[Haskell-beginners] questions on evaluation.

[Rein Henrichs]

 (blowup' n+1 xs) should be  (blowup' (n+1) xs) but otherwise this is spot
on.

On Mon, Jan 11, 2016 at 3:18 PM Peter McIlroy <pmcilroy at gmail.com> wrote:

>
> Fabian,
>
> For your problem of bang ->  baanngggg
> You can derive Rein's hairy idiom from a more straightforward technique
> for getting a recursive definition:
>
> First get a basic implementation using ++ and recursion:
>
> * add a helper function with any "state variables" as arguments.
> * Ignore concatenation while thinking about a problem.
> * Use pattern matching for recursion step.
>
> This gives the basic (naïve) implementation:
>
> blowup :: [a] -> [a]
> blowup x = concat (blowup' 1 x)
>
> -- returns (e.g.) ["b", "aa", "nnn", "gggg" ]
> blowup' a b :: Integer a => a -> [b] -> [[b]]
> blowup' n (x:xs)  = (replicate n x : blowup' n+1 xs)
> blowup' _ _ = [] -- base case
>
> You can distribute concatenation into blowup', to get recursion over
> concatenation, but this really is going the wrong direction:
>
> blowup' n (x:xs)  = (replicate n x) ++ (blowup' n+1 xs)
> blowup' _ _ = []
>
> Instead, derive Rein's hairy-looking version by (as always) ignoring
> concatenation and replacing the state variable in blowup' with zip or
> zipWith.
> observe:
> > zip [1..] "bang"
> [(1,'b'),(2,'a'),(3,'n'),(4,'g')]
>
> So blowup' n x becomes blowup' x:
>   blowup x = concat (blowup' x)
>
>   blowup' x  = map (\(a,b) replicate a b) zip [1..] x
> ==> definition of zipWith
>  blowup' x = zipWith replicate [1..] x
> ==>
> blowup x = concat (zipWith replicate [1..] x)
>
> You can always push concat down into a function by unwrapping it with
> join, getting a more efficient version that creates the result directly.
>
>   blowup x = concat (zipWith replicate [1..] x)
> ==>
>   blowup x = (join . zipWith replicate [1..] ) x
> ==>
>   blowup = join . zipWith replicate [1..]
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