[Haskell-beginners] Missing termination rule for recursive function
oscar.j.benjamin at gmail.com
Tue Nov 6 00:57:14 CET 2012
On 5 November 2012 21:27, Jay Sulzberger <jays at panix.com> wrote:
> On Mon, 5 Nov 2012, Oscar Benjamin <oscar.j.benjamin at gmail.com> wrote:
>> -- Select each item and remainder from a sequence
>> selections :: [a] -> [(a, [a])]
>> selections  = 
>> selections (x:xs) = (x,xs) : [ (y,x:ys) | (y,ys) <- selections xs ]
>> -- Permutations of a sequence
>> permutations :: [a] -> [[a]]
>> permutations xs = [ y:zs | (y,ys) <- selections xs, zs <- permutations ys
> Assuming selections is correct, we have
> permutations ["a"] ~> the list of all lists of form "a":(something that
> lies in permutations )
> So what is the value of permutations ? It is the list of all things of
> such that
> (y,ys) lies in selections xs and zs lies in permutations ys
When you rephrase in set terminology like this it becomes easier to
understand. I was thinking of it all in terms of loops before.
> where xs = . But there are no such things. And so the list of
> sll such things is the empty list .
> What is perhaps confusing is that, at this juncture, one tends to
> think that
> must really be
> but it is not.
That's exactly what confused me. It doesn't confuse me when it's laid
out like this but in the list comprehension it did.
Thanks for your help.
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