Combinatorial Game Theory  Set 2 (Game of Nim)
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.
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.
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.
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 simpleyetinteresting 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
 The player who starts first.
 The initial configuration of the piles/heaps.
In fact, we can predict the winner of the game before even playing the game !
NimSum : The cumulative XOR value of the number of coins/stones in each piles/heaps at any point of the game is called NimSum 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 NimSum at the beginning of the game is nonzero. Otherwise, if the NimSum 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
Optimal Strategy :

Couple of deductions about bitwise XOR necessary for understanding the Optimal Strategy:
 If the XOR sum of ‘n’ numbers is already zero then there is no possibility to make the XOR sum zero by single reduction of a number.
 If the XOR sum of ‘n’ numbers is nonzero then there is at least a single approach by which if you reduce a number, the XOR sum is zero.
Initially two cases could exist.
Case 1: Initial Nim Sum is zero
As we know, in this case if played optimally B wins, which means B would always prefer to have Nim sum of zero for A‘s turn.
So, as the Nim Sum is initially zero, whatever number of items A removes the new Nim Sum would be nonzero (as mentioned above). Also, as B would prefer Nim sum of zero for A‘s turn, he would then play a move so as to make the Nim Sum zero again (which is always possible, as mentioned above).
The game will run as long as there are items in any of the piles and in each of their respective turns A would make Nim sum nonzero and B would make it zero again and eventually there will be no elements left and B being the one to pick the last wins the game.
It is evident by above explanation that the optimal strategy for each player is to make the Nim Sum for his opponent zero in each of their turn, which will not be possible if it’s already zero.
Case 2: Initial Nim Sum is nonzero
Now going by the optimal approach A would make the Nim Sum to be zero now (which is possible as the initial Nim sum is nonzero, as mentioned above). Now, in B‘s turn as the nim sum is already zero whatever number B picks, the nim sum would be nonzero and A can pick a number to make the nim sum zero again. This will go as long as there are items available in any pile.
And A will be the one to pick the last item.
So, as discussed in the above cases, it should be obvious now that Optimal strategy for any player is to make the nim sum zero if it’s nonzero and if it is already zero then whatever moves the player makes now, it can be countered.
Let us apply the above theorem in the games played above. In the first game A started first and the NimSum at the beginning of the game was, 3 XOR 4 XOR 5 = 2, which is a nonzero value, and hence A won. Whereas in the second gameplay, when the initial configuration of the piles were 1, 4, and 5 and A started first, then A was destined to lose as the NimSum at the beginning of the game was 1 XOR 4 XOR 5 = 0 .
Implementation:
In the program below, we play the NimGame 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 0based 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 NimSum 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 // NimSum 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 nonempty 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 nonempty 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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Recommended Posts:
 Combinatorial Game Theory  Set 1 (Introduction)
 Combinatorial Game Theory  Set 3 (Grundy Numbers/Nimbers and Mex)
 Combinatorial Game Theory  Set 4 (Sprague  Grundy Theorem)
 The prisoner's dilemma in Game theory
 Minimax Algorithm in Game Theory  Set 1 (Introduction)
 Minimax Algorithm in Game Theory  Set 5 (Zobrist Hashing)
 Minimax Algorithm in Game Theory  Set 2 (Introduction to Evaluation Function)
 Minimax Algorithm in Game Theory  Set 4 (AlphaBeta Pruning)
 Minimax Algorithm in Game Theory  Set 3 (TicTacToe AI  Finding optimal move)
 Game Theory in Balanced Ternary Numeral System  (Moving 3k steps at a time)
 A modified game of Nim
 Variation in Nim Game
 Implementation of TicTacToe game
 Check if the game is valid or not
 Find the winner in nimgame
Improved By : prafull911