I've changed the "if minimum < 3" line with an "any (< 3)" line. This has sped up the performance to be good enough. (I assume that I have to calculate the Lev. distance of all the 9-mers in order to take the minimum. I really only care if any element has a Lev. distance less than three, so I can stop when I find the first.) The rest of this discussion is purely for fun. I've swapped "y:ys" for "ys ++ [y]", and while the output is reversed, I don't appear to be able to take the first n elements still. I haven't timed how long the program takes now to see if it blows up. Rein, I don't quite understand your answer; I may need you to be more explicit if I don't figure this out. Part of what confuses me is that the recursion takes place in merge, and the (:) is in addInto. I think your comment has given me an idea, so I'm going to take some time to play with this in the afternoon, so I'll send an update tonight. Thanks for looking at this! -- Mason On Thu, Jan 21, 2016 at 11:05 AM, Rein Henrichs wrote:

s/not/note, sorry

On Thu, Jan 21, 2016 at 10:42 AM Rein Henrichs wrote:

...

But not that doing so will cause the program to have an exponential runtime as each new ys must be repeatedly traversed to append a [y].. The alternative is to *unfold* your list by recursing on the right hand side of the (:) to add new elements.

On Thu, Jan 21, 2016 at 5:43 AM Doug McIlroy wrote:

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Each time you find another good 9-mer, you add it to the head of the list. This means that the ultimate list will be in reverse order of discovery: the first element to be printed is the last one to be found. To get output in the order it was discovered, build the output by ys++[y] rather than y:ys. _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners