Comments on: Trembling hands

By: Adam Stephanides

Adam Stephanides — Sat, 11 Sep 2004 18:29:19 +0000

“Oddly, there has recently been a comoputer proof of a forced win in a position that human players always believed to be a draw in the absence of egregious player error. It takes about 75 moves without either the advance of a pawn or the capture of a piece, and, worse, nobody yet understands the logic of the play, though it can be memorized. If the position arises in a tournament, or if the player aware of the win is about to embark on move 51, what should haoppen if the other player calles the 50-move rule.” Iirc, when such positions were first discovered, FIDE adjusted the rules of chess to take account of them, so that in those endgames it took 75 moves (or however many it was) without a capture or pawn move before a draw could be called. Later FIDE reversed itself, and currently the 50-move rule has no exceptions. I don’t know why FIDE changed its mind, but my guess would be that the inconvenience of having players fumble around endlessly (iirc, a position is now known in which it takes over 200 moves w.o. a pawn move or capture to force a win) was felt to outweigh the theoretical possibility of a player being able to execute one of the wins requiring more than 50 moves.

By: Pedro

Pedro — Fri, 10 Sep 2004 01:35:56 +0000

I pass on a paraphrased excerpt of Christian Ewerhart’s 2002 paper “Backward Induction and the Game-Theoretic Analysis of Chess”, appearing on Games and Economic Behavior 39, 206-212. The proof of the result is rather elementary. No need for advanced set theory of any sort. By a “potentially infinite chess-like game”, we mean a strictly competitive perfect information game with at most three outcomes in which each node has a finite height and a finite number of immediate successors, and in which any infinite path yields a draw. When infinite chess games are taken to be draws—which is a very reasonable assumption, is it not?—the official version of chess is a “potentially infinite chess-like game”. Mathematician L. Kalmar seems to have proved a theorem in 1929, the proof of which has as a corollary the following: Corollary: Consider any potentially infinite chess-like game G. Then, if player i cannot enforce a win, then player j can ensure at least a draw. (Say a node x is a nonwinning position for player i if there does not exist a winning strategy for i in the subgame starting at x) Proof [reworking of Kalmar’s result]: Note first that, when j is called upon to play at a position x, that is nonwinning for i, then there must exist an immediate successor node of x that is nonwinning for i. To see why, assume to the contrary, in any subgame starting at some immediate successor node of x, player i can choose a winning strategy. Then, in the subgame starting at x, a grand strategy composed of these strategies in the respective subgames will be a winning strategy for i. This, however, contradicts our assumption that x is nonwinning for i. Thus, whenever player j moves at a node x that is nonwinning for i, there is an immediate successor node that is also nonwinning for i. We can therefore define player j’s action at any such x in the searched for strategy by the requirement that it leads to an immediate successor node that is nonwinning for i. At all other nodes in G, choose any of the feasible actions for j. This defines a strategy s for player j. Now we claim that any path p generated by s and some strategy of i yields at least a draw for j. This is clear for any infinite path. Assume therefore that p is finite. We will show that the terminal node of p must be nonwinning for i, and therefore yield at least a draw for j. Note first that, by assumption, the initial node of G is nonwinning for i. Hence, it suffices to show that if x is nonwinning for i, the immediate successor node of x on p has the same property. This is clear for any x, at which j is called upon to play, because j’s strategy was precisely defined that way. But this is also true for any node x at which it is i’s turn, because, if i had available a winning strategy in a subgame starting at some immediate successor node w of x, this strategy could be complemented, in a way that i moves from x to w, to a strategy of i in the subgame starting at x. As w was arbitrary, this shows that any successor node of x is nonwinning for i. This proves the theorem.

By: Bernard Yomtov

Bernard Yomtov — Fri, 10 Sep 2004 00:17:24 +0000

dsquared, Thanks. You’re forgiven for that business about Denmark.

By: dsquared

dsquared — Thu, 09 Sep 2004 22:54:11 +0000

Bernard, basically all payoff functions that anyone might choose are “Borel functions”; it’s a description of a very wide class of functions indeed. In Marco’s post “proved for all Borel functions” just means to exclude very weird mathematical constructions which take advantage of some of the more outlandish properties of the continuum, and keep the discussion to payoffs that you could measure in dollars and cents.

By: Bernard Yomtov

Bernard Yomtov — Thu, 09 Sep 2004 22:12:52 +0000

Marco, What is a Borel payoff function? Jim Harrison, Did your random games actually go to checkmate, or did you just decide that some positions were so strong that they were wins? I would expect a set of random moves to produce a mate almost never, even if one side has overwhelming strength.

By: C.J.Colucci

C.J.Colucci — Thu, 09 Sep 2004 21:35:22 +0000

In certain positions, it is knmown that a draw is the only possible outcome, even regardless of player error. If there are only 2 kings on the board, no one can win and I can only assume that a draw would bed eclared. (This never happens because players of tournament caliber would see the inevitability of a two-king piosition well in advance of its occurring and agree to a draw.) Likewise, there are well-known positions where, short of outright grotesque player error, a draw is the only outcome. Players usually agree to draws in these circumstances. Oddly, there has recently been a comoputer proof of a forced win in a position that human players always believed to be a draw in the absence of egregious player error. It takes about 75 moves without either the advance of a pawn or the capture of a piece, and, worse, nobody yet understands the logic of the play, though it can be memorized. If the position arises in a tournament, or if the player aware of the win is about to embark on move 51, what should haoppen if the other player calles the 50-move rule. Should the other player be allowed to demonstrate the existence of the forced win? (realistically, I doubt anyone could remember a 75-move combination in game conditions if the combination does not conform to the logic used by human chess players, but assume the player is exceptionally relaxed and can demonstrate that he knows the sequence)

By: Adam Stephanides

Adam Stephanides — Thu, 09 Sep 2004 14:43:22 +0000

“Zermelo was assuming that an infinitely long game is a draw, and in that case the result that one or both players has an optimal strategy is elementary.” I spoke too soon here. I’m pretty sure it’s true if payoffs are restricted to +1, 0, or -1, as in chess; but if payoffs can be any real number, the situation is more complicated.

By: Adam Stephanides

Adam Stephanides — Thu, 09 Sep 2004 14:21:35 +0000

“I’ve always believed that the repetition rule was the same as in Go – returning the board to a state it has been in before is illegal.” No, it’s perfectly legal, but if the same state occurs three times then the game is drawn (providing a player calls it). That’s not the rule in go, either, at least not in the Japanese rules, which I believe are the most widely used rules in the U.S. It’s only illegal to play a move which returns the board to the state before your opponent’s last move. Other repetitions of positions are legal—for example, triple ko, where there are three kos on the board and the players take them in rotation—and if neither player will yield the game is declared “no result.” “Actually, Martin proved that infinite perfect-information zero-sum games are determined as long as the payoff function is Borel [Annals of Mathematics 102, 1975]. Which indeed means that Chess without drawing rules, or any other potentially infinite game that can be defined without recourse to something like the Axiom of Choice, has a winning strategy for one of the players or drawing strategies for both.” Just to clarify, Martin’s result refers to the case where a game that lasts infinitely long can be either a win or a loss (i. e. some infinite sequences of moves are defined as wins for A, and some as wins for B). Zermelo was assuming that an infinitely long game is a draw, and in that case the result that one or both players has an optimal strategy is elementary. Getting back to Martin’s case: if there are no restrictions on the payoff function, then it’s actually a consequence of the Axiom of Choice that there are some infinite games in which neither player has an optimal strategy. A good deal of work has been done on set theory without the Axiom of Choice, but with the Axiom of Determinacy, which states that a certain set of such games always has a winning strategy.

By: Mike

Mike — Thu, 09 Sep 2004 13:37:53 +0000

Cogley: How many games of chess did you win from the computer, Mr. Spock? Spock: Five in all. Cogley: May that be considered unusual? Spock: Affirmative. Cogley: Why? Spock: I personally programmed the computer for chess months ago. I gave the machine an understanding of the game equal to my own. The computer cannot make an error. Assuming that I do not either, the best that could normally be hoped for would be stalemate after stalemate, and yet I beat the machine five times. - Star Trek, "Court-Martial"

By: yabonn

yabonn — Thu, 09 Sep 2004 12:14:15 +0000

… visions of the Two Last Chessmachines, computing endlessly the One Last Game, doomed to repetition or to lose/draw a winning position…

By: Marco Vervoort

Marco Vervoort — Thu, 09 Sep 2004 11:17:27 +0000

Actually, Martin proved that infinite perfect-information zero-sum games are determined as long as the payoff function is Borel [Annals of Mathematics 102, 1975]. Which indeed means that Chess without drawing rules, or any other potentially infinite game that can be defined without recourse to something like the Axiom of Choice, has a winning strategy for one of the players or drawing strategies for both.

By: David Cantrell

David Cantrell — Thu, 09 Sep 2004 09:44:15 +0000

I’ve always believed that the repetition rule was the same as in Go – returning the board to a state it has been in before is illegal.

By: Pedro

Pedro — Thu, 09 Sep 2004 06:12:58 +0000

Chess is an infinite game. But the conclusion of Zermelo’s Theorem holds for it.

By: SickMind Fraud

SickMind Fraud — Thu, 09 Sep 2004 03:52:18 +0000

This reminds me very much of a story about Aturo Toscanini in rehersal of (I believe) a Tchaikovsky symphony with some famous orchestra or another. Essentially, he caught them playing a particular passage wrong, which apparently they had been doing for over 50 years. They had learned it wrong in the first place, and had passed the error down through the generations, until Toscanini forced them to play it correctly. Again, a matter of not checking the original materials.

By: Peter Northup

Peter Northup — Thu, 09 Sep 2004 02:48:20 +0000

I have to defend my Will Baude, my fellow Crescateer, here— First, as Adam (above) already said (but it bears repeating, because Henry’s comment reproduces the confusion): Zermelo DID prove that either white has a winning position, black has a winning position, or neither does, that is, both players can stave off defeat indefinitely (because he was ignoring the 50 move rule). It’s in the penultimate paragraph of the proof. It’s just that, as Schwalbe and Walker emphasize, this wasn’t his main concern; he was more interested in putting a bound on the number of moves it takes to win GIVEN that someone has a winning position. One might quibble with the way Baude and Morrow phrase it—in particular, for Morrow to use the phrase “winning strategy” is a bit misleading, but since in the very same sentence he clarifies by saying that he means either W wins, B wins, or both can force draw, I don’t think it’s a big gaffe. One might similarly say “tic-tac-toe has a winning strategy” meaning “a sequence of optimal play for each player” (with perhaps a connotation of “that avoids losing” tacked on); it’s hardly an abuse of language.