I need algorithm for my task

[Jussi Piitulainen]

Denis McMahon writes:
> On Thu, 06 Nov 2014 15:14:05 +1100, Chris Angelico wrote:
> > On Thu, Nov 6, 2014 at 3:00 PM, Denis McMahon wrote:
> >> def baseword(s):
> >>     """find shortest sequence which repeats to generate s"""
> >>     return s[0:["".join([s[0:x]for k in range(int(len(s)/x)+1)])[0:len
> >> (s)]for x in range(1,len(s)+1)].index(s)+1]
> > 
> > That's hardly a PEP-8 compliant line, but I can help out a bit.
> > 
> > return s[0:[(s[0:x]*(len(s)//x+1))[0:len(s)]for x in
> > range(1,len(s)+1)].index(s)+1]
> > 
> > That's still 83 characters without indentation, but it's close now.
> 
> l = len
> r = range
> 
> but technically then it's no longer a one liner.
> 
> > I love the algorithm. Took me a bit of analysis (and inspection of
> > partial results) to understand what your code's doing, but it's stupidly
> > elegant and elegantly stupid.
> 
> :)
> 
> Well yes, building that list is a stupid way to solve the problem, but I 
> can't see another way to do it in one line. It's an implementation of my 
> algorithm 3 (which I think you described) but working from the other end 
> as it were.

70-character expression, even with the spaces:

 >>> x = 'tästäkötästäkötästä'
 >>> x[:min(k for k in range(len(x) + 1) if x == (x[:k] * len(x))[:len(x)])]
 'tästäkö'