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Denote the current state of the stacks as a $k$-vector $t=(t_1,\dots,t_k)$, where $k$ counts the number of stacks, $t_i$ is the number on the top of the $i$th stack, and I assume you have sorted the $t_i$ so that $t_1 < t_2 < \dots < t_k$. Define a total order on state vectors as follows: $(t_1,\dots,t_k) \prec (u_1,\dots,u_\ell)$ if $k<\ell$, ...


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In the rules of chess, one player can demand a draw when the same position is entered three times. If your opponent can demand a draw, the value of the position is 0 in the best case. So if you look forward far enough, you see that a move leading to repetition may not be that good. (In this case, you would choose the second best move if it has a value > 0, ...


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In a game-theoretical sense, both of the moves you describe are equally good in the scenarios you describe, so the algorithm is "correct" in that it doesn't matter which move it picks -- you do not explicitly encode a preference for faster wins (or faster gains of $+100$) over slower scores. Introducing a discount factor to explicitly encode such a ...


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This is because you are not using a discount factor in your search. A discount factor $\gamma$ is a number between 0 and 1. The discount factor describes the preference of an agent for current rewards over future rewards. When $\gamma$ is close to 0, rewards in the distant future are viewed as insignificant. When $\gamma$ is 1, discounted rewards are ...


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