我在微控制器上创建了一个Tic-Tac-Toe游戏，包括一个完美的AI（完美意味着它不会丢失）。我没有使用minimax算法，只是一个具有所有可能和最佳移动的小状态机。
我现在的问题是我想实现不同的困难（简单，中等和困难）。到目前为止，人工智能将是艰难的。
所以我已经考虑过如何以最好的方式做到这一点并最终想要使用minimax算法但是以某种方式计算所有游戏位置的所有分数，这样我有时也可以选择第二好成绩而非最佳成绩。由于我不能总是在微控制器上进行所有这些计算，我想创建一个可以在我的计算机上运行的小程序，它给出了所有可能的电路板状态的数组（关于对称性等，以尽量减少使用的存储）和他们的相应分数。
为此，我首先尝试实现minimax算法本身，关于depth，以便正确计算每个状态的scores。然后它应该让我回到阵列中的所有最佳动作（现在）。但是，它似乎没有那么好用。我试图用一些printf行来调试它。这是迄今为止minimax函数以及我的主函数的代码：
static int minimax(int *board, int depth)
{
int score;
int move = -1;
int scores[9];
int nextDepth;
printf("\n----- Called Minimax, Depth: %i -----\n\n", depth);
if(depth%2 ==1){
player = -1;
} else {
player = 1;
}
printf("Player: %i\n---\n", player);
if(isWin(board) != 0){
score = (10-depth)*winningPlayer;
printf("Player %i won on depth %i\n", winningPlayer, depth);
printf("Resulting score: (10-%i)*%i = %i\nScore returned to depth %i\n---\n", depth, winningPlayer, score, depth-1);
return score;
}
score = -20;
nextDepth = depth+1;
printf("Next depth is %i\n---\n", nextDepth);
int i;
for(i=0; i<9; i++){
if(board[i] == 0) {
if(nextDepth%2 ==0) {
player = -1;
} else {
player = 1;
}
printf("Found vacant space at position %i\n", i);
printf("Set value of position %i to %i\n---\n", i, player);
board[i] = player;
int thisScore = minimax(board, nextDepth);
printf("Value of the move at position %i on next depth %i is %i\n---\n", i, nextDepth, thisScore);
scores[i] = thisScore;
if(thisScore > score){
printf("New score value is greater than the old one: %i < %i\n---\n", thisScore, score);
score = thisScore;
move = i;
g_moves[nextDepth-1] = move;
printf("Score was set to %i\n", thisScore);
printf("Remembered move %i\n---\n", move);
}
board[i] = 0;
printf("Value of position %i was reset to 0 on next depth %i\n---\n", i, nextDepth);
}
}
if(move == -1) {
printf("Game ended in a draw.\n Returned score: 0\n---\n");
return 0;
}
printf("Move at position %i was selected on next depth %i\n", move, nextDepth);
printf("Returning score of %i to depth %i\n---\n", score, depth);
return score;
}
main是：
int main(int argc, char **argv)
{
memcpy(board, initBoard, sizeof(board));
int score = 0;
int depth = getDepth(board);
score = minimax(board, depth);
printf("\n--- finished ---\n\n");
printf("Moves with the highest score: ");
int i;
for(i=0; i<9; i++){
printf("%i | ", g_moves[i]);
}
printf("\n");
printf("The score is %i\n", score);
printf("The best next board is: \n|----|----|----|\n");
for(i=0; i<3; i++){
printf("| %-2i ", board[i]);
}
printf("|\n|----|----|----|\n");
for(i=3; i<6; i++){
printf("| %-2i ", board[i]);
}
printf("|\n|----|----|----|\n");
for(i=6; i<9; i++){
printf("| %-2i ", board[i]);
}
printf("|\n|----|----|----|\n");
return 0;
}
此外，我有一些变数：
//1 = Beginning Player
//-1 = second Player
static int player;
static int winningPlayer = 0;
static int g_moves[9];
/* 0 1 2
* 3 4 5
* 6 7 8
*/
int initBoard[9] = {
0, 0, 0,
0, 0, 0,
0, 0, 0,
};
int board[9];
以及我的获胜功能：
int isWin(int *board)
{
unsigned winningBoards[8][3] = {
{board[0], board[1], board[2],},
{board[3], board[4], board[5],},
{board[6], board[7], board[8],},
{board[0], board[3], board[6],},
{board[1], board[4], board[7],},
{board[2], board[5], board[8],},
{board[0], board[4], board[8],},
{board[2], board[4], board[6],},
};
int i;
for(i=0; i<8; i++){
if( (winningBoards[i][0] != 0) &&
(winningBoards[i][0] == winningBoards[i][1]) &&
(winningBoards[i][0] == winningBoards[i][2])){
winningPlayer = winningBoards[i][0];
return winningPlayer;
}
}
return 0;
}
出于某种原因，最后一次minimax从depth 7一步一步地返回到depth 1，它会覆盖全部为0的数组g_moves，从而产生以下几行我的打印输出（只有最后70行）：
...
----- Called Minimax, Depth: 7 -----
Player: -1
---
Player 1 won on depth 7
Resulting score: (10-7)*1 = 3
Score returned to depth 6
---
Value of the move at position 2 on next depth 7 is 3
---
Value of position 2 was reset to 0 on next depth 7
---
Move at position 0 was selected on next depth 7
Returning score of 3 to depth 6
---
Value of the move at position 3 on next depth 6 is 3
---
Value of position 3 was reset to 0 on next depth 6
---
Move at position 0 was selected on next depth 6
Returning score of 3 to depth 5
---
Value of the move at position 4 on next depth 5 is 3
---
Value of position 4 was reset to 0 on next depth 5
---
Move at position 0 was selected on next depth 5
Returning score of 3 to depth 4
---
Value of the move at position 5 on next depth 4 is 3
---
Value of position 5 was reset to 0 on next depth 4
---
Move at position 0 was selected on next depth 4
Returning score of 3 to depth 3
---
Value of the move at position 6 on next depth 3 is 3
---
Value of position 6 was reset to 0 on next depth 3
---
Move at position 0 was selected on next depth 3
Returning score of 5 to depth 2
---
Value of the move at position 7 on next depth 2 is 5
---
Value of position 7 was reset to 0 on next depth 2
---
Move at position 0 was selected on next depth 2
Returning score of 5 to depth 1
---
Value of the move at position 8 on next depth 1 is 5
---
Value of position 8 was reset to 0 on next depth 1
---
Move at position 0 was selected on next depth 1
Returning score of 5 to depth 0
---
--- finished ---
Moves with the highest score: 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
The score is 5
The best next board is:
|----|----|----|
| 0 | 0 | 0 |
|----|----|----|
| 0 | 0 | 0 |
|----|----|----|
| 0 | 0 | 0 |
|----|----|----|
如果您需要任何其他信息以帮助我，如果我自己拥有它们，我很乐意将它们交给您。
提前致谢。
修改
所以我重新编写了minimax函数，因此它现在使用控制台（相应文件夹中的cmd：./NAME_OF_FILE＆gt; DEST_NAME.txt）打印.txt文件中的所有可能的电路板状态。代码如下：
int minimax(int *board, int depth)
{
g_node++;
int player;
int move = -1;
int score = -20;
int thisScore = -20;
int i;
if(isWin(board) != 0){
printf("\nNode: %i\n", g_node);
printf("Board state:");
for(i=0;i<9;i++) {
if((i%3) == 0)
printf("\n");
printf("%2i ", board[i]);
}
printf("\n");
printf("has a score of %i\n", (10-depth)*winningPlayer);
return (10-depth)*winningPlayer;
}
if(depth%2 ==1){
player = -1;
} else {
player = 1;
}
for(i=0; i<9; i++){
if(board[i] == 0){
board[i] = player;
thisScore = minimax(board, depth+1);
if(thisScore > score){
score = thisScore;
move = i;
}
board[i] = 0;
}
}
printf("\nNode: %i\n", g_node);
printf("Board state:");
for(i=0;i<9;i++) {
if((i%3) == 0)
printf("\n");
printf("%2i ", board[i]);
}
printf("\n");
if(move == -1){
printf("has a score of 0\n");
return 0;
}
printf("has a score of %i\n", score);
return score;
}
我的下一步是打印出每个动作的最大score位于相应位置的棋盘。
Example:
10 8 10
8 7 8
10 8 10
for the empty board at the beginning.
编辑2：
我现在添加了另一个名为printScoredBoards的函数，基本上应该执行我在上一次编辑中所描述的内容，但是它存在问题。
如果你的对手足够愚蠢并且因为minimax尝试所有可能性，包括那些，使用以下代码，我总是有可能在第五次移动后获胜，我得到了一个空板的所有15个得分板。
void printScoredBoards(int *board, int depth)
{
int player;
int scoredBoard[9] = {0,0,0,0,0,0,0,0,0,};
int i;
if(isWin(board) == 0){
if(depth%2 ==1){
player = -1;
} else {
player = 1;
}
for(i=0; i<9; i++){
if(board[i] == 0){
board[i] = player;
scoredBoard[i] = minimax(board, depth+1)+10;
printScoredBoards(board, depth+1);
board[i] = 0;
}
}
printf("Scored board:");
dumpTable(scoredBoard);
printf("\n");
}
}
虽然角落应该更有价值，但中心价值最低。有没有人碰巧知道解决这个问题？
编辑：我已经设置了一个新的minimax算法并将其发布在另一篇文章中。你可以在＆＃34; Linked＆＃34;中找到帖子。右边的部分或here。 现在我所做的就是在微控制器代码中实现它并创建一个函数，该函数可以从所有得分的移动中选择最佳/第二最佳移动，并且如果存在具有相同分数的多个移动则随机化它。 因此，这篇文章可以关闭。
答案 0 :(得分：0)
我认为尝试通过全面深度分析获得第二好的举动是过度的。不要通过限制minmax的深度来探索整棵树（2向前移动允许获胜，但AI仍然很强），或者只是使用随机移动来获得真正不完美的AI。