Thanks! On Tuesday, August 13, 2013 12:18:33 PM UTC-7, Norman Richards wrote: > > > > > On Mon, Aug 12, 2013 at 4:03 PM, Mark <markad...@gmail.com > <javascript:>>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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