# Predict the winner of the game | Sprague-Grundy

Given a 4×4 binary matrix. Two players A and B are playing a game, at each step a player can select any rectangle with all 1’s in it and replace all 1’s with 0. The player that cannot select any rectangle loses the game. Predict the winner off the game assuming that they both play the game optimally and A starts the game.

Examples:

Input :
0 1 1 0
0 0 0 0
0 0 0 0
0 0 0 1
Output : A
Step 1: Player A chooses the rectangle with a single one at position (1, 2), so the new matrix becomes
0 0 1 0
0 0 0 0
0 0 0 0
0 0 0 1

Step 2: Player B chooses the rectangle with a single one at position (1, 3), so the new matrix becomes
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1

Step 3: Player A chooses the rectangle with a single one at position (4, 4), so the new matrix becomes
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0

Step 4: Player B cannot move, hence A wins the game.

Input :
0 0 1 0
0 0 0 0
0 0 0 0
0 0 0 1
Output : B

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The problem can be solved using sprague-grundy theorem. The base case for Sprague-Grundy is Grundy[0] = 0, which is all the positions in the matrix are filled with 0, then B wins it, hence 0. In grundy, recursively we call grundy function with all the states that are possible.

The 4×4 matrix can be represented as a binary 16 bit number which is 65535 in int, where every bit represents the position in a matrix. Below are the steps to solve the above problem.

• Convert the matrix into int val.
• Call the recursive function with val that generates the grundy value using memoization.
• Inside the recursive function, all the grundy states can be visited by generating all possible rectangles(using four for loops).
• Check the generated rectangle, if it is a rectangle of the matrix. Then this is a state to be visited by grundy.
• To get Grundy value using MEX, please see this.
• If the recursion return 0, then player B wins, else player A wins.

Below is the implementation of the above approach

 #include using namespace std;    // Gets the max value int getMex(const unordered_set& s) {     int mex = 0;     while (s.find(mex) != s.end())         mex++;     return mex; }    // Find check if the rectangle is a part of the // the original rectangle int checkOne(int mat, int i, int j, int k, int l) {        // initially create the bitset     // of original intValue     bitset<16> m(mat);        // Check if it is a part of the rectangle     for (int x = i; x <= j; x++) {         for (int y = k; y <= l; y++) {             int pos = 15 - ((x * 4) + y);                // If not set, then not part             if (!m.test(pos)) {                 return -1;             }             m.reset(pos);         }     }        // If part of rectangle     // then convert to int again and return     int res = m.to_ullong();     return res; }    // Recursive function to get the grundy value int getGrundy(int pos, int grundy[]) {        // If state has been visited     if (grundy[pos] != -1)         return grundy[pos];        // For obtaining the MEX value     unordered_set gSet;        // Generate all the possible rectangles     for (int i = 0; i <= 3; i++) {         for (int j = i; j <= 3; j++) {             for (int k = 0; k <= 3; k++) {                 for (int l = k; l <= 3; l++) {                        // check if it is part of the original                     // rectangle, if yes then get the int value                     int res = checkOne(pos, i, j, k, l);                        // If it is a part of original matrix                     if (res != -1) {                            // Store the grundy value                         // Memorize                         grundy[res] = getGrundy(res, grundy);                            // Find MEX                         gSet.insert(grundy[res]);                     }                 }             }         }     }        // Return the MEX     return getMex(gSet); }    // Conver the matrix to INT int toInt(int matrix[4][4]) {     int h = 0;        // Traverse in the matrix     for (int i = 0; i < 4; ++i)         for (int j = 0; j < 4; ++j)             h = 2 * h + matrix[i][j];     return h; }    // Driver Code int main() {     int mat[4][4] = { { 0, 1, 1, 0 },                       { 0, 0, 0, 0 },                       { 0, 0, 0, 0 },                       { 0, 0, 0, 1 } };        // Get the int value of the matrix     int intValue = toInt(mat);        int grundy[intValue + 1];        // Initially with -1     // used for memoization     memset(grundy, -1, sizeof grundy);        // Base case     grundy[0] = 0;        // If returned value is non-zero     if (getGrundy(intValue, grundy))         cout << "Player A wins";     else         cout << "Player B wins";        return 0; }

Output:
Player A wins

Time Complexity: O(N)
Auxiliary Space: O(N)

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