Pensieve header: Non-Commutative Gaussian Elimination - Day 2.

Today. Some older challenges and a new one, then commutative Gaussian elimination, then non-commutative Gaussian elimination (NCGE), then EIWL-10-12, then even less likely, Patterns.

Topics (in no particular order). Whatever you may suggest; whatever comes to my mind; the Fibonacci numbers; the Catalan numbers; the Jones polynomial; a more efficient Jones algorithm; a riddle on spheres; Khovanov homology; Γ-calculus; the Hopf fibration; Hilbert’s 13th problem; non-commutative Gaussian elimination; free Lie algebras; the Baker-Campbell-Hausdorff formula; wacky numbers; an order 4 torus; the Schwarz Lantern; knot colourings; the Temperley-Lieb pairing; the dodecahedral link; sound experiments; barycentric subdivisions; some Peano curves; braid closures and Vogel’s algorithm; the insolubility of the quintic; phase portraits; the Mandelbrot set; shadows of the Cantor aerogel; quilt plots; some image transformations; De Bruijn graphs; the Riemann series theorem; finite type invariants and the Willerton fish; the Towers of Hanoi; Hochschild homology of (some) coalgebras; convolutions and image improvements; the 8-5-3 milk jug problem; a cow problem.

SetDirectory["C:\\drorbn\\AcademicPensieve\\Classes\\17-1750-ShamelessMathematica\\"];

An Image Manipulation Challenge

The image at http://drorbn.net/bbs/show?shot=17-1750-171016-111042.jpg is pathetic. Can you improve it? Whatever you do, should also work well with all other images at http://drorbn.net/bbs/show.php?prefix=17-1750.

Solution by Charlene Chu:

NotebookOpen[Directory[]<>"\\StudentProjects\\chu_charlene_171105_animagemanipulationchallenge.nb"];

A Graphics Challenge

The torus (S^1)×(S^1) has an order 4 symmetry. Can you draw it in such a manner that it will be manifest?

Solution by Tristan Milne:

NotebookOpen[Directory[]<>"\\StudentProjects\\Milne_Tristan_171106_TorusEmbeddings2.nb"];

The Cow Problem (Leonard)

A farmer has 19 cows, and she wishes to give them to her daughters to that the first will get 1/2, the second 1/4, and the third 1/5. This is obviously impossible. A wise woman hears of the problem and suggests: “I’ll add on one of my cows, so you’ll have 20. The first daughter will get 10, the second 5, the third 4, I’ll take back the remaining cow, and everyone is happy!”.

Problem. Are there any other quadruples like (19,2,4,5), for which the same trick will work? What are all of them?

Hint. It is sometimes better to analyze and generalize, first.

Solution by Charlene Chu:

NotebookOpen[Directory[]<>"\\StudentProjects\\chu_charlene_171105_thecowproblem.nb"];

Solution by Leonard Afeke (with help from DBN):

NotebookOpen[Directory[]<>"\\StudentProjects\\Afeke_Leonard_171102_CowProblem.nb"];

The 8-5-3 Milk Jug Problem

Challenge. Draw the state graph of this problem.

Gaussian Elimination - The Commutative Case

vs=Table[(((i+j))^5),{i,12},{j,8}];
vs // Column

MatrixRank[vs]

?Position

?FirstPosition

On to Rubik’s Cube

?Cycles

n=54; 
(Subscript[g, 1])=Cycles[{{1,18,45,28},{2,27,44,19},{3,36,43,10},{46,52,54,48},{47,49,53,51}}];
(Subscript[g, 2])=Cycles[{{7,16,39,30},{8,25,38,21},{9,34,37,12},{13,15,33,31},{14,24,32,22}}];
(Subscript[g, 3])=Cycles[{{28,31,34,48},{29,32,35,47},{30,33,36,46},{37,39,45,43},{38,42,44,40}}];
(Subscript[g, 4])=Cycles[{{1,3,9,7},{2,6,8,4},{10,54,16,13},{11,53,17,14},{12,52,18,15}}];
(Subscript[g, 5])=Cycles[{{1,13,37,46},{4,22,40,49},{7,31,43,52},{10,12,30,28},{11,21,29,19}}];
(Subscript[g, 6])=Cycles[{{3,48,39,15},{6,51,42,24},{9,54,45,33},{16,18,36,34},{17,27,35,25}}];

?PermutationProduct

a_∘b_ := PermutationProduct[a,b]

?InversePermutation

?PermutationSupport

?PermutationReplace