Savage creative solution for Numbered Triangles by veky - python coding challenges

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Savage solution in Creative category for Numbered Triangles by veky

checkio=lambda x:max([B+E+H+K+N+Q for p in[__import__("itertools").permutations]for
a,b,c,d,e,f in p(x)for A,B,C in p(a)for D,E,F in p(b)if C==D for G,H,I in p(c)if F==G
for J,K,L in p(d)if I==J for M,N,O in p(e)for P,Q,R in p(f)if(L,O,R)==(M,P,A)]+[0])

Oct. 13, 2013

Comments:

veky on Oct. 13, 2013, 5:10 a.m.
Ladies and gentlemen, I present you... the Solution. No recursion. No indexing. Branch pruning. One logical line. Don't trust the horizontal scrollbar, it fits! :-D

bryukh on Oct. 13, 2013, 9:43 a.m.
Interesting idea -- i like it, really. Could you write explicit version? I think it will be useful as good example for others.

veky on Oct. 13, 2013, 3:13 p.m.
Well, this is a nice example of "too much of a good thing". :-) You can consider it a parody of Llama's style (I hope he doesn't get angry:). <a href="http://pep8online.com/s/vLnXum82">http://pep8online.com/s/vLnXum82</a> This is obviously on one side of sweet spot. "Savage" above is obviously on another. But where is the sweet spot itself, it's hard to know. :-)

bryukh on Oct. 13, 2013, 9:25 p.m.
Oh, pep8online :-) I need update it with some new ideas (i almost forget about it). About solution -- the short form is more beatiful :-)

veky on Oct. 13, 2013, 9:57 p.m.
<blockquote> About solution -- the short form is more beatiful :-) </blockquote> That's what I'm telling you. All the time. :-D

nickie on Jan. 18, 2014, 12:38 a.m.
Here's a decryption of veky's solution, for anyone interested to study it: <pre class='brush: python'>from itertools import permutations def checkio(chips): return max([B+E+H+K+N+Q for a,b,c,d,e,f in permutations(chips) for A,B,C in permutations(a) for D,E,F in permutations(b) if C==D for G,H,I in permutations(c) if F==G for J,K,L in permutations(d) if I==J for M,N,O in permutations(e) for P,Q,R in permutations(f) if (L,O,R)==(M,P,A)] + [0]) </pre> It's a quite simple idea which assumes that there are only six chips and that each chip has three sides. Other solutions work equally well for two generalizations of the problem: (1) with more than six chips, and (2) with chips that are not triangular but have at least two sides: two that we connect and the rest that we sum in the perimeter. Notice also that this solution tries many more permutations than is necessary. For example, the starting chip 'a' could be fixed to any of the given chips (we can see any chip as the circle's starting point).

Jiyoqisv on May 27, 2014, 10:31 p.m.
<blockquote> Notice also that this solution tries many more permutations than is necessary. For example, the starting chip 'a' could be fixed to any of the given chips (we can see any chip as the circle's starting point </blockquote> That's right: <pre class='brush: python'>def checkio(chips): from itertools import permutations permutations_a = list(permutations(chips[0]))[0:3] return max([B+E+H+K+N+Q for b,c,d,e,f in permutations(chips[1:]) for A,B,C in permutations_a for D,E,F in permutations(b) if C==D for G,H,I in permutations(c) if F==G for J,K,L in permutations(d) if I==J for M,N,O in permutations(e) if L==M for P,Q,R in permutations(f) if O==P and R==A] + [0]) </pre> is as fast as most solutions proposed in speedy category. Plus, with indentation, the solution is also very clear. This one could have win in all categories !

veky on May 28, 2014, 5:22 a.m.
May I ask then, why downthumb me? If I understand you, you in fact say my solution is great, and then give me -3. I don't get it.

Jiyoqisv on May 28, 2014, 5:55 a.m.
I don't give -3 but -1. I love oneliner when they simplify the solution and make it eleguant. You produced a lot of them and I +1ed them. But here I think this is too much. For this exercise, I really like the solution of Nickie as you can see the solution as in a picture.

veky on May 28, 2014, 7:15 a.m.
LOL. As far as I'm concerned, it's the same solution. Soon we'll give thumbs for how the code looks as a picture. :-DD Ok, whatever - your opinion, your right. For -1 it doesn't matter too much, but I thought I saw -3 in history.

Fermax on April 26, 2014, 11:01 p.m.
Thanks for decrypting it, nickie! That makes it so clear. Nice use of permutations, veky!

veky on Jan. 18, 2014, 3:16 a.m.
Of course. Here, the challenge was to fit in the textbox (no scrollbars), without being unnecessarily slow. I think I made it. ;-) BTW, see also "the parody of readability" that I made here: <a href="http://pep8online.com/s/vLnXum82">http://pep8online.com/s/vLnXum82</a>

guruvyasa on April 2, 2014, 10:51 a.m.
Unbelievable!! I love the shorter version. Clear yet awewsome!

hrvoje on June 30, 2015, 12:48 a.m.
ET phone home. :)

veky on June 30, 2015, 1:54 a.m.
I lost you again. :-)

hrvoje on June 30, 2015, 7:56 a.m.
Famous line from the movie ET ...? You're obviously an alien. :)))

veky on June 30, 2015, 8:48 a.m.
I wouldn't say so. This planet being the one where Python is invented, makes me feel like my home is right here. ;-P

hrvoje on June 30, 2015, 8:57 a.m.
Now rename all occurrences of variables so they form a message :)

veky on June 30, 2015, 9 a.m.
They already do. The clearest possible one. ;-D

U.V on July 6, 2023, 9:48 a.m.
A great solution, but not the quick. In the worst (the simplest) case, a complete search is performed. <pre class='brush: python'> assert checkio([[9, 9, 9], [9, 9, 9], [9, 9, 9], [9, 9, 9], [9, 9, 9], [9, 9, 9]]) == 6 * 9 </pre> And it takes ~10 seconds

veky on July 6, 2023, 9:50 a.m.
So what? As far as I see, it is not in the speedy category. :-)

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