive. C. The slope is 0. D. The slope is negative. 3. (6 pts) Solve the inequality 6 – (7 – 3x) 9(1 + 2x). Show work. 3. ______ A. x –10/21 B. x –2/3 C. x –10/21 D. x –2/3 4. (8 pts) Which of the following equations does the graph represent? Show work or explanation. 4. ______ 5. (8 pts) What is the equation of a line having slope –6 and passing through the point (–1, 8)? Show work/explanation. 5. _______ A. y = – 6x + 2 B. y = – 6x – 8 C. y = – 6x + 9 D. y = (1/6)x + 49/6 6. (12 pts) Nicole purchased a dishwasher. 4.5% sales tax and then a $36 delivery/installation charge were added. A total of $692.26 was charged to her credit card. What was the purchase price of the dishwasher (before the tax and delivery charge)? Show algebraic work/explanation. Write a sentence to answer the question. 7. (12 pts) Solve, using substitution or elimination by addition (your choice). Show work. x + 4y = −2 3x − 8y = 9 8. (16 pts) Consider the linear equation 2x + 4y = 5. (a) Write the linear equation in slope-intercept form. (b) State the value of the slope. (c) State the y-intercept for this line. (d) Find a point on this line other than the y-intercept. (There are infinitely many right answers! Just find one of them.) 9. (14 pts) A small company makes mugs. The company has daily fixed costs of $218 per day and variable costs of $1.50 per mug produced. Mugs are sold for $6.95 each. (a) What is the cost equation? (b) What is the revenue equation? (c) How many mugs must be produced and sold each day for the company to break even? Show algebraic work to find the answer. 10. (14 pts) The Washington, DC average temperature in 1960 was 56.3 degrees. In 2012, the Washington, DC average temperature was 61.5 degrees. Let y be the Washington, DC average temperature in the year x, where x = 0 represents the year 1960. (a) Find a linear equation which could be used to predict the Washington, DC average temperature y in a given year x, where x = 0 represents the year 1960. Explain/show work. (b) Use the equation from part (a) to estimate the Washington, DC average temperature for the year 2020. Show some work. (c) Interpret the slope of the equation in part (a). What is the slope and what does it represent in the context of this application involving average temperature? Bonus: From the textbook do Section 4.2, #34	{"url":"https://courseworkhero.co.uk/math-and-week-3a/","timestamp":"2024-11-03T05:59:33Z","content_type":"text/html","content_length":"45077","record_id":"<urn:uuid:205106e3-0532-4b18-b60c-0c29da24b1bc>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00368.warc.gz"}
Igor Carboni Oliveira Tutorial at the Workshop "Randomness, Information, and Complexity" (CIRM, 2024) Probabilistic Notions of Kolmogorov Complexity Complexity Theory Through the Lens of Kolmogorov Complexity Pseudodeterministic Constructions and rK Unprovability of strong complexity lower bounds in bounded arithmetic Learning from equivalence queries and unprovability of circuit upper bounds Extracting computational hardness from learning algorithms Quantum learning algorithms imply circuit lower bounds Kolmogorov complexity, prime numbers, and complexity lower bounds Consistency of circuit lower bounds with bounded theories Towards a theory of probabilistic data representation (in Portuguese) Talk at the Workshop "Boolean Devices". Simons Institute for the Theory of Computing (UC Berkeley), 2018. Addition is exponentially harder than counting for shallow monotone circuits	{"url":"https://sites.google.com/view/igorcarbonioliveira/videos","timestamp":"2024-11-14T11:51:25Z","content_type":"text/html","content_length":"125071","record_id":"<urn:uuid:ccbe645f-1ab9-47dc-bd8b-b946aaf519e4>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00496.warc.gz"}
Basic Electrical Engineering Notes VTU Pdf \| BEE VTU Notes - Eduhub \| SWBasic Electrical Engineering Notes VTU Pdf - (BEE) -SW Here you can download the Basic Electrical Engineering Notes VTU Pdf (BEE VTU Notes) of as per VTU Syllabus. Below we have list all the links as per the modules. Basic Electrical Engineering Notes VTU Pdf \| BEE VTU Notes of Total Modules Please find the download links of Basic Electrical Engineering PDF Notes VTU \| BEE Notes VTU are listed below: Module – 1 1a. D.C.Circuits: Ohm’s Law and Kirchhoff’s Laws, analysis of series, parallel and series parallel circuits excited by independent voltage sources. Power and Energy. Illustrative examples. 1b. Electromagnetism:Review of field around a conductor, coil, magnetic flux and flux density, magneto motive force and magnetic field intensity, reluctance and permeability,definition of magnetic circuit and basic analogy between electric and magnetic circuits. Electromagnetic induction: Definition of Electromagnetic Induction, Faradays Laws, Fleming’s right hand rule, Lenz’s Law, Statically and dynamically induced emf. Concept of selfinductance,mutual inductance and coefficient of coupling. Energy stored in magnetic field.Illustrative examples. Force on current carrying conductor placed in a magnetic field, Fleming’s left hand rule. Module – 2 2a. D.C. Machines: Working principle of D.C.Machine as a generator and a motor. Types and constructional features. Types of armature windings, Emf equation of generator, relation between induced emf and terminal voltage with an enumeration of brush contact drop and drop due to armature reaction. Illustrative examples, neglecting armature reaction. Operation of D.C. motor, back emf and its significance, torque equation. Types of D.C. motors, characteristics and applications. Necessity of a starter for D.C. motor. Illustrative examples on back emf and torque. 2b. Measuring Instruments: Construction and Principle of operation of dynamometer type wattmeter and single phase induction type energy meter. Module – 3 3a.Single-phase A.C. Circuits : Generation of sinusoidal voltage, frequency of generated voltage, definition and numerical values of average value, root mean square value, form factor and peak factor of sinusoidally varying voltage and current, phasor representation of alternating quantities. Analysis, with phasor diagrams, of R, L, C, R-L, R-C and R-L-C circuits and, parallel and series- parallel circuits. Real power, reactive power, apparent power and power factor. Illustrative examples. 3b. Domestic Wiring: Service mains, meter board and distribution board. Brief discussion on concealed conduit wiring. Two-way and three-way control. Elementary discussion on Circuit protective devices: fuse and Miniature Circuit Breaker (MCB’s). Electric shock, precautions against shock–Earthing, Earth leakage circuit breaker (ELCB) and Residual current circuit breaker (RCCB). Module – 4 4a. Three Phase Circuits : Necessity and advantages of three phase systems, generation of three phase power. Definition of Phase sequence, balanced supply and balanced load. Relationship between line and phase values of balanced star and delta connections. Power in balanced threephase circuits, measurement of power by two-wattmeter method. Determination power factor using wattmeter readings. Illustrative examples. 4b. Three Phase Synchronous Generators: Principle of operation, Types and constructional features, Advantages of rotating field type alternator, Synchronous speed, Frequency of generated voltage,Emf equation. Concept of winding factor (excluding the derivation of distribution and pitch factors). Illustrative examples on emf equation. Module – 5 5a. Single Phase Transformers: Necessity of transformer, Principle of operation and construction of single-phase transformers (core and shell types). Emf equation, losses, variation losses with respect to load, efficiency, Condition for maximum efficiency, Voltage regulation and its significance (Open Circuit and Short circuit tests, equivalent circuit and phasor diagrams are excluded). Illustrative problems on emf equation and efficiency only. 5b. Three Phase Induction Motors: Principle of operation, Concept and production of rotating magnetic field, Synchronous speed, rotor speed, Slip, Frequency of the rotor induced emf, Types and Constructional features. Slip and its significance. Applications of squirrel – cage and slip – ring motors. Necessity of a starter, starting of motor using stars-delta starter. Illustrative examples on slip calculations.	{"url":"https://smartzworld.com/notes/basic-electrical-engineering-pdf-notes-vtu/","timestamp":"2024-11-12T20:35:53Z","content_type":"text/html","content_length":"106774","record_id":"<urn:uuid:dd115306-6867-4460-b4a9-774c459f562e>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00774.warc.gz"}
Introduction to Machine Learning Introduction to Machine Learning. Instructor: Prof. Sudeshna Sarkar, Department of Computer Science and Engineering, IIT Kharagpur. This course provides a concise introduction to the fundamental concepts in machine learning and popular machine learning algorithms. We will cover the standard and most popular supervised learning algorithms including linear regression, logistic regression, decision trees, k-nearest neighbour, an introduction to Bayesian learning and the naive Bayes algorithm, support vector machines and kernels and neural networks with an introduction to Deep Learning. We will also cover the basic clustering algorithms. Feature reduction methods will also be discussed. We will introduce the basics of computational learning theory. In the course we will discuss various issues related to the application of machine learning algorithms. We will discuss hypothesis space, overfitting, bias and variance, tradeoffs between representational power and learnability, evaluation strategies and cross-validation. The course will be accompanied by hands-on problem solving with programming in Python and some tutorial sessions. (from nptel.ac.in) Introduction to Machine Learning Introduction to Machine Learning. Instructor: Prof. Sudeshna Sarkar, Department of Computer Science and Engineering, IIT Kharagpur. This course provides a concise introduction to the fundamental concepts in machine learning and popular machine learning algorithms.	{"url":"http://www.infocobuild.com/education/audio-video-courses/computer-science/introduction-to-machine-learning-iit-kharagpur.html","timestamp":"2024-11-12T16:07:35Z","content_type":"text/html","content_length":"13974","record_id":"<urn:uuid:0d0ea047-4f0b-4117-8fce-f4badfd0529b>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00141.warc.gz"}
Sonic Conductance for Everyone - Fluid Power JournalSonic Conductance for Everyone Sonic Conductance for Everyone (Or Why You Should Consider Moving Away From the Cv Flow Standard in Pneumatics) Let us start by having a closer look at Cv (ANSI T3.21.3) and the implications we encounter when using this standard to determine the size of a pneumatic valve. Coefficient of flow, or simply Cv, is a standard applied to determine the flow capacity of components using water as the test fluid. It also defines a pressure drop of 1 psi (1-2%) across the test object. To eventually get an airflow rate (Q), it is necessary to apply a conversion factor. Based on its inherent properties, it is apparent that this method will not provide an accurate result. In particular, the low-pressure drop is a problem, since pneumatic circuits frequently operate with a pressure drop across the valve seat up to 15% of the inlet pressure. I know there have been countless valves selected based on the Cv data, and they probably worked just fine. The truth is, there are critical applications out there that require a more refined selection procedure, rather than chronically over-sizing components “just to be on the safe side.” When it comes to sizing, I like to say: “As small as possible and as big as necessary!” This also makes sense from an economical and ecological point of view. Sonic Conductance as defined by ISO 6358, released in 1989, uses two key parameters: (1) sonic conductance [C], which I will refer to as C-Value, and (2) critical pressure ratio [b], which I will call b-Value. • C-Value is the maximum flow capacity of any pneumatic component with an open flow path including plumbing, directional, and flow control valves. A pneumatic component has reached its maximum flow rate when the medium passing through reaches sonic flow condition. This is also referred to choked flow. The unit used is dm^3/sbar (liters per second per bar). • b-Value is the critical pressure ratio between the output and input pressure when the flow condition changes from subsonic to sonic, or vice versa. In other words, it is exactly the point where the pneumatic component has reached its maximum flow capacity. There is no unit for b-Value, since it is a ratio. Let’s examine a typical flow curve in relation to pressure, as shown in Fig. 1. You will notice the straight horizontal line on the left side of the graph. This represents the maximum flow rate of the test component. Right when this line starts to fall, we enter the subsonic flow condition. This point is the critical pressure ratio b. As the pressure ratio increases, the flow rate decreases. When we reach a pressure ratio of 1, flow has stopped (e.g., an actuator reaching its end position). With the help of Fig. 2, which represents the simplified setup of ISO 6358, it will be easy to comprehend the concept. To begin, compressed air is supplied to a pressure regulator set to 6 bar (87 psi). The compressed air flows through the test component. On the downstream side of the test component, we monitor the output pressure. As the test begins, the flow control valve is fully open (no back pressure), so we achieve maximum flow. The flow rate is measured by the flow meter, shown in our example as 100 l/min. Next we start to close our flow control valve. We will continue until we observe the flow rate dropping. In Fig. 3, with an 80% open flow control valve, we can now see that P2 has increased to 1.5 bar and the flow has just started to decline to 99.9 l/min. This is where we get the b-Value. Simply divide P2[abs] by P1[abs] [(1.5+1)/(6+1) = 0.36]. That means this test component has a rated b-Value of 0.36. Any pressure ratio below 0.36 indicates a sonic flow condition, and any ratio higher than 0.36 implies a subsonic flow condition. In order to plot the graph, we continue to reduce the flow by continuing to close the flow control valve. On the way to a fully closed flow control valve, we collect P1 and P2 data and calculate pressure ratios accordingly, just as we did before. When the flow control valve is fully closed, supply pressure and output pressure will be equal; hence, the pressure ratio will be 1 with no flow occurring. I had mentioned earlier that Cv is obtained at a pressure drop of 1 psi. If we assume a supply pressure P1 of 100 psi[abs] and a pressure drop of 1 psi, this would give us a P2 pressure of 99 psi [abs]. If we calculate the pressure ratio, we will get 0.99. I also mentioned that a 15% pressure drop across a valve is very typical for a standard pneumatic application. A 15% drop equals a pressure ratio of 0.85. If we look at Figs. 4 and 5, the problem suddenly becomes very clear. Fig. 4 shows three valves with the same Cv, but different b-Values and different maximum flow rates. Fig. 5 shows three valves with identical Cv, but different b-Values resulting in the different gradient of the flow curve. Since Cv is based on a single data point in the flow curve, it cannot provide reliable data for the remaining course of the curve. On the other hand, however, with the use of C- and b-Values, it is possible to obtain an accurate flow rate at any given pressure ratio, including the ratio when the flow enters the sonic or subsonic flow condition respectively. ISO 6358 accounts for the inner design and construction of the flow path of a pneumatic component, where Cv cannot. Now, how do we use those values to get the flow rate? I will show the scientific way and the easy way. 1. Scientific way: 1.1 Maximum flow rate in sonic condition (choked flow) 1.2 Subsonic flow rate (normal operation) Q = flow rate [l/min] C = C-Value [dm^3/sbar] P1 = Supply pressure [MPa] P2 = Output pressure [MPa] b = b-Value [-] t = Temperature [°C] 2. Let us quickly move on to the easy way before we get a headache from those formulas. 2.1 Most valve selections will be done for standard applications with the mentioned 15% pressure drop. See the chart in Fig. 6 to quickly get the flow rate. All you need to do is locate the C- and b-Values of the valve you think might work. Take the b-value of the valve you pre-selected to the chart and get the multiplier based on your system pressure. Multiply the number you obtained from the chart by the C-Value of the valve you pre-selected. Note: the chart is based on a 15% pressure drop and a standard air temperature of 20°C (68°F). C-Value: 0.8 b-Value: 0.35 System pressure: 0.6 MPa (87 psi) Multiplier from the chart: 251 Flow rate Q = 251 x 0.8 = 200 l/min 2.2 If you quickly want to know the maximum flow capacity (sonic flow) of a valve, multiply the C-Value by 420. Note: 420 is based on a system pressure of 0.6 MPa (87 psi) and a standard air temperature of 20°C (68°F). We have learned now that this standard has huge benefits. It allows you to determine the exact flow rate of a pneumatic component at any given differential pressure between input and output. It gives you the maximum flow capacity of the component and tells you at what pressure ratio this happens. Furthermore, with most major pneumatic equipment manufacturers providing the relevant data, it is possible to find the most suitable component for the application. So the next time you open up a valve specification data sheet, I hope this article encourages you to specifically look for sonic conductance C and critical pressure ratio b. About the Author: Pius Landolt, CFPPS, has worked in the automotive industry for over 10 years. He has worked in the automation industry since 2001 as an application engineer, sales engineer, and was a full-time pneumatics instructor for four years. He can be reached at 2 thoughts on “Sonic Conductance for Everyone” 1. Is C value is Cv ? Correct Please inform me ? 2. Then Sonic Conductance C [dm3/(s·bar)] = C *0.219512195 = Cv You see we are in America and we use Imperial units not metric units. What has worked for hundreds of years will work for this as well. We cannot buy a compressor based on Sonic Conductance, we buy them using SCFM at a certain pressure.	{"url":"https://fluidpowerjournal.com/sonic-conductance-for-everyone/","timestamp":"2024-11-08T11:19:15Z","content_type":"text/html","content_length":"95029","record_id":"<urn:uuid:8aea9fb9-ad9e-49ac-8250-1de630a589b4>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00769.warc.gz"}