**Prerequisites: **

Sprague Gruncy 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 and valid move is the move that will lead the current player to winning state.

**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 loose 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 move available, grundy number will same as standard nim game. (2) n with zero move available, we can reach above state and other states with zero move 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 is not a valid move, as it will lead the current player into loosing state. Therefore, g(2) = mex{0} which implies that g(2)=1.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.

**Output:**

Player 1 wins

**Time Complexity:** O(n)

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