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Python one-letter-off word solver help

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Join Date: Jul 2008

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simonches

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Python one-letter-off word solver help

Sep 21st, 2009

Hi,
I'm trying to make a program in python that solves word puzzles like this:
"nj" (the answer to that is "ok")
where each letter is off by one, for example a or c instead of b.

the program is supposed to generate a list of possible words which will be checked against a dictionary.

it's recursive, and each time it's called it's supposed to find the two possible letters for the beginning letter and call itself on the word minus what's just been solved.

 Help with Code Tags 
 python Syntax (Toggle Plain Text) 
def one(s):
        print "s:"+s
        if s == '':
            return [s]
	else:
            ans = []
	    if s[1:] == '': return ans
            print "s1:"+s[1:]
            for o in one(s[1:]):
                ans.append(prevLetter(s[0])+string.join(one(s[1:]),''))
                ans.append(nextLetter(s[0])+string.join(one(s[1:]),''))
                print "o:"+o
            return ans
def one(s):
        print "s:"+s
        if s == '':
            return [s]
	else:
            ans = []
	    if s[1:] == '': return ans
            print "s1:"+s[1:]
            for o in one(s[1:]):
                ans.append(prevLetter(s[0])+string.join(one(s[1:]),''))
                ans.append(nextLetter(s[0])+string.join(one(s[1:]),''))
                print "o:"+o
            return ans

but i just can't figure out how to make it work.
can anybody help me?

by the way, the functions nextLetter and prevLetter return the next and previous letter, respectively, for example nextLetter('b') will return 'c'

Thanks,
Simon Chester

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Join Date: Jul 2008

Posts: 1,067

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jlm699

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Knows where his Towel is

Re: Python one-letter-off word solver help

Sep 21st, 2009

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Originally Posted by simonches

but i just can't figure out how to make it work.
can anybody help me?

You were very close:

 Help with Code Tags 
 python Syntax (Toggle Plain Text) 
def one(s):
    if s == '':
        return [s]
    else:
        ans = []
        if s[1:] == '': return ans
        for o in one(s[1:]):
            ans.append(prevLetter(s[0])+o)
            ans.append(nextLetter(s[0])+o)
        return ans
def one(s):
    if s == '':
        return [s]
    else:
        ans = []
        if s[1:] == '': return ans
        for o in one(s[1:]):
            ans.append(prevLetter(s[0])+o)
            ans.append(nextLetter(s[0])+o)
        return ans

You never made use of o in your function. You were for whatever reason trying to use ''.join(x) (NOTE: that is equivalent to string.join(x, '') ).

Your function can be further simplified as such:

 Help with Code Tags 
 python Syntax (Toggle Plain Text) 
def one4(s):
    if s == '': return [s]
    else:
        return [prevLetter(s[0]) + o for o in one(s[1:])] + \
               [nextLetter(s[0]) + o for o in one(s[1:])]
def one4(s):
    if s == '': return [s]
    else:
        return [prevLetter(s[0]) + o for o in one(s[1:])] + \
               [nextLetter(s[0]) + o for o in one(s[1:])]