Finding maximum size sub-matrix of all 1's in a matrix having 1's and 0's

Question

Suppose you are given an mXn bitmap, represented by an array M[1..m,1.. n] whose entries are all 0 or 1. A all-one block is a subarray of the form M[i .. i0, j .. j0] in whi

Answer 1

Define a new matrix A wich will store in A[i,j] two values: the width and the height of the largest submatrix with the left upper corner at i,j, fill this matrix starting from the bottom right corner, by rows bottom to top. You'll find four cases: Perform these cases when given matrix at [i,j]=1

case 1: none of the right or bottom neighbour elements in the original matrix are equal to the current one, i.e: M[i,j] != M[i+1,j] and M[i,j] != M[i,j+1] being M the original matrix, in this case, the value of A[i,j] is 1x1

case 2: the neighbour element to the right is equal to the current one but the bottom one is different, the value of A[i,j].width is A[i+1,j].width+1 and A[i,j].height=1

case 3: the neighbour element to the bottom is equal but the right one is different, A[i,j].width=1, A[i,j].height=A[i,j+1].height+1

case 4: both neighbours are equal: Three rectangles are considered:

1. A[i,j].width=A[i,j+1].width+1; A[i,j].height=1;

2. A[i,j].height=A[i+1,j].height+1; a[i,j].width=1;

3. A[i,j].width = min(A[i+1,j].width+1,A[i,j+1].width) and A[i,j].height = min(A[i,j+1]+1,A[i+1,j])

The one with the max area in the above three cases will be considered to represent the rectangle at this position.

The size of the largest matrix that has the upper left corner at i,j is A[i,j].width*A[i,j].height so you can update the max value found while calculating the A[i,j]

the bottom row and the rightmost column elements are treated as if their neighbours to the bottom and to the right respectively are different.