Question

In Python 3, I am checking whether a given value is triangular, that is, it can be represented as n * (n + 1) / 2 for some positive integer n.

Can I just write:

import math

def is_triangular1(x):
    num = (1 / 2) * (math.sqrt(8 * x + 1) - 1)
    return int(num) == num

Or do I need to do check within a tolerance instead?

epsilon = 0.000000000001
def is_triangular2(x):
    num = (1 / 2) * (math.sqrt(8 * x + 1) - 1)
    return abs(int(num) - num) < epsilon

I checked that both of the functions return same results for x up to 1,000,000. But I am not sure if generally speaking int(x) == x will always correctly determine whether a number is integer, because of the cases when for example 5 is represented as 4.99999999999997 etc.

As far as I know, the second way is the correct one if I do it in C, but I am not sure about Python 3.

Answer 1

You’ll want to do the latter. In Programming in Python 3 the following example is given as the most accurate way to compare

def equal_float(a, b):
    #return abs(a - b) <= sys.float_info.epsilon
    return abs(a - b) <= chosen_value #see edit below for more info

Also, since epsilon is the “smallest difference the machine can distinguish between two floating-point numbers”, you’ll want to use <= in your function.

Edit: After reading the comments below I have looked back at the book and it specifically says “Here is a simple function for comparing floats for equality to the limit of the machines accuracy”. I believe this was just an example for comparing floats to extreme precision but the fact that error is introduced with many float calculations this should rarely if ever be used. I characterized it as the “most accurate” way to compare in my answer, which in some sense is true, but rarely what is intended when comparing floats or integers to floats. Choosing a value (ex: 0.00000000001) based on the “problem domain” of the function instead of using sys.float_info.epsilon is the correct approach.

Thanks to S.Lott and Sven Marnach for their corrections, and I apologize if I led anyone down the wrong path.