博弈论中的极小极大算法第2组（评估功能简介） _博弈论

本文概述

C ++
Java
Python3
C#

先决条件：博弈论中的极小极大算法
从上面的文章中可以看出, 每个叶节点都有一个与之关联的值。我们已经将该值存储在数组中。但是在现实世界中, 当我们创建一个程序来玩井字棋, 国际象棋, 步步高等游戏时, 我们需要实现一个函数, 该函数根据棋子在棋盘上的放置来计算棋盘的价值。此功能通常称为评估功能。有时也称为启发式函数。
评估功能对于每种类型的游戏都是唯一的。在这篇文章中, 讨论了井字游戏的评估功能。评估功能背后的基本思想是为董事会提供高价值, 如果最大化器轮到板子还是走低最小化器轮到你了。
对于这种情况, 让我们考虑 X作为最大化器和?作为最小化器.
让我们建立评估函数：
如果X在棋盘上获胜, 我们给它正值+10。

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如果O在棋盘上获胜, 我们给它负值-10。

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如果没有人获胜或游戏导致平局, 那么我们给出+0的值。

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我们可以选择除10以外的任何正/负值。为简单起见, 为简单起见, 我们选择10, 我们将使用小写字母" x"和小写字母" o"代表玩家, 并使用下划线" _"代表板上的空白。
如果我们将板表示为3×3 2D字符矩阵, 例如char board [3] [3]; 那么我们必须检查每一行, 每一列和对角线, 以检查是否有一个玩家连续获得3个。
C ++

// C++ program to compute evaluation function for // Tic Tac Toe Game. #include< stdio.h> #include< algorithm> 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 = https://www.lsbin.com/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 = https://www.lsbin.com/evaluate(board); System.out.printf("The value of this board is %d\n" , value); } }// This code is contributed by PrinciRaj1992

Python3

# 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 = https://www.lsbin.com/evaluate(board) print ("The value of this board is" , value) # This code is contributed by Rituraj Jain

// 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 = https://www.lsbin.com/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

本文的目的是了解如何为井字游戏编写简单的评估函数。在下一篇文章中, 我们将看到如何将此评估函数与minimax函数结合使用。敬请关注。
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