- A+

Category：Languages

It's easier to explain this with an example:

I want to write a function [a] -> [(a,a)] so if I get a list

`[A, B, C, D] `

I want this list to return:

`[(A, A), (A,B), (A,C), (A,D), (B,B), (B,C), (B,D), (C,C), (C,D), (D,D)] `

I came up with this code:

`function s = [(x,y) | x <- s, y <- s, x<=y] `

Which works correctly for a list of integers, but I want this to work for data types that are not instances of the Ord class. My data type derives Show and Eq. So is there a simple way to solve this problem? I'm thinking maybe by filtering the tuples from

`function s = [(x,y) | x <- s, y <- s] `

But I dont know how I can do that either.

Solution using recursion:

`f :: [a] -> [(a, a)] f [] = [] f (x:xs) = [(x, y) | y <- (x:xs)] ++ f xs `

Without recursion:

`import Data.List (tails) f' :: [a] -> [(a, a)] f' xs = concat [[(head x, y) | y <- x] | x <- tails xs] `

Without list comprehension:

`import Data.List (tails) f'' :: [a] -> [(a, a)] f'' xs = concatMap (/sl -> zip (repeat $ head sl) sl) (tails xs) `

Best is by Daniel Wagner, just use

`[(head x, y) | x <- tails xs, y <- x] `