使用 C 中的結構返回多個極小值 [score, x, y]

Question

我的代碼中的問題在於 minimax 算法，更具體地說是返回值。 我目前用 C++ 編寫的解決方案如下，使用元組、返回分數、x 和 y 位置。 我決定使用 struct 在 C 中做類似的事情。

在給定當前板子 position [3468] 的情況下，該算法會正確找到所有可能的最終狀態，但不會返回正確的下一步移動坐標。

鑒於現在輪到 Xs，我們應該從 function 中得到 [score = 0, x = 0, y = 1]。 目前返回的值為 [score = 1, x = 1, y = 0]。

我應該如何正確實現結構以使算法正常工作？

#include <stdio.h>

int s = 3; //Board size
int cnt = 0;
char player = 'X', opponent = 'O';

struct Best {
  int score, x, y;
};

void printBoard(char v[][s]);
char eval(char v[][s]);
struct Best miniMax(char Table[][s], int depth, bool isMaximising); //Problem: structs?

int main(int argc, char const *argv[]) {
  char b[3][3] = {
    {'X','_','_'},
    {'_','O','_'},
    {'_','_','_'}
  };
  printBoard(b);

  struct Best s = miniMax(b, 0, true);
  printf("Possible games: %d\n", cnt);
  cnt = 0;
  printf("Score: %d, Best move: [%d][%d]\n", s.score, s.x, s.y);

  return 0;
}
void printBoard(char v[][s]){
  for (size_t i = 0; i < s; i++){
    for (size_t j = 0; j < s; j++){
      printf("%c|", v[i][j]);
    }
    printf("\n");
  }
  printf("\n");
}
char eval(char v[][s]){
  for (size_t i = 0; i < s; i++) {
    if (v[i][0] == v[i][1] && v[i][1] == v[i][2] && v[i][0] != '_') //Row
    return v[i][0];
    if (v[0][i] == v[1][i] && v[1][i] == v[2][i] && v[0][i] != '_') //Col
    return v[0][i];
  }
  if (v[0][0] == v[1][1] && v[1][1] == v[2][2] && v[0][0] != '_') //Diag
  return v[0][0];

  if (v[0][2] == v[1][1] && v[1][1] == v[2][0] && v[0][2] != '_') //Diag2
  return v[0][2];

  for (size_t i = 0; i < s; i++) {
    for (size_t j = 0; j < s; j++) {
      if (v[i][j] == '_')
      return '\0';
    }
  }
  return '=';
}

以下是 C 中存在問題的 miniMax function 實現：

struct Best miniMax(char Table[][s], int depth, bool isMaximising) {
  struct Best v;
  char result = eval(Table);
  if (result == 'X') {
    cnt++;
    v.score = 1;
    v.x = 0;
    v.y = 0;
    return v;
  }
  if (result == 'O') {
    cnt++;
    v.score = -1;
    v.x = 0;
    v.y = 0;
    return v;
  }
  if (result == '=') {
    cnt++;
    v.score = 0;
    v.x = 0;
    v.y = 0;
    return v;
  }
  if (isMaximising) {
    int bestScore = -100;
    for (size_t i = 0; i < s; i++) {
      for (size_t j = 0; j < s; j++) {
        if (Table[i][j] == '_') {
          Table[i][j] = player;
          v = miniMax(Table, depth + 1, false);
          if (v.score > bestScore) {
            bestScore = v.score;
            v.score = bestScore;
            v.x = i;
            v.y = j;
          }
          Table[i][j] = '_';
        }
      }
    }
    return v;
  } else {
    int bestScore = 100;
    for (size_t i = 0; i < s; i++) {
      for (size_t j = 0; j < s; j++) {
        if (Table[i][j] == '_') {
          Table[i][j] = opponent;
          v = miniMax(Table, depth + 1, true);
          if (v.score < bestScore) {
            bestScore = v.score;
            v.score = bestScore;
            v.x = i;
            v.y = j;
          }
          Table[i][j] = '_';
        }
      }
    }
    return v;
  }
}

C++ 中的實現：

#include <tuple>
#include <iostream>
using namespace std;

tuple <int, int, int> Minimax(char Board[][s], int depth, bool isMaximising){
  char res = eval(Board);

  if (res == 'X') {
    cnt++;
    return {1, 0, 0};
  }
  if (res == 'O') {
    cnt++;
    return {-1, 0, 0};
  }
  if (res == '=') {
    cnt++;
    return {0, 0, 0};
  }

  if (isMaximising) {
    int bestScore = -100;
    int px = 0, py = 0;
    for (size_t i = 0; i < s; i++) {
      for (size_t j = 0; j < s; j++) {
        if (Board[i][j] == '_') {
          Board[i][j] = player;
          auto [score, x, y] = Minimax(Board, depth + 1, false);
          if (score > bestScore) {
            bestScore = score;
            px = i;
            py = j;
          }
          Board[i][j] = '_';
        }
      }
    }
    return {bestScore, px, py};
  } else {
    int bestScore = 100;
    int qx = 0, qy = 0;
    for (size_t i = 0; i < s; i++) {
      for (size_t j = 0; j < s; j++) {
        if (Board[i][j] == '_') {
          Board[i][j] = opponent;
          auto [score, x, y] = Minimax(Board, depth + 1, true);
          if (score < bestScore) {
            bestScore = score;
            qx = i;
            qy = j;
          }
          Board[i][j] = '_';
        }
      }
    }
    return {bestScore, qx, qy};
  }
}

Answer 1

基本問題是，每次 miniMax 遞歸調用自身時，它都會用返回的值覆蓋v ，從而丟失任何早期迭代中保存的最佳值。 所以你總是從最后一次測試的 position 中得到 x/y，而不是從最好的那個中得到。

添加一個新的struct Best best value，它是您在找到更好的移動時更新並返回的那個。 請注意，您也不需要bestScore local 然后 - 您可以使用best.score