Variation in Nim Game

Prerequisites: 
Sprague Grundy theorem
Grundy Numbers
Nim 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++

// CPP program for the variation
// in nim game
#include <bits/stdc++.h>
using namespace std;

// Function to return final
// grundy Number(G) of game
int 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 program
int 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

// Java program for the variation
// in nim game
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 (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

# Python3 program for the
# variation in nim game

# Function to return final
# grundy Number(G) of game
def 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 piles
n = 3

# pile with different sizes
p = [32, 49, 58]

# Function to return result of game
res = 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#

// C# program for the variation
// in nim game
using 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
// php program for the variation
// in nim game

// Function to return final
// grundy Number(G) of game
function 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
?>

Javascript

<script>

// Javascript program for the variation
// in nim game

// Function to return final
// grundy Number(G) of game
function 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 piles
let n = 3;

// Pile with different sizes
let p = [ 32, 49, 58 ];

// Function to return result of game
let res = solve(p, n);

// If G is zero
if (res == 0)
    document.write("Player 2 wins");

// If G is non zero
else
    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)