How to recognize a square in a bitmap? - algorithm

Question

I came across this interview question:
In a N x N bi-dimensional array of boolean elements, how do you determine if the values form a square?
For example:
true true true true
true false false true
true false false true
true true true true
form a square.
I figured that I have to start by checking if there is a square in the middle (if N is odd that is always true) and then recursively checking the values at the perimeter.
Is this the best way to do it or is there a better, faster, way to find out?

Answer 1

A squre could be determined by two points. Let's say the left-top point (x1,y1) and the right-bottom point (x2,y2). And, let's use 1 as true, and 0 as false.
Consider an array:
array = [None] * 5
array[0] = [1, 1, 1, 1, 0]
array[1] = [1, 0, 0, 1, 0]
array[2] = [1, 0, 0, 1, 0]
array[3] = [1, 1, 1, 1, 0]
array[4] = [1, 0, 0, 1, 0]
It's obviously that (0,0)(3,3) forms a square in this case. And we could find a property that:
A square is formed if and only if:
By adding the two row borders together, you will get a sequence of 2;
The length of the sequence is equal to the distance between to two row borders.
By adding the two column borders together, you will get a sequence of 2;
The length of the sequence is equal to the distance between to two column borders.
Exploiting the property above, you would get an algorithm:
row_segment = []
col_segment = []
for v1 in range(len(array)):
for v2 in range(v1+1, len(array)):
add_row = [array[v1][col]+array[v2][col] for col in range(len(array))]
add_col = [array[row][v1]+array[row][v2] for row in range(len(array))]
row_distance = v2-v1
row_sum = sum(add_row[:row_distance+1])
col_sum = sum(add_col[:row_distance+1])
for i in range(len(array)-row_distance):
j = i+row_distance
if row_sum == 2*(row_distance+1):
row_segment.append([v1, i, v2, j])
if col_sum == 2*(row_distance+1):
col_segment.append([i, v1, j, v2])
row_sum = row_sum - add_row[i] + add_row[j+1] if j+1 < len(array) else None
col_sum = col_sum - add_col[i] + add_col[j+1] if j+1 < len(array) else None
for i in row_segment:
if i in col_segment:
print "Square ({x1}, {y1}) ({x2}, {y2})".format(x1=i[0], y1=i[1], x2=i[2], y2=i[3])
Let's run some tests:
Test 1:
0 0 0 0 0
0 0 1 1 1
0 0 1 0 1
0 0 1 1 1
0 0 0 0 0
Square (1, 2) (3, 4)
Test 2:
0 0 0 0 0
1 1 1 1 1
1 0 1 0 1
1 1 1 0 1
0 0 1 1 1
Square (1, 0) (3, 2)
Test 3:
0 0 0 0 0
1 1 1 1 1
1 0 1 0 1
1 1 1 1 1
0 0 0 0 0
Square (1, 0) (3, 2)
Square (1, 2) (3, 4)
Test 4:
1 1 1 1 1
1 0 1 0 1
1 1 0 0 1
1 0 0 1 0
1 1 1 1 1
No squares found