Pythagorean triple in Haskell without symmetrical solutions

Question

I gotta do the Pythagorean triple in Haskell without symmetrical solutions. My try is:

terna :: Int -> [(Int,Int,Int)]
terna x = [(a,b,c)|a<-[1..x], b<-[1..x], c<-[1..x], (a^2)+(b^2) == (c^2)]

and I get as a result:

Main> terna 10
[(3,4,5),(4,3,5),(6,8,10),(8,6,10)]

As you can see, I´m getting symmetrical solutions like: (3,4,5) (4,3,5). I need to get rid of them but I don´t know how. Can anyone help me?

Answer 1

Every time you have a duplicate you have one version in which a is greater than b and one where b is greater than a. So if you want to make sure you only ever get one of them, you just need to make sure that either a is always equal to or less than b or vice versa.

One way to achieve this would be to add it as a condition to the list comprehension.

Another, more efficient way, would be to change b's generator to b <- [1..a], so it only generates values for b which are smaller or equal to a.

Speaking of efficiency: There is no need to iterate over c at all. Once you have values for a and b, you could simply calculate (a^2)+(b^2) and check whether it has a natural square root.