Introduction to Evaluation Function of Minimax Algorithm in Game Theory

Last Updated : 20 Apr, 2023

Prerequisite: Minimax Algorithm in Game Theory
As seen in the above article, each leaf node had a value associated with it. We had stored this value in an array. But in the real world when we are creating a program to play Tic-Tac-Toe, Chess, Backgammon, etc. we need to implement a function that calculates the value of the board depending on the placement of pieces on the board. This function is often known as Evaluation Function. It is sometimes also called a Heuristic Function.
The evaluation function is unique for every type of game. In this post, the evaluation function for the game Tic-Tac-Toe is discussed. The basic idea behind the evaluation function is to give a high value for a board if the maximizer turn or a low value for the board if the minimizer turn.
For this scenario let us consider X as the maximizer and O as the minimizer.
Let us build our evaluation function :

If X wins on the board we give it a positive value of +10.

evaluation_function1

If O wins on the board we give it a negative value of -10.

evaluation_function2

If no one has won or the game results in a draw then we give a value of +0.

evaluation_function3

We could have chosen any positive/negative. For the sake of simplicity, we chose 10 and shall use lowercase ‘x’ and lowercase ‘o’ to represent the players and an underscore ‘_’ to represent a blank space on the board.
If we represent our board as a 3×3 2D character matrix, like char board[3][3]; then we have to check each row, each column, and the diagonals to check if either of the players has gotten 3 in a row.

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 = 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 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
            else if 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
            else if 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
        else if 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
        else if 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 Jain
 
  

C#

   // 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
 
  

Javascript

   <script>
// Javascript 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
function evaluate(b)
{
 
    // Checking for Rows for X or O victory.
    for (let 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 (let 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
let board=[[ 'x', '_', 'o'],
        [ '_', 'x', 'o'],
        [ '_', '_', 'x']];
 
let value = evaluate(board);
document.write("The value of this board is "+ value+"<br>");
 
// This code is contributed by avanitrachhadiya2155
</script>
 
  

Output

The value of this board is 10

Time Complexity: O(max(row,col))
Auxiliary Space: O(1)

The idea of this article is to understand how to write a simple evaluation function for the game Tic-Tac-Toe. In the next article we shall see how to combine this evaluation function with the minimax function. Stay Tuned.
This article is written by Akshay L. Aradhya.