Keith Devens .com - Weblog: Short programming problem

Wednesday, January 11, 2006

Short programming problem

Here's a short programming problem: Given a list of tokens, generate a minimal regular expression that matches those (and only those) tokens. So, given ['one','two','three','four','five','twenty','tween'], the regular expression would be "f(ive|our)|one|t(hree|w(e(en|nty)|o))". I've shown it sorted, but it doesn't have to be. Also, don't worry about a '|'.

Now, an interesting problem would be how to generate a provably minimal regular expression... maybe there's already an algorithm I don't know about that does that. But all I'm trying to do here is prefix stemming.

I'll probably post some Python code later that solves the problem -- I'm just about finished writing it, but won't get to finish tonight. I'm most interested in your approaches. Who wants to try?

(I'd love to see a solution in K.)

Permalink | 5 Comments | Tags: Programming | id 7844

20:19 (utc-5)
edited: 23:31 (utc-5)

Comments

David wrote:

The Emacs Lisp function `regexp-opt' does this. So:

(regexp-opt '("one" "two" "three" "four" "five" "twenty" "tween"))

yeilds:

"f\\(?:ive\\|our\\)\\|one\\|t\\(?:hree\\|w\\(?:e\\(?:en\\|nty\\)\\|o\\)\\)"

Yeah, Elisp suffers from backslash-itis.

∴ David | 11-Jan-2006 9:38pm est | #8978

David wrote:

If you are interested in viewing the lisp code for regexp-opt, you can find it here

∴ David | 11-Jan-2006 11:13pm est | #8979

Thanks David! I had actually been thinking of updating my "I'd love to see a solution to this in K" to include "or Lisp" at the end, but you beat me to it. Though that Elisp code is more than a little hairy. Also, I noticed from your regex that I'd forgotten to include "three" in my regex, so that's been corrected.

∴ Keith | 11-Jan-2006 11:37pm est | http://keithdevens.com/ | #8980

mtve wrote:

∴ mtve | 12-Jan-2006 9:47am est | #8989

Ok, finally got around to finishing this. Here's the code (just 41 lines including the harness):

def regexify_groups(groups):

    return "|".join(regexify_terms(terms) for terms in groups)



def regexify_terms(terms):

    #'terms' must not be empty, and all must begin with the same letter

    if len(terms) == 1:

        return terms[0]



    sublist = [term[1:] for term in terms if len(term) > 1]

    grouped_terms = [group for group in group_by_first_char(sublist)]

    regex = regexify_groups(grouped_terms)



    if len(grouped_terms) > 1: #if you have multiple groups, group them

        regex = "("+regex+")"



    if len(sublist) < len(terms): #if a term got eliminated

        if len(regex) > 1: #if multiple letters,

            regex = "("+regex+")" #need parens

        regex += '?' #optional



    return terms[0][0]+regex



def group_by_first_char(lst):

    #must not be called with empty strings or an empty list

    prev = lst[0][0] #get starting first character

    temp = []

    for item in lst:

        if item[0] != prev:

            yield temp

            prev = item[0]

            temp = []

        temp.append(item)



    if temp: yield temp



if __name__ == "__main__":

    import sys

    if len(sys.argv) < 2:

        print "Usage: "+sys.argv[0]+" term1[ term2[ term3...]"

    else:

        print regexify_groups(group_by_first_char(sorted(set(sys.argv[1:]))))

There are surely more efficient ways of doing this, but I think this is pretty elegant. Though, I'd totally use Jarkko's code over this since it does more (thanks mtve).

∴ Keith | 16-Jan-2006 3:29pm est | http://keithdevens.com/ | #9002

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