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http://en.allexperts.com/q/Physics-1358/2015/6/manometer.htm | 1,493,138,832,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917120694.49/warc/CC-MAIN-20170423031200-00271-ip-10-145-167-34.ec2.internal.warc.gz | 121,666,485 | 7,090 | You are here:
Physics/manometer
Question
In simple words, can you explain to me the Working mechanism and Basic principle of a U tube Manometer?
Is a U Tube manometer used to measure fluid pressure?
Hello Rose,
The U tube Manometer measures the difference in pressure at the 2 ends of a tube bent into the letter U. It will be more clear if you bring up a web site with an illustration of a U tube Manometer. I looked at many web sites that discussed the U-tube manometer and I was surprised how many I had to look at before finding one that I thought would work well for a simple discussion. So I am not surprised that you needed my help understanding. Use this link:
http://www.phys.ubbcluj.ro/~anghels/teaching/SIS/diverse_materiale/senzori_presi
Scroll down to the section titled "U-tube manometer" with the 2 diagrams. Even this web site seems to have an error. But it is easily fixed. It seems that they got their 2 diagrams reversed in position. When it says "The diagram opposite shows a basic U-tube manometer.", they go on to say "Both ends of the tube are open". That comment refers to the 2nd diagram, which is below.
So, looking at the 2nd diagram, that is a U tube Manometer in its neutral condition. No pressure difference has been applied yet. Read the 2nd paragraph in the U tube Manometer section of the web page, remembering that it is talking about the 2nd diagram.
Now look at the 1st diagram to see how the U tube Manometer reacts when a higher pressure is applied to the left hand end of the bent tube. Note: the higher pressure is applied by connecting the top left hand end of the tube, via a hose perhaps, to an area with an unknown pressure. That is not shown to keep the diagram simple. Read the 3rd paragraph in the U tube Manometer section. As that paragraph says, the difference in the pressures at the ends of the U tube, also called Pressure differential or delta Pressure, is given by the formula
Pdif = Punk - Patm = rho*g*h
(I am spelling the Greek letter rho because of my lack of computer skills.)
Rho is the density of the liquid in the tube,
g is the acceleration due to gravity, 9.8 m/s^2,
and h, as shown in the 1st diagram, is the elevation difference between the left and right surfaces of the liquid.
Notice at the top of the 2nd page that it rearranges the formula so that if you know Patm, you can solve for the value of Punk.
I hope this helps,
Steve
Questioner's Rating
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Physics
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Steve Johnson
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Education/Credentials
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MS Electrical Engineering, North Dakota State University | 761 | 3,388 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-17 | latest | en | 0.948139 |
https://www.jiskha.com/questions/458646/change-114-8-to-a-mixed-fraction-in-simplest-form | 1,539,688,844,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510749.37/warc/CC-MAIN-20181016093012-20181016114512-00069.warc.gz | 962,298,098 | 4,715 | change 114/8 to a mixed fraction in simplest form
1. 0
1. Divide the numerator by the denominator.
114/8 = 14 1/4
posted by Ms. Sue
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More Similar Questions | 569 | 1,912 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2018-43 | latest | en | 0.886361 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/147/2/e/b/79/1/ | 1,606,252,812,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141177566.10/warc/CC-MAIN-20201124195123-20201124225123-00263.warc.gz | 746,722,135 | 83,186 | # Properties
Label 147.2.e.b.79.1 Level $147$ Weight $2$ Character 147.79 Analytic conductor $1.174$ Analytic rank $0$ Dimension $2$ CM no Inner twists $2$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$147 = 3 \cdot 7^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 147.e (of order $$3$$, degree $$2$$, not minimal)
## Newform invariants
Self dual: no Analytic conductor: $$1.17380090971$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\zeta_{6})$$ Defining polynomial: $$x^{2} - x + 1$$ Coefficient ring: $$\Z[a_1, a_2]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 21) Sato-Tate group: $\mathrm{SU}(2)[C_{3}]$
## Embedding invariants
Embedding label 79.1 Root $$0.500000 - 0.866025i$$ of defining polynomial Character $$\chi$$ $$=$$ 147.79 Dual form 147.2.e.b.67.1
## $q$-expansion
$$f(q)$$ $$=$$ $$q+(0.500000 - 0.866025i) q^{2} +(-0.500000 - 0.866025i) q^{3} +(0.500000 + 0.866025i) q^{4} +(1.00000 - 1.73205i) q^{5} -1.00000 q^{6} +3.00000 q^{8} +(-0.500000 + 0.866025i) q^{9} +O(q^{10})$$ $$q+(0.500000 - 0.866025i) q^{2} +(-0.500000 - 0.866025i) q^{3} +(0.500000 + 0.866025i) q^{4} +(1.00000 - 1.73205i) q^{5} -1.00000 q^{6} +3.00000 q^{8} +(-0.500000 + 0.866025i) q^{9} +(-1.00000 - 1.73205i) q^{10} +(-2.00000 - 3.46410i) q^{11} +(0.500000 - 0.866025i) q^{12} -2.00000 q^{13} -2.00000 q^{15} +(0.500000 - 0.866025i) q^{16} +(3.00000 + 5.19615i) q^{17} +(0.500000 + 0.866025i) q^{18} +(-2.00000 + 3.46410i) q^{19} +2.00000 q^{20} -4.00000 q^{22} +(-1.50000 - 2.59808i) q^{24} +(0.500000 + 0.866025i) q^{25} +(-1.00000 + 1.73205i) q^{26} +1.00000 q^{27} -2.00000 q^{29} +(-1.00000 + 1.73205i) q^{30} +(2.50000 + 4.33013i) q^{32} +(-2.00000 + 3.46410i) q^{33} +6.00000 q^{34} -1.00000 q^{36} +(-3.00000 + 5.19615i) q^{37} +(2.00000 + 3.46410i) q^{38} +(1.00000 + 1.73205i) q^{39} +(3.00000 - 5.19615i) q^{40} +2.00000 q^{41} -4.00000 q^{43} +(2.00000 - 3.46410i) q^{44} +(1.00000 + 1.73205i) q^{45} -1.00000 q^{48} +1.00000 q^{50} +(3.00000 - 5.19615i) q^{51} +(-1.00000 - 1.73205i) q^{52} +(-3.00000 - 5.19615i) q^{53} +(0.500000 - 0.866025i) q^{54} -8.00000 q^{55} +4.00000 q^{57} +(-1.00000 + 1.73205i) q^{58} +(-6.00000 - 10.3923i) q^{59} +(-1.00000 - 1.73205i) q^{60} +(1.00000 - 1.73205i) q^{61} +7.00000 q^{64} +(-2.00000 + 3.46410i) q^{65} +(2.00000 + 3.46410i) q^{66} +(-2.00000 - 3.46410i) q^{67} +(-3.00000 + 5.19615i) q^{68} +(-1.50000 + 2.59808i) q^{72} +(3.00000 + 5.19615i) q^{73} +(3.00000 + 5.19615i) q^{74} +(0.500000 - 0.866025i) q^{75} -4.00000 q^{76} +2.00000 q^{78} +(8.00000 - 13.8564i) q^{79} +(-1.00000 - 1.73205i) q^{80} +(-0.500000 - 0.866025i) q^{81} +(1.00000 - 1.73205i) q^{82} -12.0000 q^{83} +12.0000 q^{85} +(-2.00000 + 3.46410i) q^{86} +(1.00000 + 1.73205i) q^{87} +(-6.00000 - 10.3923i) q^{88} +(7.00000 - 12.1244i) q^{89} +2.00000 q^{90} +(4.00000 + 6.92820i) q^{95} +(2.50000 - 4.33013i) q^{96} +18.0000 q^{97} +4.00000 q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q + q^{2} - q^{3} + q^{4} + 2q^{5} - 2q^{6} + 6q^{8} - q^{9} + O(q^{10})$$ $$2q + q^{2} - q^{3} + q^{4} + 2q^{5} - 2q^{6} + 6q^{8} - q^{9} - 2q^{10} - 4q^{11} + q^{12} - 4q^{13} - 4q^{15} + q^{16} + 6q^{17} + q^{18} - 4q^{19} + 4q^{20} - 8q^{22} - 3q^{24} + q^{25} - 2q^{26} + 2q^{27} - 4q^{29} - 2q^{30} + 5q^{32} - 4q^{33} + 12q^{34} - 2q^{36} - 6q^{37} + 4q^{38} + 2q^{39} + 6q^{40} + 4q^{41} - 8q^{43} + 4q^{44} + 2q^{45} - 2q^{48} + 2q^{50} + 6q^{51} - 2q^{52} - 6q^{53} + q^{54} - 16q^{55} + 8q^{57} - 2q^{58} - 12q^{59} - 2q^{60} + 2q^{61} + 14q^{64} - 4q^{65} + 4q^{66} - 4q^{67} - 6q^{68} - 3q^{72} + 6q^{73} + 6q^{74} + q^{75} - 8q^{76} + 4q^{78} + 16q^{79} - 2q^{80} - q^{81} + 2q^{82} - 24q^{83} + 24q^{85} - 4q^{86} + 2q^{87} - 12q^{88} + 14q^{89} + 4q^{90} + 8q^{95} + 5q^{96} + 36q^{97} + 8q^{99} + O(q^{100})$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/147\mathbb{Z}\right)^\times$$.
$$n$$ $$50$$ $$52$$ $$\chi(n)$$ $$1$$ $$e\left(\frac{1}{3}\right)$$
## Coefficient data
For each $$n$$ we display the coefficients of the $$q$$-expansion $$a_n$$, the Satake parameters $$\alpha_p$$, and the Satake angles $$\theta_p = \textrm{Arg}(\alpha_p)$$.
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
$$n$$ $$a_n$$ $$a_n / n^{(k-1)/2}$$ $$\alpha_n$$ $$\theta_n$$
$$p$$ $$a_p$$ $$a_p / p^{(k-1)/2}$$ $$\alpha_p$$ $$\theta_p$$
$$2$$ 0.500000 0.866025i 0.353553 0.612372i โ0.633316 0.773893i $$-0.718307\pi$$
0.986869 + 0.161521i $$0.0516399\pi$$
$$3$$ โ0.500000 0.866025i โ0.288675 0.500000i
$$4$$ 0.500000 + 0.866025i 0.250000 + 0.433013i
$$5$$ 1.00000 1.73205i 0.447214 0.774597i โ0.550990 0.834512i $$-0.685750\pi$$
0.998203 + 0.0599153i $$0.0190830\pi$$
$$6$$ โ1.00000 โ0.408248
$$7$$ 0 0
$$8$$ 3.00000 1.06066
$$9$$ โ0.500000 + 0.866025i โ0.166667 + 0.288675i
$$10$$ โ1.00000 1.73205i โ0.316228 0.547723i
$$11$$ โ2.00000 3.46410i โ0.603023 1.04447i โ0.992361 0.123371i $$-0.960630\pi$$
0.389338 0.921095i $$-0.372704\pi$$
$$12$$ 0.500000 0.866025i 0.144338 0.250000i
$$13$$ โ2.00000 โ0.554700 โ0.277350 0.960769i $$-0.589456\pi$$
โ0.277350 + 0.960769i $$0.589456\pi$$
$$14$$ 0 0
$$15$$ โ2.00000 โ0.516398
$$16$$ 0.500000 0.866025i 0.125000 0.216506i
$$17$$ 3.00000 + 5.19615i 0.727607 + 1.26025i 0.957892 + 0.287129i $$0.0927008\pi$$
โ0.230285 + 0.973123i $$0.573966\pi$$
$$18$$ 0.500000 + 0.866025i 0.117851 + 0.204124i
$$19$$ โ2.00000 + 3.46410i โ0.458831 + 0.794719i โ0.998899 0.0469020i $$-0.985065\pi$$
0.540068 + 0.841621i $$0.318398\pi$$
$$20$$ 2.00000 0.447214
$$21$$ 0 0
$$22$$ โ4.00000 โ0.852803
$$23$$ 0 0 โ0.866025 0.500000i $$-0.833333\pi$$
0.866025 + 0.500000i $$0.166667\pi$$
$$24$$ โ1.50000 2.59808i โ0.306186 0.530330i
$$25$$ 0.500000 + 0.866025i 0.100000 + 0.173205i
$$26$$ โ1.00000 + 1.73205i โ0.196116 + 0.339683i
$$27$$ 1.00000 0.192450
$$28$$ 0 0
$$29$$ โ2.00000 โ0.371391 โ0.185695 0.982607i $$-0.559454\pi$$
โ0.185695 + 0.982607i $$0.559454\pi$$
$$30$$ โ1.00000 + 1.73205i โ0.182574 + 0.316228i
$$31$$ 0 0 0.866025 0.500000i $$-0.166667\pi$$
โ0.866025 + 0.500000i $$0.833333\pi$$
$$32$$ 2.50000 + 4.33013i 0.441942 + 0.765466i
$$33$$ โ2.00000 + 3.46410i โ0.348155 + 0.603023i
$$34$$ 6.00000 1.02899
$$35$$ 0 0
$$36$$ โ1.00000 โ0.166667
$$37$$ โ3.00000 + 5.19615i โ0.493197 + 0.854242i โ0.999969 0.00783774i $$-0.997505\pi$$
0.506772 + 0.862080i $$0.330838\pi$$
$$38$$ 2.00000 + 3.46410i 0.324443 + 0.561951i
$$39$$ 1.00000 + 1.73205i 0.160128 + 0.277350i
$$40$$ 3.00000 5.19615i 0.474342 0.821584i
$$41$$ 2.00000 0.312348 0.156174 0.987730i $$-0.450084\pi$$
0.156174 + 0.987730i $$0.450084\pi$$
$$42$$ 0 0
$$43$$ โ4.00000 โ0.609994 โ0.304997 0.952353i $$-0.598656\pi$$
โ0.304997 + 0.952353i $$0.598656\pi$$
$$44$$ 2.00000 3.46410i 0.301511 0.522233i
$$45$$ 1.00000 + 1.73205i 0.149071 + 0.258199i
$$46$$ 0 0
$$47$$ 0 0 โ0.866025 0.500000i $$-0.833333\pi$$
0.866025 + 0.500000i $$0.166667\pi$$
$$48$$ โ1.00000 โ0.144338
$$49$$ 0 0
$$50$$ 1.00000 0.141421
$$51$$ 3.00000 5.19615i 0.420084 0.727607i
$$52$$ โ1.00000 1.73205i โ0.138675 0.240192i
$$53$$ โ3.00000 5.19615i โ0.412082 0.713746i 0.583036 0.812447i $$-0.301865\pi$$
โ0.995117 + 0.0987002i $$0.968532\pi$$
$$54$$ 0.500000 0.866025i 0.0680414 0.117851i
$$55$$ โ8.00000 โ1.07872
$$56$$ 0 0
$$57$$ 4.00000 0.529813
$$58$$ โ1.00000 + 1.73205i โ0.131306 + 0.227429i
$$59$$ โ6.00000 10.3923i โ0.781133 1.35296i โ0.931282 0.364299i $$-0.881308\pi$$
0.150148 0.988663i $$-0.452025\pi$$
$$60$$ โ1.00000 1.73205i โ0.129099 0.223607i
$$61$$ 1.00000 1.73205i 0.128037 0.221766i โ0.794879 0.606768i $$-0.792466\pi$$
0.922916 + 0.385002i $$0.125799\pi$$
$$62$$ 0 0
$$63$$ 0 0
$$64$$ 7.00000 0.875000
$$65$$ โ2.00000 + 3.46410i โ0.248069 + 0.429669i
$$66$$ 2.00000 + 3.46410i 0.246183 + 0.426401i
$$67$$ โ2.00000 3.46410i โ0.244339 0.423207i 0.717607 0.696449i $$-0.245238\pi$$
โ0.961946 + 0.273241i $$0.911904\pi$$
$$68$$ โ3.00000 + 5.19615i โ0.363803 + 0.630126i
$$69$$ 0 0
$$70$$ 0 0
$$71$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$72$$ โ1.50000 + 2.59808i โ0.176777 + 0.306186i
$$73$$ 3.00000 + 5.19615i 0.351123 + 0.608164i 0.986447 0.164083i $$-0.0524664\pi$$
โ0.635323 + 0.772246i $$0.719133\pi$$
$$74$$ 3.00000 + 5.19615i 0.348743 + 0.604040i
$$75$$ 0.500000 0.866025i 0.0577350 0.100000i
$$76$$ โ4.00000 โ0.458831
$$77$$ 0 0
$$78$$ 2.00000 0.226455
$$79$$ 8.00000 13.8564i 0.900070 1.55897i 0.0726692 0.997356i $$-0.476848\pi$$
0.827401 0.561611i $$-0.189818\pi$$
$$80$$ โ1.00000 1.73205i โ0.111803 0.193649i
$$81$$ โ0.500000 0.866025i โ0.0555556 0.0962250i
$$82$$ 1.00000 1.73205i 0.110432 0.191273i
$$83$$ โ12.0000 โ1.31717 โ0.658586 0.752506i $$-0.728845\pi$$
โ0.658586 + 0.752506i $$0.728845\pi$$
$$84$$ 0 0
$$85$$ 12.0000 1.30158
$$86$$ โ2.00000 + 3.46410i โ0.215666 + 0.373544i
$$87$$ 1.00000 + 1.73205i 0.107211 + 0.185695i
$$88$$ โ6.00000 10.3923i โ0.639602 1.10782i
$$89$$ 7.00000 12.1244i 0.741999 1.28518i โ0.209585 0.977790i $$-0.567211\pi$$
0.951584 0.307389i $$-0.0994552\pi$$
$$90$$ 2.00000 0.210819
$$91$$ 0 0
$$92$$ 0 0
$$93$$ 0 0
$$94$$ 0 0
$$95$$ 4.00000 + 6.92820i 0.410391 + 0.710819i
$$96$$ 2.50000 4.33013i 0.255155 0.441942i
$$97$$ 18.0000 1.82762 0.913812 0.406138i $$-0.133125\pi$$
0.913812 + 0.406138i $$0.133125\pi$$
$$98$$ 0 0
$$99$$ 4.00000 0.402015
$$100$$ โ0.500000 + 0.866025i โ0.0500000 + 0.0866025i
$$101$$ โ7.00000 12.1244i โ0.696526 1.20642i โ0.969664 0.244443i $$-0.921395\pi$$
0.273138 0.961975i $$-0.411939\pi$$
$$102$$ โ3.00000 5.19615i โ0.297044 0.514496i
$$103$$ โ4.00000 + 6.92820i โ0.394132 + 0.682656i โ0.992990 0.118199i $$-0.962288\pi$$
0.598858 + 0.800855i $$0.295621\pi$$
$$104$$ โ6.00000 โ0.588348
$$105$$ 0 0
$$106$$ โ6.00000 โ0.582772
$$107$$ โ2.00000 + 3.46410i โ0.193347 + 0.334887i โ0.946357 0.323122i $$-0.895268\pi$$
0.753010 + 0.658009i $$0.228601\pi$$
$$108$$ 0.500000 + 0.866025i 0.0481125 + 0.0833333i
$$109$$ 9.00000 + 15.5885i 0.862044 + 1.49310i 0.869953 + 0.493135i $$0.164149\pi$$
โ0.00790932 + 0.999969i $$0.502518\pi$$
$$110$$ โ4.00000 + 6.92820i โ0.381385 + 0.660578i
$$111$$ 6.00000 0.569495
$$112$$ 0 0
$$113$$ โ14.0000 โ1.31701 โ0.658505 0.752577i $$-0.728811\pi$$
โ0.658505 + 0.752577i $$0.728811\pi$$
$$114$$ 2.00000 3.46410i 0.187317 0.324443i
$$115$$ 0 0
$$116$$ โ1.00000 1.73205i โ0.0928477 0.160817i
$$117$$ 1.00000 1.73205i 0.0924500 0.160128i
$$118$$ โ12.0000 โ1.10469
$$119$$ 0 0
$$120$$ โ6.00000 โ0.547723
$$121$$ โ2.50000 + 4.33013i โ0.227273 + 0.393648i
$$122$$ โ1.00000 1.73205i โ0.0905357 0.156813i
$$123$$ โ1.00000 1.73205i โ0.0901670 0.156174i
$$124$$ 0 0
$$125$$ 12.0000 1.07331
$$126$$ 0 0
$$127$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$128$$ โ1.50000 + 2.59808i โ0.132583 + 0.229640i
$$129$$ 2.00000 + 3.46410i 0.176090 + 0.304997i
$$130$$ 2.00000 + 3.46410i 0.175412 + 0.303822i
$$131$$ โ2.00000 + 3.46410i โ0.174741 + 0.302660i โ0.940072 0.340977i $$-0.889242\pi$$
0.765331 + 0.643637i $$0.222575\pi$$
$$132$$ โ4.00000 โ0.348155
$$133$$ 0 0
$$134$$ โ4.00000 โ0.345547
$$135$$ 1.00000 1.73205i 0.0860663 0.149071i
$$136$$ 9.00000 + 15.5885i 0.771744 + 1.33670i
$$137$$ 3.00000 + 5.19615i 0.256307 + 0.443937i 0.965250 0.261329i $$-0.0841608\pi$$
โ0.708942 + 0.705266i $$0.750827\pi$$
$$138$$ 0 0
$$139$$ 12.0000 1.01783 0.508913 0.860818i $$-0.330047\pi$$
0.508913 + 0.860818i $$0.330047\pi$$
$$140$$ 0 0
$$141$$ 0 0
$$142$$ 0 0
$$143$$ 4.00000 + 6.92820i 0.334497 + 0.579365i
$$144$$ 0.500000 + 0.866025i 0.0416667 + 0.0721688i
$$145$$ โ2.00000 + 3.46410i โ0.166091 + 0.287678i
$$146$$ 6.00000 0.496564
$$147$$ 0 0
$$148$$ โ6.00000 โ0.493197
$$149$$ โ3.00000 + 5.19615i โ0.245770 + 0.425685i โ0.962348 0.271821i $$-0.912374\pi$$
0.716578 + 0.697507i $$0.245707\pi$$
$$150$$ โ0.500000 0.866025i โ0.0408248 0.0707107i
$$151$$ โ4.00000 6.92820i โ0.325515 0.563809i 0.656101 0.754673i $$-0.272204\pi$$
โ0.981617 + 0.190864i $$0.938871\pi$$
$$152$$ โ6.00000 + 10.3923i โ0.486664 + 0.842927i
$$153$$ โ6.00000 โ0.485071
$$154$$ 0 0
$$155$$ 0 0
$$156$$ โ1.00000 + 1.73205i โ0.0800641 + 0.138675i
$$157$$ 1.00000 + 1.73205i 0.0798087 + 0.138233i 0.903167 0.429289i $$-0.141236\pi$$
โ0.823359 + 0.567521i $$0.807902\pi$$
$$158$$ โ8.00000 13.8564i โ0.636446 1.10236i
$$159$$ โ3.00000 + 5.19615i โ0.237915 + 0.412082i
$$160$$ 10.0000 0.790569
$$161$$ 0 0
$$162$$ โ1.00000 โ0.0785674
$$163$$ โ2.00000 + 3.46410i โ0.156652 + 0.271329i โ0.933659 0.358162i $$-0.883403\pi$$
0.777007 + 0.629492i $$0.216737\pi$$
$$164$$ 1.00000 + 1.73205i 0.0780869 + 0.135250i
$$165$$ 4.00000 + 6.92820i 0.311400 + 0.539360i
$$166$$ โ6.00000 + 10.3923i โ0.465690 + 0.806599i
$$167$$ โ8.00000 โ0.619059 โ0.309529 0.950890i $$-0.600171\pi$$
โ0.309529 + 0.950890i $$0.600171\pi$$
$$168$$ 0 0
$$169$$ โ9.00000 โ0.692308
$$170$$ 6.00000 10.3923i 0.460179 0.797053i
$$171$$ โ2.00000 3.46410i โ0.152944 0.264906i
$$172$$ โ2.00000 3.46410i โ0.152499 0.264135i
$$173$$ 5.00000 8.66025i 0.380143 0.658427i โ0.610939 0.791677i $$-0.709208\pi$$
0.991082 + 0.133250i $$0.0425415\pi$$
$$174$$ 2.00000 0.151620
$$175$$ 0 0
$$176$$ โ4.00000 โ0.301511
$$177$$ โ6.00000 + 10.3923i โ0.450988 + 0.781133i
$$178$$ โ7.00000 12.1244i โ0.524672 0.908759i
$$179$$ 2.00000 + 3.46410i 0.149487 + 0.258919i 0.931038 0.364922i $$-0.118904\pi$$
โ0.781551 + 0.623841i $$0.785571\pi$$
$$180$$ โ1.00000 + 1.73205i โ0.0745356 + 0.129099i
$$181$$ โ26.0000 โ1.93256 โ0.966282 0.257485i $$-0.917106\pi$$
โ0.966282 + 0.257485i $$0.917106\pi$$
$$182$$ 0 0
$$183$$ โ2.00000 โ0.147844
$$184$$ 0 0
$$185$$ 6.00000 + 10.3923i 0.441129 + 0.764057i
$$186$$ 0 0
$$187$$ 12.0000 20.7846i 0.877527 1.51992i
$$188$$ 0 0
$$189$$ 0 0
$$190$$ 8.00000 0.580381
$$191$$ 4.00000 6.92820i 0.289430 0.501307i โ0.684244 0.729253i $$-0.739868\pi$$
0.973674 + 0.227946i $$0.0732010\pi$$
$$192$$ โ3.50000 6.06218i โ0.252591 0.437500i
$$193$$ โ1.00000 1.73205i โ0.0719816 0.124676i 0.827788 0.561041i $$-0.189599\pi$$
โ0.899770 + 0.436365i $$0.856266\pi$$
$$194$$ 9.00000 15.5885i 0.646162 1.11919i
$$195$$ 4.00000 0.286446
$$196$$ 0 0
$$197$$ 22.0000 1.56744 0.783718 0.621117i $$-0.213321\pi$$
0.783718 + 0.621117i $$0.213321\pi$$
$$198$$ 2.00000 3.46410i 0.142134 0.246183i
$$199$$ โ12.0000 20.7846i โ0.850657 1.47338i โ0.880616 0.473831i $$-0.842871\pi$$
0.0299585 0.999551i $$-0.490462\pi$$
$$200$$ 1.50000 + 2.59808i 0.106066 + 0.183712i
$$201$$ โ2.00000 + 3.46410i โ0.141069 + 0.244339i
$$202$$ โ14.0000 โ0.985037
$$203$$ 0 0
$$204$$ 6.00000 0.420084
$$205$$ 2.00000 3.46410i 0.139686 0.241943i
$$206$$ 4.00000 + 6.92820i 0.278693 + 0.482711i
$$207$$ 0 0
$$208$$ โ1.00000 + 1.73205i โ0.0693375 + 0.120096i
$$209$$ 16.0000 1.10674
$$210$$ 0 0
$$211$$ 4.00000 0.275371 0.137686 0.990476i $$-0.456034\pi$$
0.137686 + 0.990476i $$0.456034\pi$$
$$212$$ 3.00000 5.19615i 0.206041 0.356873i
$$213$$ 0 0
$$214$$ 2.00000 + 3.46410i 0.136717 + 0.236801i
$$215$$ โ4.00000 + 6.92820i โ0.272798 + 0.472500i
$$216$$ 3.00000 0.204124
$$217$$ 0 0
$$218$$ 18.0000 1.21911
$$219$$ 3.00000 5.19615i 0.202721 0.351123i
$$220$$ โ4.00000 6.92820i โ0.269680 0.467099i
$$221$$ โ6.00000 10.3923i โ0.403604 0.699062i
$$222$$ 3.00000 5.19615i 0.201347 0.348743i
$$223$$ 16.0000 1.07144 0.535720 0.844396i $$-0.320040\pi$$
0.535720 + 0.844396i $$0.320040\pi$$
$$224$$ 0 0
$$225$$ โ1.00000 โ0.0666667
$$226$$ โ7.00000 + 12.1244i โ0.465633 + 0.806500i
$$227$$ 6.00000 + 10.3923i 0.398234 + 0.689761i 0.993508 0.113761i $$-0.0362899\pi$$
โ0.595274 + 0.803523i $$0.702957\pi$$
$$228$$ 2.00000 + 3.46410i 0.132453 + 0.229416i
$$229$$ 5.00000 8.66025i 0.330409 0.572286i โ0.652183 0.758062i $$-0.726147\pi$$
0.982592 + 0.185776i $$0.0594799\pi$$
$$230$$ 0 0
$$231$$ 0 0
$$232$$ โ6.00000 โ0.393919
$$233$$ 3.00000 5.19615i 0.196537 0.340411i โ0.750867 0.660454i $$-0.770364\pi$$
0.947403 + 0.320043i $$0.103697\pi$$
$$234$$ โ1.00000 1.73205i โ0.0653720 0.113228i
$$235$$ 0 0
$$236$$ 6.00000 10.3923i 0.390567 0.676481i
$$237$$ โ16.0000 โ1.03931
$$238$$ 0 0
$$239$$ 24.0000 1.55243 0.776215 0.630468i $$-0.217137\pi$$
0.776215 + 0.630468i $$0.217137\pi$$
$$240$$ โ1.00000 + 1.73205i โ0.0645497 + 0.111803i
$$241$$ โ1.00000 1.73205i โ0.0644157 0.111571i 0.832019 0.554747i $$-0.187185\pi$$
โ0.896435 + 0.443176i $$0.853852\pi$$
$$242$$ 2.50000 + 4.33013i 0.160706 + 0.278351i
$$243$$ โ0.500000 + 0.866025i โ0.0320750 + 0.0555556i
$$244$$ 2.00000 0.128037
$$245$$ 0 0
$$246$$ โ2.00000 โ0.127515
$$247$$ 4.00000 6.92820i 0.254514 0.440831i
$$248$$ 0 0
$$249$$ 6.00000 + 10.3923i 0.380235 + 0.658586i
$$250$$ 6.00000 10.3923i 0.379473 0.657267i
$$251$$ โ20.0000 โ1.26239 โ0.631194 0.775625i $$-0.717435\pi$$
โ0.631194 + 0.775625i $$0.717435\pi$$
$$252$$ 0 0
$$253$$ 0 0
$$254$$ 0 0
$$255$$ โ6.00000 10.3923i โ0.375735 0.650791i
$$256$$ 8.50000 + 14.7224i 0.531250 + 0.920152i
$$257$$ โ13.0000 + 22.5167i โ0.810918 + 1.40455i 0.101305 + 0.994855i $$0.467698\pi$$
โ0.912222 + 0.409695i $$0.865635\pi$$
$$258$$ 4.00000 0.249029
$$259$$ 0 0
$$260$$ โ4.00000 โ0.248069
$$261$$ 1.00000 1.73205i 0.0618984 0.107211i
$$262$$ 2.00000 + 3.46410i 0.123560 + 0.214013i
$$263$$ โ8.00000 13.8564i โ0.493301 0.854423i 0.506669 0.862141i $$-0.330877\pi$$
โ0.999970 + 0.00771799i $$0.997543\pi$$
$$264$$ โ6.00000 + 10.3923i โ0.369274 + 0.639602i
$$265$$ โ12.0000 โ0.737154
$$266$$ 0 0
$$267$$ โ14.0000 โ0.856786
$$268$$ 2.00000 3.46410i 0.122169 0.211604i
$$269$$ โ3.00000 5.19615i โ0.182913 0.316815i 0.759958 0.649972i $$-0.225219\pi$$
โ0.942871 + 0.333157i $$0.891886\pi$$
$$270$$ โ1.00000 1.73205i โ0.0608581 0.105409i
$$271$$ โ8.00000 + 13.8564i โ0.485965 + 0.841717i โ0.999870 0.0161307i $$-0.994865\pi$$
0.513905 + 0.857847i $$0.328199\pi$$
$$272$$ 6.00000 0.363803
$$273$$ 0 0
$$274$$ 6.00000 0.362473
$$275$$ 2.00000 3.46410i 0.120605 0.208893i
$$276$$ 0 0
$$277$$ โ11.0000 19.0526i โ0.660926 1.14476i โ0.980373 0.197153i $$-0.936830\pi$$
0.319447 0.947604i $$-0.396503\pi$$
$$278$$ 6.00000 10.3923i 0.359856 0.623289i
$$279$$ 0 0
$$280$$ 0 0
$$281$$ โ22.0000 โ1.31241 โ0.656205 0.754583i $$-0.727839\pi$$
โ0.656205 + 0.754583i $$0.727839\pi$$
$$282$$ 0 0
$$283$$ 10.0000 + 17.3205i 0.594438 + 1.02960i 0.993626 + 0.112728i $$0.0359589\pi$$
โ0.399188 + 0.916869i $$0.630708\pi$$
$$284$$ 0 0
$$285$$ 4.00000 6.92820i 0.236940 0.410391i
$$286$$ 8.00000 0.473050
$$287$$ 0 0
$$288$$ โ5.00000 โ0.294628
$$289$$ โ9.50000 + 16.4545i โ0.558824 + 0.967911i
$$290$$ 2.00000 + 3.46410i 0.117444 + 0.203419i
$$291$$ โ9.00000 15.5885i โ0.527589 0.913812i
$$292$$ โ3.00000 + 5.19615i โ0.175562 + 0.304082i
$$293$$ 14.0000 0.817889 0.408944 0.912559i $$-0.365897\pi$$
0.408944 + 0.912559i $$0.365897\pi$$
$$294$$ 0 0
$$295$$ โ24.0000 โ1.39733
$$296$$ โ9.00000 + 15.5885i โ0.523114 + 0.906061i
$$297$$ โ2.00000 3.46410i โ0.116052 0.201008i
$$298$$ 3.00000 + 5.19615i 0.173785 + 0.301005i
$$299$$ 0 0
$$300$$ 1.00000 0.0577350
$$301$$ 0 0
$$302$$ โ8.00000 โ0.460348
$$303$$ โ7.00000 + 12.1244i โ0.402139 + 0.696526i
$$304$$ 2.00000 + 3.46410i 0.114708 + 0.198680i
$$305$$ โ2.00000 3.46410i โ0.114520 0.198354i
$$306$$ โ3.00000 + 5.19615i โ0.171499 + 0.297044i
$$307$$ 4.00000 0.228292 0.114146 0.993464i $$-0.463587\pi$$
0.114146 + 0.993464i $$0.463587\pi$$
$$308$$ 0 0
$$309$$ 8.00000 0.455104
$$310$$ 0 0
$$311$$ 12.0000 + 20.7846i 0.680458 + 1.17859i 0.974841 + 0.222900i $$0.0715523\pi$$
โ0.294384 + 0.955687i $$0.595114\pi$$
$$312$$ 3.00000 + 5.19615i 0.169842 + 0.294174i
$$313$$ โ13.0000 + 22.5167i โ0.734803 + 1.27272i 0.220006 + 0.975499i $$0.429392\pi$$
โ0.954810 + 0.297218i $$0.903941\pi$$
$$314$$ 2.00000 0.112867
$$315$$ 0 0
$$316$$ 16.0000 0.900070
$$317$$ 9.00000 15.5885i 0.505490 0.875535i โ0.494489 0.869184i $$-0.664645\pi$$
0.999980 0.00635137i $$-0.00202172\pi$$
$$318$$ 3.00000 + 5.19615i 0.168232 + 0.291386i
$$319$$ 4.00000 + 6.92820i 0.223957 + 0.387905i
$$320$$ 7.00000 12.1244i 0.391312 0.677772i
$$321$$ 4.00000 0.223258
$$322$$ 0 0
$$323$$ โ24.0000 โ1.33540
$$324$$ 0.500000 0.866025i 0.0277778 0.0481125i
$$325$$ โ1.00000 1.73205i โ0.0554700 0.0960769i
$$326$$ 2.00000 + 3.46410i 0.110770 + 0.191859i
$$327$$ 9.00000 15.5885i 0.497701 0.862044i
$$328$$ 6.00000 0.331295
$$329$$ 0 0
$$330$$ 8.00000 0.440386
$$331$$ 2.00000 3.46410i 0.109930 0.190404i โ0.805812 0.592172i $$-0.798271\pi$$
0.915742 + 0.401768i $$0.131604\pi$$
$$332$$ โ6.00000 10.3923i โ0.329293 0.570352i
$$333$$ โ3.00000 5.19615i โ0.164399 0.284747i
$$334$$ โ4.00000 + 6.92820i โ0.218870 + 0.379094i
$$335$$ โ8.00000 โ0.437087
$$336$$ 0 0
$$337$$ โ14.0000 โ0.762629 โ0.381314 0.924445i $$-0.624528\pi$$
โ0.381314 + 0.924445i $$0.624528\pi$$
$$338$$ โ4.50000 + 7.79423i โ0.244768 + 0.423950i
$$339$$ 7.00000 + 12.1244i 0.380188 + 0.658505i
$$340$$ 6.00000 + 10.3923i 0.325396 + 0.563602i
$$341$$ 0 0
$$342$$ โ4.00000 โ0.216295
$$343$$ 0 0
$$344$$ โ12.0000 โ0.646997
$$345$$ 0 0
$$346$$ โ5.00000 8.66025i โ0.268802 0.465578i
$$347$$ 14.0000 + 24.2487i 0.751559 + 1.30174i 0.947067 + 0.321037i $$0.104031\pi$$
โ0.195507 + 0.980702i $$0.562635\pi$$
$$348$$ โ1.00000 + 1.73205i โ0.0536056 + 0.0928477i
$$349$$ โ2.00000 โ0.107058 โ0.0535288 0.998566i $$-0.517047\pi$$
โ0.0535288 + 0.998566i $$0.517047\pi$$
$$350$$ 0 0
$$351$$ โ2.00000 โ0.106752
$$352$$ 10.0000 17.3205i 0.533002 0.923186i
$$353$$ โ5.00000 8.66025i โ0.266123 0.460939i 0.701734 0.712439i $$-0.252409\pi$$
โ0.967857 + 0.251500i $$0.919076\pi$$
$$354$$ 6.00000 + 10.3923i 0.318896 + 0.552345i
$$355$$ 0 0
$$356$$ 14.0000 0.741999
$$357$$ 0 0
$$358$$ 4.00000 0.211407
$$359$$ โ16.0000 + 27.7128i โ0.844448 + 1.46263i 0.0416523 + 0.999132i $$0.486738\pi$$
โ0.886100 + 0.463494i $$0.846596\pi$$
$$360$$ 3.00000 + 5.19615i 0.158114 + 0.273861i
$$361$$ 1.50000 + 2.59808i 0.0789474 + 0.136741i
$$362$$ โ13.0000 + 22.5167i โ0.683265 + 1.18345i
$$363$$ 5.00000 0.262432
$$364$$ 0 0
$$365$$ 12.0000 0.628109
$$366$$ โ1.00000 + 1.73205i โ0.0522708 + 0.0905357i
$$367$$ 0 0 0.866025 0.500000i $$-0.166667\pi$$
โ0.866025 + 0.500000i $$0.833333\pi$$
$$368$$ 0 0
$$369$$ โ1.00000 + 1.73205i โ0.0520579 + 0.0901670i
$$370$$ 12.0000 0.623850
$$371$$ 0 0
$$372$$ 0 0
$$373$$ 5.00000 8.66025i 0.258890 0.448411i โ0.707055 0.707159i $$-0.749977\pi$$
0.965945 + 0.258748i $$0.0833099\pi$$
$$374$$ โ12.0000 20.7846i โ0.620505 1.07475i
$$375$$ โ6.00000 10.3923i โ0.309839 0.536656i
$$376$$ 0 0
$$377$$ 4.00000 0.206010
$$378$$ 0 0
$$379$$ 12.0000 0.616399 0.308199 0.951322i $$-0.400274\pi$$
0.308199 + 0.951322i $$0.400274\pi$$
$$380$$ โ4.00000 + 6.92820i โ0.205196 + 0.355409i
$$381$$ 0 0
$$382$$ โ4.00000 6.92820i โ0.204658 0.354478i
$$383$$ 0 0 โ0.866025 0.500000i $$-0.833333\pi$$
0.866025 + 0.500000i $$0.166667\pi$$
$$384$$ 3.00000 0.153093
$$385$$ 0 0
$$386$$ โ2.00000 โ0.101797
$$387$$ 2.00000 3.46410i 0.101666 0.176090i
$$388$$ 9.00000 + 15.5885i 0.456906 + 0.791384i
$$389$$ โ3.00000 5.19615i โ0.152106 0.263455i 0.779895 0.625910i $$-0.215272\pi$$
โ0.932002 + 0.362454i $$0.881939\pi$$
$$390$$ 2.00000 3.46410i 0.101274 0.175412i
$$391$$ 0 0
$$392$$ 0 0
$$393$$ 4.00000 0.201773
$$394$$ 11.0000 19.0526i 0.554172 0.959854i
$$395$$ โ16.0000 27.7128i โ0.805047 1.39438i
$$396$$ 2.00000 + 3.46410i 0.100504 + 0.174078i
$$397$$ 9.00000 15.5885i 0.451697 0.782362i โ0.546795 0.837267i $$-0.684152\pi$$
0.998492 + 0.0549046i $$0.0174855\pi$$
$$398$$ โ24.0000 โ1.20301
$$399$$ 0 0
$$400$$ 1.00000 0.0500000
$$401$$ 15.0000 25.9808i 0.749064 1.29742i โ0.199207 0.979957i $$-0.563837\pi$$
0.948272 0.317460i $$-0.102830\pi$$
$$402$$ 2.00000 + 3.46410i 0.0997509 + 0.172774i
$$403$$ 0 0
$$404$$ 7.00000 12.1244i 0.348263 0.603209i
$$405$$ โ2.00000 โ0.0993808
$$406$$ 0 0
$$407$$ 24.0000 1.18964
$$408$$ 9.00000 15.5885i 0.445566 0.771744i
$$409$$ 11.0000 + 19.0526i 0.543915 + 0.942088i 0.998674 + 0.0514740i $$0.0163919\pi$$
โ0.454759 + 0.890614i $$0.650275\pi$$
$$410$$ โ2.00000 3.46410i โ0.0987730 0.171080i
$$411$$ 3.00000 5.19615i 0.147979 0.256307i
$$412$$ โ8.00000 โ0.394132
$$413$$ 0 0
$$414$$ 0 0
$$415$$ โ12.0000 + 20.7846i โ0.589057 + 1.02028i
$$416$$ โ5.00000 8.66025i โ0.245145 0.424604i
$$417$$ โ6.00000 10.3923i โ0.293821 0.508913i
$$418$$ 8.00000 13.8564i 0.391293 0.677739i
$$419$$ โ12.0000 โ0.586238 โ0.293119 0.956076i $$-0.594693\pi$$
โ0.293119 + 0.956076i $$0.594693\pi$$
$$420$$ 0 0
$$421$$ 38.0000 1.85201 0.926003 0.377515i $$-0.123221\pi$$
0.926003 + 0.377515i $$0.123221\pi$$
$$422$$ 2.00000 3.46410i 0.0973585 0.168630i
$$423$$ 0 0
$$424$$ โ9.00000 15.5885i โ0.437079 0.757042i
$$425$$ โ3.00000 + 5.19615i โ0.145521 + 0.252050i
$$426$$ 0 0
$$427$$ 0 0
$$428$$ โ4.00000 โ0.193347
$$429$$ 4.00000 6.92820i 0.193122 0.334497i
$$430$$ 4.00000 + 6.92820i 0.192897 + 0.334108i
$$431$$ 12.0000 + 20.7846i 0.578020 + 1.00116i 0.995706 + 0.0925683i $$0.0295076\pi$$
โ0.417687 + 0.908591i $$0.637159\pi$$
$$432$$ 0.500000 0.866025i 0.0240563 0.0416667i
$$433$$ โ14.0000 โ0.672797 โ0.336399 0.941720i $$-0.609209\pi$$
โ0.336399 + 0.941720i $$0.609209\pi$$
$$434$$ 0 0
$$435$$ 4.00000 0.191785
$$436$$ โ9.00000 + 15.5885i โ0.431022 + 0.746552i
$$437$$ 0 0
$$438$$ โ3.00000 5.19615i โ0.143346 0.248282i
$$439$$ 12.0000 20.7846i 0.572729 0.991995i โ0.423556 0.905870i $$-0.639218\pi$$
0.996284 0.0861252i $$-0.0274485\pi$$
$$440$$ โ24.0000 โ1.14416
$$441$$ 0 0
$$442$$ โ12.0000 โ0.570782
$$443$$ โ18.0000 + 31.1769i โ0.855206 + 1.48126i 0.0212481 + 0.999774i $$0.493236\pi$$
โ0.876454 + 0.481486i $$0.840097\pi$$
$$444$$ 3.00000 + 5.19615i 0.142374 + 0.246598i
$$445$$ โ14.0000 24.2487i โ0.663664 1.14950i
$$446$$ 8.00000 13.8564i 0.378811 0.656120i
$$447$$ 6.00000 0.283790
$$448$$ 0 0
$$449$$ โ30.0000 โ1.41579 โ0.707894 0.706319i $$-0.750354\pi$$
โ0.707894 + 0.706319i $$0.750354\pi$$
$$450$$ โ0.500000 + 0.866025i โ0.0235702 + 0.0408248i
$$451$$ โ4.00000 6.92820i โ0.188353 0.326236i
$$452$$ โ7.00000 12.1244i โ0.329252 0.570282i
$$453$$ โ4.00000 + 6.92820i โ0.187936 + 0.325515i
$$454$$ 12.0000 0.563188
$$455$$ 0 0
$$456$$ 12.0000 0.561951
$$457$$ โ5.00000 + 8.66025i โ0.233890 + 0.405110i โ0.958950 0.283577i $$-0.908479\pi$$
0.725059 + 0.688686i $$0.241812\pi$$
$$458$$ โ5.00000 8.66025i โ0.233635 0.404667i
$$459$$ 3.00000 + 5.19615i 0.140028 + 0.242536i
$$460$$ 0 0
$$461$$ โ10.0000 โ0.465746 โ0.232873 0.972507i $$-0.574813\pi$$
โ0.232873 + 0.972507i $$0.574813\pi$$
$$462$$ 0 0
$$463$$ 16.0000 0.743583 0.371792 0.928316i $$-0.378744\pi$$
0.371792 + 0.928316i $$0.378744\pi$$
$$464$$ โ1.00000 + 1.73205i โ0.0464238 + 0.0804084i
$$465$$ 0 0
$$466$$ โ3.00000 5.19615i โ0.138972 0.240707i
$$467$$ โ18.0000 + 31.1769i โ0.832941 + 1.44270i 0.0627555 + 0.998029i $$0.480011\pi$$
โ0.895696 + 0.444667i $$0.853322\pi$$
$$468$$ 2.00000 0.0924500
$$469$$ 0 0
$$470$$ 0 0
$$471$$ 1.00000 1.73205i 0.0460776 0.0798087i
$$472$$ โ18.0000 31.1769i โ0.828517 1.43503i
$$473$$ 8.00000 + 13.8564i 0.367840 + 0.637118i
$$474$$ โ8.00000 + 13.8564i โ0.367452 + 0.636446i
$$475$$ โ4.00000 โ0.183533
$$476$$ 0 0
$$477$$ 6.00000 0.274721
$$478$$ 12.0000 20.7846i 0.548867 0.950666i
$$479$$ 8.00000 + 13.8564i 0.365529 + 0.633115i 0.988861 0.148842i $$-0.0475547\pi$$
โ0.623332 + 0.781958i $$0.714221\pi$$
$$480$$ โ5.00000 8.66025i โ0.228218 0.395285i
$$481$$ 6.00000 10.3923i 0.273576 0.473848i
$$482$$ โ2.00000 โ0.0910975
$$483$$ 0 0
$$484$$ โ5.00000 โ0.227273
$$485$$ 18.0000 31.1769i 0.817338 1.41567i
$$486$$ 0.500000 + 0.866025i 0.0226805 + 0.0392837i
$$487$$ 4.00000 + 6.92820i 0.181257 + 0.313947i 0.942309 0.334744i $$-0.108650\pi$$
โ0.761052 + 0.648691i $$0.775317\pi$$
$$488$$ 3.00000 5.19615i 0.135804 0.235219i
$$489$$ 4.00000 0.180886
$$490$$ 0 0
$$491$$ 20.0000 0.902587 0.451294 0.892375i $$-0.350963\pi$$
0.451294 + 0.892375i $$0.350963\pi$$
$$492$$ 1.00000 1.73205i 0.0450835 0.0780869i
$$493$$ โ6.00000 10.3923i โ0.270226 0.468046i
$$494$$ โ4.00000 6.92820i โ0.179969 0.311715i
$$495$$ 4.00000 6.92820i 0.179787 0.311400i
$$496$$ 0 0
$$497$$ 0 0
$$498$$ 12.0000 0.537733
$$499$$ โ2.00000 + 3.46410i โ0.0895323 + 0.155074i โ0.907314 0.420455i $$-0.861871\pi$$
0.817781 + 0.575529i $$0.195204\pi$$
$$500$$ 6.00000 + 10.3923i 0.268328 + 0.464758i
$$501$$ 4.00000 + 6.92820i 0.178707 + 0.309529i
$$502$$ โ10.0000 + 17.3205i โ0.446322 + 0.773052i
$$503$$ 24.0000 1.07011 0.535054 0.844818i $$-0.320291\pi$$
0.535054 + 0.844818i $$0.320291\pi$$
$$504$$ 0 0
$$505$$ โ28.0000 โ1.24598
$$506$$ 0 0
$$507$$ 4.50000 + 7.79423i 0.199852 + 0.346154i
$$508$$ 0 0
$$509$$ 5.00000 8.66025i 0.221621 0.383859i โ0.733679 0.679496i $$-0.762199\pi$$
0.955300 + 0.295637i $$0.0955319\pi$$
$$510$$ โ12.0000 โ0.531369
$$511$$ 0 0
$$512$$ 11.0000 0.486136
$$513$$ โ2.00000 + 3.46410i โ0.0883022 + 0.152944i
$$514$$ 13.0000 + 22.5167i 0.573405 + 0.993167i
$$515$$ 8.00000 + 13.8564i 0.352522 + 0.610586i
$$516$$ โ2.00000 + 3.46410i โ0.0880451 + 0.152499i
$$517$$ 0 0
$$518$$ 0 0
$$519$$ โ10.0000 โ0.438951
$$520$$ โ6.00000 + 10.3923i โ0.263117 + 0.455733i
$$521$$ โ9.00000 15.5885i โ0.394297 0.682943i 0.598714 0.800963i $$-0.295679\pi$$
โ0.993011 + 0.118020i $$0.962345\pi$$
$$522$$ โ1.00000 1.73205i โ0.0437688 0.0758098i
$$523$$ 10.0000 17.3205i 0.437269 0.757373i โ0.560208 0.828352i $$-0.689279\pi$$
0.997478 + 0.0709788i $$0.0226123\pi$$
$$524$$ โ4.00000 โ0.174741
$$525$$ 0 0
$$526$$ โ16.0000 โ0.697633
$$527$$ 0 0
$$528$$ 2.00000 + 3.46410i 0.0870388 + 0.150756i
$$529$$ 11.5000 + 19.9186i 0.500000 + 0.866025i
$$530$$ โ6.00000 + 10.3923i โ0.260623 + 0.451413i
$$531$$ 12.0000 0.520756
$$532$$ 0 0
$$533$$ โ4.00000 โ0.173259
$$534$$ โ7.00000 + 12.1244i โ0.302920 + 0.524672i
$$535$$ 4.00000 + 6.92820i 0.172935 + 0.299532i
$$536$$ โ6.00000 10.3923i โ0.259161 0.448879i
$$537$$ 2.00000 3.46410i 0.0863064 0.149487i
$$538$$ โ6.00000 โ0.258678
$$539$$ 0 0
$$540$$ 2.00000 0.0860663
$$541$$ 17.0000 29.4449i 0.730887 1.26593i โ0.225617 0.974216i $$-0.572440\pi$$
0.956504 0.291718i $$-0.0942267\pi$$
$$542$$ 8.00000 + 13.8564i 0.343629 + 0.595184i
$$543$$ 13.0000 + 22.5167i 0.557883 + 0.966282i
$$544$$ โ15.0000 + 25.9808i โ0.643120 + 1.11392i
$$545$$ 36.0000 1.54207
$$546$$ 0 0
$$547$$ 4.00000 0.171028 0.0855138 0.996337i $$-0.472747\pi$$
0.0855138 + 0.996337i $$0.472747\pi$$
$$548$$ โ3.00000 + 5.19615i โ0.128154 + 0.221969i
$$549$$ 1.00000 + 1.73205i 0.0426790 + 0.0739221i
$$550$$ โ2.00000 3.46410i โ0.0852803 0.147710i
$$551$$ 4.00000 6.92820i 0.170406 0.295151i
$$552$$ 0 0
$$553$$ 0 0
$$554$$ โ22.0000 โ0.934690
$$555$$ 6.00000 10.3923i 0.254686 0.441129i
$$556$$ 6.00000 + 10.3923i 0.254457 + 0.440732i
$$557$$ 1.00000 + 1.73205i 0.0423714 + 0.0733893i 0.886433 0.462856i $$-0.153175\pi$$
โ0.844062 + 0.536246i $$0.819842\pi$$
$$558$$ 0 0
$$559$$ 8.00000 0.338364
$$560$$ 0 0
$$561$$ โ24.0000 โ1.01328
$$562$$ โ11.0000 + 19.0526i โ0.464007 + 0.803684i
$$563$$ โ2.00000 3.46410i โ0.0842900 0.145994i 0.820798 0.571218i $$-0.193529\pi$$
โ0.905088 + 0.425223i $$0.860196\pi$$
$$564$$ 0 0
$$565$$ โ14.0000 + 24.2487i โ0.588984 + 1.02015i
$$566$$ 20.0000 0.840663
$$567$$ 0 0
$$568$$ 0 0
$$569$$ โ5.00000 + 8.66025i โ0.209611 + 0.363057i โ0.951592 0.307364i $$-0.900553\pi$$
0.741981 + 0.670421i $$0.233886\pi$$
$$570$$ โ4.00000 6.92820i โ0.167542 0.290191i
$$571$$ 2.00000 + 3.46410i 0.0836974 + 0.144968i 0.904835 0.425762i $$-0.139994\pi$$
โ0.821138 + 0.570730i $$0.806660\pi$$
$$572$$ โ4.00000 + 6.92820i โ0.167248 + 0.289683i
$$573$$ โ8.00000 โ0.334205
$$574$$ 0 0
$$575$$ 0 0
$$576$$ โ3.50000 + 6.06218i โ0.145833 + 0.252591i
$$577$$ โ17.0000 29.4449i โ0.707719 1.22581i โ0.965701 0.259656i $$-0.916391\pi$$
0.257982 0.966150i $$-0.416942\pi$$
$$578$$ 9.50000 + 16.4545i 0.395148 + 0.684416i
$$579$$ โ1.00000 + 1.73205i โ0.0415586 + 0.0719816i
$$580$$ โ4.00000 โ0.166091
$$581$$ 0 0
$$582$$ โ18.0000 โ0.746124
$$583$$ โ12.0000 + 20.7846i โ0.496989 + 0.860811i
$$584$$ 9.00000 + 15.5885i 0.372423 + 0.645055i
$$585$$ โ2.00000 3.46410i โ0.0826898 0.143223i
$$586$$ 7.00000 12.1244i 0.289167 0.500853i
$$587$$ 28.0000 1.15568 0.577842 0.816149i $$-0.303895\pi$$
0.577842 + 0.816149i $$0.303895\pi$$
$$588$$ 0 0
$$589$$ 0 0
$$590$$ โ12.0000 + 20.7846i โ0.494032 + 0.855689i
$$591$$ โ11.0000 19.0526i โ0.452480 0.783718i
$$592$$ 3.00000 + 5.19615i 0.123299 + 0.213561i
$$593$$ 3.00000 5.19615i 0.123195 0.213380i โ0.797831 0.602881i $$-0.794019\pi$$
0.921026 + 0.389501i $$0.127353\pi$$
$$594$$ โ4.00000 โ0.164122
$$595$$ 0 0
$$596$$ โ6.00000 โ0.245770
$$597$$ โ12.0000 + 20.7846i โ0.491127 + 0.850657i
$$598$$ 0 0
$$599$$ โ24.0000 41.5692i โ0.980613 1.69847i โ0.660006 0.751260i $$-0.729446\pi$$
โ0.320607 0.947212i $$-0.603887\pi$$
$$600$$ 1.50000 2.59808i 0.0612372 0.106066i
$$601$$ โ6.00000 โ0.244745 โ0.122373 0.992484i $$-0.539050\pi$$
โ0.122373 + 0.992484i $$0.539050\pi$$
$$602$$ 0 0
$$603$$ 4.00000 0.162893
$$604$$ 4.00000 6.92820i 0.162758 0.281905i
$$605$$ 5.00000 + 8.66025i 0.203279 + 0.352089i
$$606$$ 7.00000 + 12.1244i 0.284356 + 0.492518i
$$607$$ 8.00000 13.8564i 0.324710 0.562414i โ0.656744 0.754114i $$-0.728067\pi$$
0.981454 + 0.191700i $$0.0614000\pi$$
$$608$$ โ20.0000 โ0.811107
$$609$$ 0 0
$$610$$ โ4.00000 โ0.161955
$$611$$ 0 0
$$612$$ โ3.00000 5.19615i โ0.121268 0.210042i
$$613$$ 13.0000 + 22.5167i 0.525065 + 0.909439i 0.999574 + 0.0291886i $$0.00929235\pi$$
โ0.474509 + 0.880251i $$0.657374\pi$$
$$614$$ 2.00000 3.46410i 0.0807134 0.139800i
$$615$$ โ4.00000 โ0.161296
$$616$$ 0 0
$$617$$ โ6.00000 โ0.241551 โ0.120775 0.992680i $$-0.538538\pi$$
โ0.120775 + 0.992680i $$0.538538\pi$$
$$618$$ 4.00000 6.92820i 0.160904 0.278693i
$$619$$ 10.0000 + 17.3205i 0.401934 + 0.696170i 0.993959 0.109749i $$-0.0350048\pi$$
โ0.592025 + 0.805919i $$0.701671\pi$$
$$620$$ 0 0
$$621$$ 0 0
$$622$$ 24.0000 0.962312
$$623$$ 0 0
$$624$$ 2.00000 0.0800641
$$625$$ 9.50000 16.4545i 0.380000 0.658179i
$$626$$ 13.0000 + 22.5167i 0.519584 + 0.899947i
$$627$$ โ8.00000 13.8564i โ0.319489 0.553372i
$$628$$ โ1.00000 + 1.73205i โ0.0399043 + 0.0691164i
$$629$$ โ36.0000 โ1.43541
$$630$$ 0 0
$$631$$ โ40.0000 โ1.59237 โ0.796187 0.605050i $$-0.793153\pi$$
โ0.796187 + 0.605050i $$0.793153\pi$$
$$632$$ 24.0000 41.5692i 0.954669 1.65353i
$$633$$ โ2.00000 3.46410i โ0.0794929 0.137686i
$$634$$ โ9.00000 15.5885i โ0.357436 0.619097i
$$635$$ 0 0
$$636$$ โ6.00000 โ0.237915
$$637$$ 0 0
$$638$$ 8.00000 0.316723
$$639$$ 0 0
$$640$$ 3.00000 + 5.19615i 0.118585 + 0.205396i
$$641$$ โ9.00000 15.5885i โ0.355479 0.615707i 0.631721 0.775196i $$-0.282349\pi$$
โ0.987200 + 0.159489i $$0.949015\pi$$
$$642$$ 2.00000 3.46410i 0.0789337 0.136717i
$$643$$ 20.0000 0.788723 0.394362 0.918955i $$-0.370966\pi$$
0.394362 + 0.918955i $$0.370966\pi$$
$$644$$ 0 0
$$645$$ 8.00000 0.315000
$$646$$ โ12.0000 + 20.7846i โ0.472134 + 0.817760i
$$647$$ 20.0000 + 34.6410i 0.786281 + 1.36188i 0.928231 + 0.372005i $$0.121330\pi$$
โ0.141950 + 0.989874i $$0.545337\pi$$
$$648$$ โ1.50000 2.59808i โ0.0589256 0.102062i
$$649$$ โ24.0000 + 41.5692i โ0.942082 + 1.63173i
$$650$$ โ2.00000 โ0.0784465
$$651$$ 0 0
$$652$$ โ4.00000 โ0.156652
$$653$$ 9.00000 15.5885i 0.352197 0.610023i โ0.634437 0.772975i $$-0.718768\pi$$
0.986634 + 0.162951i $$0.0521013\pi$$
$$654$$ โ9.00000 15.5885i โ0.351928 0.609557i
$$655$$ 4.00000 + 6.92820i 0.156293 + 0.270707i
$$656$$ 1.00000 1.73205i 0.0390434 0.0676252i
$$657$$ โ6.00000 โ0.234082
$$658$$ 0 0
$$659$$ 12.0000 0.467454 0.233727 0.972302i $$-0.424908\pi$$
0.233727 + 0.972302i $$0.424908\pi$$
$$660$$ โ4.00000 + 6.92820i โ0.155700 + 0.269680i
$$661$$ โ11.0000 19.0526i โ0.427850 0.741059i 0.568831 0.822454i $$-0.307396\pi$$
โ0.996682 + 0.0813955i $$0.974062\pi$$
$$662$$ โ2.00000 3.46410i โ0.0777322 0.134636i
$$663$$ โ6.00000 + 10.3923i โ0.233021 + 0.403604i
$$664$$ โ36.0000 โ1.39707
$$665$$ 0 0
$$666$$ โ6.00000 โ0.232495
$$667$$ 0 0
$$668$$ โ4.00000 6.92820i โ0.154765 0.268060i
$$669$$ โ8.00000 13.8564i โ0.309298 0.535720i
$$670$$ โ4.00000 + 6.92820i โ0.154533 + 0.267660i
$$671$$ โ8.00000 โ0.308837
$$672$$ 0 0
$$673$$ 34.0000 1.31060 0.655302 0.755367i $$-0.272541\pi$$
0.655302 + 0.755367i $$0.272541\pi$$
$$674$$ โ7.00000 + 12.1244i โ0.269630 + 0.467013i
$$675$$ 0.500000 + 0.866025i 0.0192450 + 0.0333333i
$$676$$ โ4.50000 7.79423i โ0.173077 0.299778i
$$677$$ 9.00000 15.5885i 0.345898 0.599113i โ0.639618 0.768693i $$-0.720908\pi$$
0.985517 + 0.169580i $$0.0542410\pi$$
$$678$$ 14.0000 0.537667
$$679$$ 0 0
$$680$$ 36.0000 1.38054
$$681$$ 6.00000 10.3923i 0.229920 0.398234i
$$682$$ 0 0
$$683$$ 6.00000 + 10.3923i 0.229584 + 0.397650i 0.957685 0.287819i $$-0.0929302\pi$$
โ0.728101 + 0.685470i $$0.759597\pi$$
$$684$$ 2.00000 3.46410i 0.0764719 0.132453i
$$685$$ 12.0000 0.458496
$$686$$ 0 0
$$687$$ โ10.0000 โ0.381524
$$688$$ โ2.00000 + 3.46410i โ0.0762493 + 0.132068i
$$689$$ 6.00000 + 10.3923i 0.228582 + 0.395915i
$$690$$ 0 0
$$691$$ โ10.0000 + 17.3205i โ0.380418 + 0.658903i โ0.991122 0.132956i $$-0.957553\pi$$
0.610704 + 0.791859i $$0.290887\pi$$
$$692$$ 10.0000 0.380143
$$693$$ 0 0
$$694$$ 28.0000 1.06287
$$695$$ 12.0000 20.7846i 0.455186 0.788405i
$$696$$ 3.00000 + 5.19615i 0.113715 + 0.196960i
$$697$$ 6.00000 + 10.3923i 0.227266 + 0.393637i
$$698$$ โ1.00000 + 1.73205i โ0.0378506 + 0.0655591i
$$699$$ โ6.00000 โ0.226941
$$700$$ 0 0
$$701$$ 30.0000 1.13308 0.566542 0.824033i $$-0.308281\pi$$
0.566542 + 0.824033i $$0.308281\pi$$
$$702$$ โ1.00000 + 1.73205i โ0.0377426 + 0.0653720i
$$703$$ โ12.0000 20.7846i โ0.452589 0.783906i
$$704$$ โ14.0000 24.2487i โ0.527645 0.913908i
$$705$$ 0 0
$$706$$ โ10.0000 โ0.376355
$$707$$ 0 0
$$708$$ โ12.0000 โ0.450988
$$709$$ โ3.00000 + 5.19615i โ0.112667 + 0.195146i โ0.916845 0.399244i $$-0.869273\pi$$
0.804178 + 0.594389i $$0.202606\pi$$
$$710$$ 0 0
$$711$$ 8.00000 + 13.8564i 0.300023 + 0.519656i
$$712$$ 21.0000 36.3731i 0.787008 1.36314i
$$713$$ 0 0
$$714$$ 0 0
$$715$$ 16.0000 0.598366
$$716$$ โ2.00000 + 3.46410i โ0.0747435 + 0.129460i
$$717$$ โ12.0000 20.7846i โ0.448148 0.776215i
$$718$$ 16.0000 + 27.7128i 0.597115 + 1.03423i
$$719$$ 24.0000 41.5692i 0.895049 1.55027i 0.0613050 0.998119i $$-0.480474\pi$$
0.833744 0.552151i $$-0.186193\pi$$
$$720$$ 2.00000 0.0745356
$$721$$ 0 0
$$722$$ 3.00000 0.111648
$$723$$ โ1.00000 + 1.73205i โ0.0371904 + 0.0644157i
$$724$$ โ13.0000 22.5167i โ0.483141 0.836825i
$$725$$ โ1.00000 1.73205i โ0.0371391 0.0643268i
$$726$$ 2.50000 4.33013i 0.0927837 0.160706i
$$727$$ โ40.0000 โ1.48352 โ0.741759 0.670667i $$-0.766008\pi$$
โ0.741759 + 0.670667i $$0.766008\pi$$
$$728$$ 0 0
$$729$$ 1.00000 0.0370370
$$730$$ 6.00000 10.3923i 0.222070 0.384636i
$$731$$ โ12.0000 20.7846i โ0.443836 0.768747i
$$732$$ โ1.00000 1.73205i โ0.0369611 0.0640184i
$$733$$ 9.00000 15.5885i 0.332423 0.575773i โ0.650564 0.759452i $$-0.725467\pi$$
0.982986 + 0.183679i $$0.0588007\pi$$
$$734$$ 0 0
$$735$$ 0 0
$$736$$ 0 0
$$737$$ โ8.00000 + 13.8564i โ0.294684 + 0.510407i
$$738$$ 1.00000 + 1.73205i 0.0368105 + 0.0637577i
$$739$$ โ18.0000 31.1769i โ0.662141 1.14686i โ0.980052 0.198741i $$-0.936315\pi$$
0.317911 0.948120i $$-0.397019\pi$$
$$740$$ โ6.00000 + 10.3923i โ0.220564 + 0.382029i
$$741$$ โ8.00000 โ0.293887
$$742$$ 0 0
$$743$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$744$$ 0 0
$$745$$ 6.00000 + 10.3923i 0.219823 + 0.380745i
$$746$$ โ5.00000 8.66025i โ0.183063 0.317074i
$$747$$ 6.00000 10.3923i 0.219529 0.380235i
$$748$$ 24.0000 0.877527
$$749$$ 0 0
$$750$$ โ12.0000 โ0.438178
$$751$$ 16.0000 27.7128i 0.583848 1.01125i โ0.411170 0.911559i $$-0.634880\pi$$
0.995018 0.0996961i $$-0.0317870\pi$$
$$752$$ 0 0
$$753$$ 10.0000 + 17.3205i 0.364420 + 0.631194i
$$754$$ 2.00000 3.46410i 0.0728357 0.126155i
$$755$$ โ16.0000 โ0.582300
$$756$$ 0 0
$$757$$ โ10.0000 โ0.363456 โ0.181728 0.983349i $$-0.558169\pi$$
โ0.181728 + 0.983349i $$0.558169\pi$$
$$758$$ 6.00000 10.3923i 0.217930 0.377466i
$$759$$ 0 0
$$760$$ 12.0000 + 20.7846i 0.435286 + 0.753937i
$$761$$ โ9.00000 + 15.5885i โ0.326250 + 0.565081i โ0.981764 0.190101i $$-0.939118\pi$$
0.655515 + 0.755182i $$0.272452\pi$$
$$762$$ 0 0
$$763$$ 0 0
$$764$$ 8.00000 0.289430
$$765$$ โ6.00000 + 10.3923i โ0.216930 + 0.375735i
$$766$$ 0 0
$$767$$ 12.0000 + 20.7846i 0.433295 + 0.750489i
$$768$$ 8.50000 14.7224i 0.306717 0.531250i
$$769$$ 2.00000 0.0721218 0.0360609 0.999350i $$-0.488519\pi$$
0.0360609 + 0.999350i $$0.488519\pi$$
$$770$$ 0 0
$$771$$ 26.0000 0.936367
$$772$$ 1.00000 1.73205i 0.0359908 0.0623379i
$$773$$ โ7.00000 12.1244i โ0.251773 0.436083i 0.712241 0.701935i $$-0.247680\pi$$
โ0.964014 + 0.265852i $$0.914347\pi$$
$$774$$ โ2.00000 3.46410i โ0.0718885 0.124515i
$$775$$ 0 0
$$776$$ 54.0000 1.93849
$$777$$ 0 0
$$778$$ โ6.00000 โ0.215110
$$779$$ โ4.00000 + 6.92820i โ0.143315 + 0.248229i
$$780$$ 2.00000 + 3.46410i 0.0716115 + 0.124035i
$$781$$ 0 0
$$782$$ 0 0
$$783$$ โ2.00000 โ0.0714742
$$784$$ 0 0
$$785$$ 4.00000 0.142766
$$786$$ 2.00000 3.46410i 0.0713376 0.123560i
$$787$$ 22.0000 + 38.1051i 0.784215 + 1.35830i 0.929467 + 0.368906i | 22,867 | 40,298 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-50 | latest | en | 0.274732 |
http://mathoverflow.net/questions/129091/a-sampling-and-learning-question | 1,469,316,421,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823802.12/warc/CC-MAIN-20160723071023-00183-ip-10-185-27-174.ec2.internal.warc.gz | 162,348,762 | 16,267 | # A sampling and learning question
Suppose there is an oracle that returns a number $b \in \mathbb{Z}_{n}$ whenever I press the button.
We have $b = a + e$, where $a \in \mathbb{Z}_n$ is a fixed number and $e$ is sampled according to some distribution $\chi$ over $\mathbb{Z}_n$ (say, $\chi$ looks like normal distribution over $\{-(n-1)/2, ..., (n-1)/2}\}$, so the mean is $0$).
Our goal is to learn (recover) the value of $a$, given as many samples from the oracle as you wish.
I know that if the random variables are over the real $\mathbb{R}$, we can take many samples and then compute the sample mean, and then we can upper-bound the probability of failure using Chebyshev's inequality.
However, this approach doesn't seem to work since we are working in $\mathbb{Z}_{n}$, I wonder whether anyone can suggest me an approach to remove $a$.
Thank you very much! :)
-
Try the mode of the results $b$. โย Did Apr 29 '13 at 13:15
Sorry, what do you mean? Can you elaborate a bit more? Thanks. :) โย Richard Apr 29 '13 at 13:32
โย Did Apr 29 '13 at 22:09
Is $\chi$ the normal distribution or is $\chi$ just "some distribution"? โย Kelly Davis Aug 23 '13 at 22:06
The oracle produces i.i.d. samples from a fixed one of $n$ distributions $P_b$ corresponding to the $n$ possible values of $b$. The maximum likelihood estimator of $b$ is then the value that minimizes the Kullback-Leibler divergence $\text{KL}\left(P_b\big\vert\big\vert Q\right)$, where $Q$ is the empirical distribution of samples produced by the oracle. Of course, depending on the setting maximum likelihood might not be an appropriate approach.
-
As you can take โas many samples from the oracle as you wishโ, you are in a very comfortable situation. You can compute a sample mean to an arbitrary high precision. As the precision will never be perfect in practice, the last thing to do is to perform rounding to the nearest integer.
More sophisticated approaches are not needed if you can have an arbitrarily large sample.
-
If the problem is a normal-like distribution with finite variance (i.e. not the uniform distribution over $\mathbb{Z}_n$), then treated as a discrete distribution, the probability mass function will attain a unique maximum at 0. Adding $a$ to the random variable $e$ will simply shift probability mass from $j \in \mathbb{Z}_n$ to $j + a \in \mathbb{z}_n$. Thus, the new probability mass function will attain its maximum at $0 + a \mod n = a$.
Now, since you are just searching to find the maximizer of the PMF among $n$ outcomes, you can just use Hoeffding's inequality for convergence of the empirical probability of some outcome $x$ to its actual probability, combined with a union bound to get a guarantee for all $n$ outcomes. For each outcome, maintain confidence bounds on the potential region of the actual probability of the outcome. Stop once one of them emerges as the winner with your desired level of probability.
- | 740 | 2,930 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2016-30 | latest | en | 0.889466 |
https://users.obs.carnegiescience.edu/birk/COREWAR/94/HILL/fragility.red | 1,561,391,010,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-26/segments/1560627999615.68/warc/CC-MAIN-20190624150939-20190624172939-00364.warc.gz | 643,248,952 | 1,365 | ;redcode-94 ;name Fragility ;author John Metcalf ;strategy Q^2 -> stone/imp ;assert CORESIZE==8000 ; from 1997, my first attempt to create my own qscan! org qGo qf equ (qGo+1111) qd equ (227) qs equ (qd*2) qi equ (7) qGo: seq qf, qd+qf jmp qFas, }qf+2 seq qf+qs*2, qd+qf+qs*2 jmp qSet, qPnt seq qf+qs, qd+qf+qs jmp qSet, {qf+qs+2 seq qf+qs*10, qd+qf+qs*10 jmp >qSet, {qPnt seq qf+qs*4, qd+qf+qs*4 jmp >qSet, }qPnt seq qf+qs*6, qd+qf+qs*6 jmp >qSet, {qf+qs*6+2 seq qf+qs*8, qd+qf+qs*8 jmp @qSet, qPnt seq qf+qs*12, qd+qf+qs*12 djn.f @qSet, qPnt seq qf+qs*11, qd+qf+qs*11 jmp @qSet, {qPnt seq qf+qs*5, qd+qf+qs*5 jmp @qSet, }qPnt sne qf+qs*7, qd+qf+qs*7 jmp pGo, {2493 add.f *qPnt, qKil qSet:add.f @qPnt, @-1 qFas:sne qGo-1, @qKil add qDis, qKil qLoo:mov qPnt, *qKil qKil:mov qf+4, qf add qInc, qKil djn qLoo, #10 jmp pGo, {2584 qDis:dat qd, qd qInc:dat -qi, qi qPnt:dat >qTb1, >qTb2 dat 10*qs, 10*qs qTb1:dat 6*qs, 6*qs dat 4*qs, 4*qs dat 2*qs, 2*qs qTb2:dat qs, qs dat 3*qs, 3*qs for 36 dat 0, 0 rof sStep equ (-2618) iStep equ (5091) sDist equ (pGo+6294) iDist equ (sDist+4000) pGo: iBoo:mov iImp, iDist for 4 mov sStep, 1-sStep ; incremented add off, sto jmp sto, off ; incremented iPmp:spl #iImp, >-20 sub.f #-iStep-1,iJmp mov iImp, }iPmp iJmp:jmp iImp-2*(iStep+1),>iImp-3*iStep-1 iImp:mov.i #0, iStep end | 692 | 1,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-26 | latest | en | 0.078332 |
https://se.mathworks.com/matlabcentral/answers/1650405-replace-the-repeated-number-in-a-given-sequence-with-value-of-zero?s_tid=prof_contriblnk | 1,669,711,305,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710690.85/warc/CC-MAIN-20221129064123-20221129094123-00649.warc.gz | 529,104,233 | 26,261 | # replace the repeated number in a given sequence with value of zero
1 view (last 30 days)
Kundan Prasad on 15 Feb 2022
Answered: DGM on 15 Feb 2022
I want to replace the value of Z2,Z4,Z6,Z7,Z9,Z10,Z12,Z13... and so on with value of zero.
I have attached the same.
Thank you
r=22.5;
R=linspace(0,10,10);
x=repelem(R,3);
z=-(r-sqrt(r^2-R.^2));
Z=repelem(z,3);
Z(2:3:end)= Z(2:3:end)-1.04;
plot(R,z,'.');hold on
plot(x,Z,'-');
output
DGM on 15 Feb 2022
This can probably be simplified, but eh.
r=22.5;
R=linspace(0,10,10);
x=repelem(R,3);
z=-(r-sqrt(r^2-R.^2));
Z=repelem(z,3);
Z(2:3:end)= Z(2:3:end)-1.04;
plot(R,z,'.');hold on
plot(x,Z,'-');
Z0 = Z;
tol = 1E-15; % or pick some tolerance
uz = unique(Z);
for k = 1:numel(uz)
idx = find(abs(Z-uz(k))<tol);
if numel(idx)>1
Z(idx(2:end)) = 0; % get rid of the rightmost instances
%Z(idx(1:end-1)) = 0; % get rid of the leftmost instances
end
end
[Z0' Z']
ans = 30ร2
0 0 -1.0400 -1.0400 0 0 -0.0275 -0.0275 -1.0675 -1.0675 -0.0275 0 -0.1100 -0.1100 -1.1500 -1.1500 -0.1100 0 -0.2483 -0.2483
Z now only contains unique values.
If the duplicate values are created with a known period, it may suffice to directly remove them without the need for comparison.
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Start Hunting! | 522 | 1,401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2022-49 | longest | en | 0.624501 |
http://mathhelpforum.com/differential-equations/66029-differential-equations-how-form-them.html | 1,527,488,427,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794871918.99/warc/CC-MAIN-20180528044215-20180528064215-00046.warc.gz | 190,759,748 | 11,510 | # Thread: Differential equations - how to form them?
1. ## Differential equations - how to form them?
Forum,
I seek to gain an understanding of how differential equations are arrived at.
In most cases, intuition, basic knowledge of physics, or computer simulation can tell us that the unknown variable is dependent on one or more variables. How this translates into an equation is not clear to me.
I ask because I have solved hundreds of ODE/PDEs, but given a sample real-world situation, how would I even formulate the problem, before solving it, seems to be the challenge.
Best,
wirefree
2. Originally Posted by wirefree
Forum,
I seek to gain an understanding of how differential equations are arrived at.
In most cases, intuition, basic knowledge of physics, or computer simulation can tell us that the unknown variable is dependent on one or more variables. How this translates into an equation is not clear to me.
I ask because I have solved hundreds of ODE/PDEs, but given a sample real-world situation, how would I even formulate the problem, before solving it, seems to be the challenge.
Best,
wirefree
Read a book on the subject eg. Amazon.com: Modelling with Differential and Difference Equations (Australian Mathematical Society Lecture Series): Glenn Fulford, Peter Forrester, Arthur Jones: Books
3. Originally Posted by mr fantastic
Appreciate the response, mr fantastic.
I would be much obliged should you be able to provide a little foretaste of the subject. The concept accompanied by a small example would prove beneficial.
A knack for compression is the mark of brilliance.
Best regards,
wirefree
4. Originally Posted by wirefree
Appreciate the response, mr fantastic.
I would be much obliged should you be able to provide a little foretaste of the subject. The concept accompanied by a small example would prove beneficial.
A knack for compression is the mark of brilliance.
Best regards,
wirefree
A small example from fluid mechanics:
An irrotational flow is defined a flow having zero vorticity. Vorticity is defined as the curl of the velocity vector ($\displaystyle \vec{V} = (V_x,V_y,V_z)$) of a flow:
$\displaystyle \nabla \times \vec{V} = 0$
An identity from vector calculus gives:
$\displaystyle curl(grad \phi) = \nabla \times \nabla \phi = 0$
Hence, for an irrotational flow, there exists a scalar function $\displaystyle \phi$ whose gradient is equal to the velocity vector of the flow.
$\displaystyle grad \phi = \vec{V}$.
Or:
$\displaystyle (\frac{d \phi}{dx},\frac{d \phi}{dy},\frac{d \phi}{dz}) = (V_x, V_y, V_z)$.
The continuty (mass conservation) equation of an incompressible flow can be written:
$\displaystyle \nabla . \vec{V} = 0$
And hence:
$\displaystyle \nabla . (\nabla \phi) = 0$
$\displaystyle => \nabla ^2 \phi = 0$
$\displaystyle => \frac{d^2 \phi}{dx^2} + \frac{d^2 \phi}{dy^2}+\frac{d^2 \phi}{dz^2} =0$.
This is Laplace's equation, which is a linear 2nd order homogeneous differential equation. It governs ALL fluid flows which are both irrotational and incompressible.
The concept: We started with the definitions of an irrotational flow, and an incompressible flow and expressed them in terms of vector calculus using the nabla vector (which is defined as: $\displaystyle \nabla = (\frac{d}{dx},\frac{d}{dy},\frac{d}{dz})$). Combining these definitions with the identity $\displaystyle curl(grad f) = 0$, we managed to deduce a linear 2nd order homogeneous differential equation which governs the physics behind ALL flows of this type.
5. Appreciate the response, Mush. The example you illustrated was a very sophisticated one and very well formulated at that by you.
As closure, I will be persistent in reducing the thread down to the simplest of ODEs I fished in my course text, right off page 1, and request some thoughts on the nature & complexity of problem it might address and how it might have been formulated:
$\displaystyle dy/dx + 5y = e^x$ | 969 | 3,941 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2018-22 | latest | en | 0.939855 |
https://oeis.org/A338393 | 1,620,928,405,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991943.36/warc/CC-MAIN-20210513173321-20210513203321-00621.warc.gz | 458,244,709 | 3,847 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A338393 Smallest perimeter of integer-sided triangles for which there exist exactly n triangles that have an integer inradius. 0
12, 36, 60, 162, 108, 180, 228, 84, 132, 168, 210, 640, 252, 448, 504, 612, 462, 480, 396, 1050, 1008, 630, 672, 1632, 756, 792, 1380, 420, 1740, 1232, 1584, 1560, 1188, 1540, 2052, 1428, 1820, 840, 1620, 1320, 1890, 3612, 2912, 2280, 1092, 924, 2340, 2730, 3220 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Eric Weisstein's World of Mathematics, Incircle. EXAMPLE a(1) = 12 because (3,4,5) is the smallest integer-sided triangle with an integer inradius and this integer radius = 1. a(2) = 36 and the 2 corresponding triangles are (9,10,17) with r=2 and (9,12,15) with r=3. a(3) = 60 and the 3 corresponding triangles are (6,25,29) with r=2, (10,24,26) with r=4 and (15,20,25) with r=5. CROSSREFS Cf. A005044, A070201, A120062. Sequence in context: A055926 A073762 A211609 * A043140 A043920 A049598 Adjacent sequences:ย ย A338390 A338391 A338392 * A338394 A338395 A338396 KEYWORD nonn AUTHOR Bernard Schott, Oct 28 2020 EXTENSIONS More terms from Amiram Eldar, Oct 28 2020 STATUS approved
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Last modified May 13 13:53 EDT 2021. Contains 343857 sequences. (Running on oeis4.) | 580 | 1,664 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2021-21 | latest | en | 0.676431 |
http://docplayer.net/142634-Distinguishing-between-binomial-hypergeometric-and-negative-binomial-distributions.html | 1,539,939,122,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512382.62/warc/CC-MAIN-20181019082959-20181019104459-00296.warc.gz | 89,513,243 | 35,731 | # Distinguishing Between Binomial, Hypergeometric and Negative Binomial Distributions
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1 Distinguishing Between Binomial, Hypergeometric and Negative Binomial Distributions Jacqueline Wroughton Northern Kentucky University Tarah Cole Northern Kentucky University Journal of Statistics Education Volume 21, Number 1 (2013), Copyright 2013 by Jacqueline Wroughton and Tarah Cole all rights reserved. This text may be freely shared among individuals, but it may not be republished in any medium without express written consent from the authors and advance notification of the editor. Key Words: Active Learning; Binomial; Hypergeometric; Negative Binomial; Project. Abstract Recognizing the differences between three discrete distributions (Binomial, Hypergeometric and Negative Binomial) can be challenging for students. We present an activity designed to help students differentiate among these distributions. In addition, we present assessment results in the form of pre- and post-tests that were designed to assess the effectiveness of the activity. Pilot study results show promise that the activity may help students recognize the differences in these three distributions. 1. Introduction Recently, a great deal of research has focused on active learning and hands-on activities in undergraduate statistics courses (e.g., Ledolter 1995; Chance 1997; delmas, Garfield and Chance 1999; Pfaff and Weinberg 2009). Most of the activities presented have been designed for the algebra-based introductory statistics class. However, this does not mean that students in the calculus-based introductory statistics class could not also benefit from the introduction of activities designed to improve understanding of certain statistical concepts. In addition to the complexity and level of the topics in the calculus-based course, there are issues in this course which are not completely attributed to the content, but also to the make-up of the student population. Many students who take a calculus-based course are confident in their mathematical ability, unlike many in the algebra-based course. This may pose a different kind of issue as students who are confident that they know what they are doing don t show much of their work 1
3 1. To simplify the implementation of probability calculations by utilizing their probability mass functions (pmfs) and to understand the general form of a pmf. 2. To determine and utilize their specific expected value and variance. 3. To learn about the concept of modeling outcomes of a given situation through demonstration. In order to understand how students may confuse three of these distributions, we will go through each of the three distributions characteristics in more detail. 2.1 Binomial Distribution When the Binomial Distribution is introduced, it is often done so by a list of conditions that must be satisfied. These five conditions (adapted from Wackerly, Mendenhall and Scheaffer 2008) are: 1. There is a fixed number, n, of identical trials. 2. For each trial, there are only two possible outcomes (success/failure). 3. The probability of success, p, remains the same for each trial. 4. The trials are independent of each other. 5. The random variable Y = the number of successes observed for the n trials. To determine if an experiment is Binomial, one just needs to examine whether or not each of the characteristics listed above is met. If the conditions of the Binomial Distribution are satisfied, then the following pmf can be used: n y n y f ( y; n, p) p (1 p) y 0,1, 2,..., n. y In addition, the expected value and variance can be utilized: 2.2 Hypergeometric Distribution E( Y ) np Var( Y ) np(1 p). The Hypergeometric Distribution arises when sampling is performed from a finite population without replacement thus making trials dependent on each other. However, when the Hypergeometric Distribution is introduced, there is often a comparison made to the Binomial Distribution. More specifically, it is said that if n is small relative to the population size, N, then (assuming all other conditions are met) Y could be approximated by a Binomial Distribution. This case is made due to the fact that not replacing the item has a negligible effect on the conditional p. However, when this is not the case, the independence condition is no longer met and the Binomial Distribution will no longer do an efficient job at an approximation since not replacing the item will have an effect on the conditional p. Consequently, the Hypergeometric Distribution should be used instead. So, when one compares the Hypergeometric to the Binomial conditions, one will see that conditions 1, 2, 3, and 5 still hold whereas condition 4 (given in Section 2.1), independence, no longer holds. 3
4 When the Hypergeometric Distribution is of interest, the following pmf can be used: r N r y n y f ( y; N, n, r ) y [max( 0, n ( N r )), min( r, n)]. N n Where r represents the number of success items out of the N total items. In addition, the expected value and variance can be utilized: nr ( N n) nr r E ( Y ) Var ( Y ) 1. N ( N 1) N N Comparing the pmf of the Binomial Distribution to that of the Hypergeometric Distribution, one can see that they are different due to the with replacement aspect of the Binomial Distribution compared to the without replacement aspect of the Hypergeometric Distribution. In addition, the support of y looks quite different between the two. This again is due to with replacement vs. without replacement aspect between the two distributions. For example, when drawing cards from a standard 52-card deck, suppose we are interested in Y = # of clubs drawn out of n. If n = 15, the most clubs we could draw is 13. If n = 45, the fewest clubs we could draw is 6. When one compares the expected value and variance of the two distributions, they appear to be very different. However, this is not really the case. If one were to see that r/n (of the Hypergeometric Distribution) is similar to p (of the Binomial Distribution), the expected values are the same and the variances are only different by the factor of (N-n)/(N-1), where the variances are identical in n=1; the variance of the Hypergeometric is smaller for n >1. This relates back to the idea that the Hypergeometric Distribution is used when the sample size, n, is no longer small in relation to the population size, N. However, when n is small in relation to N, this factor is negligible (and thus a Binomial Distribution may be appropriate as an approximation). 2.3 Negative Binomial Distribution When the Negative Binomial Distribution is introduced, it is often compared (and contrasted) to the Binomial Distribution. It has some of the same characteristics (conditions) as the Binomial Distribution, but has two distinct differences: The value of n (the number of trials) is no longer a fixed value and the random variable Y is defined differently; here Y is the number of trials needed to obtain r successes. That is, when looking at the conditions needed for a Binomial Distribution, conditions 2, 3, and 4 still hold whereas conditions 1 and 5 (given in Section 2.1) no longer hold due to the fact that the random variable and n basically change places. The number of successes, r, now becomes fixed and the number of trials, n, becomes the random variable. When the Negative Binomial Distribution is of interest, the following pmf can be used: y 1 r y r f ( y, r, p) p (1 p) y r, r 1,.... r 1 4
5 In addition, the expected value and variance can be utilized: E Y ) r p r(1 p) Var( Y ). p ( 2 Comparing the pmf of the Negative Binomial Distribution to that of the Binomial Distribution, one can see that they look pretty similar in construction, but the placement of the y is different (and of course, the number of combinations is slightly smaller). The support of y is quite different due to how the random variable, Y, is defined. In order to achieve r successes, one must have at least r trials. The maximum number of trials needed is not known. 2.4 Activity Motivation Looking at sections as the instructor, it appears that the differences in these three distributions are easy to discern; it seems as though one should just go through these five conditions that are needed for the Binomial Distribution to be met and figure out which ones hold and which ones don t hold. However, after having taught this class for multiple semesters, it appeared that the students had a tough time determining which of these distributions a situation followed, and why it would follow that distribution. More specifically, it appeared that the students had a tendency to believe the distribution would be Binomial more than any other of these three distributions. Thus, we felt that there was a need for the students to discover what it meant for independence to be violated, as well as seeing when n is no longer a fixed value. A hands-on activity was believed to be a good method because the impact of student discovery seems to trump the best teacher explanations. 3. The Activity In order to make scenarios similar and the situation straightforward, the design of this activity only requires a standard 52-card deck. Although the set-up is very straightforward, one could adapt what the cards represent to make the activity mimic a real-life application. For example, one could design a deck of cards where each red card represented a republican vote and each black card represented a democratic vote for a given election scenario. Another example might be to have red cards represent defective items and black cards represent satisfactory items where the students could be part of the quality control component of the company. (See Appendix A for the Activity Handout) For the activity, the students encounter three distinct scenarios (one for each distribution: Negative Binomial, Binomial, Hypergeometric respectively) and for each scenario the student is told how to shuffle and replace, as well as how the random variable, Y, is defined. This process and definition is what make the distinction among the three scenarios. Below we describe each scenario in more detail. In scenario one, the student is asked to draw a single card, then replace the card, shuffle, and repeat this process until two hearts have been obtained. The random variable, Y, is defined as the number of draws to achieve two hearts. Due to the fact that the card is replaced (and cards shuffled), independence is achieved. However, since the student does not know how many times that they will have to draw, they only know when to stop by achieving the second heart, the 5
6 number of draws is the random variable, thus making scenario one the Negative Binomial Distribution. In scenario two, the student is asked to draw a single card, then replace the card, shuffle, and repeat this process five times. The random variable, Y, is defined as the number of hearts drawn in the five draws. Due to the fact that the card is replaced (and cards shuffled), independence is achieved. The student knows that they will stop after drawing a total of five cards. Thus, this scenario meets all five conditions to follow the Binomial Distribution. In scenario three, the student is asked to draw five cards from the deck all at once. The random variable, Y, is defined as the number of hearts drawn in the five cards. Due to the fact that all five cards are drawn at once, independence no longer holds. The student knows how many cards they will draw. Thus, this scenario follows the Hypergeometric Distribution. In order for the activity to achieve enough data for demonstration of other values of interest, students are asked to go through each scenario multiple times (5-10). While students wait for the rest of the class to obtain their data, they can work on calculating the expected value and standard deviation of Y for each scenario and compute a theoretical probability using the appropriate pmf. After all students have obtained their data, they are also asked to compare the theoretical probabilities that they calculated to the empirical probabilities obtained by the class. Time and applicability permitting, one may be able to incorporate some extra topics to do as a class (or by group). For example, the class may decide to graph the theoretical and empirical pmf s for the given scenarios. This would give students a nice visual demonstration of how close the observed data is to the expected. In addition, if one would have designed an unfair deck, students would get the opportunity to notice the discrepancies between the theoretical and observed values. At the end of the activity, students are given a follow-up assignment for this activity which involves an activity write-up. These write-ups are completed for each activity that is performed in the class and are similar in nature. The group is asked to include purpose, design, and analysis/results with discussion. Students are asked to complete a write-up to give them experience on how to communicate statistical results in a meaningful way as well as to give them ownership of the results. Questions used for the write-up for this activity can be found at the end of Appendix A. As students completed the activity, some did not take it all that seriously (which happens during many in-class activities), while some experienced the light bulb moment. However, having the students work together has been anecdotally observed as beneficial as the students get the chance to explain, or question, a distribution to another student in the course. We believe that there is a difference in the level of understanding when students can explain a topic to other students versus merely being able to perform it themselves. In addition, students get confirmation when they reach the same conclusion as another student. 4. Assessment Design After implementing this activity a few times, it became of interest to know the extent to which this activity was helping students distinguish among these three distributions. In order to assess 6
8 Table 1: Timeline Day Class Material Assessments Given/Collected Wednesday Lecture on Binomial and Hypergeometric Distributions Monday Wednesday Monday Lecture on Negative Binomial; Practice Problems Activity Review for Exam Homework assignment given at the end of class Checked-off homework assignment Post-test given at the beginning of class Pre-test given at the beginning of class Activity Write-up due The option to leave a question blank complicated the scoring method as a blank answer (lack of knowledge) was believed to be different than an incorrect answer (wrong knowledge). The scoring method that was utilized was adapted from the SAT scoring method (College Board 2012): Correct answer: 1 point Incorrect answer: -0.5 point Blank answer: 0 point When utilizing this method, the highest score would be 8 points and the lowest score would be -4 points. 5. Results and Discussion At the time of this activity in the class, there were 25 students who were still registered for the course, 22 of whom gave permission for their data to be used and completed the pre-test. Two of these students were not in class the day of the post-test so their results are not included. Thus, the data are reported for 20 students with results for both the pre- and post-tests (see Table 2 below). 8
9 Table 2: Individual Student Results Version PRE-TEST POST-TEST for Pre- Test correct wrong blank Score correct wrong blank Score A A A A A A A A A A B B B B B B B B B B
10 Figure 1: Pre-Test/Post-Test Change Figure 1 is a graphical representation of how students scores changed from pre-test to post-test. A black arrow indicates a student whose score increased whereas a red arrow indicates a student whose score decreased. When looking at the Table 2 and Figure 1, one can see that there were quite a few students whose pre-test score is perfect (or near perfect). That is, there were five students who got at least seven of the questions correct on the pre-test. Thus seeing that some of their post-tests scores were lower is not as shocking since they didn t have as far to go up. However, there were also many students (11) who got at least 4 of the questions incorrect on the pre-test. Overall, the majority of the incorrect answers on both versions of the tests were Binomial leading authors to believe that this is the default answer to a student in doubt. We examined the scores to determine if post-test scores were significantly higher than pre-test scores. In order to assess this, the Wilcoxon Signed-Rank Sum Test was conducted and was found to be statistically significant (p-value = ). Although the ordering of the versions was randomized between students, it was still of interest to see if there was a significant difference in the change in scores from the pre-test to the post-test depending on which version the student was given as the pre-test. This was found to not be statistically significant (p-value = ) again using the Wilcoxon Signed-Rank Sum Test. As the semester progressed I discovered that my least successful students in the class all had version B first which can partly explain the observed difference in version ordering. 10
12 Appendix A: Activity Worksheet Activity Working with Discrete Distributions Scenario 1: With your 52-card deck you will shuffle, randomly draw a card, replace it, shuffle again, randomly draw a card, replace it, We will be interested in Y = the number of draws to obtain 2 hearts. 1. Repeat this process 10 times. (Thus, you should have 10 numbers that represent the number of draws it took to get 2 hearts). Record (tally) your values on the board with the rest of the class. 2. Based on the appropriate pmf, calculate the probability of getting the 2 nd heart on the 7 th draw. 3. Based upon our sample (on the board), what proportion of the time did the 2 nd heart come on the 7 th draw? Is this close to the answer in 2? Explain! 4. For this scenario, what is the expected number of draws to get the 2 nd heart? 5. For this scenario, what is the standard deviation for the number of draws to get the 2 nd heart? Use this as well as your answer in #4 above to interpret the standard deviation. Scenario 2: With your 52-card deck you will shuffle, randomly draw a card, replace it, shuffle again, randomly draw a card, replace it We will be interested in Y = the number of hearts drawn in 5 draws. 1. Repeat this process 10 times. (Thus, you should have 10 numbers that represent the number of hearts drawn in 5 draws). Record (tally) your values on the board with the rest of the class. 2. Based on the appropriate pmf, calculate the probability of getting exactly 1 heart in 5 draws. 3. Based upon our sample (on the board), what proportion of time did we get exactly 1 heart in 5 draws? Is this close to the answer in 2? Explain! 4. For this scenario, what is the expected number of hearts in 5 draws? 5. For this scenario, what is the standard deviation of the number of hearts in 5 draws? Use this as well as your answer in #4 above to interpret the standard deviation. 12
14 Appendix B: Question Pool Note: Questions 1 8 were Version A and questions 9-16 were Version B. Each question had the following options: Binomial Hypergeometric Negative Binomial 1. At a certain manufacturing company, approximately 5% of the products are defective. We are interested in calculating the probability that the third defective is the 20 th one sampled. 2. An assembly line produces products that they put into boxes of 50. The inspector then randomly picks 3 items inside a box to test to see if they are defective. In a box containing 4 defectives, they are interested in the probability that at least one of the three items sampled is defective. 3. When rolling a pair of dice, we generally care about the sum of the two dice. We are interested in the number of rolls that we perform before we get our first sum of The ACT is a standardized test that many students take in order to enter college. It is said that 4 out of every 5 students at NKU take the ACT. We are interested in the number of students in a given class of 30 that took the ACT. 5. A husband has 7 tasks on his to-do list and a wife has 10 things on her to-do list. Five tasks are randomly picked out of these 17 tasks. We are interest in the expected number of tasks the wife will have to do. 6. A certain stoplight, when coming from the North, is green approximately 31% of the time. Over the next few days, someone comes to this light 8 times from the North. We are interested in finding the expected number of green lights the person will come to. 7. A certain radio station s phone lines are all busy approximately 98% of the time when trying to call during a contest. We are interested in finding the probability that the 5 th time that you call is the 1 st time you get through during a contest. 8. Type O blood is one of the best to be donated since it can be used for many people. Approximately 42% of people have type O blood. In a given day at a blood bank, about 120 people come in to donate. The blood bank is interested in the number of donors who are type O. 9. A police officer has found that approximately 15% of the vehicles he pulls over are from out of state. We are interested in the number of vehicles that are out of state from the next 50 vehicles that he pulls over. 14
15 10. In a capture-recapture experiment, 20 animals were captured, tagged and released. A few weeks later, a sample of 40 of these animals is captured and we are interested in the number of animals in our sample that are tagged. 11. When taking the written drivers license test, they say that about 7 out of 8 people pass the test. A test-taker is interested in the number of times they will have to take the test in order to pass. 12. There is an urn containing 15 balls, 40% of which are green. A person gets to pull three balls out at the same time. For each ball that is green, he/she wins \$200. For each ball that is not green, he/she must pay \$50. We are interested in the number of green balls pulled out of the A student is taking a true/false test that consists of 15 questions. The student has approximately a 72% chance of getting any individual question correct. We are interested in the probability that the student gets at least 9 of the 15 questions correct. 14. A company has a pool of 15 applicants (10 male, 5 female) for a particular position that has 3 current openings. They are interested in the probability that none of the positions are filled by females. 15. When rolling a pair of dice, we generally care about the sum of the two dice. We are interested in how many times we get a sum of 7 out of 30 rolls. 16. A couple likes to play darts together. However, the female is not as good at the game and only has about a 8% chance of winning any individual game. Being stubborn, the couple will play until the female wins two games. We are interested in the expected number of games the couple will need to play. Acknowledgements This material is based upon work supported by the National Science Foundation s STEM Talent Expansion Program (DUE-STEP) program under Grant No Any opinions, findings and conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of the National Science Foundation. References Aliaga, M., Cuff, C., Garfield, J., Lock, R., Utts, J. and Witmer, J. (2005), Guidelines for Assessment and Instruction in Statistics Education (GAISE): College Report. American Statistical Association. Available at: 15
16 Chance, B. L. (1997), Experiences with Authentic Assessment Techniques in an Introductory Statistics Course, Journal of Statistics Education, 5(3). Available at: College Board. (2012), How the SAT is Scored. Last accessed on March 11, 2012 from: delmas, R. C., Garfield, J., & Chance, B. (1999), A Model of Classroom Research in Action: Developing Simulation Activities to Improve Students Statistical Reasoning, Journal of Statistics Education, 7(3). Available at: Ledolter, J. (1995), Projects in Introductory Statistics Courses, The American Statistician, 49, Lunsford, M. L., Rowell, G. H., & Goodson-Espy, T. (2006), Classroom Research: Assessment of Student Understanding of Sampling Distributions of Means and the Central Limit Theorem in Post-Calculus Probability and Statistics Classes, Journal of Statistics Education, 14(3). Available at: Mills, J. D. (2002), Using Computer Simulation Methods to Teach Statistics: A Review of the Literature, Journal of Statistics Education, 10(1). Available at: Pfaff, T. J. & Weinberg, A. (2009), Do Hands-On Activities Increase Student Understanding?: A Case Study, Journal of Statistics Education, 17(3). Available at: Wackerly, D.D., Mendenhall, W., & Scheaffer, R.L. (2008), Mathematical Statistics with Applications, Belmont, CA: Thomson Brooks/Cole. Jacqueline Wroughton Nunn Drive MP 422 Highland Heights, KY Phone: (859) Tarah Cole 311 Upham Hall Oxford, Ohio Volume 22 (2013) Archive Index Data Archive Resources Editorial Board Guidelines for Authors Guidelines for Data Contributors Guidelines for Readers/Data Users Home Page Contact JSE ASA Publications 16
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### Investigating the Investigative Task: Testing for Skewness An Investigation of Different Test Statistics and their Power to Detect Skewness
Investigating the Investigative Task: Testing for Skewness An Investigation of Different Test Statistics and their Power to Detect Skewness Josh Tabor Canyon del Oro High School Journal of Statistics Education | 11,501 | 51,933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2018-43 | longest | en | 0.922784 |
https://blogs.msdn.microsoft.com/ericlippert/2003/09/15/why-does-jscript-have-rounding-errors/ | 1,561,491,598,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999946.25/warc/CC-MAIN-20190625192953-20190625214953-00177.warc.gz | 356,488,963 | 16,458 | # Why does JScript have rounding errors?
Try this in JScript:
You might expect to get 920, but in fact you get 919.9999999999999.ย What the heck is going on here?
Boy, have I ever heard this question a lot.
Well, let me answer that question with another question.ย Suppose you did a simple division, say
Would you expect an infinitely large window that said "0.33333333333..." with an infinite number of threes, or would you expect ten or so digits?ย Clearly you'd expect it to not fill your computer's entire memory with threes.ย But that means that I must ask why are you willing to accept an error of 0.00000000000333333... in the case of dividing one by three but not willing to accept a smaller error of 0.000000000001 in the case of multiplying 9.2 by 100?
The simple fact is that computer arithmetic frequently accumulates tiny errors.ย Any mathematics which results in numbers that cannot be represented by a small number of powers of two will result in errors.
Let's look at this cases a little closer.ย We're trying to multiply 9.2 by 100.0.ย 100.0 can be EXACTLY represented as a floating point number because it's an integer.ย But 9.2 can't be -- you can't represent 46 / 5 exactly in base two any more than you can represent 1 / 3 exactly in base ten.ย So when converting from the string "9.2" to the internal binary representation, a tiny error is accrued.ย However, it's not all bad -- the 64 bit binary number which represents 9.2 internally is (a) the 64 bit float closest to 9.2, and (b) the algorithm which converts back and forth between strings and binary representation will round-trip -- that binary representation will be converted back to 9.2 if you try to convert it to a string.
But now we go and throw a wrench in the works by multiplying by one hundred.ย That's going to lose the last few bits of precision because we just multiplied the accrued error by a factor of one hundred.ย The accrued error is now large enough that the string which most exactly represents the computed value is NOT "920" but rather "919.9999999999999".
The ECMAScript specification mandates that JScript display as much precision as possible when displaying floating point numbers as strings.
You may note that VBScript does not do this -- VBScript has heuristics which look for this situation and deliberately break the round-trip property in order to make this look better.ย In JScript you are guaranteed that when you convert back and forth between string and binary representations you lose no data.ย In VBScript you sometimes lose data; there are some legal floating point values in VBScript which are impossible to represent in strings accurately. In VBScript it is impossible to represent values like 919.9999999999999 precisely because they are automatically rounded!
Ultimately, the reason that this is an issue is because we as human beings see numbers like "920" as SPECIAL.ย If you multiply 3.63874692874ย by 4.2984769284 and get a result which is one-billionth of one percent off, no one cares, but when you multiply 9.2 by 100.0 and get a result which is one-billionth off, everyone yells (at me!)ย The computer doesn't know that 9.2 is more special than 3.63874692874ย ย -- it uses the same lossy algorithms for both.
All languages which use floating point arithmetic have this feature -- C++, VBScript, JScript, whatever.ย If you don't like it, either get in the habit of calling rounding functions, or don't use floating point arithmetic, use only integer arithmetic.ย (Note that VBScript supports a "currency" type which is fixed-point number not subject to rounding errors.ย It can only represent four places after the decimal point though.)
1. Abhi says:
Hi,
I used JScript in WSH for doing 108 % 100 and i got the answer as 0. i get correct answers for all the other numbers. I think this is a bug too.
2. Eric Lippert says:
Hmm, that works fine for me and I donโt recall ever fixing a bug in the mod operator. Can you send me a SMALL program that reproduces the fault along with the version numbers on JScript.DLL and CScript.exe?
Thanks,
Eric
3. Jon Kale says:
โUltimately, the reason that this is an issue is because we as human beings see numbers like "920" as SPECIAL.โ
Ermโฆ surely you mean โnumbers like "920.0" as SPECIALโ. After all, 920 differs from 920.0: oneโs in Z, while the otherโs in R, and as any fule kno, the numbers in Z have some very interesting properties which donโt generalise to R, and vice versa.
Now, this is of course a mathematically inclined pedant writing thisโฆ but even so it raises an interesting point, which is that people in the main conflate integers and rationals when they can, even when the conversion isnโt necessarily valid โ so VBS does the "least-suprise" thing whereas JS does the "right" thing.
4. Eric Lippert says:
> after all, 920 differs from 920.0:
Well, if they differ, whatโs their difference? Last I checked, 920 โ 920.0 = 0, add 920.0 to both sides and you get 920 = 920.0, QED.
As a mathematically inclined pedant with a degree in mathematics from the University of Waterloo, lemme tell ya, youโre going to have to start from the set-theoretic underpinnings of the real number system if you want to make this argument, and it is not an argument that I myself would care to advance. That 920 in Z and the 920 in R are the same number dude.
5. 2003/09/12 Whatโs Up With Hungarian Notation? 2003/09/12 Ericโs Complete Guide To BSTR Semantics 2003/09/12โฆ
6. Kaspar Karlsson says:
Maestro,
I must agree with Jon Kale.
sometimes itโs important to have an integer when it should be, and a floating point number when needed, not just because an integer is prettier.
for example, if you want to calculate prime numbers, youโll have to divide the number (x) you want to check by all integers ranging from 2 โ ceil(sqrt(x)) ย (we call that integer y), and see whatโs left (z) ย (thus executing z = x % y). if z == (int) 0, then youโll know that x isnโt prime, as it can be divided by y w times (w being an integer and positive). however, if z is not (int) 0, youโll have to check further (by increasing y). but z can be (int) 0, plus/minus this very very very small JScript rounding error margin, making it NOT (int) 0. in that case you canโt be sure wheter you have a prime number or notโฆ
of course calculating prime numbers in JScript is plain stupid when you can do that with Python several thousands times faster, but this is just an example to show that that very very very small JScript rounding error margin can make a real difference. I can think of other cases. | 1,608 | 6,571 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2019-26 | latest | en | 0.940485 |
http://wikien4.appspot.com/wiki/Divergence | 1,606,867,595,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141685797.79/warc/CC-MAIN-20201201231155-20201202021155-00534.warc.gz | 91,563,087 | 41,393 | # Divergence
The divergence of different vector fiewds. The divergence of vectors from point (x,y) eqwaws de sum of de partiaw derivative-wif-respect-to-x of de x-component and de partiaw derivative-wif-respect-to-y of de y-component at dat point: ${\dispwaystywe \nabwa \!\cdot (\madbf {V} (x,y))={\frac {\partiaw \ {\madbf {V} _{x}(x,y)}}{\partiaw {x}}}+{\frac {\partiaw \ {\madbf {V} _{y}(x,y)}}{\partiaw {y}}}}$
In vector cawcuwus, divergence is a vector operator dat operates on a vector fiewd, producing a scawar fiewd giving de qwantity of de vector fiewd's source at each point. More technicawwy, de divergence represents de vowume density of de outward fwux of a vector fiewd from an infinitesimaw vowume around a given point.
As an exampwe, consider air as it is heated or coowed. The vewocity of de air at each point defines a vector fiewd. Whiwe air is heated in a region, it expands in aww directions, and dus de vewocity fiewd points outward from dat region, uh-hah-hah-hah. The divergence of de vewocity fiewd in dat region wouwd dus have a positive vawue. Whiwe de air is coowed and dus contracting, de divergence of de vewocity has a negative vawue.
## Physicaw interpretation of divergence
In physicaw terms, de divergence of a vector fiewd is de extent to which de vector fiewd fwux behaves wike a source at a given point. It is a wocaw measure of its "outgoingness" โ de extent to which dere is more of de fiewd vectors exiting an infinitesimaw region of space dan entering it. A point at which de fwux is outgoing has positive divergence, and is often cawwed a "source" of de fiewd. A point at which de fwux is directed inward has negative divergence, and is often cawwed a "sink" of de fiewd. The greater de fwux of fiewd drough a smaww surface encwosing a given point, de greater de vawue of divergence at dat point. A point at which dere is zero fwux drough an encwosing surface has zero divergence.
The divergence of a vector fiewd is often iwwustrated using de exampwe of de vewocity fiewd of a fwuid, a wiqwid or gas. A moving gas has a vewocity, a speed and direction, at each point which can be represented by a vector, so de vewocity of de gas forms a vector fiewd. If a gas is heated, it wiww expand. This wiww cause a net motion of gas particwes outward in aww directions. Any cwosed surface in de gas wiww encwose gas which is expanding, so dere wiww be an outward fwux of gas drough de surface. So de vewocity fiewd wiww have positive divergence everywhere. Simiwarwy, if de gas is coowed, it wiww contract. There wiww be more room for gas particwes in any vowume, so de externaw pressure of de fwuid wiww cause a net fwow of gas vowume inward drough any cwosed surface. Therefore de vewocity fiewd has negative divergence everywhere. In contrast in an unheated gas wif a constant density, de gas may be moving, but de vowume rate of gas fwowing into any cwosed surface must eqwaw de vowume rate fwowing out, so de net fwux of fwuid drough any cwosed surface is zero. Thus de gas vewocity has zero divergence everywhere. A fiewd which has zero divergence everywhere is cawwed sowenoidaw.
If de fwuid is heated onwy at one point or smaww region, or a smaww tube is introduced which suppwies a source of additionaw fwuid at one point, de fwuid dere wiww expand, pushing fwuid particwes around it outward in aww directions. This wiww cause an outward vewocity fiewd droughout de fwuid, centered on de heated point. Any cwosed surface encwosing de heated point wiww have a fwux of fwuid particwes passing out of it, so dere is positive divergence at dat point. However any cwosed surface not encwosing de point wiww have a constant density of fwuid inside, so just as many fwuid particwes are entering as weaving de vowume, dus de net fwux out of de vowume is zero. Therefore de divergence at any oder point is zero.
## Definition
The divergence at a point x is de wimit of de ratio of de fwux ${\dispwaystywe \Phi }$ drough de surface Si (red arrows) to de vowume ${\dispwaystywe |V_{\text{i}}|}$ for any seqwence of cwosed regions V1, V2, V3... encwosing x dat approaches zero vowume:
${\dispwaystywe \operatorname {div} \madbf {F} =\wim _{|V_{\text{i}}|\rightarrow 0}{\Phi (S_{\text{i}}) \over |V_{\text{i}}|}}$
The divergence of a vector fiewd F(x) at a point x0 is defined as de wimit of de ratio of de surface integraw of F out of de surface of a cwosed vowume V encwosing x0 to de vowume of V, as V shrinks to zero
${\dispwaystywe \weft.\operatorname {div} \madbf {F} \right|_{\madbf {x_{0}} }=\wim _{V\rightarrow 0}{1 \over |V|}}$ ${\dispwaystywe \scriptstywe S(V)}$ ${\dispwaystywe \madbf {F} \cdot \madbf {\hat {n}} \,dS}$
where |V| is de vowume of V, S(V) is de boundary of V, and ${\dispwaystywe \madbf {\hat {n}} }$ is de outward unit normaw to dat surface. It can be shown dat de above wimit awways converges to de same vawue for any seqwence of vowumes dat contain x0 and approach zero vowume. The resuwt, div F, is a scawar function of x.
Since dis definition is coordinate-free, it shows dat de divergence is de same in any coordinate system. However it is not often used practicawwy to cawcuwate divergence; when de vector fiewd is given in a coordinate system de coordinate definitions bewow are much simpwer to use.
A vector fiewd wif zero divergence everywhere is cawwed sowenoidaw โ in which case any cwosed surface has no net fwux across it.
## Definition in coordinates
### Cartesian coordinates
In dree-dimensionaw Cartesian coordinates, de divergence of a continuouswy differentiabwe vector fiewd ${\dispwaystywe \madbf {F} =F_{x}\madbf {i} +F_{y}\madbf {j} +F_{z}\madbf {k} }$ is defined as de scawar-vawued function:
${\dispwaystywe \operatorname {div} \madbf {F} =\nabwa \cdot \madbf {F} =\weft({\frac {\partiaw }{\partiaw x}},{\frac {\partiaw }{\partiaw y}},{\frac {\partiaw }{\partiaw z}}\right)\cdot (F_{x},F_{y},F_{z})={\frac {\partiaw F_{x}}{\partiaw x}}+{\frac {\partiaw F_{y}}{\partiaw y}}+{\frac {\partiaw F_{z}}{\partiaw z}}.}$
Awdough expressed in terms of coordinates, de resuwt is invariant under rotations, as de physicaw interpretation suggests. This is because de trace of de Jacobian matrix of an N-dimensionaw vector fiewd F in N-dimensionaw space is invariant under any invertibwe winear transformation, uh-hah-hah-hah.
The common notation for de divergence โ ยท F is a convenient mnemonic, where de dot denotes an operation reminiscent of de dot product: take de components of de operator (see dew), appwy dem to de corresponding components of F, and sum de resuwts. Because appwying an operator is different from muwtipwying de components, dis is considered an abuse of notation.
### Cywindricaw coordinates
For a vector expressed in wocaw unit cywindricaw coordinates as
${\dispwaystywe \madbf {F} =\madbf {e} _{r}F_{r}+\madbf {e} _{\deta }F_{\deta }+\madbf {e} _{z}F_{z},}$
where ea is de unit vector in direction a, de divergence is[1]
${\dispwaystywe \operatorname {div} \madbf {F} =\nabwa \cdot \madbf {F} ={\frac {1}{r}}{\frac {\partiaw }{\partiaw r}}\weft(rF_{r}\right)+{\frac {1}{r}}{\frac {\partiaw F_{\deta }}{\partiaw \deta }}+{\frac {\partiaw F_{z}}{\partiaw z}}.}$
The use of wocaw coordinates is vitaw for de vawidity of de expression, uh-hah-hah-hah. If we consider x de position vector and de functions ${\dispwaystywe r(\madbf {x} )}$, ${\dispwaystywe \deta (\madbf {x} )}$, and ${\dispwaystywe z(\madbf {x} )}$, which assign de corresponding gwobaw cywindricaw coordinate to a vector, in generaw ${\dispwaystywe r(\madbf {F} (\madbf {x} ))\neq F_{r}(\madbf {x} )}$, ${\dispwaystywe \deta (\madbf {F} (\madbf {x} ))\neq F_{\deta }(\madbf {x} )}$, and ${\dispwaystywe z(\madbf {F} (\madbf {x} ))\neq F_{z}(\madbf {x} )}$. In particuwar, if we consider de identity function ${\dispwaystywe \madbf {F} (\madbf {x} )=\madbf {x} }$, we find dat:
${\dispwaystywe \deta (\madbf {F} (\madbf {x} ))=\deta \neq F_{\deta }(\madbf {x} )=0}$.
### Sphericaw coordinates
In sphericaw coordinates, wif ฮธ de angwe wif de z axis and ฯ de rotation around de z axis, and ${\dispwaystywe \madbf {F} }$ again written in wocaw unit coordinates, de divergence is[2]
${\dispwaystywe \operatorname {div} \madbf {F} =\nabwa \cdot \madbf {F} ={\frac {1}{r^{2}}}{\frac {\partiaw }{\partiaw r}}\weft(r^{2}F_{r}\right)+{\frac {1}{r\sin \deta }}{\frac {\partiaw }{\partiaw \deta }}(\sin \deta \,F_{\deta })+{\frac {1}{r\sin \deta }}{\frac {\partiaw F_{\varphi }}{\partiaw \varphi }}.}$
### Tensor fiewd
Let ${\dispwaystywe \madbf {A} }$ be continuouswy differentiabwe second-order tensor fiewd defined as fowwows:
${\dispwaystywe \madbf {A} ={\begin{bmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{bmatrix}}}$
de divergence in cartesian coordinate system is a first-order tensor fiewd[3] and can be defined in two ways:[4]
${\dispwaystywe \operatorname {div} (\madbf {A} )={\cfrac {\partiaw A_{ik}}{\partiaw x_{k}}}~\madbf {e} _{i}=A_{ik,k}~\madbf {e} _{i}={\begin{bmatrix}{\dfrac {\partiaw A_{11}}{\partiaw x_{1}}}+{\dfrac {\partiaw A_{12}}{\partiaw x_{2}}}+{\dfrac {\partiaw A_{13}}{\partiaw x_{3}}}\\{\dfrac {\partiaw A_{21}}{\partiaw x_{1}}}+{\dfrac {\partiaw A_{22}}{\partiaw x_{2}}}+{\dfrac {\partiaw A_{23}}{\partiaw x_{3}}}\\{\dfrac {\partiaw A_{31}}{\partiaw x_{1}}}+{\dfrac {\partiaw A_{32}}{\partiaw x_{2}}}+{\dfrac {\partiaw A_{33}}{\partiaw x_{3}}}\end{bmatrix}}}$
and[5][6][7][8]
${\dispwaystywe \nabwa \cdot \madbf {A} ={\cfrac {\partiaw A_{ki}}{\partiaw x_{k}}}~\madbf {e} _{i}=A_{ki,k}~\madbf {e} _{i}={\begin{bmatrix}{\dfrac {\partiaw A_{11}}{\partiaw x_{1}}}+{\dfrac {\partiaw A_{21}}{\partiaw x_{2}}}+{\dfrac {\partiaw A_{31}}{\partiaw x_{3}}}\\{\dfrac {\partiaw A_{12}}{\partiaw x_{1}}}+{\dfrac {\partiaw A_{22}}{\partiaw x_{2}}}+{\dfrac {\partiaw A_{32}}{\partiaw x_{3}}}\\{\dfrac {\partiaw A_{13}}{\partiaw x_{1}}}+{\dfrac {\partiaw A_{23}}{\partiaw x_{2}}}+{\dfrac {\partiaw A_{33}}{\partiaw x_{3}}}\\\end{bmatrix}}}$
We have
${\dispwaystywe \operatorname {div} (\madbf {A^{T}} )=\nabwa \cdot \madbf {A} }$
If tensor is symmetric ${\dispwaystywe A_{ij}=A_{ji}}$ den ${\dispwaystywe \operatorname {div} (\madbf {A} )=\nabwa \cdot \madbf {A} }$ and dis cause dat often in witerature dis two definitions (and symbows ${\dispwaystywe \madrm {div} }$ and ${\dispwaystywe \nabwa \cdot }$) are switched and interchangeabwy used (especiawwy in mechanics eqwations where tensor symmetry is assumed).
Expressions of ${\dispwaystywe \nabwa \cdot \madbf {A} }$ in cywindricaw and sphericaw coordinates are given in de articwe dew in cywindricaw and sphericaw coordinates.
### Generaw coordinates
Using Einstein notation we can consider de divergence in generaw coordinates, which we write as x1, ..., xi, ...,xn, where n is de number of dimensions of de domain, uh-hah-hah-hah. Here, de upper index refers to de number of de coordinate or component, so x2 refers to de second component, and not de qwantity x sqwared. The index variabwe i is used to refer to an arbitrary ewement, such as xi. The divergence can den be written via de Voss- Weyw formuwa,[9] as:
${\dispwaystywe \operatorname {div} (\madbf {F} )={\frac {1}{\rho }}{\frac {\partiaw \weft(\rho \,F^{i}\right)}{\partiaw x^{i}}},}$
where ${\dispwaystywe \rho }$ is de wocaw coefficient of de vowume ewement and Fi are de components of F wif respect to de wocaw unnormawized covariant basis (sometimes written as ${\dispwaystywe \madbf {e} _{i}=\partiaw \madbf {x} /\partiaw x^{i}}$). The Einstein notation impwies summation over i, since it appears as bof an upper and wower index.
The vowume coefficient ${\dispwaystywe \rho }$ is a function of position which depends on de coordinate system. In Cartesian, cywindricaw and sphericaw coordinates, using de same conventions as before, we have ${\dispwaystywe \rho =1}$, ${\dispwaystywe \rho =r}$ and ${\dispwaystywe \rho =r^{2}\sin {\deta }}$, respectivewy. The vowume can awso be expressed as ${\dispwaystywe \rho ={\sqrt {|\operatorname {det} g_{ab}|}}}$, where ${\dispwaystywe g_{ab}}$ is de metric tensor. The determinant appears because it provides de appropriate invariant definition of de vowume, given a set of vectors. Since de determinant is a scawar qwantity which doesn't depend on de indices, dese can be suppressed, writing ${\dispwaystywe \rho ={\sqrt {|\operatorname {det} g|}}}$. The absowute vawue is taken in order to handwe de generaw case where de determinant might be negative, such as in pseudo-Riemannian spaces. The reason for de sqware-root is a bit subtwe: it effectivewy avoids doubwe-counting as one goes from curved to Cartesain coordinates, and back. The vowume (de determinant) can awso be understood as de Jacobian of de transformation from Cartesian to curiviwinear coordinates, which for n = 3 gives ${\dispwaystywe \rho =\weft|{\frac {\partiaw (x,y,z)}{\partiaw (x^{1},x^{2},x^{3})}}\right|.}$
Some conventions expect aww wocaw basis ewements to be normawized to unit wengf, as was done in de previous sections. If we write ${\dispwaystywe {\hat {\madbf {e} }}_{i}}$ for de normawized basis, and ${\dispwaystywe {\hat {F}}^{i}}$ for de components of F wif respect to it, we have dat
${\dispwaystywe \madbf {F} =F^{i}\madbf {e} _{i}=F^{i}{\wVert {\madbf {e} _{i}}\rVert }{\frac {\madbf {e} _{i}}{\wVert {\madbf {e} _{i}}\rVert }}=F^{i}{\sqrt {g_{ii}}}\,{\hat {\madbf {e} }}_{i}={\hat {F}}^{i}{\hat {\madbf {e} }}_{i},}$
using one of de properties of de metric tensor. By dotting bof sides of de wast eqwawity wif de contravariant ewement ${\dispwaystywe {\hat {\madbf {e} }}^{i}}$, we can concwude dat ${\dispwaystywe F^{i}={\hat {F}}^{i}/{\sqrt {g_{ii}}}}$. After substituting, de formuwa becomes:
${\dispwaystywe \operatorname {div} (\madbf {F} )={\frac {1}{\rho }}{\frac {\partiaw \weft({\frac {\rho }{\sqrt {g_{ii}}}}{\hat {F}}^{i}\right)}{\partiaw x^{i}}}={\frac {1}{\sqrt {\operatorname {det} g}}}{\frac {\partiaw \weft({\sqrt {\frac {\operatorname {det} g}{g_{ii}}}}\,{\hat {F}}^{i}\right)}{\partiaw x^{i}}}}$.
See ยงย In curviwinear coordinates for furder discussion, uh-hah-hah-hah.
## Properties
The fowwowing properties can aww be derived from de ordinary differentiation ruwes of cawcuwus. Most importantwy, de divergence is a winear operator, i.e.,
${\dispwaystywe \operatorname {div} (a\madbf {F} +b\madbf {G} )=a\operatorname {div} \madbf {F} +b\operatorname {div} \madbf {G} }$
for aww vector fiewds F and G and aww reaw numbers a and b.
There is a product ruwe of de fowwowing type: if ฯ is a scawar-vawued function and F is a vector fiewd, den
${\dispwaystywe \operatorname {div} (\varphi \madbf {F} )=\operatorname {grad} \varphi \cdot \madbf {F} +\varphi \operatorname {div} \madbf {F} ,}$
or in more suggestive notation
${\dispwaystywe \nabwa \cdot (\varphi \madbf {F} )=(\nabwa \varphi )\cdot \madbf {F} +\varphi (\nabwa \cdot \madbf {F} ).}$
Anoder product ruwe for de cross product of two vector fiewds F and G in dree dimensions invowves de curw and reads as fowwows:
${\dispwaystywe \operatorname {div} (\madbf {F} \times \madbf {G} )=\operatorname {curw} \madbf {F} \cdot \madbf {G} -\madbf {F} \cdot \operatorname {curw} \madbf {G} ,}$
or
${\dispwaystywe \nabwa \cdot (\madbf {F} \times \madbf {G} )=(\nabwa \times \madbf {F} )\cdot \madbf {G} -\madbf {F} \cdot (\nabwa \times \madbf {G} ).}$
The Lapwacian of a scawar fiewd is de divergence of de fiewd's gradient:
${\dispwaystywe \operatorname {div} (\operatorname {grad} \varphi )=\Dewta \varphi .}$
The divergence of de curw of any vector fiewd (in dree dimensions) is eqwaw to zero:
${\dispwaystywe \nabwa \cdot (\nabwa \times \madbf {F} )=0.}$
If a vector fiewd F wif zero divergence is defined on a baww in R3, den dere exists some vector fiewd G on de baww wif F = curw G. For regions in R3 more topowogicawwy compwicated dan dis, de watter statement might be fawse (see Poincarรฉ wemma). The degree of faiwure of de truf of de statement, measured by de homowogy of de chain compwex
${\dispwaystywe \{{\text{scawar fiewds on }}U\}~{\overset {\operatorname {grad} }{\rightarrow }}~\{{\text{vector fiewds on }}U\}~{\overset {\operatorname {curw} }{\rightarrow }}~\{{\text{vector fiewds on }}U\}~{\overset {\operatorname {div} }{\rightarrow }}~\{{\text{scawar fiewds on }}U\}}$
serves as a nice qwantification of de compwicatedness of de underwying region U. These are de beginnings and main motivations of de Rham cohomowogy.
## Decomposition deorem
It can be shown dat any stationary fwux v(r) dat is twice continuouswy differentiabwe in R3 and vanishes sufficientwy fast for |r| โ โ can be decomposed uniqwewy into an irrotationaw part E(r) and a source-free part B(r). Moreover, dese parts are expwicitwy determined by de respective source densities (see above) and circuwation densities (see de articwe Curw):
For de irrotationaw part one has
${\dispwaystywe \madbf {E} =-\nabwa \Phi (\madbf {r} ),}$
wif
${\dispwaystywe \Phi (\madbf {r} )=\int _{\madbb {R} ^{3}}\,d^{3}\madbf {r} '\;{\frac {\operatorname {div} \madbf {v} (\madbf {r} ')}{4\pi \weft|\madbf {r} -\madbf {r} '\right|}}.}$
The source-free part, B, can be simiwarwy written: one onwy has to repwace de scawar potentiaw ฮฆ(r) by a vector potentiaw A(r) and de terms โโฮฆ by +โ ร A, and de source density div v by de circuwation density โ ร v.
This "decomposition deorem" is a by-product of de stationary case of ewectrodynamics. It is a speciaw case of de more generaw Hewmhowtz decomposition, which works in dimensions greater dan dree as weww.
## In arbitrary dimensions
The divergence of a vector fiewd can be defined in any number of dimensions. If
${\dispwaystywe \madbf {F} =(F_{1},F_{2},\wdots F_{n}),}$
in a Eucwidean coordinate system wif coordinates x1, x2, ..., xn, define
${\dispwaystywe \operatorname {div} \madbf {F} =\nabwa \cdot \madbf {F} ={\frac {\partiaw F_{1}}{\partiaw x_{1}}}+{\frac {\partiaw F_{2}}{\partiaw x_{2}}}+\cdots +{\frac {\partiaw F_{n}}{\partiaw x_{n}}}.}$
In de case of one dimension, F reduces to a reguwar function, and de divergence reduces to de derivative.
For any n, de divergence is a winear operator, and it satisfies de "product ruwe"
${\dispwaystywe \nabwa \cdot (\varphi \madbf {F} )=(\nabwa \varphi )\cdot \madbf {F} +\varphi (\nabwa \cdot \madbf {F} )}$
for any scawar-vawued function ฯ.
## Rewation to de exterior derivative
One can express de divergence as a particuwar case of de exterior derivative, which takes a 2-form to a 3-form in R3. Define de current two-form as
${\dispwaystywe j=F_{1}\,dy\wedge dz+F_{2}\,dz\wedge dx+F_{3}\,dx\wedge dy.}$
It measures de amount of "stuff" fwowing drough a surface per unit time in a "stuff fwuid" of density ฯ = 1 dxdydz moving wif wocaw vewocity F. Its exterior derivative dj is den given by
${\dispwaystywe dj=\weft({\frac {\partiaw F_{1}}{\partiaw x}}+{\frac {\partiaw F_{2}}{\partiaw y}}+{\frac {\partiaw F_{3}}{\partiaw z}}\right)dx\wedge dy\wedge dz=(\nabwa \cdot {\madbf {F} })\rho }$
where ${\dispwaystywe \wedge }$ is de wedge product.
Thus, de divergence of de vector fiewd F can be expressed as:
${\dispwaystywe \nabwa \cdot {\madbf {F} }={\star }d{\star }{\big (}{\madbf {F} }^{\fwat }{\big )}.}$
Here de superscript is one of de two musicaw isomorphisms, and is de Hodge star operator. When de divergence is written in dis way, de operator ${\dispwaystywe {\star }d{\star }}$ is referred to as de codifferentiaw. Working wif de current two-form and de exterior derivative is usuawwy easier dan working wif de vector fiewd and divergence, because unwike de divergence, de exterior derivative commutes wif a change of (curviwinear) coordinate system.
## In curviwinear coordinates
The appropriate expression is more compwicated in curviwinear coordinates. The divergence of a vector fiewd extends naturawwy to any differentiabwe manifowd of dimension n dat has a vowume form (or density) ฮผ, e.g. a Riemannian or Lorentzian manifowd. Generawising de construction of a two-form for a vector fiewd on R3, on such a manifowd a vector fiewd X defines an (n โ 1)-form j = iXโฮผ obtained by contracting X wif ฮผ. The divergence is den de function defined by
${\dispwaystywe dj=(\operatorname {div} X)\mu .}$
The divergence can be defined in terms of de Lie derivative as
${\dispwaystywe {\madcaw {L}}_{X}\mu =(\operatorname {div} X)\mu .}$
This means dat de divergence measures de rate of expansion of a unit of vowume (a vowume ewement)) as it fwows wif de vector fiewd.
On a pseudo-Riemannian manifowd, de divergence wif respect to de vowume can be expressed in terms of de Levi-Civita connection :
${\dispwaystywe \operatorname {div} X=\nabwa \cdot X={X^{a}}_{;a},}$
where de second expression is de contraction of de vector fiewd vawued 1-form X wif itsewf and de wast expression is de traditionaw coordinate expression from Ricci cawcuwus.
An eqwivawent expression widout using a connection is
${\dispwaystywe \operatorname {div} (X)={\frac {1}{\sqrt {|\operatorname {det} g|}}}\partiaw _{a}\weft({\sqrt {|\operatorname {det} g|}}\,X^{a}\right),}$
where g is de metric and ${\dispwaystywe \partiaw _{a}}$ denotes de partiaw derivative wif respect to coordinate xa. The sqware-root of de (absowute vawue of de determinant of de) metric appears because de divergence must be written wif de correct conception of de vowume. In curviwinear coordinates, de basis vectors are no wonger ordonormaw; de determinant encodes de correct idea of vowume in dis case. It appears twice, here, once, so dat de ${\dispwaystywe X^{a}}$ can be transformed into "fwat space" (where coordinates are actuawwy ordonormaw), and once again so dat ${\dispwaystywe \partiaw _{a}}$ is awso transformed into "fwat space", so dat finawwy, de "ordinary" divergence can be written wif de "ordinary" concept of vowume in fwat space (i.e. unit vowume, i.e. one, i.e. not written down). The sqware-root appears in de denominator, because de derivative transforms in de opposite way (contravariantwy) to de vector (which is covariant). This idea of getting to a "fwat coordinate system" where wocaw computations can be done in a conventionaw way is cawwed a viewbein. A different way to see dis is to note dat de divergence is de codifferentiaw in disguise. That is, de divergence corresponds to de expression ${\dispwaystywe \star d\star }$ wif ${\dispwaystywe d}$ de differentiaw and ${\dispwaystywe \star }$ de Hodge star. The Hodge star, by its construction, causes de vowume form to appear in aww of de right pwaces.
## The divergence of tensors
Divergence can awso be generawised to tensors. In Einstein notation, de divergence of a contravariant vector Fฮผ is given by
${\dispwaystywe \nabwa \cdot \madbf {F} =\nabwa _{\mu }F^{\mu },}$
where ฮผ denotes de covariant derivative. In dis generaw setting, de correct formuwation of de divergence is to recognize dat it is a codifferentiaw; de appropriate properties fowwow from dere.
Eqwivawentwy, some audors define de divergence of a mixed tensor by using de musicaw isomorphism : if T is a (p, q)-tensor (p for de contravariant vector and q for de covariant one), den we define de divergence of T to be de (p, q โ 1)-tensor
${\dispwaystywe (\operatorname {div} T)(Y_{1},\wdots ,Y_{q-1})={\operatorname {trace} }{\Big (}X\mapsto \sharp (\nabwa T)(X,\cdot ,Y_{1},\wdots ,Y_{q-1}){\Big )};}$
dat is, we take de trace over de first two covariant indices of de covariant derivative.[a] The ${\dispwaystywe \sharp }$ symbow refers to de musicaw isomorphism.
## Notes
1. ^ The choice of "first" covariant index of a tensor is intrinsic and depends on de ordering of de terms of de Cartesian product of vector spaces on which de tensor is given as a muwtiwinear map V ร V ร ... ร V โ R. But eqwawwy weww defined choices for de divergence couwd be made by using oder indices. Conseqwentwy, it is more naturaw to specify de divergence of T wif respect to a specified index. There are however two important speciaw cases where dis choice is essentiawwy irrewevant: wif a totawwy symmetric contravariant tensor, when every choice is eqwivawent, and wif a totawwy antisymmetric contravariant tensor (a.k.a. a k-vector), when de choice affects onwy de sign, uh-hah-hah-hah.
## Citations
1. ^ Cywindricaw coordinates at Wowfram Madworwd
2. ^ Sphericaw coordinates at Wowfram Madworwd
3. ^ Gurtin 1981, p. 30.
4. ^ "1.14 Tensor Cawcuwus I: Tensor Fiewds" (PDF). Foundations of Continuum Mechanics.
5. ^ Wiwwiam M. Deen (2016). Introduction to Chemicaw Engineering Fwuid Mechanics. Cambridge University Press. p.ย 133. ISBNย 978-1-107-12377-9.CS1 maint: uses audors parameter (wink)
6. ^ Sara Noferesti, Hassan Ghassemi, Hashem Nowruzi (15 May 2019). "Numericaw Investigation on de Effects of Obstruction and Side Ratio on Non-Newtonian Fwuid Fwow Behavior Around a Rectanguwar Barrier" (PDF): 56,59. doi:10.17512/jamcm.2019.1.05. Cite journaw reqwires |journaw= (hewp)CS1 maint: uses audors parameter (wink)
7. ^ Tasos C. Papanastasiou, Georgios C. Georgiou, Andreas N. Awexandrou (2000). Viscous Fwuid Fwow (PDF). CRC Press. p.ย 66,68. ISBNย 0-8493-1606-5.CS1 maint: uses audors parameter (wink)
8. ^ Adam Poweww (12 Apriw 2010). "The Navier-Stokes Eqwations" (PDF).
9. ^ Grinfewd, Pavew. "The Voss-Weyw Formuwa". Retrieved 9 January 2018. | 8,226 | 25,252 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 80, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2020-50 | latest | en | 0.802887 |
https://marksmath.org/classes/Fall2020Stat185/StatTalk/discussion/12/bar-plot-and-proportion | 1,610,908,617,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703513144.48/warc/CC-MAIN-20210117174558-20210117204558-00508.warc.gz | 445,115,355 | 17,240 | # Bar plot and proportion
edited August 2020
(10 points)
Using your own random data from last Friday's forum question, let's use Colab to examine the activity_level variable. Specifically:
โข Generate a bar chart for activity_level and
โข Compute the proportion of folks whose activity level is high.
Note that creative burden is higher in this lab than in the last in that the Colab link above leads to a blank notebook. Nonetheless, you can find sample code that should help in our class presentation on Categorical Data.
ยซ1
โข edited August 2020
Data Set
import pandas as pd
df.tail()
Value Counts
value_counts = df['activity_level'].value_counts()
value_counts.to_frame()
activity_level
high 41
none 31
moderate 28
Proportion of folks whose activity level is high is: 0.41
value_counts['high']/len(df)
Bar Chart
value_counts.plot.bar(figsize=(12,7), rot = 0);
โข edited August 2020
import pandas as pd
df.tail()
Value count
value_counts = df['activity_level'].value_counts()
value_counts
moderate 37
none 37
high 26
The proportion whose activity level is high is .26
Bar chart
value_counts.plot.bar(figsize=(12,7), rot = 0);
โข edited August 2020
I generated my own random data with this code:
import pandas as pd
My data:
first_name last_name age sex height weight income activity_level
0 Retha Reese 41 female 63.91 166.41 24811 moderate
1 Felicia Hamm 41 female 61.88 152.91 11829 high
2 Lauren Poindexter 22 female 69.36 219.97 4259 none
3 Erin Davis 29 female 60.52 224.19 19351 moderate
4 Minnie Bouie 20 female 69.10 140.45 12624 high
I got my value counts with the code:
value_counts = df['activity_level'].value_counts()
value_counts.to_frame()
...and got:
activity_level
none 37
high 35
moderate 28
My proportion of people with a high activity level is .35
value_counts['high']/len(df)
I got my bar chart with this code:
value_counts.plot.bar(figsize=(12,7), rot = 0);
โข edited August 2020
Here is my data set:
import pandas as pd
Here is the value counts, which are the data points for my bar chart
value_counts = df['activity_level'].value_counts()
value_counts
Here is my bar chart for activity level, obtained by
value_counts.plot.bar(figsize=(12,7), rot = 0);
I then computed the proportion of folks whose activity level is high, which is .33
value_counts['high']/len(df)
# Output: 0.33
โข edited August 2020
import pandas as pd
df.tail()
value_counts = df['activity_level'].value_counts()
value_counts
high 39
none 31
moderate 30
Name: activity_level, dtype: int64
value_counts['high']/len(df)
0.39
value_counts.plot.bar(figsize=(12,7), rot = 0);
โข Data:
import pandas as pd
df.tail()
first_name last_name age sex height weight income activity_level
95 Robert Session 20 male 71.36 175.13 2783 high
96 Christine Powers 20 female 65.73 172.88 6692 high
97 Brandon Griffin 23 male 69.00 168.12 31407 moderate
98 Mollie Donohue 39 female 68.95 174.77 15755 high
99 Manuel Hammond 32 male 68.79 146.79 238400 moderate
I also created a table for activity level:
value_counts.to_markdown = df['activity_level'].value_counts()
value_counts
none 35
high 34
moderate 31
Name: activity_level, dtype: int64
Bar Chart:
value_counts.plot.bar(figsize=(12,7), rot = 0);
The proportion of people who were at a high activity level is 0.34
value_counts['high']/len(df)
โข edited August 2020
Data Set:
import pandas as pd
df.tail()
Value Counts Code:
value_counts = df['activity_level'].value_counts()
value_counts
none 37
high 33
moderate 30
The proportion of high activity level is 0.33
value_counts['high']/len(df)
Bar Chart:
โข edited August 2020
My Data Set
import pandas as pd
df.tail()
first_name last_name age sex height weight income activity_level
95 James Theden 24 male 64.49 135.56 20767 moderate
96 Michael Williams 33 male 68.77 199.40 65051 high
97 Phillip Yamashiro 33 male 69.81 215.21 32861 none
98 Douglas Mims 47 male 72.60 172.08 102766 high
99 Beatrice Cox 23 female 61.08 153.27 73659 none
The Value Counts of My Data
value_counts = df['activity_level'].value_counts()
value_counts.to_frame()
activity_level
high 35
none 33
moderate 32
The Proportion of My Data Set
value_counts['high']/len(df)
0.35
The Bar Plot of My Data Set
value_counts.plot.bar(figsize=(12,7), rot = 0);
โข edited August 2020
Data set:
import pandas as pd
df.tail()
first_name last_name age sex height weight income activity_level
95 Cara Rogers 22 female 59.93 153.81 9555 moderate
96 Craig Brandt 29 male 65.45 169.31 28619 moderate
97 Herbert Carolina 25 male 74.51 141.63 6421 moderate
98 Patricia Guarnieri 22 female 69.52 150.90 228591 none
99 Stacy Davis 24 male 67.87 217.32 12139 moderate
Value counts:
value_counts = df['activity_level'].value_counts()
value_counts.to_frame()
activity_level
none 47
moderate 31
high 22
Proportion whose activity is high:
value_counts['high']/len(df)
0.22
Bar Chart:
value_counts.plot.bar(figsize=(12,7), rot = 0);
โข edited August 2020
Data Set:
import pandas as pd
first_name last_name age sex height weight income activity_level
0 Barbara Dolan 57 female 57.01 177.34 6631 high
1 Caren Walters 22 female 63.84 157.23 8015 high
2 Wesley Avery 39 male 66.90 204.50 2201 moderate
3 Michael Numbers 41 male 67.09 164.74 5184 none
4 Bruce Williams 37 male 67.07 180.17 9517 none
Table for the activity level:
value_counts = df['activity_level'].value_counts()
value_counts
none 37
high 33
moderate 30
Name: activity_level, dtype: int64
My Bar Chart:
value_counts.plot.bar(figsize=(12,7), rot = 0)
The proportion of people who were at a high activity level is 0.33
value_counts['high']/len(df)
โข edited August 2020
Data:
import pandas as pd
df.tail()
first_name last_name age sex height weight income activity_level
95 Richard Schmidt 40 male 69.04 184.16 425 moderate
96 Paul Fleury 29 male 64.96 206.06 11212 moderate
97 Agnes Pollard 39 female 61.78 233.80 12416 moderate
98 Diane Morrison 42 female 61.50 179.02 17823 none
99 Frances Horn 38 female 60.43 132.64 336 moderate
Value Counts:
value_counts = df['activity_level'].value_counts()
value_counts
moderate 44
high 34
none 22
Name: activity_level, dtype: int64
Proportion of people with high activity level: .34
value_counts['high']/len(df)
Bar chart:
value_counts.plot.bar(figsize=(12,7), rot = 0);
โข edited August 2020
Data:
import pandas as pd
first_name last_name age sex height weight income activity_level
0 John Griffin 53 male 70.83 101.54 7675 moderate
1 Dorthy Brown 43 female 60.14 193.71 3096 moderate
2 Mary Geiger 36 female 59.62 160.14 34931 high
3 Ann Diaz 40 female 62.30 200.72 1526 none
4 Steve Washington 39 male 70.73 145.43 1626 high
to find my specific data for 'activity level'
value_counts = df['activity_level'].value_counts()
value_counts.to_markdown()
activity_level
moderate 39
none 34
high 27
then i put the data into a bar graph
value_counts.plot.bar(figsize=(12,7), rot = 0);
and i found who was high with this and got 0.27
value_counts['high']/len(df)
โข First I gather the data:
import pandas as pd
Then I take the value counts from the activity_level variable:
value_counts = df['activity_level'].value_counts()
value_counts.to_frame()
To get the bar chart, I just enter the following code to produce one:
value_counts.plot.bar()
Finally for the proportion data, I just enter this line which returned the value 0.34
value_counts['high']/len(df)
And that's all I need to put down.
โข edited August 2020
import pandas as pd
first_name last_name age sex height weight income activity_level
0 Benjamin Davis 58 male 75.86 151.92 2977 high
1 Cody Hicks 38 male 69.02 204.74 16815 high
2 Elizabeth Hall 31 female 61.41 167.96 17624 none
3 Dean Hunt 36 male 69.04 185.04 1823 none
4 Christine Valle 40 female 62.50 112.04 4153 none
value counts
value_counts = df['activity_level'].value_counts()
value_counts
high 36
moderate 33
none 31
Name: activity_level, dtype: int64
proportion
value_counts['high']/len(df)
0.36
โข Data
import pandas as pd
first_name last_name age sex height weight income activity_level
0 Petrina Jose 37 female 63.10 182.08 22898 high
1 Amanda Abrams 28 female 65.39 126.48 13397 moderate
2 Sandra Howard 55 female 60.88 156.78 31135 moderate
3 Elmer Lim 49 male 66.48 163.43 693 moderate
4 Albert Smith 24 male 72.84 179.31 4438
Variable chart
value_counts = df['activity_level'].value_counts()
value_counts.to_frame()
activity_level
moderate 34
none 34
high 32
Bar Chart
value_counts.plot.bar(figsize=(12,7), rot = 0);
โข edited August 2020
Data:
import pandas as pd
df.tail()
first_name last_name age sex height weight income activity_level
95 Shirley Garcia 25 female 62.52 119.57 5010 none
96 Danny Hymes 26 male 73.81 153.20 57164 high
97 April Lebrecque 34 female 69.20 141.89 45640 moderate
98 Karen Babcock 21 female 61.40 172.48 44646 moderate
99 Tony Benedict 26 male 69.80 174.12 22386 high
Value Counts:
value_counts = df['activity_level'].value_counts()
value_counts.to_frame()
activity_level
moderate 35
high 34
none 31
Proportion of folks whose activity level is high is: 0.34
value_counts['high']/len(df)
0.34
Bar:
โข import dataset:
import pandas as pd
value counts for desired variable:
value_counts = df['activity_level'].value_counts()
value_counts
high 46
none 29
moderate 25
Name: activity_level, dtype: int64
finding proportion whose activity level is high:
value_counts = df['activity_level'].value_counts()
value_counts
0.46
generate bar graph:
value_counts.plot.bar(figsize=(12,7), rot = 0);
โข edited August 2020
Data Set:
import pandas as pd
first_name last_name age sex height weight income activity_level
0 David Krause 21 male 74.85 152.97 7391 moderate
1 Karen Liebsch 27 female 70.06 116.25 4694 moderate
2 Dorothy Hill 35 female 59.56 153.08 11234 high
3 Brenda Bott 26 female 66.71 181.44 105242 none
4 Maria Flournoy 50 female 67.07 189.09 7789 none
Value Counts:
value_counts = df['activity_level'].value_counts()
value_counts.to_markdown()
"High" Proportion:
value_counts['high']/len(df)
Bar Chart:
value_counts.plot.bar(figsize=(12,7), rot = 0);
โข Data:
import pandas as pd
Value Counts:
value_counts = df['activity_level'].value_counts()
value_counts.to_frame()
Proportion of those with a high activity level: 0.35
value_counts['high']/len(df)
Bar Chart:
value_counts.plot.bar(figsize=(12,7), rot = 0)
โข edited August 2020
Data Set
import pandas as pd
df.tail()
My Data
first_name last_name age sex height weight income activity_level
95 Harold Ogle 26 male 64.84 173.37 37549 high
96 Vivian Diaz 35 female 66.05 119.89 34276 high
97 Florinda Alston 35 female 68.45 186.53 8963 none
98 Bobbie Hillis 32 female 62.81 153.58 25578 moderate
99 Errol Ricketts 42 male 70.43 183.30 30 none
Value Counts
value_counts = df['activity_level'].value_counts()
value_counts.to_frame()
activity_level
moderate 35
none 33
high 32
Proportion of people's high activity levels: .32
value_counts['high']/len(df)
โข edited August 2020
Data Set:
import pandas as pd
df.tail()
first_name last_name age sex height weight income activity_level
95 Stewart Horton 29 male 68.29 180.10 1616 none
96 Catherine Maddox 34 female 65.84 187.60 7534 moderate
97 Margaret Martin 28 female 66.34 205.22 229127 high
98 Robert Armstrong 21 male 65.95 157.64 55894 none
99 Sheila Wallis 41 female 63.27 124.36 4767 moderate
Value Counts:
value_counts = df['activity_level'].value_counts()
value_counts.to_markdown()
| activity_level |\n|:---------|-----------------:|\n| high | 40 |\n| none | 31 |\n| moderate | 29 |
High Proportion:
value_counts['high']/len(df)
.4
Bar Chart:
value_counts.plot.bar(figsize=(12,7), rot = 0);
โข edited August 2020
Data Set:
import pandas as pd
df.tail()
# Output:
first_name last_name age sex height weight income activity_level
95 William Pearson 33 male 67.46 147.26 6501 none
96 Heather Stephens 39 female 67.72 138.53 70003 none
97 Sheri Lepard 58 female 62.66 172.13 27956 moderate
98 Stacie Leray 43 female 62.64 156.19 4555 moderate
99 Elizabeth Peele 41 female 63.79 175.92 19106 moderate
Value Counts:
value_counts = df['activity_level'].value_counts()
value_counts.to_frame()
activity_level
high 37
moderate 33
none 30
High Proportion:
value_counts['high']/len(df)
.37
Bar Plot:
value_counts.plot.bar(figsize=(12,7), rot = 0);
โข Data:
import pandas as pd
first_name last_name age sex height weight income activity_level
0 Crystal Menard 21 female 66.10 184.02 6515 none
1 Amy Wise 38 female 57.31 203.48 1683 none
2 Judith Monk 30 female 63.31 186.16 1156 moderate
3 Humberto Ray 50 male 70.73 211.53 1183 high
4 Claude Baker 22 male 70.68 151.08 12819 none
Value counts:
value_counts = df['activity_level'].value_counts()
value_counts.to_frame()
activity_level
none 40
high 33
moderate 27
Activity level "High":
value_counts['high']/len(df)
0.33
Bar graph:
value_counts.plot.bar(figsize=(12,7), rot = 0);
โข edited August 2020
My data set:
import pandas as pd
first_name last_name age sex height weight income activity_level
0 Carl Gerard 36 male 69.45 215.64 2663 moderate
1 Lila Johnson 49 female 61.57 175.93 25959 none
2 Brenda Mossien 40 female 61.94 188.08 14445 high
3 Gary Lafave 49 male 67.95 169.63 76683 high
4 Jefferey Johnson 23 male 71.37 190.35 33960 none
Value Counts:
value_counts = df['activity_level'].value_counts()
value_counts
high 37
moderate 33
none 30
Name: activity_level, dtype: int64
High Proportion:
value_counts['high']/len(df)
0.37
My Bar Chart:
value_counts.plot.bar(figsize=(12,7), rot = 0);
โข edited August 2020
Data set:
import pandas as pd
Value Counts:
value_counts = df['activity_level'].value_counts()
value_counts.to_frame()
moderate 38
high 35
none 27
value_counts['high']/len(df
High proportion:![]
.35
value_counts.plot.bar(figsize=(12,7), rot = 0);
โข Data:
first_name last_name age sex height weight income activity_level
95 Marilyn Parris 29 female 64.68 150.69 115310 high
96 Robert Flores 51 male 71.18 136.76 24487 moderate
97 Daniel Bohne 36 male 72.07 177.61 833 none
98 Richard Schubbe 32 male 68.57 145.97 47000 high
99 David Smith 41 male 69.95 176.52 3249 high
import pandas as pd
df.tail()
Value Count
high 36
none 33
moderate 31
value_counts = df['activity_level'].value_counts()
High proportion:
0.36
value_counts['high']/len(df)
Bar Chart
value_counts.plot.bar(figsize=(12,7), rot = 0);
โข My Data Set Is:
import pandas as pd
first_name last_name age sex height weight income activity_level
0 John Sanchez 22 male 68.97 174.75 1675 none
1 Dennis Palmer 22 female 67.04 168.24 176113 high
2 Jennifer Walker 32 female 66.34 192.27 2180 moderate
3 Shannon Clay 38 female 62.52 169.43 3165 high
4 Leona Benson 31 female 62.58 211.73 1969 moderate
My Value Counts:
value_counts = df['activity_level'].value_counts()
value_counts
none 39
high 32
moderate 29
Name: activity_level, dtype: int64
Proportion of people with high activity: .32
value_counts['high']/len(df)
Bar Chart:
value_counts.plot.bar(figsize=(12,7), rot = 0);
โข edited August 2020
My data set
import pandas as pd
df.tail()
My output
first_name last_name age sex height weight income activity_level
95 Michael Vogel 31 male 65.23 145.48 894 moderate
96 Karl Cornely 27 male 70.21 215.22 3908 moderate
97 Phyllis Hoffman 52 female 63.34 162.52 35325 none
98 Patricia Winslow 28 female 67.92 185.47 3091 high
99 Lamont Morales 40 male 72.57 173.11 4561 high
Value Counts
value_counts = df['activity_level'].value_counts()
value_counts.to_frame()
Output
activity_level
high 36
none 34
moderate 30
Proportion of those who is high-0.36
value_counts['high']/len(df)
Bar Chart
value_counts.plot.bar(figsize=(12,7), rot = 0);
โข import pandas as pd
value_counts = df['activity_level'].value_counts()
value_counts
value_counts.plot.bar(figsize=(12,7), rot = 0);
โข Data set:
import pandas as pd
df.tail()
first_name last_name age sex height weight income activity_level
95 Kathleen Allen 30 female 60.70 139.41 520 none
96 Sarah Roche 35 female 68.96 183.76 32649 high
97 Christine Griffin 29 female 63.36 159.37 7969 moderate
98 Pamela Small 30 female 67.62 174.56 5076 high
99 Betty Overton 44 female 61.65 168.26 159 high
Value Count:
value_counts = df['activity_level'].value_counts()
value_counts.to_frame()
activity_level:
moderate: 40
none: 36
high: 24
High Proportion Level:
value_counts['high']/len(df)
0.24
My Bar Chart:
value_counts.plot.bar(figsize=(12,7), rot = 0);
This discussion has been closed. | 5,708 | 18,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2021-04 | latest | en | 0.694088 |
https://www.geeksforgeeks.org/python-custom-cycle-list/ | 1,632,267,206,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057274.97/warc/CC-MAIN-20210921221605-20210922011605-00036.warc.gz | 814,202,246 | 22,118 | Related Articles
# Python | Custom Cycle list
โข Last Updated : 09 Sep, 2019
While working with Python lists, we can have a problem in which we need to perform the cycle of lists. This problem looks quite straight forward and has been discussed earlier. But sometimes, we need to perform itโs variations, hence making task challenging. We can have customizations such as elements from which cycling starts and Number of elements in cycle list. Letโs discuss solution to these variation.
Method : Using `dropwhile() + cycle() + islice()`
This task can be performed using combination of above functions. In this, the elements are dropped till K with `dropwhile()`, then the cycle can be done using `cycle()` and `islice()` is used to restrict the count of elements in list.
`# Python3 code to demonstrate working of``# Custom Cycle list``# using dropwhile() + cycle() + islice()``import` `itertools``ย `ย `# initialize list``test_list ``=` `[``3``, ``4``, ``5``, ``7``, ``1``]``ย `ย `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))``ย `ย `# initialize element for start cycle``K ``=` `7``ย `ย `# initialize size of cycle listย ``N ``=` `12``ย `ย `# Custom Cycle list``# using dropwhile() + cycle() + islice()``res ``=` `list``(itertools.islice(itertools.dropwhile(``lambda` `i: i !``=` `K, itertools.cycle(test_list)),ย N))``ย `ย `# printing result``print``(``"The cycled list is : "` `+` `str``(res))`
Output :
```The original list is : [3, 4, 5, 7, 1]
The cycled list is : [7, 1, 3, 4, 5, 7, 1, 3, 4, 5, 7, 1]
```
Attention geek! Strengthen your foundations with the Python Programming Foundation Course and learn the basics.
To begin with, your interview preparations Enhance your Data Structures concepts with the Python DS Course. And to begin with your Machine Learning Journey, join the Machine Learning โ Basic Level Course
My Personal Notes arrow_drop_up | 516 | 1,894 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-39 | latest | en | 0.757184 |
https://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Quantum_States_of_Atoms_and_Molecules_(Zielinksi_et_al.)/04._Electronic_Spectroscopy_of_Cyanine_Dyes/4.06%3A_Using_Symmetry_to_Identify_Integrals_that_are_Zero | 1,516,502,228,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889917.49/warc/CC-MAIN-20180121021136-20180121041136-00797.warc.gz | 654,156,507 | 17,355 | # 4.6: Using Symmetry to Identify Integrals that are Zero
It generally requires much work and time to evaluate integrals analytically or even numerically on a computer. Our time, computer time, and work can be saved if we can identify by inspection when integrals are zero. The determination of when integrals are zero leads to spectroscopic selection rules and provides a better understanding of them. Here we use graphs to examine properties of the wavefunctions for a particle-in-a-box to determine when the transition dipole moment integral is zero and thereby obtain the spectroscopic selection rules for this system. These considerations are completely general and can be applied to any integrals. We essentially determine whether integrals are zero or not by drawing pictures and thinking. What could be easier?
Consider the case for a transition from orbital n = 1 to orbital n = 2 of a molecule described by the particle-in-a-box model. These two wavefunctions are shown in Figure 4.6 as f1 and f2, respectively. For the curves shown on the left in the figure, we defined the box to have unit length, L = 1, and infinite potential barriers at x = 0 and x = L as we did previously, so the particle is trapped between 0 and L. For the curves shown on the right, g1 and g2, we put the origin of the coordinate system halfway between the potential barriers, i.e. at the center of the box. The barriers haven't moved and the particle hasnโt changed, but our description of the position of the barriers and the particle has changed. We now say the barriers are located at x = -L/2 and x = +L/2, and the particle is trapped between -L/2 and +L/2.
Clearly the wavefunctions in Figure $$\PageIndex{1}$$ look the same for these two choices of coordinate systems. The appearance of the wavefunctions doesnโt depend on the coordinate system we have chosen or on our labels since the wavefunctions tell us about the probability of finding the particle. This probability does not change when we change the coordinate system or relabel the axis. The names of these functions do change, however. In Figure 4.6a, they both are sine functions. In Figure 4.6b, one is a cosine function and the other is a sine function multiplied by -1.
Figure $$\PageIndex{1}$$:ย Wavefunctions for quantum state n=1 and quantum state n=2 in the two different coordinate systems.
Exercise $$\PageIndex{1}$$
Sketch $$(f_1(x))^2 \text {and} (g_1(x))^2$$. What do you observe? Sketch $$(f_2(x))^2 \text {and} (g_2(x))^2$$. What do you observe? What is the significance with respect to the probability given by both f(x) and g(x)?
We moved the origin of the coordinate system to the center of the box to take advantage of the symmetry properties of these functions. By symmetry, we mean the correspondence in form on either side of a dividing point, line, or plane. As we shall see, the analysis of the symmetry is straightforward if the origin of the coordinate systems coincides with the dividing point, line, or plane. Since the right and left halves of the box or molecule represented by the box are the same, the square of the wavefunction for x > 0 must be the same as the square of the wavefunction for x < 0. Since the box is symmetrical, the probability density, ฮจ2, for the particle distribution also must be symmetrical because there is no reason for the particle to be located preferentially on one side or the other.
The transition moment integral for the particle-in-a-boxย involves three functions ($$ฯ_f, ฯ_i$$, and x) that are multiplied together at each point x to form the integrand. These three functions for i = 1 and f = 2 are plotted on the left in Figure 4.7. The integrand is the product of these three functions and is shown on the right in the figure. The integral is the area between the integrand and the zero on the y-axis. Clearly this area and thus also the value of the integral is not zero. The integral is negative because $$ฯ_2$$ is negative for x > 0 and x is negative for x < 0. Since $$ฮผ_T โ 0$$, the transition from $$ฯ_1$$ to $$ฯ_2$$ is allowed. As we previously mentioned in this chapter, and will see again later, the absorption coefficient is proportional to the absolute square of $$ฮผ_T$$ so it is acceptable for the transition moment integral to be negative. It even could involve $$\sqrt {-1}$$. Taking the absolute square makes both negative and imaginary quantities positive.
Figure 4.7ย Functions (a) and integrand (b) of the transition moment integral for the transition from quantum state n = 1 to quantum state n = 2.
Exerciseย $$\PageIndex{2}$$
Write the expression or function for the integrand that is plotted on the right side of Figure 4.7 in terms of x, sine, and cosine functions. Use your function to explain why the integrand is 0 at x = 0 and has minima at x = + 0.25L and - 0.25L. Sketch the corresponding probability function.Where are the peaks in the probability function?
Now consider the transition moment integral for quantum state n = 1 to quantum state n = 3. In Figure $$\PageIndex{2}$$, the wavefunctions and the x operator are shown on the left side, and the integrand is shown on the right side. For this case we see that the integrand for x < 0 is the negative of the integrand for x > 0. This difference in sign means the net positive area for x > 0 is canceled by the net negative area for x < 0, so the total area and the transition moment integral are zero. We therefore conclude that the transition from n = 1 to n = 3 is forbidden.
Figure $$\PageIndex{2}$$:ย Functions (a) and integrand (b) of the transition moment integral for the transition from quantum state n = 1 to quantum state n = 3.
Exercise $$\PageIndex{3}$$
Write the expression or function for the integrand that is plotted on the right side of Figure 4.8 in terms of x and cosine functions. Use your function to explain why the integrand is zero at x = 0, why is it negative just above x = 0, and why as x goes from 0 to โ0.5, the integrand first is positive and then negative.
In spectroscopy some special terms are used to describe the symmetry properties of wavefunctions. The terms symmetric, gerade, and even describe functions like $$f(x) = x^2$$ and $$ฯ_1(x)$$ for the particle-in-a-box that have the property f(x) = f(-x), i.e. the function has the same values for x > 0 and for x < 0. The terms antisymmetric, ungerade, and odd describe functions like f(x) = x and $$ฯ_2(x)$$ for the particle-in-a-box that have the property f(x) = -f(-x), i.e. the function for x > 0 is has values that are opposite in sign compared to the function for x < 0. Gerade and ungerade are German words meaning even and odd and are abbreviated as g and u. Note that antisymmetric doesnโt mean non-symmetric.
If an integrand is u, then the integral is zero! It is zero because the contribution from x > 0 is cancelled by the contribution from x < 0, as shown by the example in Figure 4.8. An integrand will be u if the product of the functions comprising it is u. The following rules make it possible to quickly identify whether a product of two functions is u.
$g \cdot g = g, u \cdot u = g, g \cdot u = u \label {4-33}$
These rules are the same as those for multiplying +1 for g and -1 for u. The validity of these rules can be seen by examining Figures 4.7 and 4.8. If an integrand consists of more than two functions, the rules are applied to pairs of functions to obtain the symmetry of their product, and then applied to pairs of the product functions, and so forth, until one obtains the symmetry of the integrand.
Exercise $$\PageIndex{4}$$
Use Mathcad or some other software to draw graphs of $$x^2, -x^2, x^3, and -x^3$$ as a function of x. Which of these functions are g and which are u? Is the product function $$x^2 \cdot x^3 g$$ or u? How about $$x^2 \cdot -x^2 \cdot x^3 \text {and} x^2 \cdot x^3 \cdot -x^3$$?
Exercise $$\PageIndex{5}$$
Label each function in Figures 4.7 and 4.8 as g or u. Also label the integrands.
Exercise $$\PageIndex{6}$$
Use symmetry arguments to determine which of the following transitions between quantum states are allowed for the particle-in-a-box: n = 2 to 3 or n = 2 to 4.
Symmetry properties of functions allow us to identify when the transition moment integral and other integrals are zero. This symmetry-based approach to integration can be generalized and becomes even more powerful when concepts taken from mathematical Group Theory are used. With the tools of Group Theory, one can examine symmetry properties in three-dimensional space for complicated molecular structures. A group-theoretical analysis helps understand features in molecular spectra, predict products of chemical reactions, and simplify theoretical calculations of molecular structures and properties. | 2,099 | 8,745 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2018-05 | latest | en | 0.925547 |
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Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
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NCERT Exemplar CBSE Class 9 Mathematics Chapter-8: Quadrilaterals is available here. With this article, students can download the complete PDF of Chapter 8. This chapter includes various HOTS type questions based on some important properties of a quadrilateral especially a parallelogram.
Apr 4, 2017 09:50 IST
Importance of NCERT Exemplar Problems:
NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. Every year, many questions from the NCERT Exampler are directly asked in the board exams. So, itโs very important for students to practice the NCERT exemplar problems to prepare well for the board exams.
CBSE Syllabus for Class 9 Mathematics: 2017
Chapter 8 of NCERT Exemplar class 9 Maths is Quadrilaterals. It is one of the most important chapters of CBSE class 9th maths.
Diagonal of a parallelogram divides it into two congruent triangles In a parallelogram opposite sides are equal, & converse is also true In a parallelogram opposite angles are equal, & converse is also true A quadrilateral is a parallelogram if a pair of its opposite sides is parallel & equal In a parallelogram, the diagonals bisect each other & conversely In a triangle, the line segment joining the mid points of any two sides is parallel to the third side and in half of it & its converse
Most of the questions of this chapter of NCERT Exemplar are based on above mentioned topics.
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torque
Torque, moment, or moment of force is the tendency of a force to rotate an object about an axis, fulcrum, or pivot
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inclined planes
An inclined plane, also known as a ramp, is a flat supporting surface tilted at an angle, with one end higher than the other, used as an aid for raising or lowering a load
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vector
Vector graphics is the use of polygons to represent images in computer graphics
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displacement
A displacement is a vector that is the shortest distance from the initial to the final position of a point P
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9 cards in set
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displacement
A displacement is a vector that is the shortest distance from the initial to the final position of a point P
vector
Vector graphics is the use of polygons to represent images in computer graphics
inclined planes
An inclined plane, also known as a ramp, is a flat supporting surface tilted at an angle, with one end higher than the other, used as an aid for raising or lowering a load
torque
Torque, moment, or moment of force is the tendency of a force to rotate an object about an axis, fulcrum, or pivot
acceleration due to gravity
In physics, gravitational acceleration is the acceleration on an object caused by force of gravitation
Free fall
In Newtonian physics, free fall is any motion of a body where gravity is the only force acting upon it
scalar
A scalar in physics is a physical quantity that can be described by a single element of a number field such as a real number, often accompanied by units of measurement
free-body diagram
In physics a free body diagram is a graphical illustration used to visualize the applied forces, movements, and resulting reactions on a body in a steady state condition
centripetal force
A centripetal force is a force that makes a body follow a curved path
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Chapter 7 - View Definitions
### 7.1 Introduction
Consider the kinship dataset shown below. The situation here is the same as that described in Chapter 2. Art is the parent of Bob and Bea. Bob is the parent of Cal and Cam. Bea is the parent of Cat and Coe.
parent(art,bob) parent(art,bea) parent(bob,cal) parent(bob,cam) parent(bea,cat) parent(bea,coe)
Suppose now that we wanted to express information about the grandparent relation as well as the parent relation. As illustrated in the preceding chapter, we can do this by adding facts to our dataset. In this case, we would add the facts shown below. Art is the grandparent of Cal and Cam and Cat and Coe.
grandparent(art,cal) grandparent(art,cam) grandparent(art,cat) grandparent(art,coe)
Unfortunately, doing things this way is wasteful. The grandparent relation can be defined in terms of the parent relation, and so storing grandparent data as well as parent data is redundant.
A better alternative is to write rules to encode such definitions and to use these rules to compute the relations defined by these rules when needed. As we shall see in this chapter, we can write such definitions using rules similar those that we used to define goal relations in Chapter 3. For example, in the case above, rather than adding grandparent facts to our dataset, we can write the following rule, safe in the knowledge that we can use the rule to compute our grandparent data.
grandparent(X,Z) :- parent(X,Y) & parent(Y,Z)
In what follows, we distinguish two different types of relations - base relations and view relations. We define base relations by writing facts in a dataset, and we define view relations by writing rules in a ruleset. In our example, parent is a base relation, and grandparent is a view relation.
Given a dataset defining our base relations and a ruleset defining our view relations, we can use automated reasoning tools to derive facts about our view relations. For example, given the preceding facts about the parent relation and our rule defining the grandparent relation, we can compute the facts about the grandparent relation.
Using rules to define view relations has multiple advantages over encoding those relations in the form of datasets. First of all, as we have just seen, there is economy: if view relations are defined in terms of rules, we do not need to store as many facts in our datasets. Second, there is less chance of things getting out of sync, e.g. if we change the parent relation and forget to change the grandparent relation. Third, view definitions work for any number of objects; they even work for applications with infinitely many objects (e.g. the integers) without requiring infinite storage.
In this chapter, we introduce the syntax and semantics of view definitions, and we describe the important notion of stratification. In subsequent chapters, we look at many, many examples of using rules to define views. And, in Chapters 11 and 12, we look at some practical techniques for using rules to compute view relations.
### 7.2 Syntax
The syntax of view definitions is almost identical to that of queries as described in Chapter 3. The various types of constants are the same, and the notions of term and atom and literal are also the same. The main difference comes in the syntax of rules.
As before, a rule is an expression consisting of a distinguished atom, called the head and a conjunction of zero or more literals, called the body. The literals in the body are called subgoals. In what follows, we write rules as in the example shown below. Here, r(X,Y) is the head; p(X,Y) & ~q(Y) is the body; and p(X,Y) and ~q(Y) are subgoals.
r(X,Y) :- p(X,Y) & ~q(Y)
Despite these similarities, there are two important differences between query rules and rules used in view definitions. (1) In writing query rules, we use a single generic predicate (e.g. goal) in the heads of all of our query rules. By contrast, in view definitions, we use predicates for the relations we are defining (e.g. r in the example above). (2) In writing query rules, the subgoals of our rules can mention only predicates for relations described in the dataset. By contrast, in view definitions, subgoals can contain view predicates as well as predicates for base relations.
One benefit of the more flexible syntax in view definitions is that we can define multiple relations in a single set of rules. For example, the following rules define both the f relation and the m relation in terms of p and q.
f(X,Y) :- p(X,Y) & q(X) m(X,Y) :- p(X,Y) & ~q(X)
A second benefit is that we can use view relations in defining other view relations. For example, in the following rule, we use the view relation f in our definition of g.
g(X,Z) :- f(X,Y) & p(Y,Z)
A third benefit is that views can be used in their own definitions, thus allowing us to define relations recursively. For example, the following rules define a to be the transitive closure of p.
a(X,Z) :- p(X,Z) a(X,Z) :- p(X,Y) & a(Y,Z)
Unfortunately, our relaxed language allows for rulesets with some unpleasant properties. To avoid these problems, it is good practice to comply with some syntactic restrictions on our datasets and rulesets, viz. compatibility and stratification and safety (which we describe later).
A ruleset is compatible with a dataset if and only if (1) all symbols shared between the dataset and the ruleset are of the same type (symbol, constructor, predicate), (2) all constructors and predicates have the same arity, and (3) none of the predicates in the dataset appear in the heads of any rules in the ruleset.
### 7.3 Semantics
The semantics of view definitions is more complicated than the semantics of queries due to the possible occurrence of view predicates in subgoals; and, consequently, we take a slightly different approach.
To define the result of applying a set of view definitions to a dataset, we first combine the facts in a dataset with the rules defining our views into a joint set of facts and rules, hereafter called a closed logic program, and we then define the extension of that closed logic program as follows.
The Herbrand universe for a closed logic program is the set of all ground terms that can be formed from the symbols and constructors in the program. For a program without constructors, the Herbrand universe is finite (i.e. just the symbols). For a program with constructors, the Herbrand universe is infinite (i.e. the symbols and all compound terms that can be formed from those symbols).
The Herbrand base for a closed logic program is the set of all atoms that can be formed from the constants in the program. Said another way, it is the set of all facts of the form r(t1,...,tn), where r is an n-ary predicate and t1, ... , tn are ground terms.
An interpretation for a closed logic program is an arbitrary subset of the Herbrand base for the program. As with datasets, the idea here is that the factoids in the interpretation are assumed to be true, and those that are not included are assumed to be false.
A model of a closed logic program is an interpretation that satisfies the program. We define satisfaction in two steps - we first deal with the case of ground rules, and we then deal with arbitrary rules.
An interpretation ฮ satisfies a ground atom ฯ if and only if ฯ is in ฮ. ฮ satisfies a ground negation ~ฯ if and only if ฯ is not in ฮ. ฮ satisfies a ground rule ฯ :- ฯ1 & ... & ฯn if and only if ฮ satisfies ฯ whenever it satisfies ฯ1, ... , ฯn.
An instance of a rule in a closed logic program is a rule in which all variables have been consistently replaced by terms from the Herbrand universe, i.e. the set of ground terms that can be formed from the program's vocabulary. As before, consistent replacement means that, if an occurrence of a variable in a sentence is replaced by a given term, then all occurrences of that variable in that sentence are replaced by the same term.
Using the notion of instance, we can define the notion of satisfaction for arbitrary closed logic programs (with or without variables). An interpretation ฮ satisfies an arbitrary closed logic program ฮฉ if and only if ฮ satisfies every ground instance of every sentence in ฮฉ.
As an example of these concepts, consider the dataset shown below.
p(a,b) p(b,c) p(c,d) p(d,c)
And let's assume we have the following view definition.
r(X,Y) :- p(X,Y) & ~p(Y,X)
The following interpretation satisfies the closed logic program consisting of this dataset and ruleset. All of the facts in the dataset are included in the interpretation, and every conclusion that is required by our rule is included as well.
p(a,b) p(b,c) p(c,d) p(d,c) r(a,b) r(b,c)
By contrast, the following interpretations do *not* satisfy the program. The one on the left is missing conclusions from the rule; the one in the middle is missing the facts from the dataset; and the one on the right satisfies the rules but does not contain all of the facts from the dataset.
p(a,b) p(b,c) p(c,d) p(d,c)
r(a,b) r(b,c)
p(a,b) p(b,c) p(c,d) r(a,b) r(b,c) r(c,d)
On the other hand, the model shown above (the interpretation before these three non-models) is not the only interpretation that works. In general, a closed logic program can have more than one model, which means that there can be more than one way to satisfy the rules in the program. The following interpretations also satisfy our closed logic program.
p(a,b) p(b,c) p(c,d) p(d,c) r(a,b) r(b,c) r(c,d)
p(a,b) p(b,c) p(c,d) p(d,c) r(a,b) r(b,c) r(d,c)
p(a,b) p(b,c) p(c,d) p(d,c) r(a,b) r(b,c) r(c,d) r(d,c)
This seems odd in that there is no reason to include r(c,d) or r(d,c) in our interpretation. On the other hand, given our definition of satisfaction, there is no reason not to include them.
The reason that this seems wrong is that we normally want our definitions to be if and only if. We want to include among our conclusions only those facts that must be true. (1) All factoids in our dataset must be true. (2) All factoids required by our rules must be true. (3) All other factoids should be excluded.
This is the classic definition of what is know as logical entailment. A factoid is logically entailed by a closed logic program if and only if it is true in every model of the program, i.e. the set of conclusions is the intersection of all models of the program.
One way to ensure logical entailment is to take the intersection of all interpretations that satisfy our program. This guarantees that we get only those conclusions that are true in every model. For example, if we took the intersection of the three models shown above, we would get our original model.
Another approach is to concentrate on minimal models. A model ฮ of a logic program ฮฉ is minimal if and only if there is no proper subset of ฮ that is a model for ฮฉ. If there is just one minimal model of a closed logic program, then minimality guarantees logical entailment. For example, the first model given above is minimal, and every factoid in that model must be present in every model of the program.
Many closed logic programs have unique minimal models. For example, a closed logic program that does not contain any negations has one and only one minimal model. Unfortunately, closed logic programs with negation can have more than one minimal model.
One way of eliminating with ambiguities like this is to concentrate on programs that are semipositive or programs that are stratified with respect to negation. We define these types of programs and discuss their semantics in the next two sections.
### 7.4 Semipositive Programs
A semipositive program is one in which negations apply only to base relations, i.e. there are no subgoals with negated views.
The semantics of a semipositive program can be formalized by defining the result of applying the view definitions in the program to the facts in the program's dataset. We use the word extension to refer to the set of all facts that can be "deduced" in this way.
An instance of an expression (atom, literal, or rule) is one in which all variables have been consistently replaced by ground terms. For example, if we have a language with object constants a and b, then r(a) :- p(a,a), r(a) :- p(a,b), r(b) :- p(b,a), and r(b) :- p(b,b) are all instances of r(X) :- p(X,Y).
Given this notion, we define the result of the application of a single rule to a dataset as follows. Given a rule r and a dataset ฮ, let v(r,ฮ) be the set of all ฯ such that (1) ฯ is the head of an arbitrary instance of r, (2) every positive subgoal in the instance is a member of ฮ, and (3) no negative subgoal in the instance is a member of ฮ.
Using this notion, we define the result of repeatedly applying a semipositive program ฮฉ to a dataset ฮ as follows. Consider the sequence of datasets defined recursively as follows. ฮ0 = ฮ, and ฮn+1 = โชv(r0โช...โชฮn) for all r in ฮฉ. The closure of ฮฉ on ฮ is the union of the datasets in this sequence, i.e. C(ฮฉ,ฮ) = โชฮi.
To illustrate our definition, let's start with a dataset describing a small directed graph. In the sentences below, we use the edge predicate to record the arcs of one particular graph.
edge(a,b) edge(b,c) edge(c,d) edge(d,c)
Now, let's write some rules defining various relations on the nodes in our graph. Here, the relation p is true of nodes with an outgoing arc. The relation q is true of two nodes if and only if there is an edge from the first to the second or an edge from the second to the first. The relation r is true of two nodes if and only if there is an edge from the first to the second and an edge from the second to the first. The relation s is the transitive closure of the edge relation.
p(X) :- edge(X,Y) q(X,Y) :- edge(X,Y) q(X,Y) :- edge(Y,X) r(X,Y,Z) :- edge(X,Y) & edge(Y,Z) s(X,Y) :- edge(X,Y) s(X,Z) :- edge(X,Y) & s(Y,Z)
We start the computation by initializing our dataset to the edge facts listed above.
edge(a,b) edge(b,c) edge(c,d) edge(d,c)
Looking at the p rule and matching its subgoals to the data in our dataset in all possible ways, we see that we can add the following facts. In this case, every node in our graph has an outgoing edge, so there is one p fact for each node.
p(a) p(b) p(c) p(d)
Looking at the q rules and matching their subgoals to the data in our dataset in all possible ways, we see that we can add the following facts. In this case, we end up with the symmetric closure of the original graph.
q(a,b) q(b,a) q(b,c) q(c,b) q(c,d) q(d,c)
Looking at the r rule and matching the subgoals to the data in our dataset in all possible ways, we see that we can add the following facts.
r(c,d) r(d,c)
Finally, looking at the first rule for s and matching its subgoals to the data in our dataset in all possible ways, we see that we can add the following facts.
s(a,b) s(b,c) s(c,d) s(d,c)
However, we are not quite done. With the facts just added, we can use the second rule to derive the following additional data.
s(a,c) s(b,d) s(c,c) s(d,d)
Having done this, we can use the s rule again and can derive the following fact.
s(a,d)
At this point, none of the rules when applied to this collection of data produces any results that are not already in the set, and so the process terminates. The resulting collection of 25 facts is the extension of this program.
### 7.5 Stratified Programs
We say that a set of view definitions is stratified if and only if its rules can be partitioned into strata in such a way that (1) every stratum contains at least one rule, (2) the rules defining relations that appear in positive subgoals of a rule appear in the same stratum as that rule or in some lower stratum, and (3) the rules defining relations that appear in negative subgoals of a rule occur in some lower stratum (not the same stratum).
As an example, assume we have a unary relation p that is true of all of the objects in some application area, and assume that q is an arbitrary binary relation. Now, consider the ruleset shown below. The first two rules define r to be the transitive closure of q. The third rule defines s to be the complement of the transitive closure.
r(X,Y) :- q(X,Y) r(X,Z) :- q(X,Y) & r(Y,Z) s(X,Y) :- p(X) & p(Y) & ~r(X,Y)
This is a complicated ruleset, yet it is easy to see that it is stratified. The first two rules contain no negations at all, and so we can group them together in our lowest stratum. The third rule has a negated subgoal containing a relation defined in our lowest stratum, and so we put it into a stratum above this one, as shown below. This ruleset satisfies the conditions of our definition and hence it is stratified.
StratumRules
2s(X,Y) :- p(X) & p(Y) & ~r(X,Y)
1r(X,Y) :- q(X,Y)
r(X,Z) :- q(X,Y) & r(Y,Z)
By comparison, consider the following ruleset. Here, the relation r is defined in terms of p and q, and the relation s is defined in terms of r and the negation of s.
r(X,Y) :- p(X) & p(Y) & q(X,Y) s(X,Y) :- r(X,Y) & ~s(Y,X)
There is no way of dividing the rules of this ruleset into strata in a way that satisfies the definition above. Hence, the ruleset is not stratified.
The problem with unstratified rulesets is that there is a potential ambiguity. As an example, consider the rules above and assume that our dataset also included the facts p(a), p(b), q(a,b), and q(b,a). From these facts, we can conclude r(a,b) and r(b,a) are both true. So far, so good. But what can we say about s? If we take s(a,b) to be true and s(b,a) to be false, then the second rule is satisfied. If we take s(a,b) to be false and s(b,a) to be true, then the second rule is again satisfied. The upshot is that there is ambiguity about s. By concentrating exclusively on logic programs that are stratified, we avoid such ambiguities.
Although it is sometimes possible to stratify the rules in more than one way, this does not cause any problems. So long as a program is stratified with respect to negation, the definition just given produces the same extension no matter which stratification one uses.
Finally, we define the extension of a ruleset ฮฉ on dataset ฮ as follows. The definition relies on a decomposition of ฮฉ into strata ฮฉ1, ... , ฮฉk. Since there are only finitely many rules in a closed logic program and every stratum must contain at least one rule, there are only finitely many sets to consider (though the sets themselves might be infinite). With that in mind, let ฮ0 = ฮ, and let ฮn+1 = ฮn โช C(ฮฉn+1n). The extension of a program with k strata is just ฮk.
The extension of any closed logic program without constructors must be finite. Also, the extension of any non-recursive closed logic program must be finite. In both cases, it is possible to compute the extension in finite time. In fact, it is possible to show that the computation cost is polynomial in the size of the dataset.
In the case of recursive programs without constructors, the result must still be finite. However, the cost of computing the extension may be exponential in the size of the data, but the result can be computed in finite time.
For recursive programs with constructors, it is possible that the extension is infinite. In such cases, the extension is still well-defined; and, although we obviously cannot generate the entire extension in finite time, if a factoid is in the extension, it is possible to determine this in finite time.
The preceding section illustrates our method of computing extensions for semipositive programs. We now extend our example to show how to compute the extension of a stratified program.
Suppose we add the rule shown below to the program in the preceding section. The relation t here is the complement of the transitive closure of the edge relation.
t(X,Y) :- p(X) & p(Y) & ~s(X,Y)
Since this rule contains a negated relation, it would necessarily appear at a higher stratum than the s relation, and so we would not compute the conclusions until after we were done with s.
In this case, there are sixteen ways to satisfy the first two subgoals of our rule; and, as we saw in the preceding section, nine of them satisfy the s relation. The upshot is that the remaining seven facts satisfy the t relation. So, we can add these to our extension.
s(a,a) s(b,a) s(b,b) s(c,a) s(c,b) s(d,a) s(d,b)
Note that, in the presence of rules with negated subgoals, it is sometimes possible to stratify the rules in more than one way. The good news here is that does not cause any problems. So long as a program is stratified with respect to negation, the definition just given produces the same dataset no matter which stratification one uses. Consequently, there is just one extension for any safe, stratified logic program.
### Exercises
Exercise 7.1: Say whether each of the following expressions is a syntactically legal view definition.
(a) r(X,Y) :- p(X,Y) & q() (b) r(X,Y) :- p(X,Y) & ~q(Y,X) (c) ~r(X,Y) :- p(X,Y) & q(Y,X) (d) p(X,Y) & q(Y,X) :- r(X,Y) (e) p(X,Y) & ~q(Y,X) :- r(X,Y)
Exercise 7.2: Suppose we have a dataset with two symbols a and b and two unary relations p and q where all possible facts are true, i.e. the dataset is {p(a), p(b), q(a), q(b)}. Suppose we have a closed logic program consisting of this dataset and the rule r(X) :- p(X) & ~q(X).
(a) How many interpretations does this program have? (b) How many models does it have? (c) How many minimal models does it have?
Exercise 7.3: Say whether each of the following rulesets is stratified.
(a) r(X,Y) :- p(X,Y) & ~q(Y,X)r(X,Y) :- p(X,Y) & ~q(X,Y) (b) r(X,Z) :- p(X,Z) & q(X,Z)r(X,Z) :- r(X,Y) & ~r(Y,Z) (c) r(X,Z) :- p(X,Z) & ~q(X,Z)r(X,Z) :- r(X,Y) & r(Y,Z)
Exercise 7.4: What is v(r,ฮ) where r is r(X,Y) :- p(X,Y) & p(Y,X) and ฮ is the dataset shown below?
p(a,a) p(a,b) p(b,a) p(b,c)
Exercise 7.5: What is C(ฮฉ,ฮ) where ฮฉ is {r(X,Z) :- p(X,Z), r(X,Z) :- r(X,Y) & r(Y,Z)} and ฮ is the dataset shown below?
p(a,a) p(a,b) p(b,a) p(b,c)
Exercise 7.6: What is the extension of strata ฮฉ1 and ฮฉ2 on ฮ, where ฮฉ1 is {q(X) :- p(X,Y)}, where ฮฉ2 is {r(X,Y) :- p(X,Y) & ~q(Y)}, and where ฮ is the dataset shown below?
p(a,b) p(a,c) p(b,d) p(c,d) | 5,379 | 22,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2022-49 | latest | en | 0.880866 |
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1 Maths Plans Year 3
2
3 Contents Introduction Introduction 1 Using the Plans 2 Autumn 1 7 Autumn 2 21 Spring 1 43 Spring 2 59 Summer 1 75 Basic Skills 89 Progression 97 The Liverpool Maths team have developed a medium term planning document to support effective implementation of the new National Curriculum. In order to develop fluency in mathematics, children need to secure a conceptual understanding and efficiency in procedural approaches. Our materials highlight the importance of making connections between concrete materials, models and images, mathematical language, symbolic representations and prior learning. There is a key focus on the teaching sequence to ensure that children have opportunities to practise the key skills whilst building the understanding and knowledge to apply these skills into more complex application activities. For each objective, there is a breakdown which explains the key components to be addressed in the teaching and alongside this there are a series of sample questions that are pitched at an appropriate level of challenge for each year group. An additional section provides a list of key, basic skills that children must continually practise as they form the building blocks of mathematical learning. 1
4 Using the plans This is not a scheme but it is more than a medium term plan The programme of study has been split into four domains: Number Measurement Geometry Statistics These allocations serve only as a guide for the organisation of the teaching. Other factors such as term length, organisation of the daily maths lesson, prior knowledge and cross-curricular links may determine the way in which mathematics is prioritised, taught and delivered in your school. As a starting point, we have taken these domains and allocated them into five half terms: Autumn 1 Autumn 2 Spring 1 Spring 2 Summer 1 Year 3 Number - number and place value - addition and subtraction Number - multiplication and division - fractions Measurement Geometry - properties of shapes - position and direction Statistics 2
5 Using the plans Within each half term, are some new objectives and some continuous objectives: Year 3 New objectives Continuous objectives Autumn Autumn Spring Spring Summer The new objectives vary in length but cover the new learning for that half term, they will not appear again in their entirety. If the objective is in italics, it has been identified as an area that, once taught, should be re-visited and consolidated through basic skills sessions as these key skills form the building blocks of mathematical learning (see appendix 1). As before, the timings allocated and the organisation and frequency of delivery of these continuous objectives is flexible and will vary from school to school. Please note that Summer 2 has deliberately been left free for the testing period traditionally carried out at the end of summer 1. This also allows the flexibility to allocate time in Summer 2 to target specific areas identified through the assessment process as needing additional teaching time. There are 2 appendices attached: Appendix 1 - List of key basic skills with guidance notes Appendix 2 - Progression through the domains across the key stages The continuous objectives build up as you move through each half term. These objectives cover all the application aspects in mathematics. It is crucial that they are woven into the teaching continually during the year, so that once fluent in the fundamentals of mathematics, children can apply their knowledge rapidly and accurately to problem solving. 3
6 4
7 Autumn
8 6
9 YEAR 3 PROGRAMME OF STUDY DOMAIN 1 NUMBER NEW OBJECTIVES AUTUMN 1 NUMBER AND PLACE VALUE Objectives (statutory requirements) What does this mean? Example questions Notes and guidance (non-statutory) Count from 0 in multiples of 4,8,50 and 100; finding 10 or 100 more or less than a given number Recognise the place value of each digit in a three-digit number (hundreds, tens, ones) Count out loud forwards and backwards from different starting points and in steps of different sizes Be presented with any two-digit or three-digit number and be able to say the number that is 10 or 100 more or less Have an understanding of the number system up to three-digit numbers Understanding of zero as a place holder Make the links between the place value columns using apparatus to support (i.e. 100 is ten times bigger than 10) and understand the effects of multiplying by 10 and 100 Tell me all the multiples of 4 between 28 and 60 If I count in steps of 8 from zero, how many numbers will I have said by the time I get to 56? Tell me which multiples of 10 are between 386 and 421 How many multiples of 50 are there between 250 and 600? What is 10 more than 27? What is 100 less than 508? Give three digit cards (for example 3, 8, 0) can they make a number bigger than, smaller than, in between? Look at these numbers (for example 352, 405, 65, 511) tell me what the 5 digit represents in each 13 x 10 = 130 2cm x 100 = 200cm Pupils now use multiples of 2, 3, 4, 5, 8, 10, 50 and 100 They use larger numbers to at least 1000, applying partitioning related to place value using varied and increasingly complex problems, building on work in year 2 (for example, 146 = and 6, 146 = ) Using a variety of representations, including those related to measure, pupils should continue to count in ones, tens and hundreds, so that they become fluent in the order and place value of numbers to
10 Notes 8
11 Compare and order numbers up to 1000 Be able to talk about the relative size of numbers, a number bigger than, less than, between Order consecutive and non-consecutive numbers in ascending and descending order with a particular focus on crossing boundaries Repeating this with units of measure and money Present number lines in different ways and in different contexts (horizontal number line, vertical scale etc.) and place random numbers between two demarcations on a number line Using any number up to three digits, be able to round numbers to the nearest 10 and 100 Place 368 on a number line from 100 to 500 Order these numbers from smallest to largest and largest to smallest 102, 98, 101, 100, 99 Order these numbers from smallest to largest and largest to smallest 211, 193, 301, 209, 299 Order these lengths from smallest to largest and largest to smallest 101cm, 1m, 100cm, 100mm, 1m and 10cm On a number line with 300 and 500 marked, place the number 450 accurately Is 847 nearer to 800 or 900? Explain how you know Tell me all the numbers that round to 340 as the nearest 10 Tell me any three numbers that round to 700 as the nearest 100 9
12 Notes 10
13 Identify, represent and estimate numbers using different representations Have an understanding of the number system up to four-digit numbers in different contexts Children can build on place value knowledge by practising exchange (for example ten bundles of 10 for one 100) Be able to recognise and record numbers in words and figures Using apparatus such as Numicon, bundles of straws, Deines and place value counters, be able to estimate a number and then identify it Children can work with apparatus to represent numbers accurately Read and write numbers up to 1000 in numerals and in words Be able to recognise and record numbers in words and figures Listen to the numbers that I say and write them Alternate writing the figures and the words (e.g. 101, two hundred and fifteen, 300, ninety three) 11
14 Notes 12
15 NEW OBJECTIVES AUTUMN 1 ADDITION AND SUBTRACTION Add and subtract numbers mentally, including a three-digit number and ones a three-digit number and tens a three-digit number and hundreds HTU + U HTU + multiples of 10 HTU + multiples of 100 Building on knowledge of place value, identifying which digits will be changing Remember, this is a mental strategy, and although it may include informal jottings but would not be seen in books as a written calculation = = = Include similar questions for subtraction Pupils practise solving varied addition and subtraction questions. For mental calculations with two-digit numbers, the answers could exceed 100 Pupils use their understanding of place value and partitioning, and practice using columnar addition and subtraction with increasingly large numbers up to three digits to become fluent (see Mathematics Appendix 1) Add and subtract numbers with up to Teaching to be in line with school Calculation Policy Number line three digits, using formal written methods of columnar addition and subtraction Methods: Number line Expanded columnar Expanded columnar Column Column 13
16 Notes 14
17 Progression shown through: HTU + TU (no bridging) HTU + TU (bridging 10) HTU + TU (bridging 100) HTU + TU (bridging 10 and 100) HTU + HTU (no bridging) HTU + HTU (bridging 10) HTU + HTU (bridging 10 and 100) Same progression as above for subtraction Refer to the calculation sequence in the continuous objectives section to ensure children are given opportunities to apply these calculation skills 15
18 Notes 16
19 CONTINUOUS OBJECTIVES AUTUMN 1 Solve number problems and practical problems involving the ideas from number and place value Be able to answer word and reasoning problems linked to place value Emma has used these digit cards to make the number 250 How many different numbers can you make? Be able to use known facts in order to explore others: Include commutativity and inverse and other relationships between numbers (for example 4 x 8 is also 2 x 16 because one side of the multiplication is halved, the other side is doubled) Can you put all the numbers in order? If you made the number that is ten less than Emma s, which digit cards would you need? If you know that 4 x 8 = 32, how many other number facts can you tell me? 17
20 Notes 18
21 Estimate the answer to a calculation and use inverse operations to check answers Working with numbers up to three digits, ensure that children have opportunities to: Following the calculation sequence: Solve problems, including missing number problems, using number facts, place value and more complex addition and subtraction Estimate the answer Evidence the skill of addition and/or subtraction Prove the inverse using the skill of addition and/or subtraction Practice calculation skill including units of measure (m, cm, mm, kg, g, l, ml, hours, minutes and seconds) Solve missing box questions, including those where missing box represents a digit or represents a number Solve problems including those with more than one step, for numbers and measures Estimate Calculate Prove = 245 Calculate 368ml 123ml 368cm - = 245cm I have 245ml of water in one jug and 123ml in another jug, how much do I have altogether? I drink 200ml, how much is now left? Solve open-ended investigations Using the digit cards 1 to 9, make the smallest/biggest answer, an answer that is odd/even, divisible by 10 etc. 19
22 Notes 20
23 YEAR 3 PROGRAMME OF STUDY DOMAIN 1 NUMBER NEW OBJECTIVES AUTUMN 2 MULTIPLICATION AND DIVISION Objectives (statutory requirements) What does this mean? Example questions Notes and guidance (non-statutory) Recall and use multiplication and division facts for the 3, 4 and 8 multiplication tables Include chanting of multiplication tables both consecutively and non-consecutively Explore commutativity of multiplication Identify of multiples of 3, 4 and 8 Recall related division facts and explore the inverse relationship of multiplication and division Know that to multiply by 4, double and double again and that doubling this total is the same as multiplying by 8 and that the opposite is true for division 4 x 8 = This is the same as 8 x 4 = 32 is a multiple of 4 and 8 (and 2 as it is an even number) If 8 x 4 = 32, then 32 4 = 8 To find 8 x 4, double 8 and double again, for example 8, 16, 32 4) (make sure the children understand they are multiplying by 4) To find 32 4, halve 32 and halve again, for example 32, 16, 8 (make sure the children understand they are dividing by Pupils continue to practise their mental recall of multiplication tables when they are calculating mathematical statements in order to improve fluency. Through doubling, they connect the 2, 4 and 8 multiplication tables. Pupils develop efficient mental methods, for example, using commutativity and associativity (for example, = = = 240) and multiplication and division facts (for example, using 3 2 = 6, 6 3 = 2 and 2 = 6 3) to derive related facts (for example, 30 2 = 60, 60 3 = 20 and 20 = 60 3). 21
24 Notes 22
25 Write and calculate mathematical statements for multiplication and division using the multiplication tables that they know, including for two-digit numbers times onedigit numbers, using mental and progressing into formal written methods Ensure that children: Understand that multiplication is the same as repeated addition Understand that multiplication is commutative Write inverse statements Can derive and write related facts Can factorise in order to use known facts Teaching to be in line with school Calculation Policy Methods for X: Grouping on a number line to show progression from repeated addition Expanded (grid) Short Progression shown through: TU x U If 5 x 4 = 20 then 20 5 = 4 and 20 4 = 5 If 5 x 4 = 20, then 5 x 40 = 200 and 50 x 4 = 200 The factors of 20 are 1, 2, 4, 5, 10 and 20 Grouping Expanded (grid) Pupils develop reliable written methods for multiplication and division, starting with calculations of two-digit numbers by one-digit numbers and progressing to the formal written methods of short multiplication and division. Pupils solve simple problems in contexts, deciding which of the four operations to use and why. These include measuring and scaling contexts, (for example, four times as high, eight times as long etc.) and correspondence problems in which m objects are connected to n objects (for example, 3 hats and 4 coats, how many different outfits?; 12 sweets shared equally between 4 children; 4 cakes shared equally between 8 children). Methods for : Grouping on a number line to show progression from repeated subtraction Short Grouping on a number line to show links with multiplication Short Progression shown through: Grouping (repeated subtraction) TU U Refer to the calculation sequence in the continuous objectives section to ensure children are given opportunities to apply these calculation skills Grouping (using addition) Short 23
26 Notes 24
27 NEW OBJECTIVES AUTUMN 2 FRACTIONS Count up and down in tenths; recognise that tenths arise from dividing an object into 10 equal parts and dividing one-digit numbers or quantities by 10 Include different starting points, count forwards and backwards within 0 to 1, use the images as support From images, children can say what fraction is shaded Children can place fractions on a 0 1 number line and know which fractions are missing in a sequence (use fractions with the same denominator) Children understand that is the 10 same as dividing by 10 and the explicit link of fractions with division, use visual representations to support this As the children count, show images to support understanding Using different shapes that are divided into tenths, ask questions such as, How many tenths are shaded here? We have divided this shape into ten sections 1 and shaded of it, this shows that dividing by is the same as finding 10 1 Pupils connect tenths to place value, decimal measures and to division by 10 They begin to understand unit and non-unit fractions as numbers on the number line, and deduce relations between them, such as size and equivalence. They should go beyond the [0,1] interval, including relating this to measure Pupils understand the relation between unit fractions as operators (fractions of), and division by integers They continue to recognise fractions in the context of parts of a whole, numbers, measurements, a shape, and unit fractions as a division of a quantity Pupils practice adding and subtracting fractions with the same denominator through a variety of increasingly complex problems to improve fluency Children understand that when dividing a single digit by 10, the answer will always be in tenths 3 (for example 3 10 = ) This image represents three bars of chocolate 3 each divided by ten or 3 10 = 10 25
28 Notes 26
29 Recognise, find and write fractions of a discrete set of objects: unit fractions and non-unit fractions with small denominators Children understand fractions in different contexts: Fractions as part of the number system Fractions as part of a whole Fractions of a quantity Use the same fraction to illustrate this concept (e.g. 1 ) 4 Fraction as part of the number system: Children can place fractions on a number line demarcated 0-1 Fractions as part of a whole: whole shape divided into quarters 1 Fractions of a quantity: 12 divided into 4 1 groups or of 12 or When finding fractions of quantities, ensure questions include those relating fractions to measure and money. When finding non unit fractions of a quantity, children do so practically or pictorially, building on their knowledge gained when finding a unit fraction of a whole. Children should be able to answer questions such as, What is one quarter of 12cm? and What is one fifth of 1? From an image like this, children use their 1 3 knowledge that if equals 3 then must 4 4 equal 9. 27
30 Notes 28
31 Recognise and use fractions as numbers: unit fractions and nonunit fractions with small denominators Understand the place value of fractions in the number system (work with denominators 2, 3, 4, 5, 8 and 10 to build on work covered during the teaching of multiplication and division) From a set of fractions, children show knowledge of place value to position them accurately on a 0 1 number line, understanding the relationship between them Start to introduce units of measure and include numbers greater than one, for example, 2 1 Place of a metre or 1 metres on a 3 2 number line 1 Recognise and show, using diagrams, equivalent fractions with small denominators Using fraction bars (or any visual representation that shows fractions of a shape) children can identify fractions and can find pairs of equivalent fractions Children should start to explore the links between fraction families = 7 10 Build on the relationship between tenths and hundredths to show common fraction equivalents 29
32 Notes 30
33 Compare and order unit fractions, and fractions with the same denominators From a group of unit fractions with a denominator up to 10, children can compare the size of fractions and order them From a group of fractions with the same denominator, children can compare the size and order them 31
34 Notes 32
35 CONTINUOUS OBJECTIVES AUTUMN 2 Solve number problems and practical problems involving the ideas from number and place value Be able to answer word and reasoning problems linked to place value Emma has used these digit cards to make the number 250 How many different numbers can you make? Be able to use known facts in order to explore others: Can you put all the numbers in order? Include commutativity and inverse and other relationships between numbers (e.g. 4 x 8 is also 2 x 16 because one side of the multiplication is halved, the other side is doubled) If you made the number that is ten less than Emma s, which digit cards would you need? If you know that 4 x 8 = 32, how many other number facts can you tell me? Estimate the answer to a calculation and use inverse operations to check answers Solve problems, including missing number problems, using number facts, place value and more complex addition and subtraction Working with numbers up to three digits, ensure that children have opportunities to: Estimate the answer Evidence the skill of addition and/or subtraction Prove the inverse using the skill of addition and/or subtraction Practice calculation skill including units of measure (m, cm, mm, kg, g, l, ml, hours, minutes and seconds) Following the calculation sequence: Estimate Calculate Prove = 245 Calculate 368ml 123ml 33
36 Notes 34
37 Solve missing box questions, including those where missing box represents a digit or represents a number 368cm - = 245cm Solve problems including those with more than one step, for numbers and measures I have 245ml of water in one jug and 123ml in another jug, how much do I have altogether? I drink 200ml, how much is now left? Solve open-ended investigations Using the digit cards 1 to 9, make the smallest/biggest answer, an answer that is odd/even, divisible by 10 etc. Solve problems, including missing number problems, involving multiplication and division, including integer scaling problems and correspondence problems in which n objects are connected to m objects Working with numbers including up to two-digit multiplied by one-digit, ensure that children have opportunities to: Estimate the answer Evidence the skill of multiplication and/or division Prove the inverse using the skill of multiplication and/or division Practice calculation skill including units of measure (m, cm, mm, kg, g, l, ml, hrs, minutes and seconds) Following the calculation sequence: Estimate 32 x 3 Calculate 32 x 3 Prove 96 3 = 32 Calculate 32cm x 3 Solve missing box questions including those where missing boxes represents a digit or represents a number 96cm = 32cm 35
38 Notes 36
39 Solve problems including those with more than one step Three children each have 32ml of water, how much water is there altogether? = Solve open-ended investigations Solve correspondence problems (where there is a given relationship between the given variables) including finding all possibilities / combinations Use skills of doubling and halving to scale up and down to solve problems Using the digit cards 1 to 9, make the smallest/biggest answer, an answer that is odd/even etc. If there are 4 chocolate bars, how can I share them equally between 8 children? I have 3 skirts, and 5 tops, how many different outfits can I make? If 2 pizzas feed 3 children, how many pizzas are needed for 6 children? Solve problems involving fractions Building on the fraction work covered above, apply this knowledge into problem solving Which is bigger, or? Find a fraction that is bigger than, smaller than , between and, equivalent to Which two of these diagrams show fractions that are equivalent? 37
40 Notes 38
41 Shade these diagrams to show that = What fraction of this shape is shaded? = + = 1 - =
42 40
43 Spring
44 42
45 YEAR 3 PROGRAMME OF STUDY DOMAIN 2 MEASUREMENT NEW OBJECTIVES - SPRING 1 Objectives What does this mean? Example questions Notes and guidance (statutory requirements) (non-statutory) Measure, compare, add and subtract: lengths (m/cm/mm); mass (kg/g); volume/capacity (l/ml) Choosing appropriate units of measurement for the task Two of these sentences could be true, tick the two sentences that could be true: Adam s pencil is 12cm long Leah is 12 metres tall Katie s sister weighs 12kg Pupils continue to measure using the appropriate tools and units, progressing to using a wider range of measures, including comparing and using mixed units (for example, 1 kg and 200g) and simple equivalents of mixed units (for example, 5m = 500cm). Practical measuring to appropriate degrees of accuracy Record measurements in writing using correct units of measurement and compare them Knowing relationships and simple equivalents between given units for length, mass and volume/capacity Solve addition and subtraction calculations involving measure keeping the size of numbers in line with the progression outlined in the objective for addition and subtraction Jake s glass holds 12 litres of milk What would I use to measure the length of the hall? Weigh these items and write down their weight in order from smallest to largest How many metres are there in four and a half kilometres? What fraction of a litre is 500ml? In January, John was 105cm tall, he grew by 17cm, how tall is he now? The comparison of measures includes simple scaling by integers (for example, a given quantity or measure is twice as long or five times as high) and this connects to multiplication. Pupils continue to become fluent in recognising the value of coins, by adding and subtracting amounts, including mixed units, and giving change using manageable amounts. They record and p separately. The decimal recording of money is introduced formally in year 4. Pupils use both analogue and digital 12-hour clocks and record their times. In this way they become fluent in and prepared for using digital 24-hour clocks in year 4. 43
46 Notes 44
47 Start with same units of measurement progressing to different units of measurement (but not to include decimals) Compare measurements including scaling up and down Jane has 356cm of ribbon, Sally has 311cm of ribbon, how much more ribbon does Jane have than Sally? If there is 1litre 20 millilitres of water in one jug and 1litre 35 millilitres of water in another jug, how much water is there altogether? A glass holds 25ml of liquid, a jug holds five times as much liquid, how much does the jug hold? Measure the perimeter of simple 2-D shapes Perimeter is a continuous line forming the boundary of a closed geometric figure and its length can be measured Calculate a shape s perimeter by measuring its sides accurately and expressing the answer in centimetres Measurement can be by using a cm ruler accurately or a single length of string which can then be measured Use a ruler to find the perimeter of these shapes in centimetres Add and subtract amounts of money to give change, using both and p in practical contexts Solve addition and subtraction calculations keeping the size of numbers in line with the progression outlined in the objective for addition and subtraction. Pounds and pence are recorded separately (i.e. no decimal point) Start with same units of money progressing to mixed units of money If crisps cost 55p and cola costs 65p, what is the total cost? (recording the answer as 120p or 1 and 20p) 45
48 Notes 46
49 Calculate change from given amount using number line method Make sure examples are from whole pounds, using the method of counting on to find the difference A newspaper cost 70p and a chocolate bar cost 50p, John paid with a 2 coin, how much change did he get? John had 10, he spent 2 and 35p, how much money did he have left? Bridge up to 3 and then on to 10 (65p + 7 = 7 and 65p) Tell and write the time from an analogue clock, including using Roman numerals from I to XII, and 12-hour and 24-hour clocks From an analogue clock displaying either numbers 1 to 12 or Roman numerals I to XII, can read the time out loud and write it in words From a digital clock displaying 12-hour clock notation, tell and write the time Introduce the concept of a 24-hour clock linking it to 24 hours in a day Using this visual, ensure clock face is labelled with both numbers and Roman numerals and ask children to read and write the time Using this visual, children can say that the time is Six fifty five moving towards saying Five to seven 47
50 Notes 48
51 Estimate and read time with increasing accuracy to the nearest minute; record and compare time in terms of seconds, minutes, hours and o clock; use vocabulary such as am/pm, morning, afternoon, noon and midnight From a range of clock displays, children can read the time to the nearest minute When given a range of times with the same units or mixed units and using the vocabulary given, children can compare and order them From a range of clock displays, children can answer questions such as: What time is it? Is it am or pm? Which clock shows noon or midnight? Order these time durations from the shortest to the longest: 65 minutes, I hour 15 minutes, 1 ยฝ hours and fifteen minutes 2 minutes, 180 seconds, 45 seconds and 1 ยฝ minutes Know the number of seconds in a minute and the number of days in each month, year and leap year Permanent display for reference and linked to mental skills or basic skills to enable continuous practice Children can answer a range of questions and examples may be: How many seconds in two minutes? How many days in October? How many days in two leap years? Compare duration of events, for example to calculate the time taken by particular events or tasks When given the start and finish time, children can calculate how long something has taken Using this method, children can gather information to compare different time durations Katie left the house for a walk at 10:05 and returned at 10:40, for how long was she out? Mark got into the swimming pool at 3.30pm, he got out at 4.15pm, for how long was he in the pool? Who did more exercise? How many minutes more did he/she do? 49
52 Notes 50
53 CONTINUOUS OBJECTIVES SPRING 1 Solve number problems and practical problems involving the ideas from number and place value Be able to answer word and reasoning problems linked to place value Emma has used these digit cards to make the number 250 How many different numbers can you make? Be able to use known facts in order to explore others: Can you put all the numbers in order? Include commutativity and inverse and other relationships between numbers (e.g. 4 x 8 is also 2 x 16 because one side of the multiplication is halved, the other side is doubled) If you made the number that is ten less than Emma s, which digit cards would you need? If you know that 4 x 8 = 32, how many other number facts can you tell me? Estimate the answer to a calculation and use inverse operations to check answers Solve problems, including missing number problems, using number facts, place value and more complex addition and subtraction Working with numbers up to three digits, ensure that children have opportunities to: Estimate the answer Evidence the skill of addition and/or subtraction Prove the inverse using the skill of addition and/or subtraction Practice calculation skill including units of measure (m, cm, mm, kg, g, l, ml, hours, minutes and seconds) Following the calculation sequence: Estimate Calculate Prove = 245 Calculate 368ml 123ml 51
54 Notes 52
55 Solve missing box questions, including those where missing box represents a digit or represents a number 368cm - = 245cm Solve problems including those with more than one step, for numbers and measures I have 245ml of water in one jug and 123ml in another jug, how much do I have altogether? I drink 200ml, how much is now left? Solve open-ended investigations Using the digit cards 1 to 9, make the smallest/biggest answer, an answer that is odd/even, divisible by 10 etc. Solve problems, including missing number problems, involving multiplication and division, including integer scaling problems and correspondence problems in which n objects are connected to m objects Working with numbers including up to two-digit multiplied by one-digit, ensure that children have opportunities to: Estimate the answer Evidence the skill of multiplication and/or division Prove the inverse using the skill of multiplication and/or division Practice calculation skill including units of measure (m, cm, mm, kg, g, l, ml, hrs, minutes and seconds) Following the calculation sequence: Estimate 32 x 3 Calculate 32 x 3 Prove 96 3 = 32 Calculate 32cm x 3 Solve missing box questions including those where missing box represents a digit or represents a number 96cm = 32cm 53
56 Notes 54
57 Solve problems including those with more than one step Three children each have 32ml of water, how much water is there altogether? Solve open-ended investigations Using the digit cards 1 to 9, make the smallest/biggest answer, an answer that is odd/even etc. Solve correspondence problems (where there is a given relationship between the given variables) including finding all possibilities / combinations Use skills of doubling and halving to scale up and down to solve problems If there are 4 chocolate bars, how can I share them equally between 8 children? I have 3 skirts, and 5 tops, how many different outfits can I make? If 2 pizzas feed 3 children, how many pizzas are needed for 6 children? Solve problems involving fractions Building on the fraction work covered above, apply this knowledge into problem solving Which is bigger, or? Find a fraction that is bigger than, smaller than , between and, equivalent to Which two of these diagrams show fractions that are equivalent? 55
58 Notes 56
59 Shade these diagrams to show that = What fraction of this shape is shaded? = + = 1 - =
60 Notes 58
61 YEAR 3 PROGRAMME OF STUDY DOMAIN 3 GEOMETRY NEW OBJECTIVES - SPRING 2, PROPERTIES OF SHAPE Objectives What does this mean? Example questions Notes and guidance (statutory requirements) (non-statutory) Draw 2-D shapes and make 3-D shapes using modelling materials; recognise 3-D shapes in different orientations and describe them A polygon is a 2-D shape with straight sides. If all sides and angles are the same, it is a regular polygon When drawing 2-D shapes, rulers are used and lines are drawn with accuracy to a given length in cms Children work practically to construct 3-D shapes (with straws, polydron etc.) Building on the knowledge of the properties of shapes introduced in Year 2 (such as sides, edges, vertices and faces) children explore symmetry and use this knowledge to enable them to classify 2-D and 3-D shapes according to these criteria Draw a square where each side measures 4cm Children can construct shapes like this using modelling materials When presented with these shapes, children can classify them to satisfy a range of criteria Pupils knowledge of the properties of shapes is extended at this stage to symmetrical and non-symmetrical polygons and polyhedra. Pupils extend their use of the properties of shapes. They should be able to describe the properties of 2-D and 3-D shapes using accurate language, including lengths of lines and acute and obtuse for angles greater or lesser than a right angle. Pupils connect decimals and rounding to drawing and measuring straight lines in centimetres, in a variety of contexts. Recognise that angles are a property of shape or description of a turn An angle is the space (usually measured in degrees) between two intersecting lines. The angle measures the amount of turn between these lines 59
62 Notes 60
63 Children understand the definition of an angle Children understand that angles measure the amount of turn Children can identify angles in 2-D shapes Identify right angles, recognise that two right angles make a half turn, three make three quarters of a turn and four a complete turn; identify whether angles are greater than or less than a right angle Identify right angles in 2-D shapes and know that a right angle measures 90 Practically investigate turns and the right angles within them so that the children see the link between a quarter turn and a right angle Children can identify right angles from real-life photographs or the environment Through movement, children can make quarter turns, half turns, three quarter turns and full turns, match their movements to the number of right angles each represents and the corresponding measure in degrees (e.g. a half turn equals 180 ) When presented with these shapes, children can identify and mark the angles When presented with these shapes, children can identify and mark the right angles Face the window, make a half turn clock-wise: Where are you facing now? How many right angles have you turned through? How many degrees have you turned through? 61
64 Notes 62
65 When presented with a set of angles, children can classify them into bigger than, smaller than or equal to a right angle Identify horizontal and vertical lines and pairs of perpendicular and parallel lines Within 2-D drawings of shapes, children can identify horizontal and vertical lines and use this vocabulary with confidence Identify the horizontal and vertical lines in these 2-D shapes Within 2-D and 3-D shapes, children can identify perpendicular and parallel lines and use this vocabulary with confidence Identify the perpendicular and parallel lines in these pictures and photographs 63
66 Notes 64
67 CONTINUOUS OBJECTIVES SPRING 2 Solve number problems and practical problems involving the ideas from number and place value Be able to answer word and reasoning problems linked to place value Emma has used these digit cards to make the number 250 How many different numbers can you make? Be able to use known facts in order to explore others: Can you put all the numbers in order? Include commutativity and inverse and other relationships between numbers (e.g. 4 x 8 is also 2 x 16 because one side of the multiplication is halved, the other side is doubled) If you made the number that is ten less than Emma s, which digit cards would you need? If you know that 4 x 8 = 32, how many other number facts can you tell me? Estimate the answer to a calculation and use inverse operations to check answers Solve problems, including missing number problems, using number facts, place value and more complex addition and subtraction Working with numbers up to three digits, ensure that children have opportunities to: Estimate the answer Evidence the skill of addition and/or subtraction Prove the inverse using the skill of addition and/or subtraction Practice calculation skill including units of measure (m, cm, mm, kg, g, l, ml, hours, minutes and seconds) Following the calculation sequence: Estimate Calculate Prove = 245 Calculate 368ml 123ml 65
68 Notes 66
69 Solve missing box questions, including those where missing box represents a digit or represents a number 368cm - = 245cm Solve problems including those with more than one step, for numbers and measures I have 245ml of water in one jug and 123ml in another jug, how much do I have altogether? I drink 200ml, how much is now left? Solve open-ended investigations Using the digit cards 1 to 9, make the smallest/biggest answer, an answer that is odd/even, divisible by 10 etc. Solve problems, including missing number problems, involving multiplication and division, including integer scaling problems and correspondence problems in which n objects are connected to m objects Working with numbers including up to two-digit multiplied by one-digit, ensure that children have opportunities to: Estimate the answer Evidence the skill of multiplication and/or division Prove the inverse using the skill of multiplication and/or division Practice calculation skill including units of measure (m, cm, mm, kg, g, l, ml, hrs, minutes and seconds) Solve missing box questions including those where missing box represents a digit or represents a number Following the calculation sequence: Estimate 32 x 3 Calculate 32 x 3 Prove 96 3 = 32 Calculate 32cm x 3 96cm = 32cm 67
70 Notes 68
71 Solve problems including those with more than one step Solve open-ended investigations Three children each have 32ml of water, how much water is there altogether? = Using the digit cards 1 to 9, make the smallest/biggest answer, an answer that is odd/even etc. Solve correspondence problems (where there is a given relationship between the given variables) including finding all possibilities / combinations If there are 4 chocolate bars, how can I share them equally between 8 children? I have 3 skirts, and 5 tops, how many different outfits can I make? Use skills of doubling and halving to scale up and down to solve problems If 2 pizzas feed 3 children, how many pizzas are needed for 6 children? Solve problems involving fractions Building on the fraction work covered above, apply this knowledge into problem solving Which is bigger, or? Find a fraction that is bigger than, smaller than , between and, equivalent to Which two of these diagrams show fractions that are equivalent? 69
72 Notes 70
73 8 12 Shade these diagrams to show that = 2 3 What fraction of this shape is shaded? = + = 1 - =
74 72
75 Summer
76 74
77 YEAR 3 PROGRAMME OF STUDY DOMAIN 4 STATISTICS NEW OBJECTIVES - SUMMER 1 Objectives What does this mean? Example questions Notes and guidance (statutory requirements) (non-statutory) Interpret and present data using bar charts, pictograms and tables When given examples of constructed bar charts, children can identify the key features and answer simple questions (including examples where the scale is in increments of 2, 5 and 10) Using data given in a tally chart or frequency table, children can construct a bar chart with accurate labels and scaling (remember to include questions where the child is required to use scales in increments of 2, 5 and 10) Children should begin to understand which increments are the most appropriate for the data given and why How many people went into the supermarket? How many more people went into the post office than the shoe shop? Draw the missing bar in on the bar chart Pupils understand and use simple scales (for example, 2, 5, 10 units per cm) in pictograms and bar charts with increasing accuracy. They continue to interpret data presented in many contexts. When given examples of constructed pictograms, children can identify the key features and answer simple questions (including examples where one picture represents 4, 8, 50 and 100) How many girls are in the class? Children can construct a pictogram adhering to one of the above criteria, moving towards selecting own scaling There are 12 boys in the class, show this on the pictogram 75
78 Notes 76
79 Solve one-step and two-step questions such as How many more? and How many fewer? using information presented in scaled bar charts, pictograms and tables Building on understanding of bar charts, pictograms and tables, children apply these skills to answer questions How many fewer green cars than silver cars were seen? What colour was the second highest number of cars? True or false? Twice as many silver cars were seen as blue 77
80 Notes 78
81 CONTINUOUS OBJECTIVES SUMMER 1 Solve number problems and practical problems involving the ideas from number and place value Be able to answer word and reasoning problems linked to place value Emma has used these digit cards to make the number 250 How many different numbers can you make? Be able to use known facts in order to explore others: Can you put all the numbers in order? Include commutativity and inverse and other relationships between numbers (e.g. 4 x 8 is also 2 x 16 because one side of the multiplication is halved, the other side is doubled) If you made the number that is ten less than Emma s, which digit cards would you need? If you know that 4 x 8 = 32, how many other number facts can you tell me? Estimate the answer to a calculation and use inverse operations to check answers Solve problems, including missing number problems, using number facts, place value and more complex addition and subtraction Working with numbers up to three digits, ensure that children have opportunities to: Estimate the answer Evidence the skill of addition and/or subtraction Prove the inverse using the skill of addition and/or subtraction Practice calculation skill including units of measure (m, cm, mm, kg, g, l, ml, hours, minutes and seconds) Following the calculation sequence: Estimate Calculate Prove = 245 Calculate 368ml 123ml 79
82 Notes 80
83 Solve missing box questions, including those where missing box represents a digit or represents a number 368cm - = 245cm Solve problems including those with more than one step, for numbers and measures I have 245ml of water in one jug and 123ml in another jug, how much do I have altogether? I drink 200ml, how much is now left? Solve open-ended investigations Using the digit cards 1 to 9, make the smallest/biggest answer, an answer that is odd/even, divisible by 10 etc. Solve problems, including missing number problems, involving multiplication and division, including integer scaling problems and correspondence problems in which n objects are connected to m objects Working with numbers including up to two-digit multiplied by one-digit, ensure that children have opportunities to: Estimate the answer Evidence the skill of multiplication and/or division Prove the inverse using the skill of multiplication and/or division Practice calculation skill including units of measure (m, cm, mm, kg, g, l, ml, hrs, minutes and seconds) Solve missing box questions including those where missing box represents a digit or represents a number Following the calculation sequence: Estimate 32 x 3 Calculate 32 x 3 Prove 96 3 = 32 Calculate 32cm x 3 96cm = 32cm 81
84 Notes 82
85 Solve problems including those with more than one step Three children each have 32ml of water, how much water is there altogether? Solve open-ended investigations Using the digit cards 1 to 9, make the smallest/biggest answer, an answer that is odd/even etc. Solve correspondence problems (where there is a given relationship between the given variables) including finding all possibilities / combinations Use skills of doubling and halving to scale up and down to solve problems If there are 4 chocolate bars, how can I share them equally between 8 children? I have 3 skirts, and 5 tops, how many different outfits can I make? If 2 pizzas feed 3 children, how many pizzas are needed for 6 children? Solve problems involving fractions Building on the fraction work covered above, apply this knowledge into problem solving Which is bigger, or? Find a fraction that is bigger than, smaller than , between and, equivalent to Which two of these diagrams show fractions that are equivalent? 83
86 Notes 84
87 Shade these diagrams to show that = What fraction of this shape is shaded? = + = 1 - =
88 86
89 Basic Skills Appendix 1
90 88
91 YEAR 3 - BASIC SKILLS SKILLS GUIDANCE NOTES Count from zero in multiples of 4, 8, 50 and 100 using bridging strategies If children are not secure in reciting their 8 times tables they should use a as appropriate bridging strategy, e.g. ( = ) Recall multiplication facts and related division facts for 3, 4, 8 times tables Add and subtract a series of one-digit numbers Use knowledge of complements to 100 to find change from 1 Use knowledge of complements to 30 to calculate time within half an hour Find 10 or 100 more or less than a given number Read and write numbers up to 1000 Chanting forwards and backwards from different starting points as well as recalling random and non-consecutive multiplication and division facts Use skills such as number bonds, doubles, halves and near doubles e.g , and Know that there are 100 pence in one pound, use this to calculate 1 60p, 1 35p etc. Know that there are 30 minutes in half an hour, use this to calculate half an hour 10 minutes etc. Use structured apparatus such as base 10 or bundles of straws to illustrate the concept, include measures and money as context Use structured apparatus and place value grid to support conceptual understanding of place value Recognise the place value of each digit in a three-digit number What is the value of the 5 digit in these three numbers, 105, 523 and 258? Play place value games to reinforce this concept (e.g. if I add 20 to the number 523, which digit would change, what would the new digit be?) Compare and order numbers up to 1000 Comparing two three-digit numbers, children can say which is the bigger, the smaller, they also use the < and > signs. Children can order consecutive and non-consecutive numbers both forwards and backwards Partition numbers into place value columns Children can partition three-digit numbers (e.g. 364) Partition numbers in different ways 364 is and is also etc. Round any three-digit number to the nearest 10 and is approximately , is approximately
92 Notes 90
93 YEAR 3 - BASIC SKILLS SKILLS Use rounding to support estimation and calculation GUIDANCE NOTES is approximately so children can estimate the answer to be about 240 Use knowledge of place value to derive new addition and subtraction facts If I know = 15, I know = 150, = 1500 Use knowledge of inverse to derive associated addition and subtraction facts If I know = 20, then 20 5 must be 15, = 26, 26 8 = 18 and check answers Double any number between 1 and 50 and find all corresponding halves Use partitioning to double 35 so that it becomes double 30 + double 5. Halve 70 by partitioning it into 60 and 10 then halving 60, halving 10 and recombining Add and subtract mentally HTU ยฑ U, HTU ยฑ T and HTU ยฑ H Multiply any three-digit number by 10 and any two-digit number by 100 Divide any three-digit multiple of 10 by ten Children need to be secure with the skills of bridging, partitioning, doubling and know their number pairs up to ten to add and subtract mentally Understand that when multiplying a number by ten, its digits move one place to the left (as that place value column is ten times bigger) and zero is used as a place holder and when multiplying a number by 100, its digits move two places to the left and zeros are needed as place holders Understand that when dividing a number by ten, its digits move one place to the right and why zero as the place holder is no longer needed (eg = 12) Use knowledge of inverse to derive associated multiplication and division facts If I know 4 8 = 32, I know 8 x 4 = 32, 32 8 = 4, 32 4 = 8 Use known facts to derive nearby facts If I know = 16, I know = 17 If I know 5 8 = 40, I know 6 8 = 48 Use known facts to derive equivalent facts If I know = 16, I know = 16 If I know 5 8 = 40, I know = 40 91
94 Notes 92
95 YEAR 3 - BASIC SKILLS SKILLS GUIDANCE NOTES Count up and down in tenths Recall fraction pairs to 1 Children count forwards and backwards, from different starting points, consecutively and non-consecutively (e.g. ) For fractions with the same denominator, children can state the complement 2 3 to 1 (e.g. + = 1) Identify fractions greater or less than a half Children can say whether fractions such as and are more or less than a 6 6 half, they also use the < and > signs Identify equivalent fractions with small denominators Children see the links between fraction families and can say that, 4 4 and are equivalent Order fractions with the same denominator Comparing two fractions, children can say which is the bigger, the smaller, they also use the < and > signs. Children can order consecutive and non-consecutive fractions with the same denominator both forwards and backwards Tell and write the time from a 12-hour analogue clock and a clock with Roman numerals and a digital clock display Convert between money and measures including time Recognise right angles, straight angles, half and full turns and identify whether the turn is greater, less than or the same as a right angle Children can alternate between stating the time from a clock display and drawing or showing a clock display to match a given time Children can convert m to cm and cm to mm, kg to g, l to ml, hours to minutes and minutes to seconds using whole numbers as start points (i.e. no decimals) Children can identify simple angles from pictures or practical experiences they can also state the corresponding turns for these angles. Using pictures or working practically, children can compare two angles stating whether they are bigger or smaller than a right angle 93
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### PUTTERIDGE PRIMARY SCHOOL
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Progression in calculation: Abbey Park Middle School NH 2013/2014 amended for APMS by IT June 2014, Addition and Subtraction Step 1 Representations Recordings count reliably with from 1 to 20 What number
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Objectives/Activities Year R Key Vocabulary fair, sharing, the same, different, whole, equal, altogether, double, half. 1. Understanding fairness Finding equal and unequal groups. Realising when two groups
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Thousands Hundreds Tens Units Tenth Hundredth thousandth Calculations Policy Introduction This Calculations Policy has been designed to support teachers, teaching assistants and parents in the progression
### Autumn - 12 Weeks. Spring 11 Weeks. Summer 12 Weeks. Not As We Know It Limited 2014
A Year 5 Mathematician Planning of coverage and resources. Autumn - 12 Weeks Spring 11 Weeks Summer 12 Weeks TARGETS NHM YR 5 Collins 5 Abacus 5 Abacus 6 LA Prior Step NHM 4 CPM 4 Ginn 4 Number, place | 17,350 | 78,714 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2019-04 | latest | en | 0.916898 |
https://abhyas.co.in/2020/09/12/olympiad-math-preparation-class-2-number-sense-expanded-form/ | 1,680,311,261,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949694.55/warc/CC-MAIN-20230401001704-20230401031704-00534.warc.gz | 104,726,444 | 24,031 | # Olympiad Math Preparation โ Class 2
## Expanded Form
1. 8 hundred + 5 tens + 2 ones is an even number ?
2. 5 hundred + 5 tens + 7 ones is an even number ?
3. The next odd number which comes after 6 hundred + 2 tens + 3 ones __________________
4. The next even number which comes after 3 hundred + 9 tens + 8 ones is _____________________
5. 4 hundred + 8 tens = _________________________
6. Expanded form of 346
7. Which abacus represents 7 hundred + 3 tens + 2 ones ?
8. Which option is correct ?
9. Duggu has 55 balls and Dakshu has 47 balls . What is total number of balls in expanded form ? | 169 | 606 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2023-14 | longest | en | 0.822128 |
https://www.commercialmortgagerates1.com/10-1-arm-rates/ | 1,563,903,023,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529481.73/warc/CC-MAIN-20190723172209-20190723194209-00278.warc.gz | 653,230,541 | 7,953 | # 10/1 Arm Rates
### Contents
10/1 ARM โ Example โ Mortgage Calculator โ 10/1 ARM โ Example. A 10/1 ARM refers to an adjustable rate mortgage with an interest rate that is fixed for 10 years and that adjusts annually after that. In this example, we look at a 10/1 ARM for \$230,000 with a starting interest rate of 6.625%. It has a 2% cap on each adjustment. It has no floor rate and a lifetime maximum interest rate of.
Should More Borrowers Be Selecting ARMs Today? โ As an illustration, if the borrower takes the 5/1 ARM and doesnโt pay it off until year 6, he still comes out way ahead. The total cost of the ARM would actually be lower until year 9. On a 7/1 ARM, the borrower benefits if she is out of the mortgage before year 11, and on a 10/1 ARM before year 13.
What Is a 10/1 ARM? โ Financial Web โ finweb.com โ A 10/1 ARM (adjustable-rate mortgage) is often one of the best alternatives to choosing a 30-year fixed-rate mortgage. Here are the basics of the 10/1 ARM and what it can provide to you as a consumer. What Does 10/1 Mean? The 10 means that you will have 10 years of a fixed interest rate.
How high can an adjustable-rate mortgage go? โ When you start adding years until the first time the mortgage rate adjusts, you have what is called a hybrid ARM. Whether itโs a 3/1 (fixed for three years and then adjusting every one year), a 5/1, a.
10-Year ARM Mortgage Rates. A ten year adjustable rate mortgage, sometimes called a 10/1 ARM, is designed to give you the stability of fixed payments during the first 10 years of the loan, but also allows you to qualify at and pay at a lower rate of interest for the first ten years.
Now letโs discuss 10/1 ARM rates, which generally come cheaper than 30-year fixed rates. However, the interest rate may only be .125% or .25% cheaper because you get a fixed rate for a full decade before any adjustment takes place.
Fannie and Freddie impeding more affordable adjustable-rate mortgages โ and a jumbo 10/1 is at 4.25 percent. What I think: As mortgage rates ratchet up and home prices continue their skyward climb, homebuyers are obsessing about ways to bring their payments closer to.
Compare 10/1 Year ARM Mortgage Rates โ BestCashCow โ 10/1 Year ARM Mortgage Rates 2019. compare virginia 10/1 year arm Conforming Mortgage rates with a loan amount of \$250,000. Use the search box below to change the mortgage product or the loan amount. Click the lender name to view more information. Mortgage rates are updated daily.
Mortgage Center. Applying online is as easy as 1-2-3. You can apply now in as little as 20 minutes.. Apply Now or complete the application.; When you have completed the application, click submit and your information will be reviewed for approval. | 675 | 2,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-30 | latest | en | 0.935582 |
https://numberworld.info/39917 | 1,632,441,401,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057479.26/warc/CC-MAIN-20210923225758-20210924015758-00061.warc.gz | 495,657,078 | 3,808 | # Number 39917
### Properties of number 39917
Cross Sum:
Factorization:
Divisors:
1, 179, 223, 39917
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
9bed
Base 32:
16vd
sin(39917)
-0.076182627422059
cos(39917)
0.99709388087535
tan(39917)
-0.076404668490371
ln(39917)
10.594557577301
lg(39917)
4.6011578940281
sqrt(39917)
199.79239224755
Square(39917)
### Number Look Up
Look Up
39917 which is pronounced (thirty-nine thousand nine hundred seventeen) is a great number. The cross sum of 39917 is 29. If you factorisate 39917 you will get these result 179 * 223. The figure 39917 has 4 divisors ( 1, 179, 223, 39917 ) whith a sum of 40320. The number 39917 is not a prime number. 39917 is not a fibonacci number. The figure 39917 is not a Bell Number. The figure 39917 is not a Catalan Number. The convertion of 39917 to base 2 (Binary) is 1001101111101101. The convertion of 39917 to base 3 (Ternary) is 2000202102. The convertion of 39917 to base 4 (Quaternary) is 21233231. The convertion of 39917 to base 5 (Quintal) is 2234132. The convertion of 39917 to base 8 (Octal) is 115755. The convertion of 39917 to base 16 (Hexadecimal) is 9bed. The convertion of 39917 to base 32 is 16vd. The sine of 39917 is -0.076182627422059. The cosine of the number 39917 is 0.99709388087535. The tangent of the number 39917 is -0.076404668490371. The root of 39917 is 199.79239224755.
If you square 39917 you will get the following result 1593366889. The natural logarithm of 39917 is 10.594557577301 and the decimal logarithm is 4.6011578940281. I hope that you now know that 39917 is very amazing figure! | 609 | 1,748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-39 | latest | en | 0.806256 |
https://communities.sas.com/t5/Base-SAS-Programming/inverse-of-log-ratio-transformation/td-p/415906?nobounce | 1,529,506,620,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863650.42/warc/CC-MAIN-20180620143814-20180620163814-00129.warc.gz | 575,394,213 | 28,935 | Solved
Contributor
Posts: 31
# inverse of log-ratio transformation
simple issue driving me nuts!
I am trying to do a log-ratio transformation on a variable before regression. The variable is an index (so has values like 450, 560, 1200, etc.). i amย doing this in the following way:
``var1_T = log(var1/lag(var1))``
I want to bring the transformed value to its original form. So wanted to do inverse log like:
``var1_FCT_Orig = exp(var1_T);``
Not working. Can anyone tell me where I am doing this wrong?
Accepted Solutions
Solution
โ11-28-2017 08:03 PM
Super User
Posts: 23,267
## Re: inverse of log-ratio transformation
[ย Editedย ]
EXP is the reverse of LOG.
So it gives you:
``var1 / lag(var1)``
If you want Var1 then you need to multiple by the lag of VAR1 as well.
``````var1_t = log(var1 / lag(var1) );
var1_fct_orig = exp(var1_t) * lag(var1);``````
The question is, do you have the lagged value to work with as well...
eemrun wrote:
simple issue driving me nuts!
I am trying to do a log-ratio transformation on a variable before regression. The variable is an index (so has values like 450, 560, 1200, etc.). i amย doing this in the following way:
``var1_T = log(var1/lag(var1))``
I want to bring the transformed value to its original form. So wanted to do inverse log like:
``var1_FCT_Orig = exp(var1_T);``
Not working. Can anyone tell me where I am doing this wrong?
All Replies
Solution
โ11-28-2017 08:03 PM
Super User
Posts: 23,267
## Re: inverse of log-ratio transformation
[ย Editedย ]
EXP is the reverse of LOG.
So it gives you:
``var1 / lag(var1)``
If you want Var1 then you need to multiple by the lag of VAR1 as well.
``````var1_t = log(var1 / lag(var1) );
var1_fct_orig = exp(var1_t) * lag(var1);``````
The question is, do you have the lagged value to work with as well...
eemrun wrote:
simple issue driving me nuts!
I am trying to do a log-ratio transformation on a variable before regression. The variable is an index (so has values like 450, 560, 1200, etc.). i amย doing this in the following way:
``var1_T = log(var1/lag(var1))``
I want to bring the transformed value to its original form. So wanted to do inverse log like:
``var1_FCT_Orig = exp(var1_T);``
Not working. Can anyone tell me where I am doing this wrong?
Contributor
Posts: 31
## Re: inverse of log-ratio transformation
thanks! that worked perfectly.
Posts: 5,479
## Re: inverse of log-ratio transformation
You have everything you need in the transformed series, except the starting value, Example:
``````title 'Logistic Growth Curve Model of U.S. Population';
data uspop;
input pop @@;
retain year 1780;
year=year+10;
logPopRatio = log(pop/lag(pop));
label pop='U.S. Population in Thousands';
datalines;
3929 5308 7239 9638 12866 17069 23191 31443 39818 50155
62947 75994 91972 105710 122775 131669 151325 179323 203211
226542 248710
;
data usNewPop;
set uspop;
retain lagNewPop;
if _n_ = 1
then NewPop = 3929;
else NewPop = exp(logPopRatio)*lagNewPop;
lagNewPop = NewPop;
drop lagNewPop;
run;
proc print; run;``````
PG
โ This topic is solved. | 921 | 3,080 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2018-26 | latest | en | 0.863712 |
https://www.coursehero.com/file/6017359/2766-Pressure-measurements-UPD/ | 1,490,471,497,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189032.76/warc/CC-MAIN-20170322212949-00436-ip-10-233-31-227.ec2.internal.warc.gz | 900,001,579 | 172,099 | [276]6._Pressure_measurements_UPD
# [276]6._Pressure_measurements_UPD - Languages for...
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Languages for engineering communication - 6 Pressure measurement 6.1 Introduction Pressure is represented as a force per unit area and has the same units as parameter discussed previously in lecture No.4 as stress, F A ฯ = The new aspect of the same parameter which will be discussed in this lecture is the force exerted by a fluid per unit area on a containing wall. The pressure measurement devices for fluid are usually measure pressure in three different forms: absolute pressure, vacuum and gage pressure. Absolute pressure is the absolute value of the force per unit area exerted on the containing wall by a fluid. Gage pressure is the positive difference between the absolute pressure and the local atmospheric pressure. Vacuum is the negative difference between the local atmospheric pressure and the absolute pressure. Fig. 1 Relationship between pressure terms From these definitions in Fig. 1 we see that: 1. The absolute pressure is usually positive. 2. The gage pressure may be positive or negative (vacuum) 3. The vacuum is usually smaller than the local atmospheric pressure. 1
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Some common units of pressure Because a fluid pressure is result of a momentum exchange between the molecules of the fluid and containing wall, it depends on the total number of molecules striking the wall per unit time and on the average velocity v of the molecules. For an ideal gas the pressure is: 2 2 3 2 2 1 1 [ ] [ ] 3 sec m N p nmv kg Pascal m m = = = (1) where n is molecular density, molecules/unit volume; m is molecular mass and velocity v is root mean square velocity: 2 2 2 2 1 2 3 ....... 3 n rms v v v v kT v v n m + + + = = = (2) where T 0 K is the absolute temperature of the gas, k =1.3803 23 10 / J K - ร is Boltzmannโs constant, molecular mass, velocity is: 2 2 2 [ /sec] sec sec J K J Nm kgmm m m K kg kg kg kg = = = = = Since the pressure depends on collisions between the molecules, it should be dependent on the average distance passed by a molecule of radius r between collisions. The mean free path ฮป in an ideal gas is 2
2 2 8 r n ฮป ฯ = [ 3 2 m m m = ] (3) More n smaller is the mean free path. For air the Eq. (3) reduced to 5 2.27 10 [ ] T m p - = ร (4) Where 0 T K and pressure p must be in Pascal [ 2 / N m ].The mean free path decreases with increase in the gas pressure p (or the density n ) and with decrease in temperature T. The typical examples of mean free path in air at 20 C were calculated using Eq.(4): 1atm = 1.0132x 5 10 Pa - ( 29 8 1 6.564 10 atm m - = ร 1torr = 133.32 Pa ( 29 5 1 4.989 10 torr m - = ร 1 m ฮผ = 0.13332 Pa ( 29 1 0.04989 m m = 0.01 m = 1.332 3 10 Pa - ร ( 29 0.01 4.989 m m = At a standard conditions (p=1 atm) is quit small (about 0.1A), while in vacuum it reaches several meters length. 6.2 Vacuum devices with electrical output
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## This note was uploaded on 11/15/2010 for the course CHEE 638 taught by Professor Hampton during the Spring '10 term at MO Southern.
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[276]6._Pressure_measurements_UPD - Languages for...
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Ask a homework question - tutors are online | 953 | 3,573 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2017-13 | longest | en | 0.868575 |
https://us.metamath.org/mpeuni/ucnimalem.html | 1,713,051,312,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816853.44/warc/CC-MAIN-20240413211215-20240414001215-00395.warc.gz | 559,110,396 | 5,062 | Metamath Proof Explorer < Previousย ย Next > Nearby theorems Mirrorsย > ย Homeย > ย MPE Homeย > ย Th. Listย > ย ucnimalem Structured versionย ย Visualization versionย ย GIF version
Theorem ucnimalemย 22881
Description: Reformulate the ๐บ function as a mapping with one variable. (Contributed by Thierry Arnoux, 19-Nov-2017.)
Hypotheses
Ref Expression
ucnprima.1 (๐๐ โ (UnifOnโ๐))
ucnprima.2 (๐๐ โ (UnifOnโ๐))
ucnprima.3 (๐๐น โ (๐ Cnu๐))
ucnprima.4 (๐๐๐)
ucnprima.5 ๐บ = (๐ฅ๐, ๐ฆ๐ โฆ โจ(๐น๐ฅ), (๐น๐ฆ)โฉ)
Assertion
Ref Expression
ucnimalem ๐บ = (๐ โ (๐ ร ๐) โฆ โจ(๐นโ(1st๐)), (๐นโ(2nd๐))โฉ)
Distinct variable groups: ย ๐ฅ,๐,๐ฆ,๐น ย ๐,๐,๐ฅ,๐ฆ
Allowed substitution hints: ย ๐(๐ฅ,๐ฆ,๐) ย ๐(๐ฅ,๐ฆ,๐) ย ๐บ(๐ฅ,๐ฆ,๐) ย ๐(๐ฅ,๐ฆ,๐) ย ๐(๐ฅ,๐ฆ,๐) ย ๐(๐ฅ,๐ฆ,๐)
Proof of Theorem ucnimalem
StepHypRef Expression
1ย ucnprima.5 . 2 ๐บ = (๐ฅ๐, ๐ฆ๐ โฆ โจ(๐น๐ฅ), (๐น๐ฆ)โฉ)
2ย vexย 3496 . . . . . 6 ๐ฅ โ V
3ย vexย 3496 . . . . . 6 ๐ฆ โ V
42, 3op1stdย 7691 . . . . 5 (๐ = โจ๐ฅ, ๐ฆโฉ โ (1st๐) = ๐ฅ)
54fveq2dย 6667 . . . 4 (๐ = โจ๐ฅ, ๐ฆโฉ โ (๐นโ(1st๐)) = (๐น๐ฅ))
62, 3op2nddย 7692 . . . . 5 (๐ = โจ๐ฅ, ๐ฆโฉ โ (2nd๐) = ๐ฆ)
76fveq2dย 6667 . . . 4 (๐ = โจ๐ฅ, ๐ฆโฉ โ (๐นโ(2nd๐)) = (๐น๐ฆ))
85, 7opeq12dย 4803 . . 3 (๐ = โจ๐ฅ, ๐ฆโฉ โ โจ(๐นโ(1st๐)), (๐นโ(2nd๐))โฉ = โจ(๐น๐ฅ), (๐น๐ฆ)โฉ)
98mpomptย 7258 . 2 (๐ โ (๐ ร ๐) โฆ โจ(๐นโ(1st๐)), (๐นโ(2nd๐))โฉ) = (๐ฅ๐, ๐ฆ๐ โฆ โจ(๐น๐ฅ), (๐น๐ฆ)โฉ)
101, 9eqtr4iย 2845 1 ๐บ = (๐ โ (๐ ร ๐) โฆ โจ(๐นโ(1st๐)), (๐นโ(2nd๐))โฉ)
Colors of variables: wff setvar class Syntax hints: ย โ wiย 4 ย = wceqย 1530 ย โ wcelย 2107 ย โจcopย 4565 ย โฆ cmptย 5137 ย ร cxpย 5546 ย โcfvย 6348 ย (class class class)coย 7148 ย โ cmpoย 7150 ย 1st c1stย 7679 ย 2nd c2ndย 7680 ย UnifOncustย 22800 ย Cnucucnย 22876 This theorem was proved from axioms: ย ax-mpย 5 ย ax-1ย 6 ย ax-2ย 7 ย ax-3ย 8 ย ax-genย 1789 ย ax-4ย 1803 ย ax-5ย 1904 ย ax-6ย 1963 ย ax-7ย 2008 ย ax-8ย 2109 ย ax-9ย 2117 ย ax-10ย 2138 ย ax-11ย 2153 ย ax-12ย 2169 ย ax-extย 2791 ย ax-sepย 5194 ย ax-nulย 5201 ย ax-powย 5257 ย ax-prย 5320 ย ax-unย 7453 This theorem depends on definitions: ย df-biย 209 ย df-anย 399 ย df-orย 844 ย df-3anย 1083 ย df-truย 1533 ย df-exย 1774 ย df-nfย 1778 ย df-sbย 2063 ย df-moย 2616 ย df-euย 2648 ย df-clabย 2798 ย df-cleqย 2812 ย df-clelย 2891 ย df-nfcย 2961 ย df-ralย 3141 ย df-rexย 3142 ย df-rabย 3145 ย df-vย 3495 ย df-sbcย 3771 ย df-csbย 3882 ย df-difย 3937 ย df-unย 3939 ย df-inย 3941 ย df-ssย 3950 ย df-nulย 4290 ย df-ifย 4466 ย df-snย 4560 ย df-prย 4562 ย df-opย 4566 ย df-uniย 4831 ย df-iunย 4912 ย df-brย 5058 ย df-opabย 5120 ย df-mptย 5138 ย df-idย 5453 ย df-xpย 5554 ย df-relย 5555 ย df-cnvย 5556 ย df-coย 5557 ย df-dmย 5558 ย df-rnย 5559 ย df-iotaย 6307 ย df-funย 6350 ย df-fvย 6356 ย df-oprabย 7152 ย df-mpoย 7153 ย df-1stย 7681 ย df-2ndย 7682 This theorem is referenced by: ย ucnima ย 22882
Copyright terms: Public domain W3C validator | 1,629 | 2,541 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-18 | latest | en | 0.201203 |
http://icodeguru.com/Embedded/Hacker's-Delight/027.htm | 1,524,518,645,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946199.72/warc/CC-MAIN-20180423203935-20180423223935-00250.warc.gz | 152,108,917 | 3,881 | ### 2-19 Exchanging Registers
A very old trick is that of exchanging the contents of two registers without using a third [IBM]:
This works well on a two-address machine. The trick also works if is replaced by the logical operation (complement of exclusive or), and can be made to work in various ways with add's and subtract's:
Unfortunately, each of these has an instruction that is unsuitable for a two-address machine, unless the machine has "reverse subtract."
This little trick can actually be useful in the application of double buffering, in which two pointers are swapped. The first instruction can be factored out of the loop in which the swap is done (although this negates the advantage of saving a register):
#### Exchanging Corresponding Fields of Registers
The problem here is to exchange the contents of two registers x and y wherever a mask bit mi = 1, and to leave x and y unaltered wherever mi = 0. By "corresponding" fields, we mean that no shifting is required. The 1-bits of m need not be contiguous. The straightforward method is as follows:
By using "temporaries" for the four and expressions, this can be seen to require seven instructions, assuming that either m or m?/span> can be loaded with a single instruction and the machine has and not as a single instruction. If the machine is capable of executing the four (independent) and expressions in parallel, the execution time is only three cycles.
A method that is probably better (five instructions, but four cycles on a machine with unlimited instruction-level parallelism) is shown in column (a) below. It is suggested by the "three exclusive or" code for exchanging registers.
The steps in column (b) do the same exchange as that of column (a), but column (b) is useful if m does not fit in an immediate field but m?/span> does, and the machine has the equivalence instruction.
Still another method is shown in column (c) above [GLS1]. It also takes five instructions (again assuming one instruction must be used to load m into a register), but executes in only three cycles on a machine with sufficient instruction-level parallelism.
#### Exchanging Two Fields of the Same Register
Assume a register x has two fields (of the same length) that are to be swapped, without altering other bits in the register. That is, the object is to swap fields B and D, without altering fields A, C, and E, in the computer word illustrated below. The fields are separated by a shift distance k.
Straightforward code would shift D and B to their new positions, and combine the words with and and or operations, as follows:
Here, m is a mask with 1's in field D (and 0's elsewhere), and m' is a mask with 1's in fields A, C, and E. This code requires nine instructions and four cycles on a machine with unlimited instruction-level parallelism, allowing for two instructions to load the two masks.
A method that requires only seven instructions and executes in five cycles, under the same assumptions, is shown below [GLS1]. It is similar to the code in column (c) on page 39 for interchanging corresponding fields of two registers. Again, m is a mask that isolates field D.
The idea is that t1 contains B D in position D (and 0's elsewhere), and t2 contains B D in position B. This code, and the straightforward code given earlier, work correctly if B and D are "split fields"ๆขฉhat is, if the 1-bits of mask m are not contiguous.
#### Conditional Exchange
The exchange methods of the preceding two sections, which are based on exclusive or, degenerate into no-operations if the mask m is 0. Hence, they can perform an exchange of entire registers, or of corresponding fields of two registers, or of two fields of the same register, if m is set to all 1's if some condition c is true, and to all 0's if c is false. This gives branch-free code if m can be set up without branching. | 860 | 3,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-17 | latest | en | 0.947152 |
https://www.sambuz.com/doc/tutorial-on-smt-solvers-ppt-presentation-760766 | 1,695,908,096,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510412.43/warc/CC-MAIN-20230928130936-20230928160936-00705.warc.gz | 1,061,327,209 | 13,274 | # Tutorial on SMT Solvers Combinatorial Problem Solving (CPS) Enric - PowerPoint PPT Presentation
## Tutorial on SMT Solvers Combinatorial Problem Solving (CPS) Enric Rodr guez-Carbonell April 23, 2019 SMT Solvers SMT solvers take as input a (quantifier-free) first-order logic formula F over a background theory T , and return: sat (+
1. Tutorial on SMT Solvers Combinatorial Problem Solving (CPS) Enric Rodrยด ฤฑguez-Carbonell April 23, 2019
2. SMT Solvers SMT solvers take as input a (quantifier-free) first-order logic formula F โ over a background theory T , and return: sat (+ model): if F is satisfiable โ unsat : if F is unsatisfiable โ We will be using Z3 : http://z3.codeplex.com โ (developed by L. de Moura and N. Bjorner at Microsoft Research) Usage: z3 [ <options> ] <input> โ Some options: โ -stm2 : use parser for SMT-LIB 2 input format โ -st : display statistics โ -rs:<seed> : set random seed โ -h : help, shows all options โ 2 / 10
3. Input Format: SMT-LIB 2 We will be using a small subset of this language. โ For going beyond: Tutorial (standard version 2.0): โ http://smtlib.github.io/jSMTLIB/SMTLIBTutorial.pdf Full standard (standard version 2.5): โ http://smtlib.cs.uiowa.edu/papers/smt-lib-reference-v2.5-r2015-06-28.pdf 3 / 10
4. Input Format: SMT-LIB 2 First, directives. E.g., asking models to be reported: โ (set -option :produce -models true) Second, set background theory: โ (set -logic QF_LIA ) Standard theories of interest to us: โ QF_LRA : quantifier-free linear real arithmetic โ QF_LIA : quantifier-free linear integer arithmetic โ QF_RDL : quantifier-free real difference logic โ QF_IDL : quantifier-free integer difference logic โ SMT-LIB 2 does not allow to have mixed problems โ (although some solvers support it outside the standard) 4 / 10
5. Input Format: SMT-LIB 2 Third, declare variables. โ E.g., integer variable x : (declare -fun x () Int) E.g., real variable z 1 3 : (declare -fun z_1_3 () Real ) 5 / 10
6. Input Format: SMT-LIB 2 Fourth, assert formula. โ Expressions should be written in prefix form: โ ( < operator > < arg 1 > ... < arg n > ) (assert (and (or (<= (+ x 3) (* 2 u) ) (>= (+ v 4) y) (>= (+ x y z ) 2) ) (= 7 (+ (ite (and (<= x 2 ) (<= 2 (+ x 3 (- 1)))) 3 0) (ite (and (<= u 2 ) (<= 2 (+ u 3 (- 1)))) 4 0) ) ) ) ) 6 / 10
7. Input Format: SMT-LIB 2 and , or , + have arbitrary arity โ - is unary or binary โ * is binary โ ite is the if-then-else operator (like ? in C, C++, Java). โ Let a be Boolean and b , c have the same sort S . Then (ite a b c) is the expression of sort S equal to: b if a holds โ c if a does not hold โ 7 / 10
8. Input Format: SMT-LIB 2 Finally ask the SMT solver to check satisfiability ... โ (check -sat) ... and report the model โ (get -model) Anything following a ; up to an end-of-line is a comment โ 8 / 10
9. Input Format: SMT-LIB 2 (set -option :produce -models true) (set -logic QF_LIA ) (declare -fun x () Int) (declare -fun y () Int) (declare -fun z () Int) ; This is an example (declare -fun u () Int) (declare -fun v () Int) (assert (and (or (<= (+ x 3) (* 2 y) ) (>= (+ x 4) z) ) ) ) (check -sat) (get -model) 9 / 10
10. Output Format 1st line is sat or unsat โ If satisfiable, then comes a description of the solution in a model โ expression, where the value of each variable is given by: ( define โ fun < variable > () < sort > < value > ) Example: โ sat (model (define -fun y () Int 0) (define -fun x () Int (- 3)) (define -fun z () Int 2) ) 10 / 10
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https://www.raiseupwa.com/lifehacks/how-do-you-teach-numbers-to-reception/ | 1,713,836,642,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00099.warc.gz | 846,424,511 | 32,849 | Categories :
## How do you teach numbers to reception?
What your child will learn. In Reception, your child will learn to: Count reliably with numbers from 1 to 20, place them in order and say which number is one more or one less than a given number. Use quantities and objects to add and subtract 2 single-digit numbers and count on or back to find the answer.
What maths do they do in reception?
Reception maths โ your child will be:
โข Counting up to ten and beyond, using cardinal numbers.
โข Recognising the numbers 1 to 9.
โข Counting aloud in ones, twos, fives, tens.
โข Estimating a number of objects and checking by counting.
โข Matching and comparing the number of objects in two groups.
### What is a fun way to teach numbers?
20 Fun Ways to Learn How to Count
โข Mouse Counts: Counting Game.
โข One Duck Stuck: Muck Sensory Play and Counting.
โข Sort It Out: Color Sorting Game.
โข Ten Black Dots: Counting and Grouping Circles.
โข Ten Red Apples: Counting Trees.
โข Bear Counts: Counting on Paws.
โข The Doorbell Rang: Counting Cookies Math Center.
How do I teach math in Eyfs?
As well as carefully planned activities, you can explore maths in lots of different ways:
1. Using songs, stories, puzzles and games to help children explore maths concepts in a fun way.
2. Reinforcing maths vocabulary and prompting them to discuss learning.
3. Highlighting daily routines that can help support maths development.
#### Should a 4 year old be able to write numbers?
The average 4-year-old can count up to ten, although he may not get the numbers in the right order every time. One big hang-up in going higher? Those pesky numbers like 11 and 20. The irregularity of their names doesnโt make much sense to a preschooler.
How do I teach early numbers?
What teaching strategies promote early number sense?
1. Work with concrete materials and familiar ideas.
2. Compose and recompose different arrangements and representations of number.
3. Discuss and share their discoveries and solutions.
4. Investigate the realistic uses of number in their everyday world.
## What math concepts should a 4-year-old know?
COUNTING: Preschoolers and Kindergartners can often count up to ten, sometimes 20, objects. They can identify the larger of two numbers, and they make fewer mistakes when counting in that they repeat or skip numbers less frequently. By first grade, they may be able to count up to 100.
What math should a 6 year old know?
Because six-year-olds can count to higher numbers, they can also be challenged to work on higher number operations. School-aged children focus on addition and subtraction at first, and then eventually reach multiplication (in the form of skip counting) and division (in the form of equal shares).
### How do you promote math in early years?
Doing maths together at home
1. Talking about maths. It is important for children to develop specific language skills related to maths.
2. Counting. Counting is one of the first experiences of maths for young children.
3. Counting everyday.
4. Hunting for numbers.
5. Using playing cards.
6. Playing shop.
7. Playing games.
8. Playing with shapes.
What a good maths lesson looks like?
A โgood maths lessonโ will always necessarily be a part of a sequence of lessons or learning experiences which will ideally build mathematical understanding, improve fluency, build problem solving capacity and then develop mathematical reasoning skills.
#### What are some activities to build number sense?
Activities 1 Give students a number and a group of small objects to count on. 2 Play games with small objects and decks of cards where counting on is needed. 3 Play a group counting game called โAround the World.โ Say a number, and each person counts on and says the next number as you go around the circle.
What are some activities that children can do to learn numbers?
This story about some troublesome dogs encourages children to find and model doubles of different numbers. In this activity, children can practise reading numbers and counting items in order to help Owl pack for his holiday. In this task, children will learn different ways of representing the same number.
## Are there any free games for reception kids?
There are thousands of free childrenโs games online that teaches how to count, identify letters, distinguish rhyming words, and more all in an engaging virtual setting. So where should you start? Here is a list of 10 top and free numeracy games which focus on the skills your child needs in reception.
How are number bonds used in the classroom?
Number bonds are a simple but incredibly helpful concept for kids learning their math facts. They demonstrate that any number can be broken down (decomposed) into smaller parts that make up a whole, and theyโre an excellent lead-in to addition and subtraction. Here are some of our favorite activities to try in the classroom or at home. | 1,042 | 4,898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-18 | latest | en | 0.913869 |
https://www.bartleby.com/questions-and-answers/suppose-in-our-model-lifespans-increased-due-to-a-productivity-in-crease.-what-would-likely-happen-t/0359cfc3-fdb1-4a80-8785-391849e48d25 | 1,580,182,365,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251773463.72/warc/CC-MAIN-20200128030221-20200128060221-00188.warc.gz | 760,866,619 | 23,851 | # Suppose in our model, lifespans increased due to a productivity in- crease. What would likely happen to hours worked and leisure over the lifetime of a representative consumer.(a) ย Letโs consider one of two common ways of implementing this change. Suppose the increase raised both lifespans and the amount of years consumers are healthy enough to work by the same amount, how would that likely affect the measured numbered of hours work by prime aged adults within a given year? What would likely hap- pen to the retirement age?(b) ย Letโs consider the other way to implement it. Suppose the change in lifetimes came about changes in technology that delayed death but did not extend the amount of years consumers are healthy enough to work. What would likely happen to hours worked within a year? What would happen to the retirement age?(c) ย In the US, the generalized stylized fact is that the actual age of retirement is not increasing. Nor are hours worked per week. Why are both of the above predictions likely off? That is, what is our current model missing?
Question
Asked Dec 9, 2019
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1. Suppose in our model, lifespans increased due to a productivity in- crease. What would likely happen to hours worked and leisure over the lifetime of a representative consumer.
1. (a) ย Letโs consider one of two common ways of implementing this change. Suppose the increase raised both lifespans and the amount of years consumers are healthy enough to work by the same amount, how would that likely affect the measured numbered of hours work by prime aged adults within a given year? What would likely hap- pen to the retirement age?
2. (b) ย Letโs consider the other way to implement it. Suppose the change in lifetimes came about changes in technology that delayed death but did not extend the amount of years consumers are healthy enough to work. What would likely happen to hours worked within a year? What would happen to the retirement age?
3. (c) ย In the US, the generalized stylized fact is that the actual age of retirement is not increasing. Nor are hours worked per week. Why are both of the above predictions likely off? That is, what is our current model missing?
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## Expert Answer
Step 1
a.
The change in the technology when leads to an increase in the lifespan as well as the amount of years that the consumers are healthy enough to work the same amount, it would impact by increasing the work hours of the labors. The labors will have more periods in front of them to work in the economy and make money which means that they will have more chances to make money which can be used for their expenditure after the period in which they are not able to work. Thus, the prime aged workers would take more hours of work since the increased lifespan has maintained their health which helps the prime aged people to work by the same amount. This would increase the retirement age probably.
Step 2
b.
The advancement of the technology when only prolongs the period of death of the individuals and not maintains their health means that the prime aged individuals will not be having the capabilities to work in the economy. This means that they will have to live more years without the energy to work and earn. This has its impac...
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Tagged in | 744 | 3,538 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2020-05 | latest | en | 0.97605 |
http://openstudy.com/updates/510fe1c3e4b0d9aa3c47e6d1 | 1,448,996,418,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398468971.92/warc/CC-MAIN-20151124205428-00319-ip-10-71-132-137.ec2.internal.warc.gz | 166,381,579 | 11,794 | Bri_Love 2 years ago In 2005, the total waste generated in a certain country was 7.125 x 10^10 pounds. Also in 2005, the countries population was 4.48 x 10^6 people. Determain the garbage per capita (per person) in that country in the year 2005.
1. ShotGunGirl
Well the answer is the amount of trash divided by the amount of people.
2. ShotGunGirl
First, you would convert the numbers to from Scientific notation. \[7.125 x 10^{10}\] would be 71,250,000,000 \[4.48 x 10^{6}\] would be 4,480,000
3. ShotGunGirl
Then you would divide the two new numbers, trash/people \[71,250,000,000\div4,480,000\]
4. Bri_Love
5. Bri_Love
Whitch then if I round it it would be 9207.59?
6. Bri_Love
No thats wrong...
7. Bri_Love
8. ShotGunGirl
If this is hard to understand, then use easier numbers. For example If there are 20 bubbles in total and 4 bubble blowers, how many bubbles did each peson blow? Then, you might see the anser is oblviously 5. Correct? So, in your case, \[71,250,000,000\div4,480,000\] equals 15904.017851
9. ShotGunGirl
Sorry I couldn't find my calculator :p
10. Bri_Love
ok so it would be 15904.02? cause i have to round it?
11. ShotGunGirl
http://www.wolframalpha.com/input/?i=71%2C250%2C000%2C000%2F4%2C480%2C000 If you need the answer in a fraction you can find it here.
12. ShotGunGirl
Yes :)
13. Bri_Love
grr it says its wrong! Is that in scientific notation?
14. ShotGunGirl
You need and answer in scientific notation?
15. Bri_Love
yes
16. ShotGunGirl
Well after you round it to 15904.02 it would be.... hold on one sec.
17. Bri_Love
ok
18. ShotGunGirl
1.590402 x 10^4
19. ritez
In scientific notation answer will be 1.590 x 10^4
20. Bri_Love
Thank you!
21. ShotGunGirl
Anytime :D
22. Bri_Love
The answer was 1.59 x 10^4 cool thanks!
23. ShotGunGirl
Well sorry I didn't take you all the way there :( A special thanks to @ritez for helping me :D
24. ritez
@ShotGunGirl you are always welcome...enjoy! | 635 | 1,960 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2015-48 | longest | en | 0.865399 |
https://wiki.ubc.ca/Course:CPSC_320/Midterm_1_Reference_Sheet | 1,715,991,709,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057216.39/warc/CC-MAIN-20240517233122-20240518023122-00242.warc.gz | 551,342,715 | 20,076 | # Course:CPSC 320/Midterm 1 Reference Sheet
## CPSC 320 2009W2 Exam Reference Sheet
Do not remove: This reference sheet is the appendix for Midterm #1. Only the first 4 printed pages will be used; so be compact!
โข ${\displaystyle f(n)\in O(g(n))}$ iff ${\displaystyle \exists c\in \mathbf {R^{+}} ,\exists n_{0}\in \mathbf {Z^{+}} ,\forall n\in \mathbf {Z^{+}} ,n\geq n_{0}\rightarrow f(n)\leq cg(n)}$
โข ${\displaystyle f(n)\in \Omega (g(n))}$ iff ${\displaystyle g(n)\in O(f(n))}$, ${\displaystyle \exists c\in \mathbf {R^{+}} ,\exists n_{0}\in \mathbf {Z^{+}} ,\forall n\in \mathbf {Z^{+}} ,n\geq n_{0}\rightarrow 0\leq cg(n)\leq f(n)}$
โข ${\displaystyle f(n)\in \Theta (g(n))}$ iff ${\displaystyle f(n)\in O(g(n))}$ and ${\displaystyle g(n)\in O(f(n))}$, ${\displaystyle \exists c_{1},c_{2}\in \mathbf {R^{+}} ,\exists n_{0}\in \mathbf {Z^{+}} ,\forall n\in \mathbf {Z^{+}} ,n\geq n_{0}\rightarrow 0\leq c_{1}g(n)\leq f(n)\leq c_{2}g(n)}$
โข ${\displaystyle f(n)\in o(g(n))}$ iff ${\displaystyle \forall c\in \mathbf {R^{+}} ,\exists n_{0}\in \mathbf {Z^{+}} ,\forall n\in \mathbf {Z^{+}} ,n\geq n_{0}\rightarrow f(n)
โข ${\displaystyle f(n)\in \omega (g(n))}$ iff ${\displaystyle g(n)\in o(f(n))}$, ${\displaystyle \forall c\in \mathbf {R^{+}} ,\exists n_{0}\in \mathbf {Z^{+}} ,\forall n\in \mathbf {Z^{+}} ,n\geq n_{0}\rightarrow 0\leq cg(n)
Also, if ${\displaystyle \displaystyle \lim _{n\to \infty }{\frac {f(n)}{g(n)}}=}$
โข ${\displaystyle \infty }$, then ${\displaystyle f(n)\in \omega (g(n))}$
โข a non-zero constant, then ${\displaystyle f(n)\in \Theta (g(n))}$
โข ${\displaystyle 0}$, then ${\displaystyle f(n)\in o(g(n))}$
(Plus bear in mind that (1) L'Hopital's rule may be handy, and (2) the limit is not always well-defined!)
L'Hopital's Rule:
If ${\displaystyle \displaystyle \lim _{n\to \infty }{\frac {f(n)}{g(n)}}={\frac {0}{0}}}$ or ${\displaystyle \displaystyle \lim _{n\to \infty }{\frac {f(n)}{g(n)}}={\frac {\infty }{\infty }}}$, then ${\displaystyle \displaystyle \lim _{n\to \infty }{\frac {f(n)}{g(n)}}={\frac {f(n)'}{g(n)'}}}$
### Analogy to Inequalities
โข ${\displaystyle f(n)=O(g(n))}$ if and only if ${\displaystyle g(n)=\Omega (f(n))}$
โข ${\displaystyle f(n)=o(g(n))}$ if and only if ${\displaystyle g(n)=\omega (f(n))}$
It follows, where ${\displaystyle a\rightarrow f(n)}$ and ${\displaystyle b\rightarrow g(n)}$:
โข ${\displaystyle f(n)=O(g(n))\approx a\leq b}$
โข ${\displaystyle f(n)=\Omega (g(n))\approx a\geq b}$
โข ${\displaystyle f(n)=\Theta (g(n))\approx a=b}$
โข ${\displaystyle f(n)=o(g(n))\approx a
โข ${\displaystyle f(n)=\omega (g(n))\approx a>b}$
Caution: For all ${\displaystyle a,b,\in \mathbf {R} }$ exactly one must hold: ${\displaystyle a or ${\displaystyle a>b}$. Not all functions are asymptotically comparable.
### Master Theorem
If ${\displaystyle \displaystyle T(n)=aT(''n/b'')+f(n)}$ and a constant in the base case, where ${\displaystyle ''n/b''}$ can be either ${\displaystyle \lfloor n/b\rfloor }$ or ${\displaystyle \lceil n/b\rceil }$, ${\displaystyle a>=1}$ and ${\displaystyle b>1}$ then:
โข If ${\displaystyle f(n)\in O(n^{log_{b}{a-{\mathcal {E}}}})}$ for some ${\displaystyle {\mathcal {E}}>0}$, then ${\displaystyle T(n)\in \Theta (n^{log_{b}a})}$
โข Dominated by leaf cost
โข If ${\displaystyle f(n)\in \Theta (n^{log_{b}a})}$ then, ${\displaystyle T(n)\in \Theta (n^{log_{b}a}lgn)}$
โข Balanced cost
โข If ${\displaystyle f(n)\in \Omega (n^{log_{b}{a+{\mathcal {E}}}})}$ for some ${\displaystyle {\mathcal {E}}>0}$ and if ${\displaystyle \displaystyle af(''n/b'')\leq cf(n)}$ for some constant ${\displaystyle \displaystyle c<1}$ and sufficiently large ${\displaystyle n}$, then ${\displaystyle T(n)\in \Theta (f(n))}$
โข Dominated by root cost - It is important in this case to check that f(n) is well behave. More specifically,
โข For sufficient large n, a*f(n/b) <= c*f(n)
The following equations cannot be solved using the master theorem:
โข ${\displaystyle T(n)=2^{n}T\left({\frac {n}{2}}\right)+n^{n}}$
a is not a constant
${\displaystyle T(n)=2T\left({\frac {n}{2}}\right)+{\frac {n}{\log n}}}$
non-polynomial difference between f(n) and ${\displaystyle n^{\log _{b}a}}$
${\displaystyle T(n)=0.5T\left({\frac {n}{2}}\right)+n}$
a<1 cannot have less than one sub problem
${\displaystyle T(n)=64T\left({\frac {n}{8}}\right)-n^{2}\log n}$
f(n) is not positive
${\displaystyle T(n)=T\left({\frac {n}{2}}\right)+n(2-\cos n)}$
case 3 but regularity violation.
Also, Case 3 always hold when f = n^k and If ${\displaystyle f(n)\in \Omega (n^{log_{b}{a+{\mathcal {E}}}})}$
### Log Laws
For all real a > 0, b > 0, c > 0, and n,
โข ${\displaystyle a=b^{\log _{b}a}}$
โข ${\displaystyle \log _{c}ab=\log _{c}a+\log _{c}b}$
โข ${\displaystyle \log _{b}a^{n}=n\log _{b}a}$
โข ${\displaystyle \log _{b}a={\frac {\log _{c}a}{\log _{c}b}}}$
โข ${\displaystyle \log _{b}{\frac {1}{a}}=-\log _{b}a}$
โข ${\displaystyle \log _{b}a={\frac {1}{\log _{a}b}}}$
โข ${\displaystyle a^{\log _{b}c}=c^{\log _{b}a}}$
โข ${\displaystyle \log 1=0}$
### Summation
โข Arithmetic Series: ${\displaystyle \displaystyle {\sum _{k=1}^{n}(k)=1+2+\dots +n}={\frac {1}{2}}n(n+1)}$
โข Sum of Squares: ${\displaystyle \displaystyle {\sum _{k=0}^{n}(k^{2})}={\frac {n(n+1)(2n+1)}{6}}}$
โข Sum of Cubes: ${\displaystyle \displaystyle {\sum _{k=0}^{n}(k^{3})}={\frac {n^{2}(n+1)^{2}}{4}}}$
โข Geometric Series: ${\displaystyle \displaystyle {\sum _{k=0}^{n}(x^{k})=1+x+x^{2}+\dots +x^{k}}={\frac {x^{n+1}-1}{x-1}},}$ for real ${\displaystyle x\neq 1}$
โข Infinite decreasing: ${\displaystyle \displaystyle {\sum _{k=0}^{\infty }(x^{k})={\frac {1}{1-x}},for|x|<1}}$
โข Telescoping: ${\displaystyle \displaystyle {\sum _{k=1}^{n-1}(a_{k}-a_{k+1})=a_{0}-a_{n}}}$
### Exponents
โข ${\displaystyle a^{0}=1}$
โข ${\displaystyle a^{1}=a}$
โข ${\displaystyle a^{-1}={\frac {1}{a}}}$
โข ${\displaystyle (a^{m})^{n}=a^{mn}}$
โข ${\displaystyle a^{m}\times a^{n}=a^{m+n}}$
### Decision Tree-Related Notes
โข for a list of ${\displaystyle n}$ elements, there's ${\displaystyle n!}$ permutations
โข a binary tree of height ${\displaystyle d}$ has at most ${\displaystyle 2^{d}}$ leaves
โข therefore, a binary tree with at least ${\displaystyle n}$ leaves must have height at least ${\displaystyle \lceil \lg n\rceil }$
โข to get the height of a tree: the longest path (say we divide by 3/2 at each level) finishes when ${\displaystyle \left({\frac {2}{3}}\right)^{k}n=1\rightarrow k=log_{\frac {3}{2}}n}$, so the height is ${\displaystyle log_{\frac {3}{2}}n}$
โข ${\displaystyle \lg(n!)\in \Theta (n\lg n)}$, which we can establish by proving big-O and big-ฮฉ bounds separately (pumping "up" or "down" the values of the terms in the factorial and the overall number of terms as needed)
### Stirling's Approximation
โข ${\displaystyle \ln n!\sim n\ln n-n\ .}$
### Derivatives
โข ${\displaystyle {\frac {d}{dx}}a^{f(x)}=a^{f(x)}*{\frac {d(f(x))}{dx}}ln(a)}$
โข ${\displaystyle {\frac {d}{dx}}\ln(x)={\frac {1}{x}},\qquad x>0}$
โข ${\displaystyle {\frac {d}{dx}}\log _{a}(x)={\frac {1}{x\ln(a)}}}$
### The Silicon Downsย : Furlongs of Asymptotic Complexity
"NEWS FLASH: Mounties Find Silicon Downs Fixed!"
โข Constant ${\displaystyle \in O(1)}$
โข Logarithmic ${\displaystyle \in O(logn)}$ ie. ${\displaystyle log_{k}n,logn^{2}\in O(logn)}$
โข Poly-Log ${\displaystyle \in O(log^{k}n)}$
โข Linear ${\displaystyle \in O(n)}$
โข Log-Linear ${\displaystyle \in O(nlogn)}$
โข Superlinear ${\displaystyle \in O(n^{1+c})}$ where: c is a constant > 0
โข Quadratic ${\displaystyle \in O(n^{2})}$
โข Cubic ${\displaystyle \in O(n^{3})}$
โข Polynomial ${\displaystyle \in O(n^{k})}$ where: k is a constant, "tractable"
โข Exponential ${\displaystyle \in O(c^{n})}$ where: c is a constant > 1, "intractable"
### Stable Marriage Notes
Algorithm:
initialize all men in M and all women in W to unengage
while (an unengaged man with at least one woman on his preferance list remains)
do choose such a man m (element of) M
propose to the next woman w(element of)W on his preference list
if( w is unengaged)
then engage m to w
else if w prefers m to fiance m'
then break engagement of m' to w
engage m to w
cross w off m's preferance list
report set of engaged pairs as final matching.
Notes:
1. All executions of the stable marriage sort produce the same matching, and all men end up with their "best possible partner".
2. In the stable matching set each woman is paired with her worst valid partner.
3. This algorithm terminates after at most ${\displaystyle n^{2}}$ iterations of the while loop.
### Skyline Problem & Algorithm
For input B[1..n] buildings as triples (left, height, right), we get output of tuples (pos of height change, new height):
SKYLINE-ALG(B) {
If there is one building (x1,h,x2)
output (x1,h),(x2,0)
Else
divide into B1[1..floor(n/2)] and B2[floor(n/2)+1..n]
out1=SKYLINE-ALG(B1)
out2=SKYLINE-ALG(B2)
OUTPUT=MERGE(out1,out2)
}
The merge step is as follows:
MERGE(out1,out2) {
h1=h2=0, currx=0,curry=0
While out1 or out2 is not empty do {
If out1 and out2 start at the same x-coord
currx=first x-coord in out1
h1=first y-coord in out1
h2=first y-coord in out2
DELETE first entries in out1 and out2
Else If out1 has smallest first x-coord
currx=first x-coord in out1
h1=first y-coord in out1
DELETE first entry in out1
Else
currx=first x-coord in out2
h2=first y-coord in out2
DELETE first entry in out2
If max(h1,h2)!=curry
APPEND (currx,max(h1,h2)) to output
}
APPEND (currx,0) to output
}
### Proving a Loop Invariant
1. induction variable: Number of times through the loop
2. base case: Prove the invariant true before the loop starts
3. inductive hypothesis: Assume the invariant holds just before begininng some (unspecified) iteration of the loop.
4. inductive step: prove the invariant holds at the end of that iteration (right before the next iteration)
5. extra-bit: make sure the loop will eventually terminate. | 3,519 | 9,843 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 106, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-22 | latest | en | 0.323806 |
http://impoohill.com/0qK9k54V/ | 1,563,553,180,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526324.57/warc/CC-MAIN-20190719161034-20190719183034-00516.warc.gz | 75,327,297 | 16,248 | # Math Test Generator Free
Published at Saturday, May 04th 2019. by in Math Worksheet.
Step 2: Work on the Skill. Once he has met the skill, it is time for him to take baby steps on his and work on it. I will usually print off 5 sets of math worksheets on a topic. There are many great places to find these available. On the first worksheet, I let him work completely independently with no outside help at all. If he gets stuck, he gets stuck.
There are some new materials being developed now based on what we are learning about how the brain learns. These brain-friendly materials should be an improvement over what has existed. I recently bought a book by Marcia L. Tate titled โMathematics Worksheets Donโt Grow Dendrites.โ I highly recommend her book. She gives a great deal of information on alternative activities that are better for your childโs brain development and for learning.
###### Beginner Math Worksheets
Step 1: Introduce the Math Skill in Real Life. I will show my son how we use this math skill in real life. I will relate the skill to everyday. For division for example, I would preset this scenario: โYou have 25 baseball cards and want to give them equally to your five best friends. How do you do it?โ If I said to Cameron, โWhat is 25 divided by 5?โ, he would just give me a funny. Using a real world example motivates him to work for it.
It was easy because of my upbringing that ensured that math and I got acquainted long before school. My mother who was a primary grade teacher told me how she began teaching me math in different guises at home before I got to school age. Practice surely makes perfect and I am very gratefully to Mom for taking her time to familiarize me with math even as a child.
All my toys were one way or the other math related. I had puzzles, and tons of things Mom had me do as games on daily basis at home to get me ready for kindergarten! In fact, she continued guiding me towards being math friendly throughout kindergarten and first grade during which time 1st grade math worksheets was my constant companion.
Parents and teachers are aware of the importance of math as well as all of the benefits. Taken in the account how important math is, parents will do whatever it takes to help their struggling children to effectively manage math anxiety. By using worksheets, it can play a major role in helping your kids cope with these stressful. This is a good way to show our children that practicing their math skills will help them improve. Here are some of the advantages using math and worksheets.
You can find several types of sheets online and offline. You can choose among multiplication, Addition, Subtraction, Division, Geometry, Decimal, Shapes and Space worksheets.
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https://discourse.mcneel.com/t/conways-game-of-life-in-ghpython/88340 | 1,696,225,297,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510967.73/warc/CC-MAIN-20231002033129-20231002063129-00016.warc.gz | 222,712,241 | 11,199 | # Conway's Game of Life in GHPython
Hi everybody,
A couple of years ago, when I first encountered the concept of Conwayโs Game of Life, oftentimes referred to as cellular automaton, during my early architecture studies, I was kind of baffled and intrigued by it. For those of you, not familiar with architectural academia, the Game of Life, like the Vornoi diagram, particles, and other geometrical curiosities are often times experimented with to produce seemingly avant-garde design objects and/or buildings. Back then my overall CG and coding skills were still in early development, and the Game of Life remained a pretty much unrealised side-project.
Iโve revisited the concept the past weekend and am eager to share the results with you all.
For those of you not familiar, Iโm not reporting about the parlor game, created in 1860 by Milton Bradley, although that would also be a great topic, but about the conceptual Game of Life devised by British mathematician John Conway in 1970.
It is basically a zero-player game, meaning that merely the initial conditions are set, and no further input is required beyond that. The initial configuration can then be observed evolving over generations, following a rather simple, yet ingenious set of rules.
The initial setup is an infinite, orthogonal, two-dimensional grid of rectangular cells, each of which is either alive or dead (populated or unpopulated). It is often referred to as seed of the game.
At each generation, the current state of every cell is reevaluated in function of its immediate neighbourhood. The following rules apply here:
โข Any live cell with fewer than two live neighbours dies, as if by underpopulation.
โข Any live cell with two or three live neighbours lives on to the next generation.
โข Any live cell with more than three live neighbours dies, as if by overpopulation.
โข Any dead cell with exactly three live neighbours becomes a live cell, as if by reproduction.
All concerned cell states are changed simultaneously, following the above rules, in a discreet moment called the thick, making each generation purely a function of the preceding one.
Iโve devised a fully animated Grasshopper User Object that is highly customisable (and even features a custom icon).
### Inputs
โข Plane: A base plane for the Game of Life (by default a world XY-plane)
โข ExtendX: A number of grid cells in the base plane x-direction (by default 10)
โข ExtendY: Number of grid cells in the base plane y-direction (by default 10)
โข SizeX: Grid cell size in the base plane x-direction (by default 1.0)
โข SizeY: Grid cell size in the base plane y-direction (by default 1.0)
โข Max: Maximum number of generations (by default no restrictions)
โข Wrap: True to wrap the grid boundary, or False (by default True).
โข Wrapping means that grid cells that form the grid boundary look for neighbours beyond the boundary on the opposite side, basically like in the Asteroids game from 1979, where you could pass the screen boundary with you spaceship and pop out on the opposite side. This is by default turned on, because the Game of Life is meant to be infinite!
โข Wrapping can be switched off though to prevent flying geometries, when stacking individual game generations to produce a three-dimensional Game of Life in recording mode.
โข Rec: True to record and output the full Game of Life history, or False to only to output only the current generation (by default False).
โข Recording allows you to produce a three-dimensional, stacked version of the Game of Life.
โข Seed: Random seed for the initial setup of alive or dead cells
โข Run: True to iterate the Game of Life, or False to pause it
โข Reset: True to reset the Game of Life completely
### Outputs
Non-recording mode:
โข Count: The current generation count
โข Cells: A tree of grid cell outlines with branches for each grid row
โข States: A tree of current cell states (1 = alive, 0 = dead) with branches for each grid row
โข Ages: A tree of current cell ages with branches for each grid row
โข The cell age can for instance be used to color a cell.
Recording mode:
โข Count: A tree of generation counts with branches for each generation
โข Cells: A tree of grid cell outlines with branches for each generation that each include branches for each grid row
โข States: A tree of current cell states (1 = alive, 0 = dead) with branches for each generation that each include branches for each grid row
โข Ages: A tree of current cell ages with branches for each generation that each include branches for each grid row
User object:
conways_game_of_life.zip (6.7 KB)
Example files:
conways_game_of_life_example_files.zip (40.8 KB)
### Installation
In order to install the component, simply move it to the Grasshopper user objects folder (File > Special Folders > User Objects Folder).
Feel free to alter, manipulate, adapt or bastardize the Python code in any way you like. It is accessible by double clicking the component.
Please, leave feedback or report any bugs here! I would also be interested in seeing project images, if anybody uses the component to realise a design.
30 Likes
Hi ,
Thanks for your nice sharing! it helps me a lot!
however, i have a problem about how to change the density of initial points?
I try to read your code,but i am not good at python at all,
Would help me please?
Thank you very much!
Regards,
Jason
The number of initial cells in x- and y-direction is defined by the ExtendX and ExtendY inputs. If you set these parameter for instance to 10 cells in the x- and 5 cells in the y-direction, you end up with 50 cell (10 x 5) total!
You can change the size of each cell with the SizeX and SizeY inputs.
A detailed description of what each input is for is shown, when you hover your mouse over it.
2 Likes
Yeah,
My question is that, for the initial grid cells, every cell(in your class) is both 0 or 1, that means that it always has 50%density of initial cells birth.
Actually what i am trying to do is , to add sliders which can control the initial number of birth cell(10% or 30% whatever.)
I mean,maybe can add an inputs called โinputptsโ ,to use these points as the birth points.
But i am not sure whether i can modify your code to achieve this function.
If you have some advice, i will be very nice!
Thanks and regards
jason
OK, simply change line 93 of the script to this:
``````self.grid[x,y]["state"] = True if random.random() < 0.35 else False
``````
Here the initial cell state is initiated.
`random.random()` generates a random number between 0.0 and 1.0. Now, if this number is smaller than the decimal percentage 0.35, which translates to 35%, the cell will be initially alive. You can adjust this to your desired percentage.
Sure, this would also be possible, it just involves more changes to the current code. Feel free to have a go at it!
Thank you so much! big hug!
1 Like
Hello,
Prompted by this yearโs Advent Of Code challenge I have quickly implemented a 3D version of the game of life in Python in Rhino 7
``````import rhinoscriptsyntax as rs
import Rhino
from itertools import permutations
from collections import defaultdict
def conway_cubes(rounds):
rs.EnableRedraw(False)
# Example from the puzzle
state = """.#.
..#
###"""
new_cubes = set()
# Build the starting grid
for y, row in enumerate(state.splitlines()):
for x, cell in enumerate(row):
if cell == '#':
# Initialise sets and lists
old_spheres = []
neighbouring = set(permutations([1,1,1,0,0,0,-1,-1,-1],3)) # 27 neighbours
neighbouring.remove((0,0,0)) # a cube is not its own neighbour
for round in range(rounds+1):
cubes = new_cubes.copy()
rs.DeleteObjects(old_spheres)
neighbours = defaultdict(int)
old_spheres=[]
# Draw cubes and identify neighbours
for cube in cubes:
for n in neighbouring:
neighbours[(cube[0] + n[0],
cube[1] + n[1],
cube[2] + n[2],
)] += 1
new_cubes = set()
rs.Redraw()
# Apply rules for creation of cubes in next round
for location, n in neighbours.iteritems():
if location in cubes:
if 2 <= n <= 3: | 1,846 | 7,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-40 | latest | en | 0.959891 |
https://mathematica.stackexchange.com/questions/172425/error-in-integral-representation-of-appells-series?noredirect=1 | 1,721,400,955,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514908.1/warc/CC-MAIN-20240719135636-20240719165636-00308.warc.gz | 345,952,656 | 41,683 | # Error in integral representation of Appell's series
According to About the confluent versions of Appell Hypergeometric Function and Lauricella Functions the integral
$$\int_0^1t^{a-1}(1-t)^{c-a-1}(1-xt)^{-b}e^{yt}~dt$$
can be expressed as
$$\int_0^1t^{a-1}(1-t)^{c-a-1}(1-xt)^{-b}e^{yt}~dt=\dfrac{\Gamma(a)\Gamma(c-a)}{\Gamma(c)}\lim\limits_{k\to\infty}F_1\left(a,b,k,c;x,\dfrac{y}{k}\right).$$
It look like true, but gives numeric error. Look that, calculated by Mathematica:
F[a_?NumericQ, b_?NumericQ, c_?NumericQ, x_?NumericQ, y_?NumericQ] :=
NIntegrate[t^(a - 1) (1 - t)^(c - a - 1) (1 - x t)^(-b) Exp[ y t], {t, 0, 1}]
N[F[3/2, 1, 2, .4, .3], 20]
{2.8964403550198865}
G[a_?NumericQ, b_?NumericQ, c_?NumericQ, x_?NumericQ, y_?NumericQ] :=
Gamma[a] Gamma[c - a]/Gamma[c] Limit[AppellF1[a, b, k, c, x, y/k],k -> Infinity]
N[G[3/2, 1, 2, .4, .3], 20]
{2.2854650559595466}
where I was careful to ensure that $|x|<1,|y|<1$, and $\text{Re}(c)>\text{Re}(a)>0$, which are the condition of $F_1$.Note that the results are different. Furthermore, according to Wolfram Alpha
$$\lim\limits_{k\rightarrow \infty}F_1[3/2,1,k;2;0.4,\text{any/k}]=1.4549$$
So, is It a math issue or Mathematica issue?
โข Question to other users: should we include the tag bugs? Commented May 2, 2018 at 19:19
โข @AccidentalFourierTransform Did you report this bug to Mathematica? Do you know if they fixed it? Commented Aug 17, 2018 at 20:53
โข In my 13.1 you now need to Rationalize to get to your N[,20] result. N[G[3/2, 1, 2, 4/10, 3/10], 20]. And F[3/2, 1, 2, 4/10, 3/10] does not give normal symbolic form (anymore?), and thus does not allow N manipulation, it is small. WorlframAlpha gives -5+5 Sqrt[5/3] for your x = 4/10 and so does 13.1. Commented Jul 4, 2022 at 16:11
โข Oh, yes! I got why it is so "cool" now for N[F[3/2, 1, 2, 0.4, .3], 20], that it cannot even listen to N[]! See, if you will change the first NIntegrate to Integrate it will actually produce an integrand from which it is calculated, but will fail to even find the antiderivative! Wow! And then you can just change Integrate to NIntegrate back again. Another bug in there (though 2.8964403550198865 is the correct result, it is G that has a problem). NIntegrate[(E^(3 t/10) Sqrt[t])/( Sqrt[1 - t] (1 - (2 t)/5)), {t, 0, 1}, PrecisionGoal -> 20, PrecisionGoal -> 50] Commented Jul 4, 2022 at 17:15
โข @ ะะฐะปะตัะธะน ะะฐะฟะพะดะพะฒะฝะธะบะพะฒ Replacing Infinity in G by a large number, say 100 or 200, leads to a good approximation $G \simeq F$. Hance the limit is the problem. Sorry, I saw afterwards that this defect was already noticed by others. Commented Jul 1, 2023 at 13:33
There is a bug in the calculation of the large $k$ limit of AppellF1. This is easy to illustrate:
Needs["NumericalCalculus"]
wrongLimit = Limit[AppellF1[3/2, 1, k, 2, 4/10, (3/10)/k], k -> Infinity];
correctLimit = NLimit[AppellF1[3/2, 1, k, 2, 4/10, (3/10)/k], k -> Infinity];
Plot[{AppellF1[3/2, 1, k, 2, 4/10, (3/10)/k], wrongLimit, correctLimit}, {k, .5, 5}
, PlotStyle -> {Blue, Directive[Red, Dashed], Directive[Green, Dashed]}, PlotRange -> {Automatic, {0, 3}}]
Perhaps the bug should be reported. In the meantime, you can use NLimit to get the correct limit. If you use NLimit in your definition of G, you get results that consistent with those of F (up to numerical inaccuracy).
โข Thank you a lot! I didn't know 'NLimit'. Commented May 2, 2018 at 19:16
โข @user3321 I'm glad I could help :-) Commented May 2, 2018 at 19:17
โข Is it normal this code NLimit[AppellF1[3/2, -1, k,1, 16(1)^2 Sin[Pi/2]^2, -(8 (1)^2 Sin[Pi/2]^2)/k], k -> Infinity] take a long time? Commented May 9, 2018 at 20:00
โข I don't know how NLimit works internally, so I can't tell. But if you want to accelerate the process, you can use SequenceLimit[AppellF1[3/2, -1, #, 1, 16, -(8/#)] & /@ N[Range[17, 35, 2]]], which yields the approximate limit 0.065 in a few seconds on my PC. It is less reliable than NLimitthough, so use at your own risk (I plotted the function and it seems that the result is more or less good, but don't take my word for it). Commented May 9, 2018 at 20:15
โข @AccidentalFourierTransform SequenceLimit is now called NumericalMathNSequenceLimit, still being used. Why, oh, why, LOL... Commented Jul 15, 2023 at 14:31 | 1,552 | 4,246 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-30 | latest | en | 0.776707 |
https://dev.to/jethrolarson/intro-to-functional-combinators-in-javascript-4ae9 | 1,669,817,324,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710764.12/warc/CC-MAIN-20221130124353-20221130154353-00427.warc.gz | 235,560,157 | 36,918 | ## DEV Community ฯยยยฉฮฒยยฯยยยปฯยยยจฮฒยยฯยยยป is a community of 963,274 amazing developers
We're a place where coders share, stay up-to-date and grow their careers.
Jethro Larson
Posted on โข Updated on
# Intro to Functional Combinators in Javascript
Functional programming is at its core about using functions as the primary means of building programs. This means that a constructed functional program is a function. To those accustomed to OOP it might seem absurd to only use functions but with the power and simplicity of function composition and a few additional concepts you can create testable applications of any size. As a quick intro I want to focus on a couple concepts that can help build complex programs from smaller ones.
# Compose
You may remember this one from High School math class as "compose" or just "ฮยฐ". Given two functions f and g `(gฮฒยยฮฒยยฮฒยยfฮฒยย)(x) = g(f(x))`. We can implement this in JS as
``````const compose = (g, f) => (x) => g(f(x))
``````
This is a higher-order function, that is, a function that either takes a function as an argument or returns one. Compose does both, taking two functions and then returning one that applies its argument to the second function and then applies the result to the first one. Let's create a couple example functions to illustrate:
``````const inc = n => n + 1;
const half = n => n / 2;
const operate = compose(half, inc);
console.log(operate(3)) //=> 2
``````
There's an important constraint to consider; the compose function only works on functions that take one argument (unary functions). Doubly troublesome is that compose itself takes two arguments so it can't be used in a fractal manner. That wont do. Fortunatly there's a cool trick to make any function unary: currying.
# Currying
Currying is the act of converting a function that takes multiple arguments into a function that takes the first argument and returns a function that takes the next recursively until all arguments have been passed before returning the result. As an example let's refactor the `inc` function from above to be based on an `add` function:
``````const add = (n, m) => n + m;
``````
The above `add` function is a normal two-argument(binary) function but we can jam an arrow between `n` and `m` to curry it:
``````const add = n => m => n + m;
// now that its curried we can partially apply it to create
// our inc function from before
``````
# Compose revisited
Now that we know how to curry let's curry the compose function itself. This is also known as the B-combinator so let's call it that here:
``````const B = g => f => x => g(f(x));
// usage is similar to before
const operate = B(half)(inc)
``````
You may find it hard to think about what a partially-applied compose function is. I like to think of it as a program that has an adapter on it perfect to fit another program. `B(half)` is a function that will take a program and return one that divides the result by two.
One of the great places to use this is anywhere you see nested function calls:
``````const process = (arr) =>
arr.map(a =>
getUser(getFirstName(a))
);
// can be rewritten as
const process = (arr) =>
arr.map(B(getUser)(getFirstName));
``````
This is just the tip of the iceberg and I invite you to try writing these yourself and playing around with them.
John Peters โข Edited on
Thanks Jethro, really nice article, something for me to study in coming weeks for sure.
I was an OOPer for over 15 years. The main take away was the SRP principal (Single Responsibility Principal). Once I learned to assess my brand new block of code and follow that rule, it lead me to the next step. If there were any mixed responsibilities, I immediately refactored them to a new method. This ultimately allowed me to see the end result which was lots of small re-usable (functions) methods. OOP leads us to functional programming styles when we take the SRP principal seriously.
Its not really that different than the DRY principal. Reason? Because if you never repeat your code, you are forced to the functional and composition styles.
Jethro Larson
I agree. But I've also heard the assertion "functions in the small, objects in the large", so I wanted to emphasize that it is possible and potentially a good idea to use functions all the way up
John Peters โข Edited on
Yes no problem with that idea. As we build reusable parts, we have no choice but to use Composition as our style, which you referred to in this article. Nice...
If we use only composed functions it is no different than using composed classes. OOPers have been composing classes for 30 years now. The concepts are identical. So really. It's a matter of containment and how to implement.
AFAIK, Typescript compiles everything down to functions anyway.
Jethro Larson
Depending on your compile target, yeah. However, those constructor functions create objects and v8 turns objects into hidden classes so everything is everything :)
John Peters โข Edited on
I wasn't referring to V8 internals, just comparing the new Javascript Class to Functions described here:
developer.mozilla.org/en-US/docs/W...
Hassan
Thanks.
I found a resource for common combinators in javascript.
gist.github.com/Avaq/1f0636ec5c8d6...
Vuong
I just "review" code, not sure about the whole article because I'm still learning JS :P
`conat inc = add(1);`
should be
`const inc = add(1);` | 1,274 | 5,374 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2022-49 | longest | en | 0.908664 |
http://www.edhelper.com/math/geometry_right_triangles2.htm | 1,462,348,930,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860122533.7/warc/CC-MAIN-20160428161522-00100-ip-10-239-7-51.ec2.internal.warc.gz | 487,374,010 | 2,304 | Right Triangles
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High School Geometry
High School Geometry
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Name _____________________________ Date ___________________ (Key 1 - Answer ID # 0463311)
1.
A rectangle has a diagonal of 10 and length of 6. Find the area of the rectangle.
2.
Find the length of the diagonal of a rectangle with sides 15 cm and 15 cm.
3.
If s = 9 and t = 11, then r = ______.
4.
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5.
If PQ = 10, RS = 8, and PS = 5, find QR.
6.
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7.
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8.
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Select the number of pages: 1 page 2 pages 3 pages 4 pages | 452 | 1,535 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2016-18 | longest | en | 0.782214 |
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Violins Professional musicians listened to five violins being played, without seeing the instruments. One violin was a Stradivarius, and the other four were modern-day violins. When asked to pick the Stradivarius (after listening to all five), 39 got it right and 113 got it wrong.
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https://www.techylib.com/en/view/wistfultitle/chapter_17_mcgraw-hill | 1,524,540,168,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946453.89/warc/CC-MAIN-20180424022317-20180424042317-00506.warc.gz | 877,441,330 | 15,881 | # Chapter 17 - McGraw-Hill
Electronics - Devices
Nov 24, 2013ย (4 years and 5 months ago)
77 views
*
ยฉ 2014 by McGraw
-
Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or
distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or pa
rt.
*
!
!
!
!
*
Youโve come a long way! You
understand manufacturing and
machining, and are now totally
ready to study CNC!
*
*
*
Starting with
Chapter 17
, we begin the lessons on how to
manage a
programmed machine tool in a CNC
world
.
*
Itโs a bigger job than just putting parts into the setup and
hitting the green button
although thatโs where youโll
probably begin your career: operating a CNC.
*
Itโs about managing data, making setups, editing programs
and solving problems when they arise!
*
In order to manage this level of responsibility we must
study some underlying techno
-
facts.
*
Weโll first learn about
the systems
that make it work in
Chapters 17 and 18.
*
While
they
may look
like technical
beasts when
taken
one aspect at a time, the whole subject is
easy to learn
.
Thatโs
the way weโll proceed in Intro to CNC,
the third part of Machining and CNC Technology
.
The subject is broken down into individual units
of learning,
easily digested, leading to
competency!
After completing Part III youโll have the baseline
knowledge and be ready to safely and
confidently learn to setup and run your own CNC
machine.
*
*
Like all other PowerPoint sets for this text, this
presentation is not intended to teach the
subject, but rather to show why the units are
important using a sampling of what youโll be
learning
.
*
Details have been omitted but will be
explained in the textbook.
*
17.1
World Axis
Standards
17.1.1
Primary linear axes
X,Y and
Z
17.1.2
Primary rotary axes
A, B and C
17.1.3
Secondary Linear
U, V and W
17.1.4
Rules for determination
17.2
Coordinate Systems and
Points
17.2.1
Absolute and incremental values
17.2.2
Four quadrants
17.2.3
Points for geometry and reference
(A) Absolute
(B) Incremental
(C)Metric
17.2.4
Conventions in program commands
17.3
CNC Machine Motions
17.3.1
Axis moves
(A) Rapid travel
(B) Linear Interpolation
(C) Circular
17.3.2 Axis
combinations
:
-
and
3
-
D motion
17.4
Polar C
oordinates
17.4.1
Absolute and incremental polar values
17.4.2
Positive and negative direction
*
*
There are 14 standard axes defined by the
Electronics Industries Association (EIA) used for
motion and position.
*
In this text weโll study
nine
of them.
3 Primary
Linear Axes X, Y and Z
3
Primary Rotary Axes A, B and C
3
Secondary Linear Axes U,V
& W
Unless itโs a multiplexed machine with several
auxiliary rotary and linear axes, these nine are
adequate to define most of the equipment in
industry today.
However, for tomorrowโs manufacturing world,
thatโs another
question.
Machines continue to
evolve as central processors are able to handle
more and more calculations per nanosecond,
thus more functions simultaneously.
*
*
Whenever you are assigned to a new CNC
machine, the axis set must be identified as the
first order of business.
*
Here are the sets for three common machines.
*
*
Itโs easy to identify the
Z axis: itโs the spindle
or it faces the spindle
itโs the drilling axis!
*
Then apply the Right Hand Rule by pointing
your right middle finger in the positive Z
direction.
*
Your fingers and thumb then form the
orthogonal axis frame (mutually at 90
ยบ
).
*
*
First identify the Z axis.
Itโs parallel to the
spindle
axis
and brings
the work toward and
away from the spindle.
*
Pointing your middle
finger in the positive Z
direction, your index
finger and thumb form
the other positive axes.
*
*
All CNC machines use the X
-
Z or X
-
Y
-
Z frame,
with each axis
mutually perpendicular to
the
others.
*
That relationship stays the same no matter
how the axis set is rotated to suit the
machine.
*
Toward stronger or more efficient machines
manufacturers arrange the set any way
convenient, but
they donโt change the
inter
-
relationship between axes.
The set (my fingers)
remain in
the
same
orientation to each
other no matter their
world
orientation
*
*
The X axis on many turning centers, is not
parallel to the floor, it slants forward.
*
That provides easy access to the turret for
setup work, since the machine isnโt as wide as
level X axis machines.
*
Plus chips and coolants slide right off to the
catch basin below.
Z
X Slanted
This latheโs
world axis orientation
is not
level but itโs still an orthogonal set.
90ยบ
*
Whenever a machine features a rotary axis, we
identify it this way:
*
If it rotates around a line parallel to
X itโs an A axis
Y itโs B
Z itโs C
*
*
Rotary axes
capable of feed rates can
move a
cutter head in an
arc during machining.
*
Or they can move the workpiece in an arc.
*
In this
video
we see A and B auxiliary axes
moving simultaneously with X, Y and
Z
to cut
this complex turbine
blade
.
Only intelligent 5 axis CAM software can compile this program.
*
*
To determine the direction of rotary motion,
either plus or minus A,B or
C (clockwise or
CCW), we
use the Rule of Thumb.
*
Itโs based on the line about which the rotary
axis pivots, X, Y or
Z.
Point the thumb of
your right hand in
the
+ direction
of
the
axis of the
rotation
, X, Y or Z
positive direction.
Positive C
direction
Z+
X+
Positive A direction
Y+
What
motion is this?
Positive
B
*
*
CNC machines move
and locate with
reference to
the axis
origin.
*
They move to locations identified with
co
ordinates
.
*
Co
meaning working together.
*
Ordinate
meaning a single line of position.
For example, on this flat screen:
This is an ordinate
the
point to be identified lies
somewhere along it.
Lets say this line is parallel to
the X axis, and
1.750 above
the X
-
Y
origin
This line is parallel to
the Y
axis and
lies
2.50 to right of
the
X
-
Y
origin.
X2.50, Y1.750
This point
is identified with
a set of coordinates
on this
plane
.
*
*
In CNC work, the
place where the coordinates
are X0
, Y0, Z0, is known as the
Program
Reference Zero (PRZ)
*
Itโs the starting point for coordinates
*
Most coordinates in the program refer their
distance from the PRZ.
*
For example.
The tip of the drill is
at
X2.250,
Y1.00, Z
-
1.00
Relative to the PRZ
which is
on the top
corner
on this
part.
Where it All Begins
You may hear different terms for the
PRZ: โprogram zero,โ โprogram data
point,โ โthe originโ (a math term) or
others, depending on the region in
which you live. PRZ is the most
commonly used, but they all
mean the place where
X=0.0
Y=0.0
Z=You know
When we want coordinates to refer to
the PRZ we put a code in the program
G90
telling the control โthis and all
others until changed are absolute
values.โ CAM programs are almost all
absolute values.
*
*
When moving across the origin
line using
absolute coordinates, the
+/
-
values change
depending on the quarter
circle
quadrant in
which the point lies.
*
For example:
*
Point A has positive X and Y
values:
X1.54,
Y1.13
*
Point C is X
-
1.54, Y
-
1.13
*
What are the coordinates of Points
B and D?
*
*
Occasionally we encounter the need for a different kind
of coordinate.
*
They do not refer to the
PRZ,
but
rather
to their last
position.
*
Incremental coordinates are jumps from where you are to
where you wish to go next.
Incremental coordinates
are useful for hand
-
compiling small setup or tooling programs
written by the machinist at the machine. They
are also used in a limited number of
commands that weโll study later. To use
incremental coordinates in a program, the
control must read a
G91
code or be told in
some other way that they are not absolute.
Itโs as though the current position is a
mini
-
PRZ. So if I want a mill spindle to
go to the left, I write
.
G91 X
-
1.00
So when I push cycle start, it will then
move from where it is 1.0 inch to the
left
.
They are also called โrelativeโ coordinates
since they relate to the current location for
their reference.
*
*
To go to a position either
at rapid rate or at a feed
rate, we use significant
points on the part
geometry
to create
program coordinates.
Trade Tip
In
Chapter
25, you will be
drawing a part image using
Mastercam, readying it for
programming. Each line
and arc on the drawing will
be created by defining
significant points.
*
Points occur
at the ends of
lines and arcs.
*
They also
occur at the
center of arcs
and at
tangent points
where lines
join arcs.
*
*
CNC machines move their axes in five
different
ways
:
Rapid
Travel
Linear
Single
or
multi
-
axis straight
-
line
motion
Circular
motion within a single
plane
Circular/Linear
, also called
-
D
motion. Two axes
move in an arc while the third moves in a straight
line.
3
-
D
motion
Few older controls
have the ability to move in
an
arc
using
three
axes simultaneously. Most approximate
these arcs through the power of the cam software
.
Due to computer evolution with
ever
-
higher computation rates,
m
ost
new CNC controls today
can perform true 3
-
D
motion.
*
Rapid
as
fast as
the machine can
move but with
the ability to
reduce speed
through
operator
override
control.
Trade Tip
Caution!
Depending
on the power of the
CPU, your machine
will rapid in one of
two
ways
be
sure to
discover why in the
text. Older
controllers take an
unexpected route!
*
*
The next four motions all move one or more
axes at the
machining rate
specified in the
program
.
*
The differences lies in how many axes are
involved
in
a straight
line
or arc
.
*
As motions become more complex, the CPU
must handle far more calculations per second
by
interpolating
each
axesโ
drive commands.
*
*
To move axes simultaneously, to produce a
constant velocity along the line A
-
B, say at 400
inches per minute.
*
N
either
the X or Y axis
drive will
be moving at
400 IPM.
*
They will run at lower speeds that combine to
create the tool motion of 400 IPM.
Interpolating means to find an intermediate
value: in this case a feed rate value for each
axis that combines to create the programmed
rate.
*
A
B
375.87 IPM, X Axis Motion
137.81 IPM
Y Axis
*
*
The operator can override the resultant tool
motion from 0% (no movement) up to 100% or
150% on some machines.
*
When the feed rate is changed by the
machinist, the controller must change each
drive proportionately to achieve the rate.
*
*
For arc motion at feed rate, the controller is
also
interpolating,
as with linear
.
*
The difference
is that
it is constantly changing
the ratio between the axes involved, as the
curvature changes slants.
*
*
Sometimes engineering information comes not in
the form of rectangular dimensions, but rather as
the radius and angle from a starting point.
*
Those points are more easily defined using polar
coordinates
a
bolt circle for example.
*
Polar
coordinates
arenโt used inside
CAM
-
generated
programs, but they are very useful for drawing the
part geometry or when doing hand program writing
of polar entities.
Trade Tip
Using polar
coordinates
often
saves a trigonometry step during
drawing or hand program writing!
If the needed significant point is
defined in radius
-
angle, rather than
X
-
Y, why do an unnecessary
calculation? Define it with
an
R
-
A
coordinate on your geometry
drawing.
*
*
While one could
operate
a machine
without
the
knowledge
of Chapter
17,
remember that
you
are in training to
become a full journey
-
level machinist.
*
The goal is to be a manager of your CNC
machine
more than an operator.
*
That requires complete oversight
including
CNC
motions and how axis drives work,
coming up in
Chapter
18. | 3,099 | 11,700 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-17 | longest | en | 0.872179 |
https://ir.nctu.edu.tw/handle/11536/62849 | 1,582,293,566,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145529.37/warc/CC-MAIN-20200221111140-20200221141140-00436.warc.gz | 412,074,826 | 9,882 | ๆจ้ก: ๆจก็ณๆญธ็ดๅญธ็ฟๆผ็ฎๆณๅจๅนณ่ก่ฟดๅๆ็จไนๆ็จA Fuzzy Inductive Learning Algorithm for Parallel Loop Scheduling ไฝ่
: ่กๆๅTsai, Chang-Jiunๆพๆฒ้Shian-Shyong Tseng่ณ่จ็งๅญธ่ๅทฅ็จ็ ็ฉถๆ ้้ตๅญ: ๆจก็ณ้ๅ็่ซ;ๆฉๅจๅญธ็ฟๆผ็ฎๆณ;ๅนณ่ก่ฟดๅๆ็จ;็ฅ่ญๅบซ็ณป็ตฑ;Fuzzy Sets Theory;Machine Learning Algorithm;Parallel Loop Scheduling;Knowledge-Based System ๅ
ฌ้ๆฅๆ: 1997 ๆ่ฆ: ๅจๅฏฆ้็ๆ็จไธญ๏ผๆไพ็ตฆๅญธ็ฟ็ณป็ตฑ็่ณๆ้ๅธธๅ
ๅซๆจก็ณ่ณ่จ๏ผ่ๅณ็ตฑ็็ฌฆ ่ๅญธ็ฟๆผ็ฎๆณ็กๆณๆจ่ซ้้ก็่ณๆใไพๅฆ๏ผๅจๅนณ่ก็ทจ่ญฏ็ๆ็จ้ ๅไธญ๏ผๅนณ ่ก่ฟดๅๆ็จๆฏ้ๅธธ้่ฆ็๏ผๅ ็บๆฏไธๅ่ฟดๅ้ฝๅ
ๅซไธไบๅฏไปฅ่กจ็คบๅ
ถ็นๆง็ ๅฑฌๆง๏ผๅจ้ๅป๏ผๆๅๅทฒ็ถๅฏฆ้่ฃฝไฝๅบไธๅๅบๆผ็ฅ่ญๅบซๆนๆณ็ๅนณ่ก่ฟดๅๆ็จ ๆนๆณ๏ผ็ฐก็จฑ็บKPLS๏ผๅ
ถๅฏไปฅ้ๅฐไธๅ็่ฟดๅๅป้ธๆๅ้ฉ็ๅนณ่ก่ฟดๅๆ็จ๏ผ ๅฐ่ฟดๅๅ้
่ณๅค่็ๅจ็ณป็ตฑไธญไปฅ้ๅฐ้ซๅท่กๆ็๏ผๆๅๅฉ็จ้ไบ่ฟดๅ็ๅฑฌ ๆง่ฎKPLSไธญ็ๆจ่ซๆฉๆจ่ซๅบๅ้ฉ็ๆ็จๆผ็ฎๆณ๏ผไฝๆฏ๏ผ้ไบๅฑฌๆงๅ
ๅซไบๆจก ็ณ็่ณ่จ๏ผไธฆไธ้ฉๅ็จๅณ็ตฑ็็ฌฆ่ๅญธ็ฟๆผ็ฎๆณไพๆจ่ซๅ
ถๆฆๅฟต่ฆๅใๅจๆฌ็ฏ ่ซๆไธญ๏ผๆๅๅฐๆจก็ณ้ๅ็่ซๆ็จๆผAQRๅญธ็ฟๆผ็ฎๆณ๏ผ็ฐก็จฑFAQR๏ผๆญคๆผ็ฎ ๆณๅฏไปฅๅพ่จ็ทดไพๅญไธญๆญธ็ดๅบๆจก็ณ่ช่จ่ฆๅ๏ผๅ ๆญค่ขซ็จๆผ่งฃๆฑบไธ่ฟฐๅนณ่ก่ฟดๅ ๆ็จ็ๅ้ก๏ผไธฆไธๆพๅฐๆบ็ขบ็ๆจ่ซ่ฆๅ๏ผ้ไบ่ฆๅๅฏไปฅๆจ่ซๅนณ่ก่ฟดๅๆ็จ ๏ผๆญคๅคFAQRไบฆ่ขซ็จๆผ่งฃๆฑบ้ณถๅฐพ่ฑ็ๅ้กๅ้ก๏ผๅจ้ๅ
ฉๅๅฏฆ้ฉไธญ๏ผๅ
ถๅฏฆ้ฉ็ต ๆ้ฝ้ๅธธ่ฏๅฅฝใ In real applications, data provided to a learning system usually contain fuzzyinformation. The conventional symbolic learning algorithm can not infer data that contains such kind of information. For example, in the application domainsof parallelizing compilers, parallel loop scheduling is very important becauseeach loop contains some attributes that can indicate its characteristics and property. In the past few years, we have designed and implemented a parallel loop scheduling based upon knowledge based approach that is called KPLS to choose an appropriate schedule for different loop to assign loop iterations toa multiprocessor system for achieving high speedup rates. Based on these attributes mentioned above, an inference engine of KPLS is used to choose suitable scheduling algorithm. Unfortunately, we found that these attributes contain some fuzzy information, which are inapplicable to the traditional symbolic learning strategy for inferring some concept descriptions.In this thesis, we apply fuzzy set concept to AQR learning algorithm that is called FAQR. FAQR can induce fuzzy linguistic rules from fuzzy instances, is then proposed to solve the above parallel loop scheduling problem. Some promising inference rules have been found and applied to infer the choice of parallel loop scheduling.Besides, we apply the fuzzy inductive learning algorithm in IRIS Flower Classification Problem. Experimental results show that our method yields high accuracy in both different domains. URI: http://140.113.39.130/cdrfb3/record/nctu/#NT860394022http://hdl.handle.net/11536/62849 Appears in Collections: Thesis | 798 | 2,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-10 | latest | en | 0.489334 |
https://www.theserverside.com/tutorial/Java-7-and-Binary-Notation | 1,726,638,531,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651836.76/warc/CC-MAIN-20240918032902-20240918062902-00096.warc.gz | 960,746,584 | 49,376 | # New Java 7 Features: Binary Notation and Literal Variable Initialization
## Now with Java 7, you can not only use standard decimal notation, or the beloved hex notation, but you can use binary notation as well. Here's a good look at the ins and outs of binary notation, with a focus on what you'll need to know in order to pass the Java Professional certification exam from Oracle.
Having Fun with Binary Notation
Given all the fun developers have with hexadecimal notation, the visionaries at Sun and Oracle decided to expand the party by introducing the equally cryptic binary notation to the Java language. With the addition of binary notation, there are now three different ways to initialize a numeric variable, first using the standard base ten number system, secondly using hex notation, and finally using binary. Here's an example that assigns the number twelve to three different variables of type byte.
```public class BinaryLiterals {
public static void main(String args[]){
byte twelve = 12;
byte sixPlusSix = 0xC;
byte fourTimesThree = 0b1100;
System.out.println(twelve);
System.out.println(sixPlusSix);
System.out.println(fourTimesThree);
}
}
```
When the code above runs, it simply prints out the number 12 on three separate lines, demonstrating the veracity of all three notations:
12
12
12
Passing the Java 7 Professional Upgrade Exam
As part of the Oracle Certified Profession, Java 7 certification upgrade exam, currently in beta as of this writing, one of the exam objectives is to "Use binary literals." So, in this article, we're going to take a look at some of the more interesting aspects of binary notation, focussing on what you'll need to know in order to whiz by the related questions on the exam. Even if you're not planning on writing a Java cert, you'll likely find the new news on binary notation quite interesting, so you are implored to keep reading.
Taking a byte out of binary initializations
In the above code snippet, you see an example of a byte being initialized to the value of 12 using binary notation: byte fourTimesThree = 0b1100;
Now, everyone knows that a byte represents eight bits of data, so the following attempted initialization, where you assign a ten bit binary number to a byte, completely blows up with the following error: Type mismatch: cannot convert from int to byte
```byte data = 0b1100110011; // Type mismatch: cannot convert from int to byte
```
Interestingly, though not necessarily surprisingly, if the leading binary digits are zeroes, the JVM recognizes that these numbers are just placeholders, and initialization using ten bits is successful:
```byte data = 0b0000110011; //successful
```
So, as was mentioned, a byte represents eight bits of binary data, so you would expect the following to compile, seeing that it initializes eight bits to a byte:
```byte data = 0b10101010;
```
Sadly, this fails, because the byte data type only uses seven of its eight bits for holding data, while reserving the extra bit to indicate whether a number is positive or negative. That's why the range of the eight bit byte (2^8 =256) is -128 to 127, and not 0 to 255.
By the way, sometimes make the mistake of thinking the leading zero before the 'b' in byte notation (0b1100) indicates whether the number is positive or negative. That's not correct. You can toggle the positivity or negativity of a binary notation number the same way as you would with a deicmimal number: with a minus sign:
```//byte data = 1b1010101; not valid
byte data = -0b1010101; //valid
```
Now speaking of the problem of type mismatches and stuffing too many ones and zeroes into a data type, the short has the same issue as the byte, although with the short, you can initialize it to fifteen binary digits instead of eight:
``` short number = 0b111111111111111; //valid
//short number = 0b1111111111111111; not valid
```
Inconsistent treatment of ints vs. shorts and bytes
So, the eight bit byte can take seven bits, the sixteen bit short can take fifteen bits, but for some reason, you can specify a full 32 binary bits for a 32-bit int:
```int overflow = 0b10101010101010101010101010101011;
System.out.println(overflow);
```
Unfortunately, the assignment here does not work as nicely as planned. This number prints out as ย -1431655765, ย which is the negative value of the first 31 binary digits. The last digit of the binary number, which is a 1, is used to indicate that the number is negative. If the row of bits was literally translated into a number, without overflowing it into an int, it would actually have the value of 2863311531, not -1431655765.
Oh, and just so you know, there is a way to force the Java compiler to treat a binary number as a long - all you have to do is throw the letter 'L' at the end of the bits. So, take a look at what happens when you print out these two numbers that use binary notation:
```System.out.println(0b10101010101010101010101010101011); //-1431655765 using 32 signed bits
System.out.println(0b10101010101010101010101010101011L); //2863311531 using 32 unsigned bits
```
The latter example can also be useful when assigning long binary numbers to a long; it's utility goes far beyond printing out data to the console.
Isn't this int-eresting?
By default, the Java compiler likes to treat numbers as ints, but that can be a problem when you're defining numbers that fall into the exclusive range of a long. Take a look at the following code, which tries to initialize a long to a 33 bit value, a number that is well within the 64 bit range of a long, but far too large to be stuffed into a variable of type int:
```long bow = 0b101010101010101010101010101010111; //causes a compile error
```
This line of code triggers the following compile error: The literal 0b101010101010101010101010101010111 of type int is out of range.
Of course, we know that this should work, so we can force the compiler to treat our binary number as a 64 bit long, again, by appending the letter 'L'.
```//long bow = 0b101010101010101010101010101010111; fails to compile
long bow = 0b101010101010101010101010101010111L; //compiles
```
The appended 'L' can be both upper or lowercase, but since the lower case letter can be easily confused with a one, it's recommended to stick with the upper case value.
Now, since it's fine and dandy to place the letter L after a variable of type long is being initialized, you'd probably expect that the same type of notation could be used for float and double types. Well, you'd be wrong. So, while the following four lines of code are all valid:
```float f1 = 12f;
double d1 = 12d;
float f2 = 0b111;
double d2 = 0b111;
```
The following lines of code will actually blow up:
```float f3 = 0b111f;
double d3 = 0b111d;
```
Using binary notation with an 'f' or a 'd' to denote that a number is a float or a double generates the following compile time error: Syntax error on token "f", delete this token Syntax error on token "d", delete this token
And that's about it. That's everything you need to know about binary notation in order to tackle the corresponding objective in the Oracle Certified Professional for Java 7 exam. Best of luck!
Learning Resources for the Java and Java 7 Certification
Check out these other tutorials:
New Java 7 Features: Binary Notation and Literal Variable Initialization
New Java 7 Features: Numeric Underscores with Literals Tutorial
New Java 7 Features: Using String in the Switch Statement Tutorial
New Java 7 Features: The Try-with-resources Language Enhancement Tutorial
New Java 7 Features: Automatic Resource Management (ARM) and the AutoCloseable Interfact Tutorial
New Java 7 Features: Suppressed Exceptions and Try-with-resources Tutorial
Java 7 Mock Certification Exam: A Tricky OCPJP Question about ARM and Try-with-Resources
OCAJP Exam to Debuts in March 2010. OCPJP Released in June?
OCPJP & OCAJP Java 7 Exams: Oracle Drops the Training Requirement
OCAJP and OCPJP Changes for Java 7: ย New Objectives, a Format Change and a Price Hike
#### Next Steps
New to Git and distributed version control? Here are some Git examples andย Jenkins-Git integrationย tutorials designed to help you master the popular source code versioning tool.
Close | 1,963 | 8,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.884074 |
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#1
January 7th 10, 09:28 PM posted to microsoft.public.excel.misc
external usenet poster First recorded activity by ExcelBanter: Aug 2006 Posts: 102
Search formula showing results plus 100 characters
I use a formula to find text in a particular text and ask the results plus
the next 100 characters to display in another cell. I was wondering if there
was a way to show the proceeding characters. The formula i am using is:
=MID(\$G6&\$H6,SEARCH(C\$2,\$G6&\$H6),100)
#2
January 7th 10, 09:38 PM posted to microsoft.public.excel.misc
external usenet poster First recorded activity by ExcelBanter: Feb 2009 Posts: 320
Search formula showing results plus 100 characters
=LEFT(G6&H6,SEARCH(C2,C6&H6)-1)
"Deb" wrote in message
news
I use a formula to find text in a particular text and ask the results plus
the next 100 characters to display in another cell. I was wondering if
there
was a way to show the proceeding characters. The formula i am using is:
=MID(\$G6&\$H6,SEARCH(C\$2,\$G6&\$H6),100)
#3
January 7th 10, 09:51 PM posted to microsoft.public.excel.misc
external usenet poster First recorded activity by ExcelBanter: Jul 2006 Posts: 2,204
Search formula showing results plus 100 characters
Assuming that the formula you have shown is in cell D2, then a formula like
this would do the trick:
=IF(LEN(D2)0,LEFT(G\$6&H\$6,LEN(G\$6&H\$6)-LEN(D2)),"")
However, if D2 (your formula) comes up with an error because the contents of
C\$2 isn't found in G\$6 & H\$6, then the formula above will display the same
error.
"Deb" wrote:
I use a formula to find text in a particular text and ask the results plus
the next 100 characters to display in another cell. I was wondering if there
was a way to show the proceeding characters. The formula i am using is:
=MID(\$G6&\$H6,SEARCH(C\$2,\$G6&\$H6),100)
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All times are GMT +1. The time now is 05:19 AM. | 692 | 2,384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-30 | longest | en | 0.890381 |
https://www.tutorialspoint.com/armature-reaction-in-alternator-at-leading-power-factor-load | 1,713,089,498,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.25/warc/CC-MAIN-20240414095752-20240414125752-00764.warc.gz | 1,010,192,301 | 23,929 | Consider a 2-pole alternator as shown in Figure-1. Suppose that the alternator is loaded with an inductive load of zero power factor leading.. Thus, the phase currents ฯยยฮฯยยย
, ฯยยฮฯยยย and ฯยยฮฯยยฮ
will be leading their respective phase voltages ฯยยฮฯยยย
, ฯยยฮฯยยย and ฯยยฮฯยยฮ
by 90ยฐ. Figure-2 shows the phase diagram of the alternator.
Now, at time t = 0, the instantaneous values of currents and fluxes are given by,
$$\mathrm{ฯยยย_{ฯยย } = 0;\:\:ฯ_{ฯยย } = 0}$$
$$\mathrm{ฯยยย_{ฯยยย} = ฯยยฮ_{ฯยยย}\:cos 30ยฐ = \frac{\sqrt{3}}{2}ฯยยฮ_{ฯยยย};\:\:\:ฯ_{ฯยยย} = \frac{\sqrt{3}}{2}ฯ_{ฯยยย}}$$
$$\mathrm{ฯยยย_{ฯยยฮ
} = โฯยยฮ_{ฯยยย}\:cos 30ยฐ = โ\frac{\sqrt{3}}{2}ฯยยฮ_{ฯยยย};\:\:\:ฯ_{B} = โ\frac{\sqrt{3}}{2}ฯ_{ฯยยย}}$$
The space diagram of the magnetic fluxes is shown in Figure-3. By the parallelogram law, the resultant armature reaction flux is given by,
$$\mathrm{ฯยยย_{ฯยยฮฯยย } =\sqrt{ฯ^{2}_{ฯยยย} + ฯ^{2}_{B} + 2 ฯ_{ฯยยย}\:ฯ_{ฯยยฮ
}\:cos\:60ยฐ}}$$
$$\mathrm{\Rightarrow\:ฯ_{ฯยยฮฯยย } =\sqrt{(\frac{\sqrt{3}}{2}ฯ_{ฯยยย})^{2}+(\frac{\sqrt{3}}{2}ฯ_{ฯยยย})^{2}+2 \times (\frac{\sqrt{3}}{2}ฯ_{ฯยยย})\times (\frac{\sqrt{3}}{2}ฯ_{ฯยยย}) \times (\frac{1}{2})}}$$
$$\mathrm{โด\:ฯยยย_{ฯยยฮฯยย } = 1.5\:ฯ_{ฯยยย}}$$
Here, it can be seen that the direction of the armature reaction flux is in the direction of the main field flux. Hence, it will aid the main field flux. It is said to be magnetising flux.
Again, for the successive positions of the rotor, it can be seen that the armature reaction flux ฯฯยยฮฯยยย
remains constant in magnitude equal to 1.5 ฯยยยฯยยย and rotates at synchronous speed. Also, the EMFs induced in each phase of the alternator by the armature reaction flux lag the respective phase currents by 90ยฐ.
Therefore, when the alternator supplies a load at leading power factor, then the armature reaction is partly magnetising and partly cross-magnetising.ย .
Updated on: 25-Sep-2021
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http://thejubileeacademy.org/home/index.php/curriculum-2/online-home-schooling-algebra-2/ | 1,603,379,741,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107879673.14/warc/CC-MAIN-20201022141106-20201022171106-00376.warc.gz | 95,621,825 | 12,505 | Call Us Today! 484.383.3900
# Online Home Schooling Algebra 2 Course
//Online Home Schooling Algebra 2 Course
Online Home Schooling Algebra 2 Course2014-09-30T20:28:36+11:30
## Algebra 2
### Course Description
An extension of Algebra 1, Algebra 2 will review and expand skills already mastered while introducing new skills. Algebra 2 provides the basis for a continuing program in higher math, including Trigonometry and Calculus.
### eBook
This eBook covers new skills such as: the simplifying and solving of inequalities, finding the from_atxs of quadratic expressions as well as complex from_atxs, logarithms and algorithms, exponential equations, and basic trigonometric functions. Many of the topic titles are the same as the beginning algebra series, but at a level consistent with intermediate algebra courses.
[testimonial_rotator id=19515 shuffle=โtrueโ]
### Chapters
0. Important Course Information
0. Important Course Information and Answer Keys
1. Unit 1 Equations, Inequalities, and Exponents I
2. Unit 1 tests and Unit 2 Equations, Inequalities, and Exponents II
3. Unit 2 tests and Equations, Inequalities, and Exponents III
4. Unit 3 Equations, Inequalities, and Exponents IV and Unit 3 tests
5. Unit 4 Equations, Inequalities, and Exponents V
6. Unit 4 test and Unit 5 Equations, Inequalities, and Exponents VI
7. Unit 5 test and Unit 6 Equations, Inequalities, and Exponents VII
8. Unit 6 test and Unit 7 Equations, Inequalities, and Exponents VIII
9. Unit 7 tests and Unit 8 Equations, Inequalities, and Exponents IX
10. Polynomials and Factoring I and Unit 8 tests
11. Unit 9 Polynomials and Factoring II
12. Polynomials and Factoring III
13. Unit 9 tests and Unit 10 Polynomials and Factoring IV
14. Unit 11 and Units 10 & 11 tests
15. Unit 12 Polynomials and Factoring VI
16. Unit 12 test and Unit 13 Polynomials and Factoring VII
17. Unit 14 Rational Expressions and Exponents I
18. Units 13 & 14 tests and Unit 15 Rational Expressions and Exponents II
19. Unit 15 tests and Unit 16 Rational Expressions and Exponents III
20. Rational Expressions and Exponents IV
21. Unit 16 test and Units 17 and 18 Rational Expressions and Exponents V
22. Units 17 & 18 tests and Unit 19 Rational Expressions and Exponents VI
23. Unit 20 Rational Expressions and Exponents VII
24. Units 19 & 20 test and Unit 21 Radical Expressions
25. Unit 22 Radical Expressions II
26. Unit 23 Radical Expressions III
27. Units 21 โ 23 test and Unit 24 Radical Expressions V
28. Units 25 and 26 Quadradic Equations I | 682 | 2,522 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-45 | latest | en | 0.830191 |
http://www.openproblemgarden.org/op/slice_ribbon_problem_0 | 1,506,172,210,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689661.35/warc/CC-MAIN-20170923123156-20170923143156-00667.warc.gz | 530,061,955 | 4,669 | # Slice-ribbon problem
Importance: Outstanding โญโญโญโญ
Author(s): Fox, R
Subject: Topology
Keywords: cobordism knot ribbon slice
Posted by: rybu on: November 7th, 2009
\begin{conjecture} Given a knot in $S^3$ which is slice, is it a ribbon knot? % Enter your conjecture in LaTeX % You may change "conjecture" to "question" or "problem" if you prefer. \end{conjecture}
The definitions of slice and ribbon: \Def{slice knot} \Def{ribbon knot}
There is a fairly vast literature on this problem. It is closely related to the problem of determining which homology $3$-spheres bound homology $4$-balls, as both are in essence a type of $4$-dimensional cobordism problem.
% You may use many features of TeX, such as % arbitrary math (between $...$ and $$...$$) % \begin{theorem}...\end{theorem} environment, also works for question, problem, conjecture, ... % % Our special features: % Links to wikipedia: \Def {mathematics} or \Def[coloring]{Graph_coloring} % General web links: \href [The On-Line Encyclopedia of Integer Sequences]{http://www.research.att.com/~njas/sequences/}
## Bibliography
% Example: %*[B] Claude Berge, Farbung von Graphen, deren samtliche bzw. deren ungerade Kreise starr sind, Wiss. Z. Martin-Luther-Univ. Halle-Wittenberg Math.-Natur. Reihe 10 (1961), 114. % %[CRS] Maria Chudnovsky, Neil Robertson, Paul Seymour, Robin Thomas: \arxiv[The strong perfect graph theorem]{math.CO/0212070}, % Ann. of Math. (2) 164 (2006), no. 1, 51--229. \MRhref{MR2233847} % % (Put an empty line between individual entries)
*[F] Fox, R. H. Some problems in knot theory. 1962 Topology of 3-manifolds and related topics (Proc. The Univ. of Georgia Institute, 1961) pp. 168--176
[G] Gilmer, Patrick M. On the slice genus of knots. Invent. Math. 66 (1982), no. 2, 191--197.
[H] Hass, Joel. The geometry of the slice-ribbon problem. Math. Proc. Cambridge Philos. Soc. 94 (1983), no. 1, 101--108.
Ana G. Lecuona. [arXiv:0910.4601] On the Slice-Ribbon Conjecture for Montesinos knots
Brendan Owens. [arXiv:0802.2109] On slicing invariants of knots.
* indicates original appearance(s) of problem. | 640 | 2,103 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-39 | longest | en | 0.693898 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/690/2/w/b/ | 1,686,022,667,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652207.81/warc/CC-MAIN-20230606013819-20230606043819-00409.warc.gz | 919,713,135 | 63,434 | # Properties
Label 690.2.w.b Level $690$ Weight $2$ Character orbit 690.w Analytic conductor $5.510$ Analytic rank $0$ Dimension $240$ CM no Inner twists $4$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment:ย Compute space of new eigenforms
[N,k,chi] = [690,2,Mod(7,690)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(690, base_ring=CyclotomicField(44))
chi = DirichletCharacter(H, H._module([0, 11, 38]))
N = Newforms(chi, 2, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("690.7");
S:= CuspForms(chi, 2);
N := Newforms(S);
Level: $$N$$ $$=$$ $$690 = 2 \cdot 3 \cdot 5 \cdot 23$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 690.w (of order $$44$$, degree $$20$$, minimal)
## Newform invariants
comment:ย select newform
sage:ย f = N[0] # Warning: the index may be different
gp:ย f = lf[1] \\ Warning: the index may be different
Self dual: no Analytic conductor: $$5.50967773947$$ Analytic rank: $$0$$ Dimension: $$240$$ Relative dimension: $$12$$ over $$\Q(\zeta_{44})$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{44}]$
## $q$-expansion
The dimension is sufficiently large that we do not compute an algebraic $$q$$-expansion, but we have computed the trace expansion.
$$\operatorname{Tr}(f)(q) =$$ $$240 q + 24 q^{6}+O(q^{10})$$ 240 * q + 24 * q^6 $$\operatorname{Tr}(f)(q) =$$ $$240 q + 24 q^{6} - 44 q^{10} + 24 q^{16} - 44 q^{21} + 96 q^{23} + 16 q^{25} + 16 q^{26} + 44 q^{28} - 16 q^{31} + 44 q^{33} + 16 q^{35} - 24 q^{36} + 44 q^{37} - 88 q^{43} + 8 q^{46} + 96 q^{47} - 24 q^{50} - 24 q^{55} + 44 q^{57} - 16 q^{58} + 88 q^{61} + 56 q^{62} - 88 q^{65} + 132 q^{67} - 56 q^{70} + 16 q^{71} + 48 q^{73} + 24 q^{81} - 24 q^{82} + 44 q^{85} - 16 q^{87} + 44 q^{88} - 124 q^{92} + 32 q^{93} + 20 q^{95} - 24 q^{96} - 56 q^{98}+O(q^{100})$$ 240 * q + 24 * q^6 - 44 * q^10 + 24 * q^16 - 44 * q^21 + 96 * q^23 + 16 * q^25 + 16 * q^26 + 44 * q^28 - 16 * q^31 + 44 * q^33 + 16 * q^35 - 24 * q^36 + 44 * q^37 - 88 * q^43 + 8 * q^46 + 96 * q^47 - 24 * q^50 - 24 * q^55 + 44 * q^57 - 16 * q^58 + 88 * q^61 + 56 * q^62 - 88 * q^65 + 132 * q^67 - 56 * q^70 + 16 * q^71 + 48 * q^73 + 24 * q^81 - 24 * q^82 + 44 * q^85 - 16 * q^87 + 44 * q^88 - 124 * q^92 + 32 * q^93 + 20 * q^95 - 24 * q^96 - 56 * q^98
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
comment:ย embeddings in the coefficient field
gp:ย mfembed(f)
Label $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
7.1 โ0.997452 + 0.0713392i โ0.212565 + 0.977147i 0.989821 0.142315i โ1.95274 + 1.08941i 0.142315 0.989821i 1.32241 2.42181i โ0.977147 + 0.212565i โ0.909632 0.415415i 1.87005 1.22594i
7.2 โ0.997452 + 0.0713392i โ0.212565 + 0.977147i 0.989821 0.142315i โ1.89440 1.18796i 0.142315 0.989821i 0.244741 0.448210i โ0.977147 + 0.212565i โ0.909632 0.415415i 1.97432 + 1.04979i
7.3 โ0.997452 + 0.0713392i โ0.212565 + 0.977147i 0.989821 0.142315i โ0.619349 2.14858i 0.142315 0.989821i โ1.99507 + 3.65370i โ0.977147 + 0.212565i โ0.909632 0.415415i 0.771049 + 2.09892i
7.4 โ0.997452 + 0.0713392i โ0.212565 + 0.977147i 0.989821 0.142315i 1.62532 + 1.53569i 0.142315 0.989821i 1.83156 3.35425i โ0.977147 + 0.212565i โ0.909632 0.415415i โ1.73074 1.41582i
7.5 โ0.997452 + 0.0713392i โ0.212565 + 0.977147i 0.989821 0.142315i 1.87213 1.22275i 0.142315 0.989821i โ0.241465 + 0.442209i โ0.977147 + 0.212565i โ0.909632 0.415415i โ1.78013 + 1.35319i
7.6 โ0.997452 + 0.0713392i โ0.212565 + 0.977147i 0.989821 0.142315i 1.98312 + 1.03307i 0.142315 0.989821i โ0.653533 + 1.19686i โ0.977147 + 0.212565i โ0.909632 0.415415i โ2.05177 0.888964i
7.7 0.997452 0.0713392i 0.212565 0.977147i 0.989821 0.142315i โ2.18918 + 0.455502i 0.142315 0.989821i 0.172948 0.316731i 0.977147 0.212565i โ0.909632 0.415415i โ2.15111 + 0.610516i
7.8 0.997452 0.0713392i 0.212565 0.977147i 0.989821 0.142315i โ1.40926 1.73608i 0.142315 0.989821i 1.26240 2.31192i 0.977147 0.212565i โ0.909632 0.415415i โ1.52952 1.63113i
7.9 0.997452 0.0713392i 0.212565 0.977147i 0.989821 0.142315i 0.0285953 + 2.23589i 0.142315 0.989821i โ1.81898 + 3.33121i 0.977147 0.212565i โ0.909632 0.415415i 0.188029 + 2.22815i
7.10 0.997452 0.0713392i 0.212565 0.977147i 0.989821 0.142315i 0.855176 + 2.06608i 0.142315 0.989821i 1.43032 2.61943i 0.977147 0.212565i โ0.909632 0.415415i 1.00039 + 1.99981i
7.11 0.997452 0.0713392i 0.212565 0.977147i 0.989821 0.142315i 1.40256 1.74150i 0.142315 0.989821i 1.87882 3.44079i 0.977147 0.212565i โ0.909632 0.415415i 1.27475 1.83712i
7.12 0.997452 0.0713392i 0.212565 0.977147i 0.989821 0.142315i 1.93237 1.12513i 0.142315 0.989821i โ0.968092 + 1.77293i 0.977147 0.212565i โ0.909632 0.415415i 1.84719 1.26012i
37.1 โ0.877679 0.479249i 0.997452 + 0.0713392i 0.540641 + 0.841254i โ2.21203 0.326962i โ0.841254 0.540641i 0.672927 + 1.80419i โ0.0713392 0.997452i 0.989821 + 0.142315i 1.78476 + 1.34708i
37.2 โ0.877679 0.479249i 0.997452 + 0.0713392i 0.540641 + 0.841254i โ1.79792 1.32947i โ0.841254 0.540641i โ1.02771 2.75539i โ0.0713392 0.997452i 0.989821 + 0.142315i 0.940853 + 2.02850i
37.3 โ0.877679 0.479249i 0.997452 + 0.0713392i 0.540641 + 0.841254i 0.217321 2.22548i โ0.841254 0.540641i 0.154967 + 0.415484i โ0.0713392 0.997452i 0.989821 + 0.142315i โ1.25730 + 1.84911i
37.4 โ0.877679 0.479249i 0.997452 + 0.0713392i 0.540641 + 0.841254i 1.06201 + 1.96778i โ0.841254 0.540641i โ0.702461 1.88337i โ0.0713392 0.997452i 0.989821 + 0.142315i 0.0109547 2.23604i
37.5 โ0.877679 0.479249i 0.997452 + 0.0713392i 0.540641 + 0.841254i 2.06085 + 0.867694i โ0.841254 0.540641i โ1.60657 4.30737i โ0.0713392 0.997452i 0.989821 + 0.142315i โ1.39292 1.74922i
37.6 โ0.877679 0.479249i 0.997452 + 0.0713392i 0.540641 + 0.841254i 2.18586 0.471182i โ0.841254 0.540641i 1.33355 + 3.57540i โ0.0713392 0.997452i 0.989821 + 0.142315i โ2.14430 0.634025i
37.7 0.877679 + 0.479249i โ0.997452 0.0713392i 0.540641 + 0.841254i โ2.09572 0.779713i โ0.841254 0.540641i โ0.448960 1.20371i 0.0713392 + 0.997452i 0.989821 + 0.142315i โ1.46569 1.68871i
37.8 0.877679 + 0.479249i โ0.997452 0.0713392i 0.540641 + 0.841254i โ1.84527 1.26292i โ0.841254 0.540641i 1.33335 + 3.57484i 0.0713392 + 0.997452i 0.989821 + 0.142315i โ1.01431 1.99278i
See next 80 embeddings (of 240 total)
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 613.12 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
5.c odd 4 1 inner
23.d odd 22 1 inner
115.l even 44 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 690.2.w.b 240
5.c odd 4 1 inner 690.2.w.b 240
23.d odd 22 1 inner 690.2.w.b 240
115.l even 44 1 inner 690.2.w.b 240
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
690.2.w.b 240 1.a even 1 1 trivial
690.2.w.b 240 5.c odd 4 1 inner
690.2.w.b 240 23.d odd 22 1 inner
690.2.w.b 240 115.l even 44 1 inner
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$T_{7}^{240} + 88 T_{7}^{237} - 1408 T_{7}^{236} + 924 T_{7}^{235} + 3872 T_{7}^{234} - 121308 T_{7}^{233} + 966606 T_{7}^{232} - 1414952 T_{7}^{231} - 4796440 T_{7}^{230} + 77828432 T_{7}^{229} + \cdots + 55\!\cdots\!96$$ acting on $$S_{2}^{\mathrm{new}}(690, [\chi])$$. | 3,770 | 7,583 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-23 | latest | en | 0.440609 |
https://15worksheets.com/worksheet-category/division-as-repeated-subtraction/ | 1,725,815,682,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00876.warc.gz | 57,491,350 | 36,151 | # Division as Repeated Subtraction Worksheets
โข ###### Subtraction Quest
Division as repeated subtraction is a foundational concept in mathematics that helps students understand division by breaking it down into a simpler, more relatable process. Worksheets that focus on this concept are essential for young learners, typically in early elementary grades, as they begin to grasp the mechanics of division. These worksheets are designed to build a solid understanding of how division works by repeatedly subtracting the divisor from the dividend until what remains is less than the divisor. By reinforcing this method, students develop a concrete sense of division as a form of subtraction, which is crucial for their future success in more advanced math.
These worksheets introduce students to the concept of division through a process they are already familiar with: subtraction. Rather than diving straight into the abstract notion of division, these worksheets guide students to subtract the divisor from the dividend multiple times, step by step, until they can no longer subtract without going negative. The number of times they subtract represents the quotient, and whatever remains represents the remainder.
These worksheets start with simple problems to ease students into the concept and gradually increase in difficulty as their understanding deepens. The problems might involve both small and larger numbers to ensure students gain confidence with a range of division scenarios. The goal of these worksheets is not only to teach division but also to enhance studentsโ subtraction skills, which are integral to their overall mathematical development.
### Types of Problems and Skills Taught
1. Basic Repeated Subtraction Problems
The most common type of problem on these worksheets involves straightforward division where the dividend is easily divisible by the divisor, resulting in no remainder. For example, a problem might present students with 12 divided by 3. The worksheet would guide them to subtract 3 from 12 repeatedly (12 โ 3 = 9, 9 โ 3 = 6, 6 โ 3 = 3, 3 โ 3 = 0), counting the number of subtractions to determine the quotient, which in this case is 4.
Skills Taught โ This type of problem helps students understand the basic concept of division and reinforces their subtraction skills. It also teaches counting in sequences and introduces the idea that division is the opposite of multiplication.
2. Problems with Remainders
As students progress, they encounter problems where the dividend is not a multiple of the divisor, resulting in a remainder. For example, if they are asked to divide 14 by 3, they would subtract 3 from 14 until they can no longer do so without going negative. After 4 subtractions (14 โ 3 = 11, 11 โ 3 = 8, 8 โ 3 = 5, 5 โ 3 = 2), the remainder is 2, meaning the quotient is 4 with a remainder of 2.
Skills Taught โ These problems introduce students to the concept of remainders, a crucial aspect of division. They learn that not all numbers divide evenly, and they begin to understand how to interpret a remainder in the context of a division problem. This reinforces their understanding of both division and subtraction while also teaching them to deal with uneven divisions.
3. Incorporated Word Problems
Division as repeated subtraction worksheets often include word problems that require students to apply the concept to real-life scenarios. For instance, a problem might describe a situation where 20 apples are being distributed equally among 4 baskets. Students would then subtract 4 from 20 repeatedly to determine how many apples go into each basket.
Skills Taught โ Word problems help students apply mathematical concepts to real-world situations, enhancing their problem-solving abilities. They also improve reading comprehension and teach students how to translate a written problem into a mathematical equation. Moreover, by working through these problems, students learn to visualize division in a practical context, making the concept more relatable and easier to understand.
4. Visual Representations
Some worksheets use visual aids, such as number lines or pictorial representations, to illustrate the process of repeated subtraction. For example, a number line might be used to show the jumps or steps taken as a student subtracts the divisor from the dividend. If dividing 15 by 3, the student would make 5 jumps of 3 along the number line, landing on 0.
Skills Taught โ Visual representation helps students who are more visually oriented or who struggle with abstract concepts. It reinforces the idea that division is a series of equal steps or jumps, making the process more concrete. Additionally, it teaches students to use multiple strategies to solve a problem, an essential skill as they advance in mathematics.
5. Mixed Practice Problems
To ensure students fully grasp the concept, worksheets often include a variety of problems that require them to use repeated subtraction in different contexts. This might include a mix of basic problems, problems with remainders, and word problems. By encountering different types of division scenarios, students learn to apply repeated subtraction in a versatile way.
Skills Taught โ Mixed practice solidifies a studentโs understanding by challenging them to adapt the repeated subtraction method to various situations. It helps them recognize patterns and builds confidence as they see how repeated subtraction can be applied universally to all division problems, regardless of complexity.
6. Comparison with Other Division Methods
Some worksheets might include a comparison between division as repeated subtraction and other methods of division, such as long division or division by grouping. These exercises are usually introduced after students are comfortable with repeated subtraction, to show them different approaches to the same problem. For instance, students might be asked to solve a division problem using both repeated subtraction and long division and then compare the results.
Skills Taught โ This type of exercise encourages flexible thinking and helps students understand that there are multiple ways to solve a division problem. It also prepares them for more advanced division methods by showing how repeated subtraction is a stepping stone to these other techniques. Furthermore, it enhances their ability to compare and contrast different mathematical processes.
These worksheets provide a concrete method for understanding division, which can often be an abstract and challenging concept for beginners. By breaking down division into a series of subtraction steps, students can see exactly how division works, making it more accessible and less intimidating.
Whether students are working through basic problems, dealing with remainders, or tackling word problems, they must think critically about how to approach each task. This practice in logical reasoning and strategy is invaluable as they progress in their education. By including a variety of problem types, these worksheets cater to different learning styles. Visual learners benefit from the use of number lines and pictorial representations, while those who learn best through reading and writing can engage with word problems. This diverse approach ensures that all students, regardless of their preferred learning style, have the opportunity to succeed. | 1,358 | 7,377 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2024-38 | latest | en | 0.951844 |
https://www.equationsworksheets.net/balancing-chemical-equations-worksheet-1-25/ | 1,675,712,472,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500357.3/warc/CC-MAIN-20230206181343-20230206211343-00031.warc.gz | 748,507,274 | 14,868 | # Balancing Chemical Equations Worksheet 1 25
Balancing Chemical Equations Worksheet 1 25 โ The objective of Expressions and Equations Worksheets is to help your child learn more effectively and efficiently. They include interactive activities and questions based on the sequence of how operations are conducted. These worksheets are designed to make it easier for children to grasp complex concepts as well as simple concepts in a short time. Download these free resources in PDF format to aid your childโs learning and practice math-related equations. These materials are great for students who are in the 5th through 8th grades.
## Get Free Balancing Chemical Equations Worksheet 1 25
These worksheets can be utilized by students from the 5th-8th grades. These two-step word problem are constructed using decimals or fractions. Each worksheet contains ten problems. The worksheets are available on the internet as well as in print. These worksheets are a fantastic opportunity to practice rearranging equations. These worksheets are a great way to practice rearranging equations , and assist students with understanding equality and inverted operations.
These worksheets can be used by fifth- and eighth grade students. They are ideal for students who are struggling to calculate percentages. There are three kinds of problems you can choose from. It is possible to work on one-step challenges that contain decimal or whole numbers or use word-based methods to solve decimals or fractions. Each page will have 10 equations. These worksheets for Equations are suggested for students from 5th to 8th grade.
These worksheets are a great tool for practicing fraction calculations as well as other concepts that are related to algebra. You can select from a variety of different types of problems with these worksheets. You can pick a numerical or word-based challenge. It is vital to pick the type of problem, as every challenge will be unique. There are ten issues on each page, meaning theyโre excellent for students from 5th to 8th grade.
These worksheets are designed to teach students about the relationship between variables and numbers. They allow students to solve polynomial equations and to learn how to apply equations in daily life. These worksheets are an excellent way to get to know more about equations and formulas. These worksheets will help you learn about the different kinds of mathematical problems and the various symbols that are used to express them.
This worksheets are very beneficial for children in the first grade. These worksheets can teach students how to solve equations and graph. These worksheets are perfect to get used to working with polynomial variable. They also assist you to learn how to factor and simplify these variables. There are plenty of worksheets to aid children in learning equations. Working on the worksheet yourself is the best method to understand equations.
You will find a lot of worksheets for teaching quadratic equations. Each level comes with their own worksheet. The worksheets were designed to assist you in solving problems of the fourth degree. After youโve solved a stage, youโll go on to solving different kinds of equations. You can continue to tackle the same problems. As an example, you may find a problem with the same axis and an elongated number. | 624 | 3,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-06 | latest | en | 0.951746 |
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# Arithmetic and Geometric Sequences
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Use the arithmetic sequence of numbers 2, 4, 6, 8, 10... to find the following:
a) What is d, the difference between any two consecutive terms?
b) Using the formula for the nth term of an arithmetic sequence, what is 101st term?
c) Using the formula for the sum of an arithmetic sequence, what is the sum of the first 20 terms?
d) Using the formula for the sum of an arithmetic sequence, what is the sum of the first 30 terms?
e) What observation can you make about the successive partial sums of this sequence (HINT: It would be beneficial to find a few more sums like the sum of the first 2, then the first 3, etc.)?
2) Use the geometric sequence of numbers 1, 3, 9, 27, ... to find the following:
a) What is r, the ratio between 2 consecutive terms?
b) Using the formula for the nth term of a geometric sequence, what is the 10th term?
c) Using the formula for the sum of a geometric sequence, what is the sum of the first 10 terms?
3) Use the geometric sequence of numbers 1, 1/3, 1/9 , 1/27... to find the following:
a) What is r, the ratio between 2 consecutive terms?
b) Using the formula for the sum of the first n terms of a geometric sequence, what is the sum of the first 10 terms? Carry all calculations to 6 decimals on all assignments.
c) Using the formula for the sum of the first n terms of a geometric sequence, what is the sum of the first 12 terms? Carry all calculations to 6 decimals on all assignments.
d) What observation can make about the successive partial sums of this sequence? In particular, what number does it appear that the sum will always be smaller than?
4) CLASSIC PROBLEM - A traveling salesman (selling shoes) stops at a farm in the Midwest. Before he could knock on the door, he noticed an old truck on fire. He rushed over and pulled a young lady out of the flaming truck. Farmer Crane came out and gratefully thanked the traveling salesman for saving his daughter's life. Mr. Crane insisted on giving the man an award for his heroism.
So, the salesman said, "If you insist, I do not want much. Get your checkerboard and place one penny on the first square. Then place two pennies on the next square. Then place four pennies on the third square. Continue this until all 64 squares are covered with pennies." As he'd been saving pennies for over 25 years, Mr. Brown did not consider this much of an award, but soon realized he made a miscalculation on the amount of money involved.
a) How much money would Mr. Crane have to put on the 32nd square?
b) Calculate the amount of money necessary to fill the whole checkerboard (64 squares). How much money would the farmer need to give the salesman?
https://brainmass.com/math/algebra/arithmetic-and-geometric-sequences-126488
#### Solution Preview
1) Use the arithmetic series of numbers 2 , 4 , 6 , 8 ...to find the following:
a) What is d, the difference between any 2 terms?
The difference between any two terms = ( n+1 )th term - ( n )th term
Hence , difference , d = 4-2 = 6 - 4 = 8 - 6 = 2
b) Using the formula for the nth term of an arithmetic series, what is 101st term? Answer:
nth term=a+(n-1)d..............(1)
Given a=first term=1
n=101
d=2
putting above values in formula 1
nth term=1+(101-1)*2
=1+200=201
Therefore , nth term = 201
Using the formula for the sum of an arithmetic series, what is the sum of the first 20 terms?
formula for the sum of an arithmetic series=n/2(2a +(n-1)d)..............(1)
given that , n = 20
a = first term = 1
Therefore , sum = 20/2(2*1+19*2)
=10*(40)
sum = 400
2) Use the geometric sequence of numbers 1, 3, 9, 27, ... to find the following:
a) What is r, the ratio between 2 consecutive terms? | 1,003 | 3,806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2020-34 | latest | en | 0.922536 |
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# 7 THE DESIGN OF THE NEWSPACE MAP
As discussed in chapter 4, a map is a useful tool when navigating in virtual domains, and the large display also demands navigational tools. To improve reader navigation in newspace, I decided to always make a map available to the reader.
## 7.1 Scale
The scale is one of the first parameters to be set when designing a map. For the newspace map, the factors deciding the scale are the number of pages in the newspace, the size and resolution of the primary and secondary monitor, and the decimation algorithm.
The number of pages in the newspace is currently set to 9 (3x3) because:
* Presuming the same granularity of section separation as newspapers use today, a good guess is that most people will be interested in around six sections. This leaves two pages for other purposes in addition to the front page.
* Since pages in the newspace are square and the map monitor is close to square, the tightest packing is achieved using a square grid.
The primary monitor has a resolution of 2048x2048 pixels, while the secondary screen offers 1152x900. The largest map possible is then (2048x3)/900 = 6.83, but since the decimation is simpler to compute given integer decimation factors, 7:1 is scale of the pixmaps.
The corresponding number for other newspace sizes are:
pages page grid newspace ratio
4 2x2 4096x4096 5:1
9 3x3 6144x6144 7:1
12 4x3 8192x6144 8:1
16 4x4 8192x8192 10:1
20 5x4 10240x8192 10:1
Comparing how different map scales should influence the map design would be interesting extension of this project.
## 7.2 Map Metaphors
Several of the different map types discussed have been implemented as a part of the project.
The hardest decision to make was wether to create a line map (rendered), or a photo map (using image processing). See section 4.2 for a discussion of maps. The scale between the main display and the map is approximately 7:1. This means that a decimated version of an original news article pixmap is able to show the structure of the article, e.g., where the headline is, where the figures are, and how many paragraphs there are. In most cases it is not possible to read the headline. See figure 14.
Some argues that all available bandwidth in the map should be utilized to get the headline across to the user; everything else is secondary information. Following this reasoning, it makes sense to render a new pixmap for use in the map. One is then free to use a relatively large font for the headline. This solution would require much good design work.
The solution settled upon tries to take advantage of the best of both alternatives. It uses a decimated "photo" of the original and renders the headline on top of that--if necessary the headline will be relatively larger than in the original. This solution is in several ways similar to the weather map described in chapter 4. It improves the legibility of the headline fonts designed for that size, and by increasing the font size.
## 7.3 What to display
By designing a suitable map one can give users access to a lot of information about the domain it represents without actually seeing it. Applied to The Electronic Broadsheet, this means that by carefully crafting the map the reader can save both time and frustration by reducing unwanted searching.
Cartographers have for centuries made a living from selecting a set of features of an area and rendering those features into a map. Selecting the right set of features is one of the first decisions the mapmaker has to make. Later, it has to be decided how these features should be rendered. A set of standards has evolved in cartography; this eases the design process and limits the number of solutions.
Since electronic newspaper maps don't have the same traditions, a significant part of the project has been dealing with the design of the map. There is a number of information items that readers may be interested in learning about without having to look at the actual article. There is also a limited number of design tools one can use to convey a message from the surface of a computer screen to the reader.
First, identifying the information items (from now or referred to as items) is not trivial. The system claims personalization, and there is an unbounded number of items that a reader may want to see. Naturally, the data available about each article is limited, and we can also make some guesses about which items will be the most popular with the readers. The list includes:
* Time/Age
* Sections
* Importance/Urgency/Priority
* Source
* Distribution
* Size/Length
### 7.3.1 Display Tools
Identifying the design tools available to convey the information items is a little easier. A color display can only emit a limited bandwidth of electromagnetic radiation over a circumscribed 2D surface. Therefore, the design tools can be coarsely divided into two groups: chromatic and spatial.
Mackinlay [Mackinlay 86] ranks the design tools as perceptual tasks with regard to how well they perform presenting quantitative, ordeal and nominal data. See figure 15. The data have not been empirically confirmed, and should be used with care. However, we can make a number of interesting observations from it:
* Position ranks on top in all list.
* Except position, the perceived quality of the tools vary a lot from list to list, e.g., hue is ranked at position 2, 4, and 9 out of 14 entries.
* Shape scores poorly--it is considered not relevant for quantitative and ordinal presentation, and below average for nominal presentation. Still, an important school in computer vision targets shape recognition.
* Intensity (or value) is not in the diagram. The term "density" cover parts of what is known as intensity.
Presuming figure 15 is correct, one should assume that connecting an information item with the suitable tool should be easy. However, there are several problems. First, classifying an item as quantitative, ordinal or nominal is not trivial. One example: is age quantitative or ordinal in the context of a newspaper map?
Second, once the items are classified, several items might claim the same tool. This is likely to happen with position since it ranks highest for all three categories. But not all information can or should be coded through position alone, and the diagram does not answer in which order items should pick tools.
Jacobson and Bender [Jacobson, Bender 90] extends the Mackinlay diagram by devising a method of quantifying the relative expressive qualities of color combinations. They show that hue alignment, coupled with contrast of value, is strongly correlated with reading speed. Therefore, when displaying text, the contrast of value should be high, while other information can afford less contrast of value. Also, they report the off-complementary dyad when contrast of value is low to be viewed as energetic. Studies by Jacobson, Bender and Feldman [Jacobsen et al. 91] show how hue alignment and value can be combined to provide a wide rage of legibility and highlighting capabilities. The authors of the two last referenced papers have been consultants in the color selection process.
A considerable amount of time has been spent juggling the relationship between items and design tools. The following description is sorted with regard to the information items, with a description of the solution following. This is not necessarily how the solution was pursued; it was sometimes easier to take a solution (design tool) and look for a problem (information item).
### 7.3.2 Time/Age
The term "newspaper" indicates that the age of the content is an important factor for the medium. Age can be represented relative to the current time ("an article was received 14 minutes ago") , or one can use the absolute calendar as a reference ("the article was received Apr 2, 9:38pm"). The first representation is more useful in the short term, while only the second is acceptable for archives. Since newspapers by definition emphasize the short term, the first representation is most useful in this project. If the goal was to create a research tool to search through news archives, the time/date would be the natural solution.
One property of paper, and especially newspaper pulp, is that it yellows when exposed to light. Therefore, an old newspaper will acquire a yellowish tint. When a new article pops up on the map its initial background color is relatively bright. As time passes the background hue of the icon will change to yellow and the value will decrease. In its current form, the aging process is fast; the time it takes from an article is displayed until it is maximum faded is only about 20 minutes. The rate of change should be adjusted to the reader's frequency of reading. To a reader that makes a living reading news (see scenario in 5.3.1) an article or piece of mail may be old after 20 minutes, but for a casual user (section 5.3.2) a 24-hour time frame seems appropriate--this is also the newspaper time frame.
When generating colors for the map, one must keep in mind that the reader will live with the colors over a long period of time. Subtle variations in hue, saturation or intensity will be recognized and appreciated; there is no need to "scream" in order to convey the message. This is a general rule that has been enforced throughout the design process.
### 7.3.3 Sections
Newspapers indicate the subject of an article in several ways. At a coarse level of granularity articles are categorized and put into one of several sections, e.g. the daily "sports" section or the weekly "real estate" section.
In The Electronic Broadsheet, sections are currently laid out in a 3x3 two-dimensional grid with the front page in the middle. The surrounding pages each contains a section with topics similar to those found in traditional newspapers. This layout limits the number of sections to a maximum of 8. If the user wants to use some of the virtual space for other applications, the number decreases further. As discussed in section 7.1, six is a good guess for how many sections an average reader wants. However, this number may increase when huge amounts of information become available. Changing the number of section pages will change the scale of the map. Also, if the number of section pages is increased, each section will no longer border the front page.
### 7.3.4 Importance
Traditional mass-produced newspapers assign priorities to articles by guessing at what the average reader will be interested in reading. Our personalized newspaper guesses at what each individual reader will be interested in. The guess boils down to a number indicating the priority of an article at a certain time. Provided the guess is somewhat accurate it will be an important factor when the user plans the "path" through the newspaper. As time passes the importance of the article will decrease. For some types of articles this will happen very quickly, while e.g. personal mail may be important until it's read.
### 7.3.5 Source
Compared to other media, newspapers are vague in their credits and references. Author or source is sometimes listed in the first or last line of the article, but this is not sufficient.
Source information is available at two levels. At the higher level, every article can be classified with regard to the wire service they origin from. Exceptions are electronic mail and USENET, but these are naturally considered separate sources.
In the Electronic Broadsheet, a "watermark" in the map icon is used to indicate the source of an article. The colored watermark is "woven" into the pixmap as a texture. Each source has a watermark. The initial letters are used for most sources while some, among them "email", have a hand crafted watermark.
The watermark metaphor, which is one of several implemented alternatives, has turned out to be a visually attractive solution while giving easy access to a valuable piece of information.
The hues of the watermarks were selected to make them as distinct as possible while keeping the value equal. Since the number of sources (ca. 20) is greater than the number of distinct hues we can generate at the relatively low intensity chosen. Therefore, a source will not be uniquely identified by the color, but the hue will play an important part in the recognition process and help lower the cognitive load of the reader.
### 7.3.6 Distribution
A constant challenge for recipients of electronic mail is to filter out important messages from the "junk mail". Personal mail is often important, while messages form mailing lists often can be ignored. Paper-based mail recipients have the same problem, but handwritten letters are often easier to distinguish from junk mail.
Knowing the distribution of a message is a valuable clue for the user when planning what to read. Currently, the system does not tell the user about article distribution other than naming the recipients.
### 7.3.7 Size/Length
When posting messages to USENET or mailing lists it is customary to warn readers if a message is extra long. By doing so, the sender provides recipients with clues for making the decision to read an article or not. The same information is hard to find in newspapers and especially in magazines where page jumps make articles longer than they look at first glance.
### 7.3.8 Summary of Design
Figure 17 summarizes the relationships between information items and design tools
.
## 7.4 The Electronic Map--Implementation
The pages in the electronic broadsheet are laid out in a "newspace", i.e., a two dimensional virtual plane. News articles naturally lend themselves to two-dimensional representations, and a map over the newspace is provided to ease navigation. The newspace map is based on a commonly used X11 window manager.
During the last year, the X11 community has seen the introduction of several so-called "virtual window managers" . The interface is clearly based on the concept pioneered by SDMS (see section 4.3). The window managers allow X11 displays to have a virtual plane larger than the physical screen size. The user can pan the real screen over the virtual screen to view a different part of the plane. The user interface for all the virtual window managers is implemented through a map. The user can see the outline of all top-level windows in a special window that is an isomorph representation of the windows on the virtual plane. The implication of using a virtual window manager for the users is that they have more real estate to lay windows out on. Instead of using overlapping windows, the screen can pan over a large bulletin board with partial.
### 7.4.1 X11 Window Managers
One merit of the X11 Window System is that it clearly separates the different parts of a window system into different processes--e.g., the task of managing the windows and the screen space is handled by the window manager. The window manager is with few exceptions an application just like any other X11 client The window manager is given authority to control the layout of windows on the screen. Other clients indicate their preferred position and size and this will normally be granted. However, the window manager can any time decide to move or resize a client window.
One of the first window managers available, TWM released by Tom LaStrange in April 1988, gained widespread popularity by offering the same functionality and ease of use as found in earlier window systems line Xerox Star and Apple Lisa [Hรฅfjeld et al. 88] . Recognizing its de facto position among window managers, the X Consortium adapted the program and extended the functionality for release 4 of X11.
Dave Edmondson of Imperial College modified the freely available source code for TWM to add the virtual feature. The new version, called VTWM (Virtual TWM), was made available in the summer of 1990 and quickly gained popularity. It was the first free window manager with virtual features. See Appendix A for background information.
### 7.4.2 Changes to VTWM
Based on the freely available code for VTWM, the NVTWM (Newspace VTWM) was developed to manage windows in the newspace.
The current release of VTWM creates a map of the virtual plane according to the user's specifications of scale and position. If VTWM manages more than one screen, each screen has its own map. Since it was feasible for the project to use a second screen for the map (see physical setup) VTWM had to be modified to allow a map from one screen to be displayed on another.
VTWM's representation of windows in the map consists of an isomorph rectangular area (technically an X11 window), optionally equipped with a name label. The user can specify background color and a label font. While this might be sufficient information in a programming environment where windows seldom are deleted or created, it is not sufficient in a news environment where new news comes in every minute. One way of increasing the information content in the window representations is to change the background pixmap. X11 already provides the functionality for clients to indicate icon pixmaps. While the use of icons is one way to manage a limited screen area, the virtual desktop model replaces traditional icons for most users. Using the icon pixmap functionality to set the background of the window representation was a natural modification to VTWM.
7.1 - Scale
7.2 - Map Metaphors
7.3 - What to display
7.3.1 - Display Tools
7.3.2 - Time/Age
7.3.3 - Sections
7.3.4 - Importance
7.3.5 - Source
7.3.6 - Distribution
7.3.7 - Size/Length
7.3.8 - Summary of Design
7.4 - The Electronic Map--Implementation
7.4.1 - X11 Window Managers
7.4.2 - Changes to VTWM
The Electronic Broadsheet - 30 JUN 95
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Generated with CERN WebMaker | 3,791 | 17,781 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-05 | latest | en | 0.88619 |
https://bathmash.github.io/HELM/29_1_line_ints_vecs-web/29_1_line_ints_vecs-webse4.html | 1,695,897,907,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510387.77/warc/CC-MAIN-20230928095004-20230928125004-00379.warc.gz | 141,019,715 | 12,323 | ### 4 Engineering Example 1
#### 4.1 Work done moving a charge in an electric field
Introduction
If a charge, $q$ , is moved through an electric field, $\underset{ฬฒ}{E}$ , from $A$ to $B$ , then the work required is given by the line integral
$\phantom{\rule{2em}{0ex}}{W}_{AB}=-q{\int }_{A}^{B}\underset{ฬฒ}{E}\cdot \underset{ฬฒ}{dl}$
Problem in words
Compare the work done in moving a charge through the electric field around a point charge in a vacuum via two different paths.
Mathematical statement of problem
An electric field $\underset{ฬฒ}{E}$ is given by
$\begin{array}{rcll}\underset{ฬฒ}{E}& =& \frac{Q}{4\pi {\epsilon }_{0}{r}^{2}}\underset{ฬฒ}{\stackrel{ฬ}{r}}& \text{}\\ & =& \frac{Q}{4\pi {\epsilon }_{0}\left({x}^{2}+{y}^{2}+{z}^{2}\right)}ร\frac{x\underset{ฬฒ}{i}+y\underset{ฬฒ}{j}+z\underset{ฬฒ}{k}}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& \frac{Q\left(x\underset{ฬฒ}{i}+y\underset{ฬฒ}{j}+z\underset{ฬฒ}{k}\right)}{4\pi {\epsilon }_{0}{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{\frac{3}{2}}}& \text{}\end{array}$
where $\underset{ฬฒ}{r}$ is the position vector with magnitude $r$ and unit vector $\stackrel{ฬ}{\underset{ฬฒ}{r}}$ , and $\frac{1}{4\pi {ฯต}_{0}}$ is a combination of constants of proportionality, where .
Given that $Q$ = $1{0}^{-8}$ C, find the work done in bringing a charge of $q$ = $1{0}^{-10}$ C from the point $A=\left(10,10,0\right)$ to the point $B=\left(1,1,0\right)$ (where the dimensions are in metres)
1. by the direct straight line $y=x$ , $z=0$
2. by the straight line pair via $C=\left(10,1,0\right)$
Figure 4:
The path comprises two straight lines from $A=\left(10,10,0\right)$ to $B=\left(1,1,0\right)$ via $C=\left(10,1,0\right)$ (see Figure 4).
Mathematical analysis
1. Here $qโ\left(4\pi {\epsilon }_{0}\right)$ = 90 so
$\phantom{\rule{2em}{0ex}}\underset{ฬฒ}{E}=\frac{90\left[x\underset{ฬฒ}{i}+y\underset{ฬฒ}{j}\right]}{{\left({x}^{2}+{y}^{2}\right)}^{\frac{3}{2}}}$
as $z=0$ over the region of interest. The work done
$\begin{array}{rcll}{W}_{AB}& =& -q{\int }_{A}^{B}\underset{ฬฒ}{E}\cdot \underset{ฬฒ}{dl}& \text{}\\ & =& -1{0}^{-10}{\int }_{A}^{B}\frac{90}{{\left({x}^{2}+{y}^{2}\right)}^{\frac{3}{2}}}\phantom{\rule{0.3em}{0ex}}\left[x\underset{ฬฒ}{i}+y\underset{ฬฒ}{j}\right]\cdot \left[dx\underset{ฬฒ}{i}+dy\underset{ฬฒ}{j}\right]\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$
Using $\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}y=x,\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}dy=dx$
2. The first part of the path is $A$ to $C$ where $x=10$ , $dx=0$ and $y$ goes from 10 to 1.
The second part is $C$ to $B$ , where $y=1$ , $dy=0$ and $x$ goes from 10 to 1.
The sum of the two components ${W}_{AC}$ and ${W}_{CB}$ is $5.73ร1{0}^{-9}$ J. Therefore the work done over the two routes is identical.
Interpretation
In fact, the work done is independent of the route taken as the electric field $\underset{ฬฒ}{E}$ around a point charge in a vacuum is a conservative field.
##### Example 11
1. Show that $I={\int }_{\left(0,0\right)}^{\left(2,1\right)}\left\{\left(2xy+1\right)dx+\left({x}^{2}-2y\right)dy\right\}$ is independent of the path taken.
2. Find $I$ using property P1.
3. Find $I$ using property P4.
4. Find $I={\oint }_{C}\left\{\left(2xy+1\right)dx+\left({x}^{2}-2y\right)dy\right\}$ where $C$ is
1. the circle ${x}^{2}+{y}^{2}=1$
2. the square with vertices $\left(0,0\right)$ , $\left(1,0\right)$ , $\left(1,1\right)$ , $\left(0,1\right)$ .
##### Solution
1. The integral $I={\int }_{\left(0,0\right)}^{\left(2,1\right)}\left\{\left(2xy+1\right)dx+\left({x}^{2}-2y\right)dy\right\}$ may be re-written ${\int }_{C}\underset{ฬฒ}{F}\cdot \underset{ฬฒ}{dr}$ where
$\underset{ฬฒ}{F}=\left(2xy+1\right)\underset{ฬฒ}{i}+\left({x}^{2}-2y\right)\underset{ฬฒ}{j}$ .
Now $\underset{ฬฒ}{\nabla }ร\underset{ฬฒ}{F}=\left|\begin{array}{ccc}\hfill \underset{ฬฒ}{i}\hfill & \hfill \underset{ฬฒ}{j}\hfill & \hfill \underset{ฬฒ}{k}\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \frac{\partial }{\partial x}\hfill & \hfill \frac{\partial }{\partial y}\hfill & \hfill \frac{\partial }{\partial z}\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill 2xy+1\hfill & \hfill {x}^{2}-2y\hfill & \hfill 0\hfill \end{array}\right|=0\underset{ฬฒ}{i}+0\underset{ฬฒ}{j}+0\underset{ฬฒ}{k}=\underset{ฬฒ}{0}$
As $\underset{ฬฒ}{\nabla }ร\underset{ฬฒ}{F}=\underset{ฬฒ}{0}$ , $\underset{ฬฒ}{F}$ is a conservative field and $I$ is independent of the path taken between $\left(0,0\right)$ and $\left(2,1\right)$ .
2. As $I$ is independent of the path taken from $\left(0,0\right)$ to $\left(2,1\right)$ , it can be evaluated along any such path. One possibility is the straight line $y=\frac{1}{2}x$ . On this line, $dy=\frac{1}{2}dx$ . The integral $I$ becomes $\begin{array}{rcll}I& =& {\int }_{\left(0,0\right)}^{\left(2,1\right)}\left\{\left(2xy+1\right)dx+\left({x}^{2}-2y\right)dy\right\}& \text{}\\ & =& {\int }_{x=0}^{2}\left\{\left(2xร\frac{1}{2}x+1\right)dx+\left({x}^{2}-4x\right)\frac{1}{2}dx\right\}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& {\int }_{0}^{2}\left(\frac{3}{2}{x}^{2}-\frac{1}{2}x+1\right)dx& \text{}\\ & =& {\left[\frac{1}{2}{x}^{3}-\frac{1}{4}{x}^{2}+x\right]}_{0}^{2}=4-1+2-0=5& \text{}\end{array}$
3. If $\underset{ฬฒ}{F}=\underset{ฬฒ}{\nabla }\varphi$ then
$\begin{array}{ccc}\hfill \frac{\partial \varphi }{\partial x}=2xy+1\hfill & \hfill \to \hfill & \hfill \varphi ={x}^{2}y+x+f\left(y\right)\hfill \\ \hfill \hfill \\ \hfill \frac{\partial \varphi }{\partial y}={x}^{2}-2y\hfill & \hfill \to \hfill & \hfill \varphi ={x}^{2}y-{y}^{2}+g\left(x\right)\hfill \end{array}}\to \varphi ={x}^{2}y+x-{y}^{2}+C$ .
These are consistent if $\varphi ={x}^{2}y+x-{y}^{2}$ (plus a constant which may be omitted since it cancels).
So $I=\varphi \left(2,1\right)-\varphi \left(0,0\right)=\left(4+2-1\right)-0=5$
4. As $F$ is a conservative field, all integrals around a closed contour are zero.
##### Exercises
1. Determine whether the following vector fields are conservative
1. $\underset{ฬฒ}{F}=\left(x-y\right)\underset{ฬฒ}{i}+\left(x+y\right)\underset{ฬฒ}{j}$
2. $\underset{ฬฒ}{F}=3{x}^{2}{y}^{2}\underset{ฬฒ}{i}+\left(2{x}^{3}y-1\right)\underset{ฬฒ}{j}$
3. $\underset{ฬฒ}{F}=2x\underset{ฬฒ}{i}+\left(xz-2\right)\underset{ฬฒ}{j}+xy\underset{ฬฒ}{k}$
4. $\underset{ฬฒ}{F}={x}^{2}z\underset{ฬฒ}{i}+{y}^{2}z\underset{ฬฒ}{j}+\frac{1}{3}\left({x}^{3}+{y}^{3}\right)\underset{ฬฒ}{k}$
2. Consider the integral ${\int }_{C}\underset{ฬฒ}{F}\cdot \underset{ฬฒ}{dr}$ with $\underset{ฬฒ}{F}=3{x}^{2}{y}^{2}\underset{ฬฒ}{i}+\left(2{x}^{3}y-1\right)\underset{ฬฒ}{j}$ . From Exercise 1(b) $\underset{ฬฒ}{F}$ is a conservative vector field. Find a scalar field $\varphi$ so that $\underset{ฬฒ}{\nabla }\varphi =\underset{ฬฒ}{F}$ . Hence use P4 to evaluate the integral ${\int }_{C}\underset{ฬฒ}{F}\cdot \underset{ฬฒ}{dr}$ where $C$ is an integral with start-point $\left(0,0\right)$ and end point $\left(1,4\right)$ .
3. For the following conservative vector fields $\underset{ฬฒ}{F}$ , find a scalar field $\varphi$ such that $\underset{ฬฒ}{\nabla }\varphi =\underset{ฬฒ}{F}$ and hence evaluate the $I={\int }_{C}\underset{ฬฒ}{F}\cdot \underset{ฬฒ}{dr}$ for the contours $C$ indicated.
1. $\underset{ฬฒ}{F}=\left(4{x}^{3}y-2x\right)\underset{ฬฒ}{i}+\left({x}^{4}-2y\right)\underset{ฬฒ}{j}$ ; any path from $\left(0,0\right)$ to $\left(2,1\right)$ .
2. $\underset{ฬฒ}{F}=\left({e}^{x}+{y}^{3}\right)\underset{ฬฒ}{i}+\left(3x{y}^{2}\right)\underset{ฬฒ}{j}$ ; closed path starting from any point on the circle ${x}^{2}+{y}^{2}=1$ .
3. $\underset{ฬฒ}{F}=\left({y}^{2}+sinz\right)\underset{ฬฒ}{i}+2xy\underset{ฬฒ}{j}+xcosz\underset{ฬฒ}{k}$ ; any path from $\left(1,1,0\right)$ to $\left(2,0,\pi \right)$ .
4. $\underset{ฬฒ}{F}=\frac{1}{x}\underset{ฬฒ}{i}+4{y}^{3}{z}^{2}\underset{ฬฒ}{j}+2{y}^{4}z\underset{ฬฒ}{k}$ ; any path from $\left(1,1,1\right)$ to $\left(1,2,3\right)$ .
1. No,
2. Yes,
3. No,
4. Yes
1. ${x}^{3}{y}^{2}-y+C$ , $12$
1. ${x}^{4}y-{x}^{2}-{y}^{2}$ $11$ ;
2. ${e}^{x}+x{y}^{3}$ $0$ ;
3. $x{y}^{2}+xsinz$ $-1$ ;
4. $lnx+{y}^{4}{z}^{2}$ , $143$ | 3,733 | 8,517 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 112, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2023-40 | latest | en | 0.629741 |
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## Transcript
One of the simplest mathematical statements possible is 2+2=4.
While the concept is very easy to understand, when you write it down you have to use mathematical symbols which are, historically speaking, a relatively recent invention.
At one point, mathematicians were doing reasonably complicated work without the benefit of symbols at all. Something which is unthinkable today.
Learn more about mathematical symbols, where they came from, and why they exist, on this episode of Everything Everywhere Daily.
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As I mentioned in the intro, there was a time when mathematic was done without symbols. If you can imagine doing your elementary school math problems without the use of plus, minus, or equal symbols, you can realize how hard this would be.
In fact, it would be really difficult to do right now without the use of symbols.
The first people we know of who used mathematics were the ancient Babylonians and Sumerians. With their cuniform system of writing, they were able to do reasonably complicated mathematics.
Their numeral system was base-60, as opposed to ours which is base-10. The theory holds that two earlier people merged to become the Sumerians and one group had a system that was base-12 and the other had a system that was base-5. They resolved the difference by using 60, which was just 5 x 12.
They were able to solve quadratic equations, knew about square and cube roots, and had solved the Pythagorean theorem well before Pythagoras.
However, they lacked a few things. They didnโt have a zero, which is something I talked about in my episode about zero, and didnโt really have any symbolic expressions to do equations.
It wouldnโt look like algebra as we are familiar with it.
The Egyptians, Greeks, Romans, and Arabs all managed to do mathematics at some level without the use of mathematical symbols.
Algebra was actually named by Arab scholars and it literally comes from โal-jabrโ which means a reunion of broken parts.
Arab scholars probably took mathematics as far as anyone in history up until that time, but they still mostly werenโt using symbolic notation.
The last great classic Arab mathematician from the early 15th century Ab? al-Hasan ibn Al? al-Qalas?d?, used symbols, but they were just letters from the Arabic alphabet.
The symbols we know and use today werenโt created until the 15th century.
The first use of the plus sign was in 1489 by German mathematician Johannes Widmann.
The plus sign just represents the letter โtโ which was a short form of the Latin word โetโ which means โandโ.
Likewise, Widmann was the first person to use the minus sign as well. The minus sign is believed to come from a tilde which was sometimes placed over a number to represent subtraction.
In his treatise, he explicitly defined the new terms which he created. He said โwas ? ist, das ist minus, und das + ist das merโ. ย Mer being German for more.
There were other previous attempts to create symbols that did the same thing, but they didnโt catch on. The Egyptians had a symbol that could be used for addition and the mirror image of it could be used for subtraction, but it never went beyond Egypt.
Not long after, in the early 17th century, the multiplication symbol was created. This is of course just the letter โxโ.
The first use of โxโ to denote multiplication was in 1618 by Scottish mathematician John Napier in his book Mirifici Logarithmorum Canonis Descriptio.
He too explained the use of this new symbol in the book by saying, โMultiplication of species [i.e. unknowns] connects both proposed magnitudes with the symbol โinโ or ร: or ordinarily without the symbol if the magnitudes be denoted with one letter.โ
Technically, in printing, the multiplication symbol isnโt actually the letter x. It is a slightly smaller character of the same shape which is raised up.
There can be confusion when using a regular keyboard with โxโ as a multiplication symbol and using โxโ as a variable.ย Gottfried Leibniz, one of the co-inventors of calculus, disliked using an โxโ for this reason.
For that reason, a dot is also sometimes used as a multiplication symbol. This is usually more popular in Europe, and it too can be confusing because a dot is used for a special type of vector multiplication.
With the advent of computers, the asterisk * has been adopted for multiplication simply because it is in the ASCII character set.
Just as with multiplication, there are several symbols for division as well.
The earliest of the modern division symbols which we use is called the obelus. This is the straight line with a dot above and below it.
This was first used in 1659 by Swiss mathematician Johann Rahn.
Of all of the symbols Iโve mentioned, this one has been deprecated by modern mathematicians. In fact, you canโt really find in use very much at all outside of elementary school math courses and the division keys on some calculators.
Personally, I hate the obelus. I found it really confusing and I donโt think kids should be taught division using it because they will never see it again in their lives.
The preferred division symbol is called the solidus or the forward slash. This is very similar to, and conveys the same meaning, as the horizontal line used in fractions. This was a much later creation and wasnโt to represent division until 1845.
The adoption of computers only strengthened the use of the solidus over the obelus as the obelus isnโt on most keyboards.
The equal symbol has a very interesting origin story.
The equal symbol was first used in 1557 by Welsh mathematician Robert Recorde in his book The Whetstone of Witte.
In his book, he was writing equations, and over 200 times he had to write the phrase โis equal toโ. He basically got sick of writing it over and over, so he eventually created a symbol so he didnโt have to write it anymore.
He said in his book, โAnd to avoid the tedious repetition of these words: โis equal toโ I will set as I do often in work use, a pair of parallels, or duplicate lines of one [the same] length, thus: =, because no 2 things can be more equal.โ
There is a similar, lesser-used symbol, with three parallel lines simply called the triple bar. It was first used in 1801 by Carl Friedrich Gauss and it is sometimes used in logic, or in modular arithmetic.
The percent sign comes from the Italian phrase per cento. It was abbreviated as a โpโ with two zeros, and eventually, the โpโ was removed and it was just a slanted line with two zeros.
The square root symbol might have come from an Arabic letter that was used by the above-mentioned al-Qalas?d?, or possibly from a lower case letter โrโ.
The first use in 1525 just looked like a checkmark. The horizontal line on top is called a vinculum, and it was added to the checkmark symbol in 1637 by Rene Descartes to create the modern symbol we use today.
The greater than and less than symbols were created in 1631 by Englishman Thomas Harriot in his book The Analytical Arts Applied to Solving Algebraic Equations.
The infinity symbol, that being the number 8 on its side, is even older than the modern number 8, which is a Hindu-Arabic number. The earliest evidence for it goes back to the cross of St. Boniface in the 7th or 8th century.
The first use of it to signify infinity wasnโt until 1655. English clergyman John Wallis used it in his book De sectionibus conicis. There was no explanation given as to why it was selected, but one hypothesis is that it is a variant of the symbol used for the Roman Number 1000, which was the letter C, followed by an I, and then with a backward C.
The last symbol Iโll go over is that of pi.
Pi of course is just the greek letter pi. However, its use to represent the ratio of the circumference of a circle to its diameter is actually relatively recent.
The knowledge that the ratio of the circumference to the diameter of a circle was special goes back to ancient China and Egypt.
What we refer to as the number pi began the use of the Greek letter delta and pi. Pi was chosen because it was the first letter of the word โperimeterโ, and delta was the first letter of โdiameterโ.ย Englishman William Oughtred first used pi over delta in 1647.
The first use of the letter pi all by itself to represent the ratio was in 1706 by the Welsh mathematician William Jones.
There is a lot to be said about pi, but Iโll save that for a later episode, probably for next yearโs Pi day.
You might have noticed that most of these symbols, especially the main ones, all began being used over the course of a 100 years, starting in the late 15th century. Basically, once people started to use symbols, it made mathematics easier, and then more people began to adopt them as a shorthand for more things.
Mathematical symbols are still being created today as new branches of mathematics create new ideas which need to be easily expressed.
If you think about it, math symbols really arenโt that much different than emojis. It is just a way to convey a complex thought into a single character. | 2,191 | 9,915 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-50 | longest | en | 0.951285 |
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# Georgia: 4/1 - 4/30/2021
1188 replies. Last post 14 days ago by meriwetherman.
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soljoiGeorgiaUnited StatesMemberย #151550January 20, 20142590ย PostsOffline Aprilย 5,ย 2021, 12:43ย pmQuote: Originally posted by BillyBlades on April 5, 2021all unhit pairs (no repeats):00ย 11ย 22ย 33ย 44ย ย 66ย 77ย ย 99 01ย 12ย 23ย 34ย 45ย ย ย 78ย ย ย ย ย 13ย ย 35ย 46ย 57ย ย 79ย 03ย 14ย ย 36ย 47ย 58ย ย ย ย ย ย 15ย 26ย 37ย 48ย ย ย ย ย ย 27ย 38ย 49ย ย ย ย ย ย ย 28ย 39ย ย ย 07ย 18ย 29ย ย ย ย ย ย ย 19ย ย 334ย 687744438600ย 27xย congrats BB
soljoiGeorgiaUnited StatesMemberย #151550January 20, 20142590ย PostsOffline Aprilย 5,ย 2021, 12:46ย pmQuote: Originally posted by dimples123 on April 5, 2021247 208 934 776
Darryl1224QueensUnited StatesMemberย #174822May 15, 20164030ย PostsOffline Aprilย 5,ย 2021, 12:50ย pmCongrats
BillyBladesAtlanta, GAUnited StatesMemberย #163904February 13, 20158671ย PostsOffline Aprilย 5,ย 2021, 12:58ย pmQuote: Originally posted by BillyBlades on April 5, 2021all unhit pairs (no repeats):ย 11ย 22ย ย ย ย 66ย 77ย ย 99 01ย 12ย 23ย ย 45ย ย ย ย ย ย ย ย 13ย ย 35ย 46ย 57ย ย 79ย 03ย 14ย ย 36ย ย 58ย ย ย ย ย ย 15ย 26ย 37ย ย ย ย ย ย 27ย ย 49ย ย ย ย ย ย ย ย 28ย 39ย ย ย 07ย 18ย 29ย ย ย ย ย ย ย ย 19ย ย all unhit pairs (no repeats):ย 11ย 22ย ย ย ย 66ย 77ย ย 99 01ย 12ย 23ย ย 45ย ย ย ย ย ย ย ย ย 13ย ย 35ย 46ย 57ย ย 79ย 03ย 14ย ย 36ย ย 58ย ย ย ย ย ย ย 15ย 26ย 37ย ย ย ย ย ย ย 27ย ย 49ย ย ย ย ย ย ย ย 28ย 39ย ย ย 07ย 18ย 29ย ย ย ย ย ย ย ย 19ย ย ย 724Sabali
BillyBladesAtlanta, GAUnited StatesMemberย #163904February 13, 20158671ย PostsOffline Aprilย 5,ย 2021, 12:59ย pmQuote: Originally posted by BillyBlades on March 30, 2021pairs (4/01 - 4/07):01x, 05x, 06x, 07x, 16x, 18x 19x, 23x, 24x, 25x, 26x, 35x 39x, 45x, 47x, 69x, 78x, 89x24x, 47x = 724 ย Sabali
BillyBladesAtlanta, GAUnited StatesMemberย #163904February 13, 20158671ย PostsOffline Aprilย 5,ย 2021, 1:00ย pmQuote: Originally posted by JKTAYLOR1 on April 5, 2021The Infamous \$99 Wheel (Pickย 3)Number used: 6008 Numbers to play:307, 490, 543, 502, 579, 724, 618, 932724 straight! Sabali
BillyBladesAtlanta, GAUnited StatesMemberย #163904February 13, 20158671ย PostsOffline Aprilย 5,ย 2021, 1:03ย pmQuote: Originally posted by dimples123 on April 5, 2021247 208 934 776724 box Sabali
Atlanta, GA
United States
Memberย #163904
February 13, 2015
8671ย Posts
Offline
APRย 5th history:
Comboย ย Timesย Drawnย ย Expectedย ย Lastย Seenย ย Drawsย Since
0162110.32Aprย 5,ย 200611225
0181110.32Aprย 5,ย 20200
0251110.32Aprย 5,ย 200015294
0281110.32Aprย 5,ย 200213938
033155.32Aprย 5,ย 20126993
0341110.32Aprย 5,ย 200710545
0351110.32Aprย 5,ย 20182195
0361110.32Aprย 5,ย 20099184
0372110.32Aprย 5,ย 200114616
0381110.32Aprย 5,ย 20164071
0571110.32Aprย 5,ย 200313260
0891110.32Aprย 5,ย 20191098
118155.32Aprย 5,ย 20117726
1231110.32Aprย 5,ย 20201
1361110.32Aprย 5,ย 20202
1371110.32Aprย 5,ย 20191100
1381110.32Aprย 5,ย 200412580
1451110.32Aprย 5,ย 200511903
1791110.32Aprย 5,ย 20173290
188155.32Aprย 5,ย 20136264
1891110.32Aprย 5,ย 20108455
199155.32Aprย 5,ย 200313259
223155.32Aprย 5,ย 20145533
227155.32Aprย 5,ย 199915973
229155.32Aprย 5,ย 200511902
2361110.32Aprย 5,ย 200015293
2451110.32Aprย 5,ย 20173288
2472110.32Aprย 5,ย 20164072
2481110.32Aprย 5,ย 200213937
2591110.32Aprย 5,ย 20136263
2671110.32Aprย 5,ย 20182193
2693110.32Aprย 5,ย 20154803
344155.32Aprย 5,ย 200611224
3451110.32Aprย 5,ย 20173289
3471110.32Aprย 5,ย 20126994
3491110.32Aprย 5,ย 200412581
355155.32Aprย 5,ย 20117725
3571110.32Aprย 5,ย 200710544
3791110.32Aprย 5,ย 199717088
4561110.32Aprย 5,ย 20191099
4571110.32Aprย 5,ย 20182194
4581110.32Aprย 5,ย 199816651
4691110.32Aprย 5,ย 20089865
477155.32Aprย 5,ย 20089864
5781110.32Aprย 5,ย 20108456
588155.32Aprย 5,ย 199617453
668155.32Aprย 5,ย 199517819
historical repeat
Sabali
BillyBladesAtlanta, GAUnited StatesMemberย #163904February 13, 20158671ย PostsOffline Aprilย 5,ย 2021, 1:07ย pm724 key picks470, 471, 472, 473, 474, 475, 476, 477, 478, 479440, 441, 442, 443, 444, 445, 446, 447, 448, 449400, 401, 402, 403, 404, 405, 406, 407, 408, 409420, 421, 422, 423, 424, 425, 426, 427, 428, 429970, 971, 972, 973, 974, 975, 976, 977, 978, 979940, 941, 942, 943, 944, 945, 946, 947, 948, 949900, 901, 902, 903, 904, 905, 906, 907, 908, 909920, 921, 922, 923, 924, 925, 926, 927, 928, 929Sabali
Page 15 of 80 | 3,199 | 7,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-21 | latest | en | 0.454037 |
http://www.mathworks.com/help/fininst/hjmtimespec.html?requestedDomain=www.mathworks.com&nocookie=true | 1,519,368,967,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814493.90/warc/CC-MAIN-20180223055326-20180223075326-00532.warc.gz | 495,920,721 | 14,832 | # Documentation
### This is machine translation
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# hjmtimespec
Specify time structure for Heath-Jarrow-Morton interest-rate tree
## Syntax
```TimeSpec = hjmtimespec(ValuationDate,Maturity,Compounding) ```
## Arguments
`ValuationDate` Scalar date marking the pricing date and first observation in the tree. Specify as serial date number or date character vector. `Maturity ` Number of levels (depth) of the tree. A number of levels (`NLEVELS`-by-`1`) vector of dates marking the cash flow dates of the tree. Cash flows with these maturities fall on tree nodes. Maturity should be in increasing order. `Compounding` (Optional) Scalar value representing the rate at which the input zero rates were compounded when annualized. Defaultย =ย `1`. This argument determines the formula for the discount factors: `Compounding` = `1`, `2`, `3`, `4`, `6`, `12````Disc = (1 + Z/F)^(-T)```, where `F` is the compounding frequency, `Z` is the zero rate, and `T` is the time in periodic units; for example, `T = F` is one year.`Compounding` = `365` ```Disc = (1 + Z/F)^(-T)```, where `F` is the number of days in the basis year and `T` is a number of days elapsed computed by basis.`Compounding` = `โ1````Disc = exp(-T*Z)```, where `T` is time in years.
## Description
`TimeSpec = hjmtimespec(ValuationDate,Maturity,Compounding)` sets the number of levels and node times for an HJM tree and determines the mapping between dates and time for rate quoting.
`TimeSpec` is a structure specifying the time layout for `hjmtree`. The state observation dates are `[Settle; Maturity(1:end-1)]`. Because a forward rate is stored at the last observation, the tree can value cash flows out to `Maturity`.
## Examples
collapse all
This example shows how to specify an eight-period tree with semiannual nodes (every six months) and use exponential compounding to report rates.
```Compounding = -1; ValuationDate = '15-Jan-1999'; Maturity = datemnth(ValuationDate, 6*(1:8)'); TimeSpec = hjmtimespec(ValuationDate, Maturity, Compounding)```
```TimeSpec = struct with fields: FinObj: 'HJMTimeSpec' ValuationDate: 730135 Maturity: [8x1 double] Compounding: -1 Basis: 0 EndMonthRule: 1 ``` | 607 | 2,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-09 | latest | en | 0.823315 |
https://socratic.org/questions/how-do-you-solve-y-2-3-56 | 1,725,892,857,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651103.13/warc/CC-MAIN-20240909134831-20240909164831-00696.warc.gz | 514,041,252 | 5,890 | How do you solve y^2 = 3.56?
Oct 19, 2015
$y = 1.88679622641$ using a calculator
Explanation:
${y}^{2} = 3.56$
Take the square root of both sides.
$\sqrt{{y}^{2}} = \sqrt{3.56}$
$y = \sqrt{3.56}$
$y = 1.88679622641$ using a calculator | 101 | 242 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-38 | latest | en | 0.649175 |
https://www.businessinsider.in/science/health/news/an-ultramarathoner-ran-over-100-miles-in-her-house-after-her-races-were-cancelled-due-to-the-pandemic/articleshow/76202664.cms | 1,716,982,112,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059239.72/warc/CC-MAIN-20240529103929-20240529133929-00293.warc.gz | 571,292,256 | 24,304 | An ultramarathoner ran over 100 miles in her house after her races were cancelled due to the pandemic
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5. An ultramarathoner ran over 100 miles in her house after her races were cancelled due to the pandemic
An ultramarathoner ran over 100 miles in her house after her races were cancelled due to the pandemic
Gabby Landsverk
โข Preschool teacher Stephanie Northway has been running ultramarathons, or races of more than 26 miles, since 2015, logging as many as 24 events per year.
โข When her events were cancelled this year due to the pandemic, Northway and her son designed 50- and 100-mile courses in her house and backyard.
โข Since April, she's completed a total of 500 miles worth of races, and earned an award from a national trail running organization for "Most Creative Race."
As an ultramarathon runner, conditioned to endure distances of up to 100 miles, Stephanie Northway is used to being patient and staying strong in the face of adversity.
But when her slate of running events this spring was cancelled due to the pandemic, she decided to just keep moving, starting by recreating a 100 mile Tennessee ultramarathon. Rather than switching hobbies, Northway and her son, a math whiz, decided to reroute all 100 miles through their Michigan home.
"I thought, I'm trained for it, I'm ready to go, so my son and I got out the tape measure and figured out it was doable," she told Insider.
But she didn't stop there. Since April, Northway has run a total of 500 miles (four 100 mile routes, and two 50 milers) in her house and backyard.
One mile indoors equals about 106 laps around the living room, Northway calculated
First, Northway ran 100 miles by jogging through her house and around the backyard. But then she decided to seek even more of a challenge by designing the full route completely indoors.
For her follow-up home 100 miler, Northway decided to run half the race in her living room and half in a hallway and spare bedroom. She and her son got to work with a tape measure calculating the distance. They found that one mile was equal to about 50 laps from the hallway to the spare bedroom, and 106 laps around the living room.
That kind of a small loop presents some logistical challenges, Northway said. First, GPS doesn't accurately measure distance and speed on that scale, so she had to manually count her laps with tally marks and keep her own time.
Second, the tiny space makes it difficult for a runner to hit her stride, since it requires constant turning around.
"Inside my house could never get a good pace, If you go too fast, you get kind of dizzy," Northway said.
As a result, she kept about a 14 minute per mile pace throughout, which she said took a little patience and a lot of persistence.
"To stay motivated, you have to not worry about how fast you're going," she said. "It may seem like you're not getting anywhere but you are."
But the work paid off, too โ Northway's home 100 miler earned her a "Most Creative Race" award from the national trail and endurance running organization Aravaipa Running.
Altogether, Northway has put together more than half a dozen events, ranging up to 131 miles each, all of which she's recorded in detail on her blog.
In addition to the long runs, Northway also took on challenges like climbing the height of Mount Everest on her stairs and running seven marathons in as many days.
At-home aid stations and naps fuel Northway's long hour runs
Despite the challenges, there are some perks to an at-home race, Northway said.
One of the highlights has been setting up in-home aid stations (stops where runners can snack and re-hydrate) so she can refuel on favorites like Gatorade, coffee, Coca Cola, Pop Tarts and bananas.
"Your fridge is always right there, the bathroom is always right there, you can turn the TV on and watch a show while you're running," she said.
Many ultra races typically have a cut-off point for time after which runners forfeit (around 30 hours for most 100 mile races). For the first two races, Northway stuck to that goal, finishing in just under 23 hours and right around 30 hours. But since then, Northway's eased up on her endeavors, stopping for meals and even to sleep if she needs to.
"I keep the clock running and consider it all part of the race," she said.
Another benefit of running at home is a constant cheering squad โ Northway's husband and son will often hang out with her as she runs to cheer her on and keep her company, something that helps pass the long hours and miles during the race.
You don't have to do 100 miles, or even 50, to run at home
Although Northway is a fitness aficionado, she said one of her favorite parts of running is that it's very accessible, and that people can benefit from adding even a little to their daily routines.
"Right now is the perfect time to start doing more, people have more time. Just get out there and move," she said.
While running might be intimidating for some people, especially at long distances, Northway said it doesn't have to be. A good starting point is to jog half a mile, followed by a half-mile walk, and gradually increase the running time and overall distance as you get comfortable. She said it's easy that you probably think, and you may even be surprised by what you can already accomplish.
"People are probably much better than they think they are when it comes to things like this," she said.
And whether you're new to running, or tackling a marathon or more, Northway has the same mantra.
"Just keep moving forward one step at a time, that's kind of my motto," she said. "Any little bit that you can, just keep moving forward." | 1,274 | 5,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-22 | latest | en | 0.98194 |
https://gamedev.stackexchange.com/questions/31170/drawing-a-dynamic-indicator-for-a-field-of-view | 1,713,986,642,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819847.83/warc/CC-MAIN-20240424174709-20240424204709-00816.warc.gz | 230,936,534 | 41,939 | # Drawing a dynamic indicator for a field of view
My goal is to draw a dynamic indicator for a cube, which represents the field of view. The ideal indicator would look the same as in Metal Gear Solid(the blue ones):
From what I have gathered, I need an angle and a length for the indicator as such for a 50ยฐ FoV:
Now my issue is what method should I use to draw the indicator? A procedural mesh or is there a less demanding way of doing this?
I can't give you a Unity-specific answer (sorry!), but I can tell how I would solve this. I would generate a bunch of points on a circle using the blue vector.
First, what is a circle? Well, a circle looks like this:
However, graphics hardware can't really draw perfect circles. You're always going to end up with a polygon approximation:
We do this approximation by using sine and cosine functions. If you remember from trigonometry, a circle is 360 degrees or 2 pi. Sine and cosine functions take a value out of 2 pi and give you a distance either in the x (cosine) or y (sine) axis. A very useful property indeed.
In code:
x = (cos(degrees_to_radians(angle));
Now, what you want is not a circle, but a half-circle. Specifically, you want this:
This can be achieved by using an angle_start and an angle_end. We then loop over this using an angle_delta that is calculated by dividing the difference between angle_start and angle_end over the quality we want.
Another useful property of approximating circles with sine and cosine (also known as the "unit circle") is that it can be shrunk or grown using a radius value. So let's add a radius to our pseudo-code:
x = (cos(degrees_to_radians(angle)) * radius;
How can we use this for our field-of-view cone? Well, we can use the same circle but draw it at different distances:
And when I add a dotted line to connect the two half-circles, it's starting to look like a cone!
Here, we have two radii: dist_min and dist_max. Our code becomes:
x_min = (cos(degrees_to_radians(angle)) * dist_min;
However, we also need two angles: one for the current step and one for the next step. We'll call these angle_curr and angle_next. Now our code becomes:
x_curr_min = (cos(degrees_to_radians(angle_curr)) * dist_min;
Now we have four points on our half-circles. Let's store these points as triangles. Here's a diagram that should help clear up how can make triangles out of these four points:
So, what do we need to do draw a cone?
โข Determine the start angle and the end angle. This is done using the enemy's direction and field of view variable.
โข Determine the quality of our approximation. 15 steps should be good enough for most purposes.
โข Loop over the difference between start and end using the delta ((end - start) / quality).
โข Convert our angles to four points.
โข Convert the four points to two triangles.
When we combine these steps, here's what we will end up with:
In full pseudo-code:
int quality = 15;
float dist_min = 0.5f;
float dist_max = 15.f;
vec2 pos = GetEnemyPosition();
float angle_lookat = GetEnemyAngle();
float angle_fov = GetEnemyFieldOfView();
float angle_start = angle_lookat - angle_fov;
float angle_end = angle_lookat + angle_fov;
float angle_delta = (angle_end - angle_start) / quality;
float angle_curr = 0.f;
float angle_next = angle_delta;
for (int i = 0; i < quality - 1; i++)
{
vec2 sphere_curr;
vec2 sphere_next;
vec2 pos_curr_min = pos + sphere_curr * dist_min;
vec2 pos_curr_max = pos + sphere_curr * dist_max;
vec2 pos_next_min = pos + sphere_next * dist_min;
vec2 pos_next_max = pos + sphere_next * dist_max;
WriteTriangle(pos_curr_min, pos_curr_max, pos_next_max);
WriteTriangle(pos_next_max, pos_curr_min, pos_next_min);
angle_curr += angle_delta;
angle_next += angle_delta;
}
This should create the cone you want. Try playing with the parameters (dist_min, dist_max and quality) to get the effect you want. It should only be a few triangles per enemy, I don't think it's much of a problem to generate this every frame. Only consider optimizing it when it becomes a problem.
โข I tried to figure out what happens in there, hard without any comments, but apparently your WriteTriangle method draws a procedural mesh according to the angles?
โย Esa
Jun 27, 2012 at 7:21
โข I always try to make code readable without any comments (because comments lie). However, in this case I will have to make an exception. WriteTriangle is a method that takes three vec2's and saves it as a triangle in some kind of mesh structure. Internally, the mesh would store a (dynamic) list of vertices and use that to render a model. Jun 27, 2012 at 7:47
โข I have updated my answer (with pictures!) to hopefully clear things up for you. Jun 27, 2012 at 8:28
โข +1. Very detailed answer. Just lacks the texturing to get the same look as in the reference image :) Jun 27, 2012 at 8:50
โข +1 from me as well. Great answer. @bummzack I think solving gradient like texturing deserves a question on its own. Oct 15, 2017 at 9:21
Good day y'all!
I've implemented knight666's answer from pseudo-code to Unity-code (C#). Some slight changes were needed but it works like charm, just attach the script to an enemy. I don't know if there are more efficient ways to do some of the things.
One thing worth mentioning is that the number of vertices can be reduced from 4*quality to 2*quality+2, since for example triangle 2 and 3 share two vertices. This is true if the circle segment is continuous, but in my application it isn't, which is my I didn't optimize that part :)
Also note that you cannot set mesh.vertices[i] or mesh.triangles[i], which is why I used the auxiliary arrays vertices and triangles (which probably should be declared in global space for speed).
The code is sparsely commented since knight666's awesome answer and pics say it all. The biggest difference is that his answer is in right hand coordinates, whereas Unity works in left hand coordinates, so some sin & cos needed switching places as well as the order of the vertices in the triangles to get the correct normal (y axis in this case)
using UnityEngine;
using System.Collections;
public class drawFieldOfView : MonoBehaviour {
int quality = 15;
Mesh mesh;
public Material material;
float angle_fov = 40;
float dist_min = 5.0f;
float dist_max = 15.0f;
void Start()
{
mesh = new Mesh();
mesh.vertices = new Vector3[4 * quality]; // Could be of size [2 * quality + 2] if circle segment is continuous
mesh.triangles = new int[3 * 2 * quality];
Vector3[] normals = new Vector3[4 * quality];
Vector2[] uv = new Vector2[4 * quality];
for (int i = 0; i < uv.Length; i++)
uv[i] = new Vector2(0, 0);
for (int i = 0; i < normals.Length; i++)
normals[i] = new Vector3(0, 1, 0);
mesh.uv = uv;
mesh.normals = normals;
}
void Update()
{
float angle_lookat = GetEnemyAngle();
float angle_start = angle_lookat - angle_fov;
float angle_end = angle_lookat + angle_fov;
float angle_delta = (angle_end - angle_start) / quality;
float angle_curr = angle_start;
float angle_next = angle_start + angle_delta;
Vector3 pos_curr_min = Vector3.zero;
Vector3 pos_curr_max = Vector3.zero;
Vector3 pos_next_min = Vector3.zero;
Vector3 pos_next_max = Vector3.zero;
Vector3[] vertices = new Vector3[4 * quality]; // Could be of size [2 * quality + 2] if circle segment is continuous
int[] triangles = new int[3 * 2 * quality];
for (int i = 0; i < quality; i++)
{
Vector3 sphere_curr = new Vector3(
Mathf.Sin(Mathf.Deg2Rad * (angle_curr)), 0, // Left handed CW
Vector3 sphere_next = new Vector3(
pos_curr_min = transform.position + sphere_curr * dist_min;
pos_curr_max = transform.position + sphere_curr * dist_max;
pos_next_min = transform.position + sphere_next * dist_min;
pos_next_max = transform.position + sphere_next * dist_max;
int a = 4 * i;
int b = 4 * i + 1;
int c = 4 * i + 2;
int d = 4 * i + 3;
vertices[a] = pos_curr_min;
vertices[b] = pos_curr_max;
vertices[c] = pos_next_max;
vertices[d] = pos_next_min;
triangles[6 * i] = a; // Triangle1: abc
triangles[6 * i + 1] = b;
triangles[6 * i + 2] = c;
triangles[6 * i + 3] = c; // Triangle2: cda
triangles[6 * i + 4] = d;
triangles[6 * i + 5] = a;
angle_curr += angle_delta;
angle_next += angle_delta;
}
mesh.vertices = vertices;
mesh.triangles = triangles;
Graphics.DrawMesh(mesh, Vector3.zero, Quaternion.identity, material, 0);
}
float GetEnemyAngle()
{
return 90 - Mathf.Rad2Deg * Mathf.Atan2(transform.forward.z, transform.forward.x); // Left handed CW. z = angle 0, x = angle 90
}
}
8 years later and this has helped another coder. I have adjusted knight666's and Rick0r's code to work modularly with a tabletop strategy game I am working on. Though I am having trouble getting it to take a shader, it will only display the color of the shader and not the effect.
using UnityEngine;
public class FiringArcMeshGenerator : MonoBehaviour
{
/// <summary>
/// Credit to Knight666 (knight666.com) & Rickz0r
/// https://gamedev.stackexchange.com/questions/31170/drawing-a-dynamic-indicator-for-a-field-of-view
/// Code modified for modular use by ThomFoxx
/// </summary>
[SerializeField] private int _quality = 6;
private Mesh _arcMesh;
[SerializeField] private bool _isPrimary;
[SerializeField] private float _primaryAngle = 90, _auxAngle = 180;
private float _angleOfFire;
private float _minDist;
private float _maxDist;
public void MeshSetup(ShipInfo ship)
{
WeaponInfo weapon;
if (_isPrimary)
{
_angleOfFire = _primaryAngle / 2;
weapon = ship.Primary;
}
else
{
_angleOfFire = _auxAngle / 2;
_quality *= 2;
weapon = ship.Auxiliary;
}
_minDist = weapon.MinRange;
_maxDist = weapon.MaxRange;
_arcMesh = new Mesh();
_arcMesh.vertices = new Vector3[4 * _quality];
_arcMesh.triangles = new int[3 * 2 * _quality];
Vector3[] normals = new Vector3[4 * _quality];
Vector2[] uv = new Vector2[4 * _quality];
for (int i = 0; i < uv.Length; i++)
uv[i] = new Vector2(0, 0);
for (int i = 0; i < normals.Length; i++)
normals[i] = new Vector3(0, 1, 0);
_arcMesh.uv = uv;
_arcMesh.normals = normals;
GenerateMesh();
GetComponent<MeshFilter>().mesh = _arcMesh;
}
void GenerateMesh()
{
float angle_lookat = 0;
float angle_start = angle_lookat - _angleOfFire;
float angle_end = angle_lookat + _angleOfFire;
float angle_delta = (angle_end - angle_start) / _quality;
float angle_curr = angle_start;
float angle_next = angle_start + angle_delta;
Vector3 pos_curr_min = Vector3.zero;
Vector3 pos_curr_max = Vector3.zero;
Vector3 pos_next_min = Vector3.zero;
Vector3 pos_next_max = Vector3.zero;
Vector3[] vertices = new Vector3[4 * _quality];
int[] triangles = new int[3 * 2 * _quality];
for (int i = 0; i < _quality; i++)
{
Vector3 sphere_curr = new Vector3(
0,
Vector3 sphere_next = new Vector3(
0,
pos_curr_min = transform.position + sphere_curr * _minDist;
pos_curr_max = transform.position + sphere_curr * _maxDist;
pos_next_min = transform.position + sphere_next * _minDist;
pos_next_max = transform.position + sphere_next * _maxDist;
int a = 4 * i;
int b = 4 * i + 1;
int c = 4 * i + 2;
int d = 4 * i + 3;
vertices[a] = pos_curr_min;
vertices[b] = pos_curr_max;
vertices[c] = pos_next_max;
vertices[d] = pos_next_min;
triangles[6 * i] = a; // Triangle1: abc
triangles[6 * i + 1] = b;
triangles[6 * i + 2] = c;
triangles[6 * i + 3] = c; // Triangle2: cda
triangles[6 * i + 4] = d;
triangles[6 * i + 5] = a;
angle_curr += angle_delta;
angle_next += angle_delta;
}
_arcMesh.vertices = vertices;
_arcMesh.triangles = triangles;
}
}
โข "I am having trouble getting it to take a shader, it will only display the color of the shader and not the effect." โ my suspicion would be that it's because you're outputting the same uv coordinates and normal vectors for every vertex, so there are no differences in input from one part of the mesh to the other for the shader to use to produce different output. You can post a question asking how to correct that for a particular shader effect. Feb 23, 2022 at 10:20
โข @DMGregory Yeah, I've been looking over this code for the better part of the day. And while graphics coding is not normally in my wheelhouse, I have come to the same conclusion. I feel there might be a lot of overlap in this code as well. It has become my sudoku puzzle for the week. :P Feb 24, 2022 at 8:34 | 3,215 | 12,217 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-18 | latest | en | 0.933055 |
https://bestmaths.net/online/index.php/year-levels/year-9/year-9-topics/circles/exercise/ | 1,656,730,739,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103983398.56/warc/CC-MAIN-20220702010252-20220702040252-00340.warc.gz | 163,732,414 | 4,512 | ## Circles Exercise
1. Name the parts of the circle, shown by letters.
Useย ฯย = 3.14 for all of the following questions up to question 10, and give your answers correct to 3 significant figures.
2. A circle has a radius of 12 cm. What is its circumference?
3. A cirlce has a diameter of 24.8 mm. What is its circumference?
4. Find the circumference and the area of the following circles:
a. b. c.
5. Find the perimeter and area of the following shapes:
a. b. c.
6. Find the radius of a circle given that its circumference is 3.9 cm.
7. Find the diameter of a circle given that its circumference is 83 metres.
8. Find radius of a circle given that its area is 34.9 mm2
9. A car tyre has an outside diameter of 85 cm, what is its circumference?
10 A wheel is used to measure the length of a running race.ย Every time it does one revolution it is recorded on a counter.ย If the wheel has a diameter of 25 cm and it turns 15345 revolutions when measuring the course, how far, to the nearest metre, is the race? | 264 | 1,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2022-27 | latest | en | 0.928318 |
https://www.beatthegmat.com/integrated-reasoning-t114971.html | 1,527,087,013,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865679.51/warc/CC-MAIN-20180523141759-20180523161759-00192.warc.gz | 687,357,055 | 34,560 | โข 5-Day Free Trial
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## integrated reasoning
This topic has 5 member replies
abcd1111 Junior | Next Rank: 30 Posts
Joined
28 Jun 2012
Posted:
15 messages
#### integrated reasoning
Thu Jun 28, 2012 7:42 pm
I have chosen 4 . Please explain if i am wrong .
abcd1111 Junior | Next Rank: 30 Posts
Joined
28 Jun 2012
Posted:
15 messages
Thu Jun 28, 2012 9:57 pm
I think actual question has not been properly uploaded . here is the actual question :
can any one pls explain !! thanks a lot for help !!
sandeep_thaparianz Senior | Next Rank: 100 Posts
Joined
21 Mar 2012
Posted:
90 messages
Followed by:
2 members
5
Fri Jul 06, 2012 9:27 pm
question mentions min years of exp. to become a principal
for analyst it says<3 that means min =0
for associate it says 2 but can be relaxed(last point) so min again =0
for principal it says 4+ so min =4
sum all mins and we will have 4
katy_123 Junior | Next Rank: 30 Posts
Joined
24 May 2010
Posted:
17 messages
Followed by:
1 members
Fri Jul 13, 2012 7:14 am
4 yrs as analyst + 2 yrs of MBA
total 6
but it doesnt say full time MBA so I am not sure. what is the OA?
srcc25anu Master | Next Rank: 500 Posts
Joined
11 Jun 2010
Posted:
423 messages
Followed by:
2 members
86
Target GMAT Score:
720+
Mon Apr 08, 2013 5:20 am
I would also go with 6 (min 4 yrs exp + 2 yr MBA)
what's the OA?
srcc25anu Master | Next Rank: 500 Posts
Joined
11 Jun 2010
Posted:
423 messages
Followed by:
2 members
86
Target GMAT Score:
720+
Mon Apr 08, 2013 5:24 am
there was another similar question for the same information set provided about analyst, associate and partner:
John joined the bank as an Analyst when he was 22. One year later he asked for promotion to Associate. Is it possible that he will be promoted?
YES
I selected NO because 2 years is required from analyst to associate which could be relaxed if atleast 2 associates recommend the analyst strongly. No information is provided about any such reco or how much the 2 year restriction can be eased. I am really confused as to how much leeway do we have in such questions?
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See More Top Beat The GMAT Experts | 1,023 | 3,709 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2018-22 | longest | en | 0.904212 |
https://www.eurocode.us/structural-timber-design/multiple-dowel-fasteners-loaded-laterally.html | 1,560,921,898,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998913.66/warc/CC-MAIN-20190619043625-20190619065625-00449.warc.gz | 741,925,292 | 5,868 | In EC5, a number of fasteners lying along a line running parallel to the grain direction, as shown in Figure 10.14, are referred to as a row of fasteners parallel to the grain.
Where there is only a single fastener in the row, the design strength of the row per shear plane will be the design lateral load carrying capacity of the fastener per shear plane and where there are r such rows, the design strength of the connection parallel to the grain per shear plane will be:
Design lateral capacity of connection per shear plane
= r x Design lateral capacity of the fastener per shear plane
Where there is more than one fastener per row parallel to the grain the strength of the row depends on the stiffness of the fastener and the strength of the bedding material and in general the stiffer the fastener the greater the design strength of the row. Several researchers have investigated this effect, and the effective characteristic load-carrying capacity of a row of fasteners parallel to the grain, Fv ef Rk, in EC5 is:
Fv,ef>Rk = nefFv Rk (EC5, equation (8.1)) (10.42)
where Fv ef jRk is the effective characteristic lateral load carrying capacity per shear plane of one row of fasteners parallel to the grain, nef is the effective number of fasteners per shear plane in the row parallel to the grain, and Fv Rk is the characteristic lateral load carrying capacity per shear plane of the fastener type being used.
10.4.1 The effective number of fasteners
The effective number of fasteners in a connection is dependent on the type of fastener and the direction of loading relative to the grain, and is covered in the following subsections.
Stagger < d nail
D"
0 0 | 374 | 1,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-26 | latest | en | 0.910307 |
http://conversion-website.com/volume/gallon-UK-to-barrel-US-dry.html | 1,695,981,551,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510501.83/warc/CC-MAIN-20230929090526-20230929120526-00791.warc.gz | 11,456,377 | 4,572 | # Gallons (UK) to barrels (US dry) (gal to bl)
## Convert gallons (UK) to barrels (US dry)
Gallons (UK) to barrels (US dry) conversion calculator above calculates how many barrels (US dry) are in 'X' gallons (UK) (where 'X' is the number of gallons (UK) to convert to barrels (US dry)). In order to convert a value from gallons (UK) to barrels (US dry) (from gal to bl) type the number of gal to be converted to bl and then click on the 'convert' button.
## Gallons (UK) to barrels (US dry) conversion factor
1 gallon (UK) is equal to 0.039316447370998 barrels (US dry)
## Gallons (UK) to barrels (US dry) conversion formula
Volume(bl) = Volume (gal) ร 0.039316447370998
Example: Pressume there is a value of volume equal to 201 gallons (UK). How to convert them in barrels (US dry)?
Volume(bl) = 201 ( gal ) ร 0.039316447370998 ( bl / gal )
Volume(bl) = 7.9026059215706 bl or
201 gal = 7.9026059215706 bl
201 gallons (UK) equals 7.9026059215706 barrels (US dry)
## Gallons (UK) to barrels (US dry) conversion table
gallons (UK) (gal)barrels (US dry) (bl)
200.78632894741996
301.1794934211299
401.5726578948399
501.9658223685499
602.3589868422599
702.7521513159699
803.1453157896799
903.5384802633898
1003.9316447370998
1104.3248092108098
1204.7179736845198
1305.1111381582298
1405.5043026319397
1505.8974671056497
1606.2906315793597
gallons (UK) (gal)barrels (US dry) (bl)
2509.8291118427495
30011.794934211299
35013.760756579849
40015.726578948399
45017.692401316949
50019.658223685499
55021.624046054049
60023.589868422599
65025.555690791149
70027.521513159699
75029.487335528249
80031.453157896799
85033.418980265348
90035.384802633898
95037.350625002448
Versions of the gallons (UK) to barrels (US dry) conversion table. To create a gallons (UK) to barrels (US dry) conversion table for different values, click on the "Create a customized volume conversion table" button.
## Related volume conversions
Back to gallons (UK) to barrels (US dry) conversion
TableFormulaFactorConverterTop | 664 | 2,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2023-40 | latest | en | 0.555755 |
https://www.calculatoratoz.com/en/number-of-absorption-stages-by-kremser-equation-calculator/Calc-33038 | 1,656,245,313,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103205617.12/warc/CC-MAIN-20220626101442-20220626131442-00286.warc.gz | 728,495,644 | 27,353 | ## Number of Absorption Stages by Kremser Equation Solution
STEP 0: Pre-Calculation Summary
Formula Used
Number of Stages = log10(((Solute Free Mole Fraction of Gas in Inlet-(Equilibrium Constant for Mass Transfer*Solute Free Mole Fraction of Liquid in Inlet))/(Solute Free Mole Fraction of Gas in Outlet-(Equilibrium Constant for Mass Transfer*Solute Free Mole Fraction of Liquid in Inlet)))*(1-(1/Absorption Factor))+(1/Absorption Factor))/log10(Absorption Factor)
N = log10(((YN+1-(ฮฑ*X0))/(Y1-(ฮฑ*X0)))*(1-(1/A))+(1/A))/log10(A)
This formula uses 1 Functions, 6 Variables
Functions Used
log10 - Common logarithm function (base 10), log10(Number)
Variables Used
Number of Stages - The Number of Stages is defined as the Ideal number of stages required to achieve the desired separation.
Solute Free Mole Fraction of Gas in Inlet - The Solute Free Mole Fraction of Gas in Inlet is the mole fraction of the Solute in the Gas stream entering the column on solute free basis.
Equilibrium Constant for Mass Transfer - The Equilibrium constant for Mass Transfer is the proportionality constant between gas phase mole fraction and liquid phase mole fraction and could be given as the ratio between the two.
Solute Free Mole Fraction of Liquid in Inlet - The Solute Free Mole Fraction of Liquid in Inlet is the mole fraction of the solute in the solvent (liquid) in inlet of the column on solute free basis.
Solute Free Mole Fraction of Gas in Outlet - The Solute Free Mole Fraction of Gas in Outlet is the mole fraction of the solute in the exit gas stream of the column on solute free basis.
Absorption Factor - The Absorption Factor is the ratio of slopes of operating line of absorption to the equilibrium line. If equilibrium line is a curve then the absorption factor is the average at the two ends.
STEP 1: Convert Input(s) to Base Unit
Solute Free Mole Fraction of Gas in Inlet: 0.8 --> No Conversion Required
Equilibrium Constant for Mass Transfer: 1.5 --> No Conversion Required
Solute Free Mole Fraction of Liquid in Inlet: 0.01 --> No Conversion Required
Solute Free Mole Fraction of Gas in Outlet: 0.1 --> No Conversion Required
Absorption Factor: 2 --> No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
N = log10(((YN+1-(ฮฑ*X0))/(Y1-(ฮฑ*X0)))*(1-(1/A))+(1/A))/log10(A) --> log10(((0.8-(1.5*0.01))/(0.1-(1.5*0.01)))*(1-(1/2))+(1/2))/log10(2)
Evaluating ... ...
N = 2.35548065459839
STEP 3: Convert Result to Output's Unit
2.35548065459839 --> No Conversion Required
2.35548065459839 <-- Number of Stages
(Calculation completed in 00.000 seconds)
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## Credits
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DJ Sanghvi College of Engineering (DJSCE), Mumbai
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National University of Judicial Science (NUJS), Kolkata
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## < 10+ Gas Absorption Calculators
Number of Absorption Stages by Kremser Equation
Number of Stages = log10(((Solute Free Mole Fraction of Gas in Inlet-(Equilibrium Constant for Mass Transfer*Solute Free Mole Fraction of Liquid in Inlet))/(Solute Free Mole Fraction of Gas in Outlet-(Equilibrium Constant for Mass Transfer*Solute Free Mole Fraction of Liquid in Inlet)))*(1-(1/Absorption Factor))+(1/Absorption Factor))/log10(Absorption Factor) Go
Maximum Gas Rate for Absorption Column
Maximum Gas Flowrate on Solute Free Basis = Liquid Flowrate on Solute Free Basis/((Solute Free Mole Fraction of Gas in Inlet- Solute Free Mole Fraction of Gas in Outlet)/((Solute Free Mole Fraction of Gas in Inlet/Equilibrium Constant for Mass Transfer)-Solute Free Mole Fraction of Liquid in Inlet)) Go
Minimum Liquid Rate for Absorption Column
Minimum Liquid Flowrate on Solute Free Basis = Gas Flowrate on Solute Free Basis*(Solute Free Mole Fraction of Gas in Inlet-Solute Free Mole Fraction of Gas in Outlet)/((Solute Free Mole Fraction of Gas in Inlet/Equilibrium Constant for Mass Transfer)-Solute Free Mole Fraction of Liquid in Inlet) Go
Minimum Operating Line Slope for Absorption Column
Minimum Operating Line Slope of Absorption Column = (Solute Free Mole Fraction of Gas in Inlet-Solute Free Mole Fraction of Gas in Outlet)/((Solute Free Mole Fraction of Gas in Inlet/Equilibrium Constant for Mass Transfer)-Solute Free Mole Fraction of Liquid in Inlet) Go
Gas Flowrate for Absorption Column on Solute Free Basis
Gas Flowrate on Solute Free Basis = Liquid Flowrate on Solute Free Basis/((Solute Free Mole Fraction of Gas in Inlet-Solute Free Mole Fraction of Gas in Outlet)/(Solute Free Mole Fraction of Liquid in Outlet-Solute Free Mole Fraction of Liquid in Inlet)) Go
Liquid Flowrate for Absorption Column on Solute Free basis
Liquid Flowrate on Solute Free Basis = Gas Flowrate on Solute Free Basis*(Solute Free Mole Fraction of Gas in Inlet-Solute Free Mole Fraction of Gas in Outlet)/(Solute Free Mole Fraction of Liquid in Outlet-Solute Free Mole Fraction of Liquid in Inlet) Go
Number of Stages for Absorption Factor Equal to 1
Number of Stages = (Solute Free Mole Fraction of Gas in Inlet-Solute Free Mole Fraction of Gas in Outlet)/(Solute Free Mole Fraction of Gas in Outlet-(Equilibrium Constant for Mass Transfer*Solute Free Mole Fraction of Liquid in Inlet)) Go
Operating Line Slope for Absorption Column
Operating Line Slope of Absorption Column = (Solute Free Mole Fraction of Gas in Inlet-Solute Free Mole Fraction of Gas in Outlet)/(Solute Free Mole Fraction of Liquid in Outlet-Solute Free Mole Fraction of Liquid in Inlet) Go
Absorption Factor
Absorption Factor = Liquid Flowrate on Solute Free Basis/(Equilibrium Constant for Mass Transfer*Gas Flowrate on Solute Free Basis) Go
Absorption Factor based on Stripping Factor
Absorption Factor = 1/Stripping Factor Go
## Number of Absorption Stages by Kremser Equation Formula
Number of Stages = log10(((Solute Free Mole Fraction of Gas in Inlet-(Equilibrium Constant for Mass Transfer*Solute Free Mole Fraction of Liquid in Inlet))/(Solute Free Mole Fraction of Gas in Outlet-(Equilibrium Constant for Mass Transfer*Solute Free Mole Fraction of Liquid in Inlet)))*(1-(1/Absorption Factor))+(1/Absorption Factor))/log10(Absorption Factor)
N = log10(((YN+1-(ฮฑ*X0))/(Y1-(ฮฑ*X0)))*(1-(1/A))+(1/A))/log10(A)
## What is the Kremser - Souders - Brown Equation ?
In unit operation design calculations, it is useful to find out the quality of separation for a given number of stages. It is also useful to find the required number of stages if the product recovery is specified. The development was first given by Kremser in 1930, and by Souders and Brown in 1932. The resulting equations are referred to as the KSB or Kremser equations.
These equation were originally developed for gas absorption operation in a plate column, however, it is applicable to other mass transfer operations as well (for example, counter current adsorption).
## How to Calculate Number of Absorption Stages by Kremser Equation?
Number of Absorption Stages by Kremser Equation calculator uses Number of Stages = log10(((Solute Free Mole Fraction of Gas in Inlet-(Equilibrium Constant for Mass Transfer*Solute Free Mole Fraction of Liquid in Inlet))/(Solute Free Mole Fraction of Gas in Outlet-(Equilibrium Constant for Mass Transfer*Solute Free Mole Fraction of Liquid in Inlet)))*(1-(1/Absorption Factor))+(1/Absorption Factor))/log10(Absorption Factor) to calculate the Number of Stages, The Number of Absorption Stages by Kremser Equation formula is defined as the calculation for number of stages for dilute systems having straight operating line in absorption algebraically. Number of Stages is denoted by N symbol.
How to calculate Number of Absorption Stages by Kremser Equation using this online calculator? To use this online calculator for Number of Absorption Stages by Kremser Equation, enter Solute Free Mole Fraction of Gas in Inlet (YN+1), Equilibrium Constant for Mass Transfer (ฮฑ), Solute Free Mole Fraction of Liquid in Inlet (X0), Solute Free Mole Fraction of Gas in Outlet (Y1) & Absorption Factor (A) and hit the calculate button. Here is how the Number of Absorption Stages by Kremser Equation calculation can be explained with given input values -> 2.355481 = log10(((0.8-(1.5*0.01))/(0.1-(1.5*0.01)))*(1-(1/2))+(1/2))/log10(2).
### FAQ
What is Number of Absorption Stages by Kremser Equation?
The Number of Absorption Stages by Kremser Equation formula is defined as the calculation for number of stages for dilute systems having straight operating line in absorption algebraically and is represented as N = log10(((YN+1-(ฮฑ*X0))/(Y1-(ฮฑ*X0)))*(1-(1/A))+(1/A))/log10(A) or Number of Stages = log10(((Solute Free Mole Fraction of Gas in Inlet-(Equilibrium Constant for Mass Transfer*Solute Free Mole Fraction of Liquid in Inlet))/(Solute Free Mole Fraction of Gas in Outlet-(Equilibrium Constant for Mass Transfer*Solute Free Mole Fraction of Liquid in Inlet)))*(1-(1/Absorption Factor))+(1/Absorption Factor))/log10(Absorption Factor). The Solute Free Mole Fraction of Gas in Inlet is the mole fraction of the Solute in the Gas stream entering the column on solute free basis, The Equilibrium constant for Mass Transfer is the proportionality constant between gas phase mole fraction and liquid phase mole fraction and could be given as the ratio between the two, The Solute Free Mole Fraction of Liquid in Inlet is the mole fraction of the solute in the solvent (liquid) in inlet of the column on solute free basis, The Solute Free Mole Fraction of Gas in Outlet is the mole fraction of the solute in the exit gas stream of the column on solute free basis & The Absorption Factor is the ratio of slopes of operating line of absorption to the equilibrium line. If equilibrium line is a curve then the absorption factor is the average at the two ends.
How to calculate Number of Absorption Stages by Kremser Equation?
The Number of Absorption Stages by Kremser Equation formula is defined as the calculation for number of stages for dilute systems having straight operating line in absorption algebraically is calculated using Number of Stages = log10(((Solute Free Mole Fraction of Gas in Inlet-(Equilibrium Constant for Mass Transfer*Solute Free Mole Fraction of Liquid in Inlet))/(Solute Free Mole Fraction of Gas in Outlet-(Equilibrium Constant for Mass Transfer*Solute Free Mole Fraction of Liquid in Inlet)))*(1-(1/Absorption Factor))+(1/Absorption Factor))/log10(Absorption Factor). To calculate Number of Absorption Stages by Kremser Equation, you need Solute Free Mole Fraction of Gas in Inlet (YN+1), Equilibrium Constant for Mass Transfer (ฮฑ), Solute Free Mole Fraction of Liquid in Inlet (X0), Solute Free Mole Fraction of Gas in Outlet (Y1) & Absorption Factor (A). With our tool, you need to enter the respective value for Solute Free Mole Fraction of Gas in Inlet, Equilibrium Constant for Mass Transfer, Solute Free Mole Fraction of Liquid in Inlet, Solute Free Mole Fraction of Gas in Outlet & Absorption Factor and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Number of Stages?
In this formula, Number of Stages uses Solute Free Mole Fraction of Gas in Inlet, Equilibrium Constant for Mass Transfer, Solute Free Mole Fraction of Liquid in Inlet, Solute Free Mole Fraction of Gas in Outlet & Absorption Factor. We can use 1 other way(s) to calculate the same, which is/are as follows -
โข Number of Stages = (Solute Free Mole Fraction of Gas in Inlet-Solute Free Mole Fraction of Gas in Outlet)/(Solute Free Mole Fraction of Gas in Outlet-(Equilibrium Constant for Mass Transfer*Solute Free Mole Fraction of Liquid in Inlet))
Let Others Know | 2,909 | 11,745 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-27 | latest | en | 0.808209 |
https://homepages.math.uic.edu/ums/2017/ | 1,566,430,579,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316549.78/warc/CC-MAIN-20190821220456-20190822002456-00514.warc.gz | 502,522,174 | 4,527 | # Undergraduate Mathematics SymposiumSaturday, November 18, 2017University of Illinois at Chicago
Organized by David Dumas and Kevin Tucker
The Undergraduate Mathematics Symposium at UIC is an annual one-day meeting focusing on undergraduate mathematical research and education. The meeting features invited lectures by mathematical researchers, as well as contributed lectures and posters by undergraduates on their own research projects.
The 2017 UMS was held at UIC on Saturday, November 18, 2017.
### Plenary speakers
โข Christine Berkesch Zamaere (U. Minnesota)
Polynomial linear algebra and geometry
Abstract: After an example-driven crash course in algebraic geometry for projective space, I will discuss how polynomial linear algebra can explain geometry. We'll then consider a generalization of this to other spaces, which is joint work with Daniel Erman and Gregory G. Smith. A basic understanding of vector spaces over fields might be helpful in understanding this talk, but it will not be necessary.
โข Ian Morrison (Fordham U.)
โCan you play a fair game of craps with a loaded pair of dice?โ and related problems about factorization in โโฅ 0[x]
Slides (PDF)
Abstract: I hope to convince you that questions about factorizations of polynomials with non-negative real coefficients---not necessarily into irreducibles and possibly with conditions on the degrees of the factors---are worth studying by discussing several examples that have interesting answers. The question in the title arises from an equivalent formulation that asks what we can say about a finite collection of dice from knowing the probability distribution of the totals that arise when the dice are rolled. No knowledge of the game of craps will be required, but you may enjoy this classic scene [warning: adult language], from the movie โ A Bronx Taleโ set in the Belmont neighborhood that abuts Fordham.
โข Julius Ross (UIC)
Roots and critical points of random polynomials
Abstract: There are many interesting questions that one can ask about solutions to the equation p(z)=0 where p(z) = a0 + a1z + ... + anzn is a polynomial in one variable. The first is simply to ask how many solutions there are to this equation, the answer being given by the Fundamental Theorem of Algebra that states that if one works with the complex numbers then there are precisely n solutions (when counted correctly). In this talk we will investigate the relationship between solutions of the equation p(z)=0 (called roots of p) and solutions of the equation p'(z)=0 (called critical points of p). In particular we will present some work of Boris Hanin, that considers this problem for random polynomials whose degree n is very large.
### Registration form
Complete this form if you would like to attend the symposium, or to apply to present your work as a poster or in a contributed lecture.
โข Applications for posters are accepted until November 10.
โข The latest date to register to attend the symposium is November 10.
โข We are no longer accepting applications to speak at the symposium or to receive travel funding. (The deadline for these applications was October 15, 2017.)
### Student lectures and posters
Five undergraduates will be given the opportunity to give a 20 minute lecture on their research project, and a poster session will give others the chance to describe their projects. Note that the symposium does not accept group applications for lectures, but only presentations by individuals. A group of students, however, may choose to collectively submit a poster.
In addition to the poster presentation on Saturday, students will also have the opportunity to preview their posters at the weekly UIC math department tea and colloquium reception at 4:00pm on Friday, November 17.
For applications to attend or speak at the symposium that were received by October 15, limited funding for travel and lodging may be available. Faculty references for these applications should be sent by email to:
### Location
All symposium events will take place in the Science and Engineering Offices building (SEO) on UIC's East Campus. Registration, lunch and coffee will be provided in room 300, while the lectures will take place in room 636.
### Schedule of events
Plenary lectures are 50 minutes and student lectures 20 minutes; breaks of 10 minutes between talks allow for questions and discussion. A catered lunch of sandwiches and salads is provided for all symposium participants.
Friday, November 17 4:00 - 5:00pm Poster previews and math department tea in SEO 300 Saturday, November 18 8:15 - 8:50am Sign-in and coffee in SEO 300 Morning session โ Plenary Lectures โ SEO 636 9:00am Ian Morrison 10:00am Christine Berkesch Zamaere 11:00am Julius Ross 12:00pm Lunch in SEO 300 Afternoon Session 1 โ Student lectures โ SEO 636 1:00pm Zeid Ghalyoun (Southern Illinois) 1:30pm Divya Chalise (UT Arlington) 2:00pm Michael Toriyama (UIUC) 2:30pm Coffee break in SEO 300 Afternoon session 2 โ Student lectures and posters โ SEO 636 3:00pm Caitlyn Booms (Notre Dame) 3:30pm Khoa D Tran (UIUC) 4:00pm Darshan Chalise (UT Arlington) 4:30 - 5:30pm Poster Session in SEO 300
### Contact
Kevin Tucker (kftucker@uic.edu) | 1,147 | 5,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-35 | longest | en | 0.932304 |
https://academy.binance.com/pl/glossary/prisoners-dilemma | 1,721,580,187,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517747.98/warc/CC-MAIN-20240721152016-20240721182016-00807.warc.gz | 69,909,808 | 63,747 | Strona Gลรณwna
Sลowniczek
Prisoner's Dilemma
# Prisoner's Dilemma
Intermediate
Prisonerโs dilemma is an example of a situation where individual decision-makers, acting in their best interest, produce a suboptimal result for the individuals as a group. It is one of the most well-known examples in game theory.
The standardized example of prisonerโs dilemma, originally proposed by mathematicians Merrill Flood and Melvin Dresher, and then formalized by Albert W. Tucker, presents the following situation:
โข Two members of a criminal gang are arrested and interrogated in separate rooms. There are no other witnesses available, and the authorities only possess sufficient evidence to convict one of the prisoners, but only if the other prisoner testifies against them.
โข The authorities offer each prisoner a bargain. They can betray the other prisoner by testifying that the other prisoner committed the crime, or they can cooperate with the other prisoner by remaining silent.
This scenario can lead to three possible outcomes:
โข If both of them remain silent, both serve one year each.
โข If both of them betray the other, both serve two years each.
โข If one betrays the other, but the other remains silent, the prisoner who testified is set free, and the prisoner who remained silent serves three years.
Betraying the other prisoner offers a greater reward than cooperating with them, so it can be assumed that all purely rational prisoners will betray the other, resulting in the only possible outcome being both prisoners betraying each other.
Pursuing individual reward logically should lead to a better result; however in a prisonerโs dilemma, pursuing individual reward leads to a worse individual result.
Prisonerโs dilemmas occur in many aspects of the economy, but a variety of solutions have been proposed and implemented over time that favors the common good over individual incentives.
For example, in real-world situations, most interactions are repeated more than once. If a prisonerโs dilemma occurs more than once, it can be referred to as an iterated prisonerโs dilemma. In such a situation, the individual actors can implement strategies that reward cooperation over time.
Another solution is formal, institutional strategies that alter the incentives that individual decision-makers can potentially face. By having an understanding of the collective goals and the ability to enforce cooperative behavior through various sets of rules, prisonerโs dilemmas can be steered towards the more collectively beneficial outcome.
Udostฤpnij Posty
Zarejestruj konto
Wykorzystaj swojฤ
wiedzฤ w praktyce, otwierajฤ
c konto Binance juลผ dziล. | 533 | 2,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-30 | latest | en | 0.925385 |
http://www.astro.ucla.edu/%7Ewright/Stolmar_Errors.html | 1,481,401,165,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543567.64/warc/CC-MAIN-20161202170903-00299-ip-10-31-129-80.ec2.internal.warc.gz | 323,730,081 | 4,162 | # Errors in Stolmar's Cosmic Background Model
Stolmar's cosmological model uses a tired light redshift. The Earth is located in the center of a uniform density sphere of stars extending out to some maximum radius Rmax which is about 7 or 8 Hubble radii. The redshift factor 1+z is exponential with distance, and thus reaches a few thousand at the edge of the sphere. Starlight redshifted by this factor of a few thousand provides the cosmic background. However, this model does not agree with the observations and must be rejected:
## The spectrum is wrong
In his CMB page on 27 July 2001 Stolmar gave the following equation for the solid angle covered by stars in a shell with radii between R- and R+,
where A is the area of a star and n is the number density of stars. Stolmar gave this equation for the energy in the background radiation:
where the index j is the frequency in GHz. It is hard to determine exactly what is meant by this equation, since it appears to give the odd combination of energy density per unit wavelength divided by the frequency, but Stolmar also gave the equation for a blackbody:
which can be used to normalize the previous equation. Now Stolmar also assumes (1+z) = exp(HR/c). If I take the limit of infinitesimally small shells in radius, I get a normalized integral for the specific intensity of the CBR which is
where y = H R/cf is the frequency in Hz, and Ifย is in [erg/cm2/sec/sr/Hz]. Stolmar prefers to use the doubling time of photon wavelengths, Hd, instead of the Hubble constant H, but these parameters are related by H = ln(2)/Hd. This formula has three free parameters: Anc/H determines the overall intensity scaling, T* determines the frequency scaling, and ymax determines the shape of the spectrum. Note that Anc/H is the optical depth per Hubble radius and it is very small in Stolmar's model [the total fraction of the sky covered by stars out to Rmax is roughly one part per trillion] so the use of the optically thin approximation here is appropriate. Also note that Stolmar assumes that all of the power emitted by the stars at redshift z reaches us, but with an apparent temperature redshifted to T*/(1+z). Normally in tired light models one allows for the loss of photon energy by a factor of (1+z) which would change the (1+z)4 = exp(4y) in the above equations to (1+z)3 = exp(3y). The only set of these parameters that agree with the FIRAS data on the CMB is ymax < 0.00005, T* = To = 2.725ย K, and Anc/H = ymax-1. In this case the optical depth per Hubble radius is high so one has an opaque isothermal source: a blackbody. But real stars are not blackbodies, so even this limit will not actually work. When ymax is not infinitesimal, one gets a Rayleigh-Jeans low frequency tail rising to a peak corresponding approximately to a graybody with temperature exp(-ymax)T*, a Wien high frequency tail corresponding approximately to a graybody with temperature T*, and an Ifย proportional to f -1 behavior between these two tails. The low frequency tail can approximate a blackbody for an appropriate choice of Anc/H. But this model can not simultaneously fit the data both above and below the peak unless ymax is infinitesimal. The examples given by Stolmar are extremely inconsistent with the FIRAS data on the CMB, as shown in Figure 1. Stolmar's comment about this discrepancy is "The higher calculated values on the right from the peak require closer examination of the reported processing of CMBR data". In other words, blame the data for not agreeing with his theory. However, these data have been confirmed by a separate experiment: see Gush, Halpern & Wishnow (1990, PRL, 65, 537).
Figure 1: Stolmar's model with T* = 4000 and Rmax = 99ย Glyr, which deviates from the FIRAS data by up to 13,000 standard deviations. This value of Rmax and Hd = 8.468ย Glyr give ymax = 8.1 while my best match to Stolmar's curve is with ymax = 7.7. In this plot I use values read from Stolmar's graph. However, Stolmar recently changed Hd and has not updated these figures.
Since Stolmar's cosmology career on sci.astro started with the announcement of the DIRBE far IR background, it is interesting to plot his model on a much wider frequency range and compare not just to the CMB but also to the IR and optical backgrounds. This is shown in Figure 2. The long Ifย proportional to f -1 section becomes a constant when one plots f If , and this behavior is quite contrary to the data, both the detections and the upper limits.
Figure 2: Stolmar's model compared to FIRAS, DIRBE, HST, groundbased and far UV measurements of the cosmic background. Since Stolmar's graphs do not extend to such short wavelengths, I have evaluated my integral version of his equations numerically. Green curve: Stolmar's model. Black curve, FIRAS BB fit. Black IR points: Hauser et al. (1998) on the CIRB. Red points: Wright et al. (2000, 2001) on the CIRB. Magenta upper limits from lack of TeV gamma-ray absorption. Blue points: Bernstein et al. (2001) on the optical background. Black optical and UV points: Toller; Dube et al.; and Hurwitz et al.
## Stars do not last 100 billion years
The stars at the edge of the sphere that produce the peak of the CMB are radiating at a time that is 7 or 8 Hubbles times ago. That is nearly 100 billion years, and only very low mass red dwarfs last that long. But low mass M dwarfs produce very little radiation, and Stolmar's model requires a lot of radiation. On 24 July 2001 Stolmar changed his H to 160 km/sec/Mpc, which alleviates this problem but disagrees with the data on the Hubble constant.
## We must be in the center of the Universe
If we are not in the center of the Universe, a large dipole anisotropy is produced that has the spectrum of a graybody at temperature T*/exp(ymax). The observed dipole anisotropy has a different spectrum. Thus we must be nearly exactly in the center of the sphere of stars. A rough analysis of the FIRAS dipole anisotropy spectrum suggests that we must be centered within Stolmar's sphere of stars to within 1 part in 100,000 of Rmax, which means that the Milky Way could be the center, but the center of mass of the Local Group could not. Having the sphere of stars centered on the Local Supercluster is completely ruled out. | 1,509 | 6,245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2016-50 | latest | en | 0.939228 |
https://samacheerkalvi.guide/samacheer-kalvi-12th-maths-guide-chapter-1-ex-1-4/ | 1,719,133,399,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862464.38/warc/CC-MAIN-20240623064523-20240623094523-00219.warc.gz | 425,899,932 | 16,380 | Tamilnadu State Board New Syllabusย Samacheer Kalvi 12th Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.4 Textbook Questions and Answers, Notes.
Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.4
Question 1.
Solve the following systems of linear equations by Cramerโs rule:
(i) 5x โ 2y + 16 = 0, x + 3y โ 7 = 0
Solution:
(ii) $$\frac{3}{x}$$ + 2y =12, $$\frac{2}{x}$$ + 3y = 13
Solution:
Let $$\frac{1}{x}$$ = a
3a + 2b = 12
2a + 3b = 13
(iii) 3x + 3y โ z = 1 1, 2x โ y + 2z = 9, 4x + 3y + 2z = 25
Solution:
ฮ = $$\left| \begin{matrix} 3 & 3 & -1 \\ 2 & -1 & 2 \\ 4 & 3 & 2 \end{matrix} \right|$$
= 3(-2 โ 6) -3 (4 โ 8) -1(6 + 4)
= 3(-8) -3(-4) -1(10)
= -24 + 12 โ 10 = -22 โ 0
ฮx = $$\left| \begin{matrix} 11 & 3 & -1 \\ 9 & -1 & 2 \\ 25 & 3 & 2 \end{matrix} \right|$$
= 11 (-2 โ 6) โ 3(18 โ 50) โ 1(27 + 25)
= 11(-8) -3(32) -1(52)
= -88 + 96 โ 52 = -44
ฮy = $$\left| \begin{matrix} 3 & 11 & -1 \\ 2 & 9 & 2 \\ 4 & 25 & 2 \end{matrix} \right|$$
= 3(18 โ 50) โ 11(4 โ 8) โ 1(50 โ 36)
= 3(32) -11(4) -1(14)
= -96 + 44 โ 14 = -66
ฮx = $$\left| \begin{matrix} 3 & 3 & 11 \\ 2 & -1 & 9 \\ 4 & 3 & 25 \end{matrix} \right|$$
= 3(-25 โ 27) โ 3(50 โ 36) + 11(6 + 4)
= 3(-52) -3(14) + 11(10)
= -156 โ 42 + 110
= -88
Solution:
3a โ 4b โ 2c = 1 โฆโฆโฆ.. (1)
a + 2b + c = 2 โฆโฆโฆโฆโฆ (2)
2a โ 5b โ 4c = -1 โฆโฆโฆโฆ.. (3)
ฮ = $$\left| \begin{matrix} 3 & -4 & -2 \\ 1 & 2 & 1 \\ 2 & -5 & -4 \end{matrix} \right|$$
= 3(-8 + 5) + 4 (-4 โ 2) โ 2(-5 โ 4)
= 3(-3) +4(-6) -2(-9)
= -9 โ 24 + 18
= -15 โ 0
ฮa = $$\left| \begin{matrix} 1 & -4 & -2 \\ 2 & 2 & 1 \\ -1 & -5 & -4 \end{matrix} \right|$$
= 1(-8 + 5) + 4(-8 + 1) โ 2(-10 + 2)
= 1(-3) + 4(-7) โ 2(-8)
= -3 โ 28 + 16
= -15
ฮb = $$\left| \begin{matrix} 3 & 1 & -2 \\ 1 & 2 & 1 \\ 2 & -1 & -4 \end{matrix} \right|$$
= 3(-8 + 1) โ 1(-4 โ 2) โ 2(-1 โ 4)
= 3(-7) -1(-6) โ 2(-5)
= โ 21 + 6 + 10 = -5
ฮc = $$\left| \begin{matrix} 3 & -4 & 1 \\ 1 & 2 & 2 \\ 2 & -5 & -1 \end{matrix} \right|$$
= 3(-2 + 10) + 4(-1 โ 4) + 1 (-5 โ 4)
= 24 โ 20 โ 9 = -5
Question 2.
In a competitive examination, one mark is awarded for every correct answer while $$\frac{1}{4}$$ mark is deducted for every wrong answer. A student answered 100 questions and got 80 marks. How many questions did he answer correctly? (Use Cramerโs rule to solve the problem).
solution:
No. of Questions answered = 100
Let the No. of questions answered correctly be x
and the No. of questions answered wrongly be y
Here, x + y = 100 and x โ $$\frac { 1 }{ 4 }$$ y = 80
(i.e) x + y = 100 and 4x โ y = 320
correct questions = 84
wrong questions = 16.
Question 3.
A chemist has one solution which is 50% acid and another solution which is 25% acid. How much each should be mixed to make 10 litres of a 40% acid solution? (Use Cramerโs rule to solve the problem).
solution:
Let two solutions x and y
x + y = 10 โฆโฆ.. (1)
0.25 x + (0.50)y = (0.40) โฆโฆโฆ.. (2)
(2) ร 100 โ 25x + 50y = 400
(2) รท 5 โ 5x + 10y = 80 โฆโฆโฆโฆ. (3)
Question 4.
A fish tank can be filled in 10 minutes using both pumps A and B simultaneously. However, pump B can pump water in or out at the same rate. If pump B is inadvertently run in reverse, then the tank will be filled in 30 minutes. How long would it take each pump to fill the tank by itself? (Use Cramerโs rule to solve the problems).
solution:
pump A fills $$(\frac {1}{x})^{th}$$ of the tank in 1 hour.
pump B fills $$(\frac {1}{y})^{th}$$ of the tank in 1 hour.
Both can filled $$(\frac {1}{10})^{th}$$ of the tank in 1 hour.
using Cramerโs rule
Pump A takes 15 minutes
Pump B takes 30 minutes
Question 5.
A family of 3 people went out for dinner in a restaurant. The cost of two dosai, three idlies and two vadais is Rs 150. The cost of the two dosai, two idlies and four vadais is Rs 200. The cost of five dosai, four idlies and two vadais is T 250. The family has Rs 350 in hand and they ate 3 dosai and six idlies and six vadais. Will they be able to manage to pay the bill within the amount they had?
solution:
Let the Cost of dosai, Idlies and vadais be x, y, z
2x + 3y + 2z = 150
2x + 2y + 4z = 200
5x + 4y + 2z = 250
ฮ = $$\left| \begin{matrix} 2 & 3 & 2 \\ 2 & 2 & 4 \\ 5 & 4 & 2 \end{matrix} \right|$$
= 2(4 โ 16) โ 3(4 โ 20) + 2(8 โ 10)
= 2(-12) โ 3(-16) + 2(-2)
= -24 + 48 โ 4
= 20 โ 0
ฮx = $$\left| \begin{matrix} 150 & 3 & 2 \\ 200 & 2 & 4 \\ 250 & 4 & 2 \end{matrix} \right|$$
= 150(4 โ 16) โ 3(400 โ 1000) + 2(800 โ 500)
= 150(-12) โ 3(-600) + 2(300)
= -1800 + 1800 + 600
= 600
ฮy = $$\left| \begin{matrix} 2 & 150 & 2 \\ 2 & 200 & 4 \\ 5 & 250 & 2 \end{matrix} \right|$$
= 2(400 โ 1000) โ 150(4 โ 20) + 2(500 โ 1000)
= 2(-600) โ 150(-16) + 2(-500)
= -1200 + 2400 โ 1000
= 200
ฮz = $$\left| \begin{matrix} 2 & 3 & 150 \\ 2 & 2 & 200 \\ 5 & 4 & 250 \end{matrix} \right|$$
= 2(500 โ 800) โ 3(500 โ 1000) + 150(8 โ 10)
= 2(-300) โ 3(-500) + 150(-2)
= -600 + 1500 โ 300
= 600
x = Rs 30, y = Rs 10, z = Rs 30
There are 3 dosai, 6 idlies and 6 vadais
= 3x + 6y + 6z
= 3(30) + 6(10) + 6 (30)
= 90 + 60 + 180
= Rs. 330
They can eat within the amount. | 2,315 | 5,054 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-26 | latest | en | 0.344899 |
https://www.physicsforums.com/threads/what-does-self-induction-per-unit-length-signify.948954/ | 1,720,965,771,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514580.77/warc/CC-MAIN-20240714124600-20240714154600-00077.warc.gz | 846,367,413 | 18,820 | # What does self induction per unit length signify?
โข Jahnavi
In summary, self induction per unit length signifies that the EMF of the inductor (=a solenoid is an inductor, and oftentimes the core is filled with iron to greatly increase the inductance), ## \mathcal{E}_{inductor}=-L \frac{dI}{dt} ##, is greater than the other EMF's.
Jahnavi
## The Attempt at a Solution
Self inductance L = ฮผ0n2ฯr2l . This means it depends only on geometrical factors .
So , self inductance in both the cases should be equal .
But if I think in terms of L = ฮฆ/I , then Magnetic field at the center is twice that at the ends (assuming long solenoid ) . In that case flux at the center will be twice that at the ends and so will be the self inductance .
I am not sure which of the above two reasoning is correct .
#### Attachments
โข self inductance.png
7.2 KB ยท Views: 1,295
The formula you have is for the self inductance of a very long solenoid. You are correct that the magnetic field is no longer uniform at the ends of a solenoid of finite length, and is approximately (almost exactly) 1/2 at the ends, and thereby the ## -\frac{d \Phi}{dt} ## will be one-half as much per unit length near the ends as anywhere near the middle. Since inductance ## L=\frac{\Phi}{I} ## the effect on the inductance in the region near the ends is that it contributes half as much per unit length. ## \\ ## Incidentally, one can take the deivative of the last equation and write that the EMF in the inductor is ## \mathcal{E}=-\frac{d \Phi}{dt}=-L \frac{dI}{dt} ##.
Jahnavi
Do you mean SI at the center is greater ?
You are correct that the magnetic field is no longer uniform at the ends of a solenoid of finite length, and is approximately (almost exactly) 1/2 at the ends,
I think this is for a long solenoid , not a finite length solenoid .
The magnetic field of a long solenoid is ## B_z=n \mu_o I ## throughout its entire length and is very uniform with ## B_r=B_{\phi}=0 ##. ## \\ ## It is a very well-known (or somewhat well-known) problem from advanced E&M theory that the magnetic field ## B_z ## at the openings of the solenoid is ## B_z=\frac{1}{2}n \mu_o I ##. ## \\ ## For the details: This result is because the magnetic field of a magnetized cylinder (with the same dimensions as the solenoid) with uniform magnetization ## M_z ## in the z-direction has the exact same magnetic field pattern as the solenoid, (both inside and outside the cylinder), with ## B_z= M_z ## inside and anywhere near the middle region. Some very straightforward calculations using the magnetic pole method give the result that there are surface magnetic charge densities ## \sigma_m=\vec{M} \cdot \hat{n} ## at the endfaces, resulting in a contribution of ## B_z=-(1/2) M_z ## inside the cylinder near this "plane of surface charge", (comes from ## H_z=-\frac{\sigma}{2 \mu_o} ## inside the cylinder, near the endface, with an inverse square law fall-off as one moves away from the endface), and ## \vec{B}=\mu_o \vec{H}+\vec{M} ##. (Note: This result is quite basic for understanding the concept of magnetic surface concepts in the magnetic surface current methodology for computing the magnetic field of a permanent magnet. Notice how the magnetic surface current per unit length ## \vec{K}_m=\vec{M} \times \hat{n}/\mu_o ## corresponds to the solenoid current per unit length of ## nI ##).## \\ ## In any case, this is a well-known result that is presented in advanced E&M courses, and so the result can be used without going through all of the details. And your second line of reasoning in the OP is the correct one.
Last edited:
Jahnavi
OK .
Do you agree that , considering energy density = (1/2)B20 and , with the similar reasoning as in the OP , energy density at the center is greater ?
#### Attachments
โข Energy density.jpg
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Jahnavi said:
OK .
View attachment 226658
Do you agree that , considering energy density = (1/2)B20 and , with the similar reasoning as in the OP , energy density at the center is greater ?
That is correct.
Jahnavi
Thanks .
Could you explain qualitatively what does self induction per unit length signify ? I mean what does it tell you about the solenoid ? Why per unit length ?
Jahnavi said:
Thanks .
Could you explain qualitatively what does self induction per unit length signify ? I mean what does it tell you about the solenoid ?
In the differential equation for the basic series L-R-C circuit, you can write ## \mathcal{E}=L \frac{dI}{dt}+IR+\frac{Q}{C} ##, where ## \mathcal{E} ## are any additional EMF's. The EMF of the inductor (=a solenoid is an inductor, and oftentimes the core is filled with iron to greatly increase the inductance), ## \mathcal{E}_{inductor}=-L \frac{dI}{dt} ##, so we put it on the other side of the equation with a plus sign. And then the current ## I=\frac{dQ}{dt } ##, so the equation is a second order differential equation in ## Q ## . ## \\ ## A solenoid is often used to create a nearly uniform magnetic field, as with an MRI machine, but it can also be used to create an inductive circuit element. In the laboratory, we had small inductors with iron filled cores, (about 1" long and 1/4" in diameter), and I seem to recall they had an inductance of ## L \approx 100 \, \mu H ##. ## \\ ## The concept of inductance per unit length could be useful if you are trying to design an inductor, but near the ends, as this problem is showing, the effective length, and correspondingly the inductance, is reduced somewhat from what the formula gives.
Last edited:
Jahnavi
## 1. What is self inductance?
Self inductance is the ability of a conducting material to generate an electromotive force (EMF) in itself when there is a change in the current passing through it. It is measured in Henrys (H).
## 2. What is a solenoid?
A solenoid is a cylindrical coil of wire that is used to create a magnetic field when an electrical current is passed through it. It is often used in electronic devices and as a component in electric motors and generators.
## 3. How is self inductance calculated for a solenoid?
The self inductance of a solenoid can be calculated using the formula L = ฮผ0 * (N^2 * A) / l, where ฮผ0 is the permeability of free space, N is the number of turns in the solenoid, A is the cross-sectional area of the solenoid, and l is the length of the solenoid.
## 4. What factors affect the self inductance of a solenoid?
The self inductance of a solenoid is affected by the number of turns in the coil, the cross-sectional area of the coil, the length of the coil, and the permeability of the material inside the coil. Additionally, the presence of a ferromagnetic core inside the solenoid can increase its self inductance.
## 5. How does self inductance impact the behavior of a solenoid?
The self inductance of a solenoid can cause a delay in the buildup of current in the coil when a voltage is applied. This delay is known as inductive reactance and can limit the rate at which the solenoid can change its magnetic field. Additionally, self inductance can cause back EMF when the current through the solenoid is suddenly stopped, which can be damaging to electronic components.
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4K | 2,001 | 7,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-30 | latest | en | 0.92996 |
https://math.stackexchange.com/questions/2856242/inequality-sqrt-x-6-sqrt10-x-geqslant1 | 1,555,744,313,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578528702.42/warc/CC-MAIN-20190420060931-20190420082931-00173.warc.gz | 500,239,919 | 30,147 | # Inequality : $\sqrt {x} - 6 - \sqrt{10} -x \geqslant1$ [closed]
I have solved it by squaring both sides and got inequality $x \geqslant 17/2$ but after that, the solution part have concluded on the equation
$4x^2 + 289 - 68x \geqslant4(10 - x)$
How this equation is formed.
## closed as unclear what you're asking by abiessu, Karn Watcharasupat, Isaac Browne, Tyrone, NamasteJul 24 '18 at 11:51
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, itโs hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
โข The LHS is not even defined outside of the interval $[6,10]$. โย mweiss Jul 19 '18 at 4:03
โข Can you clarify the question you are asking? โย abiessu Jul 19 '18 at 4:04
โข Anyone else think the title shouldn't have been changed back? โย Sambo Jul 19 '18 at 4:32
You should have done (true on the interval $[8,10]$, where the left hand side of the inequality is positive at least) : $$\sqrt{x-6} - \sqrt{10-x} \geq 1 \iff x - 6 + (10 - x) - 2 \sqrt{(x-6)(10-x)} \geq 1 \\ \iff \frac 32 \geq \sqrt{(x-6)(10-x)} \iff 9 \geq 4(x-6)(10-x) \\ \iff 9 \geq 4(-x^2 - 60 + 16 x) \iff 4x^2 - 64x + 249 \geq 0 \\ \iff x^2 - 16x + 62.25 \geq 0 \iff (x-8)^2 \geq 1.75 \iff x \notin (8 - \sqrt{1.75},8 + \sqrt{1.75})$$
So, the answer is $8 + \sqrt{1.75} \leq x \leq 10$. Therefore, your answer is wrong, and you should use my working to verify where it is wrong. The other answer seems to point out your mistake.
Suppose we don't square immediately, but transpose , say $\sqrt{10-x}$ to the other side before squaring, then we get: $$\sqrt{x-6} \geq 1 + \sqrt{10-x} \iff x-6 \geq 1 + (10-x) + 2 \sqrt{10-x} \\ \iff x - 8.5 \geq \sqrt{10-x} \iff x^2 - 17x + 72.25 \geq 10 - x \\ \iff 4x^2 - 68x + 289 \geq 4(10 - x)$$
which is the equation given in the solution, and which will also lead to the right answer. The problem with squaring both sides immediately is that it is not true that $a \geq b \iff a^2 \geq b^2$, but it is true if $a,b$ are both positive.
You've fallen into the old trap: we don't have $(a+b)^2 = a^2 + b^2$!! As my old math teacher would say, every time you make this mistake, a kitten dies.
To clarify, you got your answer by squaring both sides and erroneously ended up with: $$(x-6)-(10-x)\geq 1$$ After rearranging, you get $x\geq 17/2$. But this is incorrect; squaring both sides will in fact give you: $$(x-6)+(10-x)-2\sqrt{x-6}\sqrt{10-x} \geq 1$$ You will need to then rearrange the equation and square again to get your solution. | 923 | 2,671 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-18 | latest | en | 0.894206 |
https://intemodino.com/en/calculator/fraction-calculator/ | 1,519,565,366,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816462.95/warc/CC-MAIN-20180225130337-20180225150337-00471.warc.gz | 685,676,745 | 4,237 | # Fraction calculator with steps
Fraction calculator with steps is the best fraction app for understanding, learning and teaching fractions. It helps you validate your calculations and understand how to calculate the result.
Fraction Calculator quickly solves various fraction problems, including addition, subtraction, multiplication and division of positive and negative fractions, improper fractions, whole numbers and mixed numbers. It enables you to add, subtract, multiply, and divide two and three fractions. The answer is provided in a reduced fraction.
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## Online fraction calculator with step by step instructions
If you are looking for a handy tool for adding, subtracting, multiplying or dividing fractions, try this online .
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Sunday, February 25, 2018 | 673 | 3,595 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-09 | longest | en | 0.885679 |
https://tszsy.l0l.in.net/emerald-eternity-band-white-gold.html | 1,582,769,683,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146643.49/warc/CC-MAIN-20200227002351-20200227032351-00333.warc.gz | 576,478,393 | 1,968 | # Two variable inequalities practice sheet
3. Graph the following inequalities on number lines: (a) x 1 (b) x > 4 (c) 1 < x < 5 Section 2 Solving Inequalities Solving inequalities is similar to solving equations as in worksheet 2.2. We may add or subtract numbers or algebraic terms from both or all sides of the inequality to isolate the variable from the rest of the expression .
Step 4 The inequality is false, so shade the half-plane above the line. Graph each inequality. 3. 2x 4y 8 4. 3x y 1 a. x-intercept a. x-intercept b. y-intercept b. y-intercept c. boundary line c. boundary line d. test 0, 0 d. test 0, 0 e. shade the line e. shade the line
Linear inequalities with two variables have infinitely many ordered pair solutions, which can be graphed by shading in the appropriate half of a rectangular coordinate plane. To graph the solution set of an inequality with two variables, first graph the boundary with a dashed or solid line depending on the inequality. Here is a set of practice problems to accompany the Linear Systems with Two Variables section of the Systems of Equations chapter of the notes for Paul Dawkins Algebra course at Lamar University. Linear inequalities with two variables have infinitely many ordered pair solutions, which can be graphed by shading in the appropriate half of a rectangular coordinate plane. To graph the solution set of an inequality with two variables, first graph the boundary with a dashed or solid line depending on the inequality. Problem solving: Solve practice problems that contain inequalities with two variables Additional Learning To learn more, review the accompanying lesson on Solving and Graphing Two-Variable ... NAME 5-6 Skills Practice DATE PERIOD 7. y > 3x 10. yโ 1 13.-2y-x>-3 Glencoe Algebra 1 Graphing Inequalities in Two Variables Match each inequality to the graph of its solution.
www.mathematicsvisionproject.org Inequalities Word Problems. Inequalities Word Problems - Displaying top 8 worksheets found for this concept.. Some of the worksheets for this concept are Inequality word problems, Absolute value word problems homework, Study guide practice unit 5 test inequalities, One step word problems, Two step word problems, 19 21 absolute value inequalities real world, N5 2, Inequalities for word problems ... Problem solving: Solve practice problems that contain inequalities with two variables Additional Learning To learn more, review the accompanying lesson on Solving and Graphing Two-Variable ... This topic covers: - Solutions to linear inequalities and systems of inequalities - Graphing linear inequalities and systems of inequalities - Linear inequalities and systems of inequalities word problems www.mathematicsvisionproject.org | 587 | 2,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2020-10 | latest | en | 0.884968 |
https://www.codevscolor.com/find-cube-root-number/ | 1,586,429,353,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371833063.93/warc/CC-MAIN-20200409091317-20200409121817-00041.warc.gz | 832,321,499 | 28,562 | # Introduction :
cbrt function is used to find out the cube root of a number in C++. In this tutorial, we will learn how to use cbrt function with an example.
## cbrt function :
cbrt is used to find the cube root of a number in C++. This is defined in cmath header file.
## Definition :
As per C++ 11 specification, cbrt is defined as below :
```double cbrt (double x);
float cbrt (float x);
long double cbrt (long double x);
double cbrt (T x);
```
The last one is for integral type. This overloaded method casts the integral type to double.
## Parameter and Return value :
This method takes a parameter of type double, float, long double, or integral type.
It returns the cubic root of the parameter.
## Example :
In this example, we will take the number as input from the user.
```#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int n;
cout << "Enter a number to find out the cubic root : " << endl;
cin >> n;
cout << "The cubic root is : " << cbrt(n) << endl;
return 0;
}```
## Sample Output :
```Enter a number to find out the cubic root :
27
The cubic root is : 3
```
Categories: C++ | 294 | 1,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-16 | longest | en | 0.625899 |
http://portraitofacreative.com/books/lectures-on-2-d-yang-mills-equivariant-cohomology-and-topological-field | 1,542,558,294,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744513.64/warc/CC-MAIN-20181118155835-20181118181835-00127.warc.gz | 270,742,905 | 9,113 | # Download Lectures on 2D Yang-Mills, Equivariant cohomology, and by Cordes, Moore, Ramgoolam. PDF
By Cordes, Moore, Ramgoolam.
Those are expository lectures reviewing A) contemporary advancements in two-dimensional Yang-Mills thought and B) the development of topological box concept Lagrangians. Topological box idea is mentioned from the perspective of infinite-dimensional differential geometry. We emphasize the unifying position of equivariant cohomology either because the underlying precept within the formula of BRST transformation legislation and as a relevant thought within the geometrical interpretation of topological box concept course integrals.
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Topics in Computational Algebra
The most goal of those lectures is first to in brief survey the elemental conยญ nection among the illustration conception of the symmetric crew Sn and the speculation of symmetric services and moment to teach how combinatorial tools that come up evidently within the conception of symmetric services result in effective algorithms to specific a number of prodยญ ucts of representations of Sn when it comes to sums of irreducible representations.
Extra info for Lectures on 2D Yang-Mills, Equivariant cohomology, and topological field theories
Sample text
Also let P be an arbitrary prime divisor on K and let f(x) (x) (x)for I j ยง3. EXTENSION AND PROJECTION OF PRIME DIVISORS 27 be the decomposition of f(x) into irreducible polynomials in KP[x]. Then P has precisely g extensions P11, ... , P. Since a finite extension is obtained by the succession of simple extensions, we can deduce the next theorem from the above theorem. 12. Let L be an arbitrary finite extension of K. Then any prime divisor P on K has at least one and at most [L : K] many extensions on L.
If we let a' = xl' v,(a') = e,, x22... x, then since aoa1 ' and a'ao 1 belong to a, we have J = (a0) = (a') = (x1)e' (x2)e2 ... In particular, for J = pj , p, = (x1) follows immediately since x, is contained in p.. pn. dnen. Then J = (X1X2... xe") implies that b = Xei -ei X 2e2 -e2 2 n 1 1 Xn"-e" and b-1 must be in a. Hence, we have e1=e1 , e2 = e2 , ... , en = en . ,n. 5). If n C = C'X 1' ... xnn , 1. PREPARATION FROM VALUATION THEORY 8 then c' E o. Let a 1 be the product of c' and x1 f for e, > 0, and let a2 be the product of x1 ' for e1 <0.
Let l = Min(vP(a,) , i = 1, ... , n). We will show that the assumption l < 0 leads to a contradiction. Choose b so that we may have vp(b) _ -l. Let f1 (x) be the polynomial in o[x] defined by b; = bai. f1(x) = b f(x) = bax" + blx"-' + ... + b, Then we have vP(bn) > 0 from the assumption. For some i, 0 < i < n, vP(bl) = 0 must hold. Hence in k[x] we have the decomposition fl(x) = xkh'(x) such that (x, h'(x)) = 1, 0 < k | 782 | 2,760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-47 | longest | en | 0.89198 |
https://polytropy.com/category/mathematics-2/exposition/ | 1,685,842,945,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649348.41/warc/CC-MAIN-20230603233121-20230604023121-00140.warc.gz | 521,179,480 | 47,777 | ## Category Archives: Exposition
Expositions of a theorem or structure of mathematics
### On Playing Zpordle
After Wordle appeared, a number of variants came out. One of the least popular may be Zpordle, or โคp-ordle. I imagine it could be more popular, if people knew it did not require advanced mathematics. It still involves numbers, to which some people declare an allergy. Nonetheless, I think Zpordle can be explained in elementary-school terms, and that is what I shall try to do here.
Screenshot of the Zpordle game won on May 27, 2022
## Preface
This attempt at exposition of Gรถdelโs Incompleteness Theorem was inspired or provoked by somebody elseโs attempt at the same thing, in a blog post that a friend directed me to. I wanted in response to set the theorem in the context of mathematics rather than computer science.
As I wrote in an email to that friend (on November 18, 2020),
Iโve written two long blog posts on Gรถdel, in 2018 and a month ago. They are about a lot of related stuff too, as is usual for my posts.
The present post is independent of those and may be simpler then they are; at any rate, the approach is different. I prepared the post by the method discussed in โLaTeX to HTML.โ Details are at the end, after the bibliography.
## Summary
Gรถdelโs Incompleteness Theorem is about the logic of mathematics. The Theorem is that the theories of certain mathematical structures cannot be completely axiomatized. This means there will always be true statements about the structures that cannot be proved as theorems from previously given axioms. To give meaning to this conclusion, we review some simple examples of mathematical theorems, and their proofs, in geometry, algebra, and logic; we also give examples of theories that can be completely axiomatized. First we look at Raymond Smullyanโs interpretation of Gรถdelโs theorem as a puzzle, and then at an analogy with the inevitable incompleteness of an English guide to English style. A general philosophical conclusion is that a science like physics cannot tell us everything there is to know, and so a doctrine such as physicalism is wrong or unverifiable.
## Puzzling
The so-called Liar Paradox can be traced to the Epistle of Paul to Titus in the Greek Bible. According to a codicil at the end of the Epistle, Titus was โordained the first bishop of the church of the Cretiansโ (Carroll and Prickett 2008NT pp.ย 265โ7). In the first of the Epistleโs three chapters, Paul writes,
12 One of themselves, even a prophet of their own, said, The Cretians are alway liars, evil beasts, slow bellies.
13 This witness is true. Wherefore rebuke them sharply, that they may be sound in the faith
In the original, Paulโs quotation of the Cretan prophet reads,
ฮฯฮฎฯฮตฯ ฮฑฮตฮฏ ฯฮตแฟฆฯฯฮฑฮน, ฮบฮฑฮบฮฌ ฮธฮทฯฮฏฮฑ, ฮณฮฑฯฯฮญฯฮตฯ แผฯฮณฮฑฮฏ.
This is the first of the Presocratic fragments that Diels and Kranz attribute to Epimenides (Diels 1960, 3B1, p.ย 32).
That a Cretan should assert that Cretans are always liars: one may take this for an absurdity or a paradox. Interpreted strictly, the assertion becomes a puzzle: is it true or false? If it were true, then, as the word of a Cretan, it would be false. Therefore:
1. It is false that all Cretans are liars.
2. Some Cretan must not be a liar.
3. The Cretan prophet is a liar.
4. Paul is a liar.
Though not mentioning Paul, Raymond Smullyan draws the first three of the four conclusions above in his book called What Is the Name of This Book? The Riddle of Dracula and Other Logical Puzzles (Smullyan 1978, para. 253, p.ย 214).
I was intrigued, as a child, by a review of Smullyanโs book in Martin Gardnerโs โMathematical Gamesโ column (Gardner 1989ch.ย 20, pp.ย 281โ92). A few years later, I was delighted to find a used copy of the book itself in a West Virginia junkshop. The book culminates in an exposition of Gรถdelโs Incompleteness Theorem.
Gรถdelโs theorem is that, in certain mathematical systems, there are true statements that are not theorems. These true statements fail to be theorems, not because the systems are too weak to prove them, but because the systems are too strong. The systems let us make so many statements that some of them inevitably end up being unprovable, even though they are true.
Some of the statements are true, because they cannot be proved. For, the mathematical systems in question let us formulate the statement, โI am not a theorem.โ If this were false, then it would be a theorem, and theorems are true; thus the statement is true.
That was an exposition of Gรถdelโs Incompleteness Theorem. Such an exposition could take one of three forms:
โข an annotation of Gรถdelโs 1931 paper, โOn formally undecidable propositions of Principia mathematica and related systems Iโ (Gรถdel 2002);
โข a revision of that paper at the same or a higher level of detail and rigor;
โข a summary treatment, not rigorous, but leaving out details.
My exposition above was of the last type. Smullyan gives such an exposition too. Before giving more detail about Gรถdelโs theorem, along with some simple examples of geometric, algebraic, and logical systems and their theorems, I want to give an exposition of Smullyanโs exposition (Gardner does this too).
## Smullyanโs Account
Smullyan asks us conceive of two infinite lists:
โข a list A1, A2, A3, of sets of counting numbers;
โข a list ฯ1, ฯ2, ฯ3, of statements of some mathematical system.
We are to make two suppositions about the system and the lists:
1. One of the listed sets comprises the serial numbers of the listed statements that are not theorems of the system.
2. For every listed setโcall it Aโ, for some listed set B, for every number n, for some number k,
โข n is in B if and only if k is in A,
โข ฯk is true if and only if n is in An.
We can write the two bulleted conclusions more compactly as
nโโโBโโโkโโโA,
ฯkโโโnโโโAn.
For every listed set A then, since the set B that is guaranteed by the second supposition is also listed, the latter set is, for some n, the set An. For this n, for some k, the supposition yields
nโโโAnโโโkโโโA,
ฯkโโโnโโโAn.
Combining the two parts of this conclusion, we can eliminate the mentions of n; thus, for every listed set A, for some k,
ฯkโโโkโโโA.
This holds in particular if A is the set mentioned in the first supposition, namely the set of numbers of statements that are not theorems. If now kโโโA, then ฯk must be a theorem, and therefore true; but then also, according to this theorem, kโโโA. In short, kโโโA implies kโโโA. Therefore indeed kโโโA. This means both that ฯk is true, and that it is not a theorem.
We can understand ฯk as saying that its own serial number belongs to the set of serial numbers of statements that are not theorems; in short, โI am not a theorem.โ
That is Gรถdelโs basic result, as derived from Smullyanโs suppositions. For Smullyan, establishing the first of his two suppositions โis quite a lengthy affair, though elementary in principleโ; moreover, the second โis really a very simple matter.โ Achieving that simplicity still takes some work, which I have glossed over.
Smullyan presents the listed statements, first as if they are knights, who always speak the truth, or knaves, who always lie. The listed sets are clubs of knights and knaves. To puzzle this out, I have introduced notation as above.
As any club has a criterion for membership, so any of the listed sets will have a definition, which can be written down. Our list of sets then is effectively a list of definitions of sets. Given any number n, we need a mathematical way of inferring what the definition of An is. Then we can form the statement nโโโAn. From this statement, we need a way to obtain a number k such that the statement is ฯk. Then, given a listed set A, we can define a set B consisting of all n such that the corresponding k is in A, and we can assume that B is also a listed set. This gives us the second supposition.
In the translation between listed sets and their serial numbers, mathematics talks about itself. I am going to liken this to how a grammar of English, written in English, implicitly talks about itself.
Smullyan gives little idea of what it means to be a mathematical theorem. Iโm not sure one can have any real understanding of Gรถdelโs theorem without knowing about mathematical proof. Thus I shall also work through some examples, from geometry, algebra, and logic itself.
One may conclude from Gรถdelโs theorem that mathematics will always be incomplete. This does not mean that no particular mathematical system can be complete. I shall give some examples that are.
## The Grammar of Gรถdel
Gรถdelโs theorem is a logical theorem about proving mathematical theorems. Again, the theorem is that there are mathematical systems so strong as to contain a statement ฯ whose meaning is precisely that of the statement, โฯ is not a theorem.โ In short then, ฯ is the statement, โI am not a theorem.โ
Mathematical statements are not normally in the first person; they do not feature anything like the pronoun I. For the effect of such a pronoun, Gรถdel makes use of an ambiguity. A statement can be either of the following:
โข something statedโcall it a meaning;
โข something spoken, or written down, and thus a string of symbols, be they sounds, words, or letters; they all come from a catalogue of some kindโa phonology, a dictionary, an alphabetโand are put together according to certain grammatical rules.
A book such as Fowlerโs Dictionary of Modern English Usage (Fowler 1926) is about statements that are in English. It also consists of such statements, and is thus about itself, even though it may never literally refer to itself.
Fowlerโs book does in fact refer to itself, in the dedication to the memory of the authorโs brother,
who shared with me the planning of this book, but did not live to share the writing.
There is also a more subtle reference, in the body of the Dictionary, where the entry Inversion starts off,
By this is meant the abandonment of the usual English order &ย the placing of the subject after the verb as in Said he, or after the auxiliary of the verb as in What did he say?Never shall we see his like again.
In addition to the three explicit examples, the whole sentence is an example of inversion, since in the usual English order the sentence would read, โBy this, the abandonment of the usual English order is meant,โ or else โThe abandonment is meant by this.โ
In his Dictionary, Fowler has included inversion because, although it has its legitimate uses,
the abuse of it ranks with Elegant Variation as one of the most repellent vices of modern writing.
Nonetheless, Fowler has already used inversion to start another entry, Battered ornaments:
On this rubbish-heap are thrown, usually by a bare cross-reference, such synonyms of the Elegant Variation kind as alma mater, daughter of Eve, sleep of the just,brother of the Angle
Thus, were the article Inversion deleted from the dictionary, the remainder could be counted as incomplete, for using an example of a construction that readers should be wary of abusing in their own writing.
No account of good English will ever be strictly complete, since a poet can always come along, to write verses such as
My father moved through dooms of love
through sames of am through haves of give,
singing each morning out of each night
my father moved through depths of height
โthus E. E. Cummings (Aiken 1963, 256), who breaks the rules in a way that is still recognizably English and even good English.
Gรถdel is the e e cummings of mathematics. In a sufficiently rich mathematical system, we can give each statement a numberโits Gรถdel numberโin such a way that, for any number n, we can make the statement,
Statement number n is not a theorem.
Call this statement ฯ(n). Denote the Gรถdel number of any expression ฯ by ฯ. Gรถdel shows how to solve the equation
ฯ(n)โโ=โn.
We shall not refer to it again, but the solution happens to be
nโ=โโฯ(โฯโ)โ,
where ฯ(k) is so defined that, for every statement ฯ(k), the statement ฯ(โฯโ) is the statement
ฯ(โฯ(โฯโ)โ).
In any case, if the number a solves the equation above, then ฯ(a) has the meaning of,
I am not a theorem.
We concluded earlier that such a statement must be true, on the basis that theorems are true. Thus the class of truths is larger than the class of theorems. This is Gรถdelโs First Incompleteness Theorem.
To prove the theorem, we need not actually know that all theorems are true; it is enough that no logical contradiction be a theorem. This presupposition is a statement ฯ. Then we can form the implication
ฯโโโฯ(a),
namely โIf ฯ is true, then so is ฯ(a),โ or โฯ implies ฯ(a)โ; and we have shown that this is a theorem. Since ฯ(a) is not a theorem, it follows that ฯ cannot be a theorem, if it is true; for there is a rule of inference whereby any statement ฮฒ must be a theorem, if each of statements ฮฑ and ฮฑโโโฮฒ is a theorem.
In short, our system cannot prove that it is not contradictory, if indeed it is not. This is Gรถdelโs Second Incompleteness Theorem.
In mathematics, one is well advised to rewrite what one reads in oneโs own words and symbolism. Smullyan did this for Gรถdel, and I did it for Smullyan. If people who do it for Gรถdel are computer scientists, such as Douglas R. Hofstadter in Gรถdel, Escher, Bach (Hofstadter 1980), then their accounts end up looking more or less odd to me as a mathematician. What I write about Gรถdel may likewise look odd to the non-mathematician.
In mathematics, we prove theorems from axioms. We choose our axioms as we wish, usually because they are satisfiedโmade trueโby some mathematical structure or structures that we want to understand. We take up, as hypotheses, statements that may be satisfied by those structures. We can be mistaken here, and that is why we want to prove our hypotheses as theorems that follow from our axioms.
## Geometry
I say our proofs are based on axioms. Not all mathematicians may be prepared to explain what axioms they are working with. โHowever,โ as Timothy Gowers observes in Mathematics: A Very Short Introduction (Gowers 2002, 41),
if somebody makes an important claim and other mathematicians find it hard to follow the proof, they will ask for clarification, and the process will then begin of dividing steps of the proof into smaller, more easily understood substeps.
Those steps and substeps are ultimately based on axioms. As Gowers has observed on the previous page, the possibility of analyzing a proof into its smallest steps
is far from obvious: in fact it was one of the great discoveries of the early 20th century, largely due to Frege, Russell, and Whitehead it means that any dispute ahout the validity of a mathematical proof can always be resolved.
If this is indeed a discovery, then it must be true; but I think it is true in the way that axioms are true. Axioms are true by fiat. That mathematical disputes can always be resolved is a conviction that guides us in our work. I think the conviction means mathematics is pacifistโpeace-makingโin principle; however, in practice, we may not always have the courage of the conviction.
The prototypical examples of mathematical axioms are the postulates of Euclidโs Elements. The first four of these make possible the following activities.
1. To connect two points with a straight line.
2. To extend a given straight line as far as we like.
3. To draw a circle with any given center and radius.
4. To know that all right angles are equal to one another.
By the first three postulates, we have a ruler and compass, along with a flat surface to use them on, and an implement to leave marks or scratches with. Euclidโs word ฮณฯฮฑฮผฮผฮฎ for line (Euclid 1883) is from the verb ฮณฯฮฌฯฯ โto scratchโ (Beekes 2010).
In Postulate 4, Euclid gives us a set square. This is not to draw right angles with; Euclid will show how to do that with ruler and compass. The existence of a set square confirms that all right angles are indeed equal.
Euclidโs first four postulates entail that in any triangle, the exterior angle at any vertex is greater than either of the two opposite interior angles. This is the 16th proposition of Book i of the Elements. Thus if triangle ABฮ is given, and side Bฮ is extended to ฮ as in Figure 1,
then
ฮฮAโ>โโ BAฮ.
This is a theorem. To prove it, we complete the figure as follows.
1. Bisect Aฮ at E, using Proposition 10, so that
AEโ=โฮE.
2. Connect BE, using Postulate 1.
3. Extend that line to Z, using Postulate 2.
4. Use Postulate 3 to ensure
BEโ=โEZ.
5. Connect Zฮ, using Postulate 1 again.
The figure complete, we continue with the argument.
1. By Proposition 15, that vertical angles are equal,
AEBโ=โโ ฮEZ.
2. From our three displayed equations, by Proposition 4, โSide Angle Side,โ
EABโ=โโ EฮZ.
3. However, by inspection,
EฮZโ<โโ Eฮฮ.
4. From this and the previous equation,
EABโ<โโ Eฮฮ.
5. Since angles EAB and Eฮฮ are respectively ฮAB and Aฮฮ, the desired conclusion follows.
Does it really follow? Consider how we can draw triangles on a globe, letting โstraight linesโ be segments of great circles. For example, we can let A and B lie on the equator, while ฮ is at the north pole, as in Figure 2.
Then in triangle ABฮ, the angles at A and B are right, while the interior angle at ฮ may be greater than that, in which case the exterior angle is less.
Thus Euclidโs Proposition i.16 fails on the globe. But then so does Postulate 1, in the sense that there is not always a unique โstraight lineโ joining two points. There is no one such line, but there are infinitely many, if the points are antipodal.
However, as we have redefined โstraight lineโ on the globe, so we may redefine โpointโ to mean pair of antipodal points, such as, in Figure 2, {A,โH}, {B,โฮ}, {ฮ,โฮ}, {E,โK}, or {Z,โฮ}. This way, any two โstraight linesโ meet at exactly one โpoint.โ
With our new conception of straight lines and points, perhaps we still violate Postulate 2, in the sense that we cannot extend a โstraight lineโ if it is already a full great circle. However, it is not clear that Euclid intended the postulate to exclude spheres. He did do research in spherical geometry (Heath 1981, 348โ49).
Euclidโs mathematics lacks the kind of precision that we need for Gรถdelโs theorem. On the other hand, we may note how, as something written down on a surface, geometry is subject to itself. The proposition that we have been considering, Euclidโs i.16, begins with an absolute phrase (a genitive absolute in Greek),
ฮ ฮฑฮฝฯแฝธฯ ฯฯฮนฮณฯฮฝฮฟฯ
ฮผฮนแพถฯ ฯแฟถฮฝ ฯฮปฮตฯ
ฯแฟถฮฝ ฯฯฮฟฯฮตฮบฮฒฮปฮทฮธฮตฮฏฯฮทฯ,
One of the sides of any triangle having been extended.
Instead of assigning a Gรถdel number to the proposition, we might consider it as itself constituting a diagram, the beginning of which is in Figure 3.
For the diagram, I have used not minuscule letters as in ฯฮฑฮฝฯฯฯ, but capitals as in ฮ ฮฮฮคฮฮฃ, because the latter are easier to consider geometrically, and they are the only letters Euclid would have known anyway. If we adapted Smullyanโs argument to the present context, his first list would consist of sets whose elements were diagrams instead of numbers. If being a theorem of Euclidโs system were a geometrical property of the theorem, considered as a diagram, then we could make a true geometrical statement that was not a theorem of the system. Alternatively, there are properties of diagrams that are not geometric.
## Physics
In origin, geometry is surveying, a measuring of the earth: in a broad sense, a part of what we now call physics. We can argue with physics in general as we did with geometry in particular. Not all truths can result from physical laws.
Thus I think there is confusion, if not contradiction, in the words of โan atheist from a Catholic background,โ Julian Baggini, in his 2020 book The Godless Gospel, in the chapter called โGood Without Godโ (Baggini 2020, 155):
Our twenty-first century, scientifically informed version of natural goodness would see it as comparable to things like love or beauty. These certainly exist, but they donโt turn up in physics textbooks or under a microscope. Most of us accept that, like all the phenomena of consciousness, they emerge from the workings of embodied, socialised human brains. We donโt yet know how this happens but we can be pretty sure that these marvellous bodies and brains are made up of nothing more than the stuff of physics.
I think one might as well say that the marvelous theorems of mathematics are made up of nothing more than letters of an alphabet; and Baggini must know that that, at least, would be ridiculous.
Baggini is presumably correct that love and beauty do not turn up as subjects in a physics book. Neither does one expect to see mathematical theorems in a book about typography. A physics book, or its contents, may be beloved and thus beautiful to the reader; and perhaps this must be so, to some extent at least, if the reader is going to get very far in physics in the first place. In โCargo Cult Scienceโ (Feynman, n.d.), Richard Feynman talks about a moral requirement for doing science, but it is rather specific:
I would like to add something thatโs not essential to the science, but something I kind of believe, which is that you should not fool the layman when youโre talking as a scientist. Iโm not trying to tell you what to do about cheating on your wife, or fooling your girlfriend, or something like that, when youโre not trying to be a scientist, but just trying to be an ordinary human being. Weโll leave those problems up to you and your rabbi. Iโm talking about a specific, extra type of integrity that is not lying, but bending over backwards to show how youโre maybe wrong, that you ought to do when acting as a scientist.
There may be a mathematical requirement for doing typography, but it is minimal, by the account of Robert Bringhurst in The Elements of Typographic Style (Bringhurst 2004, 145):
It is not in the least necessary to understand the mathematics in order to perform the actions that the math describes. People walk and ride bicycles without mathematical analyses of these complex operations The typographer likewise can construct beautiful pages without knowing the meaning of symbols like ฯ or ฯ, and indeed without ever learning to add and subtract, if he has a well-educated eye and knows which buttons to push on the calculator and keyboard.
The original suggestion of Baggini was that goodness, love, beauty โemergeโ from things that physics studies. Mathematical theorems likewise โemergeโ through application of the typesetterโs art; but evidently this tells us practically nothing about mathematics. Baggini completes the paragraph that was begun above:
Love and beauty are as real as anything that has emerged from the complex arrangements of matter, like stars, chairs or sounds. Love and beauty have the capacity to fill us with joy or break our hearts. They even have some relation to truth, since we can be mistaken about what we love, even if beauty is to some extent in the eye of the beholder. We can believe all this without insisting that either needs a transcendent source. Why canโt the same be true of goodness?
Indeed, we need not assign a transcendent source to love, beauty, goodness, or truth. We need not assign any source. To assign a source is to suggest that they already exist, somehow, before they actually do. They donโt, except in a more or less jocular way, as in the saying credited to Alfrรฉd Rรฉnyi, โdefining a mathematician as a device to turn coffee into theoremsโ (Suzuki 2009, 380, n.ย 20).
## Algebra
You know from school how to add and multiply integers, which are the so-called whole numbers, both positive and negative. They are conceived as forming a list
โฆ,โโ
โโ
3,โโ
โโ
2,โโ
โโ
1,โ0,โ1,โ2,โ3,โโฆ,
infinite in both directions. These numbers compose the set called , the letter being memorable as standing for the German Zahl โnumber.โ
Let us now create something new: a set to be called M, comprising all of the lists of four integers. We shall write a typical such list as
(a,b,c,d),
for typographical convenience, although for conceptual convenience one may think of this list as a two-by-two matrix, with (a,b) in the top row and (c,d) in the bottom. Using addition and multiplication of integers, we define multiplication of elements of M by the rule
(a,b,c,d)(x,y,z,w)โ=โ(ax+bz,ay+bw,cx+dz,cy+dw).
In both and M, multiplication is associative, meaning it satisfies the following axiom:
xโ
โyโ
โzx(yz)โ=โ(xy)z.
You can read the symbol as โfor all โ Before the upside-down A was given this meaning, other symbols were used. Gรถdel uses ฮ , citing ลukasiewicz and Tarski (Gรถdel 2002, 600, n.ย 19), who in turn cite Peirce (ลukasiewicz and Tarski 1983, 54, n.ย 2). More precisely, for our x, in the printed texts, Gรถdel uses xฮ , with slanted letter Pi after the variable, although ลukasiewicz and Tarski use โโx, with an enlarged upright Pi that extends below the line and is placed before the variable.
A sign such as , or else a combination such as x, is called a universal quantifier. ลukasiewicz and Tarski say in their footnote,
The expression โquantifierโ occurs in the work of Peirce although with a somewhat different meaning.
A challenge of reading logic is that, in its study of symbolism, it uses symbolism, whose precise meaning is important to get straight, although the symbolism and its meaning will vary from author to author.
That is true in mathematics generally, but to a less extent. Perhaps the same is true for programming.
Multiplication on is commutative, meaning it satisfies the axiom
xโ
โyxyโ=โyx.
Not so M, since for example
(1,0,1,1)(1,0,0,0)โ=โ(1,0,1,0),
while in the other order the product is different:
(1,0,0,0)(1,0,1,1)โ=โ(1,0,0,0).
Nonetheless, in both and M, there is a multiplicative identity, namely an element e satisfying the axiom
xโ
(eโ
x=xxโ
e=x),
where is to be read as โand.โ Specifically, e is 1 in and I in M, where
Iโ=โ(1,0,0,1).
The name for a set equipped with an associative operation and an identity is monoid. I donโt know when the name was invented.
Now we can observe that is a monoid with respect to addition as well as multiplication. A more precise way to say this is that each of the structures (โค,โ
,1) and (โค,+,0) is a monoid. So is (M,โ
,I).
One can make M into an additive monoid too, but our concern is with the multiplicative structure.
In as an additive monoid, each element a has an inverse, namely โ
โโ
a. Combining an element with its inverse, in either order, yields the identity. Written multiplicatively, the axiom being satisfied here is
xโ
(xโ
โโ
1x=eโงxxโ
โโ
1=e).
Neither of (โค,โ
,1) and (M,โ
,I) has an operation of inversion that satisfies this axiom. However, if we define
detโ(a,b,c,d)โ=โadโ
โโ
bc,
and if we let G consist of the elements A of M such that detโA is either 1 or โ
โโ
1, then G has inverses, since
Any structure with an associative operation that has an identity and inverses is called a group. Thus (โค,+,0,โ) and (G,โ
,I,โ
โโ
1) are groups. The subset {โ
โโ
1,โ1} of is also a group with respect to multiplication.
The identity of a monoid is โtwo-sided,โ as are inverses in a group. An associative operation may have only a one-sided identity. For example, on any set, if we define an operation by the rule
xโ
โโ
yโ=โy,
this means every element is a left identity; but none is a right identity, if there are at least two elements.
Again, in a monoid, identities are required to be two-sided. It follows that there can be only one of these, since if e and eโฒ are identities, then
eโ=โeeโฒโ=โeโฒ.
Similarly, in a group, where identities are required to be two-sided, every element has only one inverse, since if aโ
โโ
1 and a are both inverses of a, then
aโ
โโ
1โ=โaโ
โโ
1โ
โ
โ
eโ=โaโ
โโ
1โ
โ
โ
(aaโฒ)โ=โ(aโ
โโ
1a)โ
โ
โ
aโฒโ=โeโ
โ
โ
aโฒโ=โa.
If a set has an associative operation with a left identity, and each element has a left inverse, then the identity and the inverses must be two-sided, and thus the structure is a group. That is, on the basis of the three axioms
xโ
โyโ
โzx(yz)โ=โ(xy)z,
xโ
eโ
โ
โ
xโ=โx,
xxโ
โโ
1โ
โ
โ
xโ=โe,
we can prove
xxโ
โ
โ
eโ=โx,
xxโ
โ
โ
xโ
โโ
1โ=โe.
Indeed, for any a we have from the axioms
(aโ
aโ
โโ
1)(aโ
aโ
โโ
1) =โaโ
โ
โ
(aโ
โโ
1โ
โ
โ
(aโ
aโ
โโ
1)) =โaโ
โ
โ
((aโ
โโ
1โ
a)โ
โ
โ
aโ
โโ
1) =โaโ
โ
โ
(eโ
aโ
โโ
1) =โaโ
โ
โ
aโ
โโ
1,
and consequently
e =โ(aโ
aโ
โโ
1)โ
โโ
1โ
โ
โ
(aโ
aโ
โโ
1) =โ(aโ
aโ
โโ
1)โ
โโ
1โ
โ
โ
((aโ
aโ
โโ
1)(aโ
aโ
โโ
1)) =โ((aโ
aโ
โโ
1)โ
โโ
1โ
โ
โ
(aโ
aโ
โโ
1))(aโ
aโ
โโ
1) =โeโ
โ
โ
(aโ
aโ
โโ
1) =โaโ
โ
โ
aโ
โโ
1.
Thus left inverses are right inverses. Therefore
aโ
โ
โ
eโ=โaโ
โ
โ
(aโ
โโ
1a)โ=โ(aaโ
โโ
1)โ
โ
โ
aโ=โeโ
โ
โ
aโ=โa,
so the left identity is also a right identity.
We have now proved that the theory of groups has the three axioms above. The theory is not complete, since it entails neither the axiom of commutativity, nor its negation. In some groups, the operation is commutative; in some, not. The former groups are called abelian, for historical and practical reasons.
The group {1,โโ
โโ
1} mentioned above is abelian. Moreover, each element is its own inverse; that is, the group satisfies
xxโ
โโ
1โ=โx.
Every group that satisfies this axiom is abelian, since for all elements a and b of such a group,
baโ=โbaโ
โ
โ
eโ=โba(ab)(ab)โ=โb(aa)b(ab)โ=โ(bb)(ab)โ=โab.
To obtain the theory of groups in which each element is its own inverse, we do not need symbols for inverses, but can use the axioms
xโ
โyโ
โzx(yz)โ=โ(xy)z,
xโ
eโ
โ
โ
xโ=โx,
xxxโ=โe.
We can require there to be infinitely many elements of each of these groups, using additional axioms:
xโ
โyxโโ โy,
xโ
โyโ
โzโ
(xyxzyz),
and so on. Here you read as โthere exists such that,โ or โfor some โ; the backwards E used to be written as , and it or a combination such as x is an existential quantifier (ลukasiewicz and Tarski 1983, 55). Thus our theory has, for each counting number n, an axiom saying that at least nโ
+โ
1 elements exist.
The theory of infinite groups in which every element is its own inverse is complete. That assertion is a theorem, not of the theory, but about the theory. One proof of the theorem is by quantifier elimination. The idea is that, in each of the groups that we have just named, each system of equations and inequations can be so simplified that it becomes obvious whether there is a solution. For example, the existential statement
xโ
(xaxbxcxd)
is obviously true for our groups, since there are at least five elements in each of them. Thus the statement is equivalent to the equation
eโ=โe,
because this is (obviously) true. Given a more complicated equation or inequation, by using associativity, commutativity, and the axiom that everything is its own inverse, we can always
โข cancel the same letter from either side,
โข move any letter to the other side,
โข cancel repetitions of a letter from the same side.
Thus, for example, the statement
xโ
(xabxca=bxaxbaxcxcxax)
simplifies to
xโ
(x=acxb)
and then to the inequation
acโโ โb.
Replacing the constants a, b, and c with the variables y, z, and t, we simplify the statement
yโ
โzโ
โtโ
โxโ
(xyzxty=zxyxzyxtxtxyx)
to
yโ
โzโ
โtytโโ โz,
which is obviously false and thus equivalent to the inequation
eโโ โe.
In this way, every statement about the groups in question is equivalent to eโ=โe or eโโ โe. This means our axioms give us a complete theory, in the sense of allowing every statement in their language to be proved or disproved.
Might the theory be inconsistent, so that the same statement can be both proved and disproved? No, because there are infinite groups in which every element is its own inverse; in other words, the theory of such groups has at least one model. One example of such a model consists of the infinite sequences
(e1,e2,e3,โฆ),
where each of entries ei belongs to the group {1,โโ
โโ
1}. Thus the elements of the groups are sequences
(ยฑ1,ยฑ1,ยฑ1,โฆ).
Two of these are multiplied entry by entry: for example,
(1,1,โ1,โ1,โฆ)โ
โ
โ
(1,โ1,1,โ1,โฆ)โ=โ(1,โ1,โ1,1,โฆ).
The group has a subgroup, in each element of which, only finitely many entries are โ
โโ
1; this subgroup is a model of the same theory.
## Logic
We have sketched a proof of a completeness theorem. This is the theorem that every statement in a certain mathematical language (namely the language of groups), or else the negation of that statement, is a theorem of a mathematical system with certain axioms (namely the axioms of infinite groups in which every element is its own inverse).
As geometry comes from surveying, so algebra comes from counting and reckoning in general. Logic comes from reasoning, but ends up creating new mathematical objects.
As algebra is concerned with equations, so logic is concerned with formulas in general. In the logic called sentential or propositional, there is nothing simpler than a formula. One definition of the formulas of this logic is as follows.
1. Each of the variables P, Q, R, and so on is a formula.
2. The constant 0 is a formula.
3. If each of F and G is a formula, then so is (FG).
The sign is a connective. For each formula that is not just a single variable or constant, there are formulas F and G such that the original formula is (FG). That these F and G are uniquely determined by the original formula is a theorem; it would be false if we had left off the parentheses in defining formulas as above. The theorem would be true if we always wrote the formula
(FG)
with one parenthesis, as
(FโโโG
โor with no parentheses, but in a different order, as
โโ
FG.
This would be so-called Polish notation, apparently due to ลukasiewicz; it is convenient for computers, but perhaps not for human readers, and we shall not use it.
In the formula (FG), the sign displayed between F and G is the principal connective of the formula. The theorem of the existence of the principal connective of every compound formula allows us to compute, for every formula, the value 0 or 1, once we assign such a value to each variable that occurs in the formula. The computation follows the rule given by the following truth table, where the value of the whole formula (FG) is written below its principal connective.
(F G)
0 1 0
1 0 0
0 1 1
1 1 1
A formula that always takes the value 1 is a tautology. For example, (FF) is always a tautology, because its truth table is computed as follows.
(F F)
0 1 0
1 1 1
In principle, a truth table can always serve as a proof that a given formula is a tautology. However, the size of truth tables grows exponentially: if n variables occur in a formula, then its truth table will have 2n lines. An alternative method of proof is to designate
โข certain tautologies as axioms,
โข certain ways of deriving new tautologies from old ones as rules of inference.
We then obtain such a mathematical system as we have been referring to.
By Gรถdelโs theorem, some systems are strong enough to be incomplete. By contrast, propositional logic is weak enough that it can have complete systems. We describe such a system, based on work of ลukasiewicz, which builds on Frege (Church 1956, 156).
For our convenience, we simplify formulas with two conventions:
1. Outer parentheses are removed.
2. Internal parentheses are removed, when they can be reinstated by assigning priority to arrows on the right.
Our axioms are now three:
1. PโโโQโโโP, or officially (Pโโโ(QP));
2. (RQP)โโโ(RQ)โโโRโโโP;
3. ((Qโ0)โโโPโโโ0)โโโPโโโQ.
Our rules of inference are two:
1. Modus Ponens: From F and FโโโG, infer G.
2. Substitution: From any formula, infer the formula that results from substituting a particular formula, the same each time, for every occurrence of a particular variable.
Here is an example of a proof.
1. PโโโQโโโP by Axiom 1.
2. Pโโโ(PP)โโโP by Substitution in step 1.
3. (RQP)โโโ(RQ)โโโRโโโP by Axiom 2.
4. (PQP)โโโ(PQ)โโโPโโโP by Substitution in step 3.
5. (Pโ(PP)โP)โโโ(PPP)โโโPโโโP by Substitution in step 4.
6. (PPP)โโโPโโโP by Modus Ponens from step 2 and step 5.
7. PโโโPโโโP by Substitution in step 1.
8. PโโโP by Modus Ponens from step 6 and step 7.
9. FโโโF by Substitution in step 8.
Strictly, the proof consists only of the nine formulas. The numbering and the explanations help prove to us that the formulas do constitute a proof. The proof establishes FโโโF as a theorem and therefore a tautology.
We did already know the latter from a truth table. Strictly though, we saw only the truth table for PโโโP. The real truth table for FโโโF has as many lines as that for F, with twice the columns and one more. For example, here is the truth table when F is (PQ)โโโ0 (the row added to the bottom give the stages in which the columns are filled in):
((Pโ
โ
Q) โ
0) (P Q) 0
0 1 0 0 0 1 0 1 0 0 0
1 0 0 1 0 1 1 0 0 1 0
0 1 1 0 0 1 0 1 1 0 0
1 1 1 0 0 1 1 1 1 0 0
2 3 2 4 1 5 2 3 2 4 1
We need not actually do all the work of writing out such a truth table, if we already know the table for PโโโP; but this is precisely because Substitution is a legitimate rule of inference.
We could let every tautology be an axiom, just as Euclid could have given us full use of a set square from the beginning, or for that matter given us as postulates all of his propositions, or at least the โobviousโ ones. However, the point is to see how little we can get by with.
Every tautology is a theorem, because it has a proof such as the one given for FโโโF. This completeness of our system is itself a theorem: not a mathematical theorem of the system, but a logical theorem about the system.
Earlier we proved mathematical theorems about groups, as that a structure with an associative operation that has a left identity and left inverses is a group. These theorems are theorems of a system that can be as precisely defined as propositional logic. However, the quantifiers mean both that the system is more complicated and that it has nothing like truth tables. It has the notion of truth in a model: the model could, for example,
โข be the plane or sphere for a geometric system,
โข consist of integers, or matrices, or sequences, for an algebraic system.
Before the incompleteness theorems, Gรถdel proved a completeness theorem, whereby in such systems as we have just mentioned, the statements that are true in every possible model have proofs. Gรถdelโs first incompleteness theorem is then that, for some models of some systems, not every statement that is true in the model will have a proof. In particular, this is true for the counting numbers as equipped with the operations of addition and multiplication.
As we have seen, the contrary is the case for the group of sequences (ยฑ1,ยฑ1,ยฑ1,โฆ); it is also the case for the counting numbers with addition alone. Telling which models are Gรถdelian or โwild,โ and which are โtameโ in the sense of being completely axiomatizable, is the subject of ongoing research.
## References
Aiken, Conrad, ed. 1963. Twentieth-Century American Poetry. New York: Modern Library.
Baggini, Julian. 2020. The Godless Gospel: Was Jesus a Great Moral Teacher? London: Granta.
Beekes, Robert. 2010. Etymological Dictionary of Greek. Vol. 10. Leiden Indo-European Etymological Dictionary Series. Leiden: Brill.
Bringhurst, Robert. 2004. The Elements of Typographic Style. Version 3.0. Hartley &ย Marks.
Carroll, Robert, and Stephen Prickett, eds. 2008. The Bible: Authorized King James Version with Apocrypha. Oxford Worldโs Classics. Oxford.
Church, Alonzo. 1956. Introduction to Mathematical Logic. Vol. I. Princeton, N.ย J.: Princeton University Press.
Diels, Hermann, ed. 1960. Die Fragmente Der Vorsokratiker. Weidmannsche.
Euclid. 1883. Euclidis Elementa. Vol. I. Euclidis Opera Omnia. Leipzig: Teubner.
Feynman, Richard. n.d. โCargo Cult Science.โ https://calteches.library.caltech.edu/51/2/CargoCult.htm.
Fowler, H. W. 1926. A Dictionary of Modern English Usage. London: Oxford University Press.
Gardner, Martin. 1989. Penrose Tiles to Trapdoor Ciphers. New York: W. H. Freeman.
Gowers, Timothy. 2002. Mathematics: A Very Short Introduction. Very Short Introductions. Oxford: Oxford University Press.
Gรถdel, Kurt. 2002. โOn Formally Undecidable Propositions of Principia Mathematica and Related Systems I.โ In From Frege to Gรถdel: A Source Book in Mathematical Logic, 1879โ1931, edited by Jean van Heijenoort, 596โ616. Cambridge, MA: Harvard University Press.
Heath, Thomas. 1981. A History of Greek Mathematics. Vol. I. From Thales to Euclid. New York: Dover Publications.
Hofstadter, Douglas R. 1980. Gรถdel, Escher Bach: An Eternal Golden Braid. Vintage.
Smullyan, Raymond. 1978. What Is the Name of This Book? The Riddle of Dracula and Other Logical Puzzles. Englewood Cliffs, New Jersey: Prentice-Hall.
Suzuki, Jeff. 2009. Mathematics in Historical Context. Mathematical Association of America.
ลukasiewicz, J., and A. Tarski. 1983. โInvestigations into the Sentential Calculus.โ In Logic, Semantics, Metamathematics, Second, 38โ59. Indianapolis, IN: Hackett Publishing Co.
I first wrote the above for typesetting by the LaTeX program. Then I edited my tex file so that the Pandoc program would produce a readable html file. In particular, I did the following.
โข My three figures are made with pstricks and related packages. I should have been able to use \PSTtoEPS to have the diagrams saved as separate eps files, but I could not manage to do this. I made the eps files โby handโ with dvips -E, and then I converted them to jpg format (although they are not very good that way, and more work could be done here).
โข Because Pandoc would not convert the Latin letters that I was used to using, as to get
ฮฯฮฎฯฮตฯ ฮฑฮตฮฏ ฯฮตแฟฆฯฯฮฑฮน, ฮบฮฑฮบฮฌ ฮธฮทฯฮฏฮฑ, ฮณฮฑฯฯฮญฯฮตฯ แผฯฮณฮฑฮฏ
from
Kr'htes ae'i ye~ustai, kak'a jhr'ia, gast'eres >arga'i,
I figured out how to use actual Greek letters in my tex file. This took a lot of work, since (if I understand correctly) the current best solution to the problem, by means of babel (and compilation with LuaLaTeX if not XeLaTeX), has been available only in the last couple of years, but old information about possible solutions can still be found. In fact a solution that worked for me used the panglossia package, until a friend introduced me to a new friend who explained that babel was now better.
โข I changed array and align environments to tabular, introducing dollar signs as needed.
โข I changed equation environments to center (because Pandoc converts the former to span elements, the latter to div), again introducing dollar signs.
The command I used to create the html file was
pandoc --shift-heading-level-by=1 --bibliography ../../../references.bib --citeproc goedel-incompleteness.tex -o goedel-incompleteness.html
After that, what I did was
โข make the whole html file a div element with the attribute
style="text-align:justify; margin-left:10%; margin-right:10%;";
โข change the attributes
โข class="center" to style="text-align:center;",
โข class="smallcaps" to style="font-variant:small-caps;";
โข give the table elements, and also the blockquote element with the verses of Cummings, the attribute
style="font-size:90%; margin-left:auto; margin-right:auto; display:table;";
โข give the other blockquote elements the attribute style="font-size:90%;";
โข give the figure elements the attribute style="text-align:center;";
โข concerning the references,
โข give them their own sectional heading,
โข to the div element that they constitute for Pandoc, give the attribute style="text-align:left;",
โข to the div element that each of them constitutes, give the attribute style="padding-left:5%;text-indent:-5%;";
### Why It Works
The last post, โKnottedness,โ constructed Alexanderโs Horned Sphere and proved, or sketched the proof, that
โข the horned sphere itself is topologically a sphere, and in particular is simply connected, meaning
โข itโs path-connected: thereโs a path from every point to every other point;
โข loops contract to pointsโare null-homotopic;
โข the space outside of the horned sphere is not simply connected.
This is paradoxical. You would think that if any loop sitting on the horned sphere can be drawn to a point, and any loop outside the horned sphere can be made to sit on the sphere and then drawn to a point, then we ought to be able to get the loop really close to the horned sphere, and let it contract it to a point, just the way it could, if it were actually on the horned sphere.
You would think that, but you would be wrong. Continue reading
### Knottedness
If you roll out a lump of clay into a snake, then tie a string loosely around it, can you contort the ends of the snake, without actually pressing them together, so that you cannot get the string off?
You can stretch the clay into a Medusaโs head of snakes, and tangle them as you like, again without letting them touch. If you are allowed to rest the string on the surface of the clay, then you can get it off: you just slide it around and over what was an end of the original snake.
### LaTeX to HTML
This is a little about mathematics, and a little about writing for the web, but mostly about the nuts and bolts of putting mathematics on the web. I want to record how, mainly with the `pandoc` program, I have converted some mathematics from a LaTeX file into `html`. Like โComputer Recoveryโ then, this post is a laboratory notebook.
The mathematics is a proof of Dirichletโs 1837 theorem on primes in arithmetic progressions. This is the theorem that, if to some number you keep adding a number that is prime to it, there will be no end to the primes that you encounter in this way.
### Discrete Logarithms
In the fall of 2017, I created what I propose to consider as being both art and mathematics. Call the art conceptual; the mathematics, expository; here it is, as a booklet of 88 pages, size A5, in pdf format.
More precisely, the work to be considered as both art and mathematics is the middle of the three chapters that make up the booklet. The first chapter is an essay on art, ultimately considering some examples that inspire my own. The last chapter establishes the principle whereby the lists of numbers in Chapter 2 are created.
### An Exercise in Analyticย Geometry
This past spring (of 2020), when my university in Istanbul was closed (like all others in Turkey) against the spread of the novel coronavirus, I created for my students an exercise, to serve at least as a distraction for those who could find distraction in learning.
From Weeks & Adkins, Second Course in Algebra, p.ย 395
Note added, April 17, 2023: An account of the mathematics involved in the exercise would ultimately be published as: Pierce, D. (2021). โConics in Place.โ Annales Universitatis Paedagogicae Cracoviensis | Studia Ad Didacticam Mathematicae Pertinentia, 13, 127โ150.
### Ordinals
This is about the ordinal numbers, which (except for the finite ones) are less well known than the real numbers, although theoretically simpler.
The numbers of either kind compose a linear order: they can be arranged in a line, from less to greater. The orders have similarities and differences:
โข Of real numbers,
โข there is no greatest,
โข there is no least,
โข there is a countable dense set (namely the rational numbers),
โข every nonempty set with an upper bound has a least upper bound.
โข Of ordinal numbers,
โข there is no greatest,
โข every nonempty set has a least element,
โข those less than a given one compose a set,
โข every set has a least upper bound.
One can conclude in particular that the ordinals as a whole do not compose a set; they are a proper class. This is the Burali-Forti Paradox.
### Elliptical Affinity
After Descartes gave geometry the power of algebra in 1637, a purely geometrical theorem of Apollonius that is both useful and beautiful was forgotten. This is what I conclude from looking at texts from the seventeenth century on.
### Logic of Ellipticย Curves
In my 1997 doctoral dissertation, the main idea came as I was lying in bed one Sunday morning. Continue reading | 12,645 | 47,609 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2023-23 | latest | en | 0.961964 |
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# Z test for proportion of accident rate of fire trucks
In Dallas, some fire trucks were painted yellow (instead of red) to heighten their visibility. During a test period, the fleet of red fire trucks made 153,348 runs and had 20 accidents, while the fleet of yellow fire trucks made 135,035 runs and had 4 accidents. At รก = .01, did the yellow fire trucks have a significantly lower accident rate?
(a) State the hypotheses.
(b) State the decision rule and sketch it.
(c) Find the sample proportions and z test statistic.
(d) Make a decision.
(e) Find the p-value and interpret it.
(f) If statistically significant, do you think the difference is large enough to be important? If so, to whom, and why?
(g) Is the normality assumption fulfilled? Explain.
#### Solution Summary
The solution provides step by step method for the calculation of Z test for proportion of accident rate of fire trucks . Formula for the calculation and Interpretations of the results are also included. This problem is from the e-text, Applied Statistics in Business and Economics, by Doane and Seward
\$2.19 | 246 | 1,121 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2016-44 | longest | en | 0.946222 |
http://mathoverflow.net/feeds/question/2305 | 1,369,023,289,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698238192/warc/CC-MAIN-20130516095718-00044-ip-10-60-113-184.ec2.internal.warc.gz | 161,902,628 | 1,748 | How to find the Fermat Point using the construction of the tangent to ellipse? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T04:14:46Z http://mathoverflow.net/feeds/question/2305 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/2305/how-to-find-the-fermat-point-using-the-construction-of-the-tangent-to-ellipse How to find the Fermat Point using the construction of the tangent to ellipse? Vasile Moฤโขoi 2009-10-24T15:55:58Z 2012-02-02T13:47:49Z <p>Be done the triangle ABC, it is known the method to finding the point Q that minimises the sum QA+QB+QC among all points Q in the plane (The Fermat point). I want a hint for solving this problem using the construction of the tangent to ellipse. (Hadamard, Lesson de Geometrie Elementaire, II, problem no. 745).</p> http://mathoverflow.net/questions/2305/how-to-find-the-fermat-point-using-the-construction-of-the-tangent-to-ellipse/2625#2625 Answer by Philipp Lampe for How to find the Fermat Point using the construction of the tangent to ellipse? Philipp Lampe 2009-10-26T15:27:55Z 2009-10-26T15:43:11Z <p>I have the vague idea that Hadamard is referring to the construction where you erect equilateral triangles BCA', CAB' and ABC' on the sides of the triangle, as described <a href="http://www.cut-the-knot.org/Generalization/fermat%5Fpoint.shtml" rel="nofollow">here</a>. The Fermat point is the intersection of the cevians AA', BB' and CC'. It can also be constructed using the various angles of 60 resp. 120 degrees. </p> <p>In the construction of the tangent from a point P to an ellipse with foci F and F' (in the book you cite), they consider an additional point f. The correspondence should be </p> <p>F ↔ C,</p> <p>F' ↔ A,</p> <p>P ↔ B,</p> <p>f ↔ P'.</p> <p>The general philosophy behind both, I think, is to convert a sum of segments into a single segment. </p> | 544 | 1,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2013-20 | latest | en | 0.733734 |
https://www.hackmath.net/en/math-problem/24851 | 1,611,106,539,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519843.24/warc/CC-MAIN-20210119232006-20210120022006-00243.warc.gz | 817,662,221 | 14,179 | # Eight
Eight small Christmas balls with a radius of 1 cm have the same volume as one large Christmas ball. What has a bigger surface: eight small balls, or one big ball?
Correct result:
x = ย 8
#### Solution:
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
Tips to related online calculators
Tip: Our volume units converter will help you with the conversion of volume units.
#### You need to know the following knowledge to solve this word math problem:
We encourage you to watch this tutorial video on this math problem:
## Next similar math problems:
โข Plasticine ball
Plasticine balls have radius r1=85 cm, r2=60 mm, r3=59 cm, r4=86 cm, r5=20 cm, r6=76 mm, r7=81 mm, r8=25 mm, r9=19 mm, r10=14 cm. For these balls
โข Balls
Three metal balls with volumes V1=71 cm3 V2=78 cm3 and V3=64 cm3 melted into one ball. Determine it's surface area.
โข Big cube
Calculate the surface of the cube, which is composed of 64 small cubes with an edge 1 cm long.
โข Volume and area
What is the volume of a cube which has area of 361 cm2?
โข The ball
The ball has a radius of 2m. What percentage of the surface and volume is another sphere whose radius is 20% larger?
โข Sphere A2V
The surface of the sphere is 241 mm2. What is its volume?
โข Cube surface and volume
The surface of the cube is 500 cm2, how much cm3 will be its volume?
โข Equilateral cylinder
Equilateral cylinder (height = base diameter; h = 2r) has a volume of V = 199 cm3 . Calculate the surface area of the cylinder.
โข Cube into sphere
The cube has brushed a sphere as large as possible. Determine how much percent was the waste.
โข Prism X
The prism with the edges of the lengths x cm, 2x cm, and 3x cm has volume 20250 cm3. What is the area of the surface of the prism?
โข Cube surfce2volume
Calculate the volume of the cube if its surface is 150 cm2.
โข Area of a cube
Calculate the surface area of a cube if its volume is equal to 729 cubic meters.
โข Chemical parison
The blown parison (with shape of a sphere) have a volume 1.5 liters. What is its surface?
โข Iron sphere
Iron sphere has weight 100 kg and density ฯ = 7600 kg/m3. Calculate the volume, surface and diameter of the sphere. | 596 | 2,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2021-04 | longest | en | 0.910742 |
https://tapintoteenminds.com/making-connections-proportional-reasoning/ | 1,679,594,617,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945182.12/warc/CC-MAIN-20230323163125-20230323193125-00423.warc.gz | 603,342,903 | 54,435 | Select Page
# Making Connections in Proportional Reasoning
## How to Apply Proportional Reasoning Concepts to Other Strands
Through my work with the Middle Years Collaborative Inquiry Project with teams of intermediate math teachers from 29 different schools, it is clear that teaching in Ontario elementary classrooms can be tough. Teaching multiple subjects throughout the day, daily interruptions, and pressure to have adequate assessment data from each strand for report cards each term are just a few of the many pressures.
In some cases, teachers feel obliged to follow the textbook because they do not feel as confident in mathematics as they might in other areas of the curriculum. Since the textbook breaks curriculum into smaller โchunks,โ it seems logical that this would be a easy and understandable way to teach mathematics. It seems as though textbooks are created with a focus on restricting math content to only the concepts introduced from that section or unit to avoid having to โre-teachโ or โreviewโ a previously learned topic from earlier in the course.
## Lack of Retention and Ability to Apply Concepts
Unfortunately, this โchunkingโ of math concepts really restricts the possibility of making connections from one topic to another throughout the course of a term/semester or course. As math teachers, we have focused on students mastering procedures on a given day without any focus on how these skills apply to concepts learned previously or how they will apply to concepts we have yet to introduce. I think Dan Meyer sums it up very well when he said:
The maths serves the conversation, not the other way around.
If our math classes introduce interesting questions that require us to learn some math, we are more likely going to develop deeper conceptual understand and encourage retention of those concepts since they will have a memory (the question) to peg the concept to. Furthermore, if we are asking questions with context and avoid too much rote/procedural memorization, it is likely that concepts from previous (and future) math lessons will be required. It is really tough to provide โreal worldโ questions while restricting the topic of the day to โPythagorean Theorem.โ
## Making Connections in Proportional Reasoning via Baby Beats 3 Act Math Task
In the Baby Beats 3 Act Math Task I created recently, we have what seems to be a pretty simple question where students might ask: โWhat is a healthy heart rate for a baby in the womb?โ
Not only can it help with their own understanding of how they can incorporate math into real-world scenarios, but it can also prepare them for when theyโre ready to have children of their own. Pregnancy can be a scary time for both mother and father, but if they know that their child has a healthy heartbeat, all of these nerves can be put at bay. Of course, some people arenโt ready to have children at all so ask themselves the question โ what is abortion? And is it the right thing for me to do? No one can make this decision but the people involved, so itโs important to lay out all your options before making a choice. Regardless of the route you go down, math can be applied to both situations, but weโre going to be looking at the healthy heart rate of a baby in this example.
As youโll see in the video and from the screenshots below, this question can be extended well beyond a simple rate of change problem to involve many concepts from across the different math strands from 7 to 10.
## Making Connections Across Math Strands
Because I have the opportunity to work with teachers from grade 7 to 10, I have used this same problem in all four grade levels in very different ways. Many math questions could be approached in a similar manner using different questions and extensions for each level. Letโs look at a few of the possibilities with any proportional reasoning question.
## Proportional Reasoning: Rates and Equivalent Fractions
Regardless of the grade level, I like to start this problem off with some pure guesses. Regardless of the varying level of readiness from one student to the next, all students can toss out a guess based on โintuitionโ as Dan Meyer suggested in his TED Talk.
Hereโs an example of where you might want to begin with this:
Whether intentional or not, this simple minds on for this problem actually introduces some basic equivalent fraction work as students must take their initial guess and determine the number of beats per 10 second, 30 seconds, and 1 minute.
## Extending to Data Management: Tables and Graphs
After teaching for 8 years, I had introduced proportional reasoning concepts such as ratios, rates, and proportions on their own in a separate unit. Providing students with strategies such as solving proportions would be the ideal method to solve the problem. A solution for this problem would typically look like this:
Over the past semester, I have been looking at how we can make connections from one strand to the next and while it is fairly obvious now, I cannot believe I would not introduce a table and graph as a strategy to solve such a problem. I always knew this was a possibility, but wonder if I was brainwashed by the โtextbook approachโ of how a proportional reasoning problem is supposed to be solved. It seems fairly reasonable for most of my students to be able to create a table of values (or ratio table) to get a visual of what is really happening here. Discussing independent and dependent variables as well as how to create a scatter plot are skills that could be introduced during the solution process of a question like this rather than taught as a prerequisite to solve a future problem.
## Rates of Change and Unit Rates
Extending the equivalent fraction work that was carried out in the estimation portion at the beginning of the task, we can then take the change in y over the change in x to discuss how all differences in the table would yield an equivalent fraction. Students could then reduce the fraction to a unit rate that could be useful to help solve the problem more easily.
## Connections Between Decimals, Fractions, and Percents
When I have discussions with grade 7 and 8 teachers about their Decimals, Fractions, and Percents unit, I can distinctly remember how boring it was as a student and how boring it is as a teacher. Rather than introducing them separately with no context, I see proportional reasoning problems like Baby Beats as an excellent opportunity to make this connection.
As you can see in the image below, there are some big connections to be made here. Rate of change could be represented as 18 beats per 7.12 seconds, or 18/7.12 in fraction form; followed by the unit rate of 2.52 beats per 1 second, or 2.52 in decimal form; and finally, 252 beats per 100 seconds, or 252% in percent form. The important piece here is that all three representations are really the same representation, but simply different equivalent fractions.
## Linear Patterning With Initial Value and Rate of Change
What better way to begin the discussion around patterning than with a direct variation relationship, such as a proportion? The concept of initial value can be easily introduced since all proportions start at zero, leaving an equation with your dependent variable equivalent to your rate of change multiplied by the independent variable.
Why did I wait to introduce this concept until the โDirect and Partial Variationโ section in the textbook?
## Creating and Interpreting Linear Equations
Finally, we can then begin looking at the relationship between time and number of beats to create linear equations.
## Proportional Reasoning Strategies Template
Feel free to download a handy template I created that could be beneficial for any proportional reasoning questions your students may be solving. Encouraging them to use different strategies can really help them solidify their understanding of the concepts.
## Paying Attention to Proportional Reasoning
Access the Ontario Ministry of Educationโs recent publication about Proportional Reasoning and the importance from K-12. Specific examples are included at each divisional level.
How are you making connections in proportional reasoning or other topics across different math strands? Share with us!
## WANT TO LEARN HOW TO TEACH THROUGH TASK?
Download our Complete Guide to successfully implementing our Make Math Moments 3-Part Framework in your math class!
## Share With Your Learning Community:
Iโm Kyle Pearce and I am a former high school math teacher. Iโm now the K-12 Mathematics Consultant with the Greater Essex County District School Board, where I uncover creative ways to spark curiosity and fuel sense making in mathematics. Read more. | 1,728 | 8,719 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-14 | longest | en | 0.958682 |
https://www.scribd.com/document/53241220/state-transition-testing-handout | 1,553,369,280,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202924.93/warc/CC-MAIN-20190323181713-20190323203713-00446.warc.gz | 876,886,104 | 47,702 | You are on page 1of 28
BCS Oxfordshire
Thursday 6th October
State transition testing
A practical workshop in the use of a
software testing technique
Peter Quentin - QBIT
contents
โข Introduction
โข State transition testing
โข Example
โข Exercises
โข Conclusion
.. very large or infinite number of test scenarios + finite amount of time = impossible to test everything .the problem.
. the solution.. Software test techniques exist to reduce the number of tests to be run whilst still providing sufficient coverage of the system under test .
state transition testing โข Models each state a system can exist in โข Models each state transition โข Defines for each state transition โฃ start state โฃ input โฃ output โฃ finish state .
OFF .
Input: switch on OFF Output: light on ON .
OFF Input: switch off ON Output: light off .
TEST 1 TEST 2 OFF start state: start state: off on input: input: switch on switch off output: output: light on light off finish state: finish state: on off ON .
OFF FAULT ON .
OFF OFF ON FAULT FAULT OFF ON ON .
TEST STEP 1 STEP 2 STEP 3 START OFF ON OFF STATE INPUT SWITCH ON SWITCH OFF SWITCH ON OUTPUT LIGHT ON LIGHT OFF LIGHT ON FINISH ON OFF ON STATE .
display time. display date and change date โข The change mode button switches between display time and display date โข The reset button switches from display time to adjust time or display date to adjust date โข The set button returns from adjust time to display time or adjust date to display date . change time.electronic clock example โข A simple electronic clock has four modes.
reset DISPLAY ADJUST TIME TIME set change mode change mode reset DISPLAY ADJUST DATE DATE set .
DISPLAY TIME DISPLAY ADJUST DATE TIME DISPLAY ADJUST DISPLAY TIME DATE TIME DISPLAY DATE 0-switch or branch coverage .
STEP 1 STEP 2 TEST 1 START STATE DISPLAY TIME DISPLAY DATE INPUT CHANGE MODE CHANGE MODE OUTPUT DISPLAY DATE DISPLAY TIME FINISH STATE DISPLAY DATE DISPLAY TIME STEP 1 STEP 2 STEP 3 TEST 2 START STATE DISPLAY TIME DISPLAY DATE ADJUST DATE INPUT CHANGE MODE RESET SET OUTPUT DISPLAY DATE ADJUST DATE DISPLAY DATE FINISH STATE DISPLAY DATE ADJUST DATE DISPLAY DATE STEP 1 STEP 2 TEST 3 START STATE DISPLAY TIME ADJUST TIME INPUT RESET SET OUTPUT ADJUST TIME DISPLAY TIME FINISH STATE ADJUST TIME DISPLAY TIME .
DISPLAY TIME DISPLAY ADJUST DATE TIME DISPLAY ADJUST DISPLAY TIME DATE TIME ADJUST DISPLAY ADJUST DISPLAY TIME DATE TIME DATE DISPLAY DATE 1-switch or switch coverage DISPLAY TIME ADJUST DATE .
n-switch or boundary interior โฃ execute each loop n times โฃ at least twice โฃ tests each loop irrespective of the start/end point โฃ 6 loops to be tested in this case n-switch or boundary interior .
. for each exercise. โข Draw a state transition diagram โฃ draw the states โฃ mark the state transitions โฃ define input and output for each state transition โข Determine the level of coverage โฃ 1-switch/switch coverage for these exercises โข Draw a testing tree โข Define the tests ..
The first press of the button turns the toothbrush from off to speed one.electric toothbrush A two-speed electric toothbrush is operated by pressing its one button. the second press of the button turns it to speed two When the button is pressed for a third time the electric toothbrush is turned off .
OFF OFF S1 S2 SPEED SPEED 2 1 OFF S1 .
When in fast forward the play button can be pressed to enter play mode directly. When in play mode. the fast forward button can be pressed again to enter fast forward or the stop button can be used to return to play. These operations can be cancelled using the stop button. fast forward and fast play. When in fast play mode. .tape cassette player A tape player has three operations: play. the fast forward can be used to fast play. Play and fast forward are activated using the play and fast forward button respectively.
fa OFF st y fo a pl rw st op ar op d st play FAST PLAY FOR- WARD fa d st ar op st rw fo fo rw FAST st ar PLAY fa d .
0-switch or branch coverage .
1-switch or switch coverage .
input. output and finish state โฃ determine coverage level to be achieved โฃ draw testing tree โฃ define tests . โข There are many useful techniques available โฃ state transition testing is only one of the many โข A finite number of tests can be defined from an infinite number of test scenarios and a level of coverage can still be achieved โข The process for state transition testing: โฃ draw state transition diagram โฃ determine start state.. conclusions..
Peter Quentin .co.uk ..thank you very much.QBIT peter@qbit.. | 973 | 4,524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-13 | latest | en | 0.6448 |
https://www.smore.com/0e9fb-problem-solving-without-numbers | 1,582,981,726,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875149238.97/warc/CC-MAIN-20200229114448-20200229144448-00030.warc.gz | 899,885,662 | 10,778 | # The Problem with Problem Solving
## Students usually struggle with problem solving because they don't understand the problem. Let's explore strategies to take the problems out of problem solving.
Students are often presented with a story problem and they immediately start guessing what operation they should do to the numbers. I call it the "pluck and poke" method of problem solving. They pluck the numbers out, poke them in an equation and say they have a solution. The problem is the answer may not be related to the problem. The students may not understand the problem they are trying to solve.
So how do we get students to slow down and move from "answer getting" to real problem solving? We need to improve reading comprehension in math. Here are some ideas from Math in Practice: A Guide for Teachers by Susan O'Connell.
## Understand the Problem
First, students must understand the problem. To help them focus on reading and understanding, leave the numbers and the question out of the problem. Let students read, retell, and visualize the story with no numbers. In place of numbers use words such as a group or some. This helps students slow down and consider the story part of the problem. Students can focus on what is happening, rather than "how many" it is happening to.
We want to know that when children read the words, they understand the words. For many years we have asked students to underline the question, but underlining the question doesn't mean students understand what is happening in the problem. After reading the problem, students retell or restate the problem in their own words. A graphic organizer used in retelling a story, can help with retelling a math problem. Students tell what they know at the beginning of the problem, what action happened during the problem, and how the problem ends. This sounds a lot like reading comprehension strategies.
Visualizing the problem may help students understand it. Students can act out what is happening, use objects to show the problem, or use a math tool like a part-part-whole mat to represent the problem. Sometimes, gathering data into a chart or table or using a number line, will help students visualize what is happening in the story.
Multiple reads will allow students to process what they are reading in order to better understand the problem. One strategy is called Three Reads. First, students read for the gist of the story. They can paraphrase without numbers and explain the story. As students show understanding of the story, add the question and the numbers to complete the story. The second read is to understand what the problem is. In this read students are identifying what they need to solve. The third reading is for gathering the details needed to solve the problem. In this read students identify information needed to solve the problem.
Focusing on reading the story without numbers will help students view problems in a different way. They are seeking understanding of the situation and the problem. Students should be thinking about how the information they have can help solve the problem. The focus shifts from right answers to solving problems.
## Identify Necessary Information
Information needed to solve a problem is not always readily apparent. Important information may be buried in a sea of information, making it difficult to identify the most important information. Once a student knows what the question to be solved is, then he or she can identify and find the information that is helpful in solving the problem.
Sometimes students need to find conditions in the problem. A condition is data that doesn't appear with the other data, but affects the outcome of the problem. Consider this problem:
Molly's Pastry Shop baked these tasty treats:
55 apple pies
65 pumpkin pies
Molly rolled out the crusts for the pies. The apple pies each needed 2 crusts. The pumpkin pies only needed 1 crust. How many crusts did she need to roll?
The number of crusts needed for each pie is a condition that affects the solution.
Identifying missing data is also critical to identifying necessary information. Sometimes the important data is not in the problem, but can be found from the information provided in the problem. This leads to a two-step problem. Students need to think about what information will help them find the missing data, and what data is unnecessary to solve the problem.
## Develop a Plan
Here is a list of questions that will help students make a plan:
โข What is a good plan for solving this problem?
โข Should I add, subtract, multiply, or divide?
โข Should I make a table to help me see patterns or relationships?
โข Would a picture or diagram help me simplify the problem?
โข Could working backward help me find the solution?
โข Would organizing the data in a systemic way help?
โข What other strategies might help me solve this problem?
## Try the Plan
Give students time to try their plan. Let students work in partners and discuss progress with each other. If you see students becoming frustrated, jump in and ask questions that lead students to possible approaches as they share their thinking. Be careful not to guide too much. You want to honor students' opportunities for discovery of a process that leads to a solution. The goal is to promote perseverance with students as they tackle challenging problems.
## Check for Reasonableness
Throughout the problem solving process, students need to ask themselves, "Does this make sense?" When students have an answer, they need to consider if this answer makes sense for this problem. Students may use rounding to determine if the answer makes sense. If students understand the problem and know the numbers, they should be able to check for reasonableness.
## Summary
As you look at the steps to problem solving above, you may notice understanding the problem is the first essential element to successful problem solving. That's because the numbers and equations are meaningless without understanding. Deep understanding is what leads to a successful solution that goes far beyond an answer.
In the video below, the teacher never gives the numbers. The students choose their own numbers to solve the problem.
In the video above notice:
โข structure
โข engagement
โข teacher moves
How does this lesson support differentiation in the classroom?
What instructional challenges does the teacher face and overcome in the lesson?
Can this instructional approach be taken in other grade levels in a similar way? | 1,263 | 6,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-10 | latest | en | 0.965605 |
https://socratic.org/questions/how-do-you-find-the-domain-of-f-g-given-f-x-3x-4-and-g-x-5-4-x | 1,585,910,001,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370510846.12/warc/CC-MAIN-20200403092656-20200403122656-00155.warc.gz | 681,057,306 | 6,212 | # How do you find the domain of f+g given f(x)=3x + 4 and g(x) = 5/(4-x)?
Jun 15, 2016
The domain of the sum of two functions is the intersection of their domains.
#### Explanation:
First of all, let us find the domain of $f \left(x\right)$ and $g \left(x\right)$ independently:
โข $f \left(x\right)$ is a polynomial function, so its domain is $\mathbb{R}$.
โข $g \left(x\right)$ is a fractional function, so its domain is $\mathbb{R}$ excepting those points where the denominator vanishes:
$4 - x = 0 \rightarrow x = 4$
So the domain of $g \left(x\right)$ is $\mathbb{R} - \left\{4\right\}$.
Now, the domain of
$\left(f + g\right) \left(x\right) = 3 x + 4 + \frac{5}{4 - x}$
consists of those points where both $f \left(x\right)$ and $g \left(x\right)$ exist, this is the intersection of both domains. Since both domains are $\mathbb{R}$ except the second one, which excludes the value $x = 4$, the domain of the sum is:
$\text{Dom} \left(f + g\right) \left(x\right) = \mathbb{R} - \left\{4\right\}$ | 330 | 1,009 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2020-16 | latest | en | 0.769922 |
https://www.physicsforums.com/threads/find-current-for-circuit-with-capacitor-resistor-capacitor-in-series.116560/ | 1,477,372,243,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719908.93/warc/CC-MAIN-20161020183839-00460-ip-10-171-6-4.ec2.internal.warc.gz | 971,521,111 | 14,408 | # Find current for circuit with capacitor, resistor, capacitor in series
1. Apr 5, 2006
### Cyrus
I have to solve this problem for a circuit. Its a capacitor, a resistor and another capacitor all in series.
Both capacitances are the same, 1uF, and the resistor is 100k-Ohm.
I have to find the current function of time.
What I did was use KVL to get:
$$- \frac{1}{c_1} \int i(t)dt + v_1 (t_0) + \frac{1}{c_2} \int i(t)dt + v_2 (t_0) + R i(t) = 0$$
Then I took the derivative to get:
$$- \frac{1}{c_1} i(t) + R \frac {di}{dt} + \frac{1}{c_2}i(t) =0$$
Which simplifies to:
$$R( \frac{1}{c_2} - \frac {1}{c_1} )^{-1} \frac {di}{dt} +i(t) = 0$$
But the two capacitances have the same value, which means that RC-eq is zero. That's wrong....hmmmmmmmmm
Last edited: Apr 5, 2006
2. Apr 5, 2006
### barob1n
WTF. Why not change the signs of one of the capacitances. They should have the same sign. Or just let the combined capacitance equal to .5uF.
3. Apr 5, 2006
### Cyrus
Because based on the diagram, they are in the passive configuration which is why it is negative. | 374 | 1,078 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2016-44 | longest | en | 0.895749 |
https://www.encyclopediaofmath.org/index.php/Erd%C3%B6s%E2%80%93Ginzburg%E2%80%93Ziv_theorem | 1,488,267,542,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501174154.34/warc/CC-MAIN-20170219104614-00108-ip-10-171-10-108.ec2.internal.warc.gz | 794,818,830 | 10,621 | # Erdรถs-Ginzburg-Ziv theorem
EGZ theorem
If $m$ is a positive integer and $a_1,\ldots,a_{2m-1}$ is a sequence of elements from the cyclic group $\mathbb{Z}_m$, then there exists a set $I\subseteq \{1,\ldots,2m-1\}$ of cardinality $m$ such that $\sum_{i\in I}a_i=0$. This theorem was first shown in [a5].
## Related theorems.
Looking at a sequence of zeros and ones one sees that cannot be replaced by a smaller number. This motivates the definition of the ErdรถsโGinzburgโZiv constant for an arbitrary Abelian group, as follows. If is an Abelian group, then is the minimum integer such that every sequence of elements from contains a subsequence of cardinality , the order of , that adds up to . It can be proven that and equality holds if and only if . This result and the observation above led to the following two directions of investigation:
i) To find bounds, or possibly determine, for groups other than in terms of and other group invariants. Recent results in this direction were obtained by Y. Caro and Weidong Gao.
ii) To find or estimate the minimum integer such that every sequence of elements from contains a subsequence of cardinality that adds up to , provided one knows that there are at least distinct elements in the sequence. Recent results in this direction are due to A. Bialostocki, P. Dierker, Y.O. Hamidoune, and M. Lotspeich. A breakthrough in this direction was achieved after the recent proof of the long standing ErdรถsโHeilbronn conjecture (cf. also ErdรถsโHeilbronn problem). Along a different line, J.E. Olson extended the definition of the constant to non-Abelian groups and proved that still holds.
## Outline of developments.
There are several reasons why this theorem has recently (1996) drawn much attention.
Quite unexpectedly, N. Alon and M. Dubiner [a1] have shown that the theorem follows from several deeper results in algebra and number theory, establishing interesting links.
The theorem has many possible generalizations, some of which have been proved and others are easy to state as fundamental open problems, see [a4] and the conjectures below.
The theorem motivated the development of what is called a zero-sum Ramsey theory: If the sequence in the theorem consists only of the elements and , then its proof follows from the pigeon hole principle (cf. Dirichlet principle), hence the theorem can be viewed as a generalization of this principle. Consequently, since Ramsey theory (cf. also Ramsey theorem) is a development of the pigeon hole principle, there is a clear motivation to develop from the EGZ theorem a zero-sum Ramsey theory along the lines of traditional Ramsey theory. While in Ramsey theory one looks for monochromatic configurations, in zero-sum Ramsey theory the colours are elements of a group and one looks for zero-sum configurations. Zero-sum Ramsey theory generalizes many results of the traditional Ramsey theory and leaves many open problems.
## Proofs.
In [a1] five proofs for the EGZ theorem have been given. In all these proofs it is assumed that is a prime number, since the transition to a non-prime is a simple induction. The original proof is based on the CauchyโDavenport theorem from elementary additive number theory; two other proofs use the Fermat little theorem, along with a counting argument and a lemma concerning permanents, respectively. A fourth proof uses the ChevalleyโWarning theorem about zeros of polynomials over a finite field. Most interesting is the proof that uses knowledge of the Davenport constant of an Abelian -group, determined by Olson. Let be a finite Abelian group. The Davenport constant of , denoted by , is the minimal such that every sequence of elements from contains a subsequence that adds up to .
## Generalizations and analogues.
Clearly, if is cyclic of order , then . An interesting relation between and the EGZ theorem is Weidong's generalization of the EGZ theorem: Let be a sequence of elements from an Abelian group . If , then there exists a set of cardinality such that .
The following two conjectures are other possible generalizations of the EGZ theorem.
### Conjecture 1.
Let and be positive integers. If is a sequence of elements from the cyclic group , then it contains at least subsequences of elements that add up to .
If one takes in this conjecture, then the EGZ theorem follows.
### Conjecture 2.
Let be a positive integer and let be a sequence of elements from the cyclic group whose sum is . If is a sequence of elements from , then it contains a subsequence such that the sequence can be reordered such that .
If one takes , , in this conjecture, then the EGZ theorem follows.
Conjecture 1 was proven by M. Kisin for and , where and are distinct prime numbers and . In [a7], Z. Fรผredi and D.J. Kleitman confirmed Conjecture 1 asymptotically for every positive integer . Conjecture 2 can be easily proven for prime. Both conjectures illustrate the general difficulty that exists in this area for handling the non-prime case.
There are many related problems to the EGZ theorem. The following conjecture was raised by H. Harborth and some progress was made by Alon and Dubiner. It illustrates that certain problems are open, even for primes.
### Conjecture 3.
If is a prime and is a sequence of elements from the group , then there exists a subsequence of elements whose elements add up to .
A sequence containing only the elements , , , and , where each element appears times, implies that can not be replaced by a smaller number.
## Zero-sum Ramsey theory.
This theory was first introduced in [a3]. Today it includes many results on zero-sum Ramsey numbers for graphs. Recent developments by Bialostocki, P. Erdรถs, H. Lefmann, and D. Schaal deal with zero-sum solutions to systems of equations and inequalities over the integers. Surveys of this can be found in [a2] and [a4]. The following theorem from zero-sum Ramsey theory, proved in [a6] and [a8], generalizes in the sense of the EGZ theorem the folkloristic fact that in every -colouring of the edges of a complete graph there is always a monochromatic spanning tree: If each edge of the complete graph (cf. also Graph theory) is assigned an element from the cyclic group , say , then there exists a spanning tree of with edges such that .
#### References
[a1] N. Alon, M. Dubiner, "Zero-sum sets of prescribed size" D. Miklรณs (ed.) V.T. Sรณs (ed.) T. Szรถnyi (ed.) , Combinatorics, Paul Erdรถs is Eighty , Bolyai Society Mathematical Studies , 1 , Keszthely (Hungary) (1993) pp. 33โ50 [a2] A. Bialostocki, "Zero-sum trees: a survey of results and open problems" N.W. Sauer (ed.) R.E. Woodrow (ed.) B. Sands (ed.) , Finite and Infinite Combinatorics in Sets and Logic , Nato ASI Ser. , Kluwer Acad. Publ. (1993) pp. 19โ29 [a3] A. Bialostocki, P. Dierker, "Zero-sum Ramsey theorems" Congressus Numerantium , 70ย : 1 (1990) pp. 19โ130 [a4] Y. Caro, "Zero-sum problems: a survey" Discrete Math. , 152 (1996) pp. 93โ113 [a5] P. Erdรถs, A. Ginzburg, A. Ziv, "A theorem in additive number theory" Israel Research and Development Nat. Council Bull., Sect. F , 10 (1961) pp. 41โ43 [a6] Z. Fรผredi, D. Kleitman, "On zero trees" J. Graph Th. , 16 (1992) pp. 107โ120 [a7] Z. Fรผredi, D. Kleitman, "The minimal number of zero sums" D. Miklรณs (ed.) V.T. Sรณs (ed.) T. Szรถnyi (ed.) , Combinatorics, Paul Erdรถs is Eighty , Bolyai Society Mathematical Studies , 1 , Keszthely (Hungary) (1993) pp. 159โ172 [a8] L. Schrijver, P. Seymour, "A simpler proof and generalization of the zero-trees theorem" J. Combin. Th. A , 58 (1991) pp. 301โ305
How to Cite This Entry:
ErdรถsโGinzburgโZiv theorem. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Erd%C3%B6s%E2%80%93Ginzburg%E2%80%93Ziv_theorem&oldid=40192 | 1,969 | 7,735 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2017-09 | longest | en | 0.937779 |
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# Master of Finance dilemma
Author Message
Intern
Joined: 16 Jan 2011
Posts: 1
Kudos [?]: [0], given: 0
### Show Tags
02 Jun 2011, 13:19
Hello Everyone!!
I am planning to pursue a master of finance the upcoming September and I have a dilemma concerning the school I will go. I have been accepted in Schulich business School in Canada and also at Cranfield school of management in the UK and I am still expecting an answer from LSE. I live in Greece and I was working here for 4 years in Grant Thornton SA. I decided I want to study and work abroad since the economy here is not very stable. I was considering Canada because I have family there, Iโve been numerous times and I like the culture. I visited Schulich last week and I really liked the atmosphere of the University. My problem is that I donโt know where I am going to work next. I would like to try and stay in Canada but I have also considered returning to Europe (not Greece). Will it make a big difference if I go to a Canadian University and then return to Europe or the opposite. I will appreciate your opinions.
Kudos [?]: [0], given: 0
SVP
Affiliations: HEC
Joined: 28 Sep 2009
Posts: 1635
Kudos [?]: 684 [0], given: 432
Concentration: Economics, Finance
GMAT 1: 730 Q48 V44
Re: Master of Finance dilemmaย [#permalink]
### Show Tags
06 Jun 2011, 15:00
Yes, I think that it will make a difference. If you want Canada, then a solid Canadian school will of course help. But for Europe, LSE is at the top. The school is reasonably fast when it comes to application decisions, so this shouldn't pose too much of a strain on your time schedule.
_________________
Kudos [?]: 684 [0], given: 432
Re: Master of Finance dilemma ย [#permalink] 06 Jun 2011, 15:00
Display posts from previous: Sort by | 634 | 2,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-43 | latest | en | 0.934967 |
https://pwnthemole.github.io/crypto/2018/09/21/picoctfweirderrsa.html | 1,569,011,726,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574077.39/warc/CC-MAIN-20190920200607-20190920222607-00192.warc.gz | 629,114,768 | 3,298 | # pwnthem0le
## pwnthem0le is a Turin-based, hacking students group born out of CyberChallenge 2018. Read more about us!
21 September 2018
# picoCTF 2017 - weirderRSA Writeup
by matpro98
We have the public key, which consist of $n$ and $e$, $d_p=d \pmod{p-1}$ and the encrypted flag $c$.
e = 65537
n = 352758655756163603130656475864162239004344663459120398951306959672239055329877644796995008368282924624780849432051543118959312685532106237568240835778731486989439626252834661294225426875963944816709371554839452465119058016363040631618359944564550348310851045841670935254841385590882490443247265126417117450357
dp = 13530055667815347122266109008252377134325151556131892235929064596659462917644020624855537451062167377041847601387880412738836767351591511886432133011921729
c = 23428056833770750219439218340180501853506449797628734848807388355447212714387039203998085387476974936419607861041793755542930286287098871510394661091846780839592290953853536571372997807697657464569729651718518301857979495046280018444198435962234642736892075369840282923945267377104440625478468507147243879631
## Understanding the hint
The hint given is to find a multiple of $p$. Since in RSA $d \cdot e \equiv 1 \pmod{\varphi(n)}$, we have $d \cdot e \equiv 1 \pmod{(p-1)(q-1)}$, $d \cdot e \equiv 1 \pmod{p-1}$, $d \pmod{p-1} \cdot e \equiv 1 \pmod{p-1}$, $d_p \cdot e=1 \pmod{p-1}$. Thus there exist $k \in \mathbb{Z}$ such that $d_p \cdot e=1+k(p-1);\ p=\dfrac{d_p \cdot e -1}{k}+1$. From $d_p=d \pmod{p-1}$ we have $% $ and then $% $; so $% $, a reasonable value to bruteforce such $k$ keeping in mind that $p$ should be an integer that divides $n$.
## Attack
Using Python, the gmpy2 library is useful to compute the modular multiplicative inverse. The code is very simple:
import gmpy2
e=65537
n=352758655756163603130656475864162239004344663459120398951306959672239055329877644796995008368282924624780849432051543118959312685532106237568240835778731486989439626252834661294225426875963944816709371554839452465119058016363040631618359944564550348310851045841670935254841385590882490443247265126417117450357
dp=13530055667815347122266109008252377134325151556131892235929064596659462917644020624855537451062167377041847601387880412738836767351591511886432133011921729
c=23428056833770750219439218340180501853506449797628734848807388355447212714387039203998085387476974936419607861041793755542930286287098871510394661091846780839592290953853536571372997807697657464569729651718518301857979495046280018444198435962234642736892075369840282923945267377104440625478468507147243879631
stuff=dp*e-1
for k in range(1,e):
if stuff%k==0: #p should be an integer
p1=stuff//k+1
if n%p1==0: #p should divide n
p=p1
q=n//p
phi=(p-1)*(q-1)
d=gmpy2.invert(e,phi)
m=pow(c,d,n)
print(bytes.fromhex(hex(pow(c,d,n))[2:]))
and returns the flag flag{wow_leaking_dp_breaks_rsa?_47413771836}
tags: cryptoย rsa | 1,035 | 2,883 | {"found_math": true, "script_math_tex": 19, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-39 | latest | en | 0.38533 |
http://teaching.callerlab.org/basic-part-1/courtesy-turn-definition/ | 1,659,932,959,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570765.6/warc/CC-MAIN-20220808031623-20220808061623-00391.warc.gz | 50,397,366 | 8,610 | # Courtesy Turn โ Definition
Teaching Resource for COURTESY TURN
Links: ย Standard Analyze Module Teach Other Extend
CALLERLAB Program: Basic Part 1
Teaching Order: After Split Two and before Ladies Chain
Recently Taught Calls: Turn Back Family, Separate
Background: This call is part of a Ladies Chain and also part of a Right and Left Thru. Both are traditional calls that go back to the time of hoop skirts. Hence we have a call from a time of elegance and grace when it was a courtesy for the gent to help the lady turn her voluminous skirts.
Definition:
Courtesy Turn involves a turning movement with a characteristic handhold and finishes with a couple facing in. It is mainly used to define and teach calls like Ladies Chain, Chain Down the Line, Right and Left Thru, Do Paso, and Eight Chain Thru. Occasionally it is used as a call by itself.
Minimum number of dancers needed: Two
Starting formation for the minimum number needed: a Couple (at Basic and Mainstream, a Normal Couple only), or a man and a woman who are facing
Command examples:
โ Heads Pass Thru; Courtesy Turn
โ (from facing lines) Pass Thru; Tag The Line In; Turn Thru; Courtesy Turn
โ Heads Star Thru; Square Thru 3; Left Touch 1/4; Walk And Dodge; Courtesy Turn
โ Walk Around the Corner; Courtesy Turn at home
โ All 4 Ladies make a right-hand star, turn it once around; boys Courtesy Turn your girl
โ (from a completed double pass thru with ladies in the lead) Cloverleaf; four women in the center Square Thru 3; men reach out and Courtesy Turn this girl
Dance action: A couple works as a unit and turns around with the left-side dancer backing up and right-side dancer walking forward. The turning point is halfway between the dancers. Facing dancers blend into a Normal Couple as they perform this action.
The amount of turning is governed by the following rules below or may be given explicitly. If the rules are contradictory or none apply, then the amount must be given explicitly.
1. A couple that has other dancers behind it turns 180 degrees to end facing the other dancers.
2. Dancers working on the outside of the set turn to end facing the center of the set.
3. If an inactive, outside man is facing in, with an active woman coming towards him, then the couple turns to face the direction in which the inactive man had been facing.
Ending Formation: Couple facing in (to their group of 4 or the center of the set)
Timing: 4
Styling: The womanโs left hand (palm down) and manโs left hand (palm up) are joined.
Right hands are placed according to the womanโs choice. If she places her right hand behind her right hip, palm out, then the man places his right hand in hers without grasping it, leaving these hands available for the next call. If she uses her right hand to work her skirt, then the man places his right hand in the small of the back (i.e., in the center).
Comments: Courtesy Turn works best when dancers have their left hands available. For example, Square Thru 3, Courtesy Turn has good hand usage. However, when dancers already have a Couples handhold, Wheel Around, California Twirl or Partner Trade may be better choices for smoother dancing.
The turning amount can be given explicitly either by the final facing direction (e.g., โto face back inโ), or the total distance, or both (e.g., โCenters go a full turn around to face the outside 2โ).
The phrase โand A Quarter Moreโ can be used after Courtesy Turn or a call ending with Courtesy Turn (e.g., Right And Left Thru And A Quarter More). The couple turns an extra 90 degrees, generally ending in a Right-Hand Two-Faced Line.
Calls that end in a Squared Set with a Courtesy Turn (e.g., 4 Ladies Chain or Do Paso) can easily blend into a left arm turn or into an Allemande Thar. For example, โ4 Ladies Chain; Chain Them Back with a Do Pasoโ. See โAdditional Detail: Blending one call into anotherโ.
Facing Couple or Ocean Wave Rule: Neither rule applies to Courtesy Turn.
Link to Taminations: Taminations Courtesy Turn | 910 | 3,994 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-33 | latest | en | 0.905023 |
https://mathoverflow.net/questions/76038/smooth-convergence-of-minimizing-varifolds | 1,627,964,743,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154420.77/warc/CC-MAIN-20210803030201-20210803060201-00074.warc.gz | 380,849,511 | 30,813 | # Smooth convergence of minimizing varifolds
Dear all,
a question came to me when I read the paper "Complete three dimensional manifolds with positive Ricci curvature and scalar curvature/ R. Schoen and S.-T. Yau, 1982".
The question is as follows. suppose we have a sequence of complete non-compact smooth submanifolds $\Sigma_k\subset M$ and each $\Sigma_k$ is area-minimizing (in its isotopy class) with respect to any compactly supported deformation. Then by treating them as Radon measures, we can get a convergent subsequence of $\Sigma_k$, say $\Sigma_k\to\Sigma$. The convergence is of the sense of Radon measure, or is called "in the sense of varifold". As I know, it is merely stronger than Cheeger-Gromov convergence. However, in Schoen-Yau's article, they say that the convergence is smooth. (page 217, the last paragraph in the proof of Lemma 3.) I cannot see the reason for this convergence to be smooth.
Maybe they have used some other properties in that paper (I guess not), but it's difficult for me to provide all the details here.... Anyway, thank you all for any comments.
The reason for this is Allard's regularity theorem.
Roughly speaking Allard's theorem says that if near a point of the support of a stationary varifold the varifold has unit density and area close to that of the ball of the appropriate dimension (for a 2-varifold it would be area of a disk) then the support of the varifold is smooth at that point.
More precisely, there is an $\epsilon>0$ and $r_0>0$ (depending on the ambient geometry and dimension of the varifold). So that if $\Sigma$ is an stationary $m$-varifold in $M$ , a Riemannian manifold, and a point $p\in spt \Sigma$ satisfies $$\mathcal{H}^m(B_r(p)\cap \Sigma) \leq \omega_m r^m(1+\epsilon)$$ for $r\leq r_0$ then $spt \Sigma$ is smooth near $p$. Here $B_r(p)$ is the $r$-ball in $M$ and $\omega_m$ is the volume of the unit ball in $\mathbb{R}^m$.
Allard's proof is unfortunately quite technical -- a good reference is Leon Simon's (sadly) hard to find book "Lectures on Geometric Measure Theory". Luckily, for your purpose there is a simpler version with a very easy proof due to Brian White (see here for the paper).
Specifically, suppose you have instead of being a stationary varifold you know that $\Sigma$ is a smooth minimal surface. If $p$ is a point of $\Sigma$ so that $$\mathcal{H}^m(B_r(p)\cap \Sigma) \leq \omega_m r^m(1+\epsilon)$$ then one has $$|A|(p)\leq r^{-1}$$ here $|A|$ is the norm of the second fundamental form. In other words you obtain a quantitive bound on curvature.
How does this relate to your question? Well you have that $\Sigma_k$ converge to $\Sigma$ as Radon measures. Let $p\in \Sigma$ this convergence implies that $$\mathcal{H}^m(\Sigma_k\cap B_r(p))\to \mathcal{H}^m(\Sigma\cap B_r(p))$$ (it is worth noting that we are using that the surfaces are area minimizing to ensure the convergence is with multiplicity one). Now since $\Sigma$ is smooth near $p$ it is locally modelled on a flat plane. In other words, there is a scale $r$ so that $$\mathcal{H}^m(B_r(p)\cap \Sigma) \leq \omega_m r^m(1+\frac{1}{2}\epsilon)$$ hence by the convergence $$\mathcal{H}^m(B_r(p)\cap \Sigma_k) \leq \omega_m r^m(1+\epsilon)$$ and so since the $\Sigma_k$ are smooth (either a priori or by Allard's full theorem) and the point $p$ was not important we deduce that for some $\delta>0$ we have $$\sup_{B_\delta(p)\cap\Sigma_k} |A|\leq r^{-1}.$$ In other words there is a uniform curvature bound on the $\Sigma_k$. The result then follows from "Standard elliptic PDE" and the Arzela-Ascoli theorem.
โข Thanks, Rbega, your answer is extremely clear and solves my problem! Thanks a lot! โย Chih-Wei Chen Sep 22 '11 at 1:35
โข Hi Rbega, it seems this argument requires that the limit $\Sigma$ is smooth, in the context of the Schoen-Yau paper how to show this? โย Caramba Oct 4 '12 at 16:27
โข @Caramba You use the regularity theory for area-minimizing hypersurfaces. The issue Chih-Wei was asking about was why the convergence was smooth. โย Rbega Oct 4 '12 at 18:33 | 1,135 | 4,049 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-31 | longest | en | 0.917984 |
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Enough Mathematical Appetizers! :ย Fall 2002 CMSC 203 - Discrete Structures 1 Enough Mathematical Appetizers! Let us look at something more interesting: Algorithms Algorithms :ย Fall 2002 CMSC 203 - Discrete Structures 2 Algorithms What is an algorithm? An algorithm is a finite set of precise instructions for performing a computation or for solving a problem. This is a rather vague definition. You will get to know a more precise and mathematically useful definition when you attend CS420. But this one is good enough for nowโฆ Algorithms :ย Fall 2002 CMSC 203 - Discrete Structures 3 Algorithms Properties of algorithms: Input from a specified set, Output from a specified set (solution), Definiteness of every step in the computation, Correctness of output for every possible input, Finiteness of the number of calculation steps, Effectiveness of each calculation step and Generality for a class of problems. Algorithm Examples:ย Fall 2002 CMSC 203 - Discrete Structures 4 Algorithm Examples We will use a pseudocode to specify algorithms, which slightly reminds us of Basic and Pascal. Example: an algorithm that finds the maximum element in a finite sequence procedure max(a 1 , a 2 , โฆ, a n : integers) max := a 1 for i := 2 to n if max < a i then max := a i {max is the largest element} Algorithm Examples:ย Fall 2002 CMSC 203 - Discrete Structures 5 Algorithm Examples Another example: a linear search algorithm, that is, an algorithm that linearly searches a sequence for a particular element. procedure linear_search(x: integer; a 1 , a 2 , โฆ, a n : integers) i := 1 while (i ๏ฃ n and x ๏น a i ) i := i + 1 if i ๏ฃ n then location := i else location := 0 {location is the subscript of the term that equals x, or is zero if x is not found} Algorithm Examples:ย Fall 2002 CMSC 203 - Discrete Structures 6 Algorithm Examples If the terms in a sequence are ordered, a binary search algorithm is more efficient than linear search. The binary search algorithm iteratively restricts the relevant search interval until it closes in on the position of the element to be located. Algorithm Examples:ย Fall 2002 CMSC 203 - Discrete Structures 7 Algorithm Examples a c d f g h j l m o p r s u v x z binary search for the letter โjโ center element search interval Algorithm Examples:ย Fall 2002 CMSC 203 - Discrete Structures 8 Algorithm Examples a c d f g h j l m o p r s u v x z binary search for the letter โjโ center element search interval Algorithm Examples:ย Fall 2002 CMSC 203 - Discrete Structures 9 Algorithm Examples a c d f g h j l m o p r s u v x z binary search for the letter โjโ center element search interval Algorithm Examples:ย Fall 2002 CMSC 203 - Discrete Structures 10 Algorithm Examples a c d f g h j l m o p r s u v x z binary search for the letter โjโ center element search interval Algorithm Examples:ย Fall 2002 CMSC 203 - Discrete Structures 11 Algorithm Examples a c d f g h j l m o p r s u v x z binary search for the letter โjโ center element search interval found ! Algorithm Examples:ย Fall 2002 CMSC 203 - Discrete Structures 12 Algorithm Examples procedure binary_search(x: integer; a 1 , a 2 , โฆ, a n : integers) i := 1 {i is left endpoint of search interval} j := n {j is right endpoint of search interval} while (i < j) begin m := ๏ซ(i + j)/2๏ป if x > a m then i := m + 1 else j := m end if x = a i then location := i else location := 0 {location is the subscript of the term that equals x, or is zero if x is not found} Complexity:ย Fall 2002 CMSC 203 - Discrete Structures 13 Complexity In general, we are not so much interested in the time and space complexity for small inputs. For example, while the difference in time complexity between linear and binary search is meaningless for a sequence with n = 10, it is gigantic for n = 2 30 . Complexity:ย Fall 2002 CMSC 203 - Discrete Structures 14 Complexity For example, let us assume two algorithms A and B that solve the same class of problems. The time complexity of A is 5,000n, the one for B is ๏ฉ1.1 n ๏น for an input with n elements. For n = 10, A requires 50,000 steps, but B only 3, so B seems to be superior to A. For n = 1000, however, A requires 5,000,000 steps, while B requires 2.5๏10 41 steps. Complexity:ย Fall 2002 CMSC 203 - Discrete Structures 15 Complexity This means that algorithm B cannot be used for large inputs, while algorithm A is still feasible. So what is important is the growth of the complexity functions. The growth of time and space complexity with increasing input size n is a suitable measure for the comparison of algorithms. Complexity:ย Fall 2002 CMSC 203 - Discrete Structures 16 Complexity Comparison: time complexity of algorithms A and B Algorithm A Algorithm B Input Size n 10 100 1,000 1,000,000 5,000n 50,000 500,000 5,000,000 5๏10 9 ๏ฉ1.1 n ๏น 3 2.5๏10 41 13,781 4.8๏10 41392 Complexity:ย Fall 2002 CMSC 203 - Discrete Structures 17 Complexity This means that algorithm B cannot be used for large inputs, while running algorithm A is still feasible. So what is important is the growth of the complexity functions. The growth of time and space complexity with increasing input size n is a suitable measure for the comparison of algorithms. The Growth of Functions:ย Fall 2002 CMSC 203 - Discrete Structures 18 The Growth of Functions The growth of functions is usually described using the big-O notation . Definition: Let f and g be functions from the integers or the real numbers to the real numbers. We say that f(x) is O(g(x)) if there are constants C and k such that |f(x)| ๏ฃ C|g(x)| whenever x > k. The Growth of Functions:ย Fall 2002 CMSC 203 - Discrete Structures 19 The Growth of Functions When we analyze the growth of complexity functions , f(x) and g(x) are always positive. Therefore, we can simplify the big-O requirement to f(x) ๏ฃ C๏g(x) whenever x > k. If we want to show that f(x) is O(g(x)), we only need to find one pair (C, k) (which is never unique). The Growth of Functions:ย Fall 2002 CMSC 203 - Discrete Structures 20 The Growth of Functions The idea behind the big-O notation is to establish an upper boundary for the growth of a function f(x) for large x. This boundary is specified by a function g(x) that is usually much simpler than f(x). We accept the constant C in the requirement f(x) ๏ฃ C๏g(x) whenever x > k, because C does not grow with x. We are only interested in large x, so it is OK if f(x) > C๏g(x) for x ๏ฃ k. The Growth of Functions:ย Fall 2002 CMSC 203 - Discrete Structures 21 The Growth of Functions Example: Show that f(x) = x 2 + 2x + 1 is O(x 2 ). For x > 1 we have: x 2 + 2x + 1 ๏ฃ x 2 + 2x 2 + x 2 ๏ x 2 + 2x + 1 ๏ฃ 4x 2 Therefore, for C = 4 and k = 1: f(x) ๏ฃ Cx 2 whenever x > k. ๏ f(x) is O(x 2 ). The Growth of Functions:ย Fall 2002 CMSC 203 - Discrete Structures 22 The Growth of Functions Question: If f(x) is O(x 2 ), is it also O(x 3 )? Yes. x 3 grows faster than x 2 , so x 3 grows also faster than f(x). Therefore, we always have to find the smallest simple function g(x) for which f(x) is O(g(x)). The Growth of Functions:ย Fall 2002 CMSC 203 - Discrete Structures 23 The Growth of Functions โPopularโ functions g(n) are n log n, 1, 2 n , n 2 , n!, n, n 3 , log n Listed from slowest to fastest growth: 1 log n n n log n n 2 n 3 2 n n! The Growth of Functions:ย Fall 2002 CMSC 203 - Discrete Structures 24 The Growth of Functions A problem that can be solved with polynomial worst-case complexity is called tractable . Problems of higher complexity are called intractable. Problems that no algorithm can solve are called unsolvable . You will find out more about this in CS420. Useful Rules for Big-O:ย Fall 2002 CMSC 203 - Discrete Structures 25 Useful Rules for Big-O For any polynomial f(x) = a n x n + a n-1 x n-1 + โฆ + a 0 , where a 0 , a 1 , โฆ, a n are real numbers, f(x) is O(x n ). If f 1 (x) is O(g 1 (x)) and f 2 (x) is O(g 2 (x)), then (f 1 + f 2 )(x) is O(max(g 1 (x), g 2 (x))) If f 1 (x) is O(g(x)) and f 2 (x) is O(g(x)), then (f 1 + f 2 )(x) is O(g(x)). If f 1 (x) is O(g 1 (x)) and f 2 (x) is O(g 2 (x)), then (f 1 f 2 )(x) is O(g 1 (x) g 2 (x)). Complexity Examples:ย Fall 2002 CMSC 203 - Discrete Structures 26 Complexity Examples What does the following algorithm compute? procedure who_knows(a 1 , a 2 , โฆ, a n : integers) m := 0 for i := 1 to n-1 for j := i + 1 to n if |a i โ a j | > m then m := |a i โ a j | {m is the maximum difference between any two numbers in the input sequence} Comparisons: n-1 + n-2 + n-3 + โฆ + 1 = (n โ 1)n/2 = 0.5n 2 โ 0.5n Time complexity is O(n 2 ). Complexity Examples:ย Fall 2002 CMSC 203 - Discrete Structures 27 Complexity Examples Another algorithm solving the same problem: procedure max_diff(a 1 , a 2 , โฆ, a n : integers) min := a1 max := a1 for i := 2 to n if a i < min then min := a i else if a i > max then max := a i m := max - min Comparisons: 2n - 2 Time complexity is O(n).
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Big-O Notation. The symbol , pronounced "big-O of ," is one of the Landau symbols and is used to symbolically express the asymptotic behavior of a given ... | 2,929 | 10,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2016-44 | latest | en | 0.741283 |
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# Using the Decibel - Part 2: Expressing Power as an Audio Level
## 6/4/2008 2:37 PM EDT
You can receive a 15% discount on "Sound System Engineering" by going to www.elsevierdirect.com and typing in the order code 92837 when ordering, or by calling 1-800-545-2522 and mentioning 92837 when ordering. Offer expires 6/30/2008.
[Part 1 introduces the decibel and examines concepts underlying its use in sound systems.]
2.5 Expressing Power as an Audio Level
The reference power is 0.001 W (one milliwatt). When expressed as a level, this power is called 0 dBm (0 dB referenced to 1 mW).
Thus, to express a power level we need two powers - first the measured power W1 and second the reference power W2. This can be written as a power change in dB:
W1/W2 = [(E12/1)/(E22/1)][(1/R1)/(1/R2)]
= (E12/E22)(R2/R1).ย ย ย ย ย ย ย ย ย ย ย ย ย ย (2-12)
This can be written as a power level:
10log[(E12/E22)(R2/R1)] = power change in dB.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย (2-13)
or
20log(E1/E2) + 10log(R2/R1) = power change in dB.ย ย ย ย ย ย ย ย ย (2-14)
Special Circumstance
When R1 = R2 and only then:
Power level in dB = 20log(E1/E2)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย (2-15)
where,
E2 is the voltage associated with the reference power.
2.6 Conventional Practice
When calculating power level in dBm, we commonly make E2 = 0.775 V and R2 = 600 ฮฉ.Note that E2 may be any voltage and R2 any resistance so long as together they represent 0.001 W.
Levels in dB
1. The term "level" is always used for a power expressed in decibels.
2. 10log(E12/E22) = 10 log(W1/W2)
when R1 = R2
2 x 10log(E1/E2) = 20log(E1/E2)
= 10 log(W1/W2)
3. Power definitions:
Apparent power = E x I or E2/Z,
The average real or absorbed power is (E2/Z)cosฮธ,
The reactive power is (E2/Z)sinฮธ,
Power factor = cosฮธ
4. The term "gain" or "loss" always means the power gain or power loss at the system's output due to the device under test.
Practical Variations of the dBm Equations
When the reference is the audio standard, i.e., 0.77459 V and 600 ฮฉ, then:
dB level to a reference = 10log[(E12/E22)(R2/R1)]ย ย ย ย (2-16)
where,
E2 = 0.77459...V,
R2 = 600 ฮฉ
then:
R2/E22 = 1000
and 1/1000 = 0.001. Note that any E2 and R2 that result in a power of 0.001 W may be used. We can then write:
Level (in dBm) = 10log(E12/0.001R1ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย (2-17)
and
E1 = โ(0.001R1(10dBm/10))ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย (2-18)
R1 = E2/(0.001(10dBm/10))ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย (2-19)
See Fig. 2-3.
Figure 2-3. Power in dB across a load versus available input power.
For all of the values in Table 2-2 the only thing known is the voltage. The indication is not a level. The apparent level can only be true across the actual reference impedance. Finally, the presence or absence of an attenuator or other sensitivity control is not known. See Section 2.20 for explanation of VU.
The power output of Boulder Dam is said to be approximately 3,160,000,000 W. Expressed in dBm, this output would be:
10log(3.16 x 109/10-3) = 125 dBm.
Table 2-2. Root Mean Square Voltages Used as Nonstandard References
sharps_eng
5/15/2011 11:57 AM EDT
I am finding this series a useful recap. Although I started in pro-audio I was seduced by embedded processors and digital video, but audio was definitely my first love - it's so damn analog!
The thrill was that to achieve barely acceptable SNRs of -94dB when working with 26dB of headroom took everything we had - the best brains, the best parts and then some. But we did it, and othing much has changed in terms of the net quality delivered, especially in a digital world where we can mix at 96kHz and 24 to 96bits if we want but the compression artefacts in the listeners' source material will reduce the SNR to 12 bits, or worse if we include class H amplification side-effects.
Hopefully the incease in available bandwidth will increase the amount of releases on 16bit linear PCM source, just to get rid of that quantisation weebling and psychacoustic algorithmic breathing!
Hifis still can't reproduce a piano in full voice, let alone a drum kit - maybe a good system a Marshall stack heard from way off.
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โข Naz93
In summary: T(ฮฆ(x1)ฮฆ(x2)ฮฆ(x3)ฮฆ(x4))|0> = <0|:ฮฆ(x1)ฮฆ(x2)ฮฆ(x3)ฮฆ(x4):|0> + all possible contractionsNow, we're back to a vacuum expectation value, but with one less field operator. We can keep using this process until we have no more field operators left. Then we can evaluate the vacuum expectation values using Wick's theorem and put everything back together to get the final result.In summary, to evaluate the given time-ordered product using Wick's theorem, we need to rewrite it in normal ordered form and then evaluate the vacuum expectation values using
Naz93
## Homework Statement
[/B]
Consider a real free scalar field ฮฆ with mass m. Evaluate the following time-ordered product of field operators using Wick's theorem: โซd^4x <0| T(ฮฆ(x1)ฮฆ(x2)ฮฆ(x3)ฮฆ(x4)(ฮฆ(x))^4) |0>
(T denotes time ordering)
## Homework Equations
Wick's theorem: T((ฮฆ(x1)...ฮฆ(xn)) = : (ฮฆ(x1)...ฮฆ(xn)) + all possible contractions :
( : ... : denotes normal ordered product)
## The Attempt at a Solution
I'm really confused here. I've seen examples of using Wick's theorem to evaluate products like T(ฮฆ(x1)ฮฆ(x2)ฮฆ(x3)ฮฆ(x4)) , i.e. field evaluated at fixed values of x only ... but here, there is another term that's integrated over all possible x values. And I have no idea how to deal with that - I've never seen anything like that done in the lectures I've had on this stuff.
I know contractions of the form "contraction(ฮฆ(x1)ฮฆ(x2))" give Feynman propagators.
But that's all I've got...
I'm really bad at this stuff, so if someone can help explain / help me work through this from absolute basics (i.e. assuming I know pretty much nothing), that would be really appreciated!
Hello! Let me try to explain this step by step.
First, let's rewrite the time-ordered product in normal ordered form using Wick's theorem:
T(ฮฆ(x1)ฮฆ(x2)ฮฆ(x3)ฮฆ(x4)(ฮฆ(x))^4) = :ฮฆ(x1)ฮฆ(x2)ฮฆ(x3)ฮฆ(x4)(ฮฆ(x))^4: + all possible contractions
Now, let's focus on the normal ordered product. We can expand it using the definition of normal ordering:
:ฮฆ(x1)ฮฆ(x2)ฮฆ(x3)ฮฆ(x4)(ฮฆ(x))^4: = ฮฆ(x1)ฮฆ(x2)ฮฆ(x3)ฮฆ(x4)(ฮฆ(x))^4 - <0|ฮฆ(x1)ฮฆ(x2)ฮฆ(x3)ฮฆ(x4)(ฮฆ(x))^4|0>
The first term in this expansion is easy to evaluate, it's just the product of field operators at different points. But the second term involves a vacuum expectation value, which we need to evaluate using Wick's theorem.
Let's look at the second term in more detail:
<0|ฮฆ(x1)ฮฆ(x2)ฮฆ(x3)ฮฆ(x4)(ฮฆ(x))^4|0>
Using Wick's theorem, we can write this as:
<0|ฮฆ(x1)ฮฆ(x2)ฮฆ(x3)ฮฆ(x4)(ฮฆ(x))^4|0> = <0|ฮฆ(x1)ฮฆ(x2)ฮฆ(x3)ฮฆ(x4)|0> + all possible contractions
Now, let's focus on the first term on the right-hand side. We can rewrite it using the definition of vacuum expectation value:
<0|ฮฆ(x1)ฮฆ(x2)ฮฆ(x3)ฮฆ(x4)|0> = ฮฆ(x1)ฮฆ(x2)ฮฆ(x3)ฮฆ(x4) - <0|T(ฮฆ(x1)ฮฆ(x2)ฮฆ(x3)ฮฆ(x4))|0>
Notice that the second term on the right-hand side is the time-ordered product of field operators, which we can evaluate using Wick's theorem. So, let's rewrite it as:
<
## 1. What is Wick's Theorem and why is it important in evaluating time-ordered products?
Wick's Theorem is a mathematical tool used to simplify the evaluation of time-ordered products in quantum field theory. It allows us to break down complex expressions into simpler ones, making calculations more manageable. This is particularly useful in theoretical physics where time-ordered products are commonly used to describe the dynamics of quantum systems.
## 2. How does Wick's Theorem work?
Wick's Theorem states that any time-ordered product of operators can be written as a sum of normal-ordered products plus a sum of contractions between pairs of operators. Normal-ordered products have the creation operators on the left and the annihilation operators on the right, while contractions represent the overlapping terms between operators in the time-ordered product. By breaking down the time-ordered product into these simpler terms, the evaluation becomes much easier.
## 3. What are the benefits of using Wick's Theorem?
Using Wick's Theorem can significantly reduce the complexity of calculations involving time-ordered products. It also allows us to identify the most important contributions to a particular quantity, which can help us gain a better understanding of the system being studied. Additionally, Wick's Theorem provides a systematic approach to evaluating time-ordered products, making the process more efficient.
## 4. Are there any limitations to using Wick's Theorem?
While Wick's Theorem is a powerful tool, it does have some limitations. It is only applicable to time-ordered products that involve creation and annihilation operators. It also assumes that the operators are in a normal-ordered form to begin with. In some cases, the contractions may also result in infinite terms, which require further mathematical techniques to handle.
## 5. How is Wick's Theorem used in practice?
In practice, Wick's Theorem is used to evaluate time-ordered products in theoretical calculations. It is commonly used in quantum field theory, particle physics, and condensed matter physics. It is also a fundamental concept in perturbation theory, where it allows for the systematic calculation of higher-order corrections to a given quantity. Additionally, Wick's Theorem has applications in other fields such as quantum chemistry and statistical mechanics.
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Which dilutions can you make using only 10ml pipettes? | 1,636 | 7,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2017-30 | longest | en | 0.798676 |
https://itecnotes.com/electrical/electronic-what-happens-to-a-transformer-if-the-supply-side-is-shorted/ | 1,679,823,946,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945440.67/warc/CC-MAIN-20230326075911-20230326105911-00030.warc.gz | 390,413,554 | 7,968 | # Electronic โ What happens to a transformer if the supply side is shorted
power-engineeringshort-circuittransformer
Suppose I have an industrial plant, where a switchboard is fed through a transformer. The switchboard have only "passive" loads, i.e. they don't contribute with short circuit currents.
If there is a short circuit on the switchboard feeding that transformer (on the grid side), my intuition tells me that the transformer would be demagnetized (disregarding remanent flux), thus energy must go from the transformer to either the secondary or primary side. Since current can't immediately change in an inductor, I think this energy should go to the secondary side.
My simulations show this behavior. As the image below show, there is current/power flowing for about 2 ms (the results on the primary side is identical, so I haven't included it). However, I'm not sure if this is the correct physical representation or a result of inaccurate mathematics. The minimum integration step I can select is 1 ms, so it might be a result of that (I think).
The reason why I doubt the results is because the exact same current flows from the shorted bus and in to the transformer on the primary side. Investigating the results on both sides of the transformer, I see that the exact same amount of energy goes in and out after the primary side is shorted. This tells me the transformer is not demagnetized.
Now, as we all learned in kinder garden, current in must equal current out, so how could I expect something different? Well, it's not really true in all situations (you can for instance charge a capacitor). I know the energy store in a transformer is magnetic energy, not electrical.
Still, there is something here that doesn't seem right to me. I appreciate any input!
Thanks!
# Update
It is possible that my explanation has not been good enough. The short circuit I'm talking about is a fault that occurs after a certain time, 3-phase, line-line, line-line-ground or line-ground.
Here's a part of the single line diagram. I don't have any "justification" for why the delta-star connection is relevant for the analysis (I don't think it is, it's simply how the system is designed). There can be many reasons, elimination of zero sequence currents for one, but this shouldn't be relevant for this quesiton.
The diagram below is how the transformer is represented in my simulation tool. I can't "show" how the circuit is shorted. The way it is done is I add an "event": "Create a (3 phase / Line-Line / โฆ) fault after 3 seconds. Then I can see the system response as a result of this event.
I would think the general behavior of the transformer is independent of the topology, vector group etc. | 589 | 2,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-14 | latest | en | 0.950153 |
http://www.vbaexpress.com/forum/archive/index.php/t-24175.html?s=91a3cd0fa6b0ef4a8b8b3d0119356050 | 1,591,053,021,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347419639.53/warc/CC-MAIN-20200601211310-20200602001310-00460.warc.gz | 217,910,020 | 5,104 | PDA
View Full Version : Need code added to a button to list some numbers
daniels012
12-11-2008, 01:55 PM
I need code to attach to a button!
In cell A1 I enter number of teams.
In Cell A2 I enter number of games to play.
Is there a way for me to click a button and...
In columns A and B it will give me a list of teams to play each other
based on the criteria above.
In Column A will be home team and B is Visitor.
Example I enter 9 teams with 12 games to play, hit the button and it will list the games in columns A and B.
If I need to explain more, I can.
Any help on this would be great!
Michael
nst1107
12-11-2008, 02:54 PM
If I knew the first thing about how to do this manually, I'm sure I could come up with the code. Are there rules about how teams can be matched, or is it just random? And where is your list of teams?
daniels012
12-12-2008, 07:49 AM
The list of teams can be just the numbers needed.
For example:
If I enter in A1 "9" (the number of teams) and in A2 "12" (the number of games).
Also, I did not clarify! Column A represents home games and Column B is Away games.
So....
Code would somehow come up with something like:
Column A Column B
1 2
3 4
5 6
7 8
9 1
2 8
3 7
etc. etc.
It would evenly match up the teams for home and away games up to 12 games.
Does this make more sense?
Michael
MaximS
12-12-2008, 08:53 AM
Just to clarify:
Is that kind of the league creator??
Number of games - is that total of games played by each team, number of times played against each opponent, number of times played against each opponent divided by two (one game at home, one away)?
daniels012
12-12-2008, 09:15 AM
Is that kind of the league creator??
Yes it IS!
Number of games - is that total of games played by each team
Yes, each team must play 12 games.
number of times played against each opponent divided by two (one game at home, one away)?
All I want here is to even out home and away in column A and Column B respectfully.
Michael
MaximS
12-12-2008, 09:33 AM
How many teams will play in your league??
Is it always only 12 games (6 home / 6 away)
daniels012
12-12-2008, 09:59 AM
No, The value in A2 will reflect the outcome.
So if it comes out to 7 home and 6 away, that is fine. I just don't want something like 8 home and 5 away
Also, the number of teams will vary as well. The value in A1 will determine this.
Michael
daniels012
12-15-2008, 06:51 AM
Try to make them as even as possible
daniels012
12-19-2008, 01:17 PM
bump
daniels012
12-22-2008, 11:43 AM
I guess I need a sort of Do/Until it reaches the number of games.
Ideas?
Micahel
MaximS
12-23-2008, 12:19 AM
I'm working on it but it's pretty hard to find enough time before christmas :)
daniels012
12-29-2008, 07:05 AM
ok,
My appologies
daniels012
01-14-2009, 08:04 AM
ideas?
daniels012
01-14-2009, 08:40 AM
I tried this code with 4 teams in A1 and 12 games in A2:
Dim TEAM(10000) As Single
With Range("A3:B290")
.ClearContents
End With
Range("A3").Select
MY_TEAMS = Range("A1").Value
MY_GAMES = Range("A2").Value
MY_TEAM = 1
MY_TEAM_2 = 2
Do Until MY_TEAM = MY_TEAMS And MY_TEAM_2 >= MY_TEAMS
Range("A" & Rows.Count).End(xlUp).Offset(1, 0).Value = MY_TEAM
Range("A" & Rows.Count).End(xlUp).Offset(0, 1).Value = MY_TEAM_2
TEAM(MY_TEAM) = TEAM(MY_TEAM) + 1
TEAM(MY_TEAM_2) = TEAM(MY_TEAM_2) + 1
If MY_TEAM_2 = MY_TEAMS Then
MY_TEAM_2 = MY_TEAM + 1
Else
MY_TEAM_2 = MY_TEAM_2 + 1
End If
If TEAM(MY_TEAM_2) = MY_TEAMS Then
MY_TEAM_2 = MY_TEAM + 1
End If
If TEAM(MY_TEAM) = MY_GAMES Then
MY_TEAM = MY_TEAM + 1
MY_TEAM_2 = MY_TEAM + 1
End If
If MY_TEAM_2 = MY_TEAM Then MY_TEAM_2 = MY_TEAM + 1
Loop
But it does not even the teams games between Column A and Column B, also it doesn't stop it goes on forever.??
Ideas?
Michael
daniels012
01-14-2009, 02:17 PM
Ideas?
mikerickson
01-14-2009, 06:28 PM
This is a demo of a permuations class that I wrote.
The button TEGWAR (The Exiciting Game Without Any Rules), randomly assigns players to teams and then schedules those teams against each other. The number of games is not determined by the user (per se) but by the number of games needed for each team to play each opponent once.
It's not the most efficient routine. Its designed to show off a generalized permutation class and generalized methods are not as efficient as specific ones.
To get a 12 team schedule, tell the first input box that 12 people want to play and tell the second box that there is 1 person per team. An 11 week round robin schedule will be produced.
mikerickson
01-15-2009, 12:58 AM
This is a more focused routine.
The notion is:
Consider a table with people sitting across from each other
A B C
D E F
if everyone shifts clockwise one seat, except for one person (F in this example)
D A B
E C F
and follows that pattern
E D A
C B F
C E D
B A F
B C E
A D F
eventualy everyone will have sat across the table from everyone else only once. (Played against everyone.)
Futhermore, everyone will have spent (approx) half their time on one side of the table and half on the other. (50% +/- home, 50% away).
(If F switches sides of the table each time, F also has the 50/50)
That's the approach of the attached routine.
Sheet 1 shows a spreadsheet approach. Each block of formulas is copy pasted to the right.
The button on Sheet 2 calls an input box which asks how many teams and a schedule is drawn. If there are an odd number of teams, a team named "bye" is added to the league.
daniels012
01-15-2009, 08:12 AM
This is a great start. I need them to have 12 games. I used your ScheduleTeamSS file, It only provided 8 games, is there something we can incorporate to give the 12 games?
Michael
mikerickson
01-15-2009, 01:16 PM
Navigate to sheet 2, press the button and enter 12.
If you want the fixed team count spreadsheet version (sheet1) to be increased to 12, I can do that after work.
(He says as he plots a fiendishly un-editable way to make it variable sized rather than fixed team count :) )
mikerickson
01-15-2009, 05:56 PM
To do that, with Sheet1, copy F1:J8 and paste it, leaving one blank column between pasting blocks, into AD1, AJ1, AP1, etc.
Note: 12 teams with 12 games means that each team plays another team twice, but only one other team.
mikerickson
01-15-2009, 06:50 PM
I came up with a spreadsheet solution to the problem of scheduling a variable number of teams. It uses dynamic named ranges and array formulas.
Put all the team names in column A of sheet Sched, starting in row 1, no header and no blank rows.
Name : teamList
RefersTo: =OFFSET(Sched!\$A\$1,0,0,COUNTA(Sched!\$A:\$A),1)
Select C1 and define
Name: LastWeeksListH
RefersTo: =OFFSET(Sched!A\$1,0,0,ROWS(teamList),1)
Select D1 and define
Name: LastWeeksListV
ReferstTo: =OFFSET(Sched!A\$1,0,0,ROWS(teamList),1)
(LastWeeksListH and LastWeeksListV look the same, but since they use relative referencing and different cells were active when they were defined, they refer to different ranges.)
Then in C1:C20, this array formula will give you a list of the home teams, for week 1
=IF(ROW(teamList)<=ROWS(teamList)/2,LastWeeksListH,"")
In D1: D20, this will give you the Visitor teams
=IF(ROW(teamList)<=ROWS(teamList)/2,INDEX(LastWeeksListV,ROWS(teamList)-ROW(teamList)+1,1),"")
(Array formulas are confirmed with Ctrl-Shift-Enter or Cmd+Return on Mac.)
The ranges LastWeeksListH and V both refer to column A in that postion. What these formulas do is select Home / Visitor from the list of all teams 2 (or 3) columns to the left.
To get next week's games, put
=A1 in cell F1
=INDEX(A:A,ROWS(teamList),1) in F2
=A2&"" in F3 and drag down to F40.
This is the equivalent of every one sitting at the table shifting clockwise once (except for Team 1)
Copying C1: D20 and pasting into H1:I20 will pull data from column F to show the Home vs. Away for Week 2.
Copy F1:I40 and paste into K1 to get week 3 and continue for as many weeks as desired.
You can see the attached.
VB is better for the variable number of teams situation, but Excel2008 users will be glad to know that it can be done with native Excel. | 2,403 | 8,043 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2020-24 | latest | en | 0.957588 |
http://mail.gap-system.org/pipermail/forum/2016.txt | 1,501,065,071,000,000,000 | text/plain | crawl-data/CC-MAIN-2017-30/segments/1500549426133.16/warc/CC-MAIN-20170726102230-20170726122230-00452.warc.gz | 201,301,233 | 126,650 | From aniketsuniljoshi at gmail.com Mon Jan 4 07:37:01 2016 From: aniketsuniljoshi at gmail.com (Aniket Joshi) Date: Mon, 4 Jan 2016 13:07:01 +0530 Subject: [GAP Forum] Using M12 semidirect product Z_2 in GAP Message-ID: Hi, I wish to find the homology H_3(M_{12} : Z_2, Z), where M_{12} is a Mathieu group of order 95040, and M_12:Z_2 indicates the semi-direct product with Z_2. I plan to use the GroupHomology(,) command from the HAP package for GAP. How do I use the semi direct product M_12:Z_2 in GAP? Is there an inbuilt function, or would I need to compute it separately? Regards, Aniket -- Aniket S Joshi 4th year B.S.-M.S. Physics Dual Degree Roll no. PH11B001 Department of Physics, IIT Madras From dmitrii.pasechnik at cs.ox.ac.uk Mon Jan 4 15:18:45 2016 From: dmitrii.pasechnik at cs.ox.ac.uk (Dmitrii Pasechnik) Date: Mon, 4 Jan 2016 15:18:45 +0000 Subject: [GAP Forum] Using M12 semidirect product Z_2 in GAP In-Reply-To: References: Message-ID: <20160104151845.GA28704@cs.ox.ac.uk> On Mon, Jan 04, 2016 at 01:07:01PM +0530, Aniket Joshi wrote: > I wish to find the homology H_3(M_{12} : Z_2, Z), where M_{12} is a Mathieu > group of order 95040, and M_12:Z_2 indicates the semi-direct product with > Z_2. I plan to use the GroupHomology(,) command from the HAP package for > GAP. > > How do I use the semi direct product M_12:Z_2 in GAP? Is there an inbuilt > function, or would I need to compute it separately? just take assuming you have AltasRep package installed, you can just do AtlasGroup("M12:2") to get a permutation representation of M:12:2 on 24 points. HTH, Dmitrii From graham.ellis at nuigalway.ie Mon Jan 4 15:24:29 2016 From: graham.ellis at nuigalway.ie (Ellis, Grahamj) Date: Mon, 4 Jan 2016 15:24:29 +0000 Subject: [GAP Forum] Using M12 semidirect product Z_2 in GAP In-Reply-To: <20160104151845.GA28704@cs.ox.ac.uk> References: , <20160104151845.GA28704@cs.ox.ac.uk> Message-ID: And here is the computation showing that H_3(M12:2, Z) = Z_2 x z_2 x Z_12. gap> LoadPackage("atlasrep");; gap> G:=AtlasGroup("M12:2");; gap> GroupHomology(G,3); [ 2, 2, 4, 3 ] Graham School of Mathematics, Statistics & Applied Mathematics National University of Ireland, Galway University Road, Galway Ireland http://hamilton.nuigalway.ie tel: 091 493011 ________________________________________ From: forum-bounces at gap-system.org [forum-bounces at gap-system.org] on behalf of Dmitrii Pasechnik [dmitrii.pasechnik at cs.ox.ac.uk] Sent: Monday, January 4, 2016 3:18 PM To: Aniket Joshi Cc: forum at gap-system.org Subject: Re: [GAP Forum] Using M12 semidirect product Z_2 in GAP On Mon, Jan 04, 2016 at 01:07:01PM +0530, Aniket Joshi wrote: > I wish to find the homology H_3(M_{12} : Z_2, Z), where M_{12} is a Mathieu > group of order 95040, and M_12:Z_2 indicates the semi-direct product with > Z_2. I plan to use the GroupHomology(,) command from the HAP package for > GAP. > > How do I use the semi direct product M_12:Z_2 in GAP? Is there an inbuilt > function, or would I need to compute it separately? just take assuming you have AltasRep package installed, you can just do AtlasGroup("M12:2") to get a permutation representation of M:12:2 on 24 points. HTH, Dmitrii _______________________________________________ Forum mailing list Forum at mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum From sara.yaftian at gmail.com Tue Jan 5 10:50:00 2016 From: sara.yaftian at gmail.com (Sara Yaftian) Date: Tue, 5 Jan 2016 14:20:00 +0330 Subject: [GAP Forum] A question about ParGAP Message-ID: Dear Forum I have a question about ParGAP. I installed parGAP by the following command cd pargap ./configure make and then I need mpi and I installed mpi. Finally I paralleled a GAP program on Linux (Ubuntu 14.04 LTS) by the following command: mpiexe -np 5 ./pargap.sh when GAP run then I typed Read("myprogram.txt"); I am not sure my commands are correct or not because my laptop use 5 of CPU but the speed of my program has not changed. Would you please help me? I wish you have a very happy New Year 2016. Best regards Sara From axyd0000 at gmail.com Tue Jan 5 13:05:38 2016 From: axyd0000 at gmail.com (Gaurav Dhingra) Date: Tue, 5 Jan 2016 18:35:38 +0530 Subject: [GAP Forum] Meaning of \= operation in GAP Message-ID: Hi, GAP members I want to know the exact purpose for which \= i.e equality operation is used in GAP. Does \= checks for the mathematical equality or does it check for structural equality (of given objects) or some other type of equality of objects in GAP that i am not aware of? It will be very helpful if you provide the examples indicating the behavior (mathematical or structural), particularly either with FreeGroup or Permutation Groups. Thanks Gaurav Dhingra From caj21 at st-andrews.ac.uk Tue Jan 5 13:20:50 2016 From: caj21 at st-andrews.ac.uk (Christopher Jefferson) Date: Tue, 5 Jan 2016 13:20:50 +0000 Subject: [GAP Forum] Meaning of \= operation in GAP In-Reply-To: References: Message-ID: <695FAFF4-E57F-4B3D-AEF4-BC4B43000AEA@st-andrews.ac.uk> The equality operation should always compare for mathematical equality. I agree the documentation could be slightly improved in this area. For example, the following two permutation groups are equal, as they are both the complete permutation group on 4 points. gap> Group([(1,2),(2,3,4)]) = Group([(1,4),(2,3,4)]); true On 05/01/2016, 13:05, "forum-bounces at gap-system.org on behalf of Gaurav Dhingra" wrote: >Hi, GAP members > >I want to know the exact purpose for which \= i.e equality operation is >used in GAP. Does \= checks for the mathematical equality or does it >check for structural equality (of given objects) or some other type of >equality of objects in GAP that i am not aware of? >It will be very helpful if you provide the examples indicating the behavior >(mathematical or structural), particularly either with FreeGroup or >Permutation Groups. > >Thanks >Gaurav Dhingra >_______________________________________________ >Forum mailing list >Forum at mail.gap-system.org >http://mail.gap-system.org/mailman/listinfo/forum From oyvind.solberg at math.ntnu.no Tue Jan 5 15:39:43 2016 From: oyvind.solberg at math.ntnu.no (=?UTF-8?Q?=c3=98yvind_Solberg?=) Date: Tue, 5 Jan 2016 16:39:43 +0100 Subject: [GAP Forum] question concerning the GAP package qpa (how to get an inclusion mprphism?) In-Reply-To: References: Message-ID: <568BE3BF.1030902@math.ntnu.no> Dear GAP Forum and Bernhard, Suppose we are given a finite dimensional quotient A = kQ/I of a path algebra and two modules M and N over A, where we know that N is a submodule of M. If M and N are given as two representations/ modules over A and you have not told QPA how N is a submodule of M, then there is in general no way QPA can find the inclusion you are thinking about. In general a module N can be a submodule of a given module M in infinitely many ways. However, if you know by which elements N is generated by inside M, say m_1, m_2,..., m_t, then the command gap> g := SubRepresentationInclusion(M, [m1,m2,...,mt]); will produce an inclusion from a module N' isomorphic to N into M, but where the images inside M are the same. Then to get an inclusion from N you could find an isomorphism between N and N' by gap> alpha := IsomorphismOfModules( N, N' ); and then find the composition alpha*g. If you are just abstractly knowing that N is a submodule of M, then I don't know an algorithm to find an inclusion of N into M. Best regards, Oeyvind Solberg. From angelblascomunoz at gmail.com Tue Jan 5 20:36:46 2016 From: angelblascomunoz at gmail.com (Angel Blasco) Date: Tue, 5 Jan 2016 21:36:46 +0100 Subject: [GAP Forum] =?utf-8?q?Save_=E2=80=9Ctime=E2=80=9D_in_a_variable_i?= =?utf-8?q?n_GAP?= Message-ID: Hi, I'm trying to write a program in GAP to obtain the time that GAP needs to construct the groups of an order given. I'm using the command: ConstructAllGroups(order);;time; Is there anyway to save the value of "time" in a variable? I did it with: a:=ConstructAllGroups(10);;time; but only the groups were saved. Regards -- *Angel Blasco.* From hulpke at math.colostate.edu Tue Jan 5 20:43:21 2016 From: hulpke at math.colostate.edu (Alexander Hulpke) Date: Tue, 5 Jan 2016 13:43:21 -0700 Subject: [GAP Forum] =?utf-8?q?Save_=E2=80=9Ctime=E2=80=9D_in_a_variable_i?= =?utf-8?q?n_GAP?= In-Reply-To: References: Message-ID: <218F3214-39A9-44CB-93FB-B2A3158D8CFB@math.colostate.edu> > I'm trying to write a program in GAP to obtain the time that GAP needs to > construct the groups of an order given. I'm using the command: > > ConstructAllGroups(order);;time; > > Is there anyway to save the value of "time" in a variable? Yes, time is just like a variable (this is different to the time command in unix and other systems): a:=ConstructAllGroups(10);;savedtime:=time; Within a program I would rather refer to the value of Runtime(), that is: start:=Runtime(); a:=ConstructAllPiffles(-33); total:=Runtime()-start; Best, Alexander Hulpke -- Colorado State University, Department of Mathematics, Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA email: hulpke at math.colostate.edu, Phone: ++1-970-4914288 http://www.math.colostate.edu/~hulpke From sl4 at st-andrews.ac.uk Tue Jan 5 20:41:56 2016 From: sl4 at st-andrews.ac.uk (Stephen Linton) Date: Tue, 5 Jan 2016 20:41:56 +0000 Subject: [GAP Forum] =?windows-1252?q?Save_=93time=94_in_a_variable_in_GAP?= In-Reply-To: References: Message-ID: <232E9C66-442E-4F95-99A1-3242509FA4A0@st-andrews.ac.uk> ConstructAllGroups(order);; a := time; ought to work, but you might want to look into the Runtime() function to get an interface designed more for programming. Steve > On 5 Jan 2016, at 20:36, Angel Blasco wrote: > > Hi, > > I'm trying to write a program in GAP to obtain the time that GAP needs to > construct the groups of an order given. I'm using the command: > > ConstructAllGroups(order);;time; > > Is there anyway to save the value of "time" in a variable? > I did it with: > > a:=ConstructAllGroups(10);;time; > > but only the groups were saved. > > > Regards > -- > *Angel Blasco.* > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From angelblascomunoz at gmail.com Tue Jan 5 20:52:06 2016 From: angelblascomunoz at gmail.com (Angel Blasco) Date: Tue, 5 Jan 2016 21:52:06 +0100 Subject: [GAP Forum] =?utf-8?q?Save_=E2=80=9Ctime=E2=80=9D_in_a_variable_i?= =?utf-8?q?n_GAP?= In-Reply-To: <218F3214-39A9-44CB-93FB-B2A3158D8CFB@math.colostate.edu> References: <218F3214-39A9-44CB-93FB-B2A3158D8CFB@math.colostate.edu> Message-ID: thanks 2016-01-05 21:43 GMT+01:00 Alexander Hulpke : > > I'm trying to write a program in GAP to obtain the time that GAP needs to > > construct the groups of an order given. I'm using the command: > > > > ConstructAllGroups(order);;time; > > > > Is there anyway to save the value of "time" in a variable? > > Yes, time is just like a variable (this is different to the time command > in unix and other systems): > > a:=ConstructAllGroups(10);;savedtime:=time; > > Within a program I would rather refer to the value of Runtime(), that is: > > start:=Runtime(); > a:=ConstructAllPiffles(-33); > total:=Runtime()-start; > > Best, > > Alexander Hulpke > > -- Colorado State University, Department of Mathematics, > Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA > email: hulpke at math.colostate.edu, Phone: ++1-970-4914288 > http://www.math.colostate.edu/~hulpke > > > -- *Angel Blasco.* From alexk at mcs.st-and.ac.uk Tue Jan 5 21:30:14 2016 From: alexk at mcs.st-and.ac.uk (Alexander Konovalov) Date: Tue, 5 Jan 2016 21:30:14 +0000 Subject: [GAP Forum] A question about ParGAP In-Reply-To: References: Message-ID: <7CEC0812-336F-4A43-9F3D-8016F5F627A9@mcs.st-andrews.ac.uk> Dear Sara, > On 5 Jan 2016, at 10:50, Sara Yaftian wrote: > > Dear Forum > > I have a question about ParGAP. > I installed parGAP by the following command > > cd pargap > ./configure > make > > and then I need mpi and I installed mpi. > Finally I paralleled a GAP program on Linux (Ubuntu 14.04 LTS) by the > following command: > > mpiexe -np 5 ./pargap.sh Thank you for your question, and Happy New Year to you too! The above steps seem to be correct. The rest depends on what "myprogram.txt" is doing - it is quite common that when one tries to parallelise the code then the first version of the parallel code runs slower than the sequential one. Also, the phrase "I paralleled a GAP program" sounds a bit ambiguous, so just in case: what mpiexe -np 5 ./pargap.sh command is doing is that it starts several copies of GAP. One of them is called 'master', and it provides the GAP command line interface and communicates with all the others, called 'slaves'. If you will call Read("myprogram.txt"); but the code there does not use any master-slave communication, it will just perform the calculation on the master. Best wishes Alexander > > when GAP run then I typed Read("myprogram.txt"); > > I am not sure my commands are correct or not because my laptop use 5 of CPU > but the speed of my program has not changed. Would you please help me? > > I wish you have a very happy New Year 2016. > > Best regards > Sara From axyd0000 at gmail.com Wed Jan 6 05:29:40 2016 From: axyd0000 at gmail.com (Gaurav Dhingra) Date: Wed, 6 Jan 2016 10:59:40 +0530 Subject: [GAP Forum] Find the source for in GAP Message-ID: Hi GAP members I asked previously http://math.stackexchange.com/questions/1570893/which-algorithm-gap-uses-to-check-equality-of-two-groups for finding the source of an "operation" in GAP. Also there is already an answer for finding the source of a "function" in GAP. My question is how to find the source of an "Attribute" in GAP? For ex. i want to find the Algorithm for finding the stabilizer of a SymmetricGroup gap> s4:=SymmetricGroup(4); Sym( [ 1 .. 4 ] ) gap> Stabilizer; function( arg ) ... end gap> StabilizerOfExternalSet; # this is used for SymmetricGroup in particular gap> func:=ApplicableMethod(StabilizerOfExternalSet, [s4]); fail gap> It fails. Is there another method instead of using ApplicableMethod? Above i used StabilizerOfExternalSet after seeing the source code for Stabilizer. Gaurav Dhingra From sl4 at st-andrews.ac.uk Wed Jan 6 09:45:00 2016 From: sl4 at st-andrews.ac.uk (Stephen Linton) Date: Wed, 6 Jan 2016 09:45:00 +0000 Subject: [GAP Forum] Find the source for in GAP In-Reply-To: References: Message-ID: Dear Gaurav, Attributes are just like Operations, and you can use ApplicableMethod in the same way. A good example is Size: gap> g := Group((1,2,3,4,5),(1,5)(2,4)); Group([ (1,2,3,4,5), (1,5)(2,4) ]) gap> ApplicableMethod(Size,[g]); function( G ) ... end gap> Print(last); function ( G ) return SizeStabChain( StabChainMutable( G ) ); end However, there is one complication, which is that once an Attribute has been computed the value is often stored in which case the method changes to one that just recovers the stored value: gap> Size(g); 10 gap> ApplicableMethod(Size,[g]); function( object ) ... end gap> Print(last); function ( object ) <> end gap> Stabilizer, however is not itself an attribute and is rather more complicated. See: http://www.gap-system.org/Manuals/doc/ref/chap41.html#X797BD60E7ACEF1B1 and also section 41.12 a bit further down. It doesn?t make sense to ask simply about the Stabilizer of a group ? you also have to specify an object to be stabilised and an action of the group and the call goes through a number of functions that fill in missing arguments and detect special cases before eventually reaching a call of StabilizerOp or StabilizerOfExternalSet where the work is really done. Steve > On 6 Jan 2016, at 05:29, Gaurav Dhingra wrote: > > Hi GAP members > > I asked previously > http://math.stackexchange.com/questions/1570893/which-algorithm-gap-uses-to-check-equality-of-two-groups > for finding the source of an "operation" in GAP. Also there is already an > answer for finding the source of a "function" in GAP. > > My question is how to find the source of an "Attribute" in GAP? > For ex. i want to find the Algorithm for finding the stabilizer of a > SymmetricGroup > > gap> s4:=SymmetricGroup(4); > Sym( [ 1 .. 4 ] ) > gap> Stabilizer; > function( arg ) ... end > gap> StabilizerOfExternalSet; # this is used for SymmetricGroup in > particular > > gap> func:=ApplicableMethod(StabilizerOfExternalSet, [s4]); > fail > gap> > > It fails. Is there another method instead of using ApplicableMethod? > Above i used StabilizerOfExternalSet after seeing the source code for > Stabilizer. > > Gaurav Dhingra > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From sashar.sashar1360 at gmail.com Mon Jan 11 08:02:50 2016 From: sashar.sashar1360 at gmail.com (sashar sashar) Date: Mon, 11 Jan 2016 00:02:50 -0800 Subject: [GAP Forum] Does Irr(g) return the principle character as its first entry? Message-ID: Dear Forum, I am working with SmallGroups Library.I have used the function Irr(SmallGroup(n,i)) for some ns and is. I have noticed that the pincipal character 1_G is the first entry for these groups. Is this always true for all the groups in SmallGroups Library? Regards Sashar From angelblascomunoz at gmail.com Wed Jan 13 19:25:26 2016 From: angelblascomunoz at gmail.com (Angel Blasco) Date: Wed, 13 Jan 2016 20:25:26 +0100 Subject: [GAP Forum] How can I fix this problem using Cubefree package Message-ID: I'm trying to calculate the number of groups of order 815409. I'm using Cubefree package but after 2 minutes processing my pc returns me this: gap> NumberCFGroups(815409); Error, Record: '.orbit' must have an assigned value in pnt := S.orbit[1]; called from StabChainStrong( S, GeneratorsOfGroup( G ), options ); called from StabChainOp( G, rec( base := [ 1 .. LargestMovedPoint( G ) ] ) ) called from MinimalStabChain( G ) called from GeneratorsSmallest( b ) called from Set( C1 ) called from ... at line 2 of *stdin* you can 'return;' after assigning a value brk> How can I fix this problem? Regards -- *Angel Blasco.* From heiko.dietrich at monash.edu Wed Jan 13 19:51:15 2016 From: heiko.dietrich at monash.edu (Heiko Dietrich) Date: Thu, 14 Jan 2016 08:51:15 +1300 Subject: [GAP Forum] How can I fix this problem using Cubefree package Message-ID: Dear Angel, This is a known bug which is fixed in the latest version; please download version 1.15 from here and extract it in your /pkg directory of your gap installation: http://users.monash.edu/~heikod/cubefree.html (This version should actually be listed on the gap website http://www.gap-system.org/Packages/cubefree.html but it isn't -- Alex?!?) Anyway, the result of this computation is gap> n:=NumberCFGroups(815409); 415 Best, Heiko ----------------------------------------- On Thu, 14/1/16, Angel Blasco wrote: I'm trying to calculate the number of groups of order 815409. I'm using Cubefree package but after 2 minutes processing my pc returns me this: gap> NumberCFGroups(815409); Error, Record: '.orbit' must have an assigned value in pnt := S.orbit[1]; called from StabChainStrong( S, GeneratorsOfGroup( G ), options ); called from StabChainOp( G, rec( base := [ 1 .. LargestMovedPoint( G ) ] ) ) called from MinimalStabChain( G ) called from GeneratorsSmallest( b ) called from Set( C1 ) called from ... at line 2 of *stdin* you can 'return;' after assigning a value brk> How can I fix this problem? Regards -- *Angel Blasco.* _______________________________________________ Forum mailing list Forum at mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum -- *Dr HEIKO DIETRICH * *School of Mathematical Sciences* Monash University Clayton Campus 9 Rainforest Walk Monash University VIC 3800 Australia T: +61 3 9905 4771 E: heiko.dietrich at monash.edu users.monash.edu/~heikod From angelblascomunoz at gmail.com Wed Jan 13 19:56:46 2016 From: angelblascomunoz at gmail.com (Angel Blasco) Date: Wed, 13 Jan 2016 19:56:46 +0000 (UTC) Subject: [GAP Forum] How can I fix this problem using Cubefree package In-Reply-To: References: Message-ID: <6A8E708086FF1F89.F6880071-CB63-4F3B-8B93-60E183B9BB84@mail.outlook.com> Ok, i've got GAP 4.7.9 installed and the cubefree package inside is 1.13 Now i'm going to download and install version 1.15 in my pkg folder. Thanks Enviado desde Outlook Mobile On Wed, Jan 13, 2016 at 11:51 AM -0800, "Heiko Dietrich" wrote: Dear Angel, This is a known bug which is fixed in the latest version; please download version 1.15 from here and extract it in your /pkg directory of your gap installation: ?? http://users.monash.edu/~heikod/cubefree.html (This version should actually be listed on the gap website http://www.gap-system.org/Packages/cubefree.html but it isn't -- Alex?!?) Anyway, the result of this computation is gap> n:=NumberCFGroups(815409); 415 Best, Heiko ----------------------------------------- On Thu, 14/1/16, Angel Blasco wrote: I'm trying to calculate the number of groups of order 815409. I'm using Cubefree package but after 2 minutes processing my pc returns me this: gap> NumberCFGroups(815409); Error, Record: '.orbit' must have an assigned value in ? pnt := S.orbit[1]; called from StabChainStrong( S, GeneratorsOfGroup( G ), options ); called from StabChainOp( G, rec( ? ? base := [ 1 .. LargestMovedPoint( G ) ] ) ) called from MinimalStabChain( G ) called from GeneratorsSmallest( b ) called from Set( C1 ) called from ...? at line 2 of *stdin* you can 'return;' after assigning a value brk> How can I fix this problem? Regards -- *Angel Blasco.* _______________________________________________ Forum mailing list Forum at mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum -- Dr HEIKO DIETRICH ?? ? ? ? ? ? ? ? School of Mathematical Sciences Monash UniversityClayton Campus9 Rainforest Walk Monash University VIC 3800Australia T: +61 3 9905 4771? ? ? ? ? ? ? ? ? ? ??E: heiko.dietrich at monash.eduusers.monash.edu/~heikod From alexk at mcs.st-and.ac.uk Wed Jan 13 23:04:57 2016 From: alexk at mcs.st-and.ac.uk (Alexander Konovalov) Date: Wed, 13 Jan 2016 23:04:57 +0000 Subject: [GAP Forum] How can I fix this problem using Cubefree package In-Reply-To: References: Message-ID: Dear all, > On 13 Jan 2016, at 19:51, Heiko Dietrich wrote: > > Dear Angel, > > This is a known bug which is fixed in the latest version; please download version 1.15 from here and extract it in your /pkg directory of your gap installation: > > http://users.monash.edu/~heikod/cubefree.html Yes - see also my former comments under http://math.stackexchange.com/q/1607419/ > > (This version should actually be listed on the gap website http://www.gap-system.org/Packages/cubefree.html but it isn't -- Alex?!?) The packages pages on the GAP website are not dynamic - they are updated only after the GAP public release, and reflect the set of packages included in that release. Thus at the moment they match GAP 4.7.9. When I am telling package authors that the new version has been picked up, it's only the initial step. It is followed by a nightly and weekly testing cycle. It could happen that the package is incompatible with the system or other packages, and then will be put on hold. You may see the current status report on package updates at this Wiki page: https://github.com/gap-system/gap/wiki/Package-updates-status and also see how packages progress between testing stage: https://trello.com/b/CTsB9FTl/package-updates-name-version-pickup-date At the moment, Cubefree 1.15 passes tests of the release candidate for the public release of GAP 4.8, so it should be safe to use it. Best wishes Alexander > Anyway, the result of this computation is > > gap> n:=NumberCFGroups(815409); > 415 > > Best, > Heiko > > > > > ----------------------------------------- > On Thu, 14/1/16, Angel Blasco wrote: > I'm trying to calculate the number of groups of order 815409. I'm using > Cubefree package but after 2 minutes processing my pc returns me this: > > gap> NumberCFGroups(815409); > Error, Record: '.orbit' must have an assigned value in > pnt := S.orbit[1]; called from > StabChainStrong( S, GeneratorsOfGroup( G ), options ); called from > StabChainOp( G, rec( > base := [ 1 .. LargestMovedPoint( G ) ] ) ) called from > MinimalStabChain( G ) called from > GeneratorsSmallest( b ) called from > Set( C1 ) called from > ... at line 2 of *stdin* > you can 'return;' after assigning a value > brk> > > > How can I fix this problem? > > Regards > > -- > *Angel Blasco.* > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum > > -- > Dr HEIKO DIETRICH > > School of Mathematical Sciences > Monash University > Clayton Campus > 9 Rainforest Walk > Monash University VIC 3800 > Australia > > T: +61 3 9905 4771 > E: heiko.dietrich at monash.edu > users.monash.edu/~heikod -- Dr. Alexander Konovalov FSSI, Senior Research Fellow Centre for Interdisciplinary Research in Computational Algebra (CIRCA) School of Computer Science, University of St Andrews http://blogs.cs.st-andrews.ac.uk/alexk/ -- The University of St Andrews is a charity registered in Scotland:No.SC013532 From sara.yaftian at gmail.com Thu Jan 14 13:44:16 2016 From: sara.yaftian at gmail.com (Sara Yaftian) Date: Thu, 14 Jan 2016 17:14:16 +0330 Subject: [GAP Forum] A question about ParGAP In-Reply-To: <7CEC0812-336F-4A43-9F3D-8016F5F627A9@mcs.st-andrews.ac.uk> References: <7CEC0812-336F-4A43-9F3D-8016F5F627A9@mcs.st-andrews.ac.uk> Message-ID: Dear Alexander Thank you very much for your helpful message. I have a question about your last message. How to use master-salve communication? On the other hand, I need to use 5 CPU's, but MPI in ParGAP used 2 CPU's. Is there any command to solve this problem? Best regards Sara On Wed, Jan 6, 2016 at 1:00 AM, Alexander Konovalov wrote: > Dear Sara, > > > On 5 Jan 2016, at 10:50, Sara Yaftian wrote: > > > > Dear Forum > > > > I have a question about ParGAP. > > I installed parGAP by the following command > > > > cd pargap > > ./configure > > make > > > > and then I need mpi and I installed mpi. > > Finally I paralleled a GAP program on Linux (Ubuntu 14.04 LTS) by the > > following command: > > > > mpiexe -np 5 ./pargap.sh > > Thank you for your question, and Happy New Year to you too! The above > steps seem to be correct. The rest depends on what "myprogram.txt" is > doing - it is quite common that when one tries to parallelise the code > then the first version of the parallel code runs slower than the > sequential one. > > Also, the phrase "I paralleled a GAP program" sounds a bit ambiguous, so > just in case: what > > mpiexe -np 5 ./pargap.sh > > command is doing is that it starts several copies of GAP. One of them is > called 'master', and it provides the GAP command line interface and > communicates with all the others, called 'slaves'. If you will call > > Read("myprogram.txt"); > > but the code there does not use any master-slave communication, it will > just perform the calculation on the master. > > Best wishes > Alexander > > > > > > > > > > > > when GAP run then I typed Read("myprogram.txt"); > > > > I am not sure my commands are correct or not because my laptop use 5 of > CPU > > but the speed of my program has not changed. Would you please help me? > > > > I wish you have a very happy New Year 2016. > > > > Best regards > > Sara > > > From tim at algebra.uni-linz.ac.at Fri Jan 15 15:12:47 2016 From: tim at algebra.uni-linz.ac.at (Tim Boykett) Date: Fri, 15 Jan 2016 17:12:47 +0200 Subject: [GAP Forum] Shortest words Message-ID: <56990C6F.7020504@algebra.uni-linz.ac.at> Dear GAPpers, I have previously used the techniques outlined in the Rubik's Cube tutorial to find words expressing certain permutations in the generators I am using. This has worked fine so far, but I have run up against a problem where I would like to prefer some permutations and use as few as possible of others. It seems that the very old collection AbStab.g used to do this, but it is no longer part of GAP. What is the current way to do this? I am sure there is a very simple technique, but I cannot seem to find it. Best wishes, Tim From e.obrien at auckland.ac.nz Sun Jan 17 08:40:03 2016 From: e.obrien at auckland.ac.nz (Eamonn O'Brien) Date: Sun, 17 Jan 2016 21:40:03 +1300 Subject: [GAP Forum] Two jobs available at the University of Auckland Message-ID: <569B5363.7090109@auckland.ac.nz> Dear Colleagues: The Department of Mathematics at the University of Auckland invites applications for two permanent full-time positions at Lecturer / Senior Lecturer level. We expect that at least one of these will be filled at Lecturer level. We expect that the research expertise of the successful applicants will complement existing research strengths in Pure Mathematics. These include: graph theory and combinatorics; group theory; number theory; and representation theory. More details about the Department can be found at www.math.auckland.ac.nz The job advertisement, including details of the application process, is available at: https://www.math.auckland.ac.nz/en/about/news-and-events/notices/notices-2015/2015/12/job-opportunities--lecturership----senior-lecturership.html Closing date: 14 February, 2016 I am happy to provide additional information, and appreciate if you can draw the attention of your colleagues to these positions. Best wishes. Eamonn O'Brien From fima.math at gmail.com Tue Jan 19 09:50:39 2016 From: fima.math at gmail.com (Fima Bahari) Date: Tue, 19 Jan 2016 13:20:39 +0330 Subject: [GAP Forum] Hamiltoniancycle Message-ID: Dear all, By GAP, Is it possible to fine a Hamiltonian cycle in a defined graph, if it exists? Many thanks in advance From sam at Math.RWTH-Aachen.De Wed Jan 20 09:07:50 2016 From: sam at Math.RWTH-Aachen.De (Thomas Breuer) Date: Wed, 20 Jan 2016 10:07:50 +0100 Subject: [GAP Forum] Does Irr(g) return the principle character as its first entry? In-Reply-To: References: Message-ID: <20160120090747.GA1440@gemma.math.rwth-aachen.de> Dear Sashar, you asked the GAP Forum > I am working with SmallGroups Library.I have used the function > Irr(SmallGroup(n,i)) for some ns and is. I have noticed that the pincipal > character 1_G is the first entry for these groups. Is this always true for > all the groups in SmallGroups Library? No. The method that is chosen by GAP for computing the irreducible characters of a group depends on properties of this group and not on the library from which it is constructed. For example, there is a method for computing the irreducible characters of a symmetric group in its natural representation; this method uses the fact that the conjugacy classes of the group are parametrized by the cycle structures of the group elements. This method is applicable only if the group has this property and if this property is stored in the group before the call to the function Irr', and currently this method can be expected to be used in this situation. Here is a sample GAP session. gap> g:= SmallGroup( 120, 34 ); # the symmetric group on five points Group([ (1,2,3,4,5), (1,2) ]) gap> Position( Irr( g ), TrivialCharacter( g ) ); 1 gap> g:= SmallGroup( 120, 34 ); # fetch the group again Group([ (1,2,3,4,5), (1,2) ]) gap> IsNaturalSymmetricGroup( g ); # now the group stores the property true gap> Position( Irr( g ), TrivialCharacter( g ) ); 7 Note that the GAP manual states explicitly that the trivial character need not be the first entry in the list of irreducible characters. All the best, Thomas From angelblascomunoz at gmail.com Fri Jan 29 19:34:36 2016 From: angelblascomunoz at gmail.com (Angel Blasco) Date: Fri, 29 Jan 2016 20:34:36 +0100 Subject: [GAP Forum] Hamiltonian groups Message-ID: <56ABBECC.6050605@gmail.com> Is there any command in GAP to generate a Hamiltonian group of an order given? Thanks From fima.math at gmail.com Sat Jan 30 07:05:33 2016 From: fima.math at gmail.com (Fima Bahari) Date: Sat, 30 Jan 2016 10:35:33 +0330 Subject: [GAP Forum] Fwd: Hamiltoniancycle In-Reply-To: References: Message-ID: ---------- Forwarded message ---------- From: Fima Bahari Date: Tue, Jan 19, 2016 at 1:20 PM Subject: Hamiltoniancycle To: forum at mail.gap-system.org Dear all, By GAP, Is it possible to fine a Hamiltonian cycle in a defined graph, if it exists? Many thanks in advance From l.h.soicher at qmul.ac.uk Sat Jan 30 09:24:20 2016 From: l.h.soicher at qmul.ac.uk (Leonard Soicher) Date: Sat, 30 Jan 2016 09:24:20 +0000 Subject: [GAP Forum] Fwd: Hamiltoniancycle In-Reply-To: References: , , Message-ID: Dear Fima, Dear Forum, There is no GRAPE function to find a Hamiltonian cycle (when such a cycle exists) in a given graph. However, the free open-source SageMath system has a function for this. See http://doc.sagemath.org/pdf/en/reference/graphs/graphs.pdf Regards, Leonard ________________________________________ From: forum-bounces at gap-system.org on behalf of Fima Bahari Sent: 30 January 2016 07:05 To: forum at gap-system.org Subject: [GAP Forum] Fwd: Hamiltoniancycle ---------- Forwarded message ---------- From: Fima Bahari Date: Tue, Jan 19, 2016 at 1:20 PM Subject: Hamiltoniancycle To: forum at mail.gap-system.org Dear all, By GAP, Is it possible to fine a Hamiltonian cycle in a defined graph, if it exists? Many thanks in advance _______________________________________________ Forum mailing list Forum at mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum From axyd0000 at gmail.com Thu Feb 4 18:30:50 2016 From: axyd0000 at gmail.com (Gaurav Dhingra) Date: Fri, 5 Feb 2016 00:00:50 +0530 Subject: [GAP Forum] Inequality of Generators of Finitely Presented and corresponding Free Group Message-ID: <56B398DA.8060601@gmail.com> Hi, GAP forum members I am reading this particular documentation http://www.gap-system.org/Manuals/doc/ref/chap47.html , from this page i quote this sentence > That means that words in the generators of the free group are not elements of the finitely presented group. Vice versa elements of the FpGroup are not words. There is also an example showing this behavior. Now, i don't why this has been done. I mean is there is any particular reason for why GAP authors did it this way? (for defining equality between the generators). Feel free explain in detail even if it is completely some development methodology (or non-mathematical) which has been followed for this. Thanks Gaurav Dhingra From kroberts at math.uni-bielefeld.de Fri Feb 5 08:54:41 2016 From: kroberts at math.uni-bielefeld.de (Kieran Roberts) Date: Fri, 5 Feb 2016 09:54:41 +0100 Subject: [GAP Forum] Default Arguments Message-ID: Dear GAP forum: Is it possible to include default arguments in a function? Such as something like this (which doesn't work): add := function(a,b:=1) return a+b; end; So that add(2,2) = 4 and add(2) = 3. The only way I've managed to do is by writing: add := function(arg) if Size(arg) = 2 then return arg[1] + arg[2]; else return arg[1] + 1; fi; end; But if you have 2 or 3 (optional) arguments with default values this becomes cumbersome. Is there a more efficient way? Best, Kieran. From alexander.konovalov at st-andrews.ac.uk Tue Feb 9 21:57:58 2016 From: alexander.konovalov at st-andrews.ac.uk (Alexander Konovalov) Date: Tue, 9 Feb 2016 21:57:58 +0000 Subject: [GAP Forum] Default Arguments In-Reply-To: References: Message-ID: <34FDD10F-A182-4C5A-B539-1915868FBB83@st-andrews.ac.uk> Dear Kieran, > On 5 Feb 2016, at 08:54, Kieran Roberts wrote: > > Dear GAP forum: > > Is it possible to include default arguments in a function? > > Such as something like this (which doesn't work): > > add := function(a,b:=1) > return a+b; > end; > > So that add(2,2) = 4 and add(2) = 3. The only way I've managed to do > is by writing: > > add := function(arg) > if Size(arg) = 2 then > return arg[1] + arg[2]; > else > return arg[1] + 1; > fi; > end; > > But if you have 2 or 3 (optional) arguments with default values this > becomes cumbersome. Is there a more efficient way? This will a bit easier in coming soon GAP 4.8: as written at https://github.com/gap-system/gap/wiki/Changes-between-GAP-4.7-and-GAP-4.8 it will have "support for partially variadic functions to allow function expressions like function(a,b,c,x...) ... end; which would require at least three arguments and assign the first three to a, b and c and then a list containing any remaining ones to x. The former special meaning of the argument arg is still supported and is now equivalent to function(arg...), so no changes in the existing code are required." The example from above could then look like gap> add := function(a,b...) > if Length(b)=0 then > return a+1; > else > return a+b[1]; > fi; > end; function( a, b... ) ... end gap> add(2,2); 4 gap> add(2); 3 If you have more optional arguments, they would all go into the list b, but since you would be no longer obliged to use the special name 'arg', your code may be more readable because of using proper identifiers for mandatory arguments. Best wishes Alexander From rahul.kitture at gmail.com Wed Feb 10 04:34:44 2016 From: rahul.kitture at gmail.com (Rahul Kitture) Date: Wed, 10 Feb 2016 10:04:44 +0530 Subject: [GAP Forum] A question on factorization of an integer Message-ID: Dear Sir, In some computations of orders of groups, which turn out to be large, I wanted to see the prime factorization of the order. For example, I use following kind of commands: gap> g:=SymmetricGroup(10);; Size(g); 3628800 gap> Factors(3628800); [ 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 5, 5, 7 ] My question is, is there some command in GAP to get this factorization into form 2^8 * 3^4 * 5^2 * 7? Thank you. - Rahul Kitture HRI, Allahabad From rahul.kitture at gmail.com Wed Feb 10 04:34:44 2016 From: rahul.kitture at gmail.com (Rahul Kitture) Date: Wed, 10 Feb 2016 10:04:44 +0530 Subject: [GAP Forum] A question on factorization of an integer Message-ID: Dear Sir, In some computations of orders of groups, which turn out to be large, I wanted to see the prime factorization of the order. For example, I use following kind of commands: gap> g:=SymmetricGroup(10);; Size(g); 3628800 gap> Factors(3628800); [ 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 5, 5, 7 ] My question is, is there some command in GAP to get this factorization into form 2^8 * 3^4 * 5^2 * 7? Thank you. - Rahul Kitture HRI, Allahabad From maasiru at yahoo.com Wed Feb 10 06:06:43 2016 From: maasiru at yahoo.com ([Muniru Asiru] maasiru@yahoo.com) Date: Tue, 9 Feb 2016 22:06:43 -0800 Subject: [GAP Forum] A question on factorization of an integer In-Reply-To: Message-ID: <1455084403.23960.YahooMailMobile@web121505.mail.ne1.yahoo.com> Yes. Partially. Use Collected From maasiru at yahoo.com Wed Feb 10 06:06:43 2016 From: maasiru at yahoo.com ([Muniru Asiru] maasiru@yahoo.com) Date: Tue, 9 Feb 2016 22:06:43 -0800 Subject: [GAP Forum] A question on factorization of an integer In-Reply-To: Message-ID: <1455084403.23960.YahooMailMobile@web121505.mail.ne1.yahoo.com> Yes. Partially. Use Collected From benjamin.sambale at gmail.com Wed Feb 10 06:28:38 2016 From: benjamin.sambale at gmail.com (Benjamin Sambale) Date: Wed, 10 Feb 2016 07:28:38 +0100 Subject: [GAP Forum] A question on factorization of an integer In-Reply-To: References: Message-ID: <56BAD896.1060703@gmail.com> PrintFactorsInt Am 10.02.2016 um 05:34 schrieb Rahul Kitture: > Dear Sir, > In some computations of orders of groups, which turn out to be large, I > wanted to see the prime factorization of the order. For example, I use > following kind of commands: > > gap> g:=SymmetricGroup(10);; Size(g); > 3628800 > gap> Factors(3628800); > [ 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 5, 5, 7 ] > > My question is, is there some command in GAP to get this factorization into > form > > 2^8 * 3^4 * 5^2 * 7? > > > > > > Thank you. > > - Rahul Kitture > HRI, Allahabad > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum > From waghoba at gmail.com Wed Feb 10 06:28:49 2016 From: waghoba at gmail.com (Vinay Wagh) Date: Wed, 10 Feb 2016 11:58:49 +0530 Subject: [GAP Forum] A question on factorization of an integer In-Reply-To: References: Message-ID: Here is what you need... gap> PrintFactorsInt( 3628800); VInay On 10 February 2016 at 10:04, Rahul Kitture wrote: > Dear Sir, > In some computations of orders of groups, which turn out to be large, I > wanted to see the prime factorization of the order. For example, I use > following kind of commands: > > gap> g:=SymmetricGroup(10);; Size(g); > 3628800 > gap> Factors(3628800); > [ 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 5, 5, 7 ] > > My question is, is there some command in GAP to get this factorization into > form > > 2^8 * 3^4 * 5^2 * 7? > > > > > > Thank you. > > - Rahul Kitture > HRI, Allahabad > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From axyd0000 at gmail.com Wed Feb 10 06:51:02 2016 From: axyd0000 at gmail.com (Gaurav Dhingra) Date: Wed, 10 Feb 2016 12:21:02 +0530 Subject: [GAP Forum] Documentation understanding of unary Subgroup Message-ID: Hi GAP forum members, For the unary version of Subgroup, i am using it like this: gap> g:=Group((1,2,3,4),(1,2));; gap> f:=Subgroup(g) Reading in the documentation it says: > The unary version of Subgroup creates a (shell) subgroup that does not even know generators but can be used to collect information about a particular subgroup over time. I don't understand what this means, can someone explain this? In particular the statement "collect information over time", sounds very unknown to me from GAP user perspective. Thanks Gaurav From hulpke at math.colostate.edu Wed Feb 10 15:44:30 2016 From: hulpke at math.colostate.edu (Alexander Hulpke) Date: Wed, 10 Feb 2016 08:44:30 -0700 Subject: [GAP Forum] Documentation understanding of unary Subgroup In-Reply-To: References: Message-ID: <4B03061D-8626-4ED6-BCFA-AF664E9191F8@math.colostate.edu> The unary version of Subgroup? is intended to have a subgroup object that is going to be further specified in the course of an algorithm. E.g. if you do not know how to get generators of a Sylow subgroup, you could still have an object representing one and have this object store e.g. order, nilpotence etc. There is currently little, if any, functionality that uses this. Best, Alexander Hulpke -- Colorado State University, Department of Mathematics, Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA email: hulpke at math.colostate.edu, Phone: ++1-970-4914288 http://www.math.colostate.edu/~hulpke From f.k.moftakhar at gmail.com Thu Feb 11 12:45:47 2016 From: f.k.moftakhar at gmail.com (fatemeh moftakhar) Date: Thu, 11 Feb 2016 16:15:47 +0330 Subject: [GAP Forum] Error, exceeded the permitted memory Message-ID: Dear Forum I have the following problem with GAP. Would you please help me? gap> k(g52,D52); Error, exceeded the permitted memory (-o' command line option) in rep := S.identity; called from ElementsStabChain( StabChainMutable( G ) ) called from AsSSortedList( coll ) called from Elements( G ) called from ( ) called from read-eval loop at line 4 of *stdin* you can 'return;' brk> How can I increase memory? Best regards Fatemeh -- Regards; Miss Fatemeh Moftakhar PhD Candidate, Department of Pure Mathematics, Faculty of Mathematical Sciences, University of Kashan, Kashan, Iran From colva at mcs.st-and.ac.uk Thu Feb 11 15:15:47 2016 From: colva at mcs.st-and.ac.uk (Colva Roney-Dougal) Date: Thu, 11 Feb 2016 15:15:47 +0000 Subject: [GAP Forum] Groups St Andrews 2017 in Birmingham Message-ID: Dear Colleagues, The organisers of Groups St Andrews 2017 in Birmingham are pleased to be able to announce the speakers invited to the conference. Main Speakers: Michael Aschbacher (California Institute of Technology) Radha Kessar (City University, London) Pierre-Emmanuel Caprace (Universite Catholique de Louvain) Gunter Malle (TU Kaiserslautern) Plenary Speakers: Tim Burness (University of Bristol) Vincent Guirardel (Universite de Rennes 1) Harald Helfgott (University of Gottingen) Andrei Jaikin-Zapirain (Universidad Autonoma de Madrid) Donna Testerman (Ecole Polytechnique Federale de Lausanne) The arrival day for the conference is Saturday 5th August 2017 and the departure day is Sunday 13th August 2017. The talks will take place from 6th to 12th August inclusive. The conference website is available at the following link: http://www.groupsstandrews.org/2017/index.shtml We encourage those interested in attending the conference to register on the website to receive further updates on the conference. We hope that you will be able to join us in 2017. Conference Organisers: Colin Campbell, Chris Parker, Martyn Quick, Edmund Robertson & Colva Roney-Dougal From alexk at mcs.st-and.ac.uk Thu Feb 11 22:14:40 2016 From: alexk at mcs.st-and.ac.uk (Alexander Konovalov) Date: Thu, 11 Feb 2016 22:14:40 +0000 Subject: [GAP Forum] Error, exceeded the permitted memory In-Reply-To: References: Message-ID: <4443A4ED-CEE7-46A9-8AED-CF6DB49DD131@mcs.st-andrews.ac.uk> Dear Fatemeh, Please see 6.2: My calculation does not finish (or GAP runs out of memory) from the GAP F.A.Q: http://gap-system.org/Faq/faq.html#6.2 If you will need further help, I suggest to ask GAP Support: http://gap-system.org/Contacts/People/supportgroup.html Hope this helps Alexander > On 11 Feb 2016, at 12:45, fatemeh moftakhar wrote: > > Dear Forum > I have the following problem with GAP. Would you please help me? > > gap> k(g52,D52); > Error, exceeded the permitted memory (-o' command line option) in > rep := S.identity; called from > ElementsStabChain( StabChainMutable( G ) ) called from > AsSSortedList( coll ) called from > Elements( G ) called from > ( ) > called from read-eval loop at line 4 of *stdin* > you can 'return;' > brk> > > How can I increase memory? > > Best regards > Fatemeh > -- > Regards; > Miss Fatemeh Moftakhar > PhD Candidate, > Department of Pure Mathematics, > Faculty of Mathematical Sciences, > University of Kashan, Kashan, Iran From f.k.moftakhar at gmail.com Fri Feb 12 06:36:25 2016 From: f.k.moftakhar at gmail.com (fatemeh moftakhar) Date: Fri, 12 Feb 2016 10:06:25 +0330 Subject: [GAP Forum] Error, exceeded the permitted memory In-Reply-To: <4443A4ED-CEE7-46A9-8AED-CF6DB49DD131@mcs.st-andrews.ac.uk> References: <4443A4ED-CEE7-46A9-8AED-CF6DB49DD131@mcs.st-andrews.ac.uk> Message-ID: Dear Professor Konovalov Thank you very much for your response. I used the following commands and my problem is solved. ./gap.sh -m 4g and when the following error was showed, I just type return; exceeded the permitted memory (-o command line option) Your response helps me. Thanks a lot Fatemeh Moftakhar On Fri, Feb 12, 2016 at 1:44 AM, Alexander Konovalov wrote: > Dear Fatemeh, > > Please see 6.2: My calculation does not finish (or GAP runs out of memory) > from the GAP F.A.Q: http://gap-system.org/Faq/faq.html#6.2 > > If you will need further help, I suggest to ask GAP Support: > http://gap-system.org/Contacts/People/supportgroup.html > > Hope this helps > Alexander > > > > > On 11 Feb 2016, at 12:45, fatemeh moftakhar > wrote: > > > > Dear Forum > > I have the following problem with GAP. Would you please help me? > > > > gap> k(g52,D52); > > Error, exceeded the permitted memory (-o' command line option) in > > rep := S.identity; called from > > ElementsStabChain( StabChainMutable( G ) ) called from > > AsSSortedList( coll ) called from > > Elements( G ) called from > > ( ) > > called from read-eval loop at line 4 of *stdin* > > you can 'return;' > > brk> > > > > How can I increase memory? > > > > Best regards > > Fatemeh > > -- > > Regards; > > Miss Fatemeh Moftakhar > > PhD Candidate, > > Department of Pure Mathematics, > > Faculty of Mathematical Sciences, > > University of Kashan, Kashan, Iran > -- Regards; Miss Fatemeh Moftakhar PhD Candidate, Department of Pure Mathematics, Faculty of Mathematical Sciences, University of Kashan, Kashan, Iran From pjc at mcs.st-andrews.ac.uk Sat Feb 13 20:24:50 2016 From: pjc at mcs.st-andrews.ac.uk (Peter Cameron) Date: Sat, 13 Feb 2016 20:24:50 -0000 (UTC) Subject: [GAP Forum] A question about Action Message-ID: Dear GAP forum, It seems that GAP can find representatives for all systems of minimal blocks of imprimitivity for an arbitrary group action, but all blocks only for a permutation group (and there seems no easy way to get all maximal blocks). I have tried to get around this as follows. G is a group acting on O (which is actually an orbit on 2-subsets of [1..n]). I do the following: GG:=Action(G,O,OnSets); BB:=AllBlocks(GG); B:=List(BB,x->List(x,y->O[y])); This assumes that the Action maps each element of O to its position in the list, and I can't find a guarantee to this effect in the manual. Can I rely on this? Thanks for any advice. Peter Cameron. -- This email address will stop working sometime soon. Please use the address pjc20 at st-andrews.ac.uk instead. From sl4 at st-andrews.ac.uk Sat Feb 13 21:11:28 2016 From: sl4 at st-andrews.ac.uk (Stephen Linton) Date: Sat, 13 Feb 2016 21:11:28 +0000 Subject: [GAP Forum] A question about Action In-Reply-To: References: Message-ID: > On 13 Feb 2016, at 20:24, Peter Cameron wrote: > > Dear GAP forum, > > It seems that GAP can find representatives for all systems of minimal blocks > of imprimitivity for an arbitrary group action, but all blocks only for a > permutation group (and there seems no easy way to get all maximal blocks). > You can build up from minimal blocks recursively, of course. Take the action on the blocks of a minimal block system and find the minimal blocks there, they correspond to non-minimal block systems for the original action, etc. > I have tried to get around this as follows. G is a group acting on O (which > is actually an orbit on 2-subsets of [1..n]). I do the following: > > GG:=Action(G,O,OnSets); > BB:=AllBlocks(GG); > B:=List(BB,x->List(x,y->O[y])); > > This assumes that the Action maps each element of O to its position in the > list, and I can't find a guarantee to this effect in the manual. Can I rely > on this? I believe so, although this is (at least) a documentation failing). The manual does say: 40.3-2 Action homomorphisms See ActionHomomorphism (41.7-1). The calculation of images is determined by the acting function used and -for large domains- is often dominated by the search for the position of an image in a list of the domain elements. This can be improved by sorting this list if an efficient method for \< (31.11-1) to compare elements of the domain is available. which strongly suggests to me that it is retaining the given order for the domain. Steve > > Thanks for any advice. > > Peter Cameron. > > > > -- > This email address will stop working sometime soon. Please use the address > pjc20 at st-andrews.ac.uk instead. > > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From jacob.benoit.1 at gmail.com Sat Feb 13 23:57:02 2016 From: jacob.benoit.1 at gmail.com (Benoit Jacob) Date: Sat, 13 Feb 2016 18:57:02 -0500 Subject: [GAP Forum] Find rank-1 matrices in given subspace of matrices Message-ID: Hello, What would be a good approach to obtain a parametrization of the set of all rank-1 matrices in a given subspace of matrices M_n(F), where F is a finite field? It seems that what I'm looking for is equivalent to Maple's Rank1Elements() function . Thanks, Benoit From hulpke at math.colostate.edu Mon Feb 15 16:53:47 2016 From: hulpke at math.colostate.edu (Alexander Hulpke) Date: Mon, 15 Feb 2016 09:53:47 -0700 Subject: [GAP Forum] Find rank-1 matrices in given subspace of matrices In-Reply-To: References: Message-ID: > On Feb 13, 2016, at 4:57 PM, Benoit Jacob wrote: > > Hello, > > What would be a good approach to obtain a parametrization of the set of all > rank-1 matrices in a given subspace of matrices M_n(F), where F is a finite > field? If you want nxm matrices over a field k, why not pick a random nonzero vector v\in k^n and a random normed (i.e. first nonzero coefficient is one) vector w\in k^m and form the (matrix) product v * w^T. I think this gives you a perfect parameterization via parameterizing v and w. Best, Alexander Hulpke -- Colorado State University, Department of Mathematics, Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA email: hulpke at math.colostate.edu, Phone: ++1-970-4914288 http://www.math.colostate.edu/~hulpke From jacob.benoit.1 at gmail.com Mon Feb 15 16:54:46 2016 From: jacob.benoit.1 at gmail.com (Benoit Jacob) Date: Mon, 15 Feb 2016 11:54:46 -0500 Subject: [GAP Forum] Find rank-1 matrices in given subspace of matrices In-Reply-To: References: Message-ID: Hi Alexander, That would give me the set of all rank-1 matrices. I want the set of those rank-1 matrices that belong to some given linear subspace of matrices, given e.g. as the span of a finite family of matrices. Cheers, Benoit 2016-02-15 11:53 GMT-05:00 Alexander Hulpke : > > > On Feb 13, 2016, at 4:57 PM, Benoit Jacob > wrote: > > > > Hello, > > > > What would be a good approach to obtain a parametrization of the set of > all > > rank-1 matrices in a given subspace of matrices M_n(F), where F is a > finite > > field? > > If you want nxm matrices over a field k, why not pick a random nonzero > vector v\in k^n and a random normed (i.e. first nonzero coefficient is one) > vector w\in k^m and form the (matrix) product v * w^T. I think this gives > you a perfect parameterization via parameterizing v and w. > > Best, > > Alexander Hulpke > > -- Colorado State University, Department of Mathematics, > Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA > email: hulpke at math.colostate.edu, Phone: ++1-970-4914288 > http://www.math.colostate.edu/~hulpke > > > From hulpke at math.colostate.edu Mon Feb 15 21:07:24 2016 From: hulpke at math.colostate.edu (Alexander Hulpke) Date: Mon, 15 Feb 2016 14:07:24 -0700 Subject: [GAP Forum] Find rank-1 matrices in given subspace of matrices In-Reply-To: References: Message-ID: <30AF9A7E-0F60-4C5D-B63E-D55BB37EB625@math.colostate.edu> Dear Benoit, Ah ? within a subspace will give you polynomial equations (all 2x2 subdeterminants=0) in the coefficients of a linear combination, and at least in principle this can be done with Groebner bases (i.e. you get rank <=1, but rank 0 is easily eliminated.) For example (using the appended function) for the standard basis of Q^{3\times 4}: gap> b:=BasisVectors(Basis(MatrixSpace(Rationals,3,4)));; gap> e:=Rank1Equations(b); [ x_1*x_6-x_2*x_5, x_1*x_7-x_3*x_5, x_1*x_8-x_4*x_5, x_2*x_7-x_3*x_6, x_2*x_8-x_4*x_6, x_3*x_8-x_4*x_7, x_1*x_10-x_2*x_9, x_1*x_11-x_3*x_9, x_1*x_12-x_4*x_9, x_2*x_11-x_3*x_10, x_2*x_12-x_4*x_10, x_3*x_12-x_4*x_11, x_5*x_10-x_6*x_9, x_5*x_11-x_7*x_9, x_5*x_12-x_8*x_9, x_6*x_11-x_7*x_10, x_6*x_12-x_8*x_10, x_7*x_12-x_8*x_11 ] gap> ReducedGroebnerBasis(e,MonomialLexOrdering()); [ x_7*x_12-x_8*x_11, x_6*x_12-x_8*x_10, x_6*x_11-x_7*x_10, x_5*x_12-x_8*x_9, x_5*x_11-x_7*x_9, x_5*x_10-x_6*x_9, x_3*x_12-x_4*x_11, x_3*x_8-x_4*x_7, x_2*x_12-x_4*x_10, x_2*x_11-x_3*x_10, x_2*x_8-x_4*x_6, x_2*x_7-x_3*x_6, x_1*x_12-x_4*x_9, x_1*x_11-x_3*x_9, x_1*x_10-x_2*x_9, x_1*x_8-x_4*x_5, x_1*x_7-x_3*x_5, x_1*x_6-x_2*x_5 ] So x_7 =x_8*x_11/x_12 (and case for x_12=0) etc. and you can build an (ugly) parameterization from these. (Alternatively one could try to use \sum_c_i M_i=v\cdot w^T with v and w given by extra variables that are to be eliminated. This will yield the same Groebner basi after variable elimination.) Best, Alexander Rank1Equations:=function(mats) local l,f,r,vars,n,m,c,d,eqs; l:=Length(mats); f:=DefaultFieldOfMatrix(mats[1]); r:=PolynomialRing(f,l); vars:=IndeterminatesOfPolynomialRing(r); n:=Length(mats[1]); m:=Length(mats[1][1]); eqs:=[]; for c in Combinations([1..n],2) do for d in Combinations([1..m],2) do Add(eqs, Sum([1..l],x->vars[x]*mats[x][c[1]][d[1]]) *Sum([1..l],x->vars[x]*mats[x][c[2]][d[2]]) -Sum([1..l],x->vars[x]*mats[x][c[1]][d[2]]) *Sum([1..l],x->vars[x]*mats[x][c[2]][d[1]])); od; od; return eqs; end; From dmitrii.pasechnik at cs.ox.ac.uk Mon Feb 15 21:19:07 2016 From: dmitrii.pasechnik at cs.ox.ac.uk (Dima Pasechnik) Date: Mon, 15 Feb 2016 21:19:07 +0000 Subject: [GAP Forum] Find rank-1 matrices in given subspace of matrices In-Reply-To: References: Message-ID: <20160215211907.GB20273@dimpase.cs.ox.ac.uk> Hi Benoit, On Mon, Feb 15, 2016 at 11:54:46AM -0500, Benoit Jacob wrote: > That would give me the set of all rank-1 matrices. I want the set of those > rank-1 matrices that belong to some given linear subspace of matrices, > given e.g. as the span of a finite family of matrices. this is a hard problem. One can write down a system of polynomial equations specifying the matrices you are interested in. Let A=sum_k x_k A_k be the matrix of linear forms in variables x_k, and A_k span your subspace. You need to set all the 2x2 minors of A to 0, giving you a system of quadratic equations in x_k. Its solutions specify the rank 1 matrices in your subspace. It does not seem likely that there is a nice parametrisation of this set. Hope this helps, Dima > Cheers, > Benoit > > 2016-02-15 11:53 GMT-05:00 Alexander Hulpke : > > > > > > On Feb 13, 2016, at 4:57 PM, Benoit Jacob > > wrote: > > > > > > Hello, > > > > > > What would be a good approach to obtain a parametrization of the set of > > all > > > rank-1 matrices in a given subspace of matrices M_n(F), where F is a > > finite > > > field? > > > > If you want nxm matrices over a field k, why not pick a random nonzero > > vector v\in k^n and a random normed (i.e. first nonzero coefficient is one) > > vector w\in k^m and form the (matrix) product v * w^T. I think this gives > > you a perfect parameterization via parameterizing v and w. > > > > Best, > > > > Alexander Hulpke > > > > -- Colorado State University, Department of Mathematics, > > Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA > > email: hulpke at math.colostate.edu, Phone: ++1-970-4914288 > > http://www.math.colostate.edu/~hulpke > > > > > > > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From jacob.benoit.1 at gmail.com Mon Feb 15 21:53:11 2016 From: jacob.benoit.1 at gmail.com (Benoit Jacob) Date: Mon, 15 Feb 2016 16:53:11 -0500 Subject: [GAP Forum] Find rank-1 matrices in given subspace of matrices In-Reply-To: <30AF9A7E-0F60-4C5D-B63E-D55BB37EB625@math.colostate.edu> References: <30AF9A7E-0F60-4C5D-B63E-D55BB37EB625@math.colostate.edu> Message-ID: Many thanks, Alexander and Dima, for your replies! I was missing the key idea that rank one is charactized by the nullness of all 2x2 minors. Thanks also Alexander for the nice implementation! Cheers, Benoit 2016-02-15 16:07 GMT-05:00 Alexander Hulpke : > Dear Benoit, > > Ah ? within a subspace will give you polynomial equations (all 2x2 > subdeterminants=0) in the coefficients of a linear combination, and at > least in principle this can be done with Groebner bases (i.e. you get rank > <=1, but rank 0 is easily eliminated.) > > For example (using the appended function) for the standard basis of > Q^{3\times 4}: > > gap> b:=BasisVectors(Basis(MatrixSpace(Rationals,3,4)));; > gap> e:=Rank1Equations(b); > [ x_1*x_6-x_2*x_5, x_1*x_7-x_3*x_5, x_1*x_8-x_4*x_5, x_2*x_7-x_3*x_6, > x_2*x_8-x_4*x_6, x_3*x_8-x_4*x_7, x_1*x_10-x_2*x_9, x_1*x_11-x_3*x_9, > x_1*x_12-x_4*x_9, x_2*x_11-x_3*x_10, x_2*x_12-x_4*x_10, > x_3*x_12-x_4*x_11, > x_5*x_10-x_6*x_9, x_5*x_11-x_7*x_9, x_5*x_12-x_8*x_9, x_6*x_11-x_7*x_10, > x_6*x_12-x_8*x_10, x_7*x_12-x_8*x_11 ] > gap> ReducedGroebnerBasis(e,MonomialLexOrdering()); > [ x_7*x_12-x_8*x_11, x_6*x_12-x_8*x_10, x_6*x_11-x_7*x_10, > x_5*x_12-x_8*x_9, > x_5*x_11-x_7*x_9, x_5*x_10-x_6*x_9, x_3*x_12-x_4*x_11, x_3*x_8-x_4*x_7, > x_2*x_12-x_4*x_10, x_2*x_11-x_3*x_10, x_2*x_8-x_4*x_6, x_2*x_7-x_3*x_6, > x_1*x_12-x_4*x_9, x_1*x_11-x_3*x_9, x_1*x_10-x_2*x_9, x_1*x_8-x_4*x_5, > x_1*x_7-x_3*x_5, x_1*x_6-x_2*x_5 ] > > So x_7 =x_8*x_11/x_12 (and case for x_12=0) etc. and you can build an > (ugly) parameterization from these. > > (Alternatively one could try to use \sum_c_i M_i=v\cdot w^T with v and w > given by extra variables that are to be eliminated. This will yield the > same Groebner basi after variable elimination.) > > Best, > > Alexander > > > > Rank1Equations:=function(mats) > local l,f,r,vars,n,m,c,d,eqs; > l:=Length(mats); > f:=DefaultFieldOfMatrix(mats[1]); > r:=PolynomialRing(f,l); > vars:=IndeterminatesOfPolynomialRing(r); > n:=Length(mats[1]); > m:=Length(mats[1][1]); > eqs:=[]; > for c in Combinations([1..n],2) do > for d in Combinations([1..m],2) do > Add(eqs, > Sum([1..l],x->vars[x]*mats[x][c[1]][d[1]]) > *Sum([1..l],x->vars[x]*mats[x][c[2]][d[2]]) > -Sum([1..l],x->vars[x]*mats[x][c[1]][d[2]]) > *Sum([1..l],x->vars[x]*mats[x][c[2]][d[1]])); > od; > od; > return eqs; > end; > > > > > > > From axyd0000 at gmail.com Fri Feb 26 09:17:42 2016 From: axyd0000 at gmail.com (Gaurav Dhingra) Date: Fri, 26 Feb 2016 14:47:42 +0530 Subject: [GAP Forum] Using GAP online Message-ID: Hi GAP members Is there any way i can use GAP online? I know that i can use Sage online (for using GAP functionality). But is there any way i can use GAP alone? Thanks Gaurav Dhingra From dmitrii.pasechnik at cs.ox.ac.uk Fri Feb 26 09:55:26 2016 From: dmitrii.pasechnik at cs.ox.ac.uk (Dima Pasechnik) Date: Fri, 26 Feb 2016 09:55:26 +0000 Subject: [GAP Forum] Using GAP online In-Reply-To: References: Message-ID: <20160226095526.GA1586@dimpase.cs.ox.ac.uk> On Fri, Feb 26, 2016 at 02:47:42PM +0530, Gaurav Dhingra wrote: > Is there any way i can use GAP online? I know that i can use Sage > online (for using GAP functionality). you most probably refer to cloud.sagemath.com, right? > But is there any way i can use GAP alone? in a cloud.sagemath.com project, you can open a terminal and run GAP at the shell prompt. i.e. just type 'gap' at the '$' prompt and hit 'Enter'. Upon opening a terminal, you will see something like the following: (and then I started GAP there, as you see below) ??????????????????????????????????????????????????????????????????????????????????? ? Welcome to the SageMathCloud Terminal Environment ? ? ? ? Software: sage, R, ipython, gap, gp, git, latexmk, isympy, java, julia, octave, ? ? vim, emacs, nano, gcc, clang, pdflatex, xetex, node, convert, mc, htop, atop, ...? ? ??????????????????????????? ? Anaconda Python environment: anaconda3 ? Usage: type command in ? ? ... and to exit Anaconda: exit-anaconda ? then hit the return key ? ? ??????????????????????????? ? Learn about the Linux Bash terminal: http://ryanstutorials.net/linuxtutorial/ ? ? Are there any problems or is something missing? email us at help at sagemath.com ? ???????????????????????????????????????????????????????????????????????????????????? ~$ ~$gap ????????? GAP, Version 4.7.8 of 09-Jun-2015 (free software, GPL) ? GAP ? http://www.gap-system.org ????????? Architecture: x86_64-unknown-linux-gnu-gcc-default64 Libs used: gmp, readline Loading the library and packages ... Components: trans 1.0, prim 2.1, small* 1.0, id* 1.0 Packages: Alnuth 3.0.0, AtlasRep 1.5.0, AutPGrp 1.6, CTblLib 1.2.2, FactInt 1.5.3, GAPDoc 1.5.1, LAGUNA 3.7.0, Polycyclic 2.11, TomLib 1.2.5 Try '?help' for help. See also '?copyright' and '?authors' Apart from the packages listed, there is a dozen other GAP packages that are installed, such as GRAPE, which you could load using LoadPackage HTH Dima > > Thanks > Gaurav Dhingra > > > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From suskudar at gmail.com Fri Feb 26 10:20:47 2016 From: suskudar at gmail.com (=?utf-8?Q?S=C3=BCmeyra_Bedir?=) Date: Fri, 26 Feb 2016 12:20:47 +0200 Subject: [GAP Forum] Using GAP online In-Reply-To: References: Message-ID: Hi, You may try this; http://artin.nuigalway.ie/gap_calculator/gap_terminal The issue was discussed before and above was a temporary solution, S?meyra Bedir 26 ?ub 2016 tarihinde 11:17 saatinde, Gaurav Dhingra ?unlar? yazd?: > Hi GAP members > > Is there any way i can use GAP online? I know that i can use Sage online (for using GAP functionality). > But is there any way i can use GAP alone? > > Thanks > Gaurav Dhingra > > > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From alexk at mcs.st-and.ac.uk Fri Feb 26 16:42:19 2016 From: alexk at mcs.st-and.ac.uk (Alexander Konovalov) Date: Fri, 26 Feb 2016 16:42:19 +0000 Subject: [GAP Forum] GAP 4.8.2 release announcement Message-ID: <77E52CF0-802D-4178-A470-ED4F514633AD@mcs.st-andrews.ac.uk> Dear all, This is to announce the release of GAP 4.8.2, which could be downloaded from http://www.gap-system.org/Releases/ This is the first major release made from the GAP repository on GitHub (at https://github.com/gap-system/gap). An overview of the most significant ones is provided below. A more detailed version with hyperlinks is available on the GAP website here: http://www.gap-system.org/Manuals/doc/changes/chap2.htm New features: * Added support for profiling which tracks how much time in spent on each line of GAP code. This can be used to show where code is spending a long time and also check which lines of code are even executed. * Added _ability_ to install (in the library or packages) methods for accessing lists using multiple indices and indexing into lists using indices other than positive small integers. Such methods could allow you, for example, to define behaviour for expressions like m[1,2]; m[1,2,3] := x; IsBound(m["a","b",Z(7)]); Unbind(m[1][2,3]) * Added support for partially variadic functions to allow function expressions like function(a,b,c,x...) ... end; which would require at least three arguments and assign the first three to a, b and c and then a list containing any remaining ones to x. The former special meaning of the argument arg is still supported and is now equivalent to function(arg...), so no changes in the existing code are required. * Introduced CallWithTimeout and CallWithTimeoutList to call a function with a limit on the CPU time it can consume. This functionality may not be available on all systems and you should check GAPInfo.TimeoutsSupported before using this functionality. * GAP now displays the filename and line numbers of statements in backtraces when entering the break loop. Improved and extended functionality: * Method tracing shows the filename and line of function during tracing. * TraceAllMethods and UntraceAllMethods to turn on and off tracing all methods in GAP. Also, for the uniform approach UntraceImmediateMethods has been added as an equivalent of TraceImmediateMethods(false) * Performance improvements for dictionaries. * Improved methods for symmetric and alternating groups in the "natural" representations. * Package authors may optionally specify the source code repository, issue tracker and support email address for their package using new components in the PackageInfo.g file. Changed functionality: * As a preparation for the future developments to support multithreading, some language extensions from the HPC-GAP project were backported to the GAP library to help to unify the codebase of both GAP 4 and HPC-GAP. In particular, this introduces new keywords "atomic", "readonly" and "readwrite". * There was inconsistent use of the following properties of semigroups: IsGroupAsSemigroup, IsMonoidAsSemigroup, and IsSemilatticeAsSemigroup, which has been corrected. * ReadTest became obsolete and for backwards compatibility is replaced by Test with the option to compare the output up to whitespaces. * The function ErrorMayQuit, which differs from Error by not allowing execution to continue, has been renamed to ErrorNoReturn. Fixed bugs: * Problems arising when an element of an extension field was assigned into a compressed matrix. * Problems in the Fitting-free code and code inheriting PCGS. * A bug that caused a break loop in the computation of the Hall subgroup for groups having a trivial Fitting subgroup. * Including a break or continue statement in a function body but not in a loop now gives a syntax error instead of failing at run time. * GroupGeneralMappingByImages now verifies that that image of a mapping is contained in its range. This may sometimes slow down construction of mappings. If this is a problem, and you are sure the image of the mapping will be contained in the range, use GroupGeneralMappingByImagesNC instead. * Fixed a bug in caching the degree of transformation that could lead to a non-identity transformation accidentally changing its value to the identity transformation. * Fixed the problem with using Windows default browser as a help viewer using SetHelpViewer("browser");. New and updated packages since GAP 4.7.8 At the time of the release of GAP 4.7.8 there were 119 packages redistributed with GAP. New packages that have been added to the redistribution since the release of GAP 4.7.8 are: * CAP (Categories, Algorithms, Programming) package by Sebastian Gutsche, Sebastian Posur and ?ystein Skarts?terhagen, together with three associated packages GeneralizedMorphismsForCAP, LinearAlgebraForCAP] and ModulePresentationsForCAP] (all three - by Sebastian Gutsche and Sebastian Posur). * Digraphs by Jan De Beule, Julius Jonu?as, James Mitchell, Michael Torpey and Wilf Wilson, which provides functionality to work with graphs, digraphs, and multidigraphs. * FinInG package by John Bamberg, Anton Betten, Philippe Cara, Jan De Beule, Michel Lavrauw and Max Neunh?ffer for computation in Finite Incidence Geometry. * HeLP package by Andreas B?chle and Leo Margolis, which computes constraints on partial augmentations of torsion units in integral group rings using a method developed by Luthar, Passi and Hertweck. * matgrp package by Alexander Hulpke, which provides an interface to the solvable radical functionality for matrix groups, building on constructive recognition. * NormalizInterface package by Sebastian Gutsche, Max Horn and Christof S?ger, which provides a GAP interface to Normaliz, enabling direct access to the complete functionality of Normaliz, such as computations in affine monoids, vector configurations, lattice polytopes, and rational cones. * profiling package by Christopher Jefferson for transforming profiles produced by ProfileLineByLine and CoverageLineByLine into a human-readable form. * Utils package by Sebastian Gutsche, Stefan Kohl and Christopher Wensley, which provides a collection of utility functions gleaned from many packages. * XModAlg package by Zekeriya Arvasi and Alper Odabas, which provides a collection of functions for computing with crossed modules and Cat1-algebras and morphisms of these structures. In addition, a number of packages previously redistributed with GAP has been updated. We encourage all users to upgrade to GAP 4.8.2. If you need any help or would like to report any problems, please do not hesitate to contact us at support at gap-system.org or submit new issues on GitHub: https://github.com/gap-system/gap/issues Wishing you fun and success using GAP, The GAP Group From alexk at mcs.st-and.ac.uk Fri Feb 26 16:50:18 2016 From: alexk at mcs.st-and.ac.uk (Alexander Konovalov) Date: Fri, 26 Feb 2016 16:50:18 +0000 Subject: [GAP Forum] GAP 4.8.2 release announcement In-Reply-To: <77E52CF0-802D-4178-A470-ED4F514633AD@mcs.st-andrews.ac.uk> References: <77E52CF0-802D-4178-A470-ED4F514633AD@mcs.st-andrews.ac.uk> Message-ID: <0A347B6A-83DD-4D08-89F6-E39FC5903856@mcs.st-andrews.ac.uk> P.S. Only if you are using Windows: 1) We strongly advise to use the executable Windows installer for GAP: http://www.gap-system.org/pub/gap/gap48/exe/gap4r8p2_2016_02_20-18_51.exe which will update all bat-files automatically. Note that the path to the GAP directory should not contain spaces. For example, you may install it in C:\gap4r8(default), D:\gap4r8p2 or C:\Math\GAP\gap4r8, but you must not install it in a directory named like C:\Program files\gap4r8 or C:\Users\alice\My Documents\gap4r8 etc. 2) In GAP 4.8.2 for Windows, there are no compiled binaries for the Browse package and there is no readline support. We hope to fix this in GAP 4.8.3 3) There is an experimental 64-bit version of GAP for Windows, also available from http://www.gap-system.org/Releases/. As it is experimental, some packages may not work there. To install it, you need to unzip the archive and adjust *.bat files manually. We will be interested to hear any feedback about its usability. Best wishes Alexander > On 26 Feb 2016, at 16:42, Alexander Konovalov wrote: > > Dear all, > > This is to announce the release of GAP 4.8.2, which could be downloaded from > > http://www.gap-system.org/Releases/ > > This is the first major release made from the GAP repository on GitHub > (at https://github.com/gap-system/gap). An overview of the most > significant ones is provided below. A more detailed version with > hyperlinks is available on the GAP website here: > > http://www.gap-system.org/Manuals/doc/changes/chap2.htm From benjamin.sambale at gmail.com Sat Feb 27 11:04:13 2016 From: benjamin.sambale at gmail.com (Benjamin Sambale) Date: Sat, 27 Feb 2016 12:04:13 +0100 Subject: [GAP Forum] installation problems with 4.8.2 Message-ID: <56D182AD.1000508@gmail.com> Dear GAP people, I installed the new release and the compiling process seemed fine to me (using a recent arch linux version). When I started, I got the error that /usr/local/gap4r8/pkg/PolymakeInterface/PackageInfo.g is unreadable or does not exist (notice that it is a new package). However, the file does exist, but I found out that most files are owned by 50009 whoever that is (this was never a problem in previous versions). So I made this particular file readable for group users. Now I can start GAP and also the packages load, but the help system is broken: gap> ?help Error, no method found! For debugging hints type ?Recovery from NoMethodFound Error, no 1st choice method found for ReadLine' on 1 arguments I have no idea what is going on. Any ideas? Best wishes, Benjamin From alexk at mcs.st-and.ac.uk Sat Feb 27 13:09:52 2016 From: alexk at mcs.st-and.ac.uk (Alexander Konovalov) Date: Sat, 27 Feb 2016 13:09:52 +0000 Subject: [GAP Forum] installation problems with 4.8.2 In-Reply-To: <56D182AD.1000508@gmail.com> References: <56D182AD.1000508@gmail.com> Message-ID: <4E6E1B40-8968-4B57-845C-CDE10B25A5E1@mcs.st-andrews.ac.uk> Dear Benjamin, dear GAP Forum, Thank you for reporting this - we have heard this three times so far, from users installing GAP as root outside their home directories (this will not affect you if you install and run GAP as the same user). What happened is that several packages were released with files not having appropriate permissions. This will be fixed in GAP 4.8.3. In the meantime, please use the following fix to make all package files world-readable: cd gap4r8 # or whatever the name is chmod -R a+r pkg (we suggest to call it on the whole 'pkg' subdirectory rather than on individual packages involved, just in case). Apologies for any inconveniences, Alexander P.S. The help system reads information about known help books, including package manuals, so this is an instance of the same problem with permissions. Calling chmod -R a+r pkg on the whole pkg directory should fix all problems at once. > On 27 Feb 2016, at 11:04, Benjamin Sambale wrote: > > Dear GAP people, > > I installed the new release and the compiling process seemed fine to me (using a recent arch linux version). When I started, I got the error that /usr/local/gap4r8/pkg/PolymakeInterface/PackageInfo.g is unreadable or does not exist (notice that it is a new package). However, the file does exist, but I found out that most files are owned by 50009 whoever that is (this was never a problem in previous versions). So I made this particular file readable for group users. Now I can start GAP and also the packages load, but the help system is broken: > > gap> ?help > Error, no method found! For debugging hints type ?Recovery from NoMethodFound > Error, no 1st choice method found for ReadLine' on 1 arguments > > I have no idea what is going on. Any ideas? > > Best wishes, > Benjamin From benjamin.sambale at gmail.com Sat Feb 27 18:33:05 2016 From: benjamin.sambale at gmail.com (Benjamin Sambale) Date: Sat, 27 Feb 2016 19:33:05 +0100 Subject: [GAP Forum] installation problems with 4.8.2 In-Reply-To: <4E6E1B40-8968-4B57-845C-CDE10B25A5E1@mcs.st-andrews.ac.uk> References: <56D182AD.1000508@gmail.com> <4E6E1B40-8968-4B57-845C-CDE10B25A5E1@mcs.st-andrews.ac.uk> Message-ID: <56D1EBE1.7070209@gmail.com> Dear Alexander, many thanks! Problem solved. Benjamin Am 27.02.2016 um 14:09 schrieb Alexander Konovalov: > Dear Benjamin, dear GAP Forum, > > Thank you for reporting this - we have heard this three times so far, > from users installing GAP as root outside their home directories (this > will not affect you if you install and run GAP as the same user). > > What happened is that several packages were released with files not > having appropriate permissions. This will be fixed in GAP 4.8.3. In > the meantime, please use the following fix to make all package files > world-readable: > > cd gap4r8 # or whatever the name is > chmod -R a+r pkg > > (we suggest to call it on the whole 'pkg' subdirectory rather than on > individual packages involved, just in case). > > Apologies for any inconveniences, > Alexander > > P.S. The help system reads information about known help books, > including package manuals, so this is an instance of the same > problem with permissions. Calling chmod -R a+r pkg on the whole > pkg directory should fix all problems at once. > > >> On 27 Feb 2016, at 11:04, Benjamin Sambale wrote: >> >> Dear GAP people, >> >> I installed the new release and the compiling process seemed fine to me (using a recent arch linux version). When I started, I got the error that /usr/local/gap4r8/pkg/PolymakeInterface/PackageInfo.g is unreadable or does not exist (notice that it is a new package). However, the file does exist, but I found out that most files are owned by 50009 whoever that is (this was never a problem in previous versions). So I made this particular file readable for group users. Now I can start GAP and also the packages load, but the help system is broken: >> >> gap> ?help >> Error, no method found! For debugging hints type ?Recovery from NoMethodFound >> Error, no 1st choice method found for ReadLine' on 1 arguments >> >> I have no idea what is going on. Any ideas? >> >> Best wishes, >> Benjamin > > > From sara.yaftian at gmail.com Wed Mar 2 06:03:19 2016 From: sara.yaftian at gmail.com (Sara Yaftian) Date: Wed, 2 Mar 2016 09:33:19 +0330 Subject: [GAP Forum] Boolean matrix Message-ID: Dear GAP Forum I need a m by n Boolean matrix for saving memory in my works. I know A:=NullMat(m,n); is a Null matrix but I need boolean version. I wrote the following code B:=[]; for i in [1..3] do Add(B,false); od; A:=[B]; for j in [1..3] do Add(A,B); od; this is the output: gap> A; [ [ false, false, false ], [ false, false, false ], [ false, false, false ], [ false, false, false ] ] and I obtained a null Boolean matrix but if I use the following command A[2][3] := true; I have the following problem gap> A; [ [ false, false, true ], [ false, false, true ], [ false, false, true ], [ false, false, true ] ] I need just A[2][3] := true not A[i][3]. Would you please help me? Best regards Sara From gordon.royle at uwa.edu.au Wed Mar 2 07:24:40 2016 From: gordon.royle at uwa.edu.au (Gordon Royle) Date: Wed, 2 Mar 2016 15:24:40 +0800 Subject: [GAP Forum] Boolean matrix In-Reply-To: References: Message-ID: Each iteration of Add(A,B) simply adds (to A) another reference to the same copy of B, so all four rows of A refer to a single copy of B. If you change that one copy (using *any* of the references) then every reference to that copy will reflect the change. You need to actually create four genuine ?deep copies" of B if you want them to be independent. (I am also not sure whether GAP treats arrays of booleans as bit patterns which is what you would need in order to really save memory) On 2 Mar 2016, at 2:03 pm, Sara Yaftian > wrote: Dear GAP Forum I need a m by n Boolean matrix for saving memory in my works. I know A:=NullMat(m,n); is a Null matrix but I need boolean version. I wrote the following code B:=[]; for i in [1..3] do Add(B,false); od; A:=[B]; for j in [1..3] do Add(A,B); od; this is the output: gap> A; [ [ false, false, false ], [ false, false, false ], [ false, false, false ], [ false, false, false ] ] and I obtained a null Boolean matrix but if I use the following command A[2][3] := true; I have the following problem gap> A; [ [ false, false, true ], [ false, false, true ], [ false, false, true ], [ false, false, true ] ] I need just A[2][3] := true not A[i][3]. Would you please help me? Best regards Sara _______________________________________________ Forum mailing list Forum at mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum Professor Gordon Royle School of Mathematics and Statistics University of Western Australia Gordon.Royle at uwa.edu.au From alexk at mcs.st-and.ac.uk Tue Mar 8 22:34:19 2016 From: alexk at mcs.st-and.ac.uk (Alexander Konovalov) Date: Tue, 8 Mar 2016 22:34:19 +0000 Subject: [GAP Forum] Boolean matrix In-Reply-To: References: Message-ID: <36F148EA-875A-4CEB-B54A-FF36473A0FFE@mcs.st-andrews.ac.uk> Dear Gordon, Sara, dear Forum, > On 2 Mar 2016, at 07:24, Gordon Royle wrote: > > Each iteration of Add(A,B) simply adds (to A) another reference to the same copy of B, so all four rows of A refer to a single copy of B. > > If you change that one copy (using *any* of the references) then every reference to that copy will reflect the change. > > You need to actually create four genuine ?deep copies" of B if you want them to be independent. > > (I am also not sure whether GAP treats arrays of booleans as bit patterns which is what you would need in order to really save memory) GAP indeed has boolean lists (blists) which are internally represented in a compact form - see http://www.gap-system.org/Manuals/doc/ref/chap22.html for further details and some pitfalls to avoid. Best wishes Alexander From l.h.soicher at qmul.ac.uk Wed Mar 9 17:25:58 2016 From: l.h.soicher at qmul.ac.uk (Leonard Soicher) Date: Wed, 9 Mar 2016 17:25:58 +0000 Subject: [GAP Forum] Important news for users and installers of GRAPE Message-ID: Dear GAP Forum, I am pleased to announce that the new version 4.7 of GRAPE is included in the recently released GAP 4.8.2. All users and installers of GRAPE should read this email in its entirety. The main new feature of GRAPE 4.7 is that the user may run their own separately installed copy of bliss or nauty or nauty/traces rather than using the included version of nauty 2.2 included in GRAPE 4.7. How to do this is described in Chapter 1 of the GRAPE 4.7 manual. Please note that the nauty interface for GRAPE 4.7 has only been extensively tested with the included version 2.2 of nauty, and the bliss interface has only been tested with version 0.73 of bliss. The interface to bliss was originally written by Jerry James (who I thank), who modified and packaged GRAPE 4.6.1 as a Fedora package (in Fedora 22, 23 and 24) to use bliss instead of nauty, due to Fedora licensing rules. Unfortunately, there were some bugs in this initial interface, which could cause wrong results to be returned. This interface was fixed for inclusion in GRAPE 4.7 and I am pleased to see that now there are Fedora package versions of GRAPE 4.7 for Fedora 23 and 24, but I have not tried them and would appreciate any feedback you have on those Fedora package versions. However, please note that I can only guarantee support for official versions of GRAPE obtained either from http://www.maths.qmul.ac.uk/~leonard/grape/ or as part of the official GAP distribution downloaded from http://www.gap-system.org The user should test any installation of GRAPE 4.7 (and its interface to bliss or nauty or nauty/traces) using the new GRAPE testfile. In GAP, the command Test(Filename(DirectoriesPackageLibrary("grape","tst"),"testall.tst")); should return the value true. I wish you happy computing with GRAPE 4.7! Regards, Leonard From samuel.lelievre at gmail.com Thu Mar 10 06:01:15 2016 From: samuel.lelievre at gmail.com (=?UTF-8?Q?Samuel_Leli=C3=A8vre?=) Date: Thu, 10 Mar 2016 07:01:15 +0100 Subject: [GAP Forum] Important news for users and installers of GRAPE In-Reply-To: References: Message-ID: 2016-03-09 18:25 GMT+01:00 Leonard Soicher : > The interface to bliss was originally written by Jerry James (who I > thank), who modified and packaged GRAPE 4.6.1 as a Fedora package > (in Fedora 22, 23 and 24) to use bliss instead of nauty, due to Fedora > licensing rules. Note that this might evolve since the authors of Nauty released version 2.6 under an updated license, see http://pallini.di.uniroma1.it/ http://users.cecs.anu.edu.au/~bdm/nauty/ and in particular http://users.cecs.anu.edu.au/~bdm/nauty/COPYRIGHT.txt Best, Samuel From z060822400814a at rezozer.net Thu Mar 10 23:33:03 2016 From: z060822400814a at rezozer.net (z060822400814a at rezozer.net) Date: Fri, 11 Mar 2016 00:33:03 +0100 Subject: [GAP Forum] Important news for users and installers of GRAPE In-Reply-To: References: Message-ID: <56E2042F.6050604@rezozer.net> Hello Forum: On 10/03/16 07:01, Samuel Leli?vre wrote: > 2016-03-09 18:25 GMT+01:00 Leonard Soicher : > >> The interface to bliss was originally written by Jerry James (who I >> thank), who modified and packaged GRAPE 4.6.1 as a Fedora package >> (in Fedora 22, 23 and 24) to use bliss instead of nauty, due to Fedora >> licensing rules. > > Note that this might evolve since the authors of Nauty > released version 2.6 under an updated license, see > > http://pallini.di.uniroma1.it/ > http://users.cecs.anu.edu.au/~bdm/nauty/ > > and in particular > > http://users.cecs.anu.edu.au/~bdm/nauty/COPYRIGHT.txt > > Best, Samuel Thanks for the notice. I am on my way to update the nauty package for Debian [1] and to resume the packaging for GRAPE [2]. Thanks, Jerome [1] https://packages.qa.debian.org/n/nauty.html [2] https://bugs.debian.org/cgi-bin/bugreport.cgi?bug=762583 > > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum > From l.h.soicher at qmul.ac.uk Fri Mar 11 10:47:36 2016 From: l.h.soicher at qmul.ac.uk (Leonard Soicher) Date: Fri, 11 Mar 2016 10:47:36 +0000 Subject: [GAP Forum] Important news for users and installers of GRAPE In-Reply-To: <56E2042F.6050604@rezozer.net> References: , <56E2042F.6050604@rezozer.net> Message-ID: Hello GAP-Forum, I installed nauty 2.6 (r2), and ran GRAPE 4.7, setting GRAPE_NAUTY := true; GRAPE_DREADNAUT_EXE := "mypathto/nauty26r2/dreadnaut"; Initial basic tests indicate that GRAPE 4.7 works with nauty 2.6 (r2), but I will continue with further checks. I remark that I have had a problem with bliss crashing on occasion. Regards, Leonard ________________________________________ From: forum-bounces at gap-system.org on behalf of z060822400814a at rezozer.net Sent: 10 March 2016 23:33 To: forum at gap-system.org Subject: Re: [GAP Forum] Important news for users and installers of GRAPE Hello Forum: On 10/03/16 07:01, Samuel Leli?vre wrote: > 2016-03-09 18:25 GMT+01:00 Leonard Soicher : > >> The interface to bliss was originally written by Jerry James (who I >> thank), who modified and packaged GRAPE 4.6.1 as a Fedora package >> (in Fedora 22, 23 and 24) to use bliss instead of nauty, due to Fedora >> licensing rules. > > Note that this might evolve since the authors of Nauty > released version 2.6 under an updated license, see > > http://pallini.di.uniroma1.it/ > http://users.cecs.anu.edu.au/~bdm/nauty/ > > and in particular > > http://users.cecs.anu.edu.au/~bdm/nauty/COPYRIGHT.txt > > Best, Samuel Thanks for the notice. I am on my way to update the nauty package for Debian [1] and to resume the packaging for GRAPE [2]. Thanks, Jerome [1] https://packages.qa.debian.org/n/nauty.html [2] https://bugs.debian.org/cgi-bin/bugreport.cgi?bug=762583 > > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum > _______________________________________________ Forum mailing list Forum at mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum From axyd0000 at gmail.com Sat Mar 19 06:31:47 2016 From: axyd0000 at gmail.com (Gaurav Dhingra) Date: Sat, 19 Mar 2016 12:01:47 +0530 Subject: [GAP Forum] Reduction of elements of FpGroup Message-ID: <56ECF253.3090303@gmail.com> Hi GAP forum members, I am looking at the source code FpElementNFFunction (no documentation for it), in that there is a comment # first try whether the group is small'' what does "small" means in GAP terms? Is this some standard jargon used in GAP source code? If so how is this used internally? It will be helpful if you could explain this used in a little deail. I believe GAP uses somehow "small" group to refer to groups below certain "order", if this is the case. To what things it makes things useful? Thank you Gaurav From axyd0000 at gmail.com Sat Mar 19 06:33:13 2016 From: axyd0000 at gmail.com (Gaurav Dhingra) Date: Sat, 19 Mar 2016 12:03:13 +0530 Subject: [GAP Forum] Reduction of elements of FpGroup In-Reply-To: <56ECF253.3090303@gmail.com> References: <56ECF253.3090303@gmail.com> Message-ID: <56ECF2A9.2010103@gmail.com> May be here my title of mail is mis-leading. Sorry about that. On Saturday 19 March 2016 12:01 PM, Gaurav Dhingra wrote: > Hi GAP forum members, > > I am looking at the source code FpElementNFFunction (no > documentation for it), in that there is a comment > > # first try whether the group is small'' > > > what does "small" means in GAP terms? Is this some standard jargon > used in GAP source code? If so how is this used internally? It will be > helpful if you could explain this used in a little deail. > I believe GAP uses somehow "small" group to refer to groups below > certain "order", if this is the case. To what things it makes things > useful? > > Thank you > Gaurav From sl4 at st-andrews.ac.uk Sat Mar 19 13:20:11 2016 From: sl4 at st-andrews.ac.uk (Stephen Linton) Date: Sat, 19 Mar 2016 13:20:11 +0000 Subject: [GAP Forum] Reduction of elements of FpGroup In-Reply-To: <56ECF253.3090303@gmail.com> References: <56ECF253.3090303@gmail.com> Message-ID: <5291C739-3221-4470-970B-540286812427@st-andrews.ac.uk> Dear Gaurav Dhingra, Firstly, I?m afraid there is no standard meaning of ?small? in GAP. This is simply a comment by the developer of this routine to explain what the next few lines are doing. In fact we can see from the next lines that the test is whether the group can be seen to have order less than 50000. That is whether a faithful homomorphism is known, or can be found (FPFaithHom) and if so whether the image has order < 50000. This information is used to choose between two different ways of reducing the element to normal form. For ?small? groups the work is done in Factorisation which produces short words and is quick for small groups but can be very slow for larger ones. For larger groups a more sophisticated method is used, which may produce longer normal words, and requires quite a lot of setup but can be much faster for very large groups. Steve > On 19 Mar 2016, at 06:31, Gaurav Dhingra wrote: > > Hi GAP forum members, > > I am looking at the source code FpElementNFFunction (no documentation for it), in that there is a comment > > # first try whether the group is small'' > > > what does "small" means in GAP terms? Is this some standard jargon used in GAP source code? If so how is this used internally? It will be helpful if you could explain this used in a little deail. > I believe GAP uses somehow "small" group to refer to groups below certain "order", if this is the case. To what things it makes things useful? > > Thank you > Gaurav > > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From graham.ellis at nuigalway.ie Mon Mar 21 09:53:01 2016 From: graham.ellis at nuigalway.ie (Ellis, Grahamj) Date: Mon, 21 Mar 2016 09:53:01 +0000 Subject: [GAP Forum] lectureship at Galway In-Reply-To: References: <561E0F8F.3080502@gmail.com> , <56C33D7B.50403@gmail.com>, , , , , Message-ID: A permanent lectureship in mathematics at NUI Galway was advertized last Friday. See http://www.nuigalway.ie/media/humanresources/publicdocuments/Lecturer-BTB_Maths.docx . The closing date is 28 April 2016. All the best, Graham School of Mathematics, Statistics & Applied Mathematics National University of Ireland, Galway University Road, Galway Ireland http://hamilton.nuigalway.ie tel: 091 493011 From alexk at mcs.st-and.ac.uk Thu Mar 24 15:37:40 2016 From: alexk at mcs.st-and.ac.uk (Alexander Konovalov) Date: Thu, 24 Mar 2016 15:37:40 +0000 Subject: [GAP Forum] GAP 4.8.3 release announcement Message-ID: Dear GAP Forum, This is to announce the release of GAP 4.8.3, which could be downloaded from http://www.gap-system.org/Releases/ An overview of the changes introduced in GAP 4.8.3 is provided below. A more detailed version with hyperlinks to documentation is available here: http://www.gap-system.org/Manuals/doc/changes/chap2.htm New features: * New function TestPackage to run standard tests (if available) for a single package in the current GAP session. This function is also callable via make testpackage PKGNAME=pkgname to run package tests in the same settings that are used for testing GAP releases. Improved and extended functionality: * TestDirectory now prints a special status message to indicate the outcome of the test (this is convenient for automated testing). If necessary, this message may be suppressed by using the option suppressStatusMessage * Improved output of tracing methods (which may be invoked, for example, with TraceAllMethods) by displaying filename and line number in some more cases. Changed functionality: * Fixed some inconsistencies in the usage of IsGeneratorsOfSemigroup. Fixed bugs that could lead to incorrect results: * Fallback methods for conjugacy classes, that were intended only for finite groups, now use IsFinite filter to prevent them being called for infinite groups. [Reported by Gabor Horvath] Fixed bugs that could lead to break loops: * Calculating stabiliser for the alternating group caused a break loop in the case when it defers to the corresponding symmetric group. * It was not possible to use DotFileLatticeSubgroups for a trivial group. [Reported by Sergio Siccha] * A break loop while computing AutomorphismGroup for TransitiveGroup(12,269). [Reported by Ignat Soroko] * A break loop while computing conjugacy classes of PSL(6,4). [Reported by Martin Macaj] Other fixed bugs: * Fix for using Firefox as a default help viewer with SetHelpViewer. [Reported by Tom McDonough] In addition, GAP 4.8.3 distribution includes updates for 29 packages: ==================================================== Package name | Version | Date ---------------------------------------------------- ACE | 5.2 | 11/03/2016 ANUPQ | 3.1.4 | 08/03/2016 AutoDoc | 2016.03.08 | 08/03/2016 CRISP | 1.4.3 | 29/02/2016 cvec | 2.5.5 | 08/03/2016 Digraphs | 0.5 | 03/03/2016 FGA | 1.3.1 | 28/02/2016 fr | 2.3.3 | 27/02/2016 GBNP | 1.0.3 | 08/03/2016 genss | 1.6.4 | 08/03/2016 GUAVA | 3.13 | 31/01/2016 HeLP | 3.0 | 03/03/2016 IO | 4.4.6 | 08/03/2016 matgrp | 0.4 | 25/02/2015 NormalizInterface | 0.9.7 | 10/03/2016 nq | 2.5.3 | 08/03/2016 orb | 4.7.6 | 08/03/2016 Polenta | 1.3.6 | 08/03/2016 profiling | 0.5.1 | 24/02/2016 qaos | 1.2 | 25/02/2016 QPA | 1.24 | 24/02/2016 RCWA | 4.3.1 | 09/03/2016 recog | 1.2.5 | 08/03/2016 recogbase | 1.2.5 | 08/03/2016 ResClasses | 4.4.2 | 09/03/2016 Semigroups | 2.7.4 | 02/03/2016 Utils | 0.39 | 04/03/2016 XGAP | 4.24 | 28/02/2016 XMod | 2.56 | 08/03/2016 ==================================================== We encourage all users to upgrade to GAP 4.8.3. If you need any help or would like to report any problems, please do not hesitate to contact us at support at gap-system.org or submit new issues on GitHub: https://github.com/gap-system/gap/issues Wishing you fun and success using GAP, The GAP Group From z060822400814a at rezozer.net Sat Mar 26 16:08:53 2016 From: z060822400814a at rezozer.net (z060822400814a at rezozer.net) Date: Sat, 26 Mar 2016 17:08:53 +0100 Subject: [GAP Forum] Important news for users and installers of GRAPE: [DEBIAN] In-Reply-To: References: <56E2042F.6050604@rezozer.net> Message-ID: <56F6B415.8020106@rezozer.net> Hello Forum: GRAPE is now available as Debian package (DFSG-free) [1]. Note that the nauty Debian package has migrated from non-free to main (DFSG-free) [2] as nauty is now distributed under the Apache 2.0 license. Thanks, Jerome [1] https://packages.qa.debian.org/g/gap-grape.html [2] https://packages.qa.debian.org/n/nauty.html On 11/03/16 11:47, Leonard Soicher wrote: > Hello GAP-Forum, > > I installed nauty 2.6 (r2), and ran GRAPE 4.7, setting > > GRAPE_NAUTY := true; GRAPE_DREADNAUT_EXE := > "mypathto/nauty26r2/dreadnaut"; > > Initial basic tests indicate that GRAPE 4.7 works with nauty 2.6 > (r2), but I will continue with further checks. > > I remark that I have had a problem with bliss crashing on occasion. > > Regards, Leonard > > ________________________________________ From: > forum-bounces at gap-system.org on behalf > of z060822400814a at rezozer.net Sent: 10 > March 2016 23:33 To: forum at gap-system.org Subject: Re: [GAP Forum] > Important news for users and installers of GRAPE > > Hello Forum: > > On 10/03/16 07:01, Samuel Leli?vre wrote: >> 2016-03-09 18:25 GMT+01:00 Leonard Soicher >> : >> >>> The interface to bliss was originally written by Jerry James (who >>> I thank), who modified and packaged GRAPE 4.6.1 as a Fedora >>> package (in Fedora 22, 23 and 24) to use bliss instead of nauty, >>> due to Fedora licensing rules. >> >> Note that this might evolve since the authors of Nauty released >> version 2.6 under an updated license, see >> >> http://pallini.di.uniroma1.it/ >> http://users.cecs.anu.edu.au/~bdm/nauty/ >> >> and in particular >> >> http://users.cecs.anu.edu.au/~bdm/nauty/COPYRIGHT.txt >> >> Best, Samuel > > Thanks for the notice. > > I am on my way to update the nauty package for Debian [1] and to > resume the packaging for GRAPE [2]. > > Thanks, Jerome > > [1] https://packages.qa.debian.org/n/nauty.html [2] > https://bugs.debian.org/cgi-bin/bugreport.cgi?bug=762583 > > >> >> _______________________________________________ Forum mailing list >> Forum at mail.gap-system.org >> http://mail.gap-system.org/mailman/listinfo/forum >> > > _______________________________________________ Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum > From s.dikson2016 at gmail.com Mon Apr 11 08:19:55 2016 From: s.dikson2016 at gmail.com (Sara Dikson) Date: Mon, 11 Apr 2016 11:49:55 +0430 Subject: [GAP Forum] Used memory for a process Message-ID: Dear Forum, I am using a super computer for some processes. But suddenly, the GAP screen will be closed when my program is running. How can I prevent this problem? I would like to use all memory and run more than one program simultaneously. How can I do that? Best regards, Sara. From dmitrii.pasechnik at cs.ox.ac.uk Mon Apr 11 09:08:17 2016 From: dmitrii.pasechnik at cs.ox.ac.uk (Dima Pasechnik) Date: Mon, 11 Apr 2016 09:08:17 +0100 Subject: [GAP Forum] Used memory for a process In-Reply-To: References: Message-ID: <20160411080817.GA20376@dimpase.cs.ox.ac.uk> Dear Sara, On Mon, Apr 11, 2016 at 11:49:55AM +0430, Sara Dikson wrote: > I am using a super computer for some processes. But suddenly, the GAP > screen will be closed when my program is running. > How can I prevent this problem? it depends a lot on how your supercomputer is set up. Often you can only run jobs which are "big" - either memory-wise or CPU-time wise only as a batch job, and not interactively. It could well be that this is exactly what causes your issue. One way or another, typically, if you need to run a task on a remote machine (super or not) you run it using a program called screen. https://www.gnu.org/software/screen/manual/screen.html > I would like to use all memory and run more than one program > simultaneously. > How can I do that? how GAP controls the amount of memory it can use, is described in the GAP manual. Again, it is not clear how your GAP in installed, whether the amount of RAM it can use is controlled by the supercomputer scheduler, etc... More details would be needed - perhaps you should talk to your supercomputer admins, too. If you use screen, then you can have many sessions of it running, with a GAP session in each of them. HTH, Dima From f.k.moftakhar at gmail.com Sat Apr 16 12:33:19 2016 From: f.k.moftakhar at gmail.com (fatemeh moftakhar) Date: Sat, 16 Apr 2016 16:03:19 +0430 Subject: [GAP Forum] Source code of GAP Message-ID: Dear Forum The command "Print(ff)" in GAP gives the code of the "ff" written in GAP. See for example the following command in GAP: gap> Print(Combinations); function ( arg ) local combs, mset; if Length( arg ) = 1 then mset := ShallowCopy( arg[1] ); Sort( mset ); combs := CombinationsA( mset, 1, Length( mset ), [ ], 1 ); elif Length( arg ) = 2 then mset := ShallowCopy( arg[1] ); Sort( mset ); combs := CombinationsK( mset, 1, Length( mset ), arg[2], [ ], 1 ); else Error( "usage: Combinations( [, ] )" ); fi; return combs; end If I run the command "Print(CharacterTable)", then we can see the following GAP message: gap> Print(CharacterTable); Is it possible to find the GAP code for CharacterTable? If no, is it possible to find the source of the program for calculating with primitive roots of unity in GAP? finally where is the library of primitive roots of unity? For example I need to source program of GAP for computing E(8)-E(8)^3? Best regards Fatemeh Moftakhar -- Regards; Miss Fatemeh Moftakhar PhD Candidate, Department of Pure Mathematics, Faculty of Mathematical Sciences, University of Kashan, Kashan, Iran From caj21 at st-andrews.ac.uk Sat Apr 16 13:59:05 2016 From: caj21 at st-andrews.ac.uk (Christopher Jefferson) Date: Sat, 16 Apr 2016 12:59:05 +0000 Subject: [GAP Forum] Source code of GAP In-Reply-To: References: Message-ID: For operations like CharacterTable, there are several implementations in GAP. To find out what is actually called, you can (often) use 'ApplicableMethod', which accepts a function and it's arguments (as a list) For example, Print(ApplicableMethod(CharacterTable, [AlternatingGroup(5)])) shows the function for finding this character table. Unfortunately, if we try the same for primitive root, we get: gap> Print(ApplicableMethod(PrimitiveRoot, [GF(3^5)])); function ( object ) <> end This tells us this code is not written in GAP, but instead in C. On 16/04/2016, 12:33, "forum-bounces at gap-system.org on behalf of fatemeh moftakhar" wrote: >Dear Forum > >The command "Print(ff)" in GAP gives the code of the "ff" written in GAP. >See for example the following command in GAP: >gap> Print(Combinations); >function ( arg ) > local combs, mset; > if Length( arg ) = 1 then > mset := ShallowCopy( arg[1] ); > Sort( mset ); > combs := CombinationsA( mset, 1, Length( mset ), [ ], 1 ); > elif Length( arg ) = 2 then > mset := ShallowCopy( arg[1] ); > Sort( mset ); > combs := CombinationsK( mset, 1, Length( mset ), arg[2], [ ], 1 ); > else > Error( "usage: Combinations( [, ] )" ); > fi; > return combs; >end > >If I run the command "Print(CharacterTable)", then we can see the following >GAP message: > >gap> Print(CharacterTable); > > >Is it possible to find the GAP code for CharacterTable? > >If no, is it possible to find the source of the program for calculating >with primitive roots of unity in GAP? > >finally where is the library of primitive roots of unity? For example I >need to source program of GAP for computing E(8)-E(8)^3? > >Best regards >Fatemeh Moftakhar > >-- >Regards; >Miss Fatemeh Moftakhar >PhD Candidate, >Department of Pure Mathematics, >Faculty of Mathematical Sciences, >University of Kashan, Kashan, Iran >_______________________________________________ >Forum mailing list >Forum at mail.gap-system.org >http://mail.gap-system.org/mailman/listinfo/forum From sam at Math.RWTH-Aachen.De Sun Apr 17 13:34:28 2016 From: sam at Math.RWTH-Aachen.De (Thomas Breuer) Date: Sun, 17 Apr 2016 14:34:28 +0200 Subject: [GAP Forum] Source code of GAP In-Reply-To: References: Message-ID: <20160417123428.GB9297@gemma.math.rwth-aachen.de> Dear GAP Forum, probably the question is about the GAP functions that compute the irreducible characters of a given group, not about the function 'CharacterTable'. If yes then one should look for the methods for the function 'Irr', for example using 'ApplicableMethod'. Concerning the question about calculations with primitive roots of unity, I would suggest to look at the chapters Cyclotomics'' and Abelian Number Fields'' of the GAP Reference Manual. Also the paper cited as [Bre97] in this manual might be of interest. All the best, Thomas On Sat, Apr 16, 2016 at 04:03:19PM +0430, fatemeh moftakhar wrote: > Dear Forum > > The command "Print(ff)" in GAP gives the code of the "ff" written in GAP. > See for example the following command in GAP: > gap> Print(Combinations); > function ( arg ) > local combs, mset; > if Length( arg ) = 1 then > mset := ShallowCopy( arg[1] ); > Sort( mset ); > combs := CombinationsA( mset, 1, Length( mset ), [ ], 1 ); > elif Length( arg ) = 2 then > mset := ShallowCopy( arg[1] ); > Sort( mset ); > combs := CombinationsK( mset, 1, Length( mset ), arg[2], [ ], 1 ); > else > Error( "usage: Combinations( [, ] )" ); > fi; > return combs; > end > > If I run the command "Print(CharacterTable)", then we can see the following > GAP message: > > gap> Print(CharacterTable); > > > Is it possible to find the GAP code for CharacterTable? > > If no, is it possible to find the source of the program for calculating > with primitive roots of unity in GAP? > > finally where is the library of primitive roots of unity? For example I > need to source program of GAP for computing E(8)-E(8)^3? > > Best regards > Fatemeh Moftakhar From f.k.moftakhar at gmail.com Sun Apr 17 19:33:45 2016 From: f.k.moftakhar at gmail.com (fatemeh moftakhar) Date: Sun, 17 Apr 2016 23:03:45 +0430 Subject: [GAP Forum] Source code of GAP In-Reply-To: <20160417123428.GB9297@gemma.math.rwth-aachen.de> References: <20160417123428.GB9297@gemma.math.rwth-aachen.de> Message-ID: Dear Professors Stefan Kohl, Chiristopher Jefferson and Thomas Breuer Thank you very much for your useful comments. Have a nice week ahead. Bset regards Fatemeh Moftakhar On Sun, Apr 17, 2016 at 5:04 PM, Thomas Breuer wrote: > Dear GAP Forum, > > probably the question is about the GAP functions that compute > the irreducible characters of a given group, > not about the function 'CharacterTable'. > > If yes then one should look for the methods for the function 'Irr', > for example using 'ApplicableMethod'. > > Concerning the question about calculations with primitive roots > of unity, I would suggest to look at the chapters Cyclotomics'' > and Abelian Number Fields'' of the GAP Reference Manual. > Also the paper cited as [Bre97] in this manual might be of interest. > > All the best, > Thomas > > > On Sat, Apr 16, 2016 at 04:03:19PM +0430, fatemeh moftakhar wrote: > > Dear Forum > > > > The command "Print(ff)" in GAP gives the code of the "ff" written in GAP. > > See for example the following command in GAP: > > gap> Print(Combinations); > > function ( arg ) > > local combs, mset; > > if Length( arg ) = 1 then > > mset := ShallowCopy( arg[1] ); > > Sort( mset ); > > combs := CombinationsA( mset, 1, Length( mset ), [ ], 1 ); > > elif Length( arg ) = 2 then > > mset := ShallowCopy( arg[1] ); > > Sort( mset ); > > combs := CombinationsK( mset, 1, Length( mset ), arg[2], [ ], 1 > ); > > else > > Error( "usage: Combinations( [, ] )" ); > > fi; > > return combs; > > end > > > > If I run the command "Print(CharacterTable)", then we can see the > following > > GAP message: > > > > gap> Print(CharacterTable); > > > > > > Is it possible to find the GAP code for CharacterTable? > > > > If no, is it possible to find the source of the program for calculating > > with primitive roots of unity in GAP? > > > > finally where is the library of primitive roots of unity? For example I > > need to source program of GAP for computing E(8)-E(8)^3? > > > > Best regards > > Fatemeh Moftakhar > > -- Regards; Miss Fatemeh Moftakhar PhD Candidate, Department of Pure Mathematics, Faculty of Mathematical Sciences, University of Kashan, Kashan, Iran From s.dikson2016 at gmail.com Tue Apr 19 07:56:19 2016 From: s.dikson2016 at gmail.com (Sara Dikson) Date: Tue, 19 Apr 2016 11:26:19 +0430 Subject: [GAP Forum] "SimpleGroup" command Message-ID: Dear Forum I used the "SimpleGroup("J4");" command for the Janko group. But Gap can't recognize it. Also I used "AtlasGroup("J4");" but it return "fail". I try it for "J3" but it return "fail".? errorg.png ? How can I access to all simple groups? I send the error as an attachment. Best regards Sara From dmitrii.pasechnik at cs.ox.ac.uk Tue Apr 19 09:35:47 2016 From: dmitrii.pasechnik at cs.ox.ac.uk (Dima Pasechnik) Date: Tue, 19 Apr 2016 09:35:47 +0100 Subject: [GAP Forum] "SimpleGroup" command In-Reply-To: References: Message-ID: <20160419083547.GA25303@dimpase.cs.ox.ac.uk> Dear Sarah, On Tue, Apr 19, 2016 at 11:26:19AM +0430, Sara Dikson wrote: > Dear Forum > I used the "SimpleGroup("J4");" command for the Janko group. > But Gap can't recognize it. Also I used "AtlasGroup("J4");" but it return > "fail". I try it for "J3" but it return "fail".? AtlasGroup("J3") and AtlasGroup("J4") should work, provided you have the necessary pre-requisits on. As far as I know it does not quite work on Windows, as it won't be able to get the data from the net. (Or it used to be the case in the past). Although you can download the data it needs manually. > errorg.png > you did not set the permissions on this link to be public, I cannot open it. HTH, Dmitrii From arashrafi at kashanu.ac.ir Tue Apr 19 09:43:44 2016 From: arashrafi at kashanu.ac.ir (arashrafi at kashanu.ac.ir) Date: Tue, 19 Apr 2016 13:13:44 +0430 (IRDT) Subject: [GAP Forum] "SimpleGroup" command In-Reply-To: <4646042.13146.1461054974790.JavaMail.root@mail.kashanu.ac.ir> Message-ID: <24089212.13148.1461055424897.JavaMail.root@mail.kashanu.ac.ir> Dear Sara, The character table of "J4" can be computed by the following command: gap> CharacterTable("j4"); For sporadic groups, it is possible to work with character tables (and no with groups other than Mathieu groups). But, a permutation or matrix representation for some sporadic groups are recorded in ATLAS of Finite Group Representations - Version 3 (http://brauer.maths.qmul.ac.uk/Atlas/v3/spor/). You can copy a generator set for "j4" from ATLAS and then paste into GAP. But you need a good computer. I hope this help you. Regards, Alireza ----- Original Message ----- From: "Sara Dikson" To: forum at gap-system.org Sent: Monday, April 18, 2016 11:56:19 PM Subject: [GAP Forum] "SimpleGroup" command Dear Forum I used the "SimpleGroup("J4");" command for the Janko group. But Gap can't recognize it. Also I used "AtlasGroup("J4");" but it return "fail". I try it for "J3" but it return "fail".? errorg.png ? How can I access to all simple groups? I send the error as an attachment. Best regards Sara _______________________________________________ Forum mailing list Forum at mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum From alexander.konovalov at st-andrews.ac.uk Tue Apr 19 10:12:12 2016 From: alexander.konovalov at st-andrews.ac.uk (Alexander Konovalov) Date: Tue, 19 Apr 2016 09:12:12 +0000 Subject: [GAP Forum] "SimpleGroup" command In-Reply-To: <20160419083547.GA25303@dimpase.cs.ox.ac.uk> References: <20160419083547.GA25303@dimpase.cs.ox.ac.uk> Message-ID: Dear Sara, Dima, No, this should work on Windows. It uses IO package to download data from the remote host, and GAP distribution for Windows includes IO binaries. Sara, which operating system do you actually have? This is where things may go wrong for AtlasRep: 1) IO package not compiled. Then AtlasRep may try to fallback to using wget for downloads, but if that is missing too, it will fail. 2) No internet connection, or some restrictive firewall settings. 3) No write permission in the directory where AtlasRep tries to write data. It will be useful to see the error message (and it is better to copy and paste the text into email instead of providing a screenshot) for the further advice. Hope this helps Alexander > On 19 Apr 2016, at 09:35, Dima Pasechnik wrote: > > Dear Sarah, > > On Tue, Apr 19, 2016 at 11:26:19AM +0430, Sara Dikson wrote: >> Dear Forum >> I used the "SimpleGroup("J4");" command for the Janko group. >> But Gap can't recognize it. Also I used "AtlasGroup("J4");" but it return >> "fail". I try it for "J3" but it return "fail".? > > AtlasGroup("J3") and AtlasGroup("J4") should work, provided you have the necessary pre-requisits > on. As far as I know it does not quite work on Windows, as it won't be able to get the data from the net. > (Or it used to be the case in the past). > Although you can download the data it needs manually. > > >> errorg.png >> > > you did not set the permissions on this link to be public, I cannot open it. > > HTH, > Dmitrii From z060822400814a at rezozer.net Tue Apr 19 12:53:43 2016 From: z060822400814a at rezozer.net (z060822400814a at rezozer.net) Date: Tue, 19 Apr 2016 12:53:43 +0100 Subject: [GAP Forum] gapdoc: BibTeX: url Message-ID: <57161C47.8090909@rezozer.net> Hello Forum: in a bib data file with which I am dealing, there is the entry @misc{Br, author = {A.~E. Brouwer}, title = {Bounds on the minimum distance of linear codes}, howpublished = {\url{http://www.win.tue.nl/~aeb/voorlincod.html}}, year = {1997?2006}, } When this bib data file is transformed to an XML file through the following commands bibdata := ParseBibFiles("doc/dum.bib");; WriteBibXMLextFile("doc/dum_bib.xml",bibdata), I get: A. E.Brouwer Bounds on the minimum distance of linear codes \url{http://www.win.tue.nl/~aeb/voorlincod.html} 1997?2006 Is it possible to get a more appropriate URL output in the XML output ? Thanks in advance, Jerome From dmitrii.pasechnik at cs.ox.ac.uk Tue Apr 19 13:25:34 2016 From: dmitrii.pasechnik at cs.ox.ac.uk (Dima Pasechnik) Date: Tue, 19 Apr 2016 13:25:34 +0100 Subject: [GAP Forum] gapdoc: BibTeX: url In-Reply-To: <57161C47.8090909@rezozer.net> References: <57161C47.8090909@rezozer.net> Message-ID: <20160419122534.GA30120@dimpase.cs.ox.ac.uk> On Tue, Apr 19, 2016 at 12:53:43PM +0100, z060822400814a at rezozer.net wrote: > Hello Forum: > > in a bib data file with which I am dealing, there is the entry > > @misc{Br, > author = {A.~E. Brouwer}, > title = {Bounds on the minimum distance of linear codes}, > howpublished = {\url{http://www.win.tue.nl/~aeb/voorlincod.html}}, > year = {1997?2006}, > } for what it's worth, this link is obsolete. These tables now live in http://www.codetables.de/ Just in case, Dima From z060822400814a at rezozer.net Tue Apr 19 13:42:29 2016 From: z060822400814a at rezozer.net (z060822400814a at rezozer.net) Date: Tue, 19 Apr 2016 13:42:29 +0100 Subject: [GAP Forum] gapdoc: BibTeX: url In-Reply-To: <20160419122534.GA30120@dimpase.cs.ox.ac.uk> References: <57161C47.8090909@rezozer.net> <20160419122534.GA30120@dimpase.cs.ox.ac.uk> Message-ID: <571627B5.80703@rezozer.net> Hi Dima, thanks for remark. On 19/04/16 13:25, Dima Pasechnik wrote: > On Tue, Apr 19, 2016 at 12:53:43PM +0100, z060822400814a at rezozer.net wrote: >> Hello Forum: >> >> in a bib data file with which I am dealing, there is the entry >> >> @misc{Br, >> author = {A.~E. Brouwer}, >> title = {Bounds on the minimum distance of linear codes}, >> howpublished = {\url{http://www.win.tue.nl/~aeb/voorlincod.html}}, >> year = {1997?2006}, >> } > for what it's worth, this link is obsolete. These tables now live in http://www.codetables.de/ > Indeed, as some part of the full bib data file: I am not maintaining the GAP package itself, but its Debian package: once my Debian is ready, I will submit patch to the upstream maintainer of the GAP package. Whatever, for now, I want to fix some documentation issue, what is my job; revisiting and refreshing is rather under the authority of the upstream maintainer. Best wishes, Jerome > Just in case, > Dima > From rafaelv at uaeh.edu.mx Tue Apr 19 22:06:57 2016 From: rafaelv at uaeh.edu.mx (Rafael Villarroel) Date: Tue, 19 Apr 2016 16:06:57 -0500 Subject: [GAP Forum] procedure to change a permutation, not returning it Message-ID: <87a8kp1f66.fsf@uaeh.edu.mx> Hello GAP forum, Is it possible to define a function that returns, say, an integer, but changes a permutation given as argument, as a side effect. Something like this: testperm := function(s) local i; i := 1; s := s^-1; Print(s,"\n"); return i; end; but that when running it like this... gap> a:=(1,2,3); (1,2,3) gap> testperm(a); (1,3,2) 1 gap> a; (1,2,3) ...actually changes the permutation a. Thanks in advance, Rafael From caj21 at st-andrews.ac.uk Tue Apr 19 22:14:28 2016 From: caj21 at st-andrews.ac.uk (Christopher Jefferson) Date: Tue, 19 Apr 2016 21:14:28 +0000 Subject: [GAP Forum] procedure to change a permutation, not returning it In-Reply-To: <87a8kp1f66.fsf@uaeh.edu.mx> References: <87a8kp1f66.fsf@uaeh.edu.mx> Message-ID: The short answer is no I'm afraid. The slightly longer answer is while you can't do this, you can do something similar if you pass in a list containing a permutation, or a record with a permutation member, and change that. For example: testperm := function(s) local i; i := 1; s[1] := s[1]^-1; Print(s,"\n"); return i; end; gap> a:=[(1,2,3)]; [ (1,2,3) ] gap> testperm(a); [ (1,3,2) ] 1 gap> a; [ (1,3,2) ] You might think, why can I change lists? The rule is you can change members of a list (or members of a record), but not the list itself. For example, if in our function we wrote ' s := [(3,2,1)] ', we would find that didn't change a, because now we are changing s, not a member of s. Chris On 19/04/2016, 22:06, "forum-bounces at gap-system.org on behalf of Rafael Villarroel" wrote: > >Hello GAP forum, > >Is it possible to define a function that returns, say, an integer, but >changes a permutation given as argument, as a side effect. > >Something like this: > >testperm := function(s) > local i; > i := 1; > s := s^-1; > Print(s,"\n"); > return i; >end; > >but that when running it like this... > >gap> a:=(1,2,3); >(1,2,3) >gap> testperm(a); >(1,3,2) >1 >gap> a; >(1,2,3) > >...actually changes the permutation a. > >Thanks in advance, >Rafael > > >_______________________________________________ >Forum mailing list >Forum at mail.gap-system.org >http://mail.gap-system.org/mailman/listinfo/forum From wclark at mail.usf.edu Thu Apr 28 20:01:13 2016 From: wclark at mail.usf.edu (W. Edwin Clark) Date: Thu, 28 Apr 2016 15:01:13 -0400 Subject: [GAP Forum] Checking equality of elements of words in a finitely presented group Message-ID: I am trying to use GAP to establish equality of two words in a finitely presented group and am having problems. First here is a successful test case: f:=FreeGroup(5);; rel:=[(f.(5)*f.(1))^5];; g:=f/rel;; u:=GeneratorsOfGroup(g);; (u[5]*u[1])^5 = Identity(g); #testing this identity which is actually a given relation works successfully true But if I add some more relators to rel, I cannot even get equality to the identity of the first relator. Here's what I tried: f:=FreeGroup(5);; rel:=[(f.(5)*f.(1))^5];; for i in [1..4] do Add(rel,(f.(i)*f.(i+1))*(f.(5)*f.(1))^(-1)); od; rel;; g:=f/rel;; u:=GeneratorsOfGroup(g); (u[5]*u[1])^5 = Identity(g); #I gave up waiting for an answer Why is this so difficult? What I really want in this group is to show that u[1]^5*u[2]*u[4]*u[1]*u[3]*u[5] = Identity(g) Any help would be appreciated. --Edwin Clark From D.F.Holt at warwick.ac.uk Fri Apr 29 14:01:19 2016 From: D.F.Holt at warwick.ac.uk (Derek Holt) Date: Fri, 29 Apr 2016 14:01:19 +0100 Subject: [GAP Forum] Checking equality of elements of words in a finitely presented group In-Reply-To: References: Message-ID: <20160429130119.GB5543@warwick.ac.uk> Dear Edwin, dear GAP Forum You can attempt to solve problems like this using the kbmag package (which needs to be compiled). I would guess that it is also possible to do so using the built-in GAP Knuth-Bendix rewriting system functionality, but I will leave someone else to help with that, since I am not so familiar with it. In this example, we are fortunate and, although the group is infinite, we get a finite confluent rewriting system (with 1734 rewrite rules), which we can use to solve the word problem. In most examples of infinite groups defined by a finite presentation, you do not get a finite confluent rewriting system (although you still may be able to find an automatic structure), but you can still use the rewrite rules to attempt to prove that eleemnts are equal to the identity. In this situation, you cannot in general prove that an element is not equal to the identity, but you can verify that it is. Here are the commands need for this calcualtion using kbmag. LoadPackage("kbmag"); f:=FreeGroup(5);; rel:=[(f.(5)*f.(1))^5];; for i in [1..4] do Add(rel,(f.(i)*f.(i+1))*(f.(5)*f.(1))^(-1)); od; g:=f/rel;; R:=KBMAGRewritingSystem(g);; MakeConfluent(R); #true F:=FreeStructureOfRewritingSystem(R);; u := GeneratorsOfGroup(F);; ReducedWord(R, (u[5]*u[1])^5); # ReducedWord(R, u[1]^5*u[2]*u[4]*u[1]*u[3]*u[5]); # Good luck! Derek Holt. On Thu, Apr 28, 2016 at 03:01:13PM -0400, W. Edwin Clark wrote: > I am trying to use GAP to establish equality of two words in a finitely > presented group and am having problems. > > First here is a successful test case: > > f:=FreeGroup(5);; > rel:=[(f.(5)*f.(1))^5];; > g:=f/rel;; > u:=GeneratorsOfGroup(g);; > (u[5]*u[1])^5 = Identity(g); #testing this identity which is actually a > given relation works successfully > > true > > But if I add some more relators to rel, I cannot even get > equality to the identity of the first relator. Here's what I tried: > > f:=FreeGroup(5);; > rel:=[(f.(5)*f.(1))^5];; > for i in [1..4] do > Add(rel,(f.(i)*f.(i+1))*(f.(5)*f.(1))^(-1)); > od; > rel;; > g:=f/rel;; > u:=GeneratorsOfGroup(g); > > (u[5]*u[1])^5 = Identity(g); #I gave up waiting for an answer > > Why is this so difficult? > > What I really want in this group is to show that > > u[1]^5*u[2]*u[4]*u[1]*u[3]*u[5] = Identity(g) > > Any help would be appreciated. > > --Edwin Clark > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From jdm3 at st-andrews.ac.uk Fri Apr 29 09:36:01 2016 From: jdm3 at st-andrews.ac.uk (James Mitchell) Date: Fri, 29 Apr 2016 08:36:01 +0000 Subject: [GAP Forum] Checking equality of elements of words in a finitely presented group In-Reply-To: <2db1c285781d4b0f875aac4da0f4b261@AM3PR06MB1362.eurprd06.prod.outlook.com> References: <2db1c285781d4b0f875aac4da0f4b261@AM3PR06MB1362.eurprd06.prod.outlook.com> Message-ID: Dear Edwin, Probably someone else will reply with a more elaborate answer, but you might want to consult: https://en.wikipedia.org/wiki/Word_problem_for_groups#Unsolvability_of_the_uniform_word_problem A very short answer is that the word problem for finitely presented groups is unsolvable, i.e. it is known that there is no single algorithm that can for every finitely presented group say whether two words represent the same element. Best wishes, James On Thu, 28 Apr 2016 at 20:03 W. Edwin Clark wrote: > I am trying to use GAP to establish equality of two words in a finitely > presented group and am having problems. > > First here is a successful test case: > > f:=FreeGroup(5);; > rel:=[(f.(5)*f.(1))^5];; > g:=f/rel;; > u:=GeneratorsOfGroup(g);; > (u[5]*u[1])^5 = Identity(g); #testing this identity which is actually a > given relation works successfully > > true > > But if I add some more relators to rel, I cannot even get > equality to the identity of the first relator. Here's what I tried: > > f:=FreeGroup(5);; > rel:=[(f.(5)*f.(1))^5];; > for i in [1..4] do > Add(rel,(f.(i)*f.(i+1))*(f.(5)*f.(1))^(-1)); > od; > rel;; > g:=f/rel;; > u:=GeneratorsOfGroup(g); > > (u[5]*u[1])^5 = Identity(g); #I gave up waiting for an answer > > Why is this so difficult? > > What I really want in this group is to show that > > u[1]^5*u[2]*u[4]*u[1]*u[3]*u[5] = Identity(g) > > Any help would be appreciated. > > --Edwin Clark > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum > From z060822400814a at rezozer.net Fri Apr 29 20:04:00 2016 From: z060822400814a at rezozer.net (z060822400814a at rezozer.net) Date: Fri, 29 Apr 2016 19:04:00 +0000 Subject: [GAP Forum] TestPackage and arbitrary message Message-ID: <5723B020.4010400@rezozer.net> Hello Forum: I am looking for a GAP way to perform GAP package installation tests with reproducible output (as in Test files, section 7.9 in the Reference Manual): TestPackage is certainly what I am looking for. Unfortunately, TestPackage outputs arbitrary messages (or not) when the tests succeed, something as: GRAPE package: testall.tst GAP4stones: 6 where the stones number is arbitrary (as CPU time can be) --- and sometime, there is no message at all. In practice, I place the TestPackage in a Test file (see above) and I launch something as echo "QUIT_GAP(Test(\"d/t/makecheck.tst\"));" | gap -A -q -T Is there a way to silence TestPackage in case of success ? For now, I use TestDirectory : echo "QUIT_GAP(TestDirectory(\"d/t\"));" | gap -A -q -T which seems to be blind to these messages. Thanks, Jerome From caj21 at st-andrews.ac.uk Wed May 4 20:15:36 2016 From: caj21 at st-andrews.ac.uk (Christopher Jefferson) Date: Wed, 4 May 2016 19:15:36 +0000 Subject: [GAP Forum] TestPackage and arbitrary message In-Reply-To: <5723B020.4010400@rezozer.net> References: <5723B020.4010400@rezozer.net> Message-ID: Not at the moment, but this is an area where I think we need to improve our output. I would suggest writing a bug report on github.com/gap-system/gap , with a suggested new output. You could even work on a patch if you like. Chris On 29/04/2016, 20:04, "forum-bounces at gap-system.org on behalf of z060822400814a at rezozer.net" wrote: >Hello Forum: > >I am looking for a GAP way to perform GAP package installation tests with reproducible output >(as in Test files, section 7.9 in the Reference Manual): TestPackage is certainly what >I am looking for. Unfortunately, TestPackage outputs arbitrary messages (or not) when >the tests succeed, something as: > > GRAPE package: testall.tst > GAP4stones: 6 > >where the stones number is arbitrary (as CPU time can be) --- >and sometime, there is no message at all. In practice, I place the TestPackage >in a Test file (see above) and I launch something as > > echo "QUIT_GAP(Test(\"d/t/makecheck.tst\"));" | gap -A -q -T > >Is there a way to silence TestPackage in case of success ? > >For now, I use TestDirectory : > > echo "QUIT_GAP(TestDirectory(\"d/t\"));" | gap -A -q -T > >which seems to be blind to these messages. > >Thanks, >Jerome > > >_______________________________________________ >Forum mailing list >Forum at mail.gap-system.org >http://mail.gap-system.org/mailman/listinfo/forum From f.alibabaee at gmail.com Tue May 10 14:39:12 2016 From: f.alibabaee at gmail.com (fahime babaee) Date: Tue, 10 May 2016 14:39:12 +0100 Subject: [GAP Forum] Question Message-ID: Dear Forum Let$H$and$K$be finitely generated subgroups of the free group$F(n)$such that$H\subseteq K$. We Know that$K$is a free group. Now we choose basis for$K$and$H$and I want to rewrite the basis of$H$in the chosen basis of$K$. I can do this calculation with hand by using automaton of$K$but it take a long time. How can I do this with GAP? Fahime Alibabaei PhD student of Pure Mathematics(Algebra) University of Porto. From alexander.konovalov at st-andrews.ac.uk Wed May 11 21:45:06 2016 From: alexander.konovalov at st-andrews.ac.uk (Alexander Konovalov) Date: Wed, 11 May 2016 20:45:06 +0000 Subject: [GAP Forum] Question In-Reply-To: References: Message-ID: FYI: cross-posted yesterday and answered today at http://math.stackexchange.com/questions/1779212/ Alexander > On 10 May 2016, at 14:39, fahime babaee wrote: > > Dear Forum > > Let$H$and$K$be finitely generated subgroups of the free group$F(n)$> such that$H\subseteq K$. We Know that$K$is a free group. Now we choose > basis for$K$and$H$and I want to rewrite the basis of$H$in the chosen > basis of$K$. I can do this calculation with hand by using automaton of$K$> but it take a long time. How can I do this with GAP? > > Fahime Alibabaei > PhD student of Pure Mathematics(Algebra) University of Porto. From marek at mitros.org Fri May 13 15:30:49 2016 From: marek at mitros.org (Marek Mitros) Date: Fri, 13 May 2016 16:30:49 +0200 Subject: [GAP Forum] Square root of cyclotomic real number Message-ID: Dear All, I am doing following calculation and I need square root of cyclotomic number being real number. Here is example below. Is there any workaround for this ? What I am doing I obtained vector v of complex number which not unit length. In order to normalize it I perform following and I receive the error. Do I need more knowledge about cyclotomic numbers ? gap> v1:=v/Sqrt(v*ComplexConjugate(v)); Error, no method found! For debugging hints type ?Recovery from NoMethodFound Error, no 1st choice method found for Sqrt' on 1 arguments called from ( ) (...) gap> n:=E(20)-E(20)^9; E(20)-E(20)^9 gap> RealPart(n); E(20)-E(20)^9 gap> Sqrt(n); Error, no method found! For debugging hints type ?Recovery from NoMethodFound Error, no 1st choice method found for Sqrt' on 1 arguments called from Regards, Marek From hulpke at fastmail.fm Fri May 13 17:15:52 2016 From: hulpke at fastmail.fm (Alexander Hulpke) Date: Fri, 13 May 2016 10:15:52 -0600 Subject: [GAP Forum] Square root of cyclotomic real number In-Reply-To: References: Message-ID: Dear Forum, Dear Marek Mitros, Sqrt (as a short hand for the ER cyclotomic) will only work for rational numbers. If you wanted a square root of a cyclotomic number you would need to factor x^2-a over a suitable cyclotomic field. Best, Alexander Hulpke > On May 13, 2016, at 8:30 AM, Marek Mitros wrote: > > Dear All, > > I am doing following calculation and I need square root of cyclotomic > number being real number. Here is example below. Is there any workaround > for this ? > What I am doing I obtained vector v of complex number which not unit > length. In order to normalize it I perform following and I receive the > error. Do I need more knowledge about cyclotomic numbers ? > > gap> v1:=v/Sqrt(v*ComplexConjugate(v)); > > Error, no method found! For debugging hints type ?Recovery from > NoMethodFound > > Error, no 1st choice method found for Sqrt' on 1 arguments called from > > ( ) > > (...) > > gap> n:=E(20)-E(20)^9; > > E(20)-E(20)^9 > > gap> RealPart(n); > > E(20)-E(20)^9 > > gap> Sqrt(n); > > Error, no method found! For debugging hints type ?Recovery from > NoMethodFound > > Error, no 1st choice method found for Sqrt' on 1 arguments called from > > > Regards, > > Marek > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From saeidi.amin at gmail.com Sat May 14 16:25:53 2016 From: saeidi.amin at gmail.com (Amin Saeidi) Date: Sat, 14 May 2016 17:25:53 +0200 Subject: [GAP Forum] Character Table of semigroups Message-ID: Dear all, I am trying to find the character tables of some finitely presented semigroups, say for example: . It is an inverse semigroup of order 14. I can Define this semigroup in GAP 4.8.3 but I fail to find the character table: *gap> IsInverseSemigroup(S);* *true* *gap> CharacterTableOfInverseSemigroup(S);* *Error, no method found! For debugging hints type ?Recovery from NoMethodFound* *Error, no 1st choice method found for CharacterTableOfInverseSemigroup' on 1 arguments called from* *( )* * called from read-eval loop at line 87 of *stdin** *you can 'quit;' to quit to outer loop, or* *you can 'return;' to continue* I thought this command may only work for monoids, so I considered a monoid of order 15 but it didn't work either. However it works for some other monoids. For instance the command gives the character table of S if I write *S:=SymmetricInverseMonoid(3); * I will be grateful to let me know if it is a bug or I am doing something wrong. Best regards, Amin From marek at mitros.org Mon May 16 20:34:14 2016 From: marek at mitros.org (Marek Mitros) Date: Mon, 16 May 2016 21:34:14 +0200 Subject: [GAP Forum] Square root of cyclotomic real number In-Reply-To: References: Message-ID: Thank you for the answer. It is a pity, that I cannot obtain unitary matrix from the set of orthogonal columns. I must think of another method. Regards, Marek 13-05-2016 18:08 u?ytkownik "Alexander Hulpke" napisa?: > Dear Forum, Dear Marek Mitros, > > Sqrt (as a short hand for the ER cyclotomic) will only work for rational > numbers. If you wanted a square root of a cyclotomic number you would need > to factor x^2-a over a suitable cyclotomic field. > > Best, > > Alexander Hulpke > > > On May 13, 2016, at 8:30 AM, Marek Mitros wrote: > > > > Dear All, > > > > I am doing following calculation and I need square root of cyclotomic > > number being real number. Here is example below. Is there any workaround > > for this ? > > What I am doing I obtained vector v of complex number which not unit > > length. In order to normalize it I perform following and I receive the > > error. Do I need more knowledge about cyclotomic numbers ? > > > > gap> v1:=v/Sqrt(v*ComplexConjugate(v)); > > > > Error, no method found! For debugging hints type ?Recovery from > > NoMethodFound > > > > Error, no 1st choice method found for Sqrt' on 1 arguments called from > > > > ( ) > > > > (...) > > > > gap> n:=E(20)-E(20)^9; > > > > E(20)-E(20)^9 > > > > gap> RealPart(n); > > > > E(20)-E(20)^9 > > > > gap> Sqrt(n); > > > > Error, no method found! For debugging hints type ?Recovery from > > NoMethodFound > > > > Error, no 1st choice method found for Sqrt' on 1 arguments called from > > > > > > Regards, > > > > Marek > > _______________________________________________ > > Forum mailing list > > Forum at mail.gap-system.org > > http://mail.gap-system.org/mailman/listinfo/forum > > From dmitrii.pasechnik at cs.ox.ac.uk Tue May 17 12:41:13 2016 From: dmitrii.pasechnik at cs.ox.ac.uk (Dmitrii Pasechnik) Date: Tue, 17 May 2016 12:41:13 +0100 Subject: [GAP Forum] Square root of cyclotomic real number In-Reply-To: References: Message-ID: <20160517114113.GB18282@cs.ox.ac.uk> Dear Marek, Just in case, Sage(math) has no problem with computing these square roots. sage: n=libgap.eval('E(20)-E(20)^9').sage() sage: s=n.sqrt(); s # ugly, unsimplified sqrt(-(1/4*I*sqrt(5) + 1/4*sqrt(2*sqrt(5) + 10) - 1/4*I)^7 + (1/4*I*sqrt(5) + 1/4*sqrt(2*sqrt(5) + 10) - 1/4*I)^5 - (1/4*I*sqrt(5) + 1/4*sqrt(2*sqrt(5) + 10) - 1/4*I)^3 + 1/2*I*sqrt(5) + 1/2*sqrt(2*sqrt(5) + 10) - 1/2*I) sage: n.n() # looking at numerical approximations 1.90211303259031 + 1.11022302462516e-16*I sage: (s^2).n() # looking at numerical approximations 1.90211303259031 - 2.22044604925031e-16*I One way or another, you need functionality to construct and to compute in quadratic extensions of cyclotomic fields, and this might be slow, unless someone programs a way to do the square root in the cyclotomic field (my number theory is not good, but IIRC you would always be able to work in cyclotomics, perhaps of bigger degree). HTH, Dima On Mon, May 16, 2016 at 09:34:14PM +0200, Marek Mitros wrote: > Thank you for the answer. It is a pity, that I cannot obtain unitary matrix > from the set of orthogonal columns. > I must think of another method. > > Regards, > Marek > 13-05-2016 18:08 u?ytkownik "Alexander Hulpke" napisa?: > > > Dear Forum, Dear Marek Mitros, > > > > Sqrt (as a short hand for the ER cyclotomic) will only work for rational > > numbers. If you wanted a square root of a cyclotomic number you would need > > to factor x^2-a over a suitable cyclotomic field. > > > > Best, > > > > Alexander Hulpke > > > > > On May 13, 2016, at 8:30 AM, Marek Mitros wrote: > > > > > > Dear All, > > > > > > I am doing following calculation and I need square root of cyclotomic > > > number being real number. Here is example below. Is there any workaround > > > for this ? > > > What I am doing I obtained vector v of complex number which not unit > > > length. In order to normalize it I perform following and I receive the > > > error. Do I need more knowledge about cyclotomic numbers ? > > > > > > gap> v1:=v/Sqrt(v*ComplexConjugate(v)); > > > > > > Error, no method found! For debugging hints type ?Recovery from > > > NoMethodFound > > > > > > Error, no 1st choice method found for Sqrt' on 1 arguments called from > > > > > > ( ) > > > > > > (...) > > > > > > gap> n:=E(20)-E(20)^9; > > > > > > E(20)-E(20)^9 > > > > > > gap> RealPart(n); > > > > > > E(20)-E(20)^9 > > > > > > gap> Sqrt(n); > > > > > > Error, no method found! For debugging hints type ?Recovery from > > > NoMethodFound > > > > > > Error, no 1st choice method found for Sqrt' on 1 arguments called from > > > > > > > > > Regards, > > > > > > Marek > > > _______________________________________________ > > > Forum mailing list > > > Forum at mail.gap-system.org > > > http://mail.gap-system.org/mailman/listinfo/forum > > > > > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From dsxoiniana at gmail.com Tue May 17 14:04:44 2016 From: dsxoiniana at gmail.com (Dimitris Schinianakis) Date: Tue, 17 May 2016 15:04:44 +0200 Subject: [GAP Forum] Cayley diagrams for general groups Message-ID: Dear all, I'm a beginner in group theory, currently trying to solve all the material in the "A Book of abstract Algebra", by Pinter. I'd like to use Sage to verify some of the solutions but also to experiment a little bit with Cayley Diagrams. 1) Is there a way to draw a Cayley diagram in Sage by defining the group and the equations for the generators? For example let G={e,a,b,b^2,b^3,ab,ab^2,ab^3} and the generators satisfy a^2=e, b^4=e, ba=ab^2. e is of course the identity element. 2) I have a Cayley diagram with no labels, only connections and points are depicted. I need to create the group table. How can I be sure that my table is correct? Does a cayley diagram correspond always to only one group? Regards Dimitris From hulpke at fastmail.fm Tue May 17 15:59:25 2016 From: hulpke at fastmail.fm (Alexander Hulpke) Date: Tue, 17 May 2016 08:59:25 -0600 Subject: [GAP Forum] Square root of cyclotomic real number In-Reply-To: <20160517114113.GB18282@cs.ox.ac.uk> References: <20160517114113.GB18282@cs.ox.ac.uk> Message-ID: <4779F1C7-113F-4A7C-84C6-E3EBDB62ED3F@fastmail.fm> Dear Forum, Dear Dima, > One way or another, you need functionality to construct and to compute > in quadratic extensions of cyclotomic fields, and this might be slow, > unless someone programs a way to do the square root in the cyclotomic > field (my number theory is not good, but IIRC you would always be able > to work in cyclotomics, perhaps of bigger degree). Alas no, and that?s the whole difficulty. The number you construct (as root of n=E(20)-E(20)^9) > sqrt(-(1/4*I*sqrt(5) + 1/4*sqrt(2*sqrt(5) + 10) - 1/4*I)^7 + > (1/4*I*sqrt(5) + 1/4*sqrt(2*sqrt(5) + 10) - 1/4*I)^5 - (1/4*I*sqrt(5) + > 1/4*sqrt(2*sqrt(5) + 10) - 1/4*I)^3 + 1/2*I*sqrt(5) + 1/2*sqrt(2*sqrt(5) > + 10) - 1/2*I) has minimal polynomial x^8-5*x^4+5 with a nonabelian Galois group. By Kronecker-Weber, n has no square root in a cyclotomic field. (Indeed, if every square root of a cyclotomic number was cyclotomic, all 2-groups would be abelian.) While generic theory will give you a way to express such square roots in radicals, trying to do sensible arithmetic with such iterated radical expressions is rather delicate and will run into surprising problems concerning equality of different expressions (Keyword: Radial Denesting). I have not seen any method to perform such arithmetic universally, reasonably fast, and not prone to potential problems once it comes to issues such as branch cuts. Best, Alexander From wdjoyner at gmail.com Wed May 18 11:41:52 2016 From: wdjoyner at gmail.com (David Joyner) Date: Wed, 18 May 2016 06:41:52 -0400 Subject: [GAP Forum] Cayley diagrams for general groups In-Reply-To: References: Message-ID: On Tue, May 17, 2016 at 9:04 AM, Dimitris Schinianakis wrote: > Dear all, > > I'm a beginner in group theory, currently trying to solve all the material > in the "A Book of abstract Algebra", by Pinter. > > I'd like to use Sage to verify some of the solutions but also to experiment > a little bit with Cayley Diagrams. > I assume by Cayley diagram you mean Cayley graph, https://en.wikipedia.org/wiki/Cayley_graph > 1) Is there a way to draw a Cayley diagram in Sage by defining the group > and the equations for the generators? For example let > G={e,a,b,b^2,b^3,ab,ab^2,ab^3} and the generators satisfy a^2=e, b^4=e, > ba=ab^2. e is of course the identity element. > Not currently in Sage. However, in GAP, please check out the Grape and Digraphs packages at http://www.gap-system.org/Packages/packages.html The example below shows that your problem can be done in GAP using Grape. Please see the Grape manual for more options and examples. gap> f := FreeGroup( "a", "b" ); gap> g := f / [ f.1^2, f.2^3, (f.1*f.2)^5 ]; gap> CayleyGraph(g); rec( adjacencies := [ [ 5, 7, 13 ] ], group := , isGraph := true, isSimple := true, names := [ , b*a*(b*a*b)^2, a*b*a*(b*a*b)^2*a, b*a*(b*a*b)^2*b, b, b*(b*a)^2*b^2*a, b^2, (b*a)^2*b^2*a, b^2*a*(b*a*b)^2, a*(b*a*b)^2, a*(b*a*b)^2*b, (a*b)^2*b*a, a, a*b*a*(b*a*b)^2, b*a*(b*a*b)^2*a, (b*a*b)^2*a*b, b*(b*a)^2*b^2, b*a, a*(b*a*b)^2*a, a*b*(b*a)^2*b, (a*b)^2*b, b^2*a, (b*a)^2*b^2, (b*(b*a)^2)^2, a*b*a*(b*a*b)^2*b, a*b, a*b*(b*a)^2*b^2*a, a*(b*a*b)^2*a*b, a*b*(b*a)^2*b^2, a*b*a, (b*a*b)^2*a*b^2, b*(b*a)^2, b*a*b, b^2*a*b, (b*a*b^2*a)^2, (b*a)^2, a*b^2, (a*b)^3*b*a, (a*b^2*a*b)^2, (b*a*b)^2*a, b*(b*a)^2*b, b*a*b^2, ((a*b)^2*b)^2, a*b*(b*a)^2, (a*b)^2, b^2*a*b^2, (b*a)^2*b, (b*a*b^2*a)^2*b, (b*a*b)^2, (b*a*b)^2*b, b*a*b^2*a, a*b^2*a, (a*b)^3*b, (a*b^2*a*b)^2*a, a*b^2*a*b, ((a*b)^2*b)^2*a, (a*b)^2*a, (a*b^2)^2, (a*b)^3, ((a*b)^2*b)^2*a*b ], order := 60, representatives := [ 1 ], schreierVector := [ -1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1, 1, 2, 1, 2, 1, 1, 2, 2, 2 ] ) > 2) I have a Cayley diagram with no labels, only connections and points are > depicted. I need to create the group table. How can I be sure that my table > is correct? Does a cayley diagram correspond always to only one group? > No, the Cayley graph does not correspond to only one group. In fact, one can select symmetric generators for ZZ/6ZZ (an abelian group) and S_3 (a non-abelian group) such that the associated Cayley graphs are isomorphic. By the way, while the GAP Forum is a place for questions and comments of general interest about GAP, the SageMath system has it's own support resources, including the SageMath support mailing list: http://www.sagemath.org/help-groups.html" See also the related stackexchange post: http://math.stackexchange.com/questions/1789007/plot-cayley-graphs-for-generic-element-groups Hope this helps! > Regards > Dimitris > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From samuel.lelievre at gmail.com Thu May 19 03:30:01 2016 From: samuel.lelievre at gmail.com (=?UTF-8?Q?Samuel_Leli=C3=A8vre?=) Date: Thu, 19 May 2016 04:30:01 +0200 Subject: [GAP Forum] Square root of cyclotomic real number In-Reply-To: <4779F1C7-113F-4A7C-84C6-E3EBDB62ED3F@fastmail.fm> References: <20160517114113.GB18282@cs.ox.ac.uk> <4779F1C7-113F-4A7C-84C6-E3EBDB62ED3F@fastmail.fm> Message-ID: Dear forum, To complement Dima's answer, if you are using Sage, you should be aware that Sage provides two useful exact fields: - QQbar, the algebraic closure of QQ, - AA, the "algebraic real field", which is the intersection of QQbar with the reals. I would advise you to work in AA for your problem. Defining, as Dima suggested, sage: a = libgap.eval('E(20)-E(20)^9').sage() you get an element in a cyclotomic field: sage: a -zeta20^7 + zeta20^5 - zeta20^3 + 2*zeta20 sage: a.parent() Cyclotomic Field of order 20 and degree 8 Now if you take its square root naively, you end up in Sage's "Symbolic Ring" which is something to avoid in general. sage: b = a.sqrt() sage: b sqrt(-(1/4*I*sqrt(5) + 1/4*sqrt(2*sqrt(5) + 10) - 1/4*I)^7 + (1/4*I*sqrt(5) + 1/4*sqrt(2*sqrt(5) + 10) - 1/4*I)^5 - (1/4*I*sqrt(5) + 1/4*sqrt(2*sqrt(5) + 10) - 1/4*I)^3 + 1/2*I*sqrt(5) + 1/2*sqrt(2*sqrt(5) + 10) - 1/2*I) sage: b.parent() Symbolic Ring The best is to work with AA sage: AA Algebraic Real Field by moving there explicitly sage: aa = AA(a) sage: aa 1.902113032590308? sage: aa.parent() Algebraic Real Field Elements of AA are exact, and well-defined: sage: aa.minpoly() x^4 - 5*x^2 + 5 Taking square roots stays in AA sage: bb = aa.sqrt() sage: bb 1.379171139703231? sage: bb.parent() Algebraic Real Field sage: bb.minpoly() x^8 - 5*x^4 + 5 I hope this helps. Samuel From marek at mitros.org Thu May 19 06:48:05 2016 From: marek at mitros.org (Marek Mitros) Date: Thu, 19 May 2016 07:48:05 +0200 Subject: [GAP Forum] Square root of cyclotomic real number In-Reply-To: References: <20160517114113.GB18282@cs.ox.ac.uk> <4779F1C7-113F-4A7C-84C6-E3EBDB62ED3F@fastmail.fm> Message-ID: Thank you for advice. Let me explain my plan, then we can judge whether this advice work for me or it doesn't. I have group of size 5120 of complex matrices of size 16x16. In this group I listed two conjugacy classes of involutions of size 10 and 80. Next I wanted to change the basis in a way that mentioned 10 matrices are diagonal. Therefore I obtained 16 vectors as intersections of eigenspaces. Unfortunately these vectors are not unit length, so I tried to normalize them. This doesn't work in GAP, because complex numbers are cyclotomics there. Then my question appeared on GAP forum. If I could normalize those vectors, then obtained matrix A built with orthonormal hermitian vectors is unitary and I can apply x-> A^-1 x A to group elements. Can this plan be done in GAP or Sage ? Mentioned group is index 2 in involution stabilizer of Tits group ^2F4(2)'. Regards, Marek 19-05-2016 04:30 u?ytkownik "Samuel Leli?vre" napisa?: > Dear forum, > > To complement Dima's answer, if you are using Sage, > you should be aware that Sage provides two useful > exact fields: > - QQbar, the algebraic closure of QQ, > - AA, the "algebraic real field", which is the intersection > of QQbar with the reals. > > I would advise you to work in AA for your problem. > > Defining, as Dima suggested, > > sage: a = libgap.eval('E(20)-E(20)^9').sage() > > you get an element in a cyclotomic field: > > sage: a > -zeta20^7 + zeta20^5 - zeta20^3 + 2*zeta20 > sage: a.parent() > Cyclotomic Field of order 20 and degree 8 > > Now if you take its square root naively, you end up > in Sage's "Symbolic Ring" which is something to avoid > in general. > > sage: b = a.sqrt() > sage: b > sqrt(-(1/4*I*sqrt(5) + 1/4*sqrt(2*sqrt(5) + 10) - 1/4*I)^7 > + (1/4*I*sqrt(5) + 1/4*sqrt(2*sqrt(5) + 10) - 1/4*I)^5 > - (1/4*I*sqrt(5) + 1/4*sqrt(2*sqrt(5) + 10) - 1/4*I)^3 > + 1/2*I*sqrt(5) + 1/2*sqrt(2*sqrt(5) + 10) - 1/2*I) > sage: b.parent() > Symbolic Ring > > The best is to work with AA > > sage: AA > Algebraic Real Field > > by moving there explicitly > > sage: aa = AA(a) > sage: aa > 1.902113032590308? > sage: aa.parent() > Algebraic Real Field > > Elements of AA are exact, and well-defined: > > sage: aa.minpoly() > x^4 - 5*x^2 + 5 > > Taking square roots stays in AA > > sage: bb = aa.sqrt() > sage: bb > 1.379171139703231? > sage: bb.parent() > Algebraic Real Field > sage: bb.minpoly() > x^8 - 5*x^4 + 5 > > I hope this helps. > Samuel > From dmitrii.pasechnik at cs.ox.ac.uk Thu May 19 10:32:21 2016 From: dmitrii.pasechnik at cs.ox.ac.uk (Dmitrii Pasechnik) Date: Thu, 19 May 2016 10:32:21 +0100 Subject: [GAP Forum] Square root of cyclotomic real number In-Reply-To: References: <20160517114113.GB18282@cs.ox.ac.uk> <4779F1C7-113F-4A7C-84C6-E3EBDB62ED3F@fastmail.fm> Message-ID: <20160519093221.GA1757@dimpase.cs.ox.ac.uk> Dear Marek, On Thu, May 19, 2016 at 07:48:05AM +0200, Marek Mitros wrote: > Thank you for advice. Let me explain my plan, then we can judge whether > this advice work for me or it doesn't. > > I have group of size 5120 of complex matrices of size 16x16. In this group > I listed two conjugacy classes of involutions of size 10 and 80. Next I > wanted to change the basis in a way that mentioned 10 matrices are > diagonal. Therefore I obtained 16 vectors as intersections of eigenspaces. Another way to get a diagonalising transformation H would be to sum up g^*g, with g running through your size 10 conjugacy class T (and ^* denotes taking conjugate transpose). H is the matrix of a positive definite Hermitean form stabilised by the subgroup generated by T. Then you'd need to factor H=UU^* (A Cholesky factorisation), and conjugate the group by U. This is basically the standard way to show that every representation of a finite groups is unitary). Computing U needs taking square roots. However, there is also a decomposition of the form H=VDV^*, with V being lower-triangular with 1s on the diagonal, and D a diaginal matrix; such a decomposition does not need square roots, IIRC. To pass between U and V you'd need to take square roots of the elements of D. But you can conjugate your group by V, still getting a representation hopefully close to the form you want to see. (Computing V and D can be done in GAP, although you probably would need to program it yourself). > Unfortunately these vectors are not unit length, so I tried to normalize > them. This doesn't work in GAP, because complex numbers are cyclotomics > there. Then my question appeared on GAP forum. If I could normalize those > vectors, then obtained matrix A built with orthonormal hermitian vectors is > unitary and I can apply x-> A^-1 x A to group elements. using Sage, and the data type AA, as proposed by Samuel, one would see quickly whether building A is feasible in the sense that you do not get to deal with algebraic numbers of huge degree. (If you go the way I described above, you'd get to compute the Cholesky decomposition of H, something that is implemented in Sage). Converting a Sage matrix of algebraic numbers to a GAP one isn't implemented well, but should be doable. HTH, Dima > > Can this plan be done in GAP or Sage ? > > Mentioned group is index 2 in involution stabilizer of Tits group ^2F4(2)'. > > Regards, > Marek > 19-05-2016 04:30 u?ytkownik "Samuel Leli?vre" > napisa?: > > > Dear forum, > > > > To complement Dima's answer, if you are using Sage, > > you should be aware that Sage provides two useful > > exact fields: > > - QQbar, the algebraic closure of QQ, > > - AA, the "algebraic real field", which is the intersection > > of QQbar with the reals. > > > > I would advise you to work in AA for your problem. > > > > Defining, as Dima suggested, > > > > sage: a = libgap.eval('E(20)-E(20)^9').sage() > > > > you get an element in a cyclotomic field: > > > > sage: a > > -zeta20^7 + zeta20^5 - zeta20^3 + 2*zeta20 > > sage: a.parent() > > Cyclotomic Field of order 20 and degree 8 > > > > Now if you take its square root naively, you end up > > in Sage's "Symbolic Ring" which is something to avoid > > in general. > > > > sage: b = a.sqrt() > > sage: b > > sqrt(-(1/4*I*sqrt(5) + 1/4*sqrt(2*sqrt(5) + 10) - 1/4*I)^7 > > + (1/4*I*sqrt(5) + 1/4*sqrt(2*sqrt(5) + 10) - 1/4*I)^5 > > - (1/4*I*sqrt(5) + 1/4*sqrt(2*sqrt(5) + 10) - 1/4*I)^3 > > + 1/2*I*sqrt(5) + 1/2*sqrt(2*sqrt(5) + 10) - 1/2*I) > > sage: b.parent() > > Symbolic Ring > > > > The best is to work with AA > > > > sage: AA > > Algebraic Real Field > > > > by moving there explicitly > > > > sage: aa = AA(a) > > sage: aa > > 1.902113032590308? > > sage: aa.parent() > > Algebraic Real Field > > > > Elements of AA are exact, and well-defined: > > > > sage: aa.minpoly() > > x^4 - 5*x^2 + 5 > > > > Taking square roots stays in AA > > > > sage: bb = aa.sqrt() > > sage: bb > > 1.379171139703231? > > sage: bb.parent() > > Algebraic Real Field > > sage: bb.minpoly() > > x^8 - 5*x^4 + 5 > > > > I hope this helps. > > Samuel > > From primoz.moravec at fmf.uni-lj.si Thu May 19 11:52:48 2016 From: primoz.moravec at fmf.uni-lj.si (Primoz Moravec) Date: Thu, 19 May 2016 12:52:48 +0200 Subject: [GAP Forum] PhD position in Ljubljana Message-ID: <573D9B00.2010607@fmf.uni-lj.si> Dear Colleagues, could you pass this information on to potential PhD students who may consider applying to University of Ljubljana: The research program ?Algebra in Operator Theory and Financial Mathematics? at the Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia, is offering funding for a student to undertake PhD research in one of the following areas: Functional Analysis Operator Theory Linear Algebra Ring Theory Semigroup Theory Universal Algebra Algebraic Geometry Group Theory Probability and Statistics Financial Mathematics See http://www.sicris.si/public/jqm/prg.aspx?lang=eng&opdescr=search&opt=2&subopt=700&code1=prg&code2=auto&psize=10&hits=13&page=1&count=1&search_term=algebra&id=9652&slng=eng&order_by= for more details about our research interests and members of the program. We are looking for a highly motivated research student with an interest in one of the above areas. The studentship offers costs of fees for taking PhD courses at the Department of Mathematics, University of Ljubljana, and an annual stipend of about 16.320 EUR per year for 4 years. It might also be possible to fund non-EU students on an equivalent basis, so students of any nationality are encouraged to apply. Students should normally have or expect at least Masters degree in Mathematics or a related discipline by 16 September 2016. The expected start of funding is 1 October 2016. Candidates should send their cover letter, CV and academic record to primoz.moravec at fmf.uni-lj.si They should also have two recommendation letters sent to the above email address. The closing date for applications is 30th June 2016. Best wishes, Primoz. From tvonp at gmx.net Thu May 19 13:37:47 2016 From: tvonp at gmx.net (Timm von Puttkamer) Date: Thu, 19 May 2016 14:37:47 +0200 Subject: [GAP Forum] Small Cancellation Conditions Message-ID: <573DB39B.6020709@gmx.net> Dear all, Has anybody written some GAP code which decides whether a group presentation satisfies one of the common small cancellation conditions? (for example, the C'(\lambda) condition) Best, Timm From s.dikson2016 at gmail.com Thu May 19 15:44:10 2016 From: s.dikson2016 at gmail.com (Sara Dikson) Date: Thu, 19 May 2016 19:14:10 +0430 Subject: [GAP Forum] Parallel Gap Message-ID: Dear Forum, 1) Is it possible I open several terminals in ubuntu and use gap processes simultaneous? 2) I would like use "Parallel Gap" but I don't have much information about it. I would be grateful if you help me. Best regard Sara Dikson From max at quendi.de Thu May 19 17:37:25 2016 From: max at quendi.de (Max Horn) Date: Thu, 19 May 2016 18:37:25 +0200 Subject: [GAP Forum] Square root of cyclotomic real number In-Reply-To: References: <20160517114113.GB18282@cs.ox.ac.uk> <4779F1C7-113F-4A7C-84C6-E3EBDB62ED3F@fastmail.fm> Message-ID: <19BB8A57-1C64-4B15-86C2-BB908C0EC70C@quendi.de> > On 19 May 2016, at 07:48, Marek Mitros wrote: > > Thank you for advice. Let me explain my plan, then we can judge whether > this advice work for me or it doesn't. > > I have group of size 5120 of complex matrices of size 16x16. In this group > I listed two conjugacy classes of involutions of size 10 and 80. Next I > wanted to change the basis in a way that mentioned 10 matrices are > diagonal. Therefore I obtained 16 vectors as intersections of eigenspaces. > Unfortunately these vectors are not unit length, so I tried to normalize > them. You don't need to normalize them in order to construct an orthogonal basis -- that is only necessary for an orthonormal basis. So if you really just need a base change to diagonalize your matrices, that is possible without computing squares and normed vectors. Cheers, Max From rafaelv at uaeh.edu.mx Fri May 27 22:11:03 2016 From: rafaelv at uaeh.edu.mx (Rafael Villarroel) Date: Fri, 27 May 2016 16:11:03 -0500 Subject: [GAP Forum] Why do my objects do not store attributes? Message-ID: <87pos78avs.fsf@uaeh.edu.mx> Hello GAP forum I'm puzzled as to why the following code, which I think should work, does not. I have reduced the situation to the following: --------------- start tempo.gap DeclareCategory("IsMyStruct", IsObject); StrFamily := NewFamily("MyStructsFamily"); StrType := NewType(StrFamily, IsMyStruct); DeclareRepresentation("StrRep", IsComponentObjectRep and IsAttributeStoringRep, ["uno", "dos"]); CreateStr := function(u,v) return Objectify(StrType,rec(uno:=u,dos:=v)); end; InstallMethod(PrintObj,[IsMyStruct],function(S) Print("uno=",S!.uno,"dos=",S!.dos); end); InstallMethod(ViewObj,[IsMyStruct],function(S) Print("uno=",S!.uno,"dos=",S!.dos); end); DeclareAttribute("MyAtr",IsMyStruct); InstallMethod(MyAtr,[IsMyStruct],function(S) return("hello"); end); DeclareProperty("MyProp",IsMyStruct); InstallMethod(MyProp,[IsMyStruct],function(S) return false; end); --------------- end tempo.gap then, after loading in a GAP 4.8.2 session, started with "gap -r": I get: gap> Read("tempo.gap"); gap> S:=CreateStr(1,2); uno=1dos=2 gap> MyAtr(S); "hello" gap> KnownAttributesOfObject(S); [ ] why is that? However, storing of properties work? gap> MyProp(S); false gap> KnownPropertiesOfObject(S); [ "MyProp" ] Thanks in advance for any help. Rafael From caj21 at st-andrews.ac.uk Fri May 27 22:39:56 2016 From: caj21 at st-andrews.ac.uk (Christopher Jefferson) Date: Fri, 27 May 2016 21:39:56 +0000 Subject: [GAP Forum] Why do my objects do not store attributes? In-Reply-To: <87pos78avs.fsf@uaeh.edu.mx> References: <87pos78avs.fsf@uaeh.edu.mx> Message-ID: <81CF90E2-23A5-4C15-B779-1B1100C792AF@st-andrews.ac.uk> I?m not claiming this is obvious, but the definition of StrType has to have IsAttributeStoringRep: StrType := NewType(StrFamily, IsMyStruct and IsAttributeStoringRep); Chris On 27/05/2016, 22:11, "forum-bounces at gap-system.org on behalf of Rafael Villarroel" wrote: > >Hello GAP forum > >I'm puzzled as to why the following code, which I think should work, >does not. I have reduced the situation to the following: > >--------------- start tempo.gap > >DeclareCategory("IsMyStruct", IsObject); >StrFamily := NewFamily("MyStructsFamily"); >StrType := NewType(StrFamily, IsMyStruct); >DeclareRepresentation("StrRep", > IsComponentObjectRep and IsAttributeStoringRep, > ["uno", "dos"]); >CreateStr := function(u,v) > return Objectify(StrType,rec(uno:=u,dos:=v)); >end; > >InstallMethod(PrintObj,[IsMyStruct],function(S) Print("uno=",S!.uno,"dos=",S!.dos); > end); >InstallMethod(ViewObj,[IsMyStruct],function(S) Print("uno=",S!.uno,"dos=",S!.dos); > end); > >DeclareAttribute("MyAtr",IsMyStruct); >InstallMethod(MyAtr,[IsMyStruct],function(S) return("hello"); > end); > >DeclareProperty("MyProp",IsMyStruct); >InstallMethod(MyProp,[IsMyStruct],function(S) return false; > end); > >--------------- end tempo.gap > >then, after loading in a GAP 4.8.2 session, started with "gap -r": >I get: > >gap> Read("tempo.gap"); >gap> S:=CreateStr(1,2); >uno=1dos=2 >gap> MyAtr(S); >"hello" >gap> KnownAttributesOfObject(S); >[ ] > >why is that? However, storing of properties work? > >gap> MyProp(S); >false >gap> KnownPropertiesOfObject(S); >[ "MyProp" ] > >Thanks in advance for any help. > >Rafael > > > > > > > > > >_______________________________________________ >Forum mailing list >Forum at mail.gap-system.org >http://mail.gap-system.org/mailman/listinfo/forum From alexander.konovalov at st-andrews.ac.uk Fri Jun 10 16:27:31 2016 From: alexander.konovalov at st-andrews.ac.uk (Alexander Konovalov) Date: Fri, 10 Jun 2016 15:27:31 +0000 Subject: [GAP Forum] GAP 4.8.4 release announcement Message-ID: Dear GAP Forum, This is to announce the release of GAP 4.8.4, which could be downloaded from http://www.gap-system.org/Releases/ An overview of the changes introduced in GAP 4.8.4 is provided below. A more detailed version with hyperlinks to documentation is available here: http://www.gap-system.org/Manuals/doc/changes/chap2.html#X85F7B6F17D29A589 New features: * The GAP distribution now includes bin/BuildPackages.sh, a script which can be started from the pkg directory via ../bin/BuildPackages.sh and will attempt to build as many packages as possible. It replaces the InstPackages.sh script which was not a part of the GAP distribution and had to be downloaded separately from the GAP website. The new script is more robust and simplifies adding new packages with binaries, as it requires no adjustments if the new package supports the standard ./configure; make build procedure. Improved and extended functionality: * SimpleGroup now produces more informative error message in the case when AtlasGroup could not load the requested group. * An info message with suggestions to use InfoPackageLoading will now be displayed when LoadPackage returns fail (unless GAP is started with -b option). * The build system will now enable C++ support in GMP only if a working C++ compiler is detected. * More checks were added when embedding coefficient rings or rational numbers into polynomial rings in order to forbid adding polynomials in different characteristic. Fixed bugs that could lead to crashes: * Fixed the crash in --cover mode when reading files with more than 65,536 lines. Fixed bugs that could lead to incorrect results: * Fixed an error in the code for partial permutations that occurred on big-endian systems. [Reported by Bill Allombert] * Fixed the kernel method for Remove with one argument, which failed to reduce the length of a list to the position of the last bound entry. [Reported by Peter Schauenburg] Fixed bugs that could lead to break loops: * Fixed the break loop while using Factorization on permutation groups by removing some old code that relied on further caching in Factorization. [Reported by Grahame Erskine] * Fixed a problem with computation of maximal subgroups in an almost simple group. [Reported by Ramon Esteban Romero] * Added missing methods for Intersection2 when one of the arguments is an empty list. [Reported by Wilf Wilson] Other fixed bugs: * Fixed several bugs in RandomPrimitivePolynomial. [Reported by Nusa Zidaric] * Fixed several problems with Random on long lists in 64-bit GAP installations. In addition, GAP 4.8.4 distribution includes updates for 13 GAP packages: ==================================================== Package name | Version | Date ---------------------------------------------------- AtlasRep | 1.5.1 | 30/03/2016 AutomGrp | 1.3 | 28/03/2016 Carat | 2.1.6 | 23/05/2016 CRISP | 1.4.4 | 20/03/2016 float | 0.7.3 | 11/05/2016 fr | 2.3.6 | 21/04/2016 gpd | 1.43 | 16/03/2016 IRREDSOL | 1.3.1 | 20/03/2016 NormalizInterface | 0.9.8 | 07/05/2016 RCWA | 4.4.1 | 22/05/2016 ResClasses | 4.5.0 | 14/05/2016 Semigroups | 2.8.0 | 26/05/2016 Utils | 0.40 | 17/03/2016 ==================================================== We encourage all users to upgrade to GAP 4.8.4. If you need any help or would like to report any problems, please do not hesitate to contact us at support at gap-system.org or submit new issues on GitHub: https://github.com/gap-system/gap/issues Wishing you fun and success using GAP, The GAP Group From Bill.Allombert at math.u-bordeaux.fr Sat Jun 11 22:20:09 2016 From: Bill.Allombert at math.u-bordeaux.fr (Bill Allombert) Date: Sat, 11 Jun 2016 23:20:09 +0200 Subject: [GAP Forum] make testinstall fails with GAP 4r8p4 Message-ID: <20160611212009.GA31317@yellowpig> Dear GAP team, I just tried to build gap 4r8p4 and make testinstall fails: testing: /tmp/gap4r8/tst/testinstall/morpheus.tst ########> Diff in /tmp/gap4r8/tst/testinstall/morpheus.tst, line 14: # Input is: LoadPackage("autpgrp", false);; # Expected output: # But found: #I autpgrp package is not available. Check that the name is correct #I and it is present in one of the GAP root directories (see '??RootPaths') ######## testing: /tmp/gap4r8/tst/testinstall/grpmat.tst ########> Diff in /tmp/gap4r8/tst/testinstall/grpmat.tst, line 32: # Input is: Size( AutomorphismGroup( G ) ); # Expected output: 24 # But found: #I autpgrp package is not available. Check that the name is correct #I and it is present in one of the GAP root directories (see '??RootPaths') 24 ######## This is due to the extra warnings for missing packages. make testinstall is not supposed to require extra packages. Cheers, Bill. From alexander.konovalov at st-andrews.ac.uk Sat Jun 11 22:38:42 2016 From: alexander.konovalov at st-andrews.ac.uk (Alexander Konovalov) Date: Sat, 11 Jun 2016 21:38:42 +0000 Subject: [GAP Forum] make testinstall fails with GAP 4r8p4 In-Reply-To: <20160611212009.GA31317@yellowpig> References: <20160611212009.GA31317@yellowpig> Message-ID: Dear Bill, First, this problem will not affect anyone who installed GAP using the full distribution downloaded from http://www.gap-system.org/Releases/, which includes all packages currently redistributed with GAP, so I suggest to discuss it in the open GAP development mailing list (CC). > On 11 Jun 2016, at 22:20, Bill Allombert wrote: > > Dear GAP team, > > I just tried to build gap 4r8p4 and make testinstall fails: > > testing: /tmp/gap4r8/tst/testinstall/morpheus.tst > ########> Diff in /tmp/gap4r8/tst/testinstall/morpheus.tst, line 14: > # Input is: > LoadPackage("autpgrp", false);; > # Expected output: > # But found: > #I autpgrp package is not available. Check that the name is correct > #I and it is present in one of the GAP root directories (see > '??RootPaths') > ######## The call to LoadPackage has been already removed from this file in the master branch: https://github.com/gap-system/gap/blob/master/tst/testinstall/morpheus.tst so in this case neither this warning nor the one below should not appear. Note that the performance of the system without autpgrp will be worse. In particular, it significantly improves the performance of AutomorphismGroup for p-groups. There are some other packages which also improve or extend the performance, and while there is an interest of testing the core GAP system without packages, and documenting dependencies better, for practical purposes I *strongly do not recommend to use the core GAP system without packages*. Best wishes Alexander > testing: /tmp/gap4r8/tst/testinstall/grpmat.tst > ########> Diff in /tmp/gap4r8/tst/testinstall/grpmat.tst, line 32: > # Input is: > Size( AutomorphismGroup( G ) ); > # Expected output: > 24 > # But found: > #I autpgrp package is not available. Check that the name is correct > #I and it is present in one of the GAP root directories (see > '??RootPaths') > 24 > ######## > > This is due to the extra warnings for missing packages. > make testinstall is not supposed to require extra packages. > > Cheers, > Bill. From r_n_tsai at yahoo.com Tue Jun 14 04:52:58 2016 From: r_n_tsai at yahoo.com (R.N. Tsai) Date: Tue, 14 Jun 2016 03:52:58 +0000 (UTC) Subject: [GAP Forum] saving variables to a file In-Reply-To: <207889948.41401.1442357415550.JavaMail.yahoo@mail.yahoo.com> References: <1392580602.83900.YahooMailNeo@web121206.mail.ne1.yahoo.com> <207889948.41401.1442357415550.JavaMail.yahoo@mail.yahoo.com> Message-ID: <1472117324.2153157.1465876378635.JavaMail.yahoo@mail.yahoo.com> Dear GAP forum, I have same fairly large matrices that take a long time to calculate. Is there a way to save these efficiently?I read about saving the workspace but that won't help with this; I have many matrices and I want to load morethan one at a time. I tried printing to a text file but that takes a very long time because of the size of the matrices. Here's how I would have done this in Ocatave/Matlab : M1=(results of long calculations)save('M1.mat','M1);M2=...save('M2.mat','M2); so in the future I can bring up these matrices into gap like this ?: load('M1.mat','M1') load('M2.mat','M2') and now I can process both say M3=M1+M2,....in the same gap session... Any suggestions? Also, is there an efficient way to pass data between gap and Octave/Matlab? Thanks,R.N. From max at quendi.de Tue Jun 14 10:01:55 2016 From: max at quendi.de (Max Horn) Date: Tue, 14 Jun 2016 11:01:55 +0200 Subject: [GAP Forum] saving variables to a file In-Reply-To: <1472117324.2153157.1465876378635.JavaMail.yahoo@mail.yahoo.com> References: <1392580602.83900.YahooMailNeo@web121206.mail.ne1.yahoo.com> <207889948.41401.1442357415550.JavaMail.yahoo@mail.yahoo.com> <1472117324.2153157.1465876378635.JavaMail.yahoo@mail.yahoo.com> Message-ID: <555C8018-8D91-41A3-A34B-FF84D8F5AF0E@quendi.de> Dear R.N., > On 14 Jun 2016, at 05:52, R.N. Tsai wrote: > > Dear GAP forum, > I have same fairly large matrices that take a long time to calculate. Is there a way to save these efficiently?I read about saving the workspace but that won't help with this; I have many matrices and I want to load morethan one at a time. I tried printing to a text file but that takes a very long time because of the size of the matrices. To get a better feeling for your problem: How large are you matrices? Over which ring are they defined? > Here's how I would have done this in Ocatave/Matlab : > M1=(results of long calculations)save('M1.mat','M1);M2=...save('M2.mat','M2); > so in the future I can bring up these matrices into gap like this : > load('M1.mat','M1') > load('M2.mat','M2') > > and now I can process both say M3=M1+M2,....in the same gap session... > Any suggestions? > Also, is there an efficient way to pass data between gap and Octave/Matlab? One way to achieve this would be to use the IO_Pickle / IO_Unpickle functions from the GAP IO package. You can finde an example of using them here: . For details, see also Hope that helps, Max From dmitrii.pasechnik at cs.ox.ac.uk Tue Jun 14 12:19:58 2016 From: dmitrii.pasechnik at cs.ox.ac.uk (Dima Pasechnik) Date: Tue, 14 Jun 2016 12:19:58 +0100 Subject: [GAP Forum] saving variables to a file In-Reply-To: <555C8018-8D91-41A3-A34B-FF84D8F5AF0E@quendi.de> References: <1392580602.83900.YahooMailNeo@web121206.mail.ne1.yahoo.com> <207889948.41401.1442357415550.JavaMail.yahoo@mail.yahoo.com> <1472117324.2153157.1465876378635.JavaMail.yahoo@mail.yahoo.com> <555C8018-8D91-41A3-A34B-FF84D8F5AF0E@quendi.de> Message-ID: <20160614111958.GA14547@dimpase.cs.ox.ac.uk> Dear all, On Tue, Jun 14, 2016 at 11:01:55AM +0200, Max Horn wrote: > > On 14 Jun 2016, at 05:52, R.N. Tsai wrote: > > I have same fairly large matrices that take a long time to calculate. Is there a way to save these efficiently?I read about saving the workspace but that won't help with this; I have many matrices and I want to load morethan one at a time. I tried printing to a text file but that takes a very long time because of the size of the matrices. > > To get a better feeling for your problem: How large are you matrices? Over which ring are they defined? > > > Here's how I would have done this in Ocatave/Matlab : > > M1=(results of long calculations)save('M1.mat','M1);M2=...save('M2.mat','M2); > > so in the future I can bring up these matrices into gap like this : > > load('M1.mat','M1') > > load('M2.mat','M2') > > > > and now I can process both say M3=M1+M2,....in the same gap session... > > Any suggestions? > > Also, is there an efficient way to pass data between gap and Octave/Matlab? One way to pass data between GAP and Matlab (or Octave) is to use Sage(math) (www.sagemath.org) which is a Python library with interfaces both to GAP (and libGAP) and Matlab (or Octave). And in fact you can do a lot with matrices using other Sagemath parts, e.g. numpy. Another thing I can think of is creating a Matlab mex interface to libGAP, which allows you to load GAP as callable C library. See https://bitbucket.org/vbraun/libgap (not sure whether Octave has such capabilities) By the way, this would be a great tool regardless of your task at hand. HTH, Dima > > One way to achieve this would be to use the IO_Pickle / IO_Unpickle functions from the GAP IO package. You can finde an example of using them here: . For details, see also > > > Hope that helps, > Max > > > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From r_n_tsai at yahoo.com Tue Jun 14 17:20:03 2016 From: r_n_tsai at yahoo.com (R.N. Tsai) Date: Tue, 14 Jun 2016 16:20:03 +0000 (UTC) Subject: [GAP Forum] saving variables to a file In-Reply-To: <555C8018-8D91-41A3-A34B-FF84D8F5AF0E@quendi.de> References: <1392580602.83900.YahooMailNeo@web121206.mail.ne1.yahoo.com> <207889948.41401.1442357415550.JavaMail.yahoo@mail.yahoo.com> <1472117324.2153157.1465876378635.JavaMail.yahoo@mail.yahoo.com> <555C8018-8D91-41A3-A34B-FF84D8F5AF0E@quendi.de> Message-ID: <135404139.2393131.1465921203608.JavaMail.yahoo@mail.yahoo.com> Thanks for the response Max,The matrices are over GF(2), so hopefully they can be managed more efficiently than generic matrices.R.N From: Max Horn To: R.N. Tsai Cc: GAP Forum Sent: Tuesday, June 14, 2016 2:01 AM Subject: Re: [GAP Forum] saving variables to a file Dear R.N., > On 14 Jun 2016, at 05:52, R.N. Tsai wrote: > > Dear GAP forum, > I have same fairly large matrices that take a long time to calculate. Is there a way to save these efficiently?I read about saving the workspace but that won't help with this; I have many matrices and I want to load morethan one at a time. I tried printing to a text file but that takes a very long time because of the size of the matrices. To get a better feeling for your problem: How large are you matrices? Over which ring are they defined? > Here's how I would have done this in Ocatave/Matlab : > M1=(results of long calculations)save('M1.mat','M1);M2=...save('M2.mat','M2); > so in the future I can bring up these matrices into gap like this? : > load('M1.mat','M1') > load('M2.mat','M2') > > and now I can process both say M3=M1+M2,....in the same gap session... > Any suggestions? > Also, is there an efficient way to pass data between gap and Octave/Matlab? One way to achieve this would be to use the IO_Pickle / IO_Unpickle functions from the GAP IO package. You can finde an example of using them here: . For details, see also Hope that helps, Max From r_n_tsai at yahoo.com Wed Jun 15 18:02:00 2016 From: r_n_tsai at yahoo.com (R.N. Tsai) Date: Wed, 15 Jun 2016 17:02:00 +0000 (UTC) Subject: [GAP Forum] saving variables to a file In-Reply-To: <20160614111958.GA14547@dimpase.cs.ox.ac.uk> References: <1392580602.83900.YahooMailNeo@web121206.mail.ne1.yahoo.com> <207889948.41401.1442357415550.JavaMail.yahoo@mail.yahoo.com> <1472117324.2153157.1465876378635.JavaMail.yahoo@mail.yahoo.com> <555C8018-8D91-41A3-A34B-FF84D8F5AF0E@quendi.de> <20160614111958.GA14547@dimpase.cs.ox.ac.uk> Message-ID: <1626947282.3072997.1466010120929.JavaMail.yahoo@mail.yahoo.com> Dear all,I'll definitely look into using Sage and libGAP, but it also seems simple enough to add such a?capability within GAP itself (save a single variable to a specific file even if restricting the variableto be numeric...)R.N. From: Dima Pasechnik To: R.N. Tsai Cc: GAP Forum Sent: Tuesday, June 14, 2016 4:19 AM Subject: Re: [GAP Forum] saving variables to a file Dear all, On Tue, Jun 14, 2016 at 11:01:55AM +0200, Max Horn wrote: > > On 14 Jun 2016, at 05:52, R.N. Tsai wrote: > > I have same fairly large matrices that take a long time to calculate. Is there a way to save these efficiently?I read about saving the workspace but that won't help with this; I have many matrices and I want to load morethan one at a time. I tried printing to a text file but that takes a very long time because of the size of the matrices. > > To get a better feeling for your problem: How large are you matrices? Over which ring are they defined? > > > Here's how I would have done this in Ocatave/Matlab : > > M1=(results of long calculations)save('M1.mat','M1);M2=...save('M2.mat','M2); > > so in the future I can bring up these matrices into gap like this? : > > load('M1.mat','M1') > > load('M2.mat','M2') > > > > and now I can process both say M3=M1+M2,....in the same gap session... > > Any suggestions? > > Also, is there an efficient way to pass data between gap and Octave/Matlab? One way to pass data between GAP and Matlab (or Octave) is to use Sage(math) (www.sagemath.org) which is a Python library with interfaces both to GAP (and libGAP) and Matlab (or Octave). And in fact you can do a lot with matrices using other Sagemath parts, e.g. numpy. Another thing I can think of is creating a Matlab mex interface to libGAP, which allows you to load GAP as callable C library. See https://bitbucket.org/vbraun/libgap (not sure whether Octave has such capabilities) By the way, this would be a great tool regardless of your task at hand. HTH, Dima > > One way to achieve this would be to use the IO_Pickle / IO_Unpickle functions from the GAP IO package. You can finde an example of using them here: . For details, see also > > > Hope that helps, > Max > > > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From stefan at mcs.st-and.ac.uk Wed Jun 15 21:51:30 2016 From: stefan at mcs.st-and.ac.uk (Stefan Kohl) Date: Wed, 15 Jun 2016 21:51:30 +0100 (BST) Subject: [GAP Forum] saving variables to a file In-Reply-To: <135404139.2393131.1465921203608.JavaMail.yahoo@mail.yahoo.com> References: <1392580602.83900.YahooMailNeo@web121206.mail.ne1.yahoo.com> <207889948.41401.1442357415550.JavaMail.yahoo@mail.yahoo.com> <1472117324.2153157.1465876378635.JavaMail.yahoo@mail.yahoo.com> <555C8018-8D91-41A3-A34B-FF84D8F5AF0E@quendi.de> <135404139.2393131.1465921203608.JavaMail.yahoo@mail.yahoo.com> Message-ID: On Tue, June 14, 2016 5:20 pm, R.N. Tsai wrote: > Thanks for the response Max,The matrices are over GF(2), so hopefully they can be > managed more efficiently than generic matrices.R.N One thing you could do specifically for matrices over GF(2) is to save them as monochrome bitmap pictures -- this uses just one bit per entry (plus a tiny overhead for the description block for the picture): gap> M := NullMat(10000,10000,GF(2));; # just some GF(2) matrix, could be any gap> file := Filename(DirectoryTemporary(),"matrix.bmp"); # pick a file name "/tmp/tmIthaXA/matrix.bmp" gap> SaveAsBitmapPicture(M,file); # save the matrix to the file, 12.5 MB in this case gap> time; 18392 gap> N := LoadBitmapPicture(file);; # read it in again gap> time; 8320 gap> N = M; true Hope this helps, Stefan From r_n_tsai at yahoo.com Thu Jun 16 00:35:28 2016 From: r_n_tsai at yahoo.com (R.N. Tsai) Date: Wed, 15 Jun 2016 23:35:28 +0000 (UTC) Subject: [GAP Forum] saving variables to a file In-Reply-To: References: <1392580602.83900.YahooMailNeo@web121206.mail.ne1.yahoo.com> <207889948.41401.1442357415550.JavaMail.yahoo@mail.yahoo.com> <1472117324.2153157.1465876378635.JavaMail.yahoo@mail.yahoo.com> <555C8018-8D91-41A3-A34B-FF84D8F5AF0E@quendi.de> <135404139.2393131.1465921203608.JavaMail.yahoo@mail.yahoo.com> Message-ID: <1883193502.3157980.1466033728338.JavaMail.yahoo@mail.yahoo.com> This is actually perfect for GF(2) matrices (which is all I need for now).Interestingly enough if I convert the GF(2) matrix to 1 and 0's (using "Int") the bitmapfile is 24x larger...that was just an experiment so it's actually better to stay in GF(2).I should be able to process a .bmp file in matlab so this could be one way to passthe matrices between the two.Thanks for your help.R.N. From: Stefan Kohl To: R.N. Tsai Cc: Max Horn ; GAP Forum Sent: Wednesday, June 15, 2016 1:51 PM Subject: Re: [GAP Forum] saving variables to a file On Tue, June 14, 2016 5:20 pm, R.N. Tsai wrote: > Thanks for the response Max,The matrices are over GF(2), so hopefully they can be > managed more efficiently than generic matrices.R.N One thing you could do specifically for matrices over GF(2) is to save them as monochrome bitmap pictures -- this uses just one bit per entry (plus a tiny overhead for the description block for the picture): gap> M := NullMat(10000,10000,GF(2));; # just some GF(2) matrix, could be any gap> file := Filename(DirectoryTemporary(),"matrix.bmp"); # pick a file name "/tmp/tmIthaXA/matrix.bmp" gap> SaveAsBitmapPicture(M,file); # save the matrix to the file, 12.5 MB in this case gap> time; 18392 gap> N := LoadBitmapPicture(file);; # read it in again gap> time; 8320 gap> N = M; true Hope this helps, ? ? Stefan From f.k.moftakhar at gmail.com Thu Jun 16 06:21:52 2016 From: f.k.moftakhar at gmail.com (fatemeh moftakhar) Date: Wed, 15 Jun 2016 22:21:52 -0700 Subject: [GAP Forum] PartitionsSet Message-ID: Dear GAP forum I need all partitions of {2,..,21}. I mean, I want to run PartitionsSet([2..21]). I have servers with 512 gigabyte ram and my computer stopped after two days, because it needs more ram. I don't have better computer. Is there a faster algorithm? I need these partitions for computing supercharacter theories of Janko group J2. Best regards Fatemeh Moftakhar -- Regards; Miss Fatemeh Moftakhar PhD Candidate, Department of Pure Mathematics, Faculty of Mathematical Sciences, University of Kashan, Kashan, Iran From sam at Math.RWTH-Aachen.De Thu Jun 16 08:19:01 2016 From: sam at Math.RWTH-Aachen.De (Thomas Breuer) Date: Thu, 16 Jun 2016 09:19:01 +0200 Subject: [GAP Forum] PartitionsSet In-Reply-To: References: Message-ID: <20160616071900.GA15203@gemma.math.rwth-aachen.de> Dear GAP Forum, Fatemeh Moftakhar wrote > I need all partitions of {2,..,21}. I mean, I want to run > PartitionsSet([2..21]). I have servers with 512 gigabyte ram and my > computer stopped after two days, because it needs more ram. I don't have > better computer. Is there a faster algorithm? I need these partitions for > computing supercharacter theories of Janko group J2. If you really have to run over all these partitions then this will take a lot of time. Let us assume that you can process a million of them in a second, which would be quite fast. In one year (356 days with 24 hours) you can then process the following number of partitions. gap> 10^6 * 3600 * 24 * 356; 30758400000000 gap> NrPartitionsSet( [ 2 .. 21 ] ); 51724158235372 I would suggest to find criteria that allow you to exclude most of the candidates without looking at them. All the best, Thomas P.S.: Even if you really have to look at each element from a large set then keeping all of them in memory at the same time is probably not necessary. For such purposes, one can think about developing a so-called iterator for the set. This is essentially a function that knows which was the last element you have looked at, and can can be called in order to give you the next element. Such functionality exists already in GAP for several kinds of sets, for example for the set of partitions of a given number. I am not aware of iterator functionality for the set of partitions of a set, but the above numbers show that this approach would not be suitable in your situation. From f.k.moftakhar at gmail.com Thu Jun 16 08:33:21 2016 From: f.k.moftakhar at gmail.com (fatemeh moftakhar) Date: Thu, 16 Jun 2016 00:33:21 -0700 Subject: [GAP Forum] ReadLine Message-ID: Dear GAP forum Is there any way to read just a special line of a large file (that has too much data). Would you please explain with a sample code? Best regards Fatemeh Moftakhar -- Regards; Miss Fatemeh Moftakhar PhD Candidate, Department of Pure Mathematics, Faculty of Mathematical Sciences, University of Kashan, Kashan, Iran From stefan at mcs.st-and.ac.uk Fri Jun 17 00:14:07 2016 From: stefan at mcs.st-and.ac.uk (Stefan Kohl) Date: Fri, 17 Jun 2016 00:14:07 +0100 (BST) Subject: [GAP Forum] saving variables to a file In-Reply-To: <1883193502.3157980.1466033728338.JavaMail.yahoo@mail.yahoo.com> References: <1392580602.83900.YahooMailNeo@web121206.mail.ne1.yahoo.com> <207889948.41401.1442357415550.JavaMail.yahoo@mail.yahoo.com> <1472117324.2153157.1465876378635.JavaMail.yahoo@mail.yahoo.com> <555C8018-8D91-41A3-A34B-FF84D8F5AF0E@quendi.de> <135404139.2393131.1465921203608.JavaMail.yahoo@mail.yahoo.com> <1883193502.3157980.1466033728338.JavaMail.yahoo@mail.yahoo.com> Message-ID: On Thu, June 16, 2016 12:35 am, R.N. Tsai wrote: > This is actually perfect for GF(2) matrices (which is all I need for now).Interestingly > enough if I convert the GF(2) matrix to 1 and 0's (using "Int") the bitmapfile is 24x > larger... Yes -- if you take an integer matrix, you get a 24-bit color picture, where the entries of the matrix are read as 256^2 * red value + 256 * green value + blue value. Stefan > From: Stefan Kohl > To: R.N. Tsai > Cc: Max Horn ; GAP Forum > Sent: Wednesday, June 15, 2016 1:51 PM > Subject: Re: [GAP Forum] saving variables to a file > > On Tue, June 14, 2016 5:20 pm, R.N. Tsai wrote: >> Thanks for the response Max,The matrices are over GF(2), so hopefully they can be >> managed more efficiently than generic matrices.R.N > > One thing you could do specifically for matrices over GF(2) is to save them > as monochrome bitmap pictures -- this uses just one bit per entry (plus a tiny > overhead for the description block for the picture): > > gap> M := NullMat(10000,10000,GF(2));; # just some GF(2) matrix, could be any > gap> file := Filename(DirectoryTemporary(),"matrix.bmp"); # pick a file name > "/tmp/tmIthaXA/matrix.bmp" > gap> SaveAsBitmapPicture(M,file); # save the matrix to the file, 12.5 MB in this case > gap> time; > 18392 > gap> N := LoadBitmapPicture(file);; # read it in again > gap> time; > 8320 > gap> N = M; > true > > Hope this helps, > > ?? ?? Stefan > > > > > From l.h.soicher at qmul.ac.uk Tue Jun 21 09:38:55 2016 From: l.h.soicher at qmul.ac.uk (Leonard Soicher) Date: Tue, 21 Jun 2016 08:38:55 +0000 Subject: [GAP Forum] Announcing the SLA package Message-ID: Dear Forum Members, It is my pleasure to announce formally the official acceptance of the SLA package. The officially accepted version 1.1 of SLA has been publicly available in GAP starting with GAP version 4.8.2, and I apologise for the lateness of this announcement. SLA provides functions for computing with various aspects of the theory of simple Lie algebras in characteristic zero. The SLA package is authored by Willem de Graaf (University of Trento), and on behalf of the GAP Council, I thank him for this contribution to GAP. Leonard Soicher (Chair of the GAP Council) From l.h.soicher at qmul.ac.uk Tue Jun 21 10:25:56 2016 From: l.h.soicher at qmul.ac.uk (Leonard Soicher) Date: Tue, 21 Jun 2016 09:25:56 +0000 Subject: [GAP Forum] Announcing the AutomGrp package Message-ID: Dear Forum Members, It is my pleasure to announce formally the official acceptance of the AutomGrp package. The officially accepted version 1.3 of AutomGrp is now available in the newly released GAP version 4.8.4. AutomGrp provides methods for computations with groups and semigroups generated by finite automata or given by wreath recursion, as well as with their finitely generated subgroups and elements. The AutomGroup package is authored by Yevgen Muntyan (Tableau Software) and Dmytro Savchuk (University of South Florida), and on behalf of the GAP Council, I thank them for this contribution to GAP. Leonard Soicher (Chair of the GAP Council) From fawadhayat05 at gmail.com Fri Jun 24 10:05:38 2016 From: fawadhayat05 at gmail.com (Fawad Hayat) Date: Fri, 24 Jun 2016 05:05:38 -0400 Subject: [GAP Forum] FAWAD ALI M.Phil (Pure Mathematics) 3rd semester scholar From Pakistan, Message-ID: Hello sir, sir I have a problem that how to write a command in GAP the automorphism group of finite abelian group and their fixed points. Let Z_pXZ_q be cyclic group where p & q are distinct primes, Let suppose p=2 & q=3 => G:=Z_2XZ_3= {(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)} be a cyclic group, and 'd' is the divisor of the order of a group G i.e ( d/IGI =>d=1,2,3,6) how to list all the automorphisms of a group Aut(G), and how to find explicitly all those automorphisms fixing 'd' elements, where fix means f(x)=x ; for some x belongs to G & for f belongs to Aut(G). f(e)=e : by homomorphism property, i.e For d=1, A={set of all those automorphisms of Aut(G) fixing d=1 element only ( Identity element)} For d=2, B={set of all those automorphisms of Aut(G) fixing d=2 element} For d=3 C={set of all those automorphisms of Aut(G) fixing d=3 element} For d=6 D={ I : because identity is the only auto fixing all the elements of a group G } Sir, I need The command which gives me the complete calculation from GAP Thanks Sir, Regards FAWAD ALI From pakistan From fawadhayat05 at gmail.com Fri Jun 24 22:04:31 2016 From: fawadhayat05 at gmail.com (Fawad Hayat) Date: Sat, 25 Jun 2016 02:04:31 +0500 Subject: [GAP Forum] Forum Digest, Vol 151, Issue 3 In-Reply-To: References: Message-ID: Dear Forum please I need your response.. thanks FAWAD ALI On Fri, Jun 24, 2016 at 2:05 PM, wrote: > Send Forum mailing list submissions to > forum at mail.gap-system.org > > To subscribe or unsubscribe via the World Wide Web, visit > http://mail.gap-system.org/mailman/listinfo/forum > or, via email, send a message with subject or body 'help' to > forum-request at mail.gap-system.org > > You can reach the person managing the list at > forum-owner at mail.gap-system.org > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of Forum digest..." > > > Today's Topics: > > 1. PartitionsSet (fatemeh moftakhar) > 2. Re: PartitionsSet (Thomas Breuer) > 3. ReadLine (fatemeh moftakhar) > 4. Re: saving variables to a file (Stefan Kohl) > 5. Announcing the SLA package (Leonard Soicher) > 6. Announcing the AutomGrp package (Leonard Soicher) > 7. FAWAD ALI M.Phil (Pure Mathematics) 3rd semester scholar From > Pakistan, (Fawad Hayat) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Wed, 15 Jun 2016 22:21:52 -0700 > From: fatemeh moftakhar > To: forum at gap-system.org > Subject: [GAP Forum] PartitionsSet > Message-ID: > < > CAOEkHNL7a2wFTZVgiFvu+4fcGPm-bGBeuy1uRbtP_auaQEe3qw at mail.gmail.com> > Content-Type: text/plain; charset=UTF-8 > > Dear GAP forum > > I need all partitions of {2,..,21}. I mean, I want to run > PartitionsSet([2..21]). I have servers with 512 gigabyte ram and my > computer stopped after two days, because it needs more ram. I don't have > better computer. Is there a faster algorithm? I need these partitions for > computing supercharacter theories of Janko group J2. > > > > Best regards > Fatemeh Moftakhar > > > > -- > Regards; > Miss Fatemeh Moftakhar > PhD Candidate, > Department of Pure Mathematics, > Faculty of Mathematical Sciences, > University of Kashan, Kashan, Iran > > > ------------------------------ > > Message: 2 > Date: Thu, 16 Jun 2016 09:19:01 +0200 > From: Thomas Breuer > To: fatemeh moftakhar , forum at gap-system.org > Subject: Re: [GAP Forum] PartitionsSet > Message-ID: <20160616071900.GA15203 at gemma.math.rwth-aachen.de> > Content-Type: text/plain; charset=us-ascii > > Dear GAP Forum, > > Fatemeh Moftakhar wrote > > > I need all partitions of {2,..,21}. I mean, I want to run > > PartitionsSet([2..21]). I have servers with 512 gigabyte ram and my > > computer stopped after two days, because it needs more ram. I don't have > > better computer. Is there a faster algorithm? I need these partitions for > > computing supercharacter theories of Janko group J2. > > If you really have to run over all these partitions then > this will take a lot of time. > Let us assume that you can process a million of them in a second, > which would be quite fast. > In one year (356 days with 24 hours) you can then process > the following number of partitions. > > gap> 10^6 * 3600 * 24 * 356; > 30758400000000 > gap> NrPartitionsSet( [ 2 .. 21 ] ); > 51724158235372 > > I would suggest to find criteria that allow you to exclude most of the > candidates without looking at them. > > All the best, > Thomas > > P.S.: > Even if you really have to look at each element from a large set > then keeping all of them in memory at the same time is probably > not necessary. > For such purposes, one can think about developing a so-called iterator > for the set. > This is essentially a function that knows which was the last element > you have looked at, and can can be called in order to give you the > next element. > Such functionality exists already in GAP for several kinds of sets, > for example for the set of partitions of a given number. > I am not aware of iterator functionality for the set of partitions > of a set, > but the above numbers show that this approach would not be suitable > in your situation. > > > > > ------------------------------ > > Message: 3 > Date: Thu, 16 Jun 2016 00:33:21 -0700 > From: fatemeh moftakhar > To: forum at gap-system.org > Subject: [GAP Forum] ReadLine > Message-ID: > QW2paWJDhS9NUD47D6w at mail.gmail.com> > Content-Type: text/plain; charset=UTF-8 > > Dear GAP forum > > Is there any way to read just a special line of a large file (that has too > much data). Would you please explain with a sample code? > > Best regards > Fatemeh Moftakhar > > -- > Regards; > Miss Fatemeh Moftakhar > PhD Candidate, > Department of Pure Mathematics, > Faculty of Mathematical Sciences, > University of Kashan, Kashan, Iran > > > ------------------------------ > > Message: 4 > Date: Fri, 17 Jun 2016 00:14:07 +0100 (BST) > From: "Stefan Kohl" > To: "R.N. Tsai" > Cc: GAP Forum > Subject: Re: [GAP Forum] saving variables to a file > Message-ID: > Content-Type: text/plain;charset=iso-8859-1 > > On Thu, June 16, 2016 12:35 am, R.N. Tsai wrote: > > This is actually perfect for GF(2) matrices (which is all I need for > now).Interestingly > > enough if I convert the GF(2) matrix to 1 and 0's (using "Int") the > bitmapfile is 24x > > larger... > > Yes -- if you take an integer matrix, you get a 24-bit color picture, > where the > entries of the matrix are read as 256^2 * red value + 256 * green value + > blue value. > > Stefan > > > From: Stefan Kohl > > To: R.N. Tsai > > Cc: Max Horn ; GAP Forum > > Sent: Wednesday, June 15, 2016 1:51 PM > > Subject: Re: [GAP Forum] saving variables to a file > > > > On Tue, June 14, 2016 5:20 pm, R.N. Tsai wrote: > >> Thanks for the response Max,The matrices are over GF(2), so hopefully > they can be > >> managed more efficiently than generic matrices.R.N > > > > One thing you could do specifically for matrices over GF(2) is to save > them > > as monochrome bitmap pictures -- this uses just one bit per entry (plus > a tiny > > overhead for the description block for the picture): > > > > gap> M := NullMat(10000,10000,GF(2));; # just some GF(2) matrix, could > be any > > gap> file := Filename(DirectoryTemporary(),"matrix.bmp"); # pick a file > name > > "/tmp/tmIthaXA/matrix.bmp" > > gap> SaveAsBitmapPicture(M,file); # save the matrix to the file, 12.5 MB > in this case > > gap> time; > > 18392 > > gap> N := LoadBitmapPicture(file);; # read it in again > > gap> time; > > 8320 > > gap> N = M; > > true > > > > Hope this helps, > > > > ?? ?? Stefan > > > > > > > > > > > > > > > ------------------------------ > > Message: 5 > Date: Tue, 21 Jun 2016 08:38:55 +0000 > From: Leonard Soicher > To: "forum at gap-system.org" > Subject: [GAP Forum] Announcing the SLA package > Message-ID: > < > VI1PR07MB084605159C07DB25A4FAAD388E2B0 at VI1PR07MB0846.eurprd07.prod.outlook.com > > > > Content-Type: text/plain; charset="iso-8859-1" > > > Dear Forum Members, > > It is my pleasure to announce formally the official acceptance > of the SLA package. The officially accepted version 1.1 of SLA has > been publicly available in GAP starting with GAP version 4.8.2, > and I apologise for the lateness of this announcement. > > SLA provides functions for computing with various aspects of > the theory of simple Lie algebras in characteristic zero. > > The SLA package is authored by Willem de Graaf (University of Trento), > and on behalf of the GAP Council, I thank him for this contribution to GAP. > > Leonard Soicher > (Chair of the GAP Council) > > > > ------------------------------ > > Message: 6 > Date: Tue, 21 Jun 2016 09:25:56 +0000 > From: Leonard Soicher > To: "forum at gap-system.org" > Subject: [GAP Forum] Announcing the AutomGrp package > Message-ID: > < > VI1PR07MB0846F9DF39864EBF0031D7498E2B0 at VI1PR07MB0846.eurprd07.prod.outlook.com > > > > Content-Type: text/plain; charset="iso-8859-1" > > > Dear Forum Members, > > It is my pleasure to announce formally the official acceptance > of the AutomGrp package. The officially accepted version 1.3 of AutomGrp > is now available in the newly released GAP version 4.8.4. > > AutomGrp provides methods for computations with groups and semigroups > generated by finite automata or given by wreath recursion, as well as with > their > finitely generated subgroups and elements. > > The AutomGroup package is authored by Yevgen Muntyan (Tableau Software) > and Dmytro Savchuk (University of South Florida), and on behalf of the GAP > Council, > I thank them for this contribution to GAP. > > Leonard Soicher > (Chair of the GAP Council) > > > > > ------------------------------ > > Message: 7 > Date: Fri, 24 Jun 2016 05:05:38 -0400 > From: Fawad Hayat > To: forum at gap-system.org > Subject: [GAP Forum] FAWAD ALI M.Phil (Pure Mathematics) 3rd semester > scholar From Pakistan, > Message-ID: > 2LyD21qwKmVSYWWzg at mail.gmail.com> > Content-Type: text/plain; charset=UTF-8 > > Hello sir, > sir I have a problem that how to write a command in GAP the automorphism > group of finite abelian group and their fixed points. > Let Z_pXZ_q be cyclic group where p & q are distinct primes, > Let suppose p=2 & q=3 > => G:=Z_2XZ_3= {(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)} > be a cyclic group, and 'd' is the divisor of the order of a group G i.e ( > d/IGI =>d=1,2,3,6) > how to list all the automorphisms of a group Aut(G), and how to find > explicitly all those automorphisms fixing 'd' elements, where fix means > f(x)=x ; for some x belongs to G & for f belongs to Aut(G). > f(e)=e : by homomorphism property, > > i.e For d=1, > A={set of all those automorphisms of Aut(G) fixing d=1 element only ( > Identity element)} > For d=2, > B={set of all those automorphisms of Aut(G) fixing d=2 element} > For d=3 > C={set of all those automorphisms of Aut(G) fixing d=3 element} > For d=6 > D={ I : because identity is the only auto fixing all the elements of a > group G } > > Sir, I need The command which gives me the complete calculation from GAP > Thanks Sir, > Regards FAWAD ALI From pakistan > > > ------------------------------ > > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum > > > End of Forum Digest, Vol 151, Issue 3 > ************************************* > From shahmaths_problem at hotmail.com Tue Jun 28 21:07:33 2016 From: shahmaths_problem at hotmail.com (muhammad shah) Date: Wed, 29 Jun 2016 01:07:33 +0500 Subject: [GAP Forum] How to find some special automorphisms? Message-ID: Dear All, A student Fawad asked me the following question. Let G:=Z_2XZ_3= {(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)}be a cyclic group, and 'd' is the divisor of the order of a group G i.e ( d/IGI =>d=1,2,3,6)how to list all the automorphisms of a group Aut(G), and how to find explicitly all those automorphisms fixing 'd' elements, where fix means f(x)=x ; for some x belongs to G & for f belongs to Aut(G).f(e)=e : by homomorphism property, i.e For d=1,A={set of all those automorphisms of Aut(G) fixing d=1 element only ( Identity element)}For d=2,B={set of all those automorphisms of Aut(G) fixing d=2 element} For d=3C={set of all those automorphisms of Aut(G) fixing d=3 element}For d=6D={ I : because identity is the only auto fixing all the elements of a group G } I wrote a function Fawad () which I think is just a step away from what he wants. I would invite your comments on it that whether it is correct or not and what will be a more correct and efficient way in this case. Here is Fawad. Fawad:=function(G) local A,B,d,f,i,Aut; A:=[]; B:=[]; Aut:=AutomorphismGroup(G); for d in DivisorsInt(Size(G)) do for f in Aut do if Number([1..Size(G)],i -> Image(f,AsList(G)[i])=AsList(G)[i])=d then Add(A,f); Add(B,d); fi; od; od; return [B,A]; end; All the best, Muhammad Shah From shahmaths_problem at hotmail.com Thu Jun 30 15:19:19 2016 From: shahmaths_problem at hotmail.com (muhammad shah) Date: Thu, 30 Jun 2016 19:19:19 +0500 Subject: [GAP Forum] [group-pub-forum] How to find some special automorphisms? In-Reply-To: <33CB032D-C5AD-4332-A1E0-3E5CACBF9908@wisc.edu> References: , <33CB032D-C5AD-4332-A1E0-3E5CACBF9908@wisc.edu> Message-ID: I am thankful to Marty Isaacs, Witek and Petr Savicky for their valuable comments on the problem. As pointed out by Marty Isaacs, the question was looking a homework problem apparently but I was more interested in its solution in the general form and in the gap function provided there for this purpose. Petr Savicky ,in a private response to me checked the function to be correct. He provided me with a much more efficient and fast version of it. Cheers, Muhammad Shah > From: isaacs at math.wisc.edu > To: shahmaths_problem at hotmail.com > CC: group-pub-forum at lists.maths.bath.ac.uk > Date: Tue, 28 Jun 2016 20:25:39 +0000 > Subject: Re: [group-pub-forum] How to find some special automorphisms? > > Your question sounds very much like a homework > problem, so I will not give much detail in my answer. > > Your group G has just two automorphisms: the > identity and the map that carries every element > to its inverse. Obviously, the identity automorphism > fixes all elements of G, and the nonidentity > automorphism fixes exactly two elements. > > Marty Isaacs > From dpopova at fau.edu Sun Jul 3 17:48:05 2016 From: dpopova at fau.edu (Daniela Nikolova) Date: Sun, 3 Jul 2016 16:48:05 +0000 Subject: [GAP Forum] Crystallographic groups Message-ID: Hi everybody, I am Daniela - a returning user - formally from Bulgaria, now from Florida! I was wondering if there is a software in GAP concerning crystallographic groups, more specifically - I am interested in classification algorithms for 2D, and perhaps 3D? Thanks Ps: which of the two addresses above is the best address to send questions? Sent from my iPhone Sent from my iPhone From stefan at mcs.st-and.ac.uk Sun Jul 3 18:35:02 2016 From: stefan at mcs.st-and.ac.uk (Stefan Kohl) Date: Sun, 3 Jul 2016 18:35:02 +0100 (BST) Subject: [GAP Forum] Crystallographic groups In-Reply-To: References: Message-ID: On Sun, July 3, 2016 5:48 pm, Daniela Nikolova wrote: > Hi everybody, > I am Daniela - a returning user - formally from Bulgaria, now from Florida! > I was wondering if there is a software in GAP concerning crystallographic groups, more > specifically - I am interested in classification algorithms for 2D, and perhaps 3D? There are several GAP packages concerning crystallographic groups: - Cryst, http://www.gap-system.org/Packages/cryst.html - CrystCat, http://www.gap-system.org/Packages/crystcat.html - Carat, http://www.gap-system.org/Packages/carat.html Apart from this, the GAP Data Library "Integral Matrix Groups" might be relevant for you as well -- cf. http://www.gap-system.org/Datalib/imf.html. > Ps: which of the two addresses above is the best address to send questions? - For GAP questions of general interest, the GAP Forum is the best choice. - For more technical questions on GAP, there is the address support at gap-system.org. - The Group Pub Forum is for questions on group theory. Hope this helps, Stefan P.S.: Haven't you come from Florida already to Ischia Group Theory 2014? From bmenrigh at brandonenright.net Tue Jul 5 00:48:18 2016 From: bmenrigh at brandonenright.net (Brandon Enright) Date: Mon, 4 Jul 2016 23:48:18 +0000 Subject: [GAP Forum] Applying Burnside's Lemma to the 2x2x2 Rubik's Cube? Message-ID: <20160704234818.2aeaef7b@lambda> -----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 Hi GAP Forum, I'm a relatively new user of GAP and this is my first Forum post. I'm having a great time exploring various permutation puzzle groups so thank you very much to all of the developers that have helped to make such great software! A common problem that comes up on certain permutation puzzles has to do with "extra" symmetry of certain positions when the whole puzzle is reoriented in space. When this happens simple combinatoric counting arguments only produce an approximate number of distinct states. This doesn't happen for the 3x3x3 Rubik's Cube because the 6 face centers always provide a reference. My trouble is when you don't have a reference and certain positions have extra symmetry so they look identical from more than one orientation of the puzzle. If you'd like to see an image of a puzzle where this happens, I've asked about this problem (with images) at http://math.stackexchange.com/q/1822752/132892 So let me be specific. I have a 2x2x2 Rubik's Cube defined by the following 3 generators: # Turn U face TU := (1, 2, 3, 4)(5, 7, 9, 11)(6, 8, 10, 12); # Rotate whole puzzle around U face RU := (1, 2, 3, 4)(5, 7, 9, 11)(6, 8, 10, 12) (13, 15, 17, 19)(14, 16, 18, 20)(21, 22, 23, 24); # Rotate whole puzzle around L face RL := (5, 6, 14, 13)(1, 7, 22, 20)(2, 15, 21, 12) (4, 8, 23, 19)(3, 16, 24, 11)(10, 9, 17, 18); I can make a group using these generators and also a group using just the two orientation generators: # The whole 2x2x2 cube group custom222 := Group([TU, RU, RL]); # The 24 orientations of a 2x2x2 custom222o := Group([RU, RL]); gap> Size(custom222o); 24 gap> Size(custom222) / 24; 3674160 The 3674160 number is the correct number of positions for a normal 2x2x2 Rubik's cube. I had to divide the number by 24 because the cube has 24 orientations and positions that are equivalent under orientation are counted as the same position. Trouble arises when I want to look at variations of this 2x2x2 where some stickers look identical. I have defined which stickers are identical this way: samecolors := [[1, 5, 12, 16, 17, 23],[3, 7, 10, 14, 19, 21], [2, 4, 6, 8, 9, 11, 13, 15, 18, 20, 22, 24]]; Now I'd like to know how many identical positions there are using this set of stickers. I can do that with a stabilizer using OnTuplesSets which will count how many configurations of the 2x2x2 permute stickers within each set of identical stickers. If I divide the size of the whole group by the size of the identical solutions group I should get the number of distinct positions: gap> (Size(custom222) / Size(Stabilizer(custom222, samecolors, OnTuplesSets))) / 24; 1701/2 Unfortunately 1701/2 isn't the right answer. It's not even a whole number. The problem is that some positions look identical in more than one orientation so Burnside's Lemma needs to be applied. I have already done this application of Burnside's Lemma by hand to get the correct answer of 893. The answer GAP gave of 850.5 wasn't that far away. One thing I need to know for Burnside's Lemma is all the symmetries of the cube. I can enumerate this by hand but GAP can also tell me it using the custom222o group: gap> List(ConjugacyClasses(custom222o), Size); [ 1, 6, 3, 8, 6 ] 1 is the identity so the cube isn't rotated at all 6 is the six 90-degree rotations about a cube face 3 is the three 180-degree rotations about a cube face 8 is the eight 120-degree rotations about a cube corner 6 is the six 180-degree rotations about a cube edge By hand I can calculate how many positions of the custom222 group fall into each of these conjugacy classes, sum them up, and divide by 24 to get the correct results of 983. I have also written a program to use iterative-deepening BFS to enumerate all positions and it confirms that 893 is indeed correct. But how do I get GAP to give me this number? I feel like I have all the pieces but I don't know how to put them together. If I can get it working for this simple case, that will allow me to tackle much more complicated puzzles for which a manual by-hand calculation just isn't possible. Thanks in advance for any guidance or pointers for how to handle this! Best regards, Brandon -----BEGIN PGP SIGNATURE----- Version: GnuPG v2 iEYEARECAAYFAld69doACgkQqaGPzAsl94KQ+QCeMHWG5rHLUwaKXO8jvCoV5nt2 fE0AoKqEvsi9hdB87DU5IBtMruzmBfVM =KoEP -----END PGP SIGNATURE----- From r_n_tsai at yahoo.com Thu Jul 7 07:19:06 2016 From: r_n_tsai at yahoo.com (R.N. Tsai) Date: Thu, 7 Jul 2016 06:19:06 +0000 (UTC) Subject: [GAP Forum] running with grape package under windows 64 References: <1659561778.208289.1467872346589.JavaMail.yahoo.ref@mail.yahoo.com> Message-ID: <1659561778.208289.1467872346589.JavaMail.yahoo@mail.yahoo.com> Dear GAP forum,I'm doing some calculations with gap+grape package under windows.?Things work fine with the gap32 version, but grape doesn't seem to workwith gap64. The graphs I'm working with are very large and I would liketo run gap64 with 16GBytes memory. Is there a 64 bit version of grapeavailable somewhere?Thanks,R.N. From juergen.mueller at math.rwth-aachen.de Thu Jul 7 08:36:41 2016 From: juergen.mueller at math.rwth-aachen.de (Juergen Mueller) Date: Thu, 7 Jul 2016 09:36:41 +0200 Subject: [GAP Forum] Applying Burnside's Lemma to the 2x2x2 Rubik's Cube? Message-ID: <20160707073641.GA11920@localhost.localdomain> Dear Forum, I have been thinking about the problem Brandon is asking. Here is a quick computation: samecolors := [[1, 5, 12, 16, 17, 23],[3, 7, 10, 14, 19, 21], [2, 4, 6, 8, 9, 11, 13, 15, 18, 20, 22, 24]]; # this is not stricly necessary, but helps me understanding the problem lst:=[]; for i in [1..Length(samecolors)] do for j in samecolors[i] do lst[j]:=i; od; od; TU := (1, 2, 3, 4)(5, 7, 9, 11)(6, 8, 10, 12); RU := (1, 2, 3, 4)(5, 7, 9, 11)(6, 8, 10, 12) (13, 15, 17, 19)(14, 16, 18, 20)(21, 22, 23, 24); RL := (5, 6, 14, 13)(1, 7, 22, 20)(2, 15, 21, 12) (4, 8, 23, 19)(3, 16, 24, 11)(10, 9, 17, 18); G:=Group([TU,RU,RL]); # order 88179840 S:=Group([RU,RL]); # S_4 H:=Stabiliser(G,lst,Permuted); # order 4320 orb:=Orbit(G,lst,Permuted);; # length 20412 orbs:=OrbitsDomain(S,orb,Permuted);; # 894 reps:=List(orbs,Minimum);; len:=List(orbs,Length);; Collected(len); # [ [ 4, 3 ], [ 8, 3 ], [ 12, 78 ], [ 24, 810 ] ] This says that 810 of the 894 S-orbits are indeed regular, but there are quite a few shorter ones. Usually, one would like to see how the latter look like... (In the meantime, I have checked privately with Brandon, that 894 indeed is the correct answer.) Anyway, this runs in less than a second on my machine. So, although it involves computing the full orbit of the puzzle group, and the suborbits under the rotational symmetry group, it does not seem to make much sense to invoke Burnside's Lemma to just find the number of S-orbits. Of course this picture might change, when it comes to different problems... Best wishes, J?rgen From alexander.konovalov at st-andrews.ac.uk Thu Jul 7 10:46:34 2016 From: alexander.konovalov at st-andrews.ac.uk (Alexander Konovalov) Date: Thu, 7 Jul 2016 09:46:34 +0000 Subject: [GAP Forum] running with grape package under windows 64 In-Reply-To: <1659561778.208289.1467872346589.JavaMail.yahoo@mail.yahoo.com> References: <1659561778.208289.1467872346589.JavaMail.yahoo.ref@mail.yahoo.com> <1659561778.208289.1467872346589.JavaMail.yahoo@mail.yahoo.com> Message-ID: <01FDD14C-4B19-43A8-BDA8-D4CF2C99A159@st-andrews.ac.uk> > On 7 Jul 2016, at 07:19, R.N. Tsai wrote: > > Dear GAP forum,I'm doing some calculations with gap+grape package under windows. Things work fine with the gap32 version, but grape doesn't seem to workwith gap64. The graphs I'm working with are very large and I would liketo run gap64 with 16GBytes memory. Is there a 64 bit version of grapeavailable somewhere?Thanks,R.N. Dear R.N., Grape uses a stand-alone binary, so it can interact with 64-bit GAP version even being 32-bit application. The problem is that GAP looks for the binary for GRAPE using the architecture-dependent path which it assumes to be 'x86_64-unknown-cygwin-gcc-default64'. As a quick workaround, I suggest to try this: in the 'bin' directory of your GRAPE installation, you should have the directory 'i686-pc-cygwin-gcc-default32'. Copy it to the directory 'x86_64-unknown-cygwin-gcc-default64' and try to load GRAPE from 64-bit version of GAP now. Please let me know if this helps! Best wishes Alexander From r_n_tsai at yahoo.com Thu Jul 7 20:57:38 2016 From: r_n_tsai at yahoo.com (R.N. Tsai) Date: Thu, 7 Jul 2016 19:57:38 +0000 (UTC) Subject: [GAP Forum] running with grape package under windows 64 In-Reply-To: <01FDD14C-4B19-43A8-BDA8-D4CF2C99A159@st-andrews.ac.uk> References: <1659561778.208289.1467872346589.JavaMail.yahoo.ref@mail.yahoo.com> <1659561778.208289.1467872346589.JavaMail.yahoo@mail.yahoo.com> <01FDD14C-4B19-43A8-BDA8-D4CF2C99A159@st-andrews.ac.uk> Message-ID: <1002701175.502384.1467921458261.JavaMail.yahoo@mail.yahoo.com> Dear Alexander and forum, Thanks for your suggestion; it did in fact work, so now grape works with both gap32 and gap64.Performance under windows seems to get much worse as the size of the graph gets bigger. I movedto linux64 and the performance is much better there until gap runs out of memory. I have some followup questions regarding memory management in gap and graph calculations with grapeor other packages : (the moderators of the forum can move to another topic if they think that's better). (1)I'm using grape to calculate the girth of large graphs. These graphs are the "Tanner graphs" of low? ?density parity check (LDPC) codes. These type of graphs are frequently used in coding theory to analyze? ?the error correcting capability of the code. In particular, the length of the shortest cycles (girth) and? ?number of such cycles is important. I looked in the guava package which has some constructs for LDPC codes? ?("QCLDPCCodeFromGroup" for example) but nothing specific for Tanner graphs..... (2)Tanner graphs are easily described : take the parity check matrix of the code "H" of size mxn and form? ?A=(m+n)x(m+n) matrix defined as A=[0,H],[H',0]]; here H' is transpose. H is large (say 10K x 10K) binary? ?matrix (0,1 entries) but very sparse (percentage of 1's is <2%). Here's how I get "A" from "H" : ? dim:=DimensionsMat(H);m:=dim[1];n:=dim[2];? T:=H;? T1:=List([1..m],x->Concatenation(0*[1..m],T[x]));? T:=TransposedMat(H); ? T2:=List([1..n],x->Concatenation(T[x],0*[1..n]));? A:=Concatenation(T1,T2); ? This results in a large sparse matrix "A" which is >99% 0's. Is there a better way to define it? (3)With the adjacency matrix "A" I define a graph "gamma" using grape's documentation : ? ?gamma:=Graph(Group(()),[1..n+m],OnPoints,function(x,y)return A[x][y]=1;end,true); ? ?by the construction of "A", it is known to be bipartite with the partition of vertices well? ?known, namely [1..m] and [m+1..m+n] but I don't know how to use this extra knowledge to help? ?grape work with the graph more efficiently. (4)I get the girth through girth:=Girth(gamma); I would also like to get the number of cycles? ?with length=girth but I don't know how that's done. (5)Finally, is there a way to "clear memory" within gap? I noticed that if I process H1 then? ?try to process H2 in the same gap session gap reports that the memory limit was exceeded, but? ?if I exit gap and start it again with H2, then it works. I don't save anything from the calculations? ?(I just print out the results) so I don't know what's using the memory. Here's an example :? ?gap>Process(H1); #finishes ok; no variables saved...? ?gap>Process(H2); #comes back with memory problem? ?if I exit gap and start again? ?gap>Process(H2); #now this works without memory problem. Thanks for your help. R.N. From: Alexander Konovalov To: R.N. Tsai Cc: GAP Forum Sent: Thursday, July 7, 2016 2:46 AM Subject: Re: [GAP Forum] running with grape package under windows 64 > On 7 Jul 2016, at 07:19, R.N. Tsai wrote: > > Dear GAP forum,I'm doing some calculations with gap+grape package under windows. Things work fine with the gap32 version, but grape doesn't seem to workwith gap64. The graphs I'm working with are very large and I would liketo run gap64 with 16GBytes memory. Is there a 64 bit version of grapeavailable somewhere?Thanks,R.N. Dear R.N., Grape uses a stand-alone binary, so it can interact with 64-bit GAP version even being 32-bit application. The problem is that GAP looks for the binary for GRAPE using the architecture-dependent path which it assumes to be 'x86_64-unknown-cygwin-gcc-default64'. As a quick workaround, I suggest to try this: in the 'bin' directory of your GRAPE installation, you should have the directory 'i686-pc-cygwin-gcc-default32'. Copy it to the directory 'x86_64-unknown-cygwin-gcc-default64' and try to load GRAPE from 64-bit version of GAP now. Please let me know if this helps! Best wishes Alexander From rbailey at grenfell.mun.ca Thu Jul 7 21:54:43 2016 From: rbailey at grenfell.mun.ca (Bailey, Robert F.) Date: Thu, 7 Jul 2016 20:54:43 +0000 Subject: [GAP Forum] running with grape package under windows 64 In-Reply-To: <1002701175.502384.1467921458261.JavaMail.yahoo@mail.yahoo.com> References: <1659561778.208289.1467872346589.JavaMail.yahoo.ref@mail.yahoo.com> <1659561778.208289.1467872346589.JavaMail.yahoo@mail.yahoo.com> <01FDD14C-4B19-43A8-BDA8-D4CF2C99A159@st-andrews.ac.uk> <1002701175.502384.1467921458261.JavaMail.yahoo@mail.yahoo.com> Message-ID: <88dca1fa3e214a40b55b0d825694c703@CAS2.swgc.ca> Dear R.N., Some suggestions: If you are using Linux, you can override GAP's default memory allocation when you launch it (see http://www.gap-system.org/Faq/faq.html#6.2 for a discussion of how to handle memory problems). However, if the size of the adjacency matrix is the issue, you should consider an alternative way of constructing the graph in GRAPE. In particular, as you say, because the graph is bipartite its adjacency matrix will have two massive all-zero blocks. One idea would be to rewrite the adjacency function (i.e. replace the "function(x,y) return A[x][y]=1; end" part) to be based on entries of H rather than A: the edges of your graph are determined precisely from where the entries of H are 1. That way, you avoid the need to construct A. If the graph is really sparse, just listing all edges may possibly take up less memory than the matrix H anyway, and the adjacency function would only need to return [x,y] as an edge if [x,y] is in your list. Regards, Robert Bailey. ============================== Dr. Robert Bailey Division of Science (Mathematics) Grenfell Campus Memorial University of Newfoundland Corner Brook, NL A2H 6P9, Canada Office: AS 3022 Phone: +1 (709) 637-6293 Web: http://www2.grenfell.mun.ca/rbailey/ -----Original Message----- From: forum-bounces at gap-system.org [mailto:forum-bounces at gap-system.org] On Behalf Of R.N. Tsai Sent: July-07-16 5:28 PM To: Alexander Konovalov Cc: GAP Forum Subject: Re: [GAP Forum] running with grape package under windows 64 Dear Alexander and forum, Thanks for your suggestion; it did in fact work, so now grape works with both gap32 and gap64.Performance under windows seems to get much worse as the size of the graph gets bigger. I movedto linux64 and the performance is much better there until gap runs out of memory. I have some followup questions regarding memory management in gap and graph calculations with grapeor other packages : (the moderators of the forum can move to another topic if they think that's better). (1)I'm using grape to calculate the girth of large graphs. These graphs are the "Tanner graphs" of low density parity check (LDPC) codes. These type of graphs are frequently used in coding theory to analyze the error correcting capability of the code. In particular, the length of the shortest cycles (girth) and number of such cycles is important. I looked in the guava package which has some constructs for LDPC codes ("QCLDPCCodeFromGroup" for example) but nothing specific for Tanner graphs..... (2)Tanner graphs are easily described : take the parity check matrix of the code "H" of size mxn and form A=(m+n)x(m+n) matrix defined as A=[0,H],[H',0]]; here H' is transpose. H is large (say 10K x 10K) binary matrix (0,1 entries) but very sparse (percentage of 1's is <2%). Here's how I get "A" from "H" : dim:=DimensionsMat(H);m:=dim[1];n:=dim[2]; T:=H; T1:=List([1..m],x->Concatenation(0*[1..m],T[x])); T:=TransposedMat(H); T2:=List([1..n],x->Concatenation(T[x],0*[1..n])); A:=Concatenation(T1,T2); This results in a large sparse matrix "A" which is >99% 0's. Is there a better way to define it? (3)With the adjacency matrix "A" I define a graph "gamma" using grape's documentation : gamma:=Graph(Group(()),[1..n+m],OnPoints,function(x,y)return A[x][y]=1;end,true); by the construction of "A", it is known to be bipartite with the partition of vertices well known, namely [1..m] and [m+1..m+n] but I don't know how to use this extra knowledge to help grape work with the graph more efficiently. (4)I get the girth through girth:=Girth(gamma); I would also like to get the number of cycles with length=girth but I don't know how that's done. (5)Finally, is there a way to "clear memory" within gap? I noticed that if I process H1 then try to process H2 in the same gap session gap reports that the memory limit was exceeded, but if I exit gap and start it again with H2, then it works. I don't save anything from the calculations (I just print out the results) so I don't know what's using the memory. Here's an example : gap>Process(H1); #finishes ok; no variables saved... gap>Process(H2); #comes back with memory problem if I exit gap and start again gap>Process(H2); #now this works without memory problem. Thanks for your help. R.N. From: Alexander Konovalov To: R.N. Tsai Cc: GAP Forum Sent: Thursday, July 7, 2016 2:46 AM Subject: Re: [GAP Forum] running with grape package under windows 64 > On 7 Jul 2016, at 07:19, R.N. Tsai wrote: > > Dear GAP forum,I'm doing some calculations with gap+grape package under windows. Things work fine with the gap32 version, but grape doesn't seem to workwith gap64. The graphs I'm working with are very large and I would liketo run gap64 with 16GBytes memory. Is there a 64 bit version of grapeavailable somewhere?Thanks,R.N. Dear R.N., Grape uses a stand-alone binary, so it can interact with 64-bit GAP version even being 32-bit application. The problem is that GAP looks for the binary for GRAPE using the architecture-dependent path which it assumes to be 'x86_64-unknown-cygwin-gcc-default64'. As a quick workaround, I suggest to try this: in the 'bin' directory of your GRAPE installation, you should have the directory 'i686-pc-cygwin-gcc-default32'. Copy it to the directory 'x86_64-unknown-cygwin-gcc-default64' and try to load GRAPE from 64-bit version of GAP now. Please let me know if this helps! Best wishes Alexander _______________________________________________ Forum mailing list Forum at mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum This electronic communication is governed by the terms and conditions at http://www.mun.ca/cc/policies/electronic_communications_disclaimer_2011.php. From r_n_tsai at yahoo.com Fri Jul 8 03:18:19 2016 From: r_n_tsai at yahoo.com (R.N. Tsai) Date: Fri, 8 Jul 2016 02:18:19 +0000 (UTC) Subject: [GAP Forum] running with grape package under windows 64 In-Reply-To: <88dca1fa3e214a40b55b0d825694c703@CAS2.swgc.ca> References: <1659561778.208289.1467872346589.JavaMail.yahoo.ref@mail.yahoo.com> <1659561778.208289.1467872346589.JavaMail.yahoo@mail.yahoo.com> <01FDD14C-4B19-43A8-BDA8-D4CF2C99A159@st-andrews.ac.uk> <1002701175.502384.1467921458261.JavaMail.yahoo@mail.yahoo.com> <88dca1fa3e214a40b55b0d825694c703@CAS2.swgc.ca> Message-ID: <80304294.708113.1467944299535.JavaMail.yahoo@mail.yahoo.com> Dear Robert and forum, That was actually a very good suggestion. I changed the "Graph" call to one that usesthis function instead : ?tanner:=function(x,y)? if(x<=m)then?? ?if(y<=m)then? ? return false;? ?else? ? return H[x][y-m]=1;? ? fi;? else? ?if(y>m)then? ? return false;? ?else? ? return H[y][x-m]=1;? ?fi;? fi;?end; ?gamma:=Graph(Group(()),[1..n+m],OnPoints,tanner,true); ?girth:=Girth(gamma); which avoids creating the large matrix. There are no memory problems with this even for H larger than 10Kx10K.The speed is also very good even under windows64. I will need to look at larger matrices soon so any othersuggestions for efficiency are appreciated. Regards,R.N. From: "Bailey, Robert F." To: R.N. Tsai Cc: GAP Forum Sent: Thursday, July 7, 2016 1:54 PM Subject: RE: [GAP Forum] running with grape package under windows 64 Dear R.N., Some suggestions: If you are using Linux, you can override GAP's default memory allocation when you launch it (see http://www.gap-system.org/Faq/faq.html#6.2 for a discussion of how to handle memory problems). However, if the size of the adjacency matrix is the issue, you should consider an alternative way of constructing the graph in GRAPE.? In particular, as you say, because the graph is bipartite its adjacency matrix will have two massive all-zero blocks.? One idea would be to rewrite the adjacency function (i.e. replace the "function(x,y) return A[x][y]=1; end" part) to be based on entries of H rather than A: the edges of your graph are determined precisely from where the entries of H are 1.? That way, you avoid the need to construct A. If the graph is really sparse, just listing all edges may possibly take up less memory than the matrix H anyway, and the adjacency function would only need to return [x,y] as an edge if [x,y] is in your list. Regards, Robert Bailey. ============================== Dr. Robert Bailey Division of Science (Mathematics) Grenfell Campus Memorial University of Newfoundland Corner Brook, NL A2H 6P9, Canada Office: AS 3022 Phone: +1 (709) 637-6293 Web: http://www2.grenfell.mun.ca/rbailey/ -----Original Message----- From: forum-bounces at gap-system.org [mailto:forum-bounces at gap-system.org] On Behalf Of R.N. Tsai Sent: July-07-16 5:28 PM To: Alexander Konovalov Cc: GAP Forum Subject: Re: [GAP Forum] running with grape package under windows 64 Dear Alexander and forum, Thanks for your suggestion; it did in fact work, so now grape works with both gap32 and gap64.Performance under windows seems to get much worse as the size of the graph gets bigger. I movedto linux64 and the performance is much better there until gap runs out of memory. I have some followup questions regarding memory management in gap and graph calculations with grapeor other packages : (the moderators of the forum can move to another topic if they think that's better). (1)I'm using grape to calculate the girth of large graphs. These graphs are the "Tanner graphs" of low? density parity check (LDPC) codes. These type of graphs are frequently used in coding theory to analyze? the error correcting capability of the code. In particular, the length of the shortest cycles (girth) and? number of such cycles is important. I looked in the guava package which has some constructs for LDPC codes? ("QCLDPCCodeFromGroup" for example) but nothing specific for Tanner graphs..... (2)Tanner graphs are easily described : take the parity check matrix of the code "H" of size mxn and form? A=(m+n)x(m+n) matrix defined as A=[0,H],[H',0]]; here H' is transpose. H is large (say 10K x 10K) binary? matrix (0,1 entries) but very sparse (percentage of 1's is <2%). Here's how I get "A" from "H" : ? dim:=DimensionsMat(H);m:=dim[1];n:=dim[2];? T:=H;? T1:=List([1..m],x->Concatenation(0*[1..m],T[x]));? T:=TransposedMat(H); ? T2:=List([1..n],x->Concatenation(T[x],0*[1..n]));? A:=Concatenation(T1,T2); ? This results in a large sparse matrix "A" which is >99% 0's. Is there a better way to define it? (3)With the adjacency matrix "A" I define a graph "gamma" using grape's documentation : ? gamma:=Graph(Group(()),[1..n+m],OnPoints,function(x,y)return A[x][y]=1;end,true); ? by the construction of "A", it is known to be bipartite with the partition of vertices well? known, namely [1..m] and [m+1..m+n] but I don't know how to use this extra knowledge to help? grape work with the graph more efficiently. (4)I get the girth through girth:=Girth(gamma); I would also like to get the number of cycles? with length=girth but I don't know how that's done. (5)Finally, is there a way to "clear memory" within gap? I noticed that if I process H1 then? try to process H2 in the same gap session gap reports that the memory limit was exceeded, but? if I exit gap and start it again with H2, then it works. I don't save anything from the calculations? (I just print out the results) so I don't know what's using the memory. Here's an example :? gap>Process(H1); #finishes ok; no variables saved...? gap>Process(H2); #comes back with memory problem? if I exit gap and start again? gap>Process(H2); #now this works without memory problem. Thanks for your help. R.N. ? ? ? From: Alexander Konovalov To: R.N. Tsai Cc: GAP Forum Sent: Thursday, July 7, 2016 2:46 AM Subject: Re: [GAP Forum] running with grape package under windows 64 > On 7 Jul 2016, at 07:19, R.N. Tsai wrote: > > Dear GAP forum,I'm doing some calculations with gap+grape package under windows. Things work fine with the gap32 version, but grape doesn't seem to workwith gap64. The graphs I'm working with are very large and I would liketo run gap64 with 16GBytes memory. Is there a 64 bit version of grapeavailable somewhere?Thanks,R.N. Dear R.N., Grape uses a stand-alone binary, so it can interact with 64-bit GAP version even being 32-bit application. The problem is that GAP looks for the binary for GRAPE using the architecture-dependent path which it assumes to be 'x86_64-unknown-cygwin-gcc-default64'. As a quick workaround, I suggest to try this: in the 'bin' directory of your GRAPE installation, you should have the directory 'i686-pc-cygwin-gcc-default32'. Copy it to the directory 'x86_64-unknown-cygwin-gcc-default64' and try to load GRAPE from 64-bit version of GAP now. Please let me know if this helps! Best wishes Alexander _______________________________________________ Forum mailing list Forum at mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum This electronic communication is governed by the terms and conditions at http://www.mun.ca/cc/policies/electronic_communications_disclaimer_2011.php. From hutagana at gmail.com Fri Jul 15 02:26:19 2016 From: hutagana at gmail.com (Huta Gana) Date: Fri, 15 Jul 2016 08:26:19 +0700 Subject: [GAP Forum] Bijection from k^n to the extension of k of degree n Message-ID: Dear GAP forum, I would like to ask some questions relating to the construction of the bijection from k^n to the extension of k of degree n. I have used the following codes *gap> k:=GF(2^2);;gap> x:=X(k,"x");; g:=x^3+x+1;; gap> Kg:=AlgebraicExtension(k,g);;* Then, an element in Kg has a form: a+bt+bt^2 (a, b, c in *k*, g(t)=0). I want to construct the bijection T: *k*^3->Kg is define by (a,b,c)->a+bt+bt^2. As we know that *k*^3 and Kg are 3-dimesional vector space over over *k*. To consider Kg as a *k*-vector space, I use *gap> Kg:=AsVectorSpace(k,Kg);* So, we can build an isomorphism from *k*^3 to K. But *gap> Size(Basis(Kg));* *6 //not 3gap> Size(Basis(k^3));3* >From that, I can not construct an isomorphism from to *k*^3 ->Kg. *gap>f:=LeftModuleGeneralMappingByImages(k^3,Kg,Basis(k^3),Basis(Kg));Error, and must have the same length* Could you help me how to construct the the bijection T: *k*^3->Kg? Thank you very much. Best regards, Huta From lovepgroups at gmail.com Mon Jul 18 16:56:27 2016 From: lovepgroups at gmail.com (abdulhakeem alayiwola) Date: Mon, 18 Jul 2016 16:56:27 +0100 Subject: [GAP Forum] Automorphism group of Aut(D2n) In-Reply-To: References: Message-ID: Dear forum, Can anyone in the house describe the steps to find Aut(D2n) using GAP. Note that Aut(D2n) is Automorphism group of Dihedral group of order 2n. Regards. From markus.szymik at math.ntnu.no Mon Jul 18 18:10:31 2016 From: markus.szymik at math.ntnu.no (Markus Szymik) Date: Mon, 18 Jul 2016 17:10:31 +0000 Subject: [GAP Forum] Automorphism group of Aut(D2n) In-Reply-To: References: Message-ID: <92C4B70B-FD10-41B0-A1FD-4C92A8F68C16@math.ntnu.no> If you only care about the result: Aut(D_2n) is isomorphic to the semi-direct product Z/n x| (Z/n)^x for the natural action of the units on the ring. You can even take n = \infty in that formula. > On Jul 18, 2016, at 11:56 AM, abdulhakeem alayiwola wrote: > > Dear forum, > Can anyone in the house describe the steps to find Aut(D2n) using GAP. > Note that Aut(D2n) is Automorphism group of Dihedral group of order 2n. > Regards. > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From lovepgroups at gmail.com Tue Jul 19 05:30:21 2016 From: lovepgroups at gmail.com (abdulhakeem alayiwola) Date: Tue, 19 Jul 2016 05:30:21 +0100 Subject: [GAP Forum] Automorphism group of Aut(D2n) In-Reply-To: References: Message-ID: I am more interested in the procedure than the result. If anyone can help we the procedure or steps to find Aut(D2n) using GAP. I will be very glad on any hint as well. Regards On Jul 18, 2016 4:56 PM, "abdulhakeem alayiwola" wrote: > Dear forum, > Can anyone in the house describe the steps to find Aut(D2n) using GAP. > Note that Aut(D2n) is Automorphism group of Dihedral group of order 2n. > Regards. > From sven.reichard at tu-dresden.de Tue Jul 19 07:39:18 2016 From: sven.reichard at tu-dresden.de (Sven Reichard) Date: Tue, 19 Jul 2016 08:39:18 +0200 Subject: [GAP Forum] Automorphism group of Aut(D2n) In-Reply-To: References: Message-ID: <578DCB16.3000909@tu-dresden.de> Dear Abdulhakeem, I don't think GAP can give you a general answer. However, for each value of n it is straightforward: gap> n := 10; 10 gap> g := DihedralGroup(2*n); gap> aut := AutomorphismGroup(g); gap> StructureDescription(aut); "C2 x (C5 : C4)" Hope this helps, Sven. -- Institut f?r Algebra TU Dresden On 18.07.2016 17:56, abdulhakeem alayiwola wrote: > Dear forum, > Can anyone in the house describe the steps to find Aut(D2n) using GAP. > Note that Aut(D2n) is Automorphism group of Dihedral group of order 2n. > Regards. > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum > From tkohl at math.bu.edu Tue Jul 19 14:26:14 2016 From: tkohl at math.bu.edu (tkohl at math.bu.edu) Date: Tue, 19 Jul 2016 09:26:14 -0400 (EDT) Subject: [GAP Forum] Automorphism group of Aut(D2n) In-Reply-To: References: Message-ID: There are a couple of ways to do it: -------------------------------------------- n:=3;; # or whatever value you wish G:=DihedralGroup(2*n); RG:=Action(G,Elements(G),OnRight); # basically the right regular representation AutG:=Action(AutomorphismGroup(G),Elements(G)); -------------------------------------------- Alternately, you can do this: -------------------------------------------- n:=3;; G:=DihedralGroup(2*n); RG:=Action(G,Elements(G),OnRight); # basically the right regular representation HolG:=Normalizer(SymmetricGroup(2*n),RG);; AutG:=AsGroup(Filtered(Elements(HolG),g->OnPoints(1,g)=1));; -------------------------------------------- which is a bit more computationally intensive as n increases because one is filtering the elements of a group (the holomorph). However, it's nice in that it is a manifestation of one of the classic definitions of the automorphism group, namely as the set of those elements (permutations of G as a set) which normalize the right (or left) regular represenation and fix the identity, which of course the automorphism group *must* do. (Note, the ambient symmetric group these are embedded in is not Perm(G) but rather S_{2n}, so one is tacitly identifying '1' with 'e_G' the identity of G.) -Tim K. On Tue, 19 Jul 2016, abdulhakeem alayiwola wrote: > I am more interested in the procedure than the result. If anyone can help > we the procedure or steps to find Aut(D2n) using GAP. I will be very glad > on any hint as well. > Regards > > On Jul 18, 2016 4:56 PM, "abdulhakeem alayiwola" > wrote: > > > Dear forum, > > Can anyone in the house describe the steps to find Aut(D2n) using GAP. > > Note that Aut(D2n) is Automorphism group of Dihedral group of order 2n. > > Regards. > > > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum > -- From oxeimon at gmail.com Wed Jul 20 06:35:33 2016 From: oxeimon at gmail.com (Will Chen) Date: Wed, 20 Jul 2016 01:35:33 -0400 Subject: [GAP Forum] what does "." and "=" mean in StructureDescription? Message-ID: A good example is AllSmallGroups(96,191), which seems to have StructureDescription: "(C2.S4=SL(2.3).C2):C2" I understand that the colon ":" denotes semidirect product, so this is the semidirect product of something with the cyclic group of order 2. Here is C2 the normal subgroup of the semidirect product? Does the normal subgroup always appear on the right? What about the rest? Since the whole group has order 96, the stuff on the left of the ":" must have order 48, so I can only assume that the "=" means "or equivalently...". Is that correct? In this case, I'm guessing "." refers to a non-split extension? If I see G.H, which is the normal subgroup? Thanks, - Will -- William Yun Chen Ph.D. Student & Graduate Teaching Associate Department of Mathematics Pennsylvania State University, University Park, PA, 16801 chen_w at math.psu.edu oxeimon at gmail.com From max at quendi.de Wed Jul 20 09:48:20 2016 From: max at quendi.de (Max Horn) Date: Wed, 20 Jul 2016 10:48:20 +0200 Subject: [GAP Forum] what does "." and "=" mean in StructureDescription? In-Reply-To: References: Message-ID: Dear Will, > On 20 Jul 2016, at 07:35, Will Chen wrote: > > A good example is AllSmallGroups(96,191), which seems to have > StructureDescription: > > "(C2.S4=SL(2.3).C2):C2" > > I understand that the colon ":" denotes semidirect product, so this is the > semidirect product of something with the cyclic group of order 2. Correct. > > Here is C2 the normal subgroup of the semidirect product? Does the normal > subgroup always appear on the right? No, always on the left, i.e. C_2 acts on the normal subgroup on the left. > What about the rest? Since the whole group has order 96, the stuff on the > left of the ":" must have order 48, so I can only assume that the "=" means > "or equivalently...". Is that correct? Yes, correct. > > In this case, I'm guessing "." refers to a non-split extension? Correct. > If I see G.H, which is the normal subgroup? The normal subgroup is always on the left. This fits with the overall GAP convention of always considering right actions. Most of this is explained in the manual (with the exception of the "=" sign, unfortunately). If you enter "?StructureDescription" on the GAP prompt and press return, you can see the manual entry. Alternatively, look here: Cheers, Max From hutagana at gmail.com Wed Jul 20 15:33:20 2016 From: hutagana at gmail.com (Huta Gana) Date: Wed, 20 Jul 2016 21:33:20 +0700 Subject: [GAP Forum] How to compute the image of a variable under a map Message-ID: Dear GAP forum, I have run the following codes: *gap> k:=GF(2^2);;gap>T:=MappingByFunction(k,k,t->t^2);;gap> x:=Indeterminate(k,"x");gap> Image(T,x);Error, usage: Image(), Image(,), Image(,)* It appears the error because x is not an element of k *gap> x in k;* *false* Could you show me how to compute the image of a variable under a map? Thank you very much. Best regards, Huta From mathieu.dutour at gmail.com Fri Jul 22 09:54:49 2016 From: mathieu.dutour at gmail.com (Mathieu Dutour) Date: Fri, 22 Jul 2016 10:54:49 +0200 Subject: [GAP Forum] Loading problem in GAP Message-ID: Dear all, when trying to debug GAP packages installation, it is common to use SetUserPreference( "InfoPackageLoadingLevel", 4 ); However, if one does that then the error is following: Error, PrintFormattedString: function is not yet defined in ValueGlobal( "PrintFormattedString" )( s ); at /home/mathieu/opt/gap4r8p4/lib/info.gi:85 called from fun( "#I " ); at /home/mathieu/opt/gap4r8p4/lib/info.gi:93 called from fun( InfoData.LastClass, InfoData.LastLevel, arglist ); at /home/mathieu/opt/gap4r8p4/lib/info.gi:336 called from message[1] at /home/mathieu/opt/gap4r8p4/lib/package.gi:544 called from LogPackageLoadingMessage( PACKAGE_DEBUG, "finish reading file 'init.g'", info.PackageName ); at /home/mathieu/opt/gap4r8p4/lib/package.gi:1589 called from LoadPackage( pair[1], pair[2], false ) at /home/mathieu/opt/gap4r8p4/lib/ package.gi:1832 called from ... at line 890 of /home/mathieu/opt/gap4r8p4/lib/init.g Syntax warning: unbound global variable in /home/mathieu/opt/gap4r8p4/lib/init.g line 928 ColorPrompt( UserPreference( "UseColorPrompt" ) ); ^ Error, SetGasmanMessageStatus: function is not yet defined in SetGasmanMessageStatus( "none" ); at /home/mathieu/opt/gap4r8p4/lib/init.g:916 called from func( ); at /home/mathieu/opt/gap4r8p4/lib/system.g:156 called from ( ) called from read-eval loop at line 929 of /home/mathieu/opt/gap4r8p4/lib/init.g Error, Variable: 'L1_IMMUTABLE_ERROR' must have a value not in any function at line 366 of /home/mathieu/opt/gap4r8p4/lib/ obsolete.gi Syntax warning: unbound global variable in /home/mathieu/opt/gap4r8p4/lib/ obsolete.gi line 706 Print( CharacterTableDisplayLegendDefault( data ) ); ^ Packages: GAPDoc 1.5.1 Try '??help' for help. See also '?copyright', '?cite' and '?authors' gap> From ares.1995 at mail.ru Fri Jul 22 10:56:41 2016 From: ares.1995 at mail.ru (=?UTF-8?B?QXJzZW5lIEdhbHN0eWFu?=) Date: Fri, 22 Jul 2016 12:56:41 +0300 Subject: [GAP Forum] =?utf-8?q?_Counting_Latin_squares_using_a_package_=22?= =?utf-8?q?DESIGN=22=2E?= Message-ID: <1469181401.614333418@f419.i.mail.ru> ? Dear GAP forum, ? I'm interested in function SemiLatinSquareDuals(n,k) that is included in the package "DESIGN". I would like to know it is possible to use this function that to find the number of Latin squares of a given order.? ? As it is written in manual, this function " can classify semi-Latin squares with certain given properties, and return a list of their duals as block designs". Note that?(nxn)/1 semi-Latin square (that is, for k = 1)?is the same thing as a Latin square of order n. ? Thus, for k=1, function breaks set of Latin square of order n on the set non-isomorphic classes. Each record resulting list corresponds exactly to one of his class.?For n=3, it returns the following: gap> SemiLatinSquareDuals(3,1); [ rec( autSubgroup := Group([ (2,4)(3,7)(6,8), (2,3)(4,7)(5,9)(6,8), (1,5,9) ????? (2,6,7)(3,4,8), (1,4,7)(2,5,8)(3,6,9) ]), blockNumbers := [ 3 ],? ????? blockSizes := [ 3 ], blocks := [ [ 1, 5, 9 ], [ 2, 6, 7 ], [ 3, 4, 8 ] ] ??????? , isBinary := true, isBlockDesign := true, isSimple := true,? ????? pointNames := [ [ 1, 1 ], [ 1, 2 ], [ 1, 3 ], [ 2, 1 ], [ 2, 2 ],? ????????? [ 2, 3 ], [ 3, 1 ], [ 3, 2 ], [ 3, 3 ] ], r := 1,? ????? tSubsetStructure := rec( lambdas := [ 1 ], t := 1 ), v := 9 ) ] ? That is,?as I understand it, exist only one isomorphic class, i.e. the set of all Latin square of order 3 is one class. In other words, all Latin square of order 3 are pairwise isomorphic. That is, one can be obtained from the other by using composite of permutations of rows, columns and transposition operations. ? If I understand correctly, field "autSubgroup" is?permutation subgroup which is a group of automorphisms of a class of Latin squares corresponding to this record. That is, each element of this group carries permutations of rows, or(and) columns, or(and) transposition of Latin square. ? In this way, as I think, you can act on one element of this class by all elements of the group of automorphisms of this class in course and get all the elements of the class, i.e., all Latin squares of order 3. ? Field "blocks" is binary block design, which dual of Latin square (with the first row is ordered),?which belongs to the?class of Latin squares corresponding to this record. As I understand, the square can be recovered from the block design in the following way. i-th block of block design corresponding i-th symbol of Latin square and each point of block design represents number of cells of Latin square. ? In this way, ?it is possible to write a function that transforms a block design of each record in the Latin square, corresponding to it, and then acts on each obtained square of all permutations of automorphism group that corresponds to him. ? ? This is text of function "ConclusionSquares": Print("Load 'ConclusionSquares(n)'","\n"); ConclusionSquares:= function(n) local i, j, k, NumberClasses, SLSD, SubsidiaryArray, ListLatinSquares, LenList, SZ, permut, NumberRepetition, NumberRepetitionForEach, AutomSquare, AutomSquareForEach, NumberLatinSquareForEach; SLSD:= SemiLatinSquareDuals(n,1); NumberClasses:= Length(SLSD); ListLatinSquares:= [ ]; ? ?#conclusion Latin square that corresponding of records for i in [1..NumberClasses] do ? ListLatinSquares[i]:= [ ]; ? for j in [1..n] do ? ? for k in [1..n] do ? ? ? ListLatinSquares[i][SLSD[i].blocks[j][k]]:= j; ? ? od; ? od; ? SubsidiaryArray:= [ ]; ? for j in [1..n] do ? ? SubsidiaryArray[j]:= [ ]; ? ? for k in [1..n] do ? ? ? SubsidiaryArray[j][k]:= ListLatinSquares[i][n*(j-1) + k]; ? ? od; ? od; ? Print(" ",i,")","\n"); ? Display(SubsidiaryArray); ? Print("\n"); od; ? ?#count of Latin squares LenList:= NumberClasses; NumberRepetition:= 0; AutomSquare:= 0; for i in [1..NumberClasses] do ? permut:= Difference(AsList(SLSD[i].autSubgroup),[()]); ? SZ:= Size(permut); ? NumberLatinSquareForEach:= 1; ? NumberRepetitionForEach:= 0; ? AutomSquareForEach:= 0; ? for j in [1..SZ] do ? ? SubsidiaryArray:= [ ]; ? ? for k in [1..n*n] do ? ? ? SubsidiaryArray[k^permut[j]]:= ListLatinSquares[i][k]; ? ? od; ? ? if (SubsidiaryArray in ListLatinSquares) then ? ? ? NumberRepetitionForEach:= NumberRepetitionForEach + 1; ? ? ? ? #number repetitions for each record ? ? ? if (SubsidiaryArray = ListLatinSquares[i]) then ? ? ? ? AutomSquareForEach:= AutomSquareForEach + 1; ? ? #number of permutetions which not change Latin square ? ? ? fi; ? ? else ? ? ? NumberLatinSquareForEach:= NumberLatinSquareForEach + 1; ? #number different Latin square for each record ? ? ? LenList:= LenList + 1; ? ? ? ListLatinSquares[LenList]:= [ ]; ? ? ? ListLatinSquares[LenList]:= SubsidiaryArray; ? ? fi; ? od; ? Print ("[rec #",i,": Number elements in autSubgroup = ",SZ+1,"; Number Latin Squares = ",NumberLatinSquareForEach,"; Number repetition = ? ? ?",NumberRepetitionForEach,"; Number automorfisms = ",AutomSquareForEach,"]","\n","\n"); ? NumberRepetition:= NumberRepetition + NumberRepetitionForEach; ? AutomSquare:= AutomSquare + AutomSquareForEach; od; ? ?#conclusion list of Latin square for i in [NumberClasses+1..LenList] do ? SubsidiaryArray:= [ ]; ? for j in [1..n] do ? ? SubsidiaryArray[j]:= [ ]; ? ? for k in [1..n] do ? ? ? SubsidiaryArray[j][k]:= ListLatinSquares[i][n*(j-1) + k]; ? ? od; ? od; ? Print(" ",i,")","\n"); ? Display(SubsidiaryArray); od; return ["Length(ListLatinSquares)=",Length(ListLatinSquares)," NumberRepetition=",NumberRepetition," AutomSquare=",AutomSquare]; end; Function result for n=3: gap> ConclusionSquares(3);??????????? ? ?1) [ [? 1,? 2,? 3 ], ? [? 3,? 1,? 2 ], ? [? 2,? 3,? 1 ] ] [rec #1: Number elements in autSubgroup = 36; Number Latin Squares = 6; Number repetition = 30; Number automorfisms = 5] ?2) [ [? 1,? 3,? 2 ], ? [? 2,? 1,? 3 ], ? [? 3,? 2,? 1 ] ] ?3) [ [? 2,? 1,? 3 ], ? [? 3,? 2,? 1 ], ? [? 1,? 3,? 2 ] ] ?4) [ [? 3,? 1,? 2 ], ? [? 2,? 3,? 1 ], ? [? 1,? 2,? 3 ] ] ?5) [ [? 2,? 3,? 1 ], ? [? 1,? 2,? 3 ], ? [? 3,? 1,? 2 ] ] ?6) [ [? 3,? 2,? 1 ], ? [? 1,? 3,? 2 ], ? [? 2,? 1,? 3 ] ] [ "Length(ListLatinSquares)=", 6, " NumberRepetition=", 30, " AutomSquare=", 5 ] ?? So, we got exactly 6 different squares. But the number of Latin squares of order 3 is 12 pieces. Due to the fact that all first rows are different and the number is 3 !, there is an idea to mix all the rows from the second to last together. That is , multiplied by 2 !. Just get 3 ! * 2 ! = 12 . But this idea is crumbling at n=4: gap> ConclusionSquares(4); ?1) [ [? 1,? 2,? 3,? 4 ], ? [? 4,? 1,? 2,? 3 ], ? [? 3,? 4,? 1,? 2 ], ? [? 2,? 3,? 4,? 1 ] ] ?2) [ [? 1,? 2,? 3,? 4 ], ? [? 2,? 1,? 4,? 3 ], ? [? 3,? 4,? 1,? 2 ], ? [? 4,? 3,? 2,? 1 ] ] [rec #1: Number elements in autSubgroup = 64; Number Latin Squares = 8; Number repetition = 56; Number automorfisms = 7] [rec #2: Number elements in autSubgroup = 192; Number Latin Squares = 24; Number repetition = 168; Number automorfisms = 7] [ "Length(ListLatinSquares)=", 32, " NumberRepetition=", 224, " AutomSquare=", 14 ] Latin squares of order 4 exactly 576, but 32*3!=192. In this situation , of course, you can still guess as from 32 to receive 576 , because 576 divided by 32. But if n=5, then gap> ConclusionSquares(5); ?1) [ [? 1,? 2,? 3,? 4,? 5 ], ? [? 5,? 1,? 2,? 3,? 4 ], ? [? 4,? 5,? 1,? 2,? 3 ], ? [? 3,? 4,? 5,? 1,? 2 ], ? [? 2,? 3,? 4,? 5,? 1 ] ] ?2) [ [? 1,? 2,? 3,? 4,? 5 ], ? [? 3,? 1,? 5,? 2,? 4 ], ? [? 5,? 4,? 1,? 3,? 2 ], ? [? 2,? 5,? 4,? 1,? 3 ], ? [? 4,? 3,? 2,? 5,? 1 ] ] [rec #1: Number elements in autSubgroup = 200; Number Latin Squares = 20; Number repetition = 180; Number automorfisms = 9] [rec #2: Number elements in autSubgroup = 24; Number Latin Squares = 24; Number repetition = 0; Number automorfisms = 0] [ "Length(ListLatinSquares)=", 44, " NumberRepetition=", 180, " AutomSquare=", 9 ] Latin squares of order 5 exactly 161280 and 161280 is not divisible by 44. ? Therefore , from the number of Latin squares , so that you can get , it is impossible to find all the squares of a given order . That is, function SemiLatinSquareDuals does not answer the question about the number of Latin squares . ? The following questions . Am I right? ?Did I everything right and reasoned?? If so, then the function SemiLatinskuare ( n , k) is not working properly or I did not understand then what it does . ? By the way, this function has a few optional parameters: SemiLatinSquareDuals(n,k,maxmult,blockintsize, isolevel). A ll these parameters in this case, for k=1, are not interesting , because we consider only Latin squares . Thank you very much! Best regards, Arsene Galstyan ares.1995 at mail.ru From crgreathouse at gmail.com Wed Jul 6 19:39:07 2016 From: crgreathouse at gmail.com (Charles Greathouse) Date: Wed, 06 Jul 2016 18:39:07 -0000 Subject: [GAP Forum] "Error, no method found" in IsIrreducibleRingElement Message-ID: I'm having trouble getting IsIrreducibleRingElement (and IsPrime) working on SmallRing elements. For example: gap> r := SmallRing(4,4); # example ring gap> e := Elements(r)[2]; # example element b gap> IsIrreducibleRingElement(e); Error, no method found! For debugging hints type ?Recovery from NoMethodFound Error, no 1st choice method found for IsIrreducibleRingElement' on 2 arguments called from IsIrreducibleRingElement( DefaultRing( [ r ] ), r ) called from ( ) called from read-eval loop at line 3 of *stdin* you can 'quit;' to quit to outer loop, or you can 'return;' to continue brk> quit; gap> IsIrreducibleRingElement(r,e); Error, no method found! For debugging hints type ?Recovery from NoMethodFound Error, no 1st choice method found for IsIrreducibleRingElement' on 2 arguments called from ( ) called from read-eval loop at line 3 of *stdin* you can 'quit;' to quit to outer loop, or you can 'return;' to continue Switching rings, elements, versions, and operating systems hasn't helped. Am I missing something? From Alexander.Hulpke at colostate.edu Mon Jul 25 13:33:18 2016 From: Alexander.Hulpke at colostate.edu (Hulpke,Alexander) Date: Mon, 25 Jul 2016 12:33:18 +0000 Subject: [GAP Forum] "Error, no method found" in IsIrreducibleRingElement In-Reply-To: References: Message-ID: Dear Charles Greathouse, Dear Forum, On Jul 6, 2016, at 8:38 PM, Charles Greathouse > wrote: I'm having trouble getting IsIrreducibleRingElement (and IsPrime) working on SmallRing elements. For example: Switching rings, elements, versions, and operating systems hasn't helped. The built-in irreducibility tests are currently only available for the integers (and some rings of integers) and polynomial rings over fields. There is no method for the small rings, which is the reason for the error message you received. Regards, Alexander Hulpke -- Colorado State University, Department of Mathematics, Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA email: hulpke at colostate.edu, Phone: ++1-970-4914288 http://www.math.colostate.edu/~hulpke From w_becker at hotmail.com Mon Jul 25 14:45:59 2016 From: w_becker at hotmail.com (Walter Becker) Date: Mon, 25 Jul 2016 13:45:59 +0000 Subject: [GAP Forum] groups of order p^3*q^3 Message-ID: The question is how many groups are there for these orders. Note here that the Group Constructions program of Besche and Eick give for the case of p=3 and q=7 the number 215 (and actually constructs them). The Thesis of of R. Laue (see page 226) seems to give 229 (plus the 25 direct product cases of the groups of order 27 and 343 ?). The compiled list of groups available on the internet does not contain an entry for the groups of order 9261. Comments ? Walter Becker From alexander.konovalov at st-andrews.ac.uk Mon Jul 25 15:02:22 2016 From: alexander.konovalov at st-andrews.ac.uk (Alexander Konovalov) Date: Mon, 25 Jul 2016 14:02:22 +0000 Subject: [GAP Forum] groups of order p^3*q^3 In-Reply-To: References: Message-ID: Dear Walter, > On 25 Jul 2016, at 14:45, Walter Becker wrote: > > The question is how many groups are there for these > > orders. Note here that the Group Constructions program > > of Besche and Eick give for the case of p=3 and q=7 the > > number 215 (and actually constructs them). The Thesis of > > of R. Laue (see page 226) seems to give 229 (plus the > > 25 direct product cases of the groups of order 27 and 343 ?). > > > The compiled list of groups available on the internet does > > not contain an entry for the groups of order 9261. The Group Numbers reproducibility project (https://github.com/alex-konovalov/gnu) actually has an entry for gnu(9261): see https://github.com/alex-konovalov/gnu/pull/43 and it is 215. Same number obtained by using the latest GrpConst package both in GAP 4.8.3 and master branch of the development version of GAP at the time of submission. I don't have Laue's thesis so I can not comment more on the reasons for the discrepancies, but hopefully some one else could add more. Best wishes Alexander From L.H.Soicher at qmul.ac.uk Mon Jul 25 16:37:33 2016 From: L.H.Soicher at qmul.ac.uk (Leonard Soicher) Date: Mon, 25 Jul 2016 16:37:33 +0100 Subject: [GAP Forum] Counting Latin squares using a package "DESIGN". In-Reply-To: <1469181401.614333418@f419.i.mail.ru> References: <1469181401.614333418@f419.i.mail.ru> Message-ID: <20160725153732.GA32024@maths.qmul.ac.uk> Dear Arsene Galstyan, Dear GAP-Forum, For positive integers n and k, the DESIGN package function call SemiLatinSquareDuals(n,k) returns a list of the duals of a complete list of isomorphism class representatives of the (n x n)/k semi-Latin squares, such that two such squares are considered to be isomorphic if one can be obtained from the other by a combination of renaming symbols, row permutations, column permutations, and transposing. Thus, the isomorphism class of a semi-Latin square S (with symbol set [1..n*k]) whose dual is obtained from a call to SemiLatinSquareDuals(n,k) is the orbit of S under the group generated by symbol permutations, row permutations, column permutations, and transposing. I've done this as shown in the appended log-file, obtaining the correct numbers of Latin squares of orders 3, 4, and 5. There are, however, much more efficient ways to find the *number* of Latin squares of order n if you are just interested in the number rather than all the Latin squares themselves. By the way, for a block design B in the DESIGN package, the component B.autSubgroup, if bound, is guaranteed to be a subgroup of the (full) automorphism group of B. Best wishes, Leonard --------------------------------------------------------------- gap> gap> LoadPackage("design"); ----------------------------------------------------------------------------- Loading GRAPE 4.7 (GRaph Algorithms using PErmutation groups) by Leonard H. Soicher (http://www.maths.qmul.ac.uk/~leonard/). Homepage: http://www.maths.qmul.ac.uk/~leonard/grape/ ----------------------------------------------------------------------------- ----------------------------------------------------------------------------- Loading DESIGN 1.6 (The Design Package for GAP) by Leonard H. Soicher (http://www.maths.qmul.ac.uk/~leonard/). Homepage: http://www.designtheory.org/software/gap_design/ ----------------------------------------------------------------------------- true gap> gap> SemiLatinSquareFromDual := function(D) > local n,S,names,i,pt; > n:=Length(D.blocks[1]); > S:=List([1..n],x->List([1..n],y->[])); > names:=D.pointNames; > for i in [1..Length(D.blocks)] do > for pt in D.blocks[i] do > Add(S[names[pt][1]][names[pt][2]],i); > od; > od; > return S; > end; function( D ) ... end gap> gap> OnSemiLatinSquares := function(S,g) > # Returns the image T of the (n x n)/k semi-Latin square S under g, > # where g acts by permuting the symbols [1..n*k], the rows > # (labelled [n*k+1..n*k+n]) and the columns (labelled [n*k+n+1..n*k+2*n]), > # and then possibly transposing the semi-Latin square. > local n,k,i,j,newposition,T; > n:=Length(S[1]); > k:=Length(S[1][1]); > T:=List([1..n],i->[]); > for i in [1..n] do > for j in [1..n] do > newposition:=OnSets([i+n*k,j+n*k+n],g); > T[newposition[1]-n*k][newposition[2]-n*k-n]:=OnSets(S[i][j],g); > od; > od; > return T; > end; function( S, g ) ... end gap> gap> SemiLatinSquareIsomorphismClass := function(S) > # Returns the full (weak) isomorphism class of the (n x n)/k > # semi-Latin square S. > local n,k,transposer,g_symbols,g_rows,g_columns,G,i; > n:=Length(S[1]); > k:=Length(S[1][1]); > transposer:=Product(List([n*k+1..n*k+n],i->(i,i+n))); > g_symbols:=GeneratorsOfGroup(SymmetricGroup([1..n*k])); > g_rows:=GeneratorsOfGroup(SymmetricGroup([n*k+1..n*k+n])); > g_columns:=GeneratorsOfGroup(SymmetricGroup([n*k+n+1..n*k+2*n])); > G:=Group(Concatenation(g_symbols,g_rows,g_columns,[transposer])); > return Set(Orbit(G,S,OnSemiLatinSquares)); > end; function( S ) ... end gap> gap> n:=3; 3 gap> duals:=SemiLatinSquareDuals(n,1);; gap> squares:=List(duals,SemiLatinSquareFromDual); [ [ [ [ 1 ], [ 2 ], [ 3 ] ], [ [ 3 ], [ 1 ], [ 2 ] ], [ [ 2 ], [ 3 ], [ 1 ] ] ] ] gap> classes:=List(squares,SemiLatinSquareIsomorphismClass);; gap> L:=List(classes,Length); [ 12 ] gap> Sum(L); 12 gap> gap> n:=4; 4 gap> duals:=SemiLatinSquareDuals(n,1);; gap> squares:=List(duals,SemiLatinSquareFromDual); [ [ [ [ 1 ], [ 2 ], [ 3 ], [ 4 ] ], [ [ 4 ], [ 1 ], [ 2 ], [ 3 ] ], [ [ 3 ], [ 4 ], [ 1 ], [ 2 ] ], [ [ 2 ], [ 3 ], [ 4 ], [ 1 ] ] ], [ [ [ 1 ], [ 2 ], [ 3 ], [ 4 ] ], [ [ 2 ], [ 1 ], [ 4 ], [ 3 ] ], [ [ 3 ], [ 4 ], [ 1 ], [ 2 ] ], [ [ 4 ], [ 3 ], [ 2 ], [ 1 ] ] ] ] gap> classes:=List(squares,SemiLatinSquareIsomorphismClass);; gap> L:=List(classes,Length); [ 432, 144 ] gap> Sum(L); 576 gap> gap> n:=5; 5 gap> duals:=SemiLatinSquareDuals(n,1);; gap> squares:=List(duals,SemiLatinSquareFromDual); [ [ [ [ 1 ], [ 2 ], [ 3 ], [ 4 ], [ 5 ] ], [ [ 5 ], [ 1 ], [ 2 ], [ 3 ], [ 4 ] ], [ [ 4 ], [ 5 ], [ 1 ], [ 2 ], [ 3 ] ], [ [ 3 ], [ 4 ], [ 5 ], [ 1 ], [ 2 ] ], [ [ 2 ], [ 3 ], [ 4 ], [ 5 ], [ 1 ] ] ], [ [ [ 1 ], [ 2 ], [ 3 ], [ 4 ], [ 5 ] ], [ [ 3 ], [ 1 ], [ 5 ], [ 2 ], [ 4 ] ], [ [ 5 ], [ 4 ], [ 1 ], [ 3 ], [ 2 ] ], [ [ 2 ], [ 5 ], [ 4 ], [ 1 ], [ 3 ] ], [ [ 4 ], [ 3 ], [ 2 ], [ 5 ], [ 1 ] ] ] ] gap> classes:=List(squares,SemiLatinSquareIsomorphismClass);; gap> L:=List(classes,Length); [ 17280, 144000 ] gap> Sum(L); 161280 gap> On Fri, Jul 22, 2016 at 12:56:41PM +0300, Arsene Galstyan wrote: > > > ? Dear GAP forum, > > ? I'm interested in function SemiLatinSquareDuals(n,k) that is included in the package "DESIGN". I would like to know it is possible to use this function that to find the number of Latin squares of a given order.? > ? As it is written in manual, this function " can classify semi-Latin squares with certain given properties, and return a list of their duals as block designs". Note that?(nxn)/1 semi-Latin square (that is, for k = 1)?is the same thing as a Latin square of order n. > ? Thus, for k=1, function breaks set of Latin square of order n on the set non-isomorphic classes. Each record resulting list corresponds exactly to one of his class.?For n=3, it returns the following: > > gap> SemiLatinSquareDuals(3,1); > [ rec( autSubgroup := Group([ (2,4)(3,7)(6,8), (2,3)(4,7)(5,9)(6,8), (1,5,9) > ????? (2,6,7)(3,4,8), (1,4,7)(2,5,8)(3,6,9) ]), blockNumbers := [ 3 ],? > ????? blockSizes := [ 3 ], blocks := [ [ 1, 5, 9 ], [ 2, 6, 7 ], [ 3, 4, 8 ] ] > ??????? , isBinary := true, isBlockDesign := true, isSimple := true,? > ????? pointNames := [ [ 1, 1 ], [ 1, 2 ], [ 1, 3 ], [ 2, 1 ], [ 2, 2 ],? > ????????? [ 2, 3 ], [ 3, 1 ], [ 3, 2 ], [ 3, 3 ] ], r := 1,? > ????? tSubsetStructure := rec( lambdas := [ 1 ], t := 1 ), v := 9 ) ] ? > > That is,?as I understand it, exist only one isomorphic class, i.e. the set of all Latin square of order 3 is one class. In other words, all Latin square of order 3 are pairwise isomorphic. That is, one can be obtained from the other by using composite of permutations of rows, columns and transposition operations. > ? If I understand correctly, field "autSubgroup" is?permutation subgroup which is a group of automorphisms of a class of Latin squares corresponding to this record. That is, each element of this group carries permutations of rows, or(and) columns, or(and) transposition of Latin square. > ? In this way, as I think, you can act on one element of this class by all elements of the group of automorphisms of this class in course and get all the elements of the class, i.e., all Latin squares of order 3. > ? Field "blocks" is binary block design, which dual of Latin square (with the first row is ordered),?which belongs to the?class of Latin squares corresponding to this record. As I understand, the square can be recovered from the block design in the following way. i-th block of block design corresponding i-th symbol of Latin square and each point of block design represents number of cells of Latin square. > ? In this way, ?it is possible to write a function that transforms a block design of each record in the Latin square, corresponding to it, and then acts on each obtained square of all permutations of automorphism group that corresponds to him. > ? > ? This is text of function "ConclusionSquares": > > Print("Load 'ConclusionSquares(n)'","\n"); > ConclusionSquares:= function(n) > local i, j, k, NumberClasses, SLSD, SubsidiaryArray, ListLatinSquares, LenList, SZ, permut, NumberRepetition, NumberRepetitionForEach, AutomSquare, AutomSquareForEach, NumberLatinSquareForEach; > SLSD:= SemiLatinSquareDuals(n,1); > NumberClasses:= Length(SLSD); > ListLatinSquares:= [ ]; > ? ?#conclusion Latin square that corresponding of records > for i in [1..NumberClasses] do > ? ListLatinSquares[i]:= [ ]; > ? for j in [1..n] do > ? ? for k in [1..n] do > ? ? ? ListLatinSquares[i][SLSD[i].blocks[j][k]]:= j; > ? ? od; > ? od; > ? SubsidiaryArray:= [ ]; > ? for j in [1..n] do > ? ? SubsidiaryArray[j]:= [ ]; > ? ? for k in [1..n] do > ? ? ? SubsidiaryArray[j][k]:= ListLatinSquares[i][n*(j-1) + k]; > ? ? od; > ? od; > ? Print(" ",i,")","\n"); > ? Display(SubsidiaryArray); > ? Print("\n"); > od; > ? ?#count of Latin squares > LenList:= NumberClasses; > NumberRepetition:= 0; > AutomSquare:= 0; > for i in [1..NumberClasses] do > ? permut:= Difference(AsList(SLSD[i].autSubgroup),[()]); > ? SZ:= Size(permut); > ? NumberLatinSquareForEach:= 1; > ? NumberRepetitionForEach:= 0; > ? AutomSquareForEach:= 0; > ? for j in [1..SZ] do > ? ? SubsidiaryArray:= [ ]; > ? ? for k in [1..n*n] do > ? ? ? SubsidiaryArray[k^permut[j]]:= ListLatinSquares[i][k]; > ? ? od; > ? ? if (SubsidiaryArray in ListLatinSquares) then > ? ? ? NumberRepetitionForEach:= NumberRepetitionForEach + 1; ? ? ? ? #number repetitions for each record > ? ? ? if (SubsidiaryArray = ListLatinSquares[i]) then > ? ? ? ? AutomSquareForEach:= AutomSquareForEach + 1; ? ? #number of permutetions which not change Latin square > ? ? ? fi; > ? ? else > ? ? ? NumberLatinSquareForEach:= NumberLatinSquareForEach + 1; ? #number different Latin square for each record > ? ? ? LenList:= LenList + 1; > ? ? ? ListLatinSquares[LenList]:= [ ]; > ? ? ? ListLatinSquares[LenList]:= SubsidiaryArray; > ? ? fi; > ? od; > ? Print ("[rec #",i,": Number elements in autSubgroup = ",SZ+1,"; Number Latin Squares = ",NumberLatinSquareForEach,"; Number repetition = ? ? ?",NumberRepetitionForEach,"; Number automorfisms = ",AutomSquareForEach,"]","\n","\n"); > ? NumberRepetition:= NumberRepetition + NumberRepetitionForEach; > ? AutomSquare:= AutomSquare + AutomSquareForEach; > od; > ? ?#conclusion list of Latin square > for i in [NumberClasses+1..LenList] do > ? SubsidiaryArray:= [ ]; > ? for j in [1..n] do > ? ? SubsidiaryArray[j]:= [ ]; > ? ? for k in [1..n] do > ? ? ? SubsidiaryArray[j][k]:= ListLatinSquares[i][n*(j-1) + k]; > ? ? od; > ? od; > ? Print(" ",i,")","\n"); > ? Display(SubsidiaryArray); > od; > return ["Length(ListLatinSquares)=",Length(ListLatinSquares)," NumberRepetition=",NumberRepetition," AutomSquare=",AutomSquare]; > > end; > Function result for n=3: > > gap> ConclusionSquares(3);??????????? ? > ?1) > [ [? 1,? 2,? 3 ], > ? [? 3,? 1,? 2 ], > ? [? 2,? 3,? 1 ] ] > > [rec #1: Number elements in autSubgroup = 36; Number Latin Squares = 6; Number repetition = > 30; Number automorfisms = 5] > > ?2) > [ [? 1,? 3,? 2 ], > ? [? 2,? 1,? 3 ], > ? [? 3,? 2,? 1 ] ] > ?3) > [ [? 2,? 1,? 3 ], > ? [? 3,? 2,? 1 ], > ? [? 1,? 3,? 2 ] ] > ?4) > [ [? 3,? 1,? 2 ], > ? [? 2,? 3,? 1 ], > ? [? 1,? 2,? 3 ] ] > ?5) > [ [? 2,? 3,? 1 ], > ? [? 1,? 2,? 3 ], > ? [? 3,? 1,? 2 ] ] > ?6) > [ [? 3,? 2,? 1 ], > ? [? 1,? 3,? 2 ], > ? [? 2,? 1,? 3 ] ] > > [ "Length(ListLatinSquares)=", 6, " NumberRepetition=", 30, " AutomSquare=", 5 ] > > ?? So, we got exactly 6 different squares. But the number of Latin squares of order 3 is 12 pieces. Due to the fact that all first rows are different and the number is 3 !, there is an idea to mix all the rows from the second to last together. That is , multiplied by 2 !. Just get 3 ! * 2 ! = 12 . But this idea is crumbling at n=4: > > gap> ConclusionSquares(4); > ?1) > [ [? 1,? 2,? 3,? 4 ], > ? [? 4,? 1,? 2,? 3 ], > ? [? 3,? 4,? 1,? 2 ], > ? [? 2,? 3,? 4,? 1 ] ] > > ?2) > [ [? 1,? 2,? 3,? 4 ], > ? [? 2,? 1,? 4,? 3 ], > ? [? 3,? 4,? 1,? 2 ], > ? [? 4,? 3,? 2,? 1 ] ] > > [rec #1: Number elements in autSubgroup = 64; Number Latin Squares = 8; Number repetition = > 56; Number automorfisms = 7] > > [rec #2: Number elements in autSubgroup = 192; Number Latin Squares = > 24; Number repetition = 168; Number automorfisms = 7] > > [ "Length(ListLatinSquares)=", 32, " NumberRepetition=", 224, " AutomSquare=", 14 ] > > > Latin squares of order 4 exactly 576, but 32*3!=192. In this situation , of course, you can still guess as from 32 to receive 576 , because 576 divided by 32. But if n=5, then > > gap> ConclusionSquares(5); > ?1) > [ [? 1,? 2,? 3,? 4,? 5 ], > ? [? 5,? 1,? 2,? 3,? 4 ], > ? [? 4,? 5,? 1,? 2,? 3 ], > ? [? 3,? 4,? 5,? 1,? 2 ], > ? [? 2,? 3,? 4,? 5,? 1 ] ] > > ?2) > [ [? 1,? 2,? 3,? 4,? 5 ], > ? [? 3,? 1,? 5,? 2,? 4 ], > ? [? 5,? 4,? 1,? 3,? 2 ], > ? [? 2,? 5,? 4,? 1,? 3 ], > ? [? 4,? 3,? 2,? 5,? 1 ] ] > > [rec #1: Number elements in autSubgroup = 200; Number Latin Squares = > 20; Number repetition = 180; Number automorfisms = 9] > > [rec #2: Number elements in autSubgroup = 24; Number Latin Squares = 24; Number repetition = > 0; Number automorfisms = 0] > > [ "Length(ListLatinSquares)=", 44, " NumberRepetition=", 180, " AutomSquare=", 9 ] > > Latin squares of order 5 exactly 161280 and 161280 is not divisible by 44. > ? Therefore , from the number of Latin squares , so that you can get , it is impossible to find all the squares of a given order . That is, function SemiLatinSquareDuals does not answer the question about the number of Latin squares . > ? The following questions . Am I right? ?Did I everything right and reasoned?? If so, then the function SemiLatinskuare ( n , k) is not working properly or I did not understand then what it does . > ? By the way, this function has a few optional parameters: SemiLatinSquareDuals(n,k,maxmult,blockintsize, isolevel). A ll these parameters in this case, for k=1, are not interesting , because we consider only Latin squares . > > Thank you very much! > > Best regards, > > Arsene Galstyan > ares.1995 at mail.ru > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From Bill.Allombert at math.u-bordeaux.fr Mon Jul 25 18:56:08 2016 From: Bill.Allombert at math.u-bordeaux.fr (Bill Allombert) Date: Mon, 25 Jul 2016 19:56:08 +0200 Subject: [GAP Forum] "Error, no method found" in IsIrreducibleRingElement In-Reply-To: References: Message-ID: <20160725175608.GC8631@yellowpig> On Mon, Jul 25, 2016 at 12:33:18PM +0000, Hulpke,Alexander wrote: > Dear Charles Greathouse, Dear Forum, > > On Jul 6, 2016, at 8:38 PM, Charles Greathouse > wrote: > > I'm having trouble getting IsIrreducibleRingElement (and IsPrime) working > on SmallRing elements. For example: > > > Switching rings, elements, versions, and operating systems hasn't helped. > > The built-in irreducibility tests are currently only available for the > integers (and some rings of integers) and polynomial rings over > fields. There is no method for the small rings, which is the reason > for the error message you received. All elements in a finite ring are either invertible or divisors of zero so in both case are not irreducible nor prime. So it should just return false. Cheers, Bill. From sam at Math.RWTH-Aachen.De Wed Jul 27 13:39:06 2016 From: sam at Math.RWTH-Aachen.De (Thomas Breuer) Date: Wed, 27 Jul 2016 14:39:06 +0200 Subject: [GAP Forum] Loading problem in GAP In-Reply-To: References: Message-ID: <20160727123906.GA19804@gemma.math.rwth-aachen.de> Dear GAP Forum, the problem reported by Mathieu Dutour is due to a recent code change. It will be fixed in the next GAP version. (For those who are interested in the details: A Boolean expression for testing whether the function 'PrintFormattedString' can already be called had been extended in such a way that 'true' is returned before the function is available. Therefore the error happens after the function has been declared and before it has been installed.) Thanks to Mathieu for reporting this bug, and sorry for the inconveniences. All the best, Thomas On Fri, Jul 22, 2016 at 10:54:49AM +0200, Mathieu Dutour wrote: > Dear all, > > when trying to debug GAP packages installation, > it is common to use > SetUserPreference( "InfoPackageLoadingLevel", 4 ); > > However, if one does that then the error is following: > Error, PrintFormattedString: function is not yet defined in > ValueGlobal( "PrintFormattedString" )( s ); at > /home/mathieu/opt/gap4r8p4/lib/info.gi:85 called from > fun( "#I " ); at /home/mathieu/opt/gap4r8p4/lib/info.gi:93 called from > fun( InfoData.LastClass, InfoData.LastLevel, arglist > ); at /home/mathieu/opt/gap4r8p4/lib/info.gi:336 called from > message[1] at /home/mathieu/opt/gap4r8p4/lib/package.gi:544 called from [...] From s.shpectorov at bham.ac.uk Wed Jul 27 16:49:39 2016 From: s.shpectorov at bham.ac.uk (Sergey Shpectorov) Date: Wed, 27 Jul 2016 15:49:39 +0000 Subject: [GAP Forum] Thue's Lemma Message-ID: <63278B7E72610F4C87EE13DA0E155D9601DF792CC2@EX11.adf.bham.ac.uk> Hello, Does GAP have a function for Thue's Lemma: Given integers m>1, X>0, Y>0, such that X<=m References: <63278B7E72610F4C87EE13DA0E155D9601DF792CC2@EX11.adf.bham.ac.uk> Message-ID: <20160728103253.GA5949@yellowpig> On Wed, Jul 27, 2016 at 03:49:39PM +0000, Sergey Shpectorov wrote: > Hello, > > Does GAP have a function for Thue's Lemma: > > Given integers m>1, X>0, Y>0, such that X<=m x,y such that |x| > or something equivalent? I could not find it in the documentation either. This is known under various other name such as "rational reconstruction", "rational lifting" and "half extended gcd". This is available in PARI/GP as bestappr. Cheers, Bill. From frank.luebeck at math.rwth-aachen.de Thu Jul 28 13:16:38 2016 From: frank.luebeck at math.rwth-aachen.de (Frank =?iso-8859-1?Q?L=FCbeck?=) Date: Thu, 28 Jul 2016 14:16:38 +0200 Subject: [GAP Forum] Thue's Lemma In-Reply-To: <63278B7E72610F4C87EE13DA0E155D9601DF792CC2@EX11.adf.bham.ac.uk> References: <63278B7E72610F4C87EE13DA0E155D9601DF792CC2@EX11.adf.bham.ac.uk> Message-ID: <20160728121638.GA7337@localhost.localdomain> On Wed, Jul 27, 2016 at 03:49:39PM +0000, Sergey Shpectorov wrote: > Hello, > > Does GAP have a function for Thue's Lemma: > > Given integers m>1, X>0, Y>0, such that X<=m there exist integers > x,y such that |x| > or something equivalent? > > Best, > Sergey Dear Sergey, dear Forum, See the documentation of the function ?RatNumberFromModular which is contained in the EDIM package. Does this what you are looking for? (When I wrote that function I was not aware that this is known under various names.) Best regards, Frank -- /// Dr. Frank L?beck, Lehrstuhl D f?r Mathematik, Pontdriesch 14/16, \\\ 52062 Aachen, Germany /// E-mail: Frank.Luebeck at Math.RWTH-Aachen.De \\\ WWW: http://www.math.rwth-aachen.de/~Frank.Luebeck/ From s.shpectorov at bham.ac.uk Thu Jul 28 19:26:58 2016 From: s.shpectorov at bham.ac.uk (Sergey Shpectorov) Date: Thu, 28 Jul 2016 18:26:58 +0000 Subject: [GAP Forum] Thue's Lemma In-Reply-To: <20160728121638.GA7337@localhost.localdomain> References: <63278B7E72610F4C87EE13DA0E155D9601DF792CC2@EX11.adf.bham.ac.uk>, <20160728121638.GA7337@localhost.localdomain> Message-ID: <63278B7E72610F4C87EE13DA0E155D9601DF794009@EX11.adf.bham.ac.uk> Dear Frank, Yes, this is exactly what I need!! And it seems to work well for very-very long numbers! All the best, Sergey ________________________________________ From: Frank L?beck [frank.luebeck at math.rwth-aachen.de] Sent: Thursday, July 28, 2016 1:16 PM To: Sergey Shpectorov Cc: forum at gap-system.org Subject: Re: [GAP Forum] Thue's Lemma On Wed, Jul 27, 2016 at 03:49:39PM +0000, Sergey Shpectorov wrote: > Hello, > > Does GAP have a function for Thue's Lemma: > > Given integers m>1, X>0, Y>0, such that X<=m there exist integers > x,y such that |x| > or something equivalent? > > Best, > Sergey Dear Sergey, dear Forum, See the documentation of the function ?RatNumberFromModular which is contained in the EDIM package. Does this what you are looking for? (When I wrote that function I was not aware that this is known under various names.) Best regards, Frank -- /// Dr. Frank L?beck, Lehrstuhl D f?r Mathematik, Pontdriesch 14/16, \\\ 52062 Aachen, Germany /// E-mail: Frank.Luebeck at Math.RWTH-Aachen.De \\\ WWW: http://www.math.rwth-aachen.de/~Frank.Luebeck/ From waghoba at gmail.com Fri Jul 29 06:59:39 2016 From: waghoba at gmail.com (Vinay Wagh) Date: Fri, 29 Jul 2016 11:29:39 +0530 Subject: [GAP Forum] Equivalent of solve_mod in sage in GAP Message-ID: Dear Forum, I would like to solve a system of multi-variable linear congruences mod n. I found that sage offers a command called solve_mod. I will be happy to get some pointers in GAP itself, which does the same job. Thanks. -- VInay From Bill.Allombert at math.u-bordeaux.fr Fri Jul 29 13:12:39 2016 From: Bill.Allombert at math.u-bordeaux.fr (Bill Allombert) Date: Fri, 29 Jul 2016 14:12:39 +0200 Subject: [GAP Forum] Equivalent of solve_mod in sage in GAP In-Reply-To: References: Message-ID: <20160729121239.GC18983@yellowpig> On Fri, Jul 29, 2016 at 11:29:39AM +0530, Vinay Wagh wrote: > Dear Forum, > > I would like to solve a system of multi-variable linear congruences > mod n. I found that sage offers a command called solve_mod. I will be > happy to get some pointers in GAP itself, which does the same job. Such problems reduce to Hermite normal form computation, so you can use HermiteNormalFormIntegerMatTransform for that. Cheers, Bill. From juergen.mueller at math.rwth-aachen.de Sat Jul 30 10:20:58 2016 From: juergen.mueller at math.rwth-aachen.de (Juergen Mueller) Date: Sat, 30 Jul 2016 11:20:58 +0200 Subject: [GAP Forum] Thue's Lemma Message-ID: <20160730092058.GA16714@localhost.localdomain> Dear Sergey, dear Frank, just a short additional comment on this: Actually, I did not know either that this Lemma has a name attached to it, in my eyes it was part of mathematical folklore. Anyway, the standard algorithmic solution (which is the one Frank has implemented, as far as I see) is also known as Gauss's algorithm finding a shortest vector in a lattice of rank 2 in Euclidean space (applied to the lattice spanned by [n,0] and [x,1]). As such, it as a variant of the Euclidean algorithm in Z, and a predecessor of the LLL algorithm. And indeed, as it is essentially the Euclidean algorithm running, this works efficiently for HUGE numbers, as you have observed. For more details, see for example Algorithm 1.3.14 in Cohen's book A course in computational algebraic number theory'. Best wishes, J?rgen (M?ller) From s.shpectorov at bham.ac.uk Sat Jul 30 11:11:47 2016 From: s.shpectorov at bham.ac.uk (Sergey Shpectorov) Date: Sat, 30 Jul 2016 10:11:47 +0000 Subject: [GAP Forum] Thue's Lemma In-Reply-To: <20160730092058.GA16714@localhost.localdomain> References: <20160730092058.GA16714@localhost.localdomain> Message-ID: <63278B7E72610F4C87EE13DA0E155D9601DF7945CF@EX11.adf.bham.ac.uk> Dear Juergen, Thanks for providing these details! I certainly have no first-hand knowledge of the origin of this fact. I found it attributed to Thue on Wikipedia, and am just very happy that the algorithm is available in GAP! Best, Sergey ________________________________________ From: forum-bounces at gap-system.org [forum-bounces at gap-system.org] on behalf of Juergen Mueller [juergen.mueller at math.rwth-aachen.de] Sent: Saturday, July 30, 2016 10:20 AM To: forum at gap-system.org Subject: Re: [GAP Forum] Thue's Lemma Dear Sergey, dear Frank, just a short additional comment on this: Actually, I did not know either that this Lemma has a name attached to it, in my eyes it was part of mathematical folklore. Anyway, the standard algorithmic solution (which is the one Frank has implemented, as far as I see) is also known as Gauss's algorithm finding a shortest vector in a lattice of rank 2 in Euclidean space (applied to the lattice spanned by [n,0] and [x,1]). As such, it as a variant of the Euclidean algorithm in Z, and a predecessor of the LLL algorithm. And indeed, as it is essentially the Euclidean algorithm running, this works efficiently for HUGE numbers, as you have observed. For more details, see for example Algorithm 1.3.14 in Cohen's book A course in computational algebraic number theory'. Best wishes, J?rgen (M?ller) _______________________________________________ Forum mailing list Forum at mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum From axyd0000 at gmail.com Tue Aug 2 15:11:01 2016 From: axyd0000 at gmail.com (gxyd) Date: Tue, 2 Aug 2016 19:41:01 +0530 Subject: [GAP Forum] Library of Finitely Presented groups exists Message-ID: <5a6b7e56-9f8a-f779-3ec8-e177903b88bc@gmail.com> Hi forum members, I am making some programs in Python (for finitely presented groups), so I was looking examples of finitely presented for testing. Perhaps I could not find sufficient number of examples through internet search. Does there exists any database of finitely presented groups as such in GAP (with a few of their properties mentioned like order or index etc.)? I am particularly looking for groups with **small** order. Thanks Gaurav (Sent from thunderbird email client) From alexander.konovalov at st-andrews.ac.uk Wed Aug 3 14:58:04 2016 From: alexander.konovalov at st-andrews.ac.uk (Alexander Konovalov) Date: Wed, 3 Aug 2016 13:58:04 +0000 Subject: [GAP Forum] groups of order p^3*q^3 In-Reply-To: References: Message-ID: Dear Walter, dear GAP Forum, > On 25 Jul 2016, at 15:02, Alexander Konovalov wrote: > > Dear Walter, > >> On 25 Jul 2016, at 14:45, Walter Becker wrote: >> >> The question is how many groups are there for these >> orders. Note here that the Group Constructions program >> of Besche and Eick give for the case of p=3 and q=7 the >> number 215 (and actually constructs them). The Thesis of >> of R. Laue (see page 226) seems to give 229 (plus the >> 25 direct product cases of the groups of order 27 and 343 ?). >> >> The compiled list of groups available on the internet does >> not contain an entry for the groups of order 9261. > > The Group Numbers reproducibility project > (https://github.com/alex-konovalov/gnu) > actually has an entry for gnu(9261): see > > https://github.com/alex-konovalov/gnu/pull/43 > > and it is 215. Same number obtained by using the > latest GrpConst package both in GAP 4.8.3 and master > branch of the development version of GAP at the > time of submission. > > I don't have Laue's thesis so I can not comment more > on the reasons for the discrepancies, but hopefully > some one else could add more. I was pointed out (thanks!) that the reader who is not familiar with the Group Numbers reproducibility project may read my email as a confirmation of an independent count of groups of order 9261, what is not true - it was counted using the same approach used by Walter (using the GrpConst package) - what I had in mind replying to Walter was to confirm that the same number is replicated in other calculations and is actually reported somewhere on the internet. Having an independent count produced by other implementations or by purely theoretical calculations is certainly interesting. Actually, in the Group Numbers reproducibility project there are several categories (GitHub labels) for gnu(n) entries to specify different levels of their reproducibility: * replicated: the same count is obtained using the same code on a different machine, possibly with different versions of the software (but calling the same GAP functions). * reproduced: the same count is obtained using another implementation (for example, for cubefree groups one could use the GrpConst package and the Cubefree package) * agrees with theory: the count is checked against the literature Hopefully more labels of the latter two kinds will appear at some point. Best wishes Alexander From e.obrien at auckland.ac.nz Sat Aug 6 11:16:56 2016 From: e.obrien at auckland.ac.nz (Eamonn O'Brien) Date: Sat, 6 Aug 2016 22:16:56 +1200 Subject: [GAP Forum] Kalman Visiting Fellowship in Mathematics at the University of Auckland Message-ID: <57A5B918.8040609@auckland.ac.nz> Dear Colleagues: I would be grateful if you can draw the attention of your colleagues to a prestigious Visiting Fellowship available in mathematics at the University of Auckland. The purpose of the Kalman Visiting Fellowship is to enable a ?rising star? in mathematics and its applications to visit the University of Auckland. The Fellowship is for a person within 10 years of PhD. It is worth NZ$10,000, and can be spent on travel, accommodation, or other associated expenses. Full details of the Fellowship, including the simple application process, can be found at www.science.auckland.ac.nz/kalman. The closing date for applications is 5^th September 2016. I am happy to provide further information. Previous recipients of the Fellowship are Tim Burness (Bristol) and Jiawang Nie (UCSD). Best wishes. Eamonn From J.Howie at hw.ac.uk Wed Aug 10 10:33:06 2016 From: J.Howie at hw.ac.uk (Jim Howie) Date: Wed, 10 Aug 2016 10:33:06 +0100 Subject: [GAP Forum] Fwd: Job Advert in Pure Maths at Heriot Watt In-Reply-To: References: Message-ID: Dear Colleagues Apologies for multiple mailings. The Maths Department at Heriot Watt is looking to recruit a pure mathematician (at assistant/associate/full professor level). Details at https://www.hw.ac.uk/about/careers/jobs/job_SVJDNTc5MA.htm The closing date is 30 September. Please forward to anyone who may be interested. Best wishes Jim Howie ________________________________ Founded in 1821, Heriot-Watt is a leader in ideas and solutions. With campuses and students across the entire globe we span the world, delivering innovation and educational excellence in business, engineering, design and the physical, social and life sciences. The contents of this e-mail (including any attachments) are confidential. If you are not the intended recipient of this e-mail, any disclosure, copying, distribution or use of its contents is strictly prohibited, and you should please notify the sender immediately and then delete it (including any attachments) from your system. From r_n_tsai at yahoo.com Fri Aug 12 06:52:33 2016 From: r_n_tsai at yahoo.com (R.N. Tsai) Date: Fri, 12 Aug 2016 05:52:33 +0000 (UTC) Subject: [GAP Forum] Error using guava function QCLDPCCodeFromGroup References: <834423031.13493661.1470981153524.JavaMail.yahoo.ref@mail.yahoo.com> Message-ID: <834423031.13493661.1470981153524.JavaMail.yahoo@mail.yahoo.com> Dear Forum, I tried to use the?QCLDPCCodeFromGroup defined in the guava package :I used the example given in the documentation : LoadPackage("guava","0",false); C:=QCLDPCCodeFromGroup(7,2,3);? I get this error : Error, no method found! For debugging hints type ?Recovery from NoMethodFoundError, no 1st choice method found for QCLDPCCodeFromGroup' on 3 arguments called from( ) ?QCLDPCCodeFromGroup returns the expected information about it, so it seems that gapknows about the function and I do load the guava package. I'm using this version of GAP under windows.GAP 4.8.3, 19-Mar-2016, build of 2016-03-20 23:48:47 (GMTST) I got the same error with win32 and win64 versions and under linux. Thanks for your help,R.N. From ciric50 at gmail.com Tue Aug 16 20:12:24 2016 From: ciric50 at gmail.com (Bill Maier) Date: Tue, 16 Aug 2016 14:12:24 -0500 Subject: [GAP Forum] Build problem Message-ID: I'm on Linux Mint 64-bit version 18 and am trying to build a local version of GAP from the latest source code. I am a software engineer and would like to start contributing to GAP. I pulled code using "git clone https://github.com/gap-system/gap.git". Ran .configure and then make which had no problems. But then when I run bin/gap.sh, the system reports this: bmaier at mint-desktop ~/Packages/github/gap/bin $./gap.sh ????????? GAP v4.8.4-702-g80a0966 of 2016-08-16 14:06:52 (CDT) ? GAP ? http://www.gap-system.org ????????? Architecture: x86_64-pc-linux-gnu-gcc-default64 Libs used: gmp Loading the library and packages ... #I gapdoc package is not available. Check that the name is correct #I and it is present in one of the GAP root directories (see '??RootPaths') #I GAP: needed package gapdoc cannot be loaded Error, failed to load needed package gapdoc' (version >= 1.2) at /home/bmaier/Packages/github/gap/lib/package.gi:1836 called from func( ); at /home/bmaier/Packages/github/gap/lib/system.g:159 called from ( ) called from read-eval loop at /home/bmaier/Packages/github/gap/lib/init.g:770 It still actually runs most simple commands. However when I try to run the command "Factorial(8);" I get this: Error, Factorial: function is not yet defined in called from func( ); at /home/bmaier/Packages/github/gap/lib/system.g:159 called from ( ) called from read-eval loop at *errin*:2 Can someone tell me what is wrong? -Bill From caj21 at st-andrews.ac.uk Tue Aug 16 20:16:51 2016 From: caj21 at st-andrews.ac.uk (Christopher Jefferson) Date: Tue, 16 Aug 2016 19:16:51 +0000 Subject: [GAP Forum] Build problem In-Reply-To: References: Message-ID: <246BD08B-316B-43C6-959C-071BED134FCE@st-andrews.ac.uk> GAP requires the ?gapdoc? package to function, but it is not distributed with GAP (we probably should either improve the messages, or not require it, but that is a longer-term issue). To get GAP working, run either: ? make bootstrap-pkg-minimal ? to get just the packages GAP needs to start or ? make bootstrap-pkg-full ? to get a complete set of packages. On 16/08/2016, 20:12, "forum-bounces at gap-system.org on behalf of Bill Maier" wrote: I'm on Linux Mint 64-bit version 18 and am trying to build a local version of GAP from the latest source code. I am a software engineer and would like to start contributing to GAP. I pulled code using "git clone https://github.com/gap-system/gap.git". Ran .configure and then make which had no problems. But then when I run bin/gap.sh, the system reports this: bmaier at mint-desktop ~/Packages/github/gap/bin$ ./gap.sh ????????? GAP v4.8.4-702-g80a0966 of 2016-08-16 14:06:52 (CDT) ? GAP ? http://www.gap-system.org ????????? Architecture: x86_64-pc-linux-gnu-gcc-default64 Libs used: gmp Loading the library and packages ... #I gapdoc package is not available. Check that the name is correct #I and it is present in one of the GAP root directories (see '??RootPaths') #I GAP: needed package gapdoc cannot be loaded Error, failed to load needed package gapdoc' (version >= 1.2) at /home/bmaier/Packages/github/gap/lib/package.gi:1836 called from func( ); at /home/bmaier/Packages/github/gap/lib/system.g:159 called from ( ) called from read-eval loop at /home/bmaier/Packages/github/gap/lib/init.g:770 It still actually runs most simple commands. However when I try to run the command "Factorial(8);" I get this: Error, Factorial: function is not yet defined in called from func( ); at /home/bmaier/Packages/github/gap/lib/system.g:159 called from ( ) called from read-eval loop at *errin*:2 Can someone tell me what is wrong? -Bill _______________________________________________ Forum mailing list Forum at mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum From ciric50 at gmail.com Tue Aug 16 20:54:02 2016 From: ciric50 at gmail.com (Bill Maier) Date: Tue, 16 Aug 2016 14:54:02 -0500 Subject: [GAP Forum] Build problem In-Reply-To: <246BD08B-316B-43C6-959C-071BED134FCE@st-andrews.ac.uk> References: <246BD08B-316B-43C6-959C-071BED134FCE@st-andrews.ac.uk> Message-ID: That did it. Thanks! On Tue, Aug 16, 2016 at 2:16 PM, Christopher Jefferson < caj21 at st-andrews.ac.uk> wrote: > GAP requires the ?gapdoc? package to function, but it is not distributed > with GAP (we probably should either improve the messages, or not require > it, but that is a longer-term issue). > > To get GAP working, run either: > > ? make bootstrap-pkg-minimal ? to get just the packages GAP needs to start > or > ? make bootstrap-pkg-full ? to get a complete set of packages. > > > > On 16/08/2016, 20:12, "forum-bounces at gap-system.org on behalf of Bill > Maier" > wrote: > > I'm on Linux Mint 64-bit version 18 and am trying to build a local > version > of GAP from the latest > source code. I am a software engineer and would like to start > contributing > to GAP. I pulled code > using "git clone https://github.com/gap-system/gap.git". Ran > .configure and > then make > which had no problems. > > But then when I run bin/gap.sh, the system reports this: > > bmaier at mint-desktop ~/Packages/github/gap/bin $./gap.sh > ????????? GAP v4.8.4-702-g80a0966 of 2016-08-16 14:06:52 (CDT) > ? GAP ? http://www.gap-system.org > ????????? Architecture: x86_64-pc-linux-gnu-gcc-default64 > Libs used: gmp > Loading the library and packages ... > #I gapdoc package is not available. Check that the name is correct > #I and it is present in one of the GAP root directories (see > '??RootPaths') > #I GAP: needed package gapdoc cannot be loaded > Error, failed to load needed package gapdoc' (version >= 1.2) at > /home/bmaier/Packages/github/gap/lib/package.gi:1836 called from > func( ); at /home/bmaier/Packages/github/gap/lib/system.g:159 called > from > ( ) > called from read-eval loop at > /home/bmaier/Packages/github/gap/lib/init.g:770 > > It still actually runs most simple commands. However when > I try to run the command "Factorial(8);" I get this: > > Error, Factorial: function is not yet defined in > called from > func( ); at /home/bmaier/Packages/github/gap/lib/system.g:159 called > from > ( ) > called from read-eval loop at *errin*:2 > > Can someone tell me what is wrong? > > -Bill > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum > > > From dmitrii.pasechnik at cs.ox.ac.uk Thu Aug 18 13:19:36 2016 From: dmitrii.pasechnik at cs.ox.ac.uk (Dima Pasechnik) Date: Thu, 18 Aug 2016 13:19:36 +0100 Subject: [GAP Forum] Error using guava function QCLDPCCodeFromGroup In-Reply-To: <834423031.13493661.1470981153524.JavaMail.yahoo@mail.yahoo.com> References: <834423031.13493661.1470981153524.JavaMail.yahoo.ref@mail.yahoo.com> <834423031.13493661.1470981153524.JavaMail.yahoo@mail.yahoo.com> Message-ID: <20160818121936.GA9850@dimpase.cs.ox.ac.uk> Dear R.N.Tsai, I have filed this as a bug report: https://github.com/osj1961/guava/issues/23 Hope this helps, Dima On Fri, Aug 12, 2016 at 05:52:33AM +0000, R.N. Tsai wrote: > Dear Forum, > I tried to use the?QCLDPCCodeFromGroup defined in the guava package :I used the example given in the documentation : > LoadPackage("guava","0",false); > C:=QCLDPCCodeFromGroup(7,2,3);? > I get this error : > Error, no method found! For debugging hints type ?Recovery from NoMethodFoundError, no 1st choice method found for QCLDPCCodeFromGroup' on 3 arguments called from( ) > ?QCLDPCCodeFromGroup returns the expected information about it, so it seems that gapknows about the function and I do load the guava package. > I'm using this version of GAP under windows.GAP 4.8.3, 19-Mar-2016, build of 2016-03-20 23:48:47 (GMTST) > I got the same error with win32 and win64 versions and under linux. > Thanks for your help,R.N. > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From waghoba at gmail.com Tue Aug 23 05:04:54 2016 From: waghoba at gmail.com (Vinay Wagh) Date: Tue, 23 Aug 2016 09:34:54 +0530 Subject: [GAP Forum] GAP-4.8.4 or 4.8.3? Message-ID: Dear Forum, I am not sure whether this is a problem or not! I am trying to install latest GAP release -- 4.8.4 on my 64bit Ubuntu 14.04 system (from the link http://www.gap-system.org/pub/gap/gap48/tar.gz/gap4r8p4_2016_06_04-12_41.tar.gz). After the successful compilation, if I start gap, then in the banner I can see the version as 4.8.3 with build of 4th June. Am I missing something while compilation? The output --> ------------------------------------------------------ ????????? GAP 4.8.3, 19-Mar-2016, build of 2016-06-04 01:23:49 (IST) ? GAP ? http://www.gap-system.org ????????? Architecture: x86_64-pc-linux-gnu-gcc-default64 Libs used: gmp Loading the library and packages ... Components: trans 1.0, prim 2.1, small* 1.0, id* 1.0 ------------------------------------------------------ Thanks in advance VInay From dmitrii.pasechnik at cs.ox.ac.uk Tue Aug 23 11:44:03 2016 From: dmitrii.pasechnik at cs.ox.ac.uk (Dima Pasechnik) Date: Tue, 23 Aug 2016 11:44:03 +0100 Subject: [GAP Forum] GAP-4.8.4 or 4.8.3? In-Reply-To: References: Message-ID: <20160823104402.GC23452@dimpase.cs.ox.ac.uk> Dear Vinay, On Tue, Aug 23, 2016 at 09:34:54AM +0530, Vinay Wagh wrote: > I am not sure whether this is a problem or not! > > I am trying to install latest GAP release -- 4.8.4 on my 64bit Ubuntu > 14.04 system (from the link > http://www.gap-system.org/pub/gap/gap48/tar.gz/gap4r8p4_2016_06_04-12_41.tar.gz). > > After the successful compilation, if I start gap, then in the banner I > can see the version as 4.8.3 with build of 4th June. > > Am I missing something while compilation? You have more than one installation of GAP on your machine, and launching the wrong one, not the one you compiled. The one you would compile this way, say, in a directory A, can be launched by changing to A and running ./bin/gap.sh And the prompt would look like: ????????? GAP 4.8.4, 04-Jun-2016, build of 2016-08-23 11:16:47 (BST) ? GAP ? http://www.gap-system.org ????????? Architecture: x86_64-pc-linux-gnu-gcc-default64 Libs used: gmp, readline Loading the library and packages ... Components: trans 1.0, prim 2.1, small* 1.0, id* 1.0 Packages: AClib 1.2, Alnuth 3.0.0, AtlasRep 1.5.1, AutPGrp 1.6, CRISP 1.4.4, Cryst 4.1.12, CrystCat 1.1.6, CTblLib 1.2.2, FactInt 1.5.3, FGA 1.3.1, GAPDoc 1.5.1, IRREDSOL 1.3.1, LAGUNA 3.7.0, Polenta 1.3.6, Polycyclic 2.11, RadiRoot 2.7, ResClasses 4.5.0, Sophus 1.23, SpinSym 1.5, TomLib 1.2.5, Utils 0.40 Try '??help' for help. See also '?copyright', '?cite' and '?authors' HTH, Dmitrii > > The output --> > > ------------------------------------------------------ > ????????? GAP 4.8.3, 19-Mar-2016, build of 2016-06-04 01:23:49 (IST) > ? GAP ? http://www.gap-system.org > ????????? Architecture: x86_64-pc-linux-gnu-gcc-default64 > Libs used: gmp > Loading the library and packages ... > Components: trans 1.0, prim 2.1, small* 1.0, id* 1.0 > ------------------------------------------------------ > > > Thanks in advance > VInay > > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From waghoba at gmail.com Tue Aug 23 14:09:30 2016 From: waghoba at gmail.com (Vinay Wagh) Date: Tue, 23 Aug 2016 18:39:30 +0530 Subject: [GAP Forum] GAP-4.8.4 or 4.8.3? In-Reply-To: <20160823104402.GC23452@dimpase.cs.ox.ac.uk> References: <20160823104402.GC23452@dimpase.cs.ox.ac.uk> Message-ID: Not really!!! I have only one installation directory (in fact I keep overwriting every version, using a customized script). However deleting the whole directory and re-running the script solved the issue. Thanks anyway. VInay$ ./bin/gap.sh ????????? GAP 4.8.3, 19-Mar-2016, build of 2016-06-04 01:23:49 (IST) ? GAP ? http://www.gap-system.org ????????? Architecture: x86_64-pc-linux-gnu-gcc-default64 Libs used: gmp Loading the library and packages ... Components: trans 1.0, prim 2.1, small* 1.0, id* 1.0 Packages: AClib 1.2, Alnuth 3.0.0, AtlasRep 1.5.1, AutPGrp 1.6, Browse 1.8.6, CRISP 1.4.4, Cryst 4.1.12, CrystCat 1.1.6, CTblLib 1.2.2, FactInt 1.5.3, FGA 1.3.1, GAPDoc 1.5.1, IO 4.4.6, IRREDSOL 1.3.1, LAGUNA 3.7.0, Polenta 1.3.6, Polycyclic 2.11, RadiRoot 2.7, ResClasses 4.5.0, Sophus 1.23, SpinSym 1.5, TomLib 1.2.5, Utils 0.40 Try '??help' for help. See also '?copyright', '?cite' and '?authors' gap> On 23-Aug-2016 16:14, "Dima Pasechnik" wrote: > Dear Vinay, > On Tue, Aug 23, 2016 at 09:34:54AM +0530, Vinay Wagh wrote: > > I am not sure whether this is a problem or not! > > > > I am trying to install latest GAP release -- 4.8.4 on my 64bit Ubuntu > > 14.04 system (from the link > > http://www.gap-system.org/pub/gap/gap48/tar.gz/gap4r8p4_2016 > _06_04-12_41.tar.gz). > > > > After the successful compilation, if I start gap, then in the banner I > > can see the version as 4.8.3 with build of 4th June. > > > > Am I missing something while compilation? > > You have more than one installation of GAP on your machine, and > launching the wrong one, not the one you compiled. > The one you would compile this way, say, in a directory A, > can be launched by changing to A and running > ./bin/gap.sh > > And the prompt would look like: > > ????????? GAP 4.8.4, 04-Jun-2016, build of 2016-08-23 11:16:47 (BST) > ? GAP ? http://www.gap-system.org > ????????? Architecture: x86_64-pc-linux-gnu-gcc-default64 > Libs used: gmp, readline > Loading the library and packages ... > Components: trans 1.0, prim 2.1, small* 1.0, id* 1.0 > Packages: AClib 1.2, Alnuth 3.0.0, AtlasRep 1.5.1, AutPGrp 1.6, CRISP > 1.4.4, Cryst 4.1.12, > CrystCat 1.1.6, CTblLib 1.2.2, FactInt 1.5.3, FGA 1.3.1, > GAPDoc 1.5.1, IRREDSOL 1.3.1, > LAGUNA 3.7.0, Polenta 1.3.6, Polycyclic 2.11, RadiRoot 2.7, > ResClasses 4.5.0, Sophus 1.23, > SpinSym 1.5, TomLib 1.2.5, Utils 0.40 > Try '??help' for help. See also '?copyright', '?cite' and '?authors' > > HTH, > Dmitrii > > > > > The output --> > > > > ------------------------------------------------------ > > ????????? GAP 4.8.3, 19-Mar-2016, build of 2016-06-04 01:23:49 (IST) > > ? GAP ? http://www.gap-system.org > > ????????? Architecture: x86_64-pc-linux-gnu-gcc-default64 > > Libs used: gmp > > Loading the library and packages ... > > Components: trans 1.0, prim 2.1, small* 1.0, id* 1.0 > > ------------------------------------------------------ > > > > > > Thanks in advance > > VInay > > > > _______________________________________________ > > Forum mailing list > > Forum at mail.gap-system.org > > http://mail.gap-system.org/mailman/listinfo/forum > From e.obrien at auckland.ac.nz Tue Aug 23 20:59:42 2016 From: e.obrien at auckland.ac.nz (Eamonn O'Brien) Date: Wed, 24 Aug 2016 07:59:42 +1200 Subject: [GAP Forum] Kalman Visiting Fellowship in Mathematics at the University of Auckland Message-ID: <0b31ae48-93f5-525d-0dbe-315368fde5e0@auckland.ac.nz> Dear Colleagues: I would be grateful if you can draw the attention of your colleagues to a prestigious Visiting Fellowship available in mathematics at the University of Auckland. The purpose of the Kalman Visiting Fellowship is to enable a ?rising star? in mathematics and its applications to visit the University of Auckland. The Fellowship is for a person within 10 years of PhD. It is worth NZ$10,000, and can be spent on travel, accommodation, or other associated expenses. Full details of the Fellowship, including the simple application process, can be found at www.science.auckland.ac.nz/kalman. The closing date for applications is 5^th September 2016. I am happy to provide further information. Previous recipients of the Fellowship are Tim Burness (Bristol) and Jiawang Nie (UCSD). Best wishes. Eamonn From s.dikson2016 at gmail.com Sat Aug 27 15:40:49 2016 From: s.dikson2016 at gmail.com (Sara Dikson) Date: Sat, 27 Aug 2016 19:10:49 +0430 Subject: [GAP Forum] Installing gap package Message-ID: Dear Forum I would like to install "atlasrep" package in ubuntu but I couldn't! Is it possible help me to install it step by step? Best regards. Sara From dmitrii.pasechnik at cs.ox.ac.uk Sat Aug 27 16:25:40 2016 From: dmitrii.pasechnik at cs.ox.ac.uk (Dima Pasechnik) Date: Sat, 27 Aug 2016 16:25:40 +0100 Subject: [GAP Forum] Installing gap package In-Reply-To: References: Message-ID: <20160827152540.GA11195@dimpase.cs.ox.ac.uk> Dear Sara, On Sat, Aug 27, 2016 at 07:10:49PM +0430, Sara Dikson wrote: > I would like to install "atlasrep" package in ubuntu but I couldn't! > Is it possible help me to install it step by step? I take it as you use GAP installed as a Ubuntu package. Assuming you have admin rights on your machine, you simpy can install the package into /usr/share/gap/pkg/ That is, you cd to this directory and untar the package you downloaded from http://www.gap-system.org/pub/gap/gap4/tar.gz/packages/atlasrep1r5p1.tar.gz In more detail, assuming the tar file in in /tmp:$ cd /usr/share/gap/pkg/ $sudo tar xf /tmp/atlasrep1r5p1.tar.gz Finally, you will need to make AtlasRep data cache directories writable to all the users (this is not very secure step - on systems with many people using it this does not look like a good idea):$ sudo chmod 777 /usr/share/gap/pkg/atlasrep/data* There are ways to make the latter more secure, described in the package manual, if you need it. Hope this helps, Dima > Best regards. > Sara From s.dikson2016 at gmail.com Mon Aug 29 13:54:10 2016 From: s.dikson2016 at gmail.com (Sara Dikson) Date: Mon, 29 Aug 2016 17:24:10 +0430 Subject: [GAP Forum] An Error Message-ID: Dear Forum, I installed "atlasrep" package. I can not get some sporadic groups like "HS" or "McL". It returns some errors: gap> s:=AtlasGroup("HS"); #I CrcFile value of #I '/usr/share/gap/pkg/atlasrep/data/datagens/HSG1-p100B0.m1' #I does not match, ignoring this file How can I get all "AtlasGroups"? I send the error as an attachment. Best regards Sara -------------- next part -------------- gap> s:=AtlasGroup("HS"); #I CrcFile value of #I '/usr/share/gap/pkg/atlasrep/data/datagens/HSG1-p100B0.m1' #I does not match, ignoring this file fail gap> s:=AtlasGroup("HS"); #I CrcFile value of #I '/usr/share/gap/pkg/atlasrep/data/datagens/HSG1-p100B0.m1' #I does not match, ignoring this file fail gap> s:=AtlasGroup("Hs"); #I CrcFile value of #I '/usr/share/gap/pkg/atlasrep/data/datagens/HSG1-p100B0.m1' #I does not match, ignoring this file fail gap> s:=AtlasGroup("McL"); #I CrcFile value of #I '/usr/share/gap/pkg/atlasrep/data/datagens/McLG1-p275B0.m1' #I does not match, ignoring this file fail gap> LoadPackage("atlasrep"); true gap> s:=AtlasGroup("HS"); #I CrcFile value of #I '/usr/share/gap/pkg/atlasrep/data/datagens/HSG1-p100B0.m1' #I does not match, ignoring this file fail gap> s:=AtlasGroup("McL"); #I CrcFile value of #I '/usr/share/gap/pkg/atlasrep/data/datagens/McLG1-p275B0.m1' #I does not match, ignoring this file fail gap> From alexander.konovalov at st-andrews.ac.uk Mon Aug 29 13:53:51 2016 From: alexander.konovalov at st-andrews.ac.uk (Alexander Konovalov) Date: Mon, 29 Aug 2016 12:53:51 +0000 Subject: [GAP Forum] Installing gap package In-Reply-To: <20160827152540.GA11195@dimpase.cs.ox.ac.uk> References: <20160827152540.GA11195@dimpase.cs.ox.ac.uk> Message-ID: <921E0BDF-A1D8-43EE-85A7-079D969586FD@st-andrews.ac.uk> Dear Sarah, dear GAP Forum, While Dima's reply tells how to install an additional GAP package to GAP installed as an Ubuntu package, I would suggest to use a different approach, and here is why. 1) 'apt install gap' offers an alternative third-party distribution which has a different structure and at the moment does not provide the latest GAP release (it gives last years' version 4.7.9 of 29-Nov-2015 while the most recent GAP is 4.8.4, 04-Jun-2016). 2) Now the version of AtlasRep 1.5.1 was released on 30/03/2016 and was first time included in the distribution of GAP 4.8.4 released on June 4th this year. Hence, there is no guarantee that this combination was carefully tested and that it works fine. Instead, I would recommend to download GAP 4.8.4 from this page http://www.gap-system.org/Releases/ and build GAP and as many packages as you can. You don't need to have admin rights on that machine since you can install it in your home directory. For example, these commands go to your home directory, download GAP, unpack the archive, delete the archive, rename the directory to indicate the minor release of GAP, then build the GAP kernel and then build some packages: cd ~ wget http://www.gap-system.org/pub/gap/gap48/tar.bz2/gap4r8p4_2016_06_04-12_41.tar.bz2 gzip -dc gap4r8p4_2016_06_04-12_41.tar.bz2 | tar xpv rm gap4r8p4_2016_06_04-12_41.tar.bz2 mv gap4r8 gap4r8p4 cd gap4r8p4 ./configure make cd pkg ../bin/BuildPackages.sh After that you will be able to start GAP with ~/gap4r8p4/bin/gap.sh Remember that 'gap' will still call the version installed as an Ubuntu package, so be careful. If you have admin rights, you can remove GAP Ubuntu package, and you can also copy ~/gap4r8p4/bin/gap.sh to /usr/local/bin/gap and be able to start it just by calling 'gap' Hope this helps Alexander P.S. One could of course argue that GAP packages provide PackageInfo.g file which documents dependencies, and that AtlasRep states "GAP version: >= 4.5". Even if this is the case for AtlasRep, in general this is not reliable. The following code summarises dependencies: l:=List(RecNames(GAPInfo.PackagesInfo),r->GAPInfo.PackagesInfo.(r)[1].Dependencies.GAP);; for s in l do RemoveCharacters(s,">= ");od; Collected(l); and for packages from GAP 4.8.4 it shows the following: [ [ "4.3", 5 ], [ "4.3fix4", 1 ], [ "4.4", 29 ], [ "4.4.10", 1 ], [ "4.4.6", 1 ], [ "4.4.9", 1 ], [ "4.5", 28 ], [ "4.5.0", 1 ], [ "4.5.3", 4 ], [ "4.5.5", 1 ], [ "4.5.6", 1 ], [ "4.6", 7 ], [ "4.6.3", 1 ], [ "4.7", 33 ], [ "4.7.4", 2 ], [ "4.7.6", 2 ], [ "4.7.8", 1 ], [ "4.8", 3 ], [ "4.8.0", 1 ], [ "4.8.1", 1 ], [ "4.8.2", 4 ], [ "4.8.3", 3 ] ] However, if you will look at release years, then you will see that actually many packages were released in last two years: gap> Collected(List(RecNames(GAPInfo.PackagesInfo),r-> > SplitString(GAPInfo.PackagesInfo.(r)[1].Date,"/")[3])); [ [ "2003", 1 ], [ "2004", 2 ], [ "2006", 1 ], [ "2007", 1 ], [ "2008", 1 ], [ "2009", 1 ], [ "2010", 1 ], [ "2011", 9 ], [ "2012", 11 ], [ "2013", 14 ], [ "2014", 9 ], [ "2015", 31 ], [ "2016", 49 ] ] and I seriously doubt that someone still tests their packages with each previous major release of GAP. May I use this rant to remind to package authors to update dependencies when you're preparing the new version of the package - for example, to the major release of GAP under which you're developing and testing the package. On 27 Aug 2016, at 16:25, Dima Pasechnik wrote: > > Dear Sara, > On Sat, Aug 27, 2016 at 07:10:49PM +0430, Sara Dikson wrote: >> I would like to install "atlasrep" package in ubuntu but I couldn't! >> Is it possible help me to install it step by step? > > I take it as you use GAP installed as a Ubuntu package. > > Assuming you have admin rights on your machine, you simpy can install > the package into /usr/share/gap/pkg/ > That is, you cd to this directory and untar the package you downloaded > from > http://www.gap-system.org/pub/gap/gap4/tar.gz/packages/atlasrep1r5p1.tar.gz > > In more detail, assuming the tar file in in /tmp: > $cd /usr/share/gap/pkg/ >$ sudo tar xf /tmp/atlasrep1r5p1.tar.gz > > Finally, you will need to make AtlasRep data cache directories writable > to all the users (this is not very secure step - on systems > with many people using it this does not look like a good idea): > > sudo chmod 777 /usr/share/gap/pkg/atlasrep/data* > > There are ways to make the latter more secure, described in the package > manual, if you need it. > > Hope this helps, > Dima From dmitrii.pasechnik at cs.ox.ac.uk Mon Aug 29 15:52:48 2016 From: dmitrii.pasechnik at cs.ox.ac.uk (dmitrii.pasechnik at cs.ox.ac.uk) Date: Mon, 29 Aug 2016 14:52:48 +0000 Subject: [GAP Forum] An Error In-Reply-To: References: Message-ID: <20160829145248.GA4675@hilbert.localdomain> Dear Sara, On Mon, Aug 29, 2016 at 05:24:10PM +0430, Sara Dikson wrote: try removing these files (such as /usr/share/gap/pkg/atlasrep/data/datagens/HSG1-p100B0.m1) and re-running gap> s:=AtlasGroup("HS"); etc. Crc errors simply indicate that files got corrupted while you were downloading them (the package does get these files from the internet) Hope this helps, Dima > Dear Forum, > I installed "atlasrep" package. I can not get some sporadic groups like > "HS" or "McL". It returns some errors: > > gap> s:=AtlasGroup("HS"); > #I CrcFile value of > #I '/usr/share/gap/pkg/atlasrep/data/datagens/HSG1-p100B0.m1' > #I does not match, ignoring this file > > How can I get all "AtlasGroups"? > > I send the error as an attachment. > Best regards > > Sara > gap> s:=AtlasGroup("HS"); > #I CrcFile value of > #I '/usr/share/gap/pkg/atlasrep/data/datagens/HSG1-p100B0.m1' > #I does not match, ignoring this file > fail > gap> s:=AtlasGroup("HS"); > #I CrcFile value of > #I '/usr/share/gap/pkg/atlasrep/data/datagens/HSG1-p100B0.m1' > #I does not match, ignoring this file > fail > gap> s:=AtlasGroup("Hs"); > #I CrcFile value of > #I '/usr/share/gap/pkg/atlasrep/data/datagens/HSG1-p100B0.m1' > #I does not match, ignoring this file > fail > gap> s:=AtlasGroup("McL"); > #I CrcFile value of > #I '/usr/share/gap/pkg/atlasrep/data/datagens/McLG1-p275B0.m1' > #I does not match, ignoring this file > fail > gap> LoadPackage("atlasrep"); > true > gap> s:=AtlasGroup("HS"); > #I CrcFile value of > #I '/usr/share/gap/pkg/atlasrep/data/datagens/HSG1-p100B0.m1' > #I does not match, ignoring this file > fail > gap> s:=AtlasGroup("McL"); > #I CrcFile value of > #I '/usr/share/gap/pkg/atlasrep/data/datagens/McLG1-p275B0.m1' > #I does not match, ignoring this file > fail > gap> > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From harshaarora.2008 at gmail.com Tue Aug 30 11:51:53 2016 From: harshaarora.2008 at gmail.com (Harsha Arora) Date: Tue, 30 Aug 2016 16:21:53 +0530 Subject: [GAP Forum] Automorphism orbits Message-ID: Dear sir Is there any command to get automorphism orbits of a group? Harsha Arora From harshaarora.2008 at gmail.com Tue Aug 30 11:54:11 2016 From: harshaarora.2008 at gmail.com (Harsha Arora) Date: Tue, 30 Aug 2016 16:24:11 +0530 Subject: [GAP Forum] Fwd: Automorphism orbits In-Reply-To: References: Message-ID: Dear sir Is there any command to get automorphism orbits of a group? Harsha Arora Dear sir Is there any command to get automorphism orbits of a group? Harsha Arora From r_n_tsai at yahoo.com Wed Aug 31 18:32:27 2016 From: r_n_tsai at yahoo.com (R.N. Tsai) Date: Wed, 31 Aug 2016 17:32:27 +0000 (UTC) Subject: [GAP Forum] formatted "Print" References: <2117109665.2701117.1472664747330.ref@mail.yahoo.com> Message-ID: <2117109665.2701117.1472664747330@mail.yahoo.com> Dear GAP forum, Is there a way to do a formatted print in GAP? Something equivalent to this for example : ?%6d or %-6d ?to print an alignedfixed width 6 bit integer....(these are c examples but are used in other languages too). Thanks,R.N. From Bill.Allombert at math.u-bordeaux.fr Thu Sep 1 08:52:07 2016 From: Bill.Allombert at math.u-bordeaux.fr (Bill Allombert) Date: Thu, 1 Sep 2016 09:52:07 +0200 Subject: [GAP Forum] Installing gap package In-Reply-To: <20160827152540.GA11195@dimpase.cs.ox.ac.uk> References: <20160827152540.GA11195@dimpase.cs.ox.ac.uk> Message-ID: <20160901075207.GD15090@yellowpig> On Sat, Aug 27, 2016 at 04:25:40PM +0100, Dima Pasechnik wrote: > Dear Sara, > On Sat, Aug 27, 2016 at 07:10:49PM +0430, Sara Dikson wrote: > > I would like to install "atlasrep" package in ubuntu but I couldn't! > > Is it possible help me to install it step by step? > > I take it as you use GAP installed as a Ubuntu package. Recent versions of Ubuntu should have a package 'gap-atlasrep', in which case you can just do: sudo apt-get install gap-atlasrep > Assuming you have admin rights on your machine, you simpy can install > the package into /usr/share/gap/pkg/ It is better to put it in /usr/local/share/gap/pkg/ or even inHOME/gap/pkg/ if you do not have admin rights. The gap script will still find it. But otherwise your instruction are good. Cheers, Bill. From alexander.konovalov at st-andrews.ac.uk Thu Sep 1 11:38:08 2016 From: alexander.konovalov at st-andrews.ac.uk (Alexander Konovalov) Date: Thu, 1 Sep 2016 10:38:08 +0000 Subject: [GAP Forum] formatted "Print" In-Reply-To: <2117109665.2701117.1472664747330@mail.yahoo.com> References: <2117109665.2701117.1472664747330.ref@mail.yahoo.com> <2117109665.2701117.1472664747330@mail.yahoo.com> Message-ID: <7195FD5A-18E0-4657-80A2-6787FB850D65@st-andrews.ac.uk> Dear R.N., > On 31 Aug 2016, at 18:32, R.N. Tsai wrote: > > Dear GAP forum, > Is there a way to do a formatted print in GAP? > Something equivalent to this for example : %6d or %-6d to print an alignedfixed width 6 bit integer....(these are c examples but are used in other languages too). > Thanks,R.N. Not that I am aware of (maybe utility function in some package though?) but you can easily achieve this using String, or PrintString or ViewString - dependently on the objects that you want to output this way. For example, this is a simple-minded attempt to add some number of spaces in front of an integer (it uses undocumented feature that ListWithIdenticalEntries returns an empty list when its 1st argument is negative): gap> PrintFormatted:=function(d,n) > local t,s; > t:=String(d); > s:=ListWithIdenticalEntries(n-Length(t),' '); > Print(s,t); > end; function( d, n ) ... end gap> for i in [1..16] do PrintFormatted(2^i,4);Print("\n");od; 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 gap> Hope this helps Alexander From sam at Math.RWTH-Aachen.De Thu Sep 1 13:06:00 2016 From: sam at Math.RWTH-Aachen.De (Thomas Breuer) Date: Thu, 1 Sep 2016 14:06:00 +0200 Subject: [GAP Forum] formatted "Print" In-Reply-To: <2117109665.2701117.1472664747330@mail.yahoo.com> References: <2117109665.2701117.1472664747330.ref@mail.yahoo.com> <2117109665.2701117.1472664747330@mail.yahoo.com> Message-ID: <20160901120600.GA1336@gemma.math.rwth-aachen.de> Dear GAP Forum, concerning the question by R.N. Tsai > Is there a way to do a formatted print in GAP? > Something equivalent to this for example : ?%6d or %-6d ?to print an alignedfixed width 6 bit integer....(these are c examples but are used in other languages too). I would suggest the 2-argument version of 'String', which creates left or right aligned strings, depending on and fills up with whitespace. Here is an example. gap> for i in [ 1 .. 6 ] do Print( String( 10^i, 6 ), "\n" ); od; 10 100 1000 10000 100000 1000000 gap> for i in [ 1 .. 6 ] do Print( String( 10^i, -6 ), ".\n" ); od; 10 . 100 . 1000 . 10000 . 100000. 1000000. All the best, Thomas From rafael at rgug.ch Thu Sep 1 16:14:14 2016 From: rafael at rgug.ch (Rafael Guglielmetti) Date: Thu, 01 Sep 2016 17:14:14 +0200 Subject: [GAP Forum] CoxIter - Computing invariants of hyperbolic Coxeter groups Message-ID: <1472742854-da909a52df580bae35c24089c091f22d@rgug.ch> Dear GAP users, I few months ago, I released CoxIter, a program to compute invariants of hyperbolic Coxeter groups (Euler characteristic, f-vector of the associated polyhedron, growth series and growth rate, cocompactness and cofiniteness test). More information can be found here: [https://coxiter.rgug.ch/](https://coxiter.rgug.ch/) and [https://github.com/rgugliel/CoxIter](https://github.com/rgugliel/CoxIter) More recently, it was suggested to create a GAP package with CoxIter. I have a preliminary version of the package and before I wrap this and write a proper documentation, I would like some feedback from GAP users/contributors to know whether the interface is nice to use, if the names of the? function are standard, if the "architecture" of the package is not stupid, ... The package, together with an example can be found here: [https://github.com/rgugliel/CoxIterGAP](https://github.com/rgugliel/CoxIterGAP) I would be happy to get any feedback. Also, if you get trouble building/using it, feel free to contact me. Thanks a lot. Best, Rafael From caj21 at st-andrews.ac.uk Thu Sep 1 17:21:59 2016 From: caj21 at st-andrews.ac.uk (Christopher Jefferson) Date: Thu, 1 Sep 2016 16:21:59 +0000 Subject: [GAP Forum] CoxIter - Computing invariants of hyperbolic Coxeter groups In-Reply-To: <1472742854-da909a52df580bae35c24089c091f22d@rgug.ch> References: <1472742854-da909a52df580bae35c24089c091f22d@rgug.ch> Message-ID: <0EFF06C7-609B-4EDC-BDC2-5E215CBCA37E@st-andrews.ac.uk> Hi, I installed the package on a mac, and checked it built / ran correctly. Everything seemed reasonable. I can?t really comment on the maths, but some minor comments: * There are a few practice functions (CoxIter_Example, CoxIter_Compute) which I assume you don?t need, that need cleaning up. * Be sure to handle the program failing (for example, if you pass it excessively large integers). * In CoxIterCompute, just do ?SetCofinite(ci, EvalString(data[2]))?. This avoids another function call when you want these values later, and also sets ?HasCofinite? to true. So for example Cofinite could be: InstallMethod( Cofinite, "for hyperbolic Coxeter groups", [IsCoxIter and IsCoxIterRep], function(obj) CoxIterCompute(obj); if HasCofinite(obj) then return Cofinite(obj); else return fail; fi; end); On 01/09/2016, 16:14, "forum-bounces at gap-system.org on behalf of Rafael Guglielmetti" wrote: Dear GAP users, I few months ago, I released CoxIter, a program to compute invariants of hyperbolic Coxeter groups (Euler characteristic, f-vector of the associated polyhedron, growth series and growth rate, cocompactness and cofiniteness test). More information can be found here: [https://coxiter.rgug.ch/](https://coxiter.rgug.ch/) and [https://github.com/rgugliel/CoxIter](https://github.com/rgugliel/CoxIter) More recently, it was suggested to create a GAP package with CoxIter. I have a preliminary version of the package and before I wrap this and write a proper documentation, I would like some feedback from GAP users/contributors to know whether the interface is nice to use, if the names of the function are standard, if the "architecture" of the package is not stupid, ... The package, together with an example can be found here: [https://github.com/rgugliel/CoxIterGAP](https://github.com/rgugliel/CoxIterGAP) I would be happy to get any feedback. Also, if you get trouble building/using it, feel free to contact me. Thanks a lot. Best, Rafael _______________________________________________ Forum mailing list Forum at mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum From r_n_tsai at yahoo.com Thu Sep 1 21:25:35 2016 From: r_n_tsai at yahoo.com (R.N. Tsai) Date: Thu, 1 Sep 2016 20:25:35 +0000 (UTC) Subject: [GAP Forum] formatted "Print" In-Reply-To: <20160901120600.GA1336@gemma.math.rwth-aachen.de> References: <2117109665.2701117.1472664747330.ref@mail.yahoo.com> <2117109665.2701117.1472664747330@mail.yahoo.com> <20160901120600.GA1336@gemma.math.rwth-aachen.de> Message-ID: <716774194.3422590.1472761535576@mail.yahoo.com> Dear GAP Forum and Thomas,This is exactly what I need; same functionality as??%6d or %-6d in an even more elegant way.Thanks,R.N.? From: Thomas Breuer To: forum at gap-system.org Sent: Thursday, September 1, 2016 5:06 AM Subject: Re: [GAP Forum] formatted "Print" Dear GAP Forum, concerning the question by R.N. Tsai > Is there a way to do a formatted print in GAP? > Something equivalent to this for example : ?%6d or %-6d ?to print an alignedfixed width 6 bit integer....(these are c examples but are used in other languages too). I would suggest the 2-argument version of 'String', which creates left or right aligned strings, depending on and fills up with whitespace. Here is an example. ? ? gap> for i in [ 1 .. 6 ] do Print( String( 10^i, 6 ), "\n" ); od; ? ? ? ? 10 ? ? ? 100 ? ? ? 1000 ? ? 10000 ? ? 100000 ? ? 1000000 ? ? gap> for i in [ 1 .. 6 ] do Print( String( 10^i, -6 ), ".\n" ); od; ? ? 10? ? . ? ? 100? . ? ? 1000? . ? ? 10000 . ? ? 100000. ? ? 1000000. All the best, Thomas _______________________________________________ Forum mailing list Forum at mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum From rafael at rgug.ch Fri Sep 2 08:13:08 2016 From: rafael at rgug.ch (Rafael Guglielmetti) Date: Fri, 2 Sep 2016 09:13:08 +0200 Subject: [GAP Forum] CoxIter - Computing invariants of hyperbolic Coxeter groups In-Reply-To: <0EFF06C7-609B-4EDC-BDC2-5E215CBCA37E@st-andrews.ac.uk> References: <1472742854-da909a52df580bae35c24089c091f22d@rgug.ch> <0EFF06C7-609B-4EDC-BDC2-5E215CBCA37E@st-andrews.ac.uk> Message-ID: <72dd6bfb-bb38-8afd-58b1-39fe2c1e25fe@rgug.ch> Dear Christopher, Thank your for your feedback, I will implement your comments. Best, Rafael On 01/09/16 18:21, Christopher Jefferson wrote: > Hi, > > I installed the package on a mac, and checked it built / ran correctly. Everything seemed reasonable. I can?t really comment on the maths, but some minor comments: > > * There are a few practice functions (CoxIter_Example, CoxIter_Compute) which I assume you don?t need, that need cleaning up. > > * Be sure to handle the program failing (for example, if you pass it excessively large integers). > > * In CoxIterCompute, just do ?SetCofinite(ci, EvalString(data[2]))?. This avoids another function call when you want these values later, and also sets ?HasCofinite? to true. > > So for example Cofinite could be: > > InstallMethod( Cofinite, > "for hyperbolic Coxeter groups", > [IsCoxIter and IsCoxIterRep], > function(obj) > CoxIterCompute(obj); > if HasCofinite(obj) then > return Cofinite(obj); > else > return fail; > fi; > end); > > > > On 01/09/2016, 16:14, "forum-bounces at gap-system.org on behalf of Rafael Guglielmetti" wrote: > > > Dear GAP users, > I few months ago, I released CoxIter, a program to compute invariants of hyperbolic Coxeter groups (Euler characteristic, f-vector of the associated polyhedron, growth series and growth rate, cocompactness and cofiniteness test). > More information can be found here: [https://coxiter.rgug.ch/](https://coxiter.rgug.ch/) and [https://github.com/rgugliel/CoxIter](https://github.com/rgugliel/CoxIter) > > More recently, it was suggested to create a GAP package with CoxIter. > > I have a preliminary version of the package and before I wrap this and write a proper documentation, I would like some feedback from GAP users/contributors to know whether the interface is nice to use, if the names of the function are standard, if the "architecture" of the package is not stupid, ... > > The package, together with an example can be found here: [https://github.com/rgugliel/CoxIterGAP](https://github.com/rgugliel/CoxIterGAP) > > I would be happy to get any feedback. Also, if you get trouble > building/using it, feel free to contact me. > > Thanks a lot. > > Best, > > Rafael > > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum > > > > From s.dikson2016 at gmail.com Fri Sep 2 18:32:07 2016 From: s.dikson2016 at gmail.com (Sara Dikson) Date: Fri, 2 Sep 2016 22:02:07 +0430 Subject: [GAP Forum] A command Message-ID: Dear Forum I use "QuoInt" command for elements of a list. Gap returns me a list with some empty spaces like following example: gap> L:=[2,23,10,15]; [ 2, 23, 10, 15 ] gap> M:=[];; gap> for i in L do > M[i]:=QuoInt(465372-Sum(L),i)+3; > od; M; [ , 232664,,,,,,,, 46535,,,,, 31024,,,,,,,, 20234 ] Although I can compact M by "Compacted(M)", I want a command for "for" loop which does not create any empty spaces. Is it possible? I mean I need such a command that calculate ''QuoInt'' for five elements of list "L" and no more. I need it because I think the empty spaces for a list with large integer takes up a lot of memory. Best regards Sara From rm43 at evansville.edu Fri Sep 2 19:04:16 2016 From: rm43 at evansville.edu (Robert Morse) Date: Fri, 2 Sep 2016 13:04:16 -0500 Subject: [GAP Forum] A command Message-ID: Dear Sara, You are using the elements in L as indices into the list M. Just Add the values to M: gap> L:=[2,23,10,15];; gap> M:=[];; gap> for i in L do > Add(M,QuoInt(465372-Sum(L),i)+3); > od; gap> M; [ 232664, 20234, 46535, 31024 ] Robert F. Morse On Fri, Sep 2, 2016 at 12:32 PM, Sara Dikson wrote: > Dear Forum > I use "QuoInt" command for elements of a list. > Gap returns me a list with some empty spaces like following example: > > gap> L:=[2,23,10,15]; > > [ 2, 23, 10, 15 ] > > gap> M:=[];; > > gap> for i in L do > >> M[i]:=QuoInt(465372-Sum(L),i)+3; > >> od; M; > > [ , 232664,,,,,,,, 46535,,,,, 31024,,,,,,,, 20234 ] > > Although I can compact M by "Compacted(M)", I want a command for "for" loop > which does not create any empty spaces. Is it possible? > > I mean I need such a command that calculate ''QuoInt'' for five elements of > list "L" and no more. > > I need it because I think the empty spaces for a list with large > integer takes up a lot of memory. > > > Best regards > > Sara > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From s.dikson2016 at gmail.com Sat Sep 3 16:04:42 2016 From: s.dikson2016 at gmail.com (Sara Dikson) Date: Sat, 3 Sep 2016 19:34:42 +0430 Subject: [GAP Forum] Building Variables In-Reply-To: References: Message-ID: Dear Forum I have a diophantine equation with many variables. (e. g. 70 variables) I use alphabet but I need more. How can I build some variables such that I use them at "for loops"? Best regards Sara From Alexander.Hulpke at colostate.edu Sun Sep 4 15:34:35 2016 From: Alexander.Hulpke at colostate.edu (Hulpke,Alexander) Date: Sun, 4 Sep 2016 14:34:35 +0000 Subject: [GAP Forum] Building Variables In-Reply-To: References: Message-ID: <9CEC8746-8888-4DBC-ABFE-2F61FC1060BA@colostate.edu> Dear Forum, Dear Ms. Dikson, Dear Forum I have a diophantine equation with many variables. (e. g. 70 variables) I use alphabet but I need more. How can I build some variables such that I use them at "for loops?? Variables can be longer than one letter, so you can easily create more than 26. Regards, Alexander Hulpke -- Colorado State University, Department of Mathematics, Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA email: hulpke at colostate.edu, Phone: ++1-970-4914288 http://www.math.colostate.edu/~hulpke From r_n_tsai at yahoo.com Sun Sep 4 18:08:05 2016 From: r_n_tsai at yahoo.com (R.N. Tsai) Date: Sun, 4 Sep 2016 17:08:05 +0000 (UTC) Subject: [GAP Forum] Building Variables In-Reply-To: References: Message-ID: <33327773.1122710.1473008885133@mail.yahoo.com> Dear GAP Forum and Sara, I use this sometimes : A:=List([1..71],k->Concatenation("a",String(k))); then A[1]...A[71] are the variables... R.N. From: Sara Dikson To: forum at gap-system.org Sent: Saturday, September 3, 2016 8:04 AM Subject: [GAP Forum] Building Variables Dear Forum I have a diophantine equation with many variables. (e. g. 70 variables) I use alphabet but I need more. How can I build some variables such that I use them at "for loops"? Best regards Sara _______________________________________________ Forum mailing list Forum at mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum From argentina.ale at gmail.com Mon Sep 5 00:16:31 2016 From: argentina.ale at gmail.com (Alejandra Alderete) Date: Sun, 4 Sep 2016 20:16:31 -0300 Subject: [GAP Forum] regular subgroups of the Symmetric Group S_24 isomorphic to S_4. Message-ID: Dear Forum, Can you help me?? I need to work with groups of hight order and my computer isn't potency o capacity , I don't know. I am sending the algorithm. I need to find the regular subgroups of the Symmetric Group S_24 isomorphic to S_4. gap> s4 := SymmetricGroup (4); gap> s24 := SymmetricGroup (24); gap> homo := AllHomomorphisms (s4, s24);; gap> inj :=Filtered (homo, function (v) return IsInjective (v)= true ; end); gap> img := List (inj, x-> Image (x, s4)); gap> reg := Filtered (img, function (v) return IsRegular(v)= true ; end); best regards Alejandra From dmitrii.pasechnik at cs.ox.ac.uk Mon Sep 5 09:16:46 2016 From: dmitrii.pasechnik at cs.ox.ac.uk (dmitrii.pasechnik at cs.ox.ac.uk) Date: Mon, 5 Sep 2016 08:16:46 +0000 Subject: [GAP Forum] regular subgroups of the Symmetric Group S_24 isomorphic to S_4. In-Reply-To: References: Message-ID: <20160905081645.GB12146@hilbert.localdomain> Dear Alejandra, On Sun, Sep 04, 2016 at 08:16:31PM -0300, Alejandra Alderete wrote: > > I need to work with groups of hight order and my computer isn't potency o > capacity , I don't know. I am sending the algorithm. I need to find the > regular subgroups of the Symmetric Group S_24 isomorphic to S_4. > > gap> s4 := SymmetricGroup (4); > gap> s24 := SymmetricGroup (24); > gap> homo := AllHomomorphisms (s4, s24);; > gap> inj :=Filtered (homo, function (v) return IsInjective (v)= true ; end); > gap> img := List (inj, x-> Image (x, s4)); > gap> reg := Filtered (img, function (v) return IsRegular(v)= true ; end); This would be a hopelessly long list. There are 24!/576 such subgroups in S_24. On the other hand constructing one copy is very easy, just let S_4 act on itself. gap> S4:=SymmetricGroup(4); Sym( [ 1 .. 4 ] ) gap> Action(S4,Elements(S4),OnRight); Group([ (1,10,17,19)(2,9,18,20)(3,12,14,21)(4,11,13,22)(5,7,16,23)(6,8,15,24), (1,7)(2,8)(3,9)(4,10)(5,11)(6,12)(13,15)(14,16)(17,18) (19,21)(20,22)(23,24) ]) Hope this helps, Dmitrii. > > > > best regards > > Alejandra From alexander.konovalov at st-andrews.ac.uk Fri Sep 9 09:54:20 2016 From: alexander.konovalov at st-andrews.ac.uk (Alexander Konovalov) Date: Fri, 9 Sep 2016 08:54:20 +0000 Subject: [GAP Forum] 2nd CoDiMa Training School in Computational Discrete Mathematics (Edinburgh 17-21 Oct 2016) Message-ID: <78776151-5387-4A17-9C90-4AA0EE02EADE@st-andrews.ac.uk> -------------------------------------------------------------------- CoDiMa Training School in Computational Discrete Mathematics International Centre for Mathematical Sciences, Edinburgh 17-21 October 2016 http://www.codima.ac.uk/school2016/ -------------------------------------------------------------------- This is the announcement of the Second CoDiMa Training School in Computational Discrete Mathematics. CoDiMa is the Collaborative Computational Project (CCP) in the area of Computational Discrete Mathematics, supported by the EPSRC (EP/M022641/1). It is centred on GAP and SageMath - open source software systems which are widely used for research and teaching in abstract algebra, number theory, cryptography, combinatorics, graph theory, coding theory, optimisation and search, among other areas. CoDiMa supports a number of activities to support users, extenders and developers of these systems and encourage best practice in their use. One of them is the Training School in Computational Discrete Mathematics, intended for PhD students and researchers from UK institutions. It will start with the 2-days hands-on Software Carpentry workshop covering basic concepts and tools, including working with the command line, version control and task automation, and the GAP Software Carpentry lesson, followed by the introduction to SageMath system and further lectures and exercise classes on a selection of topics in computational discrete mathematics. Lecturers * John Cremona (Warwick) * Mike Croucher (Sheffield) * Christopher Jefferson (St Andrews) * Alexander Konovalov (St Andrews) * Steve Linton (St Andrews) * Markus Pfeiffer (St Andrews) * Leighton Pritchard (James Hutton Institute, Dundee) * Viviane Pons (Universit? Paris-Sud) * Alexey Tarutin (University of Edinburgh) * Wilf Wilson (St Andrews) Registration Participation in the Training School is free of charge, but attendees need to be registered in advance as the number of places is limited. To register, please proceed to the Software Carpentry workshop page: https://widdowquinn.github.io/2016-10-17-edinburgh/ where you will find further information and the link to the registration form on Eventbrite. By registering there, you will automatically register for the whole week. Financial Support A limited financial support to cover travel expenses and stay in Edinburgh is available to PhD students from UK Universities who are coming for the whole duration of the Training School (from Monday until Friday). Students wishing to apply for the support should register online and then ask their PhD supervisor to email to contact at codima.ac.uk a recommendation that they should take part. Travel and accommodation Attendees are asked to make their own arrangements for travel and accommodation, following the suggestions published on the School's homepage at http://www.codima.ac.uk/school2016/ Please contact the organisers with any questions or suggestions: * Alexander Konovalov: alexander.konovalov at st-andrews.ac.uk You may also follow CoDiMa on Twitter: https://twitter.com/codima_project From f.k.moftakhar at gmail.com Sat Sep 10 09:19:58 2016 From: f.k.moftakhar at gmail.com (fatemeh moftakhar) Date: Sat, 10 Sep 2016 12:49:58 +0430 Subject: [GAP Forum] Frobenius group Message-ID: Dear forum I need the list of all Frobenius groups up to order 1000. Is there any commend or package in GAP to do this? Best regards Fatemeh Moftakhar -- Regards; Miss Fatemeh Moftakhar PhD Candidate, Department of Pure Mathematics, Faculty of Mathematical Sciences, University of Kashan, Kashan, Iran From D.F.Holt at warwick.ac.uk Sun Sep 11 20:47:33 2016 From: D.F.Holt at warwick.ac.uk (Holt, Derek) Date: Sun, 11 Sep 2016 19:47:33 +0000 Subject: [GAP Forum] Frobenius group Message-ID: Dear Fatemeh, dear Forum, I would suggest the following approach, which I think will work, although there might be other ways of doing it. Consider each integer n from 1 to 1000 in turn. For each such n find all expressions of n as a product n = dh, where d,h > 1 and h divides d-1. There could be more than such expression e.g. 42 = 21 x 2 = 7 x 6. You can discard all cases such as 30 = 6 x 5, where d is twice an odd number because a Frobenius kernel cannot have twice odd order. Now, if n has one or more expressions as dh then using the small groups library in GAP, consider each group G of order n in turn, and for each such G consider all possible factorisations n = dh. First check that G has a normal nilpotent subgroup K of order d. If it does not, then you proceed to the next group or the next factorisation. To do that, you could find all Sylow p-subgroups for p dividing d, and check that they are all normal in G and, if so, take K to be the subgroup generated by these Sylow subgroups. K is a candidate for the Frobenius kernel of G. Before going further you could use the fact that if h is even then K must be abelian and if h is divisible by 3 then K must have nilpotency class at most 2. If not, you can rule out this case. Assuming K exists, it must have a unique conjugacy class of subgroups H of order h. There is a function in GAP to find complements, which you could use to find H, but in many cases h will be a prime power, in which case you can just take a Sylow subgroup. H is a candidate for the Frobenius complement. Finally, you need to check whether H really is a Frobenius complement. To do that, you could compute the permutation action on the cosets of H and see if that is a Frobenius group. Another way, which might be quicker, is to find the subgroups P of H of prime order. If H is a Frobenius complement, then there is a single conjugacy class of such subgroups for each prime dividing |H|. If so, then check that C_K(P) = 1 for all such P. If so, then G = KH is a Frobenius group. I would guess that n = 768 with k = 256, h = 3, might be the hardest case, because there are over a million groups of order 768, but it should still work if you are patient. I hope this helps. Derek Holt. On Sat, Sep 10, 2016 at 12:49:58PM +0430, fatemeh moftakhar wrote: > Dear forum > I need the list of all Frobenius groups up to order 1000. Is there any > command or package in GAP to do this? > > Best regards > Fatemeh Moftakhar > > -- > Regards; > Miss Fatemeh Moftakhar > PhD Candidate, > Department of Pure Mathematics, > Faculty of Mathematical Sciences, > University of Kashan, Kashan, Iran > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From e.obrien at auckland.ac.nz Sun Sep 11 21:28:05 2016 From: e.obrien at auckland.ac.nz (Eamonn O'Brien) Date: Mon, 12 Sep 2016 08:28:05 +1200 Subject: [GAP Forum] Frobenius group In-Reply-To: References: Message-ID: <38d59be9-d28e-31ae-3842-08104e181fa1@auckland.ac.nz> Dear Fatemeh: Lemma 1 of Bertram's 1991 paper in the Israel Journal of Math, vol 75 (1991) pp. 243-245 offers a characterisation of Frobenius groups which is easy to implement and computationally fast. In summary it states: G is a group with a proper normal subgroup N satisfying k(G) = k(G/N) + (k(N) - 1)/[G:N], if and only if G is a Frobenius group with kernel N (in which case N is the Fitting subgroup of G). Here k(H) denotes the number of conjugacy classes of a group H. Best wishes. Eamonn From alexander.konovalov at st-andrews.ac.uk Tue Sep 27 15:24:13 2016 From: alexander.konovalov at st-andrews.ac.uk (Alexander Konovalov) Date: Tue, 27 Sep 2016 14:24:13 +0000 Subject: [GAP Forum] GAP 4.8.5 release announcement Message-ID: Dear GAP Forum, This is to announce the release of GAP 4.8.5, which could be downloaded from http://www.gap-system.org/Releases/ An overview of the changes introduced in GAP 4.8.5 is provided below. A more detailed version with hyperlinks to documentation is available here: http://www.gap-system.org/Manuals/doc/changes/chap2.html#X84E1F4618181F6A3 Improved and extended functionality: * The error messages produced when an unexpected fail is returned were made more clear by explicitly telling that the result should not be boolean or fail (before it only said "not a boolean"). * For consistency, both NrTransitiveGroups and TransitiveGroup now disallow the transitive group of degree 1. Fixed bugs that could lead to incorrect results: * A bug in the code for algebraic field extensions over non-prime fields that may cause, for example, a list of all elements of the extension not being a duplicate-free. [Reported by Huta Gana] * So far, FileString only wrote files of sizes less than 2G and did not indicate an error in case of larger strings. Now strings of any length can be written, and in the case of a failure the corresponding system error is shown. Fixed bugs that could lead to break loops: * NaturalHomomorphismByIdeal was not reducing monomials before forming a quotient ring, causing a break loop on some inputs. [Reported by Dmytro Savchuk] * A bug in DefaultInfoHandler a break loop on startup with the setting SetUserPreference( "InfoPackageLoadingLevel", 4 ). [Reported by Mathieu Dutour] * The Iterator for permutation groups was broken when the StabChainMutable of the group was not reduced, which can reasonably happen as the result of various algorithms. In addition, GAP 4.8.5 distribution includes updates for 5 GAP packages, namely Cubefree, Digraphs, float, loops and SglPPow. We encourage all users to upgrade to GAP 4.8.5. Just in case, a common pitfall during GAP installation on Linux and OS X is to compile only the GAP kernel, but not GAP packages. This procedure had changed recently (in GAP 4.8.4) so just in case let me remind you that after running ./configure make in the GAP root directory you need to change to the 'pkg' subdirectory and call ../bin/BuildPackages.sh to run the script which will build most of the packages that require compilation (provided sufficiently many libraries, headers and tools are available). If you need any help or would like to report any problems, please do not hesitate to contact us at support at gap-system.org or submit new issues on GitHub: https://github.com/gap-system/gap/issues Wishing you fun and success using GAP, The GAP Group From L.H.Soicher at qmul.ac.uk Wed Sep 28 12:00:48 2016 From: L.H.Soicher at qmul.ac.uk (Leonard Soicher) Date: Wed, 28 Sep 2016 12:00:48 +0100 Subject: [GAP Forum] Announcing the SglPPow Package Message-ID: <20160928110048.GA28638@maths.qmul.ac.uk> Dear Forum Members, It is my pleasure to announce the officially accepted SglPPow (version 2.0) package for GAP, now available in the newly released GAP 4.8.5. The SglPPow package contains an extension of the SmallGroups Library and adds the groups of order 3^8 and p^7 for primes p > 11 to the library. The package webpage is: http://www.gap-system.org/Packages/sglppow.html The SglPPow package is authored by Michael Vaughan-Lee (Oxford) and Bettina Eick (Braunschweig), and on behalf of the GAP Council, I thank them for this contribution to GAP. Leonard Soicher (Chair of the GAP Council) From jalaeeyan at gmail.com Thu Sep 29 09:04:04 2016 From: jalaeeyan at gmail.com (taleea jalaeeyan) Date: Thu, 29 Sep 2016 11:34:04 +0330 Subject: [GAP Forum] quastion Message-ID: Hi there Is Gap work on windows or linux?is there diference ? is c-nilpotent multiplier defined for Gap as default? Can Gap calculate the c-nilpotent multiplier of a group by it's presentation? Is there any sample program of Gap to calculate c-nilpotent multiplier of group? Best wishes From alexander.konovalov at st-andrews.ac.uk Fri Sep 30 15:23:15 2016 From: alexander.konovalov at st-andrews.ac.uk (Alexander Konovalov) Date: Fri, 30 Sep 2016 14:23:15 +0000 Subject: [GAP Forum] quastion In-Reply-To: References: Message-ID: <008706C8-C3AF-41E7-8C53-2BE13ACCC3AD@st-andrews.ac.uk> Dear Taleea Jalaeeyan, > On 29 Sep 2016, at 09:04, taleea jalaeeyan wrote: > > Hi there > Is Gap work on windows or linux?is there diference ? GAP works on Windows, Linux, and OS X - archives for different operating systems are at http://www.gap-system.org/Releases/index.html On Windows, some specialised packages are not available, but that will be more than sufficient to start to use GAP. > is c-nilpotent multiplier defined for Gap as default? Not as far as I am aware of (only some functionality to deal with the Schur multiplier) > Can Gap calculate the c-nilpotent multiplier of a group by it's > presentation? > Is there any sample program of Gap to calculate c-nilpotent multiplier of > group? Again, I don't know. It may be helpful to provide a reference for its definition in case someone else may be able to give further advice. Hope this helps Alexander From graham.ellis at nuigalway.ie Fri Sep 30 15:53:46 2016 From: graham.ellis at nuigalway.ie (Ellis, Grahamj) Date: Fri, 30 Sep 2016 14:53:46 +0000 Subject: [GAP Forum] quastion In-Reply-To: <008706C8-C3AF-41E7-8C53-2BE13ACCC3AD@st-andrews.ac.uk> References: , <008706C8-C3AF-41E7-8C53-2BE13ACCC3AD@st-andrews.ac.uk> Message-ID: Hi Taleea, A very naive implementation of the nilpotent multiplier M^(c)(G) is available under the name BaerInvariant(G,c). The following commands calculate the nilpotent multiplier for the dihedral group G of order 64 and c=3. The multiplier is the finite abelian group Z_2+Z_2+Z_8 . gap> LoadPackage("HAP");; gap> c:=3;;BaerInvariant(DihedralGroup(64),c); [ 2, 2, 8 ] See file:///home/graham/pkg/Hap1.11/www/SideLinks/About/aboutSchurMultiplier.html for some more details and calculations on infinite groups. All the best, Graham School of Mathematics, Statistics & Applied Mathematics National University of Ireland, Galway University Road, Galway Ireland http://hamilton.nuigalway.ie tel: 091 493011 ________________________________________ From: forum-bounces at gap-system.org [forum-bounces at gap-system.org] on behalf of Alexander Konovalov [alexander.konovalov at st-andrews.ac.uk] Sent: Friday, September 30, 2016 3:23 PM To: taleea jalaeeyan Cc: GAP Forum Subject: Re: [GAP Forum] quastion Dear Taleea Jalaeeyan, > On 29 Sep 2016, at 09:04, taleea jalaeeyan wrote: > > Hi there > Is Gap work on windows or linux?is there diference ? GAP works on Windows, Linux, and OS X - archives for different operating systems are at http://www.gap-system.org/Releases/index.html On Windows, some specialised packages are not available, but that will be more than sufficient to start to use GAP. > is c-nilpotent multiplier defined for Gap as default? Not as far as I am aware of (only some functionality to deal with the Schur multiplier) > Can Gap calculate the c-nilpotent multiplier of a group by it's > presentation? > Is there any sample program of Gap to calculate c-nilpotent multiplier of > group? Again, I don't know. It may be helpful to provide a reference for its definition in case someone else may be able to give further advice. Hope this helps Alexander _______________________________________________ Forum mailing list Forum at mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum From aodabas at ogu.edu.tr Tue Oct 4 10:31:49 2016 From: aodabas at ogu.edu.tr (=?iso-8859-9?Q?Alper_Odaba=FE?=) Date: Tue, 4 Oct 2016 12:31:49 +0300 Subject: [GAP Forum] free module Message-ID: <00f601d21e22$24f14a40$6ed3dec0$@ogu.edu.tr> Dear forum, By linear algebra, the choice of an (ordered) basis for a free module of finite rank m yields an isomorphism to Z^{1 x m}, the module whose entries are row matrices with m columns. In GAP, how to get matrices from the group algebra GroupRing(GF(2),CyclicGroup(3)) Regards, Alper From Bill.Allombert at math.u-bordeaux.fr Tue Oct 4 13:33:03 2016 From: Bill.Allombert at math.u-bordeaux.fr (Bill Allombert) Date: Tue, 4 Oct 2016 14:33:03 +0200 Subject: [GAP Forum] free module In-Reply-To: <00f601d21e22$24f14a40$6ed3dec0$@ogu.edu.tr> References: <00f601d21e22$24f14a40$6ed3dec0$@ogu.edu.tr> Message-ID: <20161004123303.GB23686@yellowpig> On Tue, Oct 04, 2016 at 12:31:49PM +0300, Alper Odaba? wrote: > Dear forum, > > By linear algebra, the choice of an (ordered) basis for a free module of > finite rank m yields an isomorphism to Z^{1 x m}, the module whose entries > are row matrices with m columns. > > In GAP, how to get matrices from the group algebra > GroupRing(GF(2),CyclicGroup(3)) M:=IdentityMat(3,GF(2)); Cheers, Bill. From max at quendi.de Tue Oct 4 15:01:01 2016 From: max at quendi.de (Max Horn) Date: Tue, 4 Oct 2016 16:01:01 +0200 Subject: [GAP Forum] free module In-Reply-To: <00f601d21e22$24f14a40$6ed3dec0$@ogu.edu.tr> References: <00f601d21e22$24f14a40$6ed3dec0@ogu.edu.tr> Message-ID: Dear Alper, > On 04 Oct 2016, at 11:31, Alper Odaba? wrote: > > Dear forum, > > > > By linear algebra, the choice of an (ordered) basis for a free module of > finite rank m yields an isomorphism to Z^{1 x m}, the module whose entries > are row matrices with m columns. > > > > In GAP, how to get matrices from the group algebra > GroupRing(GF(2),CyclicGroup(3)) Your question is a bit ambiguous, I'll interpret it as follows: Given a basis B of an algebra A, how can I express an element x as a matrix over that basis? Answer: Using the command AdjointMatrix. Here is an example: gap> R:=GroupRing(GF(2),CyclicGroup(3)); gap> x:= R.1 + R.1^2; gap> B:=Basis(R); CanonicalBasis( ) gap> AsList(B); # let's see which basis GAP picked...: [ (Z(2)^0)* of ..., (Z(2)^0)*f1, (Z(2)^0)*f1^2 ] gap> mat:=AdjointMatrix(Basis(R), x); [ [ 0*Z(2), Z(2)^0, Z(2)^0 ], [ Z(2)^0, 0*Z(2), Z(2)^0 ], [ Z(2)^0, Z(2)^0, 0*Z(2) ] ] gap> Display(mat); . 1 1 1 . 1 1 1 . Hope that helps, Max From oxeimon at gmail.com Wed Oct 12 23:34:16 2016 From: oxeimon at gmail.com (Will Chen) Date: Wed, 12 Oct 2016 18:34:16 -0400 Subject: [GAP Forum] opening a saved workspace generated on a windows machine on a mac Message-ID: So, I've got a weird situation. My mac is is rather slow, but I carry it everywhere. I have a much faster windows PC, but it's a desktop I keep at home. I'd like to be able to do some computations on my windows PC, save the workspace, and use it on my mac. The first time I tried this (all using the latest version of GAP), it seems my windows install was 32 bit (even though the PC itself has a 64bit cpu), whereas my mac install was 64bit, and hence gap complained. Next, I installed the 64bit "experimental" version of GAP on my windows PC, saved a workspace, but upon trying to open it on my mac, I get the error: #W dlopen() error: dlopen(/proc/cygdrive/C/gap4r8_x86_64/pkg/io-4.4.6/bin/x86_64-unknown-cygwin-gcc-default64/io.so, 9): image not found Failed to load needed dynamic module /proc/cygdrive/C/gap4r8_x86_64/pkg/io-4.4.6/bin/x86_64-unknown-cygwin-gcc-default64/io.so, error code 1 Is this circumventable? or are the mac and PC versions fundamentally incompatible? Thanks, - Will -- William Chen Member Institute for Advanced Study, Princeton, NJ, 08540 oxeimon at gmail.com From caj21 at st-andrews.ac.uk Thu Oct 13 10:14:51 2016 From: caj21 at st-andrews.ac.uk (Christopher Jefferson) Date: Thu, 13 Oct 2016 09:14:51 +0000 Subject: [GAP Forum] opening a saved workspace generated on a windows machine on a mac In-Reply-To: References: Message-ID: <07716CC7-682D-419C-873F-A2D718AE3765@st-andrews.ac.uk> Yes, In fact, even moving a saved workspace from one mac to another mac can break, if the versions of GAP or any package are different, or if the operating system version is different. Chris > On 12 Oct 2016, at 23:34, Will Chen wrote: > > So, I've got a weird situation. My mac is is rather slow, but I carry it > everywhere. I have a much faster windows PC, but it's a desktop I keep at > home. > > I'd like to be able to do some computations on my windows PC, save the > workspace, and use it on my mac. > > The first time I tried this (all using the latest version of GAP), it seems > my windows install was 32 bit (even though the PC itself has a 64bit cpu), > whereas my mac install was 64bit, and hence gap complained. > > Next, I installed the 64bit "experimental" version of GAP on my windows PC, > saved a workspace, but upon trying to open it on my mac, I get the error: > > #W dlopen() error: > dlopen(/proc/cygdrive/C/gap4r8_x86_64/pkg/io-4.4.6/bin/x86_64-unknown-cygwin-gcc-default64/io.so, > 9): image not found > > Failed to load needed dynamic module > /proc/cygdrive/C/gap4r8_x86_64/pkg/io-4.4.6/bin/x86_64-unknown-cygwin-gcc-default64/io.so, > error code 1 > > Is this circumventable? or are the mac and PC versions fundamentally > incompatible? > > Thanks, > > - Will > > -- > > William Chen > Member > Institute for Advanced Study, > Princeton, NJ, 08540 > oxeimon at gmail.com > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From Alexander.Hulpke at colostate.edu Fri Oct 14 22:08:09 2016 From: Alexander.Hulpke at colostate.edu (Hulpke,Alexander) Date: Fri, 14 Oct 2016 21:08:09 +0000 Subject: [GAP Forum] Postdoctoral position at Colorado State University Message-ID: Dear GAP-Forum, posted here, as I am looking for applicants with interest in Computational Group Theory, in particular involving GAP. Faculty with related interests are Wilson and Hulpke; there also could be interest overlapping with Combinatorics (Penttila, Betten) or Number Theory (Achter, Pries). Applications will be evaluated starting November 1. All the best, Alexander Hulpke The Department of Mathematics at Colorado State University seeks applications for at least two postdoctoral researchers. These are expected to be 12-month postdoctoral (non-tenure track) appointments for a possible three year period, and include a small annual travel stipend, beginning in the fall semester of 2017. The teaching load will be based on department need and funding source, and can range from 0-3 courses per academics year. A detailed description of the department and information on specific faculty research programs are available on the department web site at: http://www.math.colostate.edu/. The department is seeking candidates with excellent research potential and a strong commitment to teaching, who will contribute to and benefit from the rich environment of the department. Members of the department create, communicate, and apply mathematics of the highest caliber through activities such as internationally recognized research and graduate education, award winning undergraduate programs, and extensive outreach to local schools. Roughly 2/3 of the department?s faculty are supported by federal research funding. Applicants are expected to have a Ph.D. in mathematics or related science area by the date of hire. One to two years of teaching experience at the university level or equivalent is also preferred with a record of teaching a range of mathematics courses. In their cover letter, applicants should identify a faculty member or research group of interest for collaboration, and explain how their research can both benefit from, and enhance, our research endeavors. A complete job description can be found at: https://jobs.colostate.edu/postings/33464, and applications must be submitted through the Colorado State University jobs site. For full consideration, applications should include: a cover letter, current curriculum vitae, teaching statement, research statement, and four professional reference letters, including one addressing teaching. CSU is an EO/EA/AA employer. Colorado State University conducts background checks on all final candidates. Review of applications will begin on November 1, 2016, and applications will be considered until the positions are filled. -- Colorado State University, Department of Mathematics, Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA email: hulpke at colostate.edu, Phone: ++1-970-4914288 http://www.math.colostate.edu/~hulpke From rafael at rgug.ch Mon Oct 17 08:04:32 2016 From: rafael at rgug.ch (Rafael Guglielmetti) Date: Mon, 17 Oct 2016 09:04:32 +0200 Subject: [GAP Forum] GAPDoc and images Message-ID: <59c9226a-aeac-e24a-c1a2-7c7fcfb3729f@rgug.ch> Dear forum, Is it possible to includes images in the doc of a package with GAP doc? Especially, I would be interested to know whether it is possible to provide the image in different formats, for example: * a pdf to be included in the manual.pdf * a jpg for the .html * a text to replace the image for the other cases (the equivalent of the HTML alt property) Thanks a lot. Best, Rafael From maasiru at yahoo.com Wed Oct 19 11:11:11 2016 From: maasiru at yahoo.com ([Muniru Asiru] maasiru@yahoo.com) Date: Wed, 19 Oct 2016 10:11:11 +0000 (UTC) Subject: [GAP Forum] Factoring large integers References: <23705114.2742923.1476871871235.ref@mail.yahoo.com> Message-ID: <23705114.2742923.1476871871235@mail.yahoo.com> ?Dear Forum; I need to factor large integers.? One of then is d:=208792137511016848023422421216659133913113038868420856362546192700228823547384086468768314408434778574357732094492450898877804848025959323284721735578041792204431293469827064868376517438126765900138542676486099702859181721290457656418761796076484288301990972494617227238953673093108532474354655441019362615154103484247530007425892848695540897;; I tested d for being prime or not by using IsPrime(d); which returns false which confirms that d is composite. ? I tried PartialFactorization(d,6); which returns d.? Also Factors(d);? did not return a result in 15 minutes. ? How do I find the factors of d using GAP? Dr. Muniru A. Asiru, Department of Mathematics & Statistics, The Federal Polytechnic, P.M.B. 55, Bida, Niger State, Nigeria. OR Dr. Muniru A. Asiru, P.O.Box. 294, Bida, Niger State, Nigeria. Email Addresses: maasiru at yahoo.com, remiasiru at yahoo.com, Mobile Phone number: +2348034271010 From stefan at mcs.st-and.ac.uk Wed Oct 19 11:42:07 2016 From: stefan at mcs.st-and.ac.uk (Stefan Kohl) Date: Wed, 19 Oct 2016 11:42:07 +0100 (BST) Subject: [GAP Forum] Factoring large integers In-Reply-To: <23705114.2742923.1476871871235@mail.yahoo.com> References: <23705114.2742923.1476871871235.ref@mail.yahoo.com> <23705114.2742923.1476871871235@mail.yahoo.com> Message-ID: On Wed, October 19, 2016 11:11 am, [Muniru Asiru] maasiru at yahoo.com wrote: > > I need to factor large integers.?? One of then is > d:=208792137511016848023422421216659133913113038868420856362546192700228823547384086468768314408434778574357732094492450898877804848025959323284721735578041792204431293469827064868376517438126765900138542676486099702859181721290457656418761796076484288301990972494617227238953673093108532474354655441019362615154103484247530007425892848695540897;; > I tested d for being prime or not by using IsPrime(d); which returns false which > confirms that d is composite. ?? > I tried PartialFactorization(d,6); which returns d.?? Also Factors(d);?? did not return > a result in 15 minutes. ?? > How do I find the factors of d using GAP? If the second-largest factor is 'relatively small', you can factor your number with GAP / FactInt if you just wait 'some longer'. How long it will take depends very much on the size of the factors and on good or bad luck. However if your number has two factors of roughly comparable size, you will need a dedicated program implementing the GNFS (generalized number field sieve) and either *A LOT* of patience or a supercomputer (note that your number is greater than a 1024-bit RSA modulus). Best regards, Stefan ----------------------------------------------------------------------------- http://www.gap-system.org/DevelopersPages/StefanKohl/ ----------------------------------------------------------------------------- From alexander.konovalov at st-andrews.ac.uk Wed Oct 19 21:34:08 2016 From: alexander.konovalov at st-andrews.ac.uk (Alexander Konovalov) Date: Wed, 19 Oct 2016 20:34:08 +0000 Subject: [GAP Forum] GAPDoc and images In-Reply-To: <59c9226a-aeac-e24a-c1a2-7c7fcfb3729f@rgug.ch> References: <59c9226a-aeac-e24a-c1a2-7c7fcfb3729f@rgug.ch> Message-ID: Dear Rafael, > On 17 Oct 2016, at 08:04, Rafael Guglielmetti wrote: > > Dear forum, > Is it possible to includes images in the doc of a package with GAP doc? Especially, I would be interested to know whether it is possible to provide the image in different formats, for example: > > * a pdf to be included in the manual.pdf > * a jpg for the .html > * a text to replace the image for the other cases (the equivalent of > the HTML alt property) Of course this is possible - see e.g. this file from the GAP SCSCP package: https://github.com/gap-packages/scscp/blob/335545909d9c37bb24e8f755baab811de05a548e/tracing/tracing.g which has two images in HTML and PDF versions and just one lime of text in the text version: ## \centerline{\resizebox{150mm}{!}{\includegraphics{img/quillen.pdf}}} ## <img src="img/quillen.jpg" align="left" /> ## \vspace{10pt}\centerline{\resizebox{150mm}{!}{\includegraphics{img/euler.pdf}}} ## <img src="img/euler.jpg" align="left" /> ## /See diagrams in HTML and PDF versions of the manual/ Hope this is exactly what you have in mind. Best wishes Alexander From r_n_tsai at yahoo.com Thu Oct 20 06:06:27 2016 From: r_n_tsai at yahoo.com (R.N. Tsai) Date: Thu, 20 Oct 2016 05:06:27 +0000 (UTC) Subject: [GAP Forum] GUAVA and matrix standard forms References: <1317772781.88296.1476939987758.ref@mail.yahoo.com> Message-ID: <1317772781.88296.1476939987758@mail.yahoo.com> The GUAVA command PutStandardForm can be used to get standard forms for matrices over GF(2) : : ?perm:=PutStandardForm(A,true); ? # to get mat ~ [ I | B ] ?perm:=PutStandardForm(At,false); # to get mat ~ [B | I ] "perm" can be used to get a permutation matrix (P) so that M*A*P^-1 has the desired form.The problem is that the matrix "M" is not provided and I need that for my calculations.Is there a way to get that from GUAVA or from other commands??I looked at using SemiEchelonMatTransformation but wasn't able to work out the details.These are commonly used standard forms, so I'm hoping the details have been workedout already within GAP or a package. I would also like to be able to handle matrices that arenot full rank (PutStandardForm assumes full rank).Thanks for your help.R.N.? From nuesken at bit.uni-bonn.de Mon Oct 24 15:46:03 2016 From: nuesken at bit.uni-bonn.de (=?UTF-8?Q?Michael_N=c3=bcsken?=) Date: Mon, 24 Oct 2016 16:46:03 +0200 Subject: [GAP Forum] Factoring large integers In-Reply-To: References: <23705114.2742923.1476871871235.ref@mail.yahoo.com> <23705114.2742923.1476871871235@mail.yahoo.com> Message-ID: Just for fun, I had MuPAD (of 2004) running for 120h on one 2.4GHz kernel on your d: Am 19.10.2016 um 12:42 schrieb Stefan Kohl: >> d:=208792137511016848023422421216659133913113038868420856362546192700228823547384086468768314408434778574357732094492450898877804848025959323284721735578041792204431293469827064868376517438126765900138542676486099702859181721290457656418761796076484288301990972494617227238953673093108532474354655441019362615154103484247530007425892848695540897;; and it produced no answer. That system only implements elliptic curve factoring, but it would probably have finished if all but one prime factors have at most 60-80 bit... Well, still, the runtime of even the best known factoring algorithms (ECM, GNFS) is exponential. Enjoy, |\ /| Michael N?sken, b-it cryptography, | \/ | Room 1.22, Dahlmannstr. 2, 53113 Bonn, | \ | ++49/228/2699-214, ++49/228/2619334, | \| . From f.alibabaee at gmail.com Tue Nov 8 08:43:37 2016 From: f.alibabaee at gmail.com (fahime babaee) Date: Tue, 8 Nov 2016 08:43:37 +0000 Subject: [GAP Forum] egg-box Message-ID: Dear forum How can I draw the egg box of D class of a semigroup with GAP. I used function DrawDClasses(S,"DClasses"), but it dose not work and it ask me to install gsviwer but I already have gsviwer on my system. Sincerely From jdm3 at st-andrews.ac.uk Tue Nov 8 09:35:26 2016 From: jdm3 at st-andrews.ac.uk (James Mitchell) Date: Tue, 08 Nov 2016 09:35:26 +0000 Subject: [GAP Forum] egg-box In-Reply-To: <0d4123e370be4426b6353be2b2438371@AM3PR06MB1362.eurprd06.prod.outlook.com> References: <0d4123e370be4426b6353be2b2438371@AM3PR06MB1362.eurprd06.prod.outlook.com> Message-ID: Dear Fahima Babaee, If you use the Semigroups package for GAP: https://gap-packages.github.io/Semigroups/ you can draw the D-class diagram of a finite semigroup by doing: Splash(DotDClasses(S)); As long as you have graphviz installed on your computer. If you have graphviz installed and the above command does not work (it should, but just in case it does not), then you can write the output of DotDClasses(S) to a file by doing: FileString("name-of-a-file.dot", DotDClasses(S)); and then process the resulting file in the terminal by doing: dot -Tpdf name-of-a-file > name-of-a-file.pdf to produce a pdf file. Best wishes, James On Tue, 8 Nov 2016 at 08:46 fahime babaee wrote: > Dear forum > > How can I draw the egg box of D class of a semigroup with GAP. I used > function DrawDClasses(S,"DClasses"), but it dose not work and it ask me to > install gsviwer but I already have gsviwer on my system. > > Sincerely > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum > From nkhodyunya at gmail.com Thu Nov 10 15:19:57 2016 From: nkhodyunya at gmail.com (Nikolay Hodyunya) Date: Thu, 10 Nov 2016 22:19:57 +0700 Subject: [GAP Forum] Root systems representation in GAP Message-ID: Hello all, Could someone explain me please how does the GAP represent root systems? I'd like to have representation as in Bourbaki IV-VI chapters (tables at the end of the book). For example, any root of A_n is of the form e_i - e_j. For A_3 positive roots subsystem GAP gives me these vectors: [ [ 2, -1, 0 ], [ -1, 2, -1 ], [ 0, -1, 2 ], [ 1, 1, -1 ], [ -1, 1, 1 ], [ 1, 0, 1 ] ]. What do they mean? Thanks. From willem.degraaf at unitn.it Thu Nov 10 17:50:27 2016 From: willem.degraaf at unitn.it (Willem Adriaan De Graaf) Date: Thu, 10 Nov 2016 18:50:27 +0100 Subject: [GAP Forum] Root systems representation in GAP In-Reply-To: References: Message-ID: Dear Nikolay Hodyunya, > For A_3 positive roots subsystem GAP gives me these vectors: [ [ 2, > -1, 0 ], [ -1, 2, -1 ], [ 0, -1, 2 ], [ 1, 1, -1 ], [ -1, 1, 1 ], [ 1, > 0, 1 ] ]. What do they mean? Of course, there are many ways to represent the roots of a root system. In this case, the roots record the eigenvalues of the elements of the Cartan subalgebra, which are contained in ChevalleyBasis(L)[3] (assuming your Lie algebra is denoted L). If h1, h2, h3 are these elements, and x is an element of the second root space (for example), then h1*x = -x, h2*x = 2*x, h3*x = -x. In a GAP session: gap> L:= SimpleLieAlgebra("A",3,Rationals); gap> ch:= ChevalleyBasis(L); [ [ v.1, v.2, v.3, v.4, v.5, v.6 ], [ v.7, v.8, v.9, v.10, v.11, v.12 ], [ v.13, v.14, v.15 ] ] gap> ch[3]*ch[1][2]; [ (-1)*v.2, (2)*v.2, (-1)*v.2 ] Best wishes, Willem de Graaf On 10 November 2016 at 16:19, Nikolay Hodyunya wrote: > Hello all, > > Could someone explain me please how does the GAP represent root > systems? I'd like to have representation as in Bourbaki IV-VI chapters > (tables at the end of the book). For example, any root of A_n is of > the form e_i - e_j. > > For A_3 positive roots subsystem GAP gives me these vectors: [ [ 2, > -1, 0 ], [ -1, 2, -1 ], [ 0, -1, 2 ], [ 1, 1, -1 ], [ -1, 1, 1 ], [ 1, > 0, 1 ] ]. What do they mean? > > Thanks. > > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum > From alexander.konovalov at st-andrews.ac.uk Mon Nov 14 11:34:42 2016 From: alexander.konovalov at st-andrews.ac.uk (Alexander Konovalov) Date: Mon, 14 Nov 2016 11:34:42 +0000 Subject: [GAP Forum] GAP 4.8.6 release announcement Message-ID: <2DA61C73-33CC-4F5A-9BDD-1FC86B43CCCB@st-andrews.ac.uk> Dear GAP Forum, This is to announce the release of GAP 4.8.6, which could be downloaded from http://www.gap-system.org/Releases/ The only two changes in the core GAP system made in this release are: * the fix for the regression in the GAP kernel code introduced in GAP 4.8.5 and breaking StringFile's ability to work with compressed files. [Reported by Bill Allombert] * adding references about how to use the filter argument to the documentation of the Basic Groups library, and also adding an example using the IsPermGroup filter to the documentation of DihedralGroup. [Submitted by Sergio Siccha]. For the corresponding changes in the source code, see the GAP 4.8.6 milestone on GitHub: https://github.com/gap-system/gap/milestone/12?closed=1 Furthermore, GAP 4.8.6 distribution includes updates for 15 packages: cvec | 2.5.6 | 08/11/2016 gpd | 1.45 | 02/11/2016 idrel | 2.34 | 20/10/2016 json | 1.1.0 | 01/11/2016 kan | 1.27 | 20/10/2016 LieRing | 2.3 | 01/11/2016 loops | 3.3.0 | 26/10/2016 Polenta | 1.3.7 | 09/11/2016 profiling | 1.1.0 | 01/11/2016 QPA | 1.25 | 21/10/2016 SLA | 1.2 | 01/11/2016 TomLib | 1.2.6 | 07/11/2016 Utils | 0.43 | 20/10/2016 XGAP | 4.26 | 06/11/2016 XMod | 2.58 | 02/11/2016 We encourage all users to upgrade to GAP 4.8.6. Just in case, a common pitfall during GAP installation on Linux and OS X is to compile only the GAP kernel, but not GAP packages. This procedure had changed recently (in GAP 4.8.4) so just in case let me remind you that after running ./configure make in the GAP root directory you need to change to the 'pkg' subdirectory and call ../bin/BuildPackages.sh to run the script which will build most of the packages that require compilation (provided sufficiently many libraries, headers and tools are available). If you need any help or would like to report any problems, please do not hesitate to contact us at support at gap-system.org, or submit new issues on GitHub: https://github.com/gap-system/gap/issues There is also a tag 'gap' for questions about GAP at the Mathematics Q&A site: http://math.stackexchange.com/tags/gap/info In addition, you may find some GAP related news on Twitter: http://twitter.com/gap_system (you may also read the GAP Twitter feed at http://www.gap-system.org). Wishing you fun and success using GAP, Alexander Konovalov on behalf of the GAP Group From rbailey at grenfell.mun.ca Tue Nov 15 19:59:48 2016 From: rbailey at grenfell.mun.ca (Bailey, Robert F.) Date: Tue, 15 Nov 2016 19:59:48 +0000 Subject: [GAP Forum] 3.M_{22}:2 on 990 points Message-ID: Dear forum, The group 3.M_{22}:2 has a rank-9 imprimitive permutation representation of degree 990. (This is the full automorphism group of the distance-transitive Ivanov-Ivanov-Faradjev graph.) I would like to construct this group in GAP. The www ATLAS gives a matrix representation of this group in characteristic 2; however, applying "IsomorphismPermGroup" to this matrix group gives a group of degree 693. Does anyone have a suggestion for how to obtain the degree 990 representation? Thanks, Robert Bailey. ============================== Dr. Robert Bailey School of Science and Environment (Mathematics) Grenfell Campus Memorial University of Newfoundland Corner Brook, NL A2H 6P9, Canada Office: AS 3022 Phone: +1 (709) 637-6293 Web: http://www2.grenfell.mun.ca/rbailey/ This electronic communication is governed by the terms and conditions at http://www.mun.ca/cc/contact/policies/elect_communications_disclaimer_2011.php From juergen.mueller at math.rwth-aachen.de Tue Nov 15 20:50:25 2016 From: juergen.mueller at math.rwth-aachen.de (Juergen Mueller) Date: Tue, 15 Nov 2016 21:50:25 +0100 Subject: [GAP Forum] 3.M_{22}:2 on 990 points Message-ID: <20161115205025.GA32290@localhost.localdomain> Dear Robert, dear Forum, Enclosed please find the generators for the representation you are asking for. Would you like to know how to construct them? Best wishes, J?rgen -------------- next part -------------- gens:= [ ( 3, 27)( 4, 21)( 5, 24)( 6, 23)( 7, 36)( 8, 35)( 9, 22)( 10, 38) ( 11, 37)( 12, 30)( 13, 31)( 14, 32)( 15, 33)( 16, 34)( 17, 28)( 18, 29) ( 19, 25)( 20, 26)( 39,209)( 40,206)( 41,204)( 42,205)( 43,202)( 44,186) ( 45,208)( 46,203)( 47,207)( 49,201)( 50,210)( 51,212)( 52,211)( 53,213) ( 54,215)( 55,214)( 56,197)( 57,113)( 58,333)( 59,295)( 60,298)( 62,332) ( 63,135)( 64,329)( 65,104)( 66,328)( 67,330)( 68,108)( 69,331)( 71,224) ( 72,227)( 73,344)( 75,345)( 76,346)( 77,240)( 78,342)( 79, 96)( 80,347) ( 81,283)( 82, 91)( 83,343)( 84,279)( 86,341)( 87,138)( 88,433)( 89,432) ( 90,431)( 92,434)( 93,182)( 94,147)( 95,436)( 97,435)( 98,353)(100,383) (101,386)(102,188)(103,126)(106,384)(107,385)(109,362)(110,258)(111,388) (112,387)(114,120)(115,118)(119,124)(125,535)(127,531)(128,532)(129,291) (130,539)(131,533)(133,538)(134,530)(136,334)(137,534)(139,537)(141,536) (142,562)(143,554)(144,566)(145,568)(146,565)(148,317)(149,567)(150,564) (151,461)(152,561)(153,175)(154,311)(155,563)(156,598)(157,596)(158,595) (159,524)(160,599)(161,318)(162,597)(163,397)(164,594)(165,601)(166,399) (167,600)(168,602)(169,604)(170,603)(171,550)(172,624)(173,622)(174,572) (176,627)(177,243)(178,625)(179,483)(180,623)(181,248)(183,223)(184,439) (185,626)(187,593)(189,514)(190,661)(191,659)(192,590)(193,660)(194,644) (195,657)(196,364)(198,656)(199,658)(200,360)(217,697)(219,349)(220,698) (221,699)(222,700)(225,686)(226,696)(228,416)(229,693)(230,504)(231,708) (232,288)(233,503)(234,709)(235,522)(236,390)(237,617)(238,583)(239,695) (241,710)(242,715)(244,719)(245,718)(246,415)(247,681)(249,716)(250,717) (251,441)(252,500)(253,271)(254,274)(256,735)(257,737)(259,678)(260,736) (261,417)(262,738)(263,739)(264,378)(265,744)(266,743)(267,389)(268,527) (269,587)(270,463)(272,729)(273,742)(275,509)(276,636)(277,752)(278,340) (280,757)(281,448)(282,756)(284,759)(285,758)(286,767)(289,768)(290,361) (292,370)(293,357)(294,337)(296,790)(297,702)(299,788)(300,791)(301,782) (302,541)(303,309)(304,789)(305,471)(306,424)(307,582)(308,557)(312,795) (313,606)(314,629)(315,785)(316,796)(319,382)(320,635)(321,570)(322,814) (323,815)(324,649)(325,813)(327,365)(335,438)(336,437)(338,440)(339,380) (351,828)(352,827)(354,826)(355,543)(356,818)(358,775)(359,833)(363,724) (366,838)(367,839)(368,619)(369,425)(372,840)(373,651)(374,426)(375,712) (376,810)(377,529)(379,473)(381,843)(391,409)(392,429)(394,584)(400,792) (401,774)(402,765)(403,723)(404,517)(405,647)(406,457)(407,411)(408,418) (410,462)(412,544)(413,523)(414,846)(419,847)(420,849)(421,848)(422,586) (423,850)(427,852)(428,451)(430,851)(442,855)(444,501)(446,856)(449,746) (450,553)(452,477)(453,857)(454,472)(456,666)(458,860)(459,858)(460,859) (464,475)(465,476)(466,474)(469,470)(478,748)(479,499)(481,864)(482,865) (484,863)(485,808)(486,832)(487,831)(488,866)(489,867)(490,869)(491,542) (492,747)(493,868)(494,687)(495,505)(496,577)(497,854)(498,769)(502,706) (507,870)(508,662)(510,665)(511,871)(512,872)(513,725)(515,740)(516,633) (518,547)(519,875)(520,874)(521,873)(525,646)(526,776)(528,579)(540,675) (545,655)(546,892)(548,894)(549,778)(551,893)(552,896)(555,895)(556,614) (558,897)(559,608)(560,611)(569,923)(571,750)(573,690)(574,924)(575,668) (576,707)(578,925)(580,677)(581,820)(585,877)(588,926)(589,927)(591,929) (592,928)(605,671)(607,615)(609,672)(610,939)(612,786)(613,784)(616,933) (618,940)(620,653)(621,777)(628,733)(630,803)(631,669)(632,952)(634,905) (637,834)(638,691)(639,772)(640,680)(641,953)(642,902)(643,754)(645,730) (652,816)(654,817)(663,961)(664,948)(667,960)(670,806)(673,909)(674,921) (676,904)(679,899)(683,911)(684,949)(685,962)(688,721)(689,898)(692,835) (694,882)(701,915)(703,914)(704,918)(711,881)(713,773)(714,727)(722,910) (726,944)(728,880)(731,805)(732,876)(734,801)(741,807)(745,950)(749,781) (751,955)(753,842)(755,959)(760,907)(761,942)(762,766)(763,793)(764,943) (770,956)(771,964)(779,947)(780,920)(783,888)(787,937)(794,967)(797,984) (798,979)(799,891)(800,980)(802,931)(804,932)(809,957)(811,821)(812,883) (819,983)(829,941)(830,987)(836,988)(841,981)(844,919)(845,938)(853,922) (861,890)(862,889)(878,885)(879,965)(884,913)(886,969)(887,934)(900,989) (901,975)(903,972)(906,951)(908,970)(912,958)(916,990)(917,966)(930,935) (936,977)(945,968)(946,963)(971,985)(973,978)(976,986), ( 1,875,428,522)( 2,942, 7,357)( 3, 23,607,271)( 4,907,606,986) ( 5,210,137,820)( 6,491,466,415)( 8,532,171,422)( 9,474,528,547) ( 10,742,400,360)( 11,170,461,493)( 12,700,301,175)( 13, 37,485,253) ( 14,118,472,665)( 15,102,352,966)( 16,151,605,518)( 17,214,140,894) ( 18,656,608,903)( 19,766,356,486)( 20,386,294,780)( 21, 30,426,329) ( 22,549,597,342)( 24,203,553,708)( 25,278,209,749)( 26,121,702,620) ( 27,961,617,751)( 28,199,477,402)( 29,454,439,225)( 31,506,746,406) ( 32,380,389,217)( 33,470)( 34,911,604,750)( 35,559, 94,498)( 36,361) ( 38,828,857,124)( 39,231,266,685)( 40,144,936,938)( 41,263,476,401) ( 42,316,248,459)( 43,669,917,945)( 44,890,929,588)( 45,543,832,623) ( 46,686,274, 82)( 47,626,789,927)( 48,970,967,362)( 49,152,701, 86) ( 50,435)( 51,696,529,197)( 52,792)( 53,647)( 54,590,712,332)( 55,673) ( 56,383,740,146)( 57,455,937, 97)( 58, 74,815,178)( 59,198) ( 60,216,782,419)( 61,899)( 62,679,448,614)( 63,760)( 64,265,667,185) ( 65,974,525,520)( 66,290,444,599)( 67,398,539,895)( 68,218,889,295) ( 69,100,705,958)( 70,744,479,378)( 71,123,948,241)( 72,519) ( 73,648,381,421)( 75,213,139,897)( 76,784,920,947)( 77,226,810,297) ( 78,211,136,806)( 79,898,595,965)( 80,829,354,450)( 81,494,202,157) ( 83,192,773,664)( 84,560,568,460)( 85,434,173,721)( 87,830) ( 88,822,194,642)( 89,322,752,778)( 90,465,132,941)( 91,382,716,483) ( 92,724,796,602)( 93, 96,148,572)( 95,187,730,624)( 98,189,330,633) ( 99,882,405,120)(101,723)(103,643,675,116)(104,814,505,387) (105,433,417,699)(106,114,302,781)(107,418,844,823)(108,904,288,359) (109,404,399,507)(110,182,678,550)(111,317,704,871)(112,344,649,180) (113,659,769,364)(115,657,188,345)(117,156,234,526)(119,508) (122,254,131,524)(125,794,129,451)(126,641,355,375)(127,793,287,579) (128,230,273,591)(130,953,353,181)(133,516,340,304)(134,521,349,377) (135,463,348,743)(138,416,350,915)(141,765,351,977)(142,160,627,809) (143,912,636,145)(147,155,511,163)(149,256,270,509)(150,763,358,475) (153,981,839,770)(154,928,619,223)(158,298,169,243)(159,449,695,489) (161,715,681,343)(162,694,512,373)(164,975,500,825)(165,990,943,219) (166,177,537,960)(167,654,279,570)(168,308,289,638)(172,391,697,393) (174,767,671,851)(176,707,341,326)(179,394)(183,824,776,523) (184,692,787,886)(186,982,873,452)(190,320,339,501)(191,555,631,852) (193,413,325,533)(195,280,908,881)(196,574,594,865)(200,795,596,621) (201,567,768,818)(204,563,534,275)(205,808,392,545)(206,488,764,821) (207,305,250,639)(212,987,504,819)(215,469,429,311)(220,653,799,901) (221,538,277,651)(222,536)(224,661,854,257)(227,365,918,884) (228,655,363,410)(229,457,613,848)(232,337,313,515)(233,645,650,688) (235,950,390,616)(237,249,610,267)(238,582,310,309)(239,629,962,438) (240,813,939,885)(242,372,575,644)(244,612,589,514)(245,368) (246,931,868,930)(247,805,956,468)(251,734)(252,628,968,682)(255,892) (258,446,668,395)(259,577,713,328)(260,726,385,396)(261,412,622,414) (262,333,347,635)(264,303,921,269)(268,831,922,866)(272,738,670,736) (276,338)(281,324,497,334)(282,492)(283,291,510,978)(284,564,836,735) (285,867,728,462)(286,576,757,902)(292,535,403,487)(293,660,603,900) (296,964,443,940)(299,913,637,336)(300,859,640,714)(306,473)(307,453) (312,874,835,583)(314,924,955,584)(315,709,569,409)(318,916,833,456) (319,891,909,727)(321,495,691,646)(323,592,496,774)(327,847,683,513) (331,580)(335,798,926,834)(366,872,585,464)(369,677,548,925) (370,827,849,379)(371,424,427,554)(374,693,862,952)(376,698,615,946) (384,484,562,480)(388,551)(397,447,775,407)(408,609,856,423) (411,689,944,634)(420,684)(430,552,759,893)(431,762)(432,672,840,771) (436,785,858,618)(437,674,755,719)(440,853,983,949)(441,541,556,561) (442,880,720,879)(445,877,722,878)(458,593,797,737)(467,779,804,860) (471,600,557,565)(478,729)(481,587)(482,527)(499,745)(502,690,896,969) (503,861,863,663)(517,611,906,963)(530,571,652,959)(531,676,754,841) (540,581,843,747)(542,905,761,887)(544,923,546,850)(558,566,864,662) (573,954,812,725)(578,951,718,807)(586,658,687,601)(598,817,739,772) (625,801)(630,988,680,711)(632,753,984,914)(666,741,783,703)(706,845) (710,802)(717,826,786,919)(731,838,980,989)(732,957)(733,758,932,934) (748,816,933,888)(756,791,976,788)(777,800,883,837)(790,971) (803,972,979,869)(811,842,973,870)(846,985)(855,935,910,876) ]; From dmitrii.pasechnik at cs.ox.ac.uk Tue Nov 15 20:51:17 2016 From: dmitrii.pasechnik at cs.ox.ac.uk (dmitrii.pasechnik at cs.ox.ac.uk) Date: Tue, 15 Nov 2016 20:51:17 +0000 Subject: [GAP Forum] 3.M_{22}:2 on 990 points In-Reply-To: References: Message-ID: <20161115205117.GA10335@localhost.localdomain> Dear forum, Dear Robert, On Tue, Nov 15, 2016 at 07:59:48PM +0000, Bailey, Robert F. wrote: > The group 3.M_{22}:2 has a rank-9 imprimitive permutation representation of degree 990. (This is the full automorphism group of the distance-transitive Ivanov-Ivanov-Faradjev graph.) I would like to construct this group in GAP. > > The www ATLAS gives a matrix representation of this group in characteristic 2; however, applying "IsomorphismPermGroup" to this matrix group gives a group of degree 693. > > Does anyone have a suggestion for how to obtain the degree 990 representation? There is a diagram geometry with this group that has 990 points and 693 lines; more concretely, there is a degree 7 graph of girth 5 on the 990 vertices that is invariant under your group G, such that every 2-path lies in a unique Petersen subgraph; there are 693 these subgraphs. Each vertex is in 7 such subgraphs. Dually, for the permutation representation of degree 693 there is a G-invariant graph of degree 30, such each vertex lies in 10 maximal cliques of size 7; there are 990 such cliques, and the action on them will give you the desired action. I would use GRAPE to construct the graph on 693 vertices, and find a 7-clique there. Also the following might help: the maximum possible intersection of two such 7-cliques is in a 3-clique, and there are 15 such special 3-cliques on each of the 693 vertices- they correspond to the edges of your graph on 990 vertices. Hope this helps. I'd be happy to provide more details, if needed. Dima > Thanks, > Robert Bailey. > > ============================== > Dr. Robert Bailey > School of Science and Environment (Mathematics) > Grenfell Campus > Memorial University of Newfoundland > Corner Brook, NL A2H 6P9, Canada > > Office: AS 3022 > Phone: +1 (709) 637-6293 > Web: http://www2.grenfell.mun.ca/rbailey/ > > This electronic communication is governed by the terms and conditions at http://www.mun.ca/cc/contact/policies/elect_communications_disclaimer_2011.php > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From rbailey at grenfell.mun.ca Tue Nov 15 20:51:33 2016 From: rbailey at grenfell.mun.ca (Bailey, Robert F.) Date: Tue, 15 Nov 2016 20:51:33 +0000 Subject: [GAP Forum] 3.M_{22}:2 on 990 points In-Reply-To: <973975C0-1E1C-4ED0-8CB4-A773CB2DA91C@arcor.de> References: <973975C0-1E1C-4ED0-8CB4-A773CB2DA91C@arcor.de> Message-ID: Dear J?rgen, Thank you very much for the quick response. It would be useful to know how you obtained them! Best wishes, Robert. -----Original Message----- From: Juergen Mueller [mailto:juergen.m.mueller at arcor.de] Sent: November-15-16 5:17 PM To: Bailey, Robert F. Cc: forum at gap-system.org Subject: Re: [GAP Forum] 3.M_{22}:2 on 990 points Dar Robert, Enclosed please find the generators for the representation you are asking for. Would you like to know how to construct them? Best wishes, J?rgen This electronic communication is governed by the terms and conditions at http://www.mun.ca/cc/contact/policies/elect_communications_disclaimer_2011.php From juergen.mueller at math.rwth-aachen.de Tue Nov 15 21:16:36 2016 From: juergen.mueller at math.rwth-aachen.de (Juergen Mueller) Date: Tue, 15 Nov 2016 22:16:36 +0100 Subject: [GAP Forum] 3.M_{22}:2 on 990 points Message-ID: <20161115211636.GA32437@localhost.localdomain> Dear Robert, here is essentially how I did it: (The generators this produces are different from the ones I sent earlier, because the details of the constructions differ.) ##### # finds generators of 2x2^3:L3(2) inside M22.2 slp:=AtlasStraightLineProgram("M22.2",5).program; # 3.M22.2 in dimension 12, std.gens. lifting those of M22.2 gens:=AtlasGenerators("3.M22.2",1).generators; # a subgroup of 3.(2x2^3:L3(2)) projecting onto 2x2^3:L3(2) hgens:=ResultOfStraightLineProgram(slp,gens); # indeed 2x2^3:L3(2) g:=Group(gens); h:=Group(hgens); Size(h); # go over to perm.rep, actually on 693 points iso:=IsomorphismPermGroup(g); LargestMovedPointPerm(gg); gg:=Image(iso,g); hh:=Image(iso,h); # action on cosets cos:=RightCosets(gg,hh); Length(cos); act:=Action(gg,cos,OnRight); LargestMovedPointPerm(act); ##### Best wishes, J?rgen From dmitrii.pasechnik at cs.ox.ac.uk Tue Nov 15 21:27:54 2016 From: dmitrii.pasechnik at cs.ox.ac.uk (dmitrii.pasechnik at cs.ox.ac.uk) Date: Tue, 15 Nov 2016 21:27:54 +0000 Subject: [GAP Forum] 3.M_{22}:2 on 990 points In-Reply-To: <20161115211636.GA32437@localhost.localdomain> References: <20161115211636.GA32437@localhost.localdomain> Message-ID: <20161115212754.GA11837@localhost.localdomain> On Tue, Nov 15, 2016 at 10:16:36PM +0100, Juergen Mueller wrote: > Dear Robert, > > here is essentially how I did it: > > (The generators this produces are different from the ones > I sent earlier, because the details of the constructions differ.) > > ##### > > # finds generators of 2x2^3:L3(2) inside M22.2 > slp:=AtlasStraightLineProgram("M22.2",5).program; > > # 3.M22.2 in dimension 12, std.gens. lifting those of M22.2 > gens:=AtlasGenerators("3.M22.2",1).generators; > > # a subgroup of 3.(2x2^3:L3(2)) projecting onto 2x2^3:L3(2) > hgens:=ResultOfStraightLineProgram(slp,gens); > > # indeed 2x2^3:L3(2) > g:=Group(gens); h:=Group(hgens); Size(h); > > # go over to perm.rep, actually on 693 points > iso:=IsomorphismPermGroup(g); LargestMovedPointPerm(gg); > gg:=Image(iso,g); hh:=Image(iso,h); > > # action on cosets > cos:=RightCosets(gg,hh); Length(cos); > act:=Action(gg,cos,OnRight); LargestMovedPointPerm(act); That's more or less what I outlined in my message - 2x2^3:L3(2) is the stabiliser of a 7-clique in the 693-vertex graph. It's computationally faster to create the orbit of 990 7-subsets, but OK, that mattered a lot 25+ years ago, when that graph was constructed :-) Just in case, Dima > > ##### > > Best wishes, J?rgen > > > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From frank.luebeck at math.rwth-aachen.de Tue Nov 15 21:57:00 2016 From: frank.luebeck at math.rwth-aachen.de (Frank =?iso-8859-1?Q?L=FCbeck?=) Date: Tue, 15 Nov 2016 22:57:00 +0100 Subject: [GAP Forum] 3.M_{22}:2 on 990 points In-Reply-To: References: Message-ID: <20161115215700.GA31064@localhost.localdomain> On Tue, Nov 15, 2016 at 07:59:48PM +0000, Bailey, Robert F. wrote: > Dear forum, > > The group 3.M_{22}:2 has a rank-9 imprimitive permutation representation > of degree 990. (This is the full automorphism group of the > distance-transitive Ivanov-Ivanov-Faradjev graph.) I would like to > construct this group in GAP. > > The www ATLAS gives a matrix representation of this group in > characteristic 2; however, applying "IsomorphismPermGroup" to this matrix > group gives a group of degree 693. > > Does anyone have a suggestion for how to obtain the degree 990 representation? > > Thanks, > Robert Bailey. Dear Robert, dear Forum, [When I wanted to send this reply I have seen that J?rgen has already sent a very similar solution. Nevertheless, I send my as well because it uses less knowledge about the relation the representations of M22.2 and 3.M22.2 in the ATLAS and some other GAP commands. Here it is:] The www ATLAS does not only have a matrix representation for your group but it also contains information for generating its maximal subgroups. Use, for example BrowseAtlasInfo to find this out and to get some information on the maximal subgroups. Lets take the smallest maximal subgroup which may contain a group of the right order which may be the stabilizer of a point in the permutation representation you are looking for: gap> info:=BrowseAtlasInfo("3.M22.2"); # select the 7 maximal subgroups [ rec( groupname := "3.M22.2", identifier := [ "3.M22.2", "M22d2G1-max1W1", 1 ], size := 1330560, standardization := 1 ), rec( groupname := "3.M22.2", identifier := [ "3.M22.2", "M22d2G1-max2W1", 1 ], size := 120960, standardization := 1 ), rec( groupname := "3.M22.2", identifier := [ "3.M22.2", "M22d2G1-max3W1", 1 ], size := 34560, standardization := 1 ), rec( groupname := "3.M22.2", identifier := [ "3.M22.2", "M22d2G1-max4W1", 1 ], size := 11520, standardization := 1 ), rec( groupname := "3.M22.2", identifier := [ "3.M22.2", [ "M22d2G1-max5W1", "M22.2" ], 1 ], size := 8064, standardization := 1 ), rec( groupname := "3.M22.2", identifier := [ "3.M22.2", "M22d2G1-max6W1", 1 ], size := 4320, standardization := 1 ), rec( groupname := "3.M22.2", identifier := [ "3.M22.2", [ "M22d2G1-max7W1", "M22.2" ], 1 ], size := 3960, standardization := 1 ) ] gap> g := AtlasGroup("3.M22.2"); gap> q := Size(g)/990; 2688 gap> Filtered([1..7], i-> info[i].size mod q = 0); [ 1, 2, 5 ] gap> m := AtlasSubgroup(g, 5);; gap> maxes := List(ConjugacyClassesMaximalSubgroups(m), Representative);; gap> u := First(maxes, x-> Size(x) = q); gap> act := FactorCosetAction(g,u); CompositionMapping( , ) gap> niceg := Image(act); gap> LargestMovedPoint(niceg); 990 gap> st := Stabilizer(niceg, 1); gap> Length(Orbits(st, [1..990])); 9 Is niceg the permutation group you are looking for? With best regards, Frank -- /// Dr. Frank L?beck, Lehrstuhl D f?r Mathematik, Pontdriesch 14/16, \\\ 52062 Aachen, Germany /// E-mail: Frank.Luebeck at Math.RWTH-Aachen.De \\\ WWW: http://www.math.rwth-aachen.de/~Frank.Luebeck/ From rbailey at grenfell.mun.ca Tue Nov 15 22:20:01 2016 From: rbailey at grenfell.mun.ca (Bailey, Robert F.) Date: Tue, 15 Nov 2016 22:20:01 +0000 Subject: [GAP Forum] 3.M_{22}:2 on 990 points In-Reply-To: <20161115215700.GA31064@localhost.localdomain> References: <20161115215700.GA31064@localhost.localdomain> Message-ID: <9dc9135f4d38486d87045cafbe15b68a@CAS1.swgc.ca> Dear Frank (and everyone else), Thanks for the responses. The group "niceg" in Frank's email does appear to be the group I want (although the "Browse" packing isn't working in my GAP installation so I can't test it for sure), but I now have the group I was in need of! Best wishes, Robert. ============================== Dr. Robert Bailey School of Science and Environment (Mathematics) Grenfell Campus Memorial University of Newfoundland Corner Brook, NL A2H 6P9, Canada Office: AS 3022 Phone: +1 (709) 637-6293 Web: http://www2.grenfell.mun.ca/rbailey/ -----Original Message----- From: Frank L?beck [mailto:frank.luebeck at math.rwth-aachen.de] Sent: November-15-16 6:27 PM To: Bailey, Robert F. ; forum at gap-system.org Subject: Re: [GAP Forum] 3.M_{22}:2 on 990 points On Tue, Nov 15, 2016 at 07:59:48PM +0000, Bailey, Robert F. wrote: > Dear forum, > > The group 3.M_{22}:2 has a rank-9 imprimitive permutation representation > of degree 990. (This is the full automorphism group of the > distance-transitive Ivanov-Ivanov-Faradjev graph.) I would like to > construct this group in GAP. > > The www ATLAS gives a matrix representation of this group in > characteristic 2; however, applying "IsomorphismPermGroup" to this matrix > group gives a group of degree 693. > > Does anyone have a suggestion for how to obtain the degree 990 representation? > > Thanks, > Robert Bailey. Dear Robert, dear Forum, [When I wanted to send this reply I have seen that J?rgen has already sent a very similar solution. Nevertheless, I send my as well because it uses less knowledge about the relation the representations of M22.2 and 3.M22.2 in the ATLAS and some other GAP commands. Here it is:] The www ATLAS does not only have a matrix representation for your group but it also contains information for generating its maximal subgroups. Use, for example BrowseAtlasInfo to find this out and to get some information on the maximal subgroups. Lets take the smallest maximal subgroup which may contain a group of the right order which may be the stabilizer of a point in the permutation representation you are looking for: gap> info:=BrowseAtlasInfo("3.M22.2"); # select the 7 maximal subgroups [ rec( groupname := "3.M22.2", identifier := [ "3.M22.2", "M22d2G1-max1W1", 1 ], size := 1330560, standardization := 1 ), rec( groupname := "3.M22.2", identifier := [ "3.M22.2", "M22d2G1-max2W1", 1 ], size := 120960, standardization := 1 ), rec( groupname := "3.M22.2", identifier := [ "3.M22.2", "M22d2G1-max3W1", 1 ], size := 34560, standardization := 1 ), rec( groupname := "3.M22.2", identifier := [ "3.M22.2", "M22d2G1-max4W1", 1 ], size := 11520, standardization := 1 ), rec( groupname := "3.M22.2", identifier := [ "3.M22.2", [ "M22d2G1-max5W1", "M22.2" ], 1 ], size := 8064, standardization := 1 ), rec( groupname := "3.M22.2", identifier := [ "3.M22.2", "M22d2G1-max6W1", 1 ], size := 4320, standardization := 1 ), rec( groupname := "3.M22.2", identifier := [ "3.M22.2", [ "M22d2G1-max7W1", "M22.2" ], 1 ], size := 3960, standardization := 1 ) ] gap> g := AtlasGroup("3.M22.2"); gap> q := Size(g)/990; 2688 gap> Filtered([1..7], i-> info[i].size mod q = 0); [ 1, 2, 5 ] gap> m := AtlasSubgroup(g, 5);; gap> maxes := List(ConjugacyClassesMaximalSubgroups(m), Representative);; gap> u := First(maxes, x-> Size(x) = q); gap> act := FactorCosetAction(g,u); CompositionMapping( , ) gap> niceg := Image(act); gap> LargestMovedPoint(niceg); 990 gap> st := Stabilizer(niceg, 1); gap> Length(Orbits(st, [1..990])); 9 Is niceg the permutation group you are looking for? With best regards, Frank -- /// Dr. Frank L?beck, Lehrstuhl D f?r Mathematik, Pontdriesch 14/16, \\\ 52062 Aachen, Germany /// E-mail: Frank.Luebeck at Math.RWTH-Aachen.De \\\ WWW: http://www.math.rwth-aachen.de/~Frank.Luebeck/ This electronic communication is governed by the terms and conditions at http://www.mun.ca/cc/contact/policies/elect_communications_disclaimer_2011.php From dmitrii.pasechnik at cs.ox.ac.uk Wed Nov 16 09:12:37 2016 From: dmitrii.pasechnik at cs.ox.ac.uk (dmitrii.pasechnik at cs.ox.ac.uk) Date: Wed, 16 Nov 2016 09:12:37 +0000 Subject: [GAP Forum] 3.M_{22}:2 on 990 points In-Reply-To: <20161115205117.GA10335@localhost.localdomain> References: <20161115205117.GA10335@localhost.localdomain> Message-ID: <20161116091237.GB16752@localhost.localdomain> Dear forum, Dear Robert, for what it's worth, here is what I described, in GAP, takes an instant to run: LoadPackage("grape"); g0:=AtlasGroup("3.M22.2"); g693:=Image(IsomorphismPermGroup(g)); h:=Stabilizer(g693,1); oo:=Orbits(h,[1..693]); e:=[1,First(oo,x->Length(x)=30)[1]]; gamma:=NullGraph(g693); AddEdgeOrbit(gamma,e); c7:=Orbit(g693,CompleteSubgraphs(gamma,7)[1],OnSets); iif:=NullGraph(Action(g693,c7,OnSets)); AddEdgeOrbit(iif, List(Filtered(c7,x->IsSubset(x,e)),x->Position(c7,x))); # Now we call GlobalParameters(iif); # and see that it is # [ [ 0, 0, 7 ], [ 1, 0, 6 ], [ 1, 2, 4 ], [ 1, 2, 4 ], # [ 2, 1, 4 ], [ 4, 2, 1 ], [ 4, 2, 1 ], [ 6, 0, 1 ], [ 7, 0, 0 ] ] # - the parameters of Ivanov-Ivanov-Faradjev graph Cheers, Dima On Tue, Nov 15, 2016 at 08:51:17PM +0000, dmitrii.pasechnik at cs.ox.ac.uk wrote: > > On Tue, Nov 15, 2016 at 07:59:48PM +0000, Bailey, Robert F. wrote: > > The group 3.M_{22}:2 has a rank-9 imprimitive permutation representation of degree 990. (This is the full automorphism group of the distance-transitive Ivanov-Ivanov-Faradjev graph.) I would like to construct this group in GAP. > > > > The www ATLAS gives a matrix representation of this group in characteristic 2; however, applying "IsomorphismPermGroup" to this matrix group gives a group of degree 693. > > > > Does anyone have a suggestion for how to obtain the degree 990 representation? > > There is a diagram geometry with this group that has 990 points and 693 > lines; more concretely, there is a degree 7 graph of girth 5 on the 990 vertices > that is invariant under your group G, such that every 2-path lies in a > unique Petersen subgraph; there are 693 these subgraphs. > Each vertex is in 7 such subgraphs. > Dually, for the permutation representation of degree 693 there is a > G-invariant graph of degree 30, such each vertex lies in 10 maximal > cliques of size 7; there are 990 such cliques, and the action on them > will give you the desired action. > > I would use GRAPE to construct the graph on 693 vertices, and find a > 7-clique there. Also the following might help: > the maximum possible intersection of two such 7-cliques > is in a 3-clique, and there are 15 such special 3-cliques on each of the > 693 vertices- they correspond to the edges of your graph on 990 > vertices. > > Hope this helps. > I'd be happy to provide more details, if needed. > Dima > > > > > Thanks, > > Robert Bailey. > > > > ============================== > > Dr. Robert Bailey > > School of Science and Environment (Mathematics) > > Grenfell Campus > > Memorial University of Newfoundland > > Corner Brook, NL A2H 6P9, Canada > > > > Office: AS 3022 > > Phone: +1 (709) 637-6293 > > Web: http://www2.grenfell.mun.ca/rbailey/ > > > > This electronic communication is governed by the terms and conditions at http://www.mun.ca/cc/contact/policies/elect_communications_disclaimer_2011.php > > _______________________________________________ > > Forum mailing list > > Forum at mail.gap-system.org > > http://mail.gap-system.org/mailman/listinfo/forum From lovepgroups at gmail.com Wed Nov 16 10:07:05 2016 From: lovepgroups at gmail.com (abdulhakeem alayiwola) Date: Wed, 16 Nov 2016 11:07:05 +0100 Subject: [GAP Forum] XGAP Message-ID: Dear forum, I have tried to install XGAP but i have not been successful. I have tried the procedure available on GAP home page but yet unsuccessful, may be i didn't get language correct. Example:What is C Compiler? or how to configure the C Part. Can someone give me a simple guide on how to install XGAP as a first time user? regards From jalaeeyan at gmail.com Sun Nov 20 12:42:47 2016 From: jalaeeyan at gmail.com (taleea jalaeeyan) Date: Sun, 20 Nov 2016 16:12:47 +0330 Subject: [GAP Forum] question Message-ID: Thank you gap for answering me. 1- how can i define commutators of wight more than 3 to gap? 2- is the concept of Quadratic Non Residue (mod p) is defined for gap? how can i define it for a presentation of a group? how can i define the following presentation in gap \langle x?, ?y |x^{p^{2}}=y^{p}=1?, ?[x?, ?y?, ?x]=1?, ?[x?, ?y?, ?y]=x^{np}?, ?[x?, ?y?, ?y?, ?y]=1 \rangle when n is quadratic non residue (mod p)? From w_becker at hotmail.com Sun Nov 20 15:05:34 2016 From: w_becker at hotmail.com (Walter Becker) Date: Sun, 20 Nov 2016 15:05:34 +0000 Subject: [GAP Forum] how to get groups from GrpConst Message-ID: If you run the Group Extension program for getting cyclic group extension as follows gap> g:=SmallGroup(11^4,11); gap> CyclicSplitExtensions(g,5); rec( both := [ 8192199993 ], down := [ ], up := [ rec( code := 298568409859696239999, order := 73205 ), rec( code := 5856355022849648239999, order := 73205 ), rec( code := 5891490574296688239999, order := 73205 ), rec( code := 11282584185874032239999, order := 73205 ), rec( code := 94119189388603040990846639999, order := 73205 ), rec( code := 94538261508559871198846639999, order := 73205 ), rec( code := 94106975037650140219006639999, order := 73205 ), rec( code := 94464989257903016937086639999, order := 73205 ), rec( code := 94128831738093798622846639999, order := 73205 ), rec( code := 94535689507097728059006639999, order := 73205 ), rec( code := 94162902789490357798526639999, order := 73205 ), rec( code := 94479775706899598272126639999, order := 73205 ), rec( code := 94282447553251495872126639999, order := 73205 ), rec( code := 94138474087584556254846639999, order := 73205 ), rec( code := 94760017161234627233406639999, order := 73205 ), rec( code := 94945130963929828861566639999, order := 73205 ), rec( code := 94842931761382239127166639999, order := 73205 ), rec( code := 94925846361529851020926639999, order := 73205 ), rec( code := 94794076622846695605886639999, order := 73205 ) ] ) how does one get or construct the explicit groups that are found here. In this case it might be relatively easy given the 11 - group here but in other cases it may not be as easy to construct these groups. Commnets ??? Walter Becker From z060822400814a at rezozer.net Sun Nov 20 15:21:01 2016 From: z060822400814a at rezozer.net (z060822400814a at rezozer.net) Date: Sun, 20 Nov 2016 15:21:01 +0000 Subject: [GAP Forum] How to list Warning messages ? Message-ID: <4d42c9b1-c2fc-3ab0-21f6-4c43aac788f6@rezozer.net> Hello Forum, let assume that I am running a restored GAP session: is there an easy way to list the #W[arning] messages ? Thanks in advance, Jerome From stefan at mcs.st-and.ac.uk Sun Nov 20 20:26:48 2016 From: stefan at mcs.st-and.ac.uk (Stefan Kohl) Date: Sun, 20 Nov 2016 20:26:48 -0000 (UTC) Subject: [GAP Forum] how to get groups from GrpConst In-Reply-To: References: Message-ID: On Sun, November 20, 2016 3:05 pm, Walter Becker wrote: > If you run the Group Extension program for getting > cyclic group extension as follows > > gap> g:=SmallGroup(11^4,11); > > gap> CyclicSplitExtensions(g,5); > rec( both := [ 8192199993 ], down := [ ], > up := [ rec( code := 298568409859696239999, order := 73205 ), rec( code := > 5856355022849648239999, order := 73205 ), > rec( code := 5891490574296688239999, order := 73205 ), rec( code := > 11282584185874032239999, order := 73205 ), > rec( code := 94119189388603040990846639999, order := 73205 ), > rec( code := 94538261508559871198846639999, order := 73205 ), > rec( code := 94106975037650140219006639999, order := 73205 ), > rec( code := 94464989257903016937086639999, order := 73205 ), > rec( code := 94128831738093798622846639999, order := 73205 ), > rec( code := 94535689507097728059006639999, order := 73205 ), > rec( code := 94162902789490357798526639999, order := 73205 ), > rec( code := 94479775706899598272126639999, order := 73205 ), > rec( code := 94282447553251495872126639999, order := 73205 ), > rec( code := 94138474087584556254846639999, order := 73205 ), > rec( code := 94760017161234627233406639999, order := 73205 ), > rec( code := 94945130963929828861566639999, order := 73205 ), > rec( code := 94842931761382239127166639999, order := 73205 ), > rec( code := 94925846361529851020926639999, order := 73205 ), > rec( code := 94794076622846695605886639999, order := 73205 ) ] ) > > how does one get or construct the explicit groups that > are found here. Just do gap> g:=SmallGroup(11^4,11); gap> extcodes := CyclicSplitExtensions(g,5);; gap> extgrps := List(extcodes.up,r->PcGroupCode(r.code,r.order)); [ , , , , , , , , , , , , , , , , , , ] Hope this helps, Stefan Kohl ----------------------------------------------------------------------------- http://www.gap-system.org/DevelopersPages/StefanKohl/ ----------------------------------------------------------------------------- From alexander.konovalov at st-andrews.ac.uk Tue Nov 22 14:09:11 2016 From: alexander.konovalov at st-andrews.ac.uk (Alexander Konovalov) Date: Tue, 22 Nov 2016 14:09:11 +0000 Subject: [GAP Forum] How to list Warning messages ? In-Reply-To: <4d42c9b1-c2fc-3ab0-21f6-4c43aac788f6@rezozer.net> References: <4d42c9b1-c2fc-3ab0-21f6-4c43aac788f6@rezozer.net> Message-ID: <92A605DF-A04A-42C0-8945-7F1D222EF7B3@st-andrews.ac.uk> Hello Jerome, > On 20 Nov 2016, at 15:21, z060822400814a at rezozer.net wrote: > > Hello Forum, > > let assume that I am running a restored GAP session: > is there an easy way to list the #W[arning] messages ? > > Thanks in advance, > Jerome If you mean accessing all warnings that were issued before the workspace was saved, then I think in general the answer is 'no'. The only kind of warnings you may restore are package loading warnings, which may be displayed using DisplayPackageLoadingLog. This happens because they are saved in GAPInfo.PackageLoadingMessages. Best wishes Alexander From alexander.konovalov at st-andrews.ac.uk Tue Nov 22 14:25:43 2016 From: alexander.konovalov at st-andrews.ac.uk (Alexander Konovalov) Date: Tue, 22 Nov 2016 14:25:43 +0000 Subject: [GAP Forum] XGAP In-Reply-To: References: Message-ID: Dear Abdulhakeem Alayiwola, As you wrote to me, you have Windows and you are interested in lattices of subgroups. You can't install XGAP on Windows since it requires UNIX environment. For example, it works on Linux and on OS X (needs XQuartz - https://www.xquartz.org/). For Windows machine, one suggestion could be to install VirtualBox (https://www.virtualbox.org/) and establish Linux virtual machine. Hopefully the functionality you want will eventually appear in the GAP Jupyter interface (http://opendreamkit.org/activities/2016-08-03-gap-docker-jupyter/), but it's not there yet. However note that under Windows you should be able to compute the lattice of subgroups using LatticeSubgroups and then explore it from GAP. Strictly speaking, XGAP is needed to *visualise* the lattice of subgroups, but not to *compute* it. There is also a GAP function DotFileLatticeSubgroups that produces an input file for GraphViz (http://www.graphviz.org) which you will hopefully be able to used under Windows. See this example: http://math.stackexchange.com/questions/1737962/subgroup-lattice-of-ut3-3/1737985#1737985 Hope this helps Alexander > On 16 Nov 2016, at 10:07, abdulhakeem alayiwola wrote: > > Dear forum, > I have tried to install XGAP but i have not been successful. I have tried > the procedure available on GAP home page but yet unsuccessful, may be i > didn't get language correct. Example:What is C Compiler? or how to > configure the C Part. Can someone give me a simple guide on how to install > XGAP as a first time user? > > regards From alexander.konovalov at st-andrews.ac.uk Tue Nov 22 14:44:03 2016 From: alexander.konovalov at st-andrews.ac.uk (Alexander Konovalov) Date: Tue, 22 Nov 2016 14:44:03 +0000 Subject: [GAP Forum] question In-Reply-To: References: Message-ID: <48487AEE-E961-4DEA-81A8-F216C7BB9E0C@st-andrews.ac.uk> Dear Taleea Jalaeeyan, > On 20 Nov 2016, at 12:42, taleea jalaeeyan wrote: > > Thank you gap for answering me. > 1- how can i define commutators of wight more than 3 to gap? There is an undocumented function LeftNormedComm, so ?[x?, ?y?, ?y?, ?y] is ?[[[x?, ?y?], ?y?], ?y] > 2- is the concept of Quadratic Non Residue (mod p) is defined for gap? Yes - see Legendre, http://www.gap-system.org/Manuals/doc/ref/chap15.html#X81464ABF7F10E544 > how can i define it for a presentation of a group? You will have to construct a group below for some fixed given n and p. For example, gap> Legendre(3,5); -1 so 3 is a quadratic non-residue modulo 5. Now > how can i define the following presentation in gap > \langle x?, ?y |x^{p^{2}}=y^{p}=1?, ?[x?, ?y?, ?x]=1?, ?[x?, ?y?, > ?y]=x^{np}?, ?[x?, ?y?, ?y?, ?y]=1 \rangle > when n is quadratic non residue (mod p)? gap> x:=f.1; x gap> y:=f.2; y gap> n:=3;p:=5; 3 5 gap> rels:=[ > x^(p^2), y^(p), > LeftNormedComm([x,y,x]), > LeftNormedComm([x,y,y])*x^(-n*p), > LeftNormedComm([x,y,y,y]) ]; [ x^25, y^5, y^-1*x^-1*y*x^-1*y^-1*x*y*x, y^-1*x^-1*y*x*y^-1*x^-1*y^-1*x*y^2*x^-15, y^-2*x^-1*y*x*y*x^-1*y^-1*x*y^-1*x^-1*y*x*y^-1*x^-1*y^-1*x*y^3 ] gap> g:=f/rels; gap> Size(g); 625 gap> IdGroup(g); [ 625, 10 ] Hope this helps Alexander From saad1225 at gmail.com Wed Nov 23 23:30:55 2016 From: saad1225 at gmail.com (saad khalid) Date: Wed, 23 Nov 2016 17:30:55 -0600 Subject: [GAP Forum] Difficulty Understanding how to Generate Group and Compute Molien Series Message-ID: Hello everyone! I'm coming from using Macaulay2 (and not knowing much of anything about representation theory) to generate groups and their corresponding Molien series. In Macaulay2, I would create a field with some qth root of unity. For example, if I wanted q = 8, I would use: K = toField(QQ[zet]/(zet^4 + 1)) Then, I would pick some matrix I wanted to use to generate the group: A = matrix{{zet^(a1),0,0},{0,zet^(a2),0},{0,0,zet^(a3)}} #a1, a2, a3 are what I would pick. Notice that this is diagonal(all the matrices I'm dealing with are diagonal). Then, I would generate the group for A in the field K, and compute its molienseries: B = generateGroup({A},K) molienSeries B I would like to do this process in GAP, but I'm having trouble figuring out how to generate the group. In actuality, I don't need the group at all, I'm only interested in the final molienseries representation. However, to generate the molien series using GAP, I think I need to know the character of the group. I believe I have sort of an understanding of what that means. At least, in my case since they're just diagonal matrices, the character of each matrix in the group is just the trace of that matrix. Even if that is the case, I still don't know how to put that in GAP. Say I had q = 4 and picked a1 = 1, a2 = 2, a3 = 3; what should I input into GAP as the character of this group for it to generate the Molien series? If there isn't some easy way to just figure out and type in the character of the group, I assume that I'll have to create the group in GAP and then use GAP to get its character. I followed the examples given in the GAP documentation, but I couldn't figure out how to apply that to my situation. I just don't know how to translate what I did in Macaulay2 to GAP. I would like to learn representation theory in the future, and I hope that I will learn some by working with GAP. Until that happens though, I would like to apologize for how novice my questions here are. Thank you for taking the time to read my post! I would really appreciate any help you can give. -Saad From max at quendi.de Thu Nov 24 14:09:51 2016 From: max at quendi.de (Max Horn) Date: Thu, 24 Nov 2016 15:09:51 +0100 Subject: [GAP Forum] Difficulty Understanding how to Generate Group and Compute Molien Series In-Reply-To: References: Message-ID: <21315E51-7304-4565-B625-E7C0CE5CF1C6@quendi.de> Hi Saad, > On 24 Nov 2016, at 00:30, saad khalid wrote: > > Hello everyone! > > I'm coming from using Macaulay2 (and not knowing much of anything about > representation theory) to generate groups and their corresponding Molien > series. In Macaulay2, I would create a field with some qth root of unity. > For example, if I wanted q = 8, I would use: > K = toField(QQ[zet]/(zet^4 + 1)) > > Then, I would pick some matrix I wanted to use to generate the group: > > A = matrix{{zet^(a1),0,0},{0,zet^(a2),0},{0,0,zet^(a3)}} #a1, a2, a3 are > what I would pick. In GAP, you can get a primitive 8-th root of unity by E(8), and the field it generates by CF(8). So you could write your matrix like this: z := E(8); A := DiagonalMat([z^a1,z^a2,z^a3]); > > Notice that this is diagonal(all the matrices I'm dealing with are > diagonal). > > Then, I would generate the group for A in the field K, and compute its > molienseries: > B = generateGroup({A},K) > molienSeries B The equivalent to the first command in GAP: B := Group( A ); As to the second command, the corresponding GAP function is MolienSeries(), but it takes a character of a group, not a (matrix) group. But as you say, you can easily convert your matrix representation into a character as follows: chi := Character(B, List(ConjugacyClasses(B), c -> TraceMat(Representative(c)))); Then we can apply MolienSeries, and get something like this (for a1=1, a2=2, a3=3): gap> MolienSeries(chi); ( 1-z-z^2+2*z^3+2*z^4-2*z^5+z^7 ) / ( (1-z^8)*(1-z^2)*(1-z) ) Hope that helps, Max From vicorso at doctor.upv.es Sat Nov 26 15:17:07 2016 From: vicorso at doctor.upv.es (=?utf-8?b?VsOtY3Rvcg==?= Manuel Ortiz Sotomayor) Date: Sat, 26 Nov 2016 16:17:07 +0100 Subject: [GAP Forum] Conjugacy classes Alternating group degree 125 Message-ID: <20161126161707.Horde.ybLatfVWluRJkfXUSfzRRzC@webmail.upv.es> Let G:=AlternatingGroup(125) be the Alternating group of degree 125, and let Q:=SylowSubgroup(G, 5) be a Sylow 5-subgroup of G. I want to compute, for each element x of Q, the distinct G-conjugacy class sizes, that is, the distinct values of Size(ConjugacyClass(G, x)) (obviously, computing the distinct values of Centralizer(G, x) for all x in Q) would be the same). Needless to say that, I always get out of memory when I run over all the elements of Q. I had tried the following: compute the upper central series of Q (L:=UpperCentralSeriesOfGroup(Q)) and, for some "intermediate" normal subgroup N in that chain, to decompose Q in right cosets on N, in order to make a disjoint union of the elements of Q that is more manageable. However, I still have problems of memory because either I have so many transversals or the order of N is also too large. Any idea? Thanks in advance. From Alexander.Hulpke at colostate.edu Sat Nov 26 17:18:09 2016 From: Alexander.Hulpke at colostate.edu (Hulpke,Alexander) Date: Sat, 26 Nov 2016 17:18:09 +0000 Subject: [GAP Forum] Conjugacy classes Alternating group degree 125 In-Reply-To: <20161126161707.Horde.ybLatfVWluRJkfXUSfzRRzC@webmail.upv.es> References: <20161126161707.Horde.ybLatfVWluRJkfXUSfzRRzC@webmail.upv.es> Message-ID: Dear GAP-Forum, On Nov 26, 2016, at 8:17 AM, V?ctor Manuel Ortiz Sotomayor > wrote: Let G:=AlternatingGroup(125) be the Alternating group of degree 125, and let Q:=SylowSubgroup(G, 5) be a Sylow 5-subgroup of G. I want to compute, for each element x of Q, the distinct G-conjugacy class sizes, that is, the distinct values of Size(ConjugacyClass(G, x)) (obviously, computing the distinct values of Centralizer(G, x) for all x in Q) would be the same). Needless to say that, I always get out of memory when I run over all the elements of Q. I had tried the following: compute the upper central series of Q (L:=UpperCentralSeriesOfGroup(Q)) and, for some "intermediate" normal subgroup N in that chain, to decompose Q in right cosets on N, in order to make a disjoint union of the elements of Q that is more manageable. However, I still have problems of memory because either I have so many transversals or the order of N is also too large. Any idea? My recommendation would be to take the code for computing conjugacy classes (which is in the file claspcgs.gi) and centralizers (which works down over subsequently larger factor groups) which currently works breadth-first to convert it to a depth-first approach, not storing all classes, but deleting them once the size of the class is known. Caveat: Even the factor of Q modulo the 6th term in the lower central series (this is less than sqrt(|Q|), so only a small bit) already has almost 2 million conjugacy classes, so the whole group could easily have 10^10 classes or more. which puts the feasibility of such an enumerative approach into doubt. Best, Alexander Hulpke -- Colorado State University, Department of Mathematics, Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA email: hulpke at colostate.edu, Phone: ++1-970-4914288 http://www.math.colostate.edu/~hulpke From benjamin.sambale at gmail.com Sat Nov 26 19:11:37 2016 From: benjamin.sambale at gmail.com (Benjamin Sambale) Date: Sat, 26 Nov 2016 20:11:37 +0100 Subject: [GAP Forum] Conjugacy classes Alternating group degree 125 In-Reply-To: <20161126161707.Horde.ybLatfVWluRJkfXUSfzRRzC@webmail.upv.es> References: <20161126161707.Horde.ybLatfVWluRJkfXUSfzRRzC@webmail.upv.es> Message-ID: <93bc9cd8-8f34-6b5f-f8f1-12b54343c768@gmail.com> If I understand the question correctly, there is no need to use a computer. The conjugacy classes of the alternating groups are well-known and there sizes can be computed easily. But maybe this is not the point of the question. Best, Benjamin Am 26.11.2016 um 16:17 schrieb V?ctor Manuel Ortiz Sotomayor: > Let G:=AlternatingGroup(125) be the Alternating group of degree 125, > and let Q:=SylowSubgroup(G, 5) be a Sylow 5-subgroup of G. > > I want to compute, for each element x of Q, the distinct G-conjugacy > class sizes, that is, the distinct values of Size(ConjugacyClass(G, > x)) (obviously, computing the distinct values of Centralizer(G, x) for > all x in Q) would be the same). > > Needless to say that, I always get out of memory when I run over all > the elements of Q. I had tried the following: compute the upper > central series of Q (L:=UpperCentralSeriesOfGroup(Q)) and, for some > "intermediate" normal subgroup N in that chain, to decompose Q in > right cosets on N, in order to make a disjoint union of the elements > of Q that is more manageable. However, I still have problems of memory > because either I have so many transversals or the order of N is also > too large. Any idea? > > Thanks in advance. > > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum > From vicorso at doctor.upv.es Sun Nov 27 15:26:31 2016 From: vicorso at doctor.upv.es (VMOS) Date: Sun, 27 Nov 2016 16:26:31 +0100 Subject: [GAP Forum] Conjugacy classes Alternating group degree 125 Message-ID: First of all, thank you so much por your invaluable help. I'm a young student and I didn't know that "the conjugacy classes of the alternating groups were?well-known and there sizes could be computed easily". Anyway, I hope that this kind of computations with many conjugacy classes might be interesting in other frameworks. I'm sorry if the ease of my question has bothered someone. Best, VMOS -------- Mensaje original -------- De: V?ctor Manuel Ortiz Sotomayor Fecha:27/11/2016 15:44 (GMT+01:00) Para: benjamin.sambale at gmail.com,Alexander.Hulpke at colostate.edu Cc: Forum at mail.gap-system.org Asunto: [GAP Forum] Conjugacy classes Alternating group degree 125 ------------------------------------------------------------------------- > My recommendation would be to take the code for computing conjugacy > classes (which is in the file claspcgs.gi) and centralizers (which > works down over subsequently larger factor groups) which currently > works breadth-first to convert it to a depth-first approach, not > storing all classes, but deleting them once the size of the class is > known. > Caveat: Even the factor of Q modulo the 6th term in the lower > central series (this is less than sqrt(|Q|), so only a small bit) > already has almost 2 million conjugacy classes, so the whole group > could easily have 10^10 classes or more. which puts the feasibility > of such an enumerative approach into doubt. > Best, > Alexander Hulpke -------------------------------------------------------------------------- >> If I understand the question correctly, there is no need to use a >> computer. The conjugacy classes of the alternating groups are >> well-known and there sizes can be computed easily. But maybe this >> is not the point of the question. >> Best, >> Benjamin -------------------------------------------------------------------------- >>> Am 26.11.2016 um 16:17 schrieb V?ctor Manuel Ortiz Sotomayor: >>> Let G:=AlternatingGroup(125) be the Alternating group of degree >>> 125, and let Q:=SylowSubgroup(G, 5) be a Sylow 5-subgroup of G. >>> >>> I want to compute, for each element x of Q, the distinct >>> G-conjugacy class sizes, that is, the distinct values of >>> Size(ConjugacyClass(G, x)) (obviously, computing the distinct >>> values of Centralizer(G, x) for all x in Q) would be the same). >>> >>> Needless to say that, I always get out of memory when I run over >>> all the elements of Q. I had tried the following: compute the >>> upper central series of Q (L:=UpperCentralSeriesOfGroup(Q)) and, >>> for some "intermediate" normal subgroup N in that chain, to >>> decompose Q in right cosets on N, in order to make a disjoint >>> union of the elements of Q that is more manageable. However, I >>> still have problems of memory because either I have so many >>> transversals or the order of N is also too large. Any idea? >>> >>> Thanks in advance. From mazurov at math.nsc.ru Sat Dec 3 05:24:26 2016 From: mazurov at math.nsc.ru (Victor D. Mazurov) Date: Sat, 3 Dec 2016 12:24:26 +0700 Subject: [GAP Forum] tensor product of representations Message-ID: Dear forum, How can I get a homomorphism from given representation of finite group to the another one? Example: By Atlas of FGR, Matrices ? [[0,1,0,0], [1,1,0,0], [0,0,0,1], [0,0,1,1] ]*Z(2) , [[0,0,1,0], [0,1,1,0], [0,1,1,1], [1,1,1,0] ]*Z(2) generate? a 4-dimensional representation U of alternating group A_8 over a field of order 2 and matrices [[0,1,0,0,0,0], [1,1,0,0,0,0], [1,1,1,0,0,0], [0,0,0,1,0,0], [0,0,0,0,1,0], [0,0,0,0,0,1] ]*Z(2) , [[1,1,0,0,0,0], [0,0,1,0,0,0], [0,0,0,1,0,0], [0,0,0,0,1,0], [0,0,0,0,0,1], [1,0,1,0,1,0] ]*Z(2) ?generate a 6-dimensional representation V of A_8 over a field of order 2?. How can I calculate H=Hom(U\otimes U,V) and, if H\ne 0, a homomorphism of U\otimes U onto V? Best wishes, Victor Mazurov -- Victor Danilovich Mazurov Institute of Mathematics Novosibirsk 630090 Russia From Alexander.Hulpke at colostate.edu Sat Dec 3 16:07:07 2016 From: Alexander.Hulpke at colostate.edu (Hulpke,Alexander) Date: Sat, 3 Dec 2016 16:07:07 +0000 Subject: [GAP Forum] tensor product of representations In-Reply-To: References: Message-ID: <52EF6EE4-26BA-4542-97B5-7F241AE38C32@colostate.edu> Dear Forum, Dear Victor Mazurov, > On Dec 2, 2016, at 10:24 PM, Victor D. Mazurov wrote: > > Dear forum, > > How can I get a homomorphism from given representation of finite group to > the another one? > > Example: By Atlas of FGR, > Matrices > [?] > generate? a 4-dimensional representation U of alternating group A_8 over a > field of order 2 and > matrices > > [?] > > ?generate a 6-dimensional representation V of A_8 over a field of order > 2?. > If you get matrices from the online ATLAS, you are in luck in that they are always given on the same generators, that is isomorphisms will simply map the one generating set to the other. For example, you could use hom:=GroupHomomorphismByImages(U,V,GeneratorsOfGroup(U),GeneratorsOfGroup(V)); to construct such an isomorphism. You can apply it with Image(how,elm) on elements or subgroups. > How can I calculate H=Hom(U\otimes U,V) and, if H\ne 0, a homomorphism of > U\otimes U onto V? Do you mean by U\otimes U the tensor-square representation? If so, you do the same (with generators still fitting) gap> tens:=List(GeneratorsOfGroup(U), > x->KroneckerProduct(x,x)); gap> A:=Group(tens); gap> hom:=GroupHomomorphismByImages(A,V,GeneratorsOfGroup(A),GeneratorsOfGroup( If the generators do not agree, you would have to do an explicit homomorphism search. E.g. (forcing different generators: gap> B:=Group(Random(U),Random(U));Size(B); 20160 gap> IsomorphismGroups(B,V); CompositionMapping( [ (2,9)(4,11)(6,13)(8,15), (2,7,6,10,12)(3,11,8,4,13)(5,16,15,9,14) ] -> [ , ], ) All the best, Alexander Hulpke -- Colorado State University, Department of Mathematics, Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA email: hulpke at colostate.edu, Phone: ++1-970-4914288 http://www.math.colostate.edu/~hulpke From mazurov at math.nsc.ru Sat Dec 3 23:44:04 2016 From: mazurov at math.nsc.ru (Victor D. Mazurov) Date: Sun, 4 Dec 2016 06:44:04 +0700 Subject: [GAP Forum] tensor product of representations In-Reply-To: <52EF6EE4-26BA-4542-97B5-7F241AE38C32@colostate.edu> References: <52EF6EE4-26BA-4542-97B5-7F241AE38C32@colostate.edu> Message-ID: Dear Alexander, Thank you for your comprehensive help. Best wishes, Victor 2016-12-03 23:07 GMT+07:00 Hulpke,Alexander : > Dear Forum, Dear Victor Mazurov, > > > > On Dec 2, 2016, at 10:24 PM, Victor D. Mazurov > wrote: > > > > Dear forum, > > > > How can I get a homomorphism from given representation of finite group to > > the another one? > > > > Example: By Atlas of FGR, > > Matrices > > [?] > > > generate? a 4-dimensional representation U of alternating group A_8 over > a > > field of order 2 and > > matrices > > > > [?] > > > > ?generate a 6-dimensional representation V of A_8 over a field of order > > 2?. > > > > If you get matrices from the online ATLAS, you are in luck in that they > are always given on the same generators, that is isomorphisms will simply > map the one generating set to the other. For example, you could use > > hom:=GroupHomomorphismByImages(U,V,GeneratorsOfGroup(U), > GeneratorsOfGroup(V)); > > to construct such an isomorphism. You can apply it with Image(how,elm) > on elements or subgroups. > > > How can I calculate H=Hom(U\otimes U,V) and, if H\ne 0, a homomorphism of > > U\otimes U onto V? > > Do you mean by U\otimes U the tensor-square representation? If so, you do > the same (with generators still fitting) > > gap> tens:=List(GeneratorsOfGroup(U), > > x->KroneckerProduct(x,x)); > gap> A:=Group(tens); > gap> hom:=GroupHomomorphismByImages(A,V,GeneratorsOfGroup(A), > GeneratorsOfGroup(\$ > > If the generators do not agree, you would have to do an explicit > homomorphism search. E.g. (forcing different generators: > > gap> B:=Group(Random(U),Random(U));Size(B); > > 20160 > gap> IsomorphismGroups(B,V); > CompositionMapping( > [ (2,9)(4,11)(6,13)(8,15), (2,7,6,10,12)(3,11,8,4,13)(5,16,15,9,14) ] -> > [ , ], > ) > > All the best, > > Alexander Hulpke > > > > -- Colorado State University, Department of Mathematics, > Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA > email: hulpke at colostate.edu, Phone: ++1-970-4914288 > http://www.math.colostate.edu/~hulpke > > > -- Victor Danilovich Mazurov Institute of Mathematics Novosibirsk 630090 Russia From qijiayue14 at mails.ucas.ac.cn Sat Dec 17 09:56:34 2016 From: qijiayue14 at mails.ucas.ac.cn (=?GBK?B?xuu8ztTD?=) Date: Sat, 17 Dec 2016 17:56:34 +0800 (GMT+08:00) Subject: [GAP Forum] Consults on GAP Message-ID: <1065d8f.20f7d.1590c362f8e.Coremail.qijiayue14@mails.ucas.ac.cn> Dear Sir/Madam, Em.. here is a question I have about GAP: Is there any operations like "AllSmallGroupsOfCharacterDegrees([...])"? How do we find the certain groups in GAP database(the SmallGroup Id will be fine) holding the given character degrees? Looking forward to your reply!Thanks a lot! Best regards, Jiayue From qijiayue14 at mails.ucas.ac.cn Sat Dec 17 11:17:41 2016 From: qijiayue14 at mails.ucas.ac.cn (=?GBK?B?xuu8ztTD?=) Date: Sat, 17 Dec 2016 19:17:41 +0800 (GMT+08:00) Subject: [GAP Forum] How to express the group algebra over the complex field C[G]? Message-ID: <15f1046.210e5.1590c807040.Coremail.qijiayue14@mails.ucas.ac.cn> Dear Sir/Madam, Sorry I'm asking again... as the title says, I wonder how could I express the group algebra C[G] over the complex field C in GAP? I tried several but they didn't work... I'd appreciate it a lot if you could help me with this! Looking forward to your reply! Best regards, Jiayue From Alexander.Hulpke at colostate.edu Sat Dec 17 15:58:42 2016 From: Alexander.Hulpke at colostate.edu (Hulpke,Alexander) Date: Sat, 17 Dec 2016 15:58:42 +0000 Subject: [GAP Forum] Consults on GAP In-Reply-To: <1065d8f.20f7d.1590c362f8e.Coremail.qijiayue14@mails.ucas.ac.cn> References: <1065d8f.20f7d.1590c362f8e.Coremail.qijiayue14@mails.ucas.ac.cn> Message-ID: <46D19297-644D-44B2-85B5-25F1F88C2F9E@colostate.edu> Dear Forum, Jiayue asked: > Em.. here is a question I have about GAP: > > Is there any operations like "AllSmallGroupsOfCharacterDegrees([...])"? > > How do we find the certain groups in GAP database(the SmallGroup Id will be fine) > holding the given character degrees? The operation AllSmallGroups? can select for groups satisfying certain functions, e.g the result of the function CharacterDegrees?. For example, to get all groups of order 32 with 4 characters of degree and 16 linear characters, one could use: AllSmallGroups(Size,32,CharacterDegrees,[[[1,16],[2,4]]]); (Note that, as the result is a list, we need to wrap it in a list once more to avoid GAP interpreting it as a collection of values, alternatively one could use it as: AllSmallGroups(Size,32,x->CharacterDegrees(x)=[[1,16],[2,4]],true); Note that the character degrees are not stored precomputed in the library, so GAP will have to compute them on the fly for all groups in the otherwise specified range. So doing this test for order 256 could take a long time. If you do not care about multiplicities, you could take the degrees only with a small list operation, that is: occurringDegrees:=g->List(CharacterDegrees(g),x->x[1]); AllSmallGroups(Size,32,occurringDegrees,[[1,2]]); would return the groups of order 32 with characters only of degree 1 and 2. Regards, Alexander Hulpke -- Colorado State University, Department of Mathematics, Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA email: hulpke at colostate.edu, Phone: ++1-970-4914288 http://www.math.colostate.edu/~hulpke From qijiayue14 at mails.ucas.ac.cn Sat Dec 17 16:17:58 2016 From: qijiayue14 at mails.ucas.ac.cn (=?utf-8?B?6b2Q5ZiJ5oKm?=) Date: Sun, 18 Dec 2016 00:17:58 +0800 (GMT+08:00) Subject: [GAP Forum] Consults on GAP In-Reply-To: <46D19297-644D-44B2-85B5-25F1F88C2F9E@colostate.edu> References: <1065d8f.20f7d.1590c362f8e.Coremail.qijiayue14@mails.ucas.ac.cn> <46D19297-644D-44B2-85B5-25F1F88C2F9E@colostate.edu> Message-ID: <188c1af.2168c.1590d935c54.Coremail.qijiayue14@mails.ucas.ac.cn> Dear Alexander, Thank you so much for helping and I think my problem of this character degree stuff solved now! Best wishes, Jiayue ?2016-12-17 23:58:42,?????? > Dear Forum, > > Jiayue asked: > > > Em.. here is a question I have about GAP: > > > > Is there any operations like "AllSmallGroupsOfCharacterDegrees([...])"? > > > > How do we find the certain groups in GAP database(the SmallGroup Id will be fine) > > holding the given character degrees? > > > The operation AllSmallGroups? can select for groups satisfying certain functions, e.g the result of the function CharacterDegrees?. For example, to get all groups of order 32 with 4 characters of degree and 16 linear characters, one could use: > > AllSmallGroups(Size,32,CharacterDegrees,[[[1,16],[2,4]]]); > > (Note that, as the result is a list, we need to wrap it in a list once more to avoid GAP interpreting it as a collection of values, alternatively one could use it as: > > AllSmallGroups(Size,32,x->CharacterDegrees(x)=[[1,16],[2,4]],true); > > Note that the character degrees are not stored precomputed in the library, so GAP will have to compute them on the fly for all groups in the otherwise specified range. So doing this test for order 256 could take a long time. > > If you do not care about multiplicities, you could take the degrees only with a small list operation, that is: > > occurringDegrees:=g->List(CharacterDegrees(g),x->x[1]); > AllSmallGroups(Size,32,occurringDegrees,[[1,2]]); > > would return the groups of order 32 with characters only of degree 1 and 2. > > Regards, > > Alexander Hulpke > > -- Colorado State University, Department of Mathematics, > Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA > email: hulpke at colostate.edu, Phone: ++1-970-4914288 > http://www.math.colostate.edu/~hulpke > > From gryakj at gmail.com Mon Dec 19 01:55:50 2016 From: gryakj at gmail.com (Jonathan Gryak) Date: Sun, 18 Dec 2016 20:55:50 -0500 Subject: [GAP Forum] Alternative Methods for Constructing Non-Virtually Nilpotent Polycyclic Groups in GAP Message-ID: Hello Forum Members, The GAP package Polycyclic contains the function MaximalOrderByUnitsPcpGroup that constructs a polycyclic group that is infinite and non-virtually nilpotent. These groups are of the form O(F) \rtimes U(F), where F is an algebraic number field and O(F) and U(F) are respectively the maximal order and unit group of F, as discussed in the chapter on polycyclic group in the Handbook of Computational Group Theory . Is there another method for constructing infinite, non-nilpotent polycyclic groups that doesn't rely on the split extension method above? And can this method be implemented in GAP, say via a finite presentation? Thanks in advance, Jonathan From max at quendi.de Mon Dec 19 14:44:48 2016 From: max at quendi.de (Max Horn) Date: Mon, 19 Dec 2016 15:44:48 +0100 Subject: [GAP Forum] Alternative Methods for Constructing Non-Virtually Nilpotent Polycyclic Groups in GAP In-Reply-To: References: Message-ID: Dear Jonathan, > On 19 Dec 2016, at 02:55, Jonathan Gryak wrote: > > Hello Forum Members, > The GAP package Polycyclic contains the function MaximalOrderByUnitsPcpGroup > > that constructs a polycyclic group that is infinite and non-virtually > nilpotent. These groups are of the form O(F) \rtimes U(F), where F is an > algebraic number field and O(F) and U(F) are respectively the maximal order > and unit group of F, as discussed in the chapter on polycyclic group > in the Handbook > of Computational Group Theory > > . > > Is there another method for constructing infinite, non-nilpotent polycyclic > groups that doesn't rely on the split extension method above? And can this > method be implemented in GAP, say via a finite presentation? This is the primary purpose of the Polycyclic package you already are using: It allows you to work with arbitrary Polycyclic groups. You can do so either by constructing a suitable collector directly (as described in Chapter 3 of the Polycyclic manual), or else by defining a suitable finitely presented group, then using the function IsomorphismPcpGroupFromFpGroupWithPcPres (see chapter 5 of the Polycyclic manual). Regards, Max From qijiayue14 at mails.ucas.ac.cn Wed Dec 21 11:40:42 2016 From: qijiayue14 at mails.ucas.ac.cn (=?GBK?B?xuu8ztTD?=) Date: Wed, 21 Dec 2016 19:40:42 +0800 (GMT+08:00) Subject: [GAP Forum] AlternatingGroup(4) and its generators Message-ID: <1c235dc.540c.159212ef484.Coremail.qijiayue14@mails.ucas.ac.cn> Dear Forum members, When I type this in GAP: gap>G:=SmallGroup(12,3); gap>GeneratorsOfGroup(G); the result is [f1,f2,f3],but actually the group G here is AlternatingGroup(4) and I wonder how could I know here what f1,f2,f3 exactly means by permutations respectively? How could I know what are they in A4? Looking forward to any reply! Thanks a lot! Jiayue From qijiayue14 at mails.ucas.ac.cn Wed Dec 21 11:41:25 2016 From: qijiayue14 at mails.ucas.ac.cn (=?GBK?B?xuu8ztTD?=) Date: Wed, 21 Dec 2016 19:41:25 +0800 (GMT+08:00) Subject: [GAP Forum] AlternatingGroup(4) and its generators Message-ID: Dear Forum members, When I type this in GAP: gap>G:=SmallGroup(12,3); gap>GeneratorsOfGroup(G); the result is [f1,f2,f3],but actually the group G here is AlternatingGroup(4) and I wonder how could I know here what f1,f2,f3 exactly means by permutations respectively? How could I know what are they in A4? Looking forward to any reply! Thanks a lot! Jiayue From sven.reichard at tu-dresden.de Wed Dec 21 12:12:41 2016 From: sven.reichard at tu-dresden.de (Sven Reichard) Date: Wed, 21 Dec 2016 13:12:41 +0100 Subject: [GAP Forum] AlternatingGroup(4) and its generators In-Reply-To: <1c235dc.540c.159212ef484.Coremail.qijiayue14@mails.ucas.ac.cn> References: <1c235dc.540c.159212ef484.Coremail.qijiayue14@mails.ucas.ac.cn> Message-ID: Dear Jiayue, to be precise, G is isomorphic to the alternating group. You can find such an isomorphism and determine the images of the generators as follows: gap> G := SmallGroup(12,3); gap> StructureDescription(G); "A4" gap> a4 := AlternatingGroup(4); Alt( [ 1 .. 4 ] ) gap> iso := IsomorphismGroups(G, a4); [ f1, f2, f3 ] -> [ (2,4,3), (1,3)(2,4), (1,2)(3,4) ] gap> List(GeneratorsOfGroup(G), g -> Image(iso, g)); [ (2,4,3), (1,3)(2,4), (1,2)(3,4) ] HTH, Sven On 21.12.2016 12:40, ??? wrote: > > Dear Forum members, > > > When I type this in GAP: > > > gap>G:=SmallGroup(12,3); > gap>GeneratorsOfGroup(G); > the result is [f1,f2,f3],but actually the group G here is AlternatingGroup(4) and I wonder > how could I know here what f1,f2,f3 exactly means by permutations respectively? How > could I know what are they in A4? > > > Looking forward to any reply! > > > Thanks a lot! > > > Jiayue > > > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum > From qijiayue14 at mails.ucas.ac.cn Wed Dec 21 12:35:51 2016 From: qijiayue14 at mails.ucas.ac.cn (=?GBK?B?xuu8ztTD?=) Date: Wed, 21 Dec 2016 20:35:51 +0800 (GMT+08:00) Subject: [GAP Forum] AlternatingGroup(4) and its generators Message-ID: <4108c5.5654.159216172a8.Coremail.qijiayue14@mails.ucas.ac.cn> Dear Forum members, When I type this in GAP: gap>G:=SmallGroup(12,3); gap>GeneratorsOfGroup(G); the result is [f1,f2,f3],but actually the group G here is AlternatingGroup(4) and I wonder how could I know here what f1,f2,f3 exactly means by permutations respectively? How could I know what are they in A4? Looking forward to any reply! Thanks a lot! Jiayue From qijiayue14 at mails.ucas.ac.cn Wed Dec 21 13:12:23 2016 From: qijiayue14 at mails.ucas.ac.cn (=?utf-8?B?6b2Q5ZiJ5oKm?=) Date: Wed, 21 Dec 2016 21:12:23 +0800 (GMT+08:00) Subject: [GAP Forum] AlternatingGroup(4) and its generators In-Reply-To: References: <1c235dc.540c.159212ef484.Coremail.qijiayue14@mails.ucas.ac.cn> Message-ID: <61ac49.587b.1592182e3d6.Coremail.qijiayue14@mails.ucas.ac.cn> Dear Sven, Now I see the way to find out what f1,f2,f3 exactly are in the A4 group! Thank you so much for your help! Best regards, Jiayue > -----????----- > ???: "Sven Reichard" > ????: 2016?12?21? ??? > ???: forum at gap-system.org > ??: > ??: Re: [GAP Forum] AlternatingGroup(4) and its generators > > Dear Jiayue, > > to be precise, G is isomorphic to the alternating group. You can find > such an isomorphism and determine the images of the generators as follows: > > gap> G := SmallGroup(12,3); > > gap> StructureDescription(G); > "A4" > gap> a4 := AlternatingGroup(4); > Alt( [ 1 .. 4 ] ) > gap> iso := IsomorphismGroups(G, a4); > [ f1, f2, f3 ] -> [ (2,4,3), (1,3)(2,4), (1,2)(3,4) ] > gap> List(GeneratorsOfGroup(G), g -> Image(iso, g)); > [ (2,4,3), (1,3)(2,4), (1,2)(3,4) ] > > HTH, > Sven > > On 21.12.2016 12:40, ??? wrote: > > > > Dear Forum members, > > > > > > When I type this in GAP: > > > > > > gap>G:=SmallGroup(12,3); > > gap>GeneratorsOfGroup(G); > > the result is [f1,f2,f3],but actually the group G here is AlternatingGroup(4) and I wonder > > how could I know here what f1,f2,f3 exactly means by permutations respectively? How > > could I know what are they in A4? > > > > > > Looking forward to any reply! > > > > > > Thanks a lot! > > > > > > Jiayue > > > > > > _______________________________________________ > > Forum mailing list > > Forum at mail.gap-system.org > > http://mail.gap-system.org/mailman/listinfo/forum > > > > _______________________________________________ > Forum mailing list > Forum at mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum From qijiayue14 at mails.ucas.ac.cn Sat Dec 24 07:15:38 2016 From: qijiayue14 at mails.ucas.ac.cn (=?GBK?B?xuu8ztTD?=) Date: Sat, 24 Dec 2016 15:15:38 +0800 (GMT+08:00) Subject: [GAP Forum] consulting on how to parallel my GAP program Message-ID: <1fef795.127c2.1592faf5ca9.Coremail.qijiayue14@mails.ucas.ac.cn> Dear Sir/Madam, First of all Merry Christmas! As my GAP program needs a lot of calculation and I'm thinking about paralleling it so that I could finish my calculation sooner, hopefully. Is there such service or some certain way that can help people realize this in the GAP system itself? If not, what do you think I should do? My program is a final function who needs two parameters for input and it invokes other functions written before already. The output will be a subset triple of the group satisfying a certain property and a blank if the group doesn't fit the property. Looking forward to your reply! Best regards, Jiayue From z060822400814a at rezozer.net Wed Dec 7 15:58:31 2016 From: z060822400814a at rezozer.net (Jerome BENOIT) Date: Wed, 07 Dec 2016 15:58:31 -0000 Subject: [GAP Forum] GAP package: braid Message-ID: <89e479a9-b2b6-5a04-6940-f27669546315@rezozer.net> Hello Forum, I have just noticed that Sage[Math] tries to pre-loads the GAP package `braid', which is not present in the GAP webpage. I guess it is an old GAP package. Can anybody confirm this ? Does anybody knows whether or not this GAP pachage has been replaced/supersedeed ? Thanks in advance, Jerome From surinder.kaur at iitrpr.ac.in Mon Nov 21 14:38:54 2016 From: surinder.kaur at iitrpr.ac.in (Surinder Kaur) Date: Mon, 21 Nov 2016 14:38:54 -0000 Subject: [GAP Forum] GAP query Message-ID: LAGUNA package gives information about group rings, FG, only when G is a p-group and F is a field of characteristic p. Can I find the unit groups of modular group rings where p divides the order of the group (means group is not necessarily p-group)? -- *Regards* *Surinder Kaur* *Research scholar * *Department of Mathematics * *IIT Ropar* | 134,415 | 442,702 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-30 | latest | en | 0.716402 |
https://www.numerade.com/questions/for-each-of-the-following-vectors-describe-the-opposite-vector-a-an-airplane-flies-due-north-at-400-/ | 1,627,251,761,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151866.98/warc/CC-MAIN-20210725205752-20210725235752-00055.warc.gz | 950,999,776 | 32,868 | ๐ Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning Numerade Educator ### Problem 8 Easy Difficulty # For each of the following vectors, describe the opposite vector.a. an airplane flies due north at$400 \mathrm{km} / \mathrm{h}$b. a car travels in a northeasterly direction at$70 \mathrm{km} / \mathrm{h}$c. a bicyclist pedals in a northwesterly direction at$30 \mathrm{km} / \mathrm{h}$d. a boat travels due west at$25 \mathrm{km} / \mathrm{h}$### Answer ## a.$400 \mathrm{~km} / \mathrm{h},$due southb.$70 \mathrm{~km} / \mathrm{h},$southwesterlyc.$30 \mathrm{~km} / \mathrm{h}$southeasterlyd.$25 \mathrm{~km} / \mathrm{h},\$ due east
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### Video Transcript
in this problem, we have been asked to describe the vector which is the opposite of the victor, given. Now in the first problem, we have an airplane flying due north at 400 km/h. Now the opposite victor will have the same magnitude but the opposite direction, so the magnitude will be 400 km/h. And since the question has a vector which flies due north, the opposite vector will be in the direction opposite to north, so that will be south, so the opposite vector B 400 kilometers per hour. View south. And the second problem, a car travels in a northeasterly direction at 70 km/h, so the opposite victor will have the same magnitude, which is 70 km/h, and the direction will be opposite to the North Easterly direction, which will be the southwesterly direction, So 70 km/h in the south westerly direction. Now in the third problem, we have a vice cyclist which pedals in a northwesterly direction at 30 km/h. The opposite victor will have the same magnitude of 30 km/h, and the direction will be opposite to the direction of northwesterly, which will be southeasterly. And in the last problem, we have a boat which travels due west at 25 km/h. The opposite victor will have the same magnitude of 25 km/h and the direction of the opposite to the direction of West, so that will be east, so the opposite vector will be 25 km/h. View east.
University of North Bengal
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No Related Subtopics | 577 | 2,205 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-31 | latest | en | 0.883474 |
https://www.airmilescalculator.com/distance/brc-to-vdm/ | 1,632,869,323,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060908.47/warc/CC-MAIN-20210928214438-20210929004438-00269.warc.gz | 644,628,754 | 28,041 | # Distance between San Carlos de Bariloche (BRC) and Viedma (VDM)
Flight distance from San Carlos de Bariloche to Viedma (San Carlos De Bariloche Airport โ Gobernador Edgardo Castello Airport) is 427 miles / 687 kilometers / 371 nautical miles. Estimated flight time is 1 hour 18 minutes.
Driving distance from San Carlos de Bariloche (BRC) to Viedma (VDM) is 623 miles / 1002 kilometers and travel time by car is about 12 hours 7 minutes.
## Map of flight path and driving directions from San Carlos de Bariloche to Viedma.
Shortest flight path between San Carlos De Bariloche Airport (BRC) and Gobernador Edgardo Castello Airport (VDM).
## How far is Viedma from San Carlos de Bariloche?
There are several ways to calculate distances between San Carlos de Bariloche and Viedma. Here are two common methods:
Vincenty's formula (applied above)
โข 426.668 miles
โข 686.656 kilometers
โข 370.765 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earthโs surface, using an ellipsoidal model of the earth.
Haversine formula
โข 425.580 miles
โข 684.905 kilometers
โข 369.819 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance โ the shortest distance between two points).
## Airport information
A San Carlos De Bariloche Airport
City: San Carlos de Bariloche
Country: Argentina
IATA Code: BRC
ICAO Code: SAZS
Coordinates: 41ยฐ9โฒ4โณS, 71ยฐ9โฒ27โณW
City: Viedma
Country: Argentina
IATA Code: VDM
ICAO Code: SAVV
Coordinates: 40ยฐ52โฒ9โณS, 63ยฐ0โฒ1โณW
## Time difference and current local times
There is no time difference between San Carlos de Bariloche and Viedma.
-03
-03
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 88 kg (194 pounds).
## Frequent Flyer Miles Calculator
San Carlos de Bariloche (BRC) โ Viedma (VDM).
Distance:
427
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
427
Round trip? | 547 | 2,001 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-39 | latest | en | 0.71721 |
https://www.physicsforums.com/threads/solving-point-distributions-for-negotiation-simulation-w-thresholds.1039163/ | 1,721,560,033,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517663.24/warc/CC-MAIN-20240721091006-20240721121006-00428.warc.gz | 794,860,882 | 17,830 | # Solving Point Distributions for Negotiation Simulation w/Thresholds
โข MHB
โข Winums2
In summary, to create a negotiation simulation with an agreement rate of only 20%, it is important to define the parameters and goals, create a matrix with point values and thresholds, and manipulate the values based on the parties' values and priorities. This may require some trial and error, but with careful consideration, it is possible to achieve the desired outcome.
Winums2
I'm trying to create a negotiation simulation and don't know how to find point allocations for each party that will allow agreement in only about 20% of the cases. Wondering whether anyone here would solve the problem for me.
The point allocations I came up with allow agreement only a few times.
The table linked to below shows 6 questions (job title, club membership, parking, salary, vacation, number of employees).
For each question there are several possible answers (ex. Job title: vp, mgr, asst mgr, exec) (ex. Club membership: no, lunch only, All services).
For each possible answer, each party (ER and EE) assigns different point values (ex. for the Job Title question answer vp: ER assigns 0 pints, EE assigns 200 points)
To reach agreement, each party must accumulate more points than their threshold. To allow agreement I set ERโs threshold at 1300 and EEโs threshold at 1100.
The thresholds and point values Iโve used allow few agreements. One is shown in columns 5 and 6. Another is shown in columns 7 and 8.
The task is to find new thresholds for ER and EE and new sets of numbers for columns 2 and 3 that allows more agreements between ER and EE. Ideally:
a. About 20% of the possible combinations are to allow agreement by having both parties beat their threshold. The other 80% of possible combinations will prevent agreement by having only one or no party beat their threshold.
b. At least one issue is of little value to ER but enough value to the EE to allow the points for EE to offset lost points on another question.
c. At least one issue is of little value to EE but enough value to the ER to allow the points for ER to offset lost points on another question.
If you need to add another question or change the number of answers in particular questions, that would be fine.https://docs.google.com/document/d/1cVtzvuSFEmzWkt47IxjFXs946L6cy7rd2jFK5gDXI-Q/edit?usp=sharing
Hello,
my advice would be to approach this problem using a mathematical and logical approach rather than relying on someone else to solve it for you. First, you need to clearly define the parameters and goals of your negotiation simulation. This includes identifying the parties involved, the issues at hand, and the desired outcome (in this case, reaching agreement in only about 20% of the cases).
Once you have a clear understanding of the simulation, you can start by creating a matrix or table similar to the one provided in the forum post. The number of questions and possible answers can be adjusted as needed to meet your desired criteria. Next, assign point values for each answer and threshold values for each party. It is important to ensure that the threshold values are achievable, meaning that it is possible for both parties to reach their respective thresholds in at least 20% of the possible combinations.
To increase the chances of agreement in only 20% of the cases, you can manipulate the point values and thresholds based on the value and importance of each issue to each party. For example, if an issue is of high value to one party but of low value to the other, you can assign higher point values for that issue to the party that values it more. This will create a situation where one party may be more likely to reach their threshold, while the other party may not, resulting in agreement in only 20% of the cases.
Additionally, you can also introduce the concept of trade-offs or compromises, where one party may be willing to sacrifice points on one issue in exchange for gaining points on another issue. This can be achieved by assigning point values that are close in value for certain answers, making it more tempting for a party to compromise.
In conclusion, finding the perfect set of point allocations and thresholds may require some trial and error, but by using a logical and mathematical approach and manipulating the values based on the parties' values and priorities, you should be able to create a negotiation simulation that meets your desired criteria. Good luck!
## 1. What is a point distribution in a negotiation simulation?
A point distribution in a negotiation simulation refers to the allocation of points to different issues or items that are being negotiated. These points represent the value or importance placed on each issue and can be used as a bargaining tool to reach a mutually beneficial agreement.
## 2. How do thresholds factor into point distributions for negotiation simulations?
Thresholds are predetermined limits or minimum requirements that must be met in order for a party to agree to a certain point distribution. These thresholds may be used to protect a party's bottom line or to signal when a negotiation may be at a standstill.
## 3. How are point distributions determined in a negotiation simulation?
Point distributions are typically determined through a combination of research, analysis, and negotiation strategy. This may involve gathering information about the issues being negotiated, assessing the value of each issue, and considering the other party's interests and preferences.
## 4. What are some common challenges when solving point distributions for negotiation simulations?
Some common challenges when solving point distributions for negotiation simulations include accurately assessing the value or importance of each issue, understanding the other party's perspective and priorities, and finding a balance between reaching a mutually beneficial agreement and protecting one's own interests.
## 5. Can point distributions be changed during a negotiation simulation?
Yes, point distributions can be changed during a negotiation simulation if both parties agree to it. This may occur if new information is presented or if one party reevaluates their priorities. However, any changes should be approached carefully and with consideration for the overall negotiation strategy.
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3K | 1,432 | 6,848 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-30 | latest | en | 0.919166 |
https://www.physicsforums.com/threads/equation-of-a-parabola.353714/ | 1,527,386,127,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867977.85/warc/CC-MAIN-20180527004958-20180527024958-00395.warc.gz | 815,318,901 | 15,582 | # Homework Help: Equation of a parabola
1. Nov 11, 2009
### Jshua Monkoe
1. The problem statement, all variables and given/known data
i am given the equation of a parabola to be y=2x^2-2x+3 and asked to sketch the parabola
2. Relevant equations
y=1/2(l-p)(x-k)^2+(l+p)/2
(l+p)/2=h
vertex is at (k,h)
equation of the directrix is y=p
distance[(k,h) to y=p]=distance[(x,y) to (k,h)]
3. The attempt at a solution
(1) completing the squares for y=2x^2-2x+3 i get
y=2(x-1/2)^2+2
=>k=1/2 &
h=2 &
a=2
.'. vertex is at (1,2)
(2) (l+p)/2=h=2 ........eqn1
1/(2(l-p))=a=2....eqn2
making l the subject in eqn.1 i get
l=4-p
substituting in eqn2 i get
1/(2(4-p-p))=2
=>1/(8-4p)=2
=>1=16-8p
.'.p=-15/8~-1.08
=>eqn of directrix is y=-1
(3)given the vertex is at (1,2)
F is at (1,3) taking (1,-1) from the directrix
from (2,4) to F is sq. root of 2 units
from (2,4) to (1,-1) is sq. root of 26 units
AND THIS ISNT A PARABOLA
PLEASE, WHERE HAVE I WENT WRONG BUDDIES?!?
2. Nov 11, 2009
### tiny-tim
Welcome to PF!
Hi Jshua Monkoe! Welcome to PF!
(try using the X2 tag just above the Reply box )
Nooooo
3. Nov 11, 2009
### Staff: Mentor
You have y = 2(x2 - x + ?) + 3. Remember that whatever you add in to complete the square is multiplied by 2, so you'll need to take that into account and subtract the same amount from three to keep your expressions equal.
4. Nov 11, 2009
### Anakin_k
y=2x2-2x+3
When you complete the square, you must factor out the leading co-efficient, which in this case is 2 from the terms with the variable. The 3 however will stay outside of the brackets.
y=2(x2-x)+3
y=2(x2-x+0.25-0.25)+3
Bring out the negative 0.25 after multiplying it by the leading co-efficient that was factored out but leave the positive 0.25 inside the brackets. Now simply.
You should be able to take it from there. | 662 | 1,822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-22 | latest | en | 0.880641 |
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# Events & Promotions
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# Eligibility
new topic post reply Update application status
Author Message
Intern
Joined: 24 Feb 2011
Posts: 5
Kudos [?]: [0], given: 3
### Show Tags
28 Mar 2011, 05:23
hi every one,
being a aspirant of GMAT,want to know the eligibility and i want to get in to the top 10 b-school in the world and my score card as follows,
10th 83.84
puc 74
engineering avg(till 5th sem) 66
thank you
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Retired Moderator
Status: Darden Class of 2013
Joined: 28 Jul 2009
Posts: 1834
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Schools: University of Virginia
### Show Tags
28 Mar 2011, 11:14
Hi there, it's impossible to tell you your potential with such little information. Please provide a bit more about your extracurricular activity, your GMAT range (or even better, take the GMAT first so you have a solid score), your job, your post-MBA goals. Feel free to read up on what other people wrote to see the level of detail required.
_________________
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Intern
Joined: 24 Feb 2011
Posts: 5
Kudos [?]: [0], given: 3
### Show Tags
29 Mar 2011, 04:45
Kudos [?]: [0], given: 3
Manager
Status: Planning to retake.
Affiliations: Alpha Psi Omega
Joined: 25 Oct 2010
Posts: 89
Kudos [?]: 31 [1], given: 14
Concentration: General Management, Entrepreneurship
GMAT 1: 650 Q42 V37
GRE 1: 1310 Q630 V680
GPA: 3.16
### Show Tags
30 Mar 2011, 03:03
1
KUDOS
I mean this gently.
Based on your post, I am not sure if your current English level is good enough to get into an elite b-school.
_________________
Each moment of time ought to be put to proper use, either in business, in improving the mind, in the innocent and necessary relaxations and entertainments of life, or in the care of the moral and religious part of our nature.
-William Andrus Alcott
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Manager
Status: Preparing for GMAT !!!
Joined: 09 Apr 2010
Posts: 124
Kudos [?]: 20 [1], given: 41
### Show Tags
30 Mar 2011, 03:44
1
KUDOS
Just to add to the above posts, your high school scores don't matter much at this level. It is more important that you have your UG GPA & GMAT scores, compare them against the 80% median of your target schools. Then ask the above question by posting your work experience, extra curricular activities, leadership etc. and most importantly your target job/company and other details.
I would suggest you that you take your GMAT first.
_________________
If you like my post, consider giving me Kudos !!
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Intern
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Posts: 5
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### Show Tags
30 Mar 2011, 04:17
thanks all.
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Intern
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Posts: 5
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### Show Tags
30 Mar 2011, 04:31
is my english good?
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Manager
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### Show Tags
30 Mar 2011, 16:59
2
KUDOS
amarnathbh wrote:
is my english good?
It's not easy to say based on a couple of posts. Added to that, in an online forum, the purpose is to get the message across. Instead of worrying, just be confident and focus on gmat preparation and let it determine your english level.
_________________
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Re: Eligibility ย [#permalink] 30 Mar 2011, 16:59
Display posts from previous: Sort by
# Eligibility
new topic post reply Update application status
Powered by phpBB ยฉ phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMATยฎ test is a registered trademark of the Graduate Management Admission Councilยฎ, and this site has neither been reviewed nor endorsed by GMACยฎ. | 1,242 | 4,280 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-51 | latest | en | 0.881114 |
https://discuss.tlapl.us/msg03075.html | 1,624,075,920,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487643380.40/warc/CC-MAIN-20210619020602-20210619050602-00485.warc.gz | 206,300,004 | 5,396 | [tlaplus] Re: Help with a TLAPS proof for a refinement involving records (and a Proof Decomposition bug)
Dearย Leslie,
Now I understand "SV!<".
So, the real bug here is that the Toolbox (Versionย 1.6.0 of 10 July 2019) reports a parse error on maxBal[p] SV!< b; see the attached .png file (I failed to insert it in the text here.)
See also the attached TLA+ Specs files (SimpleVoting.tla, Record.tla) for the line numbers referred to in the error message.
Best regards,
hengxin
On Tuesday, August 20, 2019 at 12:59:21 AM UTC+8, Leslie Lamport wrote:
Stephan wrote:
please feel free to post a bug report about the malfunctioning proof decomposition.
I do not see any malfunction of the proof decomposition command.ย The command is expanding the
definition ofย ย SV!IncreaseMaxBal(p, b).ย The definition of IncreaseMaxBal in module SimpleVoting
uses the operator < (imported from the Naturals module), which is imported by the INSTANCE statement
as the operator SV!< .ย Since the module containing the theorem also imports the Naturals module,
the operator SV!< happens to equal the operator < of the current module.ย Please replace < by SV!<
in the proof and check that it still works.ย If it doesn't, please report this as a TLAPS bug.
Because the definition of < is imported into the instantiated module from a standard module, and
that same definition is imported into the instantiating module from the same standard module, a
user would almost always prefer that the Decompose Proof command replace SV!< by < .ย This should
be straightforward to implement and we would be grateful if someone would volunteer to do it.
Leslie
On Wednesday, August 14, 2019 at 9:26:22 PM UTC-7, Hengfeng Wei wrote:
Dear All,
I have some difficutlty in proving a refinement mapping involving records.
SimpleVoting.tla:
It maintains for each participant a maxBal which is a natural number.
In IncreaseMaxBal(p, b), maxBal[p] is increased to a larger value b.
---------------------------- MODULE SimpleVoting ----------------------------
EXTENDS
Naturals
-----------------------------------------------------------------------------
CONSTANT
Participant
VARIABLE maxBal
TypeOK == maxBal \in [Participant -> Nat]
-----------------------------------------------------------------------------
Init == maxBal = [p \in Participant |-> 0]
IncreaseMaxBal(p, b) ==
/\ maxBal[p] < b
/
\ maxBal' = [maxBal EXCEPT ![p] = b]
-----------------------------------------------------------------------------
Next == \E p \in Participant, b \in Nat : IncreaseMaxBal(p, b)
Spec == Init /\ [][Next]_maxBal
=============================================================================
Record.tla:
It maintains a 2D array state. state[p][q] is the State of q from the view of p and a State is a record:
State == [maxBal : Nat, maxVBal : Nat].
In Prepare(p, b): state[p][p].maxBal is increased to a larger value b.
------------------------------- MODULE Record -------------------------------
EXTENDS
Naturals, TLAPS
---------------------------------------------------------------------------
CONSTANTS
Participant ย \* the set of partipants
VARIABLES state
\* state[p][q]: the state of q \in Participant from the view of p \in Participant
State == [maxBal: Nat, maxVBal: Nat]
TypeOK == state \in [Participant -> [Participant -> State]]
---------------------------------------------------------------------------
InitState == [maxBal |-> 0, maxVBal |-> 0]
Init == state = [p \in Participant |-> [q \in Participant |-> InitState]]
Prepare(p, b) ==
/\ state[p][p].maxBal < b
/
\ state' = [state EXCEPT ![p][p].maxBal = b]
---------------------------------------------------------------------------
Next == \E p \in Participant, b \in Nat : Prepare(p, b)
Spec == Init /\ [][Next]_state
---------------------------------------------------------------------------
I want to show that Record refines SimpleVoting under the refinement mapping:
maxBal == [p \in Participant |-> state[p][p].maxBal]
SV
== INSTANCE SimpleVoting
However, step <4>2 fails (more details below). (In addition, the proof decomposition (Ctrl+G, Ctrl+D) of <3> generates a buggy formula "maxBal[p]ย SV!<ย b" in <4>1, which should be "maxBal[p]ย <ย b".)
THEOREM Spec => SV!Spec
<1>1. Init => SV!Init
BY DEF
Init, SV!Init, maxBal, InitState
<1>2. [Next]_state => [SV!Next]_maxBal
<2>1. UNCHANGED state => UNCHANGED maxBal
BY DEF maxBal
<2>2. Next => SV!Next
<3> ASSUME NEW p \in Participant, NEW b \in Nat,
Prepare(p, b)
PROVE ย SV
!IncreaseMaxBal(p, b) \* SV!Next
<4>1. maxBal[p] < b \* Wrong decomposition here: maxBal[p] SV!< b
BY DEF
Prepare, maxBal
<4>2. maxBal' = [maxBal EXCEPT ![p] = b]
BY DEF Prepare, maxBal \* Failed here!!!
<4>3. QED
BY <4>1, <4>2 DEF SV!IncreaseMaxBal
<3>1. QED
BY DEF Next, SV!Next
<2>3. QED
BY <2>1, <2>2
<1>3. QED
BY <1>1, <1>2, PTL DEF SV!Spec, Spec
The obligation at <4>2 is as follows. Isn't the state' = [state EXCEPT ![p] = ...] in the assumption the same as the conclusion [p_1 \in Participant |-> state[p_1][p_1].maxBal]' ...? What is wrong with my proof? How should I proceed?
ASSUME NEW CONSTANT Participant,
NEW VARIABLE state
,
NEW CONSTANT p
\in Participant,
NEW CONSTANT b
\in Nat,
/\ state[p][p].maxBal < b
/
\ state'
= [state EXCEPT
![p] = [state[p] EXCEPT
![p] = [state[p][p] EXCEPT !.maxBal = b]]]
PROVE ย [p_1 \in Participant |-> state[p_1][p_1].maxBal]'
= [[p_1 \in Participant |-> state[p_1][p_1].maxBal] EXCEPT ![p] = b]
Best regards,
hengxin
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https://www.nagwa.com/en/worksheets/856191502592/ | 1,680,015,049,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948867.32/warc/CC-MAIN-20230328135732-20230328165732-00159.warc.gz | 1,013,880,770 | 10,982 | # Lesson Worksheet: Properties of Operations on Vectors Mathematics
In this worksheet, we will practice using the properties of addition and multiplication on vectors.
Q1:
What is the addition property that shows that ?
โข ACommutative property
โข BAssociative property
โข DDistributive property
Q2:
Which of the following properties of vector operations is incorrect?
โข A
โข B
โข C
โข D
โข E
Q3:
Let and .
What are the components of ?
โข A
โข B
โข C
โข D
โข E
What are the components of ?
โข A
โข B
โข C
โข D
โข E
Q4:
Does the vector sum have a solution?
โข ANo
โข BYes
Q5:
Complete the following: .
โข A
โข B
โข C
โข D
โข E
Q6:
Given that and , find .
โข A
โข B
โข C
โข D
โข E
Q7:
Complete the following: .
โข A
โข B
โข C
โข D
โข E
Q8:
Given that , find .
โข A
โข B
โข C
โข D
โข E
Q9:
What is the property that shows that ?
โข BCommutative property
โข EAssociative property
Q10:
Consider that , , and .
Find .
โข A
โข B
โข C
โข D
โข E
Find .
โข A
โข B
โข C
โข D
โข E
Does equal ?
โข AYes
โข BNo | 327 | 982 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-14 | longest | en | 0.702496 |
frankverspoor.com | 1,563,937,631,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195530250.98/warc/CC-MAIN-20190724020454-20190724042454-00001.warc.gz | 62,435,573 | 13,098 | Source of inspiration:
Pythagorean theorem wiki / Stelling van Pythagoras (Dutch) wiki
In mathematics, the Pythagorean theorem, also known as Pythagorasโs theorem, is a relation in Euclidean geometry among the three sides of a right triangle. It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. The theorem can be written as an equation relating the lengths of the sides a, b and c, often called the โPythagorean equationโ: a2 + b2 = c2 , where c represents the length of the hypotenuse and a and b the lengths of the triangleโs other two sides.
47th problem of Euclid
โThe proposition is especially important in architecture. Builders have since ancient times used the theorem in constructing buildings by a process known as โsquaring a room.โ As the theorem states that 3 squared + 4 squared = 5 squared, a builder starts by marking a spot and drawing a line, say line A. This line is given the value of 3. The builder then marks another point, say point B and draws a line from it at a right angle to line A, and it is given the value of 4. The distance between line A and B is then measured, and if the distance between A and B is 5, then the room is squared. By inverting the process, a โsquaredโ (or rectangle) room can be obtained.
Engineers who tunnel from both sides through a mountain use the 47th problem to get the two shafts to meet in the center. The surveyor who wants to know how high a mountain may be ascertains the answer through the 47th problem. The astronomer who calculates the distance of the sun, the moon, the planets, and who fixes โthe duration of times and seasons, years, and cycles,โ depends upon the 47th problem for his results. The navigator traveling the trackless seas uses the 47th problem in determining his latitude, his longitude, and his true time. Eclipses are predicted, tides are specified as to height and time of occurrence, land is surveyed, roads run, shafts dug, bridges built, with the 47th problem to show the way.โ
wiki | 484 | 2,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2019-30 | latest | en | 0.956959 |
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