Re: [math-fun] A knight's surprise?
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 On 2014-04-13 09:09, Fred Lunnon wrote:
Yeah, that's much more direct than my argument! 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
Dear all, There's a lovely chess problem by Noam Elkies, which, if I recall correctly, has 16 variations, each being 4 moves with the same knight. I'll copy this to Noam, to give him a chance to confirm or deny, and perhaps remind us of the problem. R. On Sun, 13 Apr 2014, 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
On 2014-04-13 09:09, Fred Lunnon wrote:
Yeah, that's much more direct than my argument! 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.
--------
Corollary: (2)
Proof: Trivial.
--------
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).
--------
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
math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
rkg <rkg@cpsc.ucalgary.ca> wrote:
There's a lovely chess problem by Noam Elkies, which, if I recall correctly, has 16 variations, each being 4 moves with the same knight. I'll copy this to Noam, to give him a chance to confirm or deny, and perhaps remind us of the problem. R.
Thanks. The problem indeed involves a Knight tesseract, though the number of *variations* is 4! = 24, not 2^4 = 16. Sorry I didn't answer sooner; the problem is about 10 years old and builds on earlier work, and I was worried about sorting out the attributions correctly. But it turns out that I have in my a write-up from the same month when the problem was composed; I append it below. Note that "S" = knight (from German "Springer"; chess problemists usually denote Knight by S, reserving N for Nightrider <http://en.wikipedia.org/wiki/Nightrider_(chess)>, whose move extends the Knight's as the Queen's move extends the King's). NDE 11/2004, after Jelliss/Beasley +---a---b---c---d---e---f---g---h---+ | | 8 . . . . . . . . 8 | | 7 . . . P . . . . 7 | | 6 -P K . . . . . . 6 | | 5 . P . . . . . -P 5 | | 4 . . . . . . P . 4 | | 3 . -P P -P -P . . . 3 | | 2 . P -P -K -P . . . 2 | | 1 . -B R . R -S . . 1 | | +---a---b---c---d---e---f---g---h---+ #4 8 + 10 White Kb6 Rc1 Re1 Pb2 Pd7 Pb5 Pg4 Pc3 Black Kd2 Pb3 Pc2 Pd3 Pe2 Pe3 Bb1 Pa6 Ph5 Sf1 1 d8S! Fourfold Fleck (see below --NDE). Black's four thematic defenses Ba2,axb5,hxg4,Sf3 separate White's threats, each of which is a threefold Fleck; in each case, the remaining three Black defenses separate White's threats, each of which is a double threat with dual avoidance by Black's remaining two defenses: 1...Ba2 2 Sf7! 2...axb5 3 Sg5 hxg4(Sh2)/Sg3 4 Se4/Sf3# 2...hxg4 3 Sd6 axb5/Sg3 4 Se4/Sc4# 2...Sg3 3 Se5 axb5/hxg4 4 Sf3/Sc4# 1...axb5 2 Se6! 2...Ba2 3 Sg5 hxg4(Sh2)/Sg3 4 Se4/f3# 2...hxg4 3 Sc5 Ba2/Sg3 4 Se4/Sxb3# 2...Sg3 3 Sd4 Ba2/hxg4 4 Sf3/Sxb3# 1...Sg3 2 Sc6! 2...Ba2 3 Se5 axb5/hxg4 4 Sf3/Sc4# 2...axb5 3 Sd4 Ba2/hxg4 4 Sf3/Sxb3# 2...hxg4 3 Sa5 Ba2/axb5 4 Sc4/Sxb3# 1...hxg4 2 Sb7! 2...Ba2 3 Sd6 axb5/Sg3 4 Se4/Sc4# 2...axb5 3 Sc5 Ba2/Sg3 4 Se4/Sxb3# 2...Sg3 3 Sa5 Ba2/axb5 4 Sc4/Sxb3# Whaat makes this all work is the observation, apparently first made by George Jelliss [or Tylor?], that on an empty board there are 24 four-move Knight paths from d8 to d2, forming a projection of the 4-dimensional hypercube onto the chessboard: each of the four downwards directions must be used once. Each of the Black defenses covers one of the squares b3,c4,e4,f3 adjacent to d2 on this hypercube, and is answered by the Knight moving to the complementary 3-dimensional cube, where Black's remaining three defenses are answered in the same way. Thus in each thematic line ...Ba2 is always answered by a move in the ESE direction, ...axb5 by a SSE move, ...Sg3 by SSW, and ...hxg4 by WSW. This scheme was first shown by John Beasley in 2004 (_Variant Chess_) as a fairy problem with only two men: White Sc1, Black Ke8, mate in 4. The fairy rule is that bK may not move, but at each turn Black chooses a square (other than e8 and the square occupied by the White Knight) and "mines" it, permanently removing it from the chessboard. The solution is 1 Se2! Mb7/Mc6/Me6/Mf7 2 Sf3/e4/c4/b3! etc. In the orthodox #4 setting above, the non-thematic defense ...Sh2 of f3 is weaker than hxg4 because from h2 the Knight can no longer defend e4 in one move. After 1...Sh2 White plays 2 Sb7 or Se6, but not 2 Sf7? Sf3! 3 Sd6 Sxe1! (4 Sc4(e4)+ Kxc1). Fortunately after either 2 Sb7(e6) Sf3 3 Sa5/c5 Sxe1 4 Sxb3 is still checkmate. The problem is verified sound by Popeye version 3.41, in a matter of seconds -- during which it prints lines totaling about 550 half-moves! It was thus still something of a challenge to verify that each of the 24 thematic variations is dual-free. === "Fleck" is a problem theme where White's move makes n>=3 or more threats, for each of which there is at least one Black move that singles out that threat as the unique continuation by defeating the other other n-1 threats (so in particular White still has a unique mating continuation after each *thematic* Black move, and each of the Fleck threats is thematic). In problems of mate in 3 or more moves it's quite rare to have n>3. Sincerely, --Noam D. Elkies
participants (3)
-
Bill Gosper -
elkies@math.harvard.edu -
rkg