(Parentheses) creative solution for Colder-Warmer by veky - python coding challenges

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(Parentheses) solution in Creative category for Colder-Warmer by veky

checkio=lambda steps: {(x,y) for x in range(10) for y in range(10) if all(
    (lambda t: t//abs(t) if t else 0)((x-x1)**2+(y-y1)**2-(x-x2)**2-(y-y2)**2)==s
        and (x,y) not in ((x1,y1),(x2,y2))
    for (x1,y1,_),(x2,y2,s) in zip(steps,steps[1:]))}.pop()

Sept. 13, 2013

Comments:

veky on Sept. 13, 2013, 1:37 p.m.
For all those who wondered how my "golf course" looks like in the zeroth iteration. I intentionally didn't "compress" any of this code, this is what I wrote directly. Compressing the above is left as an excercise to the reader. :-D

bryukh on Sept. 16, 2013, 10:36 p.m.
How are think, 12 steps is enough or too much?

veky on Sept. 17, 2013, 6:07 a.m.
Well, my solution above is kind of random (because of pop), and really naive, but I've run it many times and it finds solution in 8 steps or less. Why would you change it to 12?

bryukh on Sept. 17, 2013, 7:29 a.m.
<blockquote> Why would you change it to 12? </blockquote> No, it's already 12.

veky on Sept. 17, 2013, 3:03 p.m.
Oh well. It was 8 when I started solving this task. :-D

bryukh on Sept. 17, 2013, 3:11 p.m.
No, it was 10. Here a proof :) <a href="https://github.com/Bryukh-Checkio-Tasks/checkio-task-colder-warmer/commit/5fc2f9478e585faa2802c72e5a86318830c38f90">https://github.com/Bryukh-Checkio-Tasks/checkio-task-colder-warmer/commit/5fc2f9478e585faa2802c72e5a86318830c38f90</a>#diff-dce0fd9e42fae5f30ddef593650b60bf

veky on Sept. 17, 2013, 5:55 p.m.
Ok, sorry. I don't know why I thought it was 8. Anyway, it seems 10 is more than enough.

ysenko on Dec. 23, 2013, 10:59 a.m.
Really cool and not readable at all :)

veky on Dec. 23, 2013, 4:23 p.m.
Err... thank you, I guess. :-D

veky on Dec. 24, 2013, 3:55 a.m.
I felt a bit guilty about this, so I wrote another solution. It's basically the same algorithm (fetch a point that satisfies all conditions), but it's deterministic and flowery. ;-D See if it helps you to understand this.

Ardillo95 on Oct. 8, 2016, 6:24 a.m.
Wow I apreciated this so much :D

terryyin on Aug. 13, 2014, 7:36 a.m.
Yes, pop is a bit risky. Because I found if you always use the smallest x, y in the set it will take more than 12 steps in some cases. Probably taking the middle one of the set is better? Then you cannot put it on one line:P

veky on Aug. 13, 2014, 6:20 p.m.
Who, me? I can put everything in one line. I'm the <a href="http://www.checkio.org/forum/post/1485/vekyization-challenge/">Chuck Norris</a> of Python oneliners. :-D BTW. That same problem you mentioned doesn't convince e.g. people who think quicksort is worthy of its name. :-P Because, really, what's the chance of that happening? :-)

terryyin on Aug. 14, 2014, 1:21 a.m.
Ok Chuck Norris:-) Say I have an unnamed list: [rand(10) for _ in range(rand(10))], how can I get the item in the middle of this list in one line?

veky on Aug. 14, 2014, 4:53 a.m.
Well, "item in the middle" isn't really defined if the list is of even length, and it's even less defined if the list is empty. But assuming it isn't (and assuming "from random import randrange as rand", which is probably what you meant), does <pre class='brush: python'>(lambda l: l[len(l)//2])([rand(10) for _ in range(rand(10))]) </pre> satisfy you? You see, the names are ephemeral. :-D

terryyin on Aug. 14, 2014, 5:47 a.m.
Orz Nice:-) Thank you!

veky on Aug. 14, 2014, 1:09 p.m.
What's interesting is that I used that same trick above, in line 2. If you just read it a bit more carefully... ;-P

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