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I'm currently looking into computability logic. Japaridze explain that a game !P v P like !Chess v Chess is always winnable thanks to the copycat strategy (http://www.csc.villanova.edu/~japaridz/CL/3.html#copycat). However, what guarantees that the environment will play the same moves? One game could be e4 e5, but the other one e4 Nf6, and then the machine won't know what to do, right?

I'm probably not getting something, but I don't know what. Could someone help me?

Thanks.

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    $\begingroup$ Actually, you can guarantee to get either one win and one loss, or two draws. And I vaguely remember this in a movie where a young lady accepted very high bets to draw even against two chess masters. $\endgroup$
    – gnasher729
    Jul 12, 2018 at 22:20

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The setting in the link you gave (where is it referred to as $chess\lor \neg chess$) is that the player is playing on two boards with different colors, and has to win in at least one.

In that case you can win by mimicking your opponent as follows. Let $b_{\text{white}},b_{\text{black}}$ be the boards on which you play white/black correspondingly. Whenever the opponent makes a move $x$ on $b_{\text{black}}$, you play $x$ on $b_{\text{white}}$, similarly when your opponent makes a move $x$ on $b_{white}$ your play $x$ on $b_{black}$. The result is that you are playing the same position on both boards, only with different colors.

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    $\begingroup$ For this to work, you need the opponent to make the very first move. What if the opponent refuses to move, and asks you to make the very first move? This point isn't really clarified in the link, which attempts very hard to avoid definitions. $\endgroup$
    – Yuval Filmus
    Jul 12, 2018 at 19:00
  • $\begingroup$ Indeed, for the entire duration of the game I can only make a move on one board after my opponent has played on the other. This doesn't really define a strategy in the standard game theoretic sense, and requires some different formalization. $\endgroup$
    – Ariel
    Jul 12, 2018 at 19:07
  • $\begingroup$ This point is explained in Part II of this paper, but the explanation is too long for me understand. $\endgroup$
    – Yuval Filmus
    Jul 12, 2018 at 19:29
  • $\begingroup$ @YuvalFilmus The reason is that, in this formalization of games, it's not only final positions that are designated as either winning or losing; intermediate positions are considered to be winning or losing, too. This means that as long as both boards have the same state, Top is winning on one board and losing on the other, and therefore winning overall. If Bottom refuses to move, then a winning strategy for Top is to also refuse to move. Whenever Bottom does make a move, Top makes the corresponding move on the other board, and thus returns to a winning position. $\endgroup$
    – Tanner Swett
    Jul 12, 2018 at 22:24
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    $\begingroup$ @YuvalFilmus I'm not sure what you mean by the opponent refusing to move. In chess, White always moves first and it is not possible to pass. If one of the opponents simply refuses to play according to the setup, then we're out of the realm of game theory and into Interpersonal Skills. $\endgroup$
    – David Richerby
    Jul 13, 2018 at 11:50
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I'm not sure this observation is worth a full answer.

The way to think about this strategy, is that you are trying to guarantee a win on one of two boards. You accomplish this, by forcing two opponents to play each other, with you acting as the intermediary. One one player moves, you duplicate the identical move on the opposite board. So the two players are playing each other, but on different boards.

If this doesn't make sense, I'd recommend setting this up with a very simple game (if you know the game of Nim, that would be a good choice) and see how this plays out.

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