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Haskell function application and currying

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( 5 months ago )

I am always interested in learning new languages, a fact that keeps me on my toes and makes me (I believe) a better programmer. My attempts at conquering Haskell come and go - twice so far - and I decided it was time to try again. 3rd time's the charm, right?

Nope. I re-read my old notes... and get disappointed :-(

The problem that made me lose faith last time, was an easy one: permutations of integers. i.e. from a list of integers, to a list of lists - a list of their permutations:

[int] -> [[int]]

This is in fact a generic problem, so replacing 'int' above with 'a', would still apply.

From my notes:

I code it first on my own, I succeed. Hurrah!

I send my solution to a good friend of mine - Haskell guru, it usually helps to learn from gurus - and he sends me this, which I am told, "expresses the true power of the language, the use of generic facilities to code your needs". All for it, I recently drank the kool-aid, let's go:

permute :: [a] -> [[a]]
permute = foldr (concatMap.ins) [[]]
   where ins x []     = [[x]]
         ins x (y:ys) = (x:y:ys):[ y:res | res <- ins x ys]

Hmm. Let's break this down:

bash$ cat b.hs
ins x []     = [[x]]
ins x (y:ys) = (x:y:ys):[ y:res | res <- ins x ys]

bash$ ghci
Prelude> :load b.hs
[1 of 1] Compiling Main             ( b.hs, interpreted )
Ok, modules loaded: Main.

*Main> ins 1 [2,3]

OK, so far, so good. Took me a minute to understand the second line of "ins", but OK: It places the 1st arg in all possible positions in the list. Cool.

Now, to understand the foldr and concatMap. in "Real world Haskell", the DOT was explained...

(f . g) x just another syntax for...

f (g x) 

And in the code the guru sent, DOT was used from a foldr, with the "ins" function as the fold "collapse":

*Main> let g=concatMap . ins
*Main> g 1 [[2,3]]

OK, since I want to understand how the DOT is used by the guru, I try the equivalent expression according to the DOT definition, (f . g) x = f (g x) ...

*Main> concatMap 


( 5 months ago )

(f . g) x = f (g x)

This is true. You concluded from that that

(f . g) x y = f (g x y)

must also be true, but that is not the case. In fact, the following is true:

(f . g) x y = f (g x) y

which is not the same.

Why is this true? Well (f . g) x y is the same as ((f . g) x) y and since we know that (f . g) x = f (g x) we can reduce that to (f (g x)) y, which is again the same as f (g x) y.

So (concatMap . ins) 1 [[2,3]] is equivalent to concatMap (ins 1) [[2,3]]. There is no magic going on here.

Another way to approach this is via the types:

. has the type (b -> c) -> (a -> b) -> a -> cconcatMap has the type (x -> [y]) -> [x] -> [y]ins has the type t -> [t] -> [[t]]. So if we use concatMap as the b -> c argument and ins as the a -> b argument, then a becomes tb becomes [t] -> [[t]] and c becomes [[t]] -> [[t]] (with x = [t] and y = [t]).

So the type of concatMap . ins is t -> [[t]] -> [[t]], which means a function taking a whatever and a list of lists (of whatevers) and returning a list of lists (of the same type).

what's your interest