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Subject: Re: [Haskell-beginners] another list comprehesion error From: d@vidplace.com Date: Thu, 21 Jul 2011 15:19:54 -0400 CC: beginners@haskell.org To: rwobben@hotmail.com
On Jul 21, 2011, at 2:47 PM, Roelof Wobben wrote:
I could make this with a for next loop and a if then if you like it that way ?
I though it would be trivial, and perhaps it is. This seems to be a case where the list comprehensions give quite a bit of help.
First of all, here is your function with all the syntax and semantic bugs fixed. Please compare it closely to your version.
roelof :: Int -> [(Int,Int,Int)] roelof n = [(x, y, z) | x<-[1..n], y<- [1..n], z<-[1..n], x^2+y^2 == z^2 ]
Oke, I see that x<[1..n] part is now in the middle. That I find quite confusing. When I have made a list compresshion where I must print out a text I look like this ; roelof n x = [x | y <- [1..n]] So I thought the syntax of a list compression would be [output | filter <- input] But the answer is now [ output | input <- filter] This is very confusing for me.
Here's a version using only functions without any of the syntactic sugar of list comprehensions.
triples xs ys zs = concatMap (\x -> concatMap (\y -> (map (\z -> (x,y,z)) zs)) ys) xs roelof' n = filter (\(x,y,z) -> x^2+y^2 == z^2) $ triples [1..n] [1..n] [1..n]
It's quite unattractive. Please someone tell me there is a better way.
I thought of doing something like this in pseudo code : For teller = 1 to n for teller2 = 1 to n antwoord = (teller1 ^2 + teller^2)^ 0.5 if antwoord == int(antwoord) then print x,y,z Roelof