On Mon, Aug 12, 2013 at 4:03 PM, Mark <markaddle...@gmail.com> wrote:
>
> At run level 6, I get all the permutations of [1 2 3], just as expected.
> However, at 7, the program does not terminate and I'd like to understand
> why. I feel like I need to constrain the relation between o-h and o better
> but I'm not sure what else to say about it.
>
Looking at this, it's clear that this will never terminate. Your recursion
is unbounded on the rembero. When you get to rembero, your l and o-h are
both always fresh. (rembero :anyting (lvar) (lvar)) will produce an
infinite set of results.
After you've generated all your 6 results, remberallo will of course fail
on everything rembero produces, but you've got nothing that stops rembero.
Now, consider putting the remberallo BEFORE the rembero:
(defne remberallo [s l o]
([() l l])
([[h . r] _ _]
(fresh [o-h]
(remberallo r o-h o)
(rembero h l o-h))))
Now, imagine you are on a final remberallo call where h is some number and
r is the empty list. The recursive remberallo will will only produce one
answer, unifying o-h and o. Now o-h is no longer fresh and your rembero
will have enough information to terminate.
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