Variation in Nim Game

15 Jun, 2022
Prerequisites: Sprague Gruncy theorem Grundy NumbersNim is a famous game in which two players take turns removing items from distinct piles. During each turn, a player must remove one or more items from a single, non-empty pile. The winner of the game is whichever player removes the last item from the last non-empty pile. 
Now, For each non-empty pile, either player can remove zero items from that pile and have it count as their move; however, this move can only be performed once per pile by either player. Given the number of items in each pile, determine who will win the game; Player 1 or player 2. If player 1 starts the game and both plays optimally.
Examples: 
Input : 3 [18, 47, 34]
Output : Player 2 wins
G = g(18)^g(47)^g(34) = (17)^(48)^(33) = 0
Grundy number(G), for this game is zero.
Player 2 wins. 

Input : 3 [32, 49, 58]
Output : Player 1 wins
G = g(31)^g(50)^g(57) = (17)^(48)^(33) = 20
Grundy number(G), for this game is non-zero.
Player 1 wins. 
Approach: Grundy number for each pile is calculated based on the number of stones.To compensate the zero move we will have to modify grundy values we used in standard nim game. If pile size is odd; grundy number is size+1 and if pile size is even; grundy number is size-1. We XOR all the grundy number values to check if final Grundy number(G) of game is non zero or not to decide who is winner of game.
Explanation: Grundy number of a state is the smallest positive integer that cannot be reached in one valid move. So, we need to calculate mex value for each n, bottom up wise so that we can induce the grundy number for each n. where n is the pile size.
Winning state: A tuple of values from where the current player will win the game no matter what opponent does. (If G!=0) Losing state: A tuple of values from where the current player will lose the game no matter what opponent does. (If G=0)  
For a given pile size n, we have two states:
(1) n with no zero moves available, 
grundy number will same as standard nim game.
(2) n with zero moves available, we can
reach above state and other states with 
zero moves remaining. 

For, n = 0, g(0) = 0, empty pile

For, n = 1, we can reach two states:
(1) n = 0 (zero move not used)
(2) n = 1 (zero move used)   
Therefore, g(1) = mex{0, 1} which implies
that g(1)=2.

For, n = 2, we can reach :
(1) n = 0 (zero move not used) state 
because this is a valid move.
(2) n = 1 (zero move not used) is a valid
 move, whose grundy number is 2.
Therefore, g(2) = mex{0,2} which implies 
that g(2)=1. 
note that n=1 (zero move used) is not a valid
move.

If we try to build a solution bottom-up
like this, it turns out that if n is even, 
the grundy number is n - 1 and when it is odd,
the grundy is n + 1.
Below is the implementation of above approach: 
C++
Java
Python3
C#
PHP
Javascript
// CPP program for the variation// in nim game#include <bits/stdc++.h>using namespace std; // Function to return final// grundy Number(G) of gameint solve(int p[], int n){    int G = 0;    for (int i = 0; i < n; i++) {         // if pile size is odd        if (p[i] & 1)                        // We XOR pile size+1            G ^= (p[i] + 1);         else // if pile size is even             // We XOR pile size-1            G ^= (p[i] - 1);    }    return G;} // driver programint main(){    // Game with 3 piles    int n = 3;     // pile with different sizes    int p[3] = { 32, 49, 58 };     // Function to return result of game    int res = solve(p, n);     if (res == 0) // if G is zero        cout << "Player 2 wins";    else // if G is non zero        cout << "Player 1 wins";     return 0;}
// Java program for the variation// in nim gameclass GFG {         // Function to return final    // grundy Number(G) of game    static int solve(int p[], int n)    {                 int G = 0;        for (int i = 0; i < n; i++) {                 // if pile size is odd            if (p[i]%2!=0)                                 // We XOR pile size+1                G ^= (p[i] + 1);                 else // if pile size is even                     // We XOR pile size-1                G ^= (p[i] - 1);        }                 return G;    }         //Driver code    public static void main (String[] args)    {                 // Game with 3 piles        int n = 3;             // pile with different sizes        int p[] = { 32, 49, 58 };             // Function to return result of game        int res = solve(p, n);             if (res == 0) // if G is zero            System.out.print("Player 2 wins");        else // if G is non zero            System.out.print("Player 1 wins");    }} // This code is contributed by Anant Agarwal.
# Python3 program for the# variation in nim game # Function to return final# grundy Number(G) of gamedef solve(p, n):    G = 0    for i in range(n):         # if pile size is odd        if (p[i] % 2 != 0):                         # We XOR pile size+1            G ^= (p[i] + 1)                 # if pile size is even        else:             # We XOR pile size-1            G ^= (p[i] - 1)         return G # Driver code # Game with 3 pilesn = 3 # pile with different sizesp = [32, 49, 58] # Function to return result of gameres = solve(p, n) if (res == 0): # if G is zero    print("Player 2 wins")     else: # if G is non zero    print("Player 1 wins")     # This code is contributed by Anant Agarwal.
// C# program for the variation// in nim gameusing System;class GFG {     // Function to return final    // grundy Number(G) of game    static int solve(int[] p, int n)    {         int G = 0;        for (int i = 0; i < n; i++) {             // if pile size is odd            if (p[i] % 2 != 0)                 // We XOR pile size+1                G ^= (p[i] + 1);             else // if pile size is even                 // We XOR pile size-1                G ^= (p[i] - 1);        }         return G;    }     // Driver code    public static void Main()    {         // Game with 3 piles        int n = 3;         // pile with different sizes        int[] p = { 32, 49, 58 };         // Function to return result of game        int res = solve(p, n);         if (res == 0) // if G is zero            Console.WriteLine("Player 2 wins");                     else // if G is non zero            Console.WriteLine("Player 1 wins");    }} // This code is contributed by vt_m.
<?php// php program for the variation// in nim game // Function to return final// grundy Number(G) of gamefunction solve($p,$n){    $G = 0;    for ($i = 0; $i < $n; $i++)    {         // if pile size is odd        if ($p[$i] & 1)                         // We XOR pile size+1            $G ^= ($p[$i] + 1);         else // if pile size is even             // We XOR pile size-1            $G ^= ($p[$i] - 1);    }    return $G;}     // Driver Code    // Game with 3 piles    $n = 3;     // pile with different sizes    $p= array( 32, 49, 58 );     // Function to return result of game    $res = solve($p, $n);     if ($res == 0) // if G is zero        echo "Player 2 wins";    else // if G is non zero        echo "Player 1 wins"; // This code is contributed by mits?>
<script> // Javascript program for the variation// in nim game // Function to return final// grundy Number(G) of gamefunction solve(p, n){    let G = 0;    for(let i = 0; i < n; i++)    {                 // If pile size is odd        if (p[i] % 2 != 0)                           // We XOR pile size+1            G ^= (p[i] + 1);                     // If pile size is even        else               // We XOR pile size-1            G ^= (p[i] - 1);    }    return G;}   // Driver code // Game with 3 pileslet n = 3; // Pile with different sizeslet p = [ 32, 49, 58 ]; // Function to return result of gamelet res = solve(p, n); // If G is zeroif (res == 0)    document.write("Player 2 wins");     // If G is non zeroelse    document.write("Player 1 wins");     // This code is contributed by sanjoy_62 </script>
Output: 
Player 1 wins
Time Complexity: O(n)Space Complexity: O(n) 
Mithun Kumar
Akanksha_Rai
ad_1608
sanjoy_62
AshokJaiswal
ranjanrohit840
Bit Magic
Competitive Programming
Game Theory
Technical Scripter
Bit Magic
Game Theory
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.