[EMAIL PROTECTED] writes:
> George Sakkis:
>> A faster algorithm is to create a 'key' for each word, defined as the
>> tuple of ord differences (modulo 26) of consecutive characters.
>
> Very nice solution, it uses the same strategy used to find anagrams,
> where keys are
> "".join(sorted(word))
> Such general strategy to look for a possible invariant key for the
> subsets of the required solutions is quite useful.
Or even:
def get_key(word):
initial = ord(word[0])
return tuple((ord(x) - initial) % 26 for x in word[1:])
--
Arnaud
--
http://mail.python.org/mailman/listinfo/python-list
