Piperoll answers level 94 code#

If we define an almost equilateral triangle as (a,a,c) then c is either a+1 or a-1. The height of the (a,a,c) triangle can be computed using a right triangle formed of half the base (c/2) with a as the hypothenuse. Pythagoreanįor this to produce an integer surface, h and c/2 must be integers and (h,c/2,a) must be a pythagorean triple.īased on this we can go through possible (even) values of c while advancing a corresponding h sequentially.

If abs(2*b-c)=1: # (c,c,2*b) almost equilateral triangle If abs(2*a-c)=1: # (c,c,2*a) almost equilateral triangle Yield m*m-n*n,2*m*n,c # return a,b,c triple If c >= k: n=m continue # skip remaining n when c >= k While h*h =m: n,m = m%2,m+1 # n reached m, advance m, reset n As c increases, so will h which will allow us to simply match h^2 with a^2 - (c/2)^2 to detect pythagorean triples. This is more than 2x faster than the previous approach. The 'unnecessary' numbers are due to floating point precision and rounding error. This is clearly not an integer, yet area % 1 = 0 returns True. You could try using Decimal to increase precision. Or instead, you could use the theorem that an integer is either a perfect square or its square root is an irrational number. For the area to be integral the term inside the square root must be a perfect square.

Piperoll answers level 94 code#

Using a function is_square to check if the term is a square, your code can be corrected and slightly optimized like this: from math import isqrt b must also be an even number, otherwise a**2 - (b/2)**2 cannot be an integer.