Re: Simple Permutations in Python and Ruby

Aldric — Thu, 30 Oct 2008 16:30:11 -0000

*chuckle* I tried to look at simple output from the function (I'll be honest, I was doing Project Euler, too) to figure out how it worked and I realized, from the output, it was nesting lists. I also figured you could do the .flatten but I think that adds quite a bit of time to the computation. Your original bugged version found the answer to the Euler problem in 24.8 seconds or so, and I didn't bother waiting past a minute for the .flatten.

On the other hand, using
[element].concat (val)

got me a result in ~25 seconds (only two runs, so not conclusive). Since this is Ruby, this should work on most anything that's enumerable.. Even sets (just tested it), however worthless that may be on a set.

This, I must admit, is by far the most elegant permutation implementation I've found on the web. Do you mind if I keep it, add your name/handle/URL to the source and reuse / distribute as I see fit (Not that I expect to find too many people who need to permute in Ruby/Python, but who knows) ?

And, to answer, yes, this was very helpful. It's really the "plain" permutation system, but quite ruby-ified with a flavor of LISP ;-)

Thanks for your explanation!

Re: Simple Permutations in Python and Ruby

Aldric — Thu, 30 Oct 2008 12:30:01 -0000

Okay - funny, I've been looking for a good way to do permutations (had no clue how to even begin) and found your code (I'm teaching myself Ruby). It taught me quite a bit about co-routines and what on earth yield could be used for.. But I'm a little curious about this bit :
permutations(li.select() {|n| n != element}) \
{|val| yield([element] << val)}

Now, what this is saying is ...
run the permutation function on every element in the list except the current one, and run that against ... What, exactly? What is "yield(|element << val)" ? I can't find a good way to put this in english..

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Re: Simple Permutations in Python and Ruby

Re: Simple Permutations in Python and Ruby