Sebastian,
This program is clever, but I like more your original one:
fib | head x =?= 1
& head (tail x) =?= 1
& tail (tail x) =?= zipWith (+) x (tail x) = x where x free
since it more clearly shows the conditions that you impose on the output. I changed the syntax of the equality operator, since I think that you want some kind of equality less strict than the standard one and different from (=:<=) which is intended for a different use.
Laziness desirable. The problem is how much lazy you want to be (or can be) to ensure the conditions. Section 3 of the functional patterns paper shows that this problem is difficult and subtle.
Sergio
On Mon, 14 Dec 2015, at 07:49, Sebastian Hanowski <sebastian_hanowski_at_gmx.de> wrote:
> On Thu, Dec 10, 2015 at 10:53:00PM +0100, Michael Hanus wrote:
> > As I wrote before, the second argument of the non-strict unification
> > operator (=:<=) must be a pattern variable in order to stay in the
> > well-understood framework of functional patterns. Since this is not the
> > case in your definition, it might compute something as intended,
> > but the general declarative semantics is not clear (to me).
>
> Challenge accepted ;)
>
> fibaux fib xs | (1, (1, tail (tail fib))) =:<= xs = fib
>
> fibs = fibaux xs (head xs, (head (tail xs), zipWith (+) xs (tail xs)))
> where xs free
>
>
> And if non-linearity of patterns isn't an issue...
>
> fibaux fib (1, (1, tail (tail fib))) = fib
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Received on Di Dez 15 2015 - 22:26:48 CET