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Predict the winner of the game | Sprague-Grundy

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  • Difficulty Level : Hard
  • Last Updated : 14 Jun, 2022

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 of 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

Approach: The problem can be solved using the 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 

CPP




#include <bits/stdc++.h>
using namespace std;
 
// Gets the max value
int getMex(const unordered_set<int>& 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<int> 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);
}
 
// Convert 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(N2), we are using recursion which will cost us O(81*N) and we are also using nested loops to traverse the matrix which will cost us O(N*N) time.

Auxiliary Space: O(N2), we are using extra space for the array grundy which will be of size N*N in the worst case.


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