## Monday, January 2, 2012

### Maximum size square sub-matrix with all 1s

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
```
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]:
(1) First copy the first row and first column as it is from M[][] to S[][]
(2) And for the remaining entries as mentioned do the following:

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

So to sum up:
```If M[i][j] is 1 then
S[i][j] = min(S[i][j-1], S[i-1][j], S[i-1][j-1]) + 1
Else /*If M[i][j] is 0*/
S[i][j] = 0
```

(3) Find the maximum entry in S[][] and use it to construct the maximum size square sub-matrix.

Example - 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.

```#define ROW 10
#define COL 10

void 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] = M[i];

/* Set first row of S[][]*/
for(j = 0; j < COL; j++)
S[j] = M[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; 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.