# Project Euler: Problem 9

Published: Thu 06 January 2011

In blog.

Happy New Year Ladies(?) and Geeks

Trying to get the year off to a good start, I'm posting up my solution for the ninth Project Euler problem. I'm not going to spend a lot of time talking about it, because the solution is a pretty easy one. Here is the code in Python:

 ```1 2 3 4 5 6 7``` ```#!/usr/bin/python3 """ python solution for project euler problem #9. """ print( [a*b*(1000 - b -a) for a in range(1,500+1) for b in range(1,500+1) \ if a * a + b * b == ((1000 -b -a) * (1000 - b - a))][0]) ```

And here is the code in Haskell:

 ```1 2 3 4 5``` ```module Main where main :: IO() main = print . head \$ [a*b*(1000-b-a) | a <- [1..500] , b <- [1..500], a ^ 2 + b ^ 2 == (1000 - b - a) ^ 2] ```

See, it really is that simple. There really isn't anything interesting between to the solutions, but I would like to make a quick note on the luxury of being able to use “head” in Haskell to simplify the whole process. In the Python solution, the answer is generated twice, that's just the nature of the algorithm, and to just get one number, I just ask for the first item in the list. Thanks to Haskell's lazy evaluation, I only have to calculate the answer once, and I think this may be reflected in the run times.

So now the part that I know everyone loves to read the most, Times:

python-2.6.6 : .165s

python-3.1.2 : .110s