Check if matrix A can be converted to B by changing parity of corner elements of any submatrix in Python

PythonServer Side ProgrammingProgramming

Suppose we have two N X M binary matrices A and B. In a single operation, we can select a sub-matrix (at least 2x2) and convert the parity of the corner elements (flip bits). Finally, we have to check whether the matrix A can be converted to B by performing any number of operations or not.

So, if the input is like

100
101
100


then the output will be True as we can perform the operation on the top left square sub-matrix of size (2x2) on mat1 to get mat2.

To solve this, we will follow these steps −

  • row := row count of mat1
  • column := column count of mat1
  • for i in range 1 to row - 1, do
    • for j in range 1 to column - 1, do
      • if mat1[i, j] is not same as mat2[i, j], then
        • mat1[i, j] := mat1[i, j] XOR 1
        • mat1[0, 0] := mat1[0, 0] XOR 1
        • mat1[0, j] := mat1[0, j] XOR 1
        • mat1[i, 0] := mat1[i, 0] XOR 1
  • for i in range 0 to row - 1, do
    • for j in range 0 to column - 1, do
      • if mat1[i, j] is not same as mat2[i, j], then
        • return False
  • return True

Example

Let us see the following implementation to get better understanding −

 Live Demo

def solve(mat1, mat2):
   row = len(mat1)
   column = len(mat1[0])
   for i in range(1, row):
      for j in range(1, column):
         if mat1[i][j] != mat2[i][j]:
            mat1[i][j] ^= 1
            mat1[0][0] ^= 1
            mat1[0][j] ^= 1
            mat1[i][0] ^= 1
   for i in range(row):
      for j in range(column):
         if mat1[i][j] != mat2[i][j]:
            return False
   return True
mat1 = [
         [1, 0, 0],
         [1, 0, 1],
         [1, 0, 0]]
mat2 = [
         [0, 1, 0],
         [0, 1, 1],
         [1, 0, 0]]
print(solve(mat1, mat2))

Input

[
   [1, 0, 0],
   [1, 0, 1],
   [1, 0, 0]],
[
   [0, 1, 0],
   [0, 1, 1],
   [1, 0, 0]]

Output

True
raja
Published on 19-Jan-2021 09:21:26
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