Part 1 of this series on using floats in python can be found here

This is a short story of a problem where using floats was just a bad idea in the first place.

The Motivation

I was asked to check whether the points of a triangle determined a right triangle or not. The points of the triangles all lied at nonnegative integer coordinates, with one of them being located at the origin, (0,0).

triangle

My solution boiled down to calculating the slopes of each side of the triangle and determining if any pair of them were perpendicular.

>>> p1, p2 = (0,0), (1,1) # first line segment
>>> q1, q2 = (1,1), (1,2) # second line segment
>>> slope1, slope2 = slope(p1,p2), slope(q1,q2)
>>> are_perp(slope1, slope2)
False

My first attempt tried using floats as an intermediary representation of the slope of these segments…

def slope(a,b):
	"""
	return slope from a = (x1,y1) to b = (x2,y2) as a decimal
	a,b assumed to be distinct points
	"""
	x1,y1 = a
	x2,y2 = b
	if x1 == x2:
		return float('inf')
	else:
		return 1.*(y2-y1)/(x2-x1)

def are_perp(m1,m2):
	"""
	return whether m1,m2 as slopes are perpendicular (inputs are decimals)
	"""
	if m2 == 0:
		return m1 == float('inf') or m1 == -float('inf')
	else:
		return -1./m2 == m1

Which passes some sanity checks…

>>> origin, p,q = (0,0), (9,10), (-10,9)
>>> m1, m2 = slope(origin,p), slope(origin,q)
>>> are_perp(m1, m2)
True

However, there were some triangles with segments that were almost perpendicular, which didn’t bode well with this use of floats…

The Problem

>>> m1 = slope((3,11),(0,0))
>>> m2 = slope((3,11),(14,8))
>>> are_perp(m1, m2)
False

Yet, these segments are perpendicular because the ratios are negative reciprocals m1 == 11/3. and m2 == -3/11..

A Solution

In part 1 I made use of an arbitrary precision library, decimal to solve a problem caused by imprecise floating-point arithmetic in python. Unfortunately, even using arbirtrary precision here will not solve this problem. The standard answer (at least in Python) for checking equality of floats is actuall to check if they are close enough, which is time-consuming to define properly…

This is a good case where it would be nice to avoid dealing with decimals in the first place, and in fact this problem lends it self easily to another representation: fractions (as tuples of integers).

That is, we represent our slopes not as decimals but as tuples of integers so m1 = 11/3. becomes m1 == (11, 3) (numerator, denomenator). And then I could confirm perpendicularity without ever entering the world of decimals…

def slope(a,b):
	"""
	return the slope from point a to b as a tuple (numerator, denominator)
	a,b assumed to be distinct points
	"""
	x1,y1 = a
	x2,y2 = b
	return (y2-y1, x2-x1)

def are_perp(m1,m2):
	"""
	check whether m1, m2 as slopes are perpendicular (inputs as tuples)
	"""
	n1,d1 = m1
	n2,d2 = m2
	return (n1 * n2) + (d1 * d2) == 0

This required refactoring other functions to correspond to this representation, but complexity avoided in dealing with floats is well worth the extra work with tuples…