先决条件：博弈论中的极小极大算法
如上一篇文章所示，每个叶节点都有一个与之关联的值。我们已经将这个值存储在一个数组中。但在现实世界中，当我们创建一个程序来玩井字游戏、国际象棋、西洋双陆棋等时，我们需要实现一个函数，根据棋子在棋盘上的位置计算棋盘的价值。这个函数通常被称为评估函数。它有时也称为启发式函数。
对于每种类型的游戏，评估函数都是独一无二的。在这篇文章中，讨论了 Tic-Tac-Toe 游戏的评估函数。评估函数背后的基本思想是，如果轮到最大化者，则为棋盘提供高值，如果轮到最小者，则为棋盘提供低值。
对于这种情况，让我们将X视为最大化器，将O视为最小化器。
让我们构建我们的评估函数：
- 如果 X 在棋盘上获胜，我们给它一个正值 +10。
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- 如果 O 在棋盘上获胜，我们给它一个负值 -10。
_1.jpg)
- 如果没有人赢或比赛结果为平局，则我们给出 +0 的值。
_2.jpg)
我们可以选择除 10 以外的任何正/负值。为了简单起见，我们选择 10 为简单起见，我们将使用小写的“x”和小写的“o”来代表玩家和下划线“_”表示板上的空白区域。
如果我们将板表示为 3×3 2D字符矩阵，如 char board[3][3]；然后我们必须检查每一行、每一列和对角线，以检查是否有玩家连续获得 3 个。
C++
// C++ program to compute evaluation function for
// Tic Tac Toe Game.
#include
#include
using namespace std;
// Returns a value based on who is winning
// b[3][3] is the Tic-Tac-Toe board
int evaluate(char b[3][3])
{
// Checking for Rows for X or O victory.
for (int row = 0; row<3; row++)
{
if (b[row][0]==b[row][1] && b[row][1]==b[row][2])
{
if (b[row][0]=='x')
return +10;
else if (b[row][0]=='o')
return -10;
}
}
// Checking for Columns for X or O victory.
for (int col = 0; col<3; col++)
{
if (b[0][col]==b[1][col] && b[1][col]==b[2][col])
{
if (b[0][col]=='x')
return +10;
else if (b[0][col]=='o')
return -10;
}
}
// Checking for Diagonals for X or O victory.
if (b[0][0]==b[1][1] && b[1][1]==b[2][2])
{
if (b[0][0]=='x')
return +10;
else if (b[0][0]=='o')
return -10;
}
if (b[0][2]==b[1][1] && b[1][1]==b[2][0])
{
if (b[0][2]=='x')
return +10;
else if (b[0][2]=='o')
return -10;
}
// Else if none of them have won then return 0
return 0;
}
// Driver code
int main()
{
char board[3][3] =
{
{ 'x', '_', 'o'},
{ '_', 'x', 'o'},
{ '_', '_', 'x'}
};
int value = evaluate(board);
printf("The value of this board is %d\n", value);
return 0;
} Java
// Java program to compute evaluation function for
// Tic Tac Toe Game.
class GFG
{
// Returns a value based on who is winning
// b[3][3] is the Tic-Tac-Toe board
static int evaluate(char b[][])
{
// Checking for Rows for X or O victory.
for (int row = 0; row < 3; row++)
{
if (b[row][0] == b[row][1] && b[row][1] == b[row][2])
{
if (b[row][0] == 'x')
return +10;
else if (b[row][0] == 'o')
return -10;
}
}
// Checking for Columns for X or O victory.
for (int col = 0; col < 3; col++)
{
if (b[0][col] == b[1][col] && b[1][col] == b[2][col])
{
if (b[0][col] == 'x')
return +10;
else if (b[0][col] == 'o')
return -10;
}
}
// Checking for Diagonals for X or O victory.
if (b[0][0] == b[1][1] && b[1][1] == b[2][2])
{
if (b[0][0] == 'x')
return +10;
else if (b[0][0] == 'o')
return -10;
}
if (b[0][2] == b[1][1] && b[1][1] == b[2][0])
{
if (b[0][2] == 'x')
return +10;
else if (b[0][2] == 'o')
return -10;
}
// Else if none of them have won then return 0
return 0;
}
// Driver code
public static void main(String[] args)
{
char board[][] =
{
{ 'x', '_', 'o'},
{ '_', 'x', 'o'},
{ '_', '_', 'x'}
};
int value = evaluate(board);
System.out.printf("The value of this board is %d\n", value);
}
}
// This code is contributed by PrinciRaj1992Python3
# Python3 program to compute evaluation
# function for Tic Tac Toe Game.
# Returns a value based on who is winning
# b[3][3] is the Tic-Tac-Toe board
def evaluate(b):
# Checking for Rows for X or O victory.
for row in range(0, 3):
if b[row][0] == b[row][1] and b[row][1] == b[row][2]:
if b[row][0] == 'x':
return 10
elif b[row][0] == 'o':
return -10
# Checking for Columns for X or O victory.
for col in range(0, 3):
if b[0][col] == b[1][col] and b[1][col] == b[2][col]:
if b[0][col]=='x':
return 10
elif b[0][col] == 'o':
return -10
# Checking for Diagonals for X or O victory.
if b[0][0] == b[1][1] and b[1][1] == b[2][2]:
if b[0][0] == 'x':
return 10
elif b[0][0] == 'o':
return -10
if b[0][2] == b[1][1] and b[1][1] == b[2][0]:
if b[0][2] == 'x':
return 10
elif b[0][2] == 'o':
return -10
# Else if none of them have won then return 0
return 0
# Driver code
if __name__ == "__main__":
board = [['x', '_', 'o'],
['_', 'x', 'o'],
['_', '_', 'x']]
value = evaluate(board)
print("The value of this board is", value)
# This code is contributed by Rituraj JainC#
// C# program to compute evaluation function for
// Tic Tac Toe Game.
using System;
class GFG
{
// Returns a value based on who is winning
// b[3,3] is the Tic-Tac-Toe board
static int evaluate(char [,]b)
{
// Checking for Rows for X or O victory.
for (int row = 0; row < 3; row++)
{
if (b[row, 0] == b[row, 1] && b[row, 1] == b[row, 2])
{
if (b[row, 0] == 'x')
return +10;
else if (b[row, 0] == 'o')
return -10;
}
}
// Checking for Columns for X or O victory.
for (int col = 0; col < 3; col++)
{
if (b[0, col] == b[1, col] && b[1, col] == b[2, col])
{
if (b[0, col] == 'x')
return +10;
else if (b[0, col] == 'o')
return -10;
}
}
// Checking for Diagonals for X or O victory.
if (b[0, 0] == b[1, 1] && b[1, 1] == b[2, 2])
{
if (b[0, 0] == 'x')
return +10;
else if (b[0, 0] == 'o')
return -10;
}
if (b[0, 2] == b[1, 1] && b[1, 1] == b[2, 0])
{
if (b[0, 2] == 'x')
return +10;
else if (b[0, 2] == 'o')
return -10;
}
// Else if none of them have won then return 0
return 0;
}
// Driver code
public static void Main(String[] args)
{
char [,]board =
{
{ 'x', '_', 'o'},
{ '_', 'x', 'o'},
{ '_', '_', 'x'}
};
int value = evaluate(board);
Console.Write("The value of this board is {0}\n", value);
}
}
// This code is contributed by Rajput-Ji输出 ：
The value of this board is 10这篇文章的想法是了解如何为游戏 Tic-Tac-Toe 编写一个简单的评估函数。在下一篇文章中，我们将看到如何将这个评估函数与极大极小函数结合起来。敬请关注。
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