[英]unbeatable Tic Tac Toe
我正在嘗試為一個輔助項目制作一個無與倫比的井字游戲,但我做對了(我實際上可以諷刺地擊敗它)。 它實際上是 MiniMax 算法的一個實現; 我帶着這個代碼
#include <iostream>
#include <string>
using namespace std;
struct Move
{
int line, columns;
};
//Return the number of remainings turn based on the number of lest boxes
int remainingsTurns(char grid[3][3])
{
int remainingTurn = 0;
for (int k = 0; k < 3; k++)
{
for (int i = 0; i < 3; i++)
{
if (grid[k][i] == ' ')
{
remainingTurn++;
}
}
}
return remainingTurn;
}
//Print the grid on screen
void printGrid(char grid[3][3])
{
for (int k = 0; k < 3; k++)
{
for (int i = 0; i < 3; i++)
{
cout << "| " << grid[k][i] << " ";
}
cout << "|" << endl;
}
}
//Give a value to the board
int evaluateBoard(char grid[3][3])
{
//Check the board for lines
for (int k = 0; k < 3; k++)
{
if (grid[k][0] == grid[k][1] && grid[k][1] == grid[k][2])
{
if (grid[k][0] == 'x')
{
return +10;
}
else if (grid[k][0] == 'o')
{
return -10;
}
}
}
//Check the board for columns
for (int k = 0; k < 3; k++)
{
if (grid[0][k] == grid[1][k] && grid[1][k] == grid[2][k])
{
if (grid[0][k] == 'x')
{
return +10;
}
else if (grid[0][k] == 'o')
{
return -10;
}
}
}
//Check the board for diagonals
if (grid[0][0] == grid[1][1] && grid[0][0] == grid[2][2])
{
if (grid[0][0] == 'x')
{
return +10;
}
else if (grid[0][0] == 'o')
{
return -10;
}
}
if (grid[0][2] == grid[1][1] && grid[0][2] == grid[2][0])
{
if (grid[0][0] == 'x')
{
return +10;
}
else if (grid[0][0] == 'o')
{
return -10;
}
}
//if no ictory return 0
return 0;
}
// MiniMax algorithm
int miniMax(char grid[3][3], int turn, bool maxMove)
{
int score = evaluateBoard(grid);
if (score == 10)
{
return score;
}
if (score == -10)
{
return score;
}
//Check if the game is a tie
if (remainingsTurns(grid) == 0)
{
return 0;
}
if (maxMove)
{
int best = -1000;
for (int k = 0; k < 3; k++)
{
for (int i = 0; i < 3; i++)
{
if (grid[k][i] == ' ')
{
grid[k][i] = 'x';
best = max(best, miniMax(grid, turn + 1, !maxMove));
grid[k][i] = ' ';
}
}
}
return best;
}
else
{
int best = 1000;
for (int k = 0; k < 3; k++)
{
for (int i = 0; i < 3; i++)
{
if (grid[k][i] == ' ')
{
grid[k][i] = 'o';
best = min(best, miniMax(grid, turn + 1, !maxMove));
grid[k][i] = ' ';
}
}
}
return best;
}
}
Move playerMov(char grid[3][3])
{
Move playerMove;
int input = -1;
cout << "Enter the column you want to play (1, 2 or 3)" << endl;
cin >> input;
if (input == 1 || input == 2 || input == 3)
{
playerMove.columns = input-1;
input = -1;
}
else
{
cout << "Error, enter a valid number!" << endl;
playerMov(grid);
}
cout << "Enter the line you want to play (1, 2 or 3)" << endl;
cin >> input;
if (input == 1 || input == 2 || input == 3)
{
playerMove.line = input-1;
input = -1;
}
else
{
cout << "Error, enter a valid number!" << endl;
playerMov(grid);
}
return playerMove;
}
//return the best move using the MiniMax
Move findMove(char grid[3][3])
{
int best = -1000;
Move move;
move.line = -1;
move.columns = -1;
//Check all move to find if this move is the best possible move
for (int k = 0; k < 3; k++)
{
for (int i = 0; i < 3; i++)
{
if (grid[k][i] == ' ')
{
grid[k][i] = 'x';
int moveValue = miniMax(grid, 0, false);
grid[k][i] = ' ';
if (moveValue > best)
{
move.line = k;
move.columns = i;
}
}
}
}
return move;
}
int main()
{
char grid[3][3];
int turn = 0;
Move playerMove, algoMove;
for (int k = 0; k < 3; k++)
{
for (int i = 0; i < 3; i++)
{
grid[k][i] = ' ';
}
}
cout << "Welcome to the unbeatable Tic Tac Toe !" << endl;
do
{
printGrid(grid);
playerMove = playerMov(grid);
grid[playerMove.line][playerMove.columns] = 'o';
Move computerMove = findMove(grid);
grid[computerMove.line][computerMove.columns] = 'x';
} while (remainingsTurns(grid) > 0);
cout << endl;
}
但是算法的動作似乎不對,它總是選擇右下角,我不明白為什么......這個實現很大程度上受到了 Geek for Geek 這篇文章的啟發,我試圖竊取算法,但我無法讓它適合單人游戲。 我想念什么?
您的代碼中有多個錯誤:
grid[0][0]
在grid[0][2] == grid[1][1] && grid[0][2] == grid[2][0]
的情況下函數evaluateBoard
。 它應該是grid[0][2]
。playerMov(grid);
在函數playerMov
應該return playerMov(grid);
. 否則,重新輸入的“正確”值將被刪除,並且(部分)未初始化的playerMov
將被返回。best
到moveValue
時moveValue > best
的功能findMove
。 (這應該是您問題的原因)
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.