Prove the following assertion: For every game tree, the utility obtained by max using minimax decisions against a suboptimal min will never be lower than the utility obtained playing against an optimal min. Can you come up with a game tree in which max can do still better using a suboptimal strategy against a suboptimal min?
Player $A$ moves first. The two players take turns moving, and each player must move his token to an open adjacent space in either direction. If the opponent occupies an adjacent space, then a player may jump over the opponent to the next open space if any. (For example, if $A$ is on 3 and $B$ is on 2, then $A$ may move back to 1.) The game ends when one player reaches the opposite end of the board. If player $A$ reaches space 4 first, then the value of the game to $A$ is $+1$; if player $B$ reaches space 1 first, then the value of the game to $A$ is $-1$.
line-game4-figure
The starting position of a simple game.

Prove the following assertion: For every game tree, the utility obtained by max using minimax decisions against a suboptimal min will never be lower than the utility obtained playing against an optimal min. Can you come up with a game tree in which max can do still better using a suboptimal strategy against a suboptimal min?
Player $A$ moves first. The two players take turns moving, and each player must move his token to an open adjacent space in either direction. If the opponent occupies an adjacent space, then a player may jump over the opponent to the next open space if any. (For example, if $A$ is on 3 and $B$ is on 2, then $A$ may move back to 1.) The game ends when one player reaches the opposite end of the board. If player $A$ reaches space 4 first, then the value of the game to $A$ is $+1$; if player $B$ reaches space 1 first, then the value of the game to $A$ is $-1$.

line-game4-figure
The starting position of a simple game.





Submit Solution

Your Display Name
Email
Solution