Comments on: Why Haskell is a joy to learn

By: dannyo152

dannyo152 — Fri, 24 Apr 2009 04:09:56 +0000

Oh dear, not only did my indents not make it through, the last line of the function disappeared.

The last line of the function is:

sum [ i | i <- [ 1 .. lim - 1 ], iOkay i ]

By: dannyo152

dannyo152 — Fri, 24 Apr 2009 04:05:27 +0000

I think one of swell things about Haskell is if you can’t find the wheel, it’s not hard to reinvent one.

So, just for giggles, let’s take the problem and make the limit and primes and their count variable:

sumPrimeMultsBeforeLimit :: [Integer] -> Integer -> Integer
sumPrimeMultsBeforeLimit primeList lim =
let
–a function that tests a number and base
modTestGood base val = mod val base == 0
–we’ll make a list of partial functions from the primes
–we’ll also say if the primes are an empty list, all the
— numbers are good and we get the sum of the numbers from 1 to n-1
— the type of testList is [Integer->Bool]
testList = if primeList == []
then [(\ i -> True )]
else map (\ k -> modTestGood k ) primeList
— we use or on a List of True and False results
— created by mapping a function that completes
— modTestGood
iOkay i = or ( map (\ f -> f i ) testList )
in
–sum the list comprehension use the iOkay filter
sum [ i | i sumPrimeMultsBeforeLimit [3,5] 1000
233168
*Vendor> seansFn
233168

Before I click “Post” a prayer to the CSS gods for a sensible layout.

By: sean

sean — Fri, 17 Apr 2009 17:38:35 +0000

I love all the discussion actually. I knew my solution was far from ideal when I posted it, especially given that I have a knowledge of the haskell Prelude that tends toward 0. So it’s been wonderfully informative to see the ongoing comments.

By: Jade NB

Jade NB — Fri, 17 Apr 2009 17:04:10 +0000

I think that you’ve just shown that, while Haskell is a joy to learn, it’s not always a joy to show others your Haskell. Go for something incomprehensible first, so that it takes a while to come up with alternate idioms for it.

By: josh

josh — Fri, 17 Apr 2009 16:38:10 +0000

What’s unclear about (a `isMultOf`) `any` [3,5]? It’s practically english.

By: Vasyl Pasternak

Vasyl Pasternak — Fri, 17 Apr 2009 13:27:26 +0000

Another way to do it:

sum_mult_3 = sum [3,6..1000]

sum_mult_5 = sum [5,10..1000]

sum_mult_15 = sum [15,30..1000]

result = sum_mult_3 + sum_mult_5 – sum_mult_15

all these lines could be reduced to:

sum $ map sum [[3,6..1000],[5,10..1000],[(-15),(-30)..(-1000)]]

but this is less understandable

By: Matteo

Matteo — Fri, 17 Apr 2009 12:40:33 +0000

What about this?

sum [ a | a <- [1..999], any ((==) 0 . (a `mod`)) [3,5] ]

I’m a beginner, so if it is any wrong please tell me.

By: remi

remi — Fri, 17 Apr 2009 12:24:56 +0000

I would argue the last two attempts are much less clear than f.ex. Stephen Johnson’s expression..

By: exlevan

exlevan — Fri, 17 Apr 2009 07:04:29 +0000

Without isMultOf:
sum [a | a <- [1..999], or $ map ((== 0) . mod a) [3,5]]

By: josh

josh — Fri, 17 Apr 2009 00:28:36 +0000

Less repetitive:

a `isMultOf` b = a `mod` b == 0

sum [a | a <- [1..999], (a `isMultOf`) `any` [3,5]]