can someone help me answer part b c and d? i really dont understand what to do.

https://www.chegg.com/homework-help/questions-and-answers/consider-two-player-line-game-shown--players-start-visualized-positions-player-moves-first-q10845895 link to graphs

1 2 3 4 Consider the following two-player game (players A and B): A B The starting position of the simple game is shown on the right: 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 or forward to 4.) 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. (a) Draw the complete game tree using the following conventions: Annotate each terminal state with its game value in a circle. Treat loop states as terminal states. Since it is not clear how to assign values to loop states, annotate each with a question mark in a circle. Loop states are states that already appear on their path to the root at a level in which it is the same player's turn to move. (b) Now mark each node with its backed-up minimax value (also in a circle). Explain how you handled the question mark values and why. (c) Explain why the standard minimax algorithm would fail on this game tree and briefly sketch how you might fix it, drawing on your answer to part (b) Does your modified algorithm give optimal decisions for all games with loops? (d) Which player has a winning strategy, and what does it look like? And what can be said about the general case, when instead of a 4-square game, we consider an n-square game for n > 2? 1 2 3 4 Consider the following two-player game (players A and B): A B The starting position of the simple game is shown on the right: 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 or forward to 4.) 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. (a) Draw the complete game tree using the following conventions: Annotate each terminal state with its game value in a circle. Treat loop states as terminal states. Since it is not clear how to assign values to loop states, annotate each with a question mark in a circle. Loop states are states that already appear on their path to the root at a level in which it is the same player's turn to move. (b) Now mark each node with its backed-up minimax value (also in a circle). Explain how you handled the question mark values and why. (c) Explain why the standard minimax algorithm would fail on this game tree and briefly sketch how you might fix it, drawing on your answer to part (b) Does your modified algorithm give optimal decisions for all games with loops? (d) Which player has a winning strategy, and what does it look like? And what can be said about the general case, when instead of a 4-square game, we consider an n-square game for n > 2