The very first problem is rather simple:

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

And a potential solution is haskell is could be to define the following:

isMultOf x n = mod n x == 0
 
sumList xs = foldr (+) 0 xs

After which the answer could be obtained with a simple:

sumList [a | a <- [1..1000], isMultOf 3 a || 
                             isMultOf 5 a, 
                             a < 1000]

And we have a list comprehension. Beautiful, and immediately expressive of exactly what’s happening for such a simple problem. The amazing thing is that properly written haskell maintains this same clarity and expressivity in even large scale programs. Sometimes it’s the little things in a language that makes you really appreciate it.

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