A = 5
B = 3
abs(A - B)

import math
from Graphics import Point

A = Point(1, 3)
B = Point(2, 5)
math.sqrt( (A.x - B.x) ** 2 + (A.y - B.y) ** 2 )

def distance(s1, s2):                                # s1 is string1, s2 is string2
    if len(s1) == 0:                                 # base case, nothing left to do
        return len(s2)                               # except return length of other
    elif len(s2) == 0:                               # base case, nothing left to do
        return len(s1)                               # except return length of other
    if s1[0] == s2[0]:                               # same
        return distance(s1[1:], s2[1:]) + 0          #   return 0 + rest
    else:                                            # difference! Take minimum of:
        return min( distance(s1[1:], s2[1:]) + 1,    #   edit, return 1 + rest
                    distance(s1, s2[1:]) + 1,        #   insert, return 1 + rest
                    distance(s1[1:], s2) + 1)        #   delete, return 1 + rest


distance("apple", "apple")

distance("apple", "aple")

distance("abcdefghijklmnopqrstuvwxyz", "abcdefghijklmnopqrstuvwxyzA")

distance("abcdefghijklmnopqrstuvwxyz", "Aabcdefghijklmnopqrstuvwxyz")

def distance1(s1, s2):
    if len(s1) == 0 or len(s2) == 0: return len(s2) + len(s1)
    cost = 0 if (s1[0] == s2[0]) else 1
    return min( distance1(s1[1:], s2[1:]) + cost,
                distance1(s1, s2[1:]) + 1,
                distance1(s1[1:], s2) + 1)


def memoize(f):
    cache = {}
    def m(*args):
        if args not in cache:
            cache[args] = f(*args)
        return cache[args]
    return m


@memoize
def distance(s1, s2):
    if len(s1) == 0 or len(s2) == 0: return len(s2) + len(s1)
    cost = 0 if (s1[0] == s2[0]) else 1
    return min( distance(s1[1:], s2[1:]) + cost,
                distance(s1, s2[1:]) + 1,
                distance(s1[1:], s2) + 1)


distance("abcdefghijklmnopqrstuvwxyz", "Aabcdefghijklmnopqrstuvwxyz")

def distance(s1, s2):
    if len(s1) == 0: return 0
    cost = 0 if s1[0] == s2[0] else 1
    return cost + distance(s1[1:], s2[1:])

def distance(s1, s2):
    cost = 0
    for i in range(len(s1)):
        cost += 0 if s1[i] == s2[i] else 1
    return cost