one line uncategorized solution for Disposable Teleports by Apua - python coding challenges

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one line solution in Uncategorized category for Disposable Teleports by Apua

def checkio(teleports_string):
    #return any route from 1 to 1 over all points
    #Don't remove checkio function

    return checkio.__setattr__('f', lambda E={(int(a),int(b)) for a,b in teleports_string.split(',')},k=1,path=[]: checkio.f.__setattr__('ans',''.join(map(str,[1]+path))) or True if len(set(path))==8 and path[-1]==1 else any(checkio.f({e for e in E if e!=(k,n) and e!=(n,k)},n,path+[n]) for n in {e[0] if e[1]==k else e[1] for e in E if k in e})) or checkio.f() and checkio.f.ans
    
if __name__ == "__main__":
    print(checkio("12,23,34,45,56,67,78,81")) #"123456781"
    print(checkio("12,28,87,71,13,14,34,35,45,46,63,65")) #"1365417821"
    print(checkio("12,15,16,23,24,28,83,85,86,87,71,74,56")) #"12382478561"
    print(checkio("13,14,23,25,34,35,47,56,58,76,68")) #"132586741"

Aug. 10, 2013

Comments:

veky on Aug. 11, 2013, 4:40 a.m.
Nice contortions you're going through just to get it all in one line. I even learned something (locals can't be changed live, but function dicts are fair game:). It can be made much shorter if you want, though. For example, e[0] if e[1]==k else e[1] can be written as e[e[0]==k]. Also, {e for e in E if e!=(k,n) and e!=(n,k)} is just E-{(k,n),(n,k)} and so on. But if you just wanted to cram it in one line (without using eval, which would be trivial:), nice work. BTW, what happened to your (un)liking my compression of Altarfinch's solution? Couldn't make up your mind? :-)

Apua on Aug. 12, 2013, 1:23 p.m.
Yes, I am sleepy at that time, and I found the original algorithm is written by Altarfinch, so I turn the 3 'liking' to him, and want to reading your solutions (I found the solution "Shortest?" you design with exceptions, it's awesome) again on the next day.

veky on Aug. 12, 2013, 2:55 p.m.
Thanks for liking my "Shortest?". I really believed it can't be made much shorter until I saw what Altarfinch did, so I had to try. :-) As for the idea (you might be interested to know), I just wondered, I have this deep recursion, but as soon as a solution is found, it has to bubble up all those stack frames and report itself to the caller of recursion. Well, that sounds like exactly what exceptions do, right? First I tried to use StopIteration as a nice pun:-), but then, looking at the condition, I wondered... is there any way way to test a condition and raise an exception if it is met? Why ofcourse, assert the inverted condition! :-)

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