Combinatorial Game Theory | Set 2 (Game of Nim)

3.2

We strongly recommend to refer below article as a prerequisite of this.

Combinatorial Game Theory | Set 1 (Introduction)

In this post, Game of Nim is discussed. The Game of Nim is described by the following rules-

Given a number of piles in which each pile contains some numbers of stones/coins. In each turn, a player can choose only one pile and remove any number of stones (at least one) from that pile. The player who cannot move is considered to lose the game (i.e., one who take the last stone is the winner).

For example, consider that there are two players- A and B, and initially there are three piles of coins initially having 3, 4, 5 coins in each of them as shown below. We assume that first move is made by A. See the below figure for clear understanding of the whole game play.

Set2_Nim_Game_start_with_A

A Won the match (Note: A made the first move)

So was A having a strong expertise in this game ? or he/she was having some edge over B by starting first ?

Let us now play again, with the same configuration of the piles as above but this time B starting first instead of A.

gameofnim11

B Won the match (Note: B made the first move)

By the above figure, it must be clear that the game depends on one important factor – Who starts the game first ?

So does the player who starts first will win everytime ?
Let us again play the game, starting from A , and this time with a different initial configuration of piles. The piles have 1, 4, 5 coins initially.

Will A win again as he has started first ? Let us see.
gameofnim4

A made the first move, but lost the Game.

So, the result is clear. A has lost. But how? We know that this game depends heavily on which player starts first. Thus, there must be another factor which dominates the result of this simple-yet-interesting game. That factor is the initial configuration of the heaps/piles. This time the initial configuration was different from the previous one.

So, we can conclude that this game depends on two factors-

  1. The player who starts first.
  2. The initial configuration of the piles/heaps.

In fact, we can predict the winner of the game before even playing the game !

Nim-Sum : The cumulative XOR value of the number of coins/stones in each piles/heaps at any point of the game is called Nim-Sum at that point.

“If both A and B play optimally (i.e- they don’t make any mistakes), then the player starting first is guaranteed to win if the Nim-Sum at the beginning of the game is non-zero. Otherwise, if the Nim-Sum evaluates to zero, then player A will lose definitely.”

For the proof of the above theorem, see- https://en.wikipedia.org/wiki/Nim#Proof_of_the_winning_formula

Let us apply the above theorem in the games played above. In the first game A started first and the Nim-Sum at the beginning of the game was, 3 XOR 4 XOR 5 = 2, which is a non-zero value, and hence A won. Whereas in the second game-play, when the initial configuration of the piles were 1, 4, and 5 and A started first, then A was destined to lose as the Nim-Sum at the beginning of the game wasv1 XOR 4 XOR 5 = 0 .

Implementation:

In the program below, we play the Nim-Game between computer and human(user)
The below program uses two functions
knowWinnerBeforePlaying() : : Tells the result before playing.
playGame() : plays the full game and finally declare the winner. The function playGame() doesn’t takes input from the human(user), instead it uses a rand() function to randomly pick up a pile and randomly remove any number of stones from the picked pile.

The below program can be modified to take input from the user by removing the rand() function and inserting cin or scanf() functions.

/* A C program to implement Game of Nim. The program
   assumes that both players are playing optimally */
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

#define COMPUTER 1
#define HUMAN 2

/* A Structure to hold the two parameters of a move

 A move has two parameters-
  1) pile_index = The index of pile from which stone is
  				  going to be removed
  2) stones_removed = Number of stones removed from the
  					  pile indexed = pile_index  */
struct move
{
    int pile_index;
    int stones_removed;
};

/*
 piles[] -> Array having the initial count of stones/coins
			in each piles before the game has started.
 n       -> Number of piles

 The piles[] are having 0-based indexing*/

// A C function to output the current game state.
void showPiles (int piles[], int n)
{
    int i;
    printf ("Current Game Status -> ");
    for (i=0; i<n; i++)
        printf ("%d ", piles[i]);
    printf("\n");
    return;
}

// A C function that returns True if game has ended and
// False if game is not yet over
bool gameOver(int piles[], int n)
{
    int i;
    for (i=0; i<n; i++)
        if (piles[i]!=0)
            return (false);

    return (true);
}

// A C function to declare the winner of the game
void declareWinner(int whoseTurn)
{
    if (whoseTurn == COMPUTER)
        printf ("\nHUMAN won\n\n");
    else
        printf("\nCOMPUTER won\n\n");
    return;
}

// A C function to calculate the Nim-Sum at any point
// of the game.
int calculateNimSum(int piles[], int n)
{
    int i, nimsum = piles[0];
    for (i=1; i<n; i++)
        nimsum = nimsum ^ piles[i];
    return(nimsum);
}

// A C function to make moves of the Nim Game
void makeMove(int piles[], int n, struct move * moves)
{
    int i, nim_sum = calculateNimSum(piles, n);

    // The player having the current turn is on a winning
    // position. So he/she/it play optimally and tries to make
    // Nim-Sum as 0
    if (nim_sum != 0)
    {
        for (i=0; i<n; i++)
        {
            // If this is not an illegal move
            // then make this move.
            if ((piles[i] ^ nim_sum) < piles[i])
            {
                (*moves).pile_index = i;
                (*moves).stones_removed =
                                 piles[i]-(piles[i]^nim_sum);
                piles[i] = (piles[i] ^ nim_sum);
                break;
            }
        }
    }

    // The player having the current turn is on losing
    // position, so he/she/it can only wait for the opponent
    // to make a mistake(which doesn't happen in this program
    // as both players are playing optimally). He randomly
    // choose a non-empty pile and randomly removes few stones
    // from it. If the opponent doesn't make a mistake,then it
    // doesn't matter which pile this player chooses, as he is
    // destined to lose this game.

    // If you want to input yourself then remove the rand()
    // functions and modify the code to take inputs.
    // But remember, you still won't be able to change your
    // fate/prediction.
    else
    {
        // Create an array to hold indices of non-empty piles
        int non_zero_indices[n], count;

        for (i=0, count=0; i<n; i++)
            if (piles[i] > 0)
                non_zero_indices [count++] = i;

        (*moves).pile_index = (rand() % (count));
        (*moves).stones_removed =
                 1 + (rand() % (piles[(*moves).pile_index]));
        piles[(*moves).pile_index] =
         piles[(*moves).pile_index] - (*moves).stones_removed;

        if (piles[(*moves).pile_index] < 0)
            piles[(*moves).pile_index]=0;
    }
    return;
}

// A C function to play the Game of Nim
void playGame(int piles[], int n, int whoseTurn)
{
    printf("\nGAME STARTS\n\n");
    struct move moves;

    while (gameOver (piles, n) == false)
    {
        showPiles(piles, n);

        makeMove(piles, n, &moves);

        if (whoseTurn == COMPUTER)
        {
            printf("COMPUTER removes %d stones from pile "
                   "at index %d\n", moves.stones_removed,
                   moves.pile_index);
            whoseTurn = HUMAN;
        }
        else
        {
            printf("HUMAN removes %d stones from pile at "
                   "index %d\n", moves.stones_removed,
                   moves.pile_index);
            whoseTurn = COMPUTER;
        }
    }

    showPiles(piles, n);
    declareWinner(whoseTurn);
    return;
}

void knowWinnerBeforePlaying(int piles[], int n,
                             int whoseTurn)
{
    printf("Prediction before playing the game -> ");

    if (calculateNimSum(piles, n) !=0)
    {
        if (whoseTurn == COMPUTER)
            printf("COMPUTER will win\n");
        else
            printf("HUMAN will win\n");
    }
    else
    {
        if (whoseTurn == COMPUTER)
            printf("HUMAN will win\n");
        else
            printf("COMPUTER will win\n");
    }

    return;
}

// Driver program to test above functions
int main()
{
    // Test Case 1
    int piles[] = {3, 4, 5};
    int n = sizeof(piles)/sizeof(piles[0]);

    // We will predict the results before playing
    // The COMPUTER starts first
    knowWinnerBeforePlaying(piles, n, COMPUTER);

    // Let us play the game with COMPUTER starting first
    // and check whether our prediction was right or not
    playGame(piles, n, COMPUTER);

    /*
    Test Case 2
    int piles[] = {3, 4, 7};
    int n = sizeof(piles)/sizeof(piles[0]);

    // We will predict the results before playing
    // The HUMAN(You) starts first
    knowWinnerBeforePlaying (piles, n, COMPUTER);

    // Let us play the game with COMPUTER starting first
    // and check whether our prediction was right or not
    playGame (piles, n, HUMAN);    */

    return(0);
}

Output : May be different on different runs as random numbers are used to decide next move (for the loosing player).

Prediction before playing the game -> COMPUTER will win

GAME STARTS

Current Game Status -> 3 4 5 
COMPUTER removes 2 stones from pile at index 0
Current Game Status -> 1 4 5 
HUMAN removes 3 stones from pile at index 1
Current Game Status -> 1 1 5 
COMPUTER removes 5 stones from pile at index 2
Current Game Status -> 1 1 0 
HUMAN removes 1 stones from pile at index 1
Current Game Status -> 1 0 0 
COMPUTER removes 1 stones from pile at index 0
Current Game Status -> 0 0 0 

COMPUTER won

References :
https://en.wikipedia.org/wiki/Nim

This article is contributed by Rachit Belwariar. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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