Dynamic programming - Largest square block

Question

I need to find the largest square of 1's in a giant file full of 1's and 0's. I know i have to use dynamic programming. I am storing it in a 2D array. Any help with the algorithm to find the largest square would be great, thanks!

example input:

1 0 1 0 1 0 1 0 1 1 1 1 0 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 

answer:

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 

My code so far:

int Square (Sq[int x][int y]) {    if (Sq[x][y]) == 0) {        return 0;    }    else {        return 1+MIN( Sq(X-1,Y), Sq(X,Y-1), Sq(X-1,Y-1) );    } } 

(assuming values already entered into the array)

int main() {     int Sq[5][6]; //5,6 = bottom right conner     int X = Square(Sq[5][6]); } 

How do I go on from there?

Answer 1

Here is a sketch of the solution:

For each of the cells we will keep a counter of how big a square can be made using that cell as top left. Clearly all cells with 0 will have 0 as the count.

Start iterating from bottom right cell and go to bottom left, then go to one row up and repeat.

At each scan do this:

1. If the cell has 0 assign count=0
2. If the cell has 1 and is an edge cell (bottom or right edge only), assign count=1
3. For all other cells, check the count of the cell on its right, right-below, and below. Take the min of them and add 1 and assign that to the count. Keep a global max_count variable to keep track of the max count so far.

At the end of traversing the matrix, max_count will have the desired value.

Complexity is no more that the cost of traversal of the matrix.

This is how the matrix will look like after the traversal. Values in parentheses are the counts, i.e. biggest square that can be made using the cell as top left.

1(1) 0(0) 1(1) 0(0) 1(1) 0(0) 1(1) 0(0) 1(4) 1(3) 1(2) 1(1) 0(0) 1(1) 1(3) 1(3) 1(2) 1(1) 0(0) 0(0) 1(2) 1(2) 1(2) 1(1) 1(1) 1(1) 1(1) 1(1) 1(1) 1(1) 

Implementation in Python

def max_size(mat, ZERO=0):     """Find the largest square of ZERO's in the matrix `mat`."""     nrows, ncols = len(mat), (len(mat[0]) if mat else 0)     if not (nrows and ncols): return 0 # empty matrix or rows     counts = [[0]*ncols for _ in xrange(nrows)]     for i in reversed(xrange(nrows)):     # for each row         assert len(mat[i]) == ncols # matrix must be rectangular         for j in reversed(xrange(ncols)): # for each element in the row             if mat[i][j] != ZERO:                 counts[i][j] = (1 + min(                     counts[i][j+1],  # east                     counts[i+1][j],  # south                     counts[i+1][j+1] # south-east                     )) if i < (nrows - 1) and j < (ncols - 1) else 1 # edges     return max(c for rows in counts for c in rows) 

Answer 2

LSBRA(X,Y) means "Largest Square with Bottom-Right At X,Y"

Pseudocode:

LSBRA(X,Y):    if (x,y) == 0:        0    else:        1+MIN( LSBRA(X-1,Y), LSBRA(X,Y-1), LSBRA(X-1,Y-1) ) 

(For edge cells, you can skip the MIN part and just return 1 if (x,y) is not 0.)

Work diagonally through the grid in "waves", like the following:

    0 1 2 3 4   +---------- 0 | 1 2 3 4 5 1 | 2 3 4 5 6 2 | 3 4 5 6 7 3 | 4 5 6 7 8 

or alternatively, work through left-to-right, top-to-bottom, as long as you fill in edge cells.

    0 1 2 3 4   +---------- 0 | 1 2 3 4 5 1 | 6 7 8 9 . 2 | . . . . . 3 | . . . . . 

That way you'll never run into a computation where you haven't previously computed the necessary data - so all of the LSBRA() "calls" are actually just table lookups of your previous computation results (hence the dynamic programming aspect).

Why it works

In order to have a square with a bottom-right at X,Y - it must contain the overlapping squares of one less dimension that touch each of the other 3 corners. In other words, to have

XXXX XXXX XXXX XXXX 

you must also have...

XXX.    .XXX    ....    .... XXX.    .XXX    XXX.    .... XXX.    .XXX    XXX.    .... ....    ....    XXX.    ...X 

As long as you have those 3 (each of the LSBRA checks) N-size squares plus the current square is also "occupied", you will have an (N+1)-size square.