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Minimax Tic-Tac-Toe

Pick the perfect tic-tac-toe move by searching the whole game tree with minimax.

このチャレンジで学べること

  • Search every possible future move
  • Score positions from the current player's view
  • Pick the optimal move recursively

問題

翻訳を表示

Minimax is how game-playing programs choose moves: assume both players play perfectly, score every finished game, and back the scores up the game tree. The player to move picks the child with the highest score for itself.

The board is a 9-character string read row by row (indices 0-8), with "X", "O" or "." for an empty cell. X always moves first, so it is X's turn when both have the same number of marks and O's turn otherwise.

Score a finished game from the point of view of the player to move: +1 for a win, 0 for a draw, -1 for a loss. Return the empty cell whose move has the best minimax score; if several are equally good, return the lowest index.

Implement best_move(board).

Input

board: A valid position where nobody has won yet and at least one cell is empty.

Output

Return the index (0-8) of the best move.

制約

board has exactly 9 characters, each X, O or .

The position is reachable in a real game, nobody has won, and at least one cell is empty

例

例 1
入力:
board = "XX.OO...."

期待される出力:
expected = 2
入力:
std::string board = "XX.OO....";

期待される出力:
long long expected = 2;
入力:
string board = "XX.OO....";

期待される出力:
long expected = 2;

解説

X and O have two marks each, so X moves. Playing 2 completes the top row and wins at once (+1).

例 2
入力:
board = "X..OO...X"

期待される出力:
expected = 5
入力:
std::string board = "X..OO...X";

期待される出力:
long long expected = 5;
入力:
string board = "X..OO...X";

期待される出力:
long expected = 5;

解説

X moves. O threatens to complete the middle row at 5. Every other move lets O win (-1), while blocking at 5 leads to a draw (0), so the answer is 5.

参考資料

この課題はCodeUtility用に調整されています。上記の説明と制約に従ってください。

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