On 4/14/14, Adam P. Goucher <apgoucher@gmx.com> wrote:
Fred Lunnon wrote:
Yeah, that's much more direct than my argument! WFL
Thanks. What was your intended proof?
I computed the symmetry group and noticed it is the hyperoctahedral group in n = 4 dimensions. Then used the theorem that the n-cube skeleton has genus 1 + (n-4)2^(n-3). A touch of BFI (brute force and ignorance)! On 4/14/14, Bill Gosper <billgosper@gmail.com> wrote:
If you draw the 16 ways to slide, vs the 8 ways to hop, you get something like http://www.log24.com/log/pix10A/100619-FourDiamondTesseract.gif --rwg
Er, I cannot make sense of this diagram at all. WFL
Bill Gosper wrote:
Years ago Scott Kim told me that when shown the rules of chess as a child, he exclaimed "Ooh, a tesseract!" when they came to the knight. --rwg
Adam>Why, exactly? As far as I can see, the tesseract only appears when one imposes that the knight inhabits a 4-by-4 toroidal grid.
He was only a kid.
[Massive crop]
I blame global warming. [What were you growing --- was it legal?] WFL
On 4/13/14, Adam P. Goucher <apgoucher@gmx.com> wrote:
Very nice discovery!
Lemma: The graph is isomorphic to the 4-by-4 toroidal grid graph.
Proof: draw the graph in the way you defined (which is on the 4-by-4 torus, but has lots of edges crossing). Then, translate all of the black vertices by (2,2) whilst preserving the positions of the white vertices. Then every vertex is connected to its four neighbours and nothing else.
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Corollary: (2)
Proof: Trivial.
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Corollary: (1)
Proof: The graph is isomorphic to the product C4 x C4 (where C4 is the cycle graph on four vertices), so is isomorphic to the product K2 x K2 x K2 x K2 (since C4 is isomorphic to K2 x K2). Consequently, the graph is isomorphic to the skeleton of the tesseract (four-dimensional hypercube).
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The last isomorphism I used (C4 x C4 == K2 x K2 x K2 x K2) may be familiar to any electronic engineers amongst you, on the basis that it allows Karnaugh maps to work.
Sincerely,
Adam P. Goucher
----- Original Message ----- From: Fred Lunnon Sent: 04/13/14 12:04 PM To: math-fun Subject: [math-fun] A knight's surprise?
Consider the graph whose vertices & edges represent squares & knight moves on a 4x4 chessboard tiling the torus.
(1) Find a spatial embedding with the full symmetry of the graph.
(2) Deduce that the graph can be embedded (without crossings) on the torus.
Fred Lunnon