Counting Squares

Count how many squares are there in the following figure.

Easy validation by Python using Bruteforce, checking every possible pair of points (lower-left, and upper right)

#!/usr/bin/env python
# https://helloacm.com

def count(pts):
    pts.sort()
    cnt = 0
    sz = len(pts)
    for x in xrange(sz):
        u = pts[x]
        for y in xrange(x + 1, sz):
            v = pts[y]
            if v[0] - u[0] == v[1] - u[1]:
                cnt += 1
                print u, v
    return cnt

def compute():
    pts = []
    for x in xrange(5):
        for y in xrange(5):
            pts.append((x, y))
    cnt1 = count(pts)    
    pts = []
    for x in xrange(3):
        for y in xrange(3):
            pts.append((1.5 + x * 0.5, 0.5 + y * 0.5))
    cnt2 = count(pts)
    pts = []
    for x in xrange(3):
        for y in xrange(3):
            pts.append((1.5 + x * 0.5, 2.5 + y * 0.5))
    cnt3 = count(pts)    
    print "total = ", cnt1 + cnt2 + cnt3

if __name__ == "__main__":
    compute()

The Output is:

(0, 0) (1, 1)
(0, 0) (2, 2)
(0, 0) (3, 3)
(0, 0) (4, 4)
(0, 1) (1, 2)
(0, 1) (2, 3)
(0, 1) (3, 4)
(0, 2) (1, 3)
(0, 2) (2, 4)
(0, 3) (1, 4)
(1, 0) (2, 1)
(1, 0) (3, 2)
(1, 0) (4, 3)
(1, 1) (2, 2)
(1, 1) (3, 3)
(1, 1) (4, 4)
(1, 2) (2, 3)
(1, 2) (3, 4)
(1, 3) (2, 4)
(2, 0) (3, 1)
(2, 0) (4, 2)
(2, 1) (3, 2)
(2, 1) (4, 3)
(2, 2) (3, 3)
(2, 2) (4, 4)
(2, 3) (3, 4)
(3, 0) (4, 1)
(3, 1) (4, 2)
(3, 2) (4, 3)
(3, 3) (4, 4)
(1.5, 0.5) (2.0, 1.0)
(1.5, 0.5) (2.5, 1.5)
(1.5, 1.0) (2.0, 1.5)
(2.0, 0.5) (2.5, 1.0)
(2.0, 1.0) (2.5, 1.5)
(1.5, 2.5) (2.0, 3.0)
(1.5, 2.5) (2.5, 3.5)
(1.5, 3.0) (2.0, 3.5)
(2.0, 2.5) (2.5, 3.0)
(2.0, 3.0) (2.5, 3.5)
total =  40

–EOF (The Ultimate Computing & Technology Blog) —