Question: Given a matrix consisting only 0s and 1s, find the maximum size square sub-matrix with all 1s.

Example: Consider the below matrix.

`   0  1  1  0  1   1  1  0  1  0   0  1  1  1  0   1  1  1  1  0   1  1  1  1  1   0  0  0  0  0`
The maximum square sub-matrix with all '1' bits is from (2,1) to (4,3)
`    1  1  1    1  1  1    1  1  1`
Answer: This is a classic Dynamic Programming problem. Lets calculate the maximum size square sub-matrix as we traverse the original matrix M[][]. We will use a auxiliary matrix S[][] of same size for memoization. S[i][j] represents size of the square sub-matrix with all 1s including M[i][j]. 'i' and 'j' will be the last row and column respectively in square sub-matrix.

How to calculate S[i][j]:
We should note that if M[i][j] is '0' then S[i][j] will obviously be '0'. If M[i][j] is '1' then S[i][j] depends on earlier values.

If M[i][j] is '1' then it will contribute to the all 1s square sub-matrix ending at either M[i][j-1] or M[i-1][j] or M[i-1][j-1]. If we visualize the conditions then, we will see:
S[i][j] = min(S[i][j-1], S[i-1][j], S[i-1][j-1]) + 1

How did we arrive at above relationship?
Note if we include M[i][j] in earlier calculated sub-matrix then we are adding S[i][j] elements from ith row and jth columns. They all should be '1' if we wanna include M[i][j]. On visualizing with some examples, readers will analyze why, minimum of 3 neighbors is taken.

For the given M[][] in above example, constructed S[][] would be:
`   0  1  1  0  1   1  1  0  1  0   0  1  1  1  0   1  1  2  2  0   1  2  2  3  1   0  0  0  0  0`
The value of maximum entry in above matrix is 3 and coordinates of the entry are (4, 3). Using the maximum value and its coordinates, we can find out the required sub-matrix.

Code:
```#define ROW 10#define COL 10void FindMaxSubSquare(bool M[ROW][COL], int &max_i, int &max_j, int &size){    int i,j;    int S[ROW][COL];     /* Set first column of S[][]*/    for(i = 0; i < ROW; i++)        S[i][0] = M[i][0];     /* Set first row of S[][]*/    for(j = 0; j < COL; j++)        S[0][j] = M[0][j];     /* Construct other entries of S[][]*/    for(i = 1; i < ROW; i++)    {        for(j = 1; j < COL; j++)        {            if(M[i][j] == 1)                S[i][j] = min(S[i][j-1], S[i-1][j], S[i-1][j-1]) + 1;            else                S[i][j] = 0;        }    }      /* Find the maximum entry, and indexes of maximum entry in S[][] */    size = S[0][0]; max_i = 0; max_j = 0;    for(i = 0; i < ROW; i++)    {        for(j = 0; j < COL; j++)        {            if(size < S[i][j])            {                size = S[i][j];                max_i = i;                max_j = j;            }        }    }         return}
```

Complexity:
Time Complexity: O(m*n) where m is number of rows and n is number of columns in the given matrix.
Space Complexity: O(m*n) where m is number of rows and n is number of columns in the given matrix.

Note: Part of this post is taken from geeksforgeeks.