Chat minigolf

Sp3000

14:49

room topic changed to Chat minigolf: Current problem: checkio.org/mission/brackets (no tags)

room topic changed to Chat minigolf: Codegolf mini-puzzles! (no tags)

room topic changed to Chat minigolf: Participate in codegolf mini-puzzles together! (no tags)

Current problem: checkio.org/mission/brackets . Function or program okay (function need not be named "checkio"). Return truthy/falsy value.

Hello! I'm going to get coding :P

FryAmTheEggman

15:06

Should we work in the same language? (Also I'm a bit busy, won't be super responsive :p )

Sp3000

Ahaha well initially what happened was grc, xnor and I were golfing down the same few Python solutions. I'm not sure if everyone knows Python though

FryAmTheEggman

I know it well enough ;p

Sp3000

But I thought if someone didn't know Python did, say, CJam, then people who didn't know CJam could learn a bit of that

def B(s):
 S=[""]
 for c in s:
  if c in "{([":S+=[{"{":"}","(":")","[":"]"}[c]]
  if c in "})]":
   if c != S[-1]:return 0
   S.pop()
 return len(S)<2

Ahaha that's so ungolfed - is there something better than a dict for Python?

FryAmTheEggman

I was trying

Using a string with both sets of brackets

And checking if .find() % 3 was the same

Sp3000

Oooh interesting

FryAmTheEggman

15:20

It's pretty long and I don't think it works with something like "(asdf" but... :P

oh, no formatting :/

def c(s,S=[],b='{[(}])'):
    for C in s:
        if C in b:S+=C
    while S:
        if b.find(S[-1])%3==b.find(S[-2])%3:
            S.pop();S.pop()
        else:return 0
    return 1

Sp3000

Hmm I've always wondered - does PPCG have some sort of rule that your function should work find on consecutive runs?

Because otherwise the mutable default argument might be a problem

FryAmTheEggman

Huh, I have no idea :P

Sp3000

Oh wait you're using += ... I think that creates a new list?

FryAmTheEggman

I don't think it changes anything here, though (size length)

Sp3000

Hm k

FryAmTheEggman

15:24

Also it does break on second run :P

Sp3000

Ahaha damn

FryAmTheEggman

Actually it's behaviour is quite funny

Sp3000

I think you could probably do something like if b[b.find(S[-1])%3]==S[-2]:

FryAmTheEggman

Since the list is emptied when it succeeds, it works if the previous answer was true :P

Also that's smart

Sp3000

:P so you only need to clear the list before the next run! :D

FryAmTheEggman

15:27

Hrm, not working...

Sp3000

Well that makes this 144:

def B(s):
 S=[""];Q=")]}([{"
 for c in s:
  if c in"{([":S+=[Q[Q.find(c)%3]]
  if c in"})]":
   if S[-1]!=c:return 0
   S.pop()
 return len(S)<2

FryAmTheEggman

Oh, I had bug >_<

Sp3000

Ah D:

FryAmTheEggman

I forgot I was running the check at the end, not as I got them :P

Sp3000

:P

FryAmTheEggman

15:34

Also, what's up with the len(S)<2?

Sp3000

There's an empty string at the base of the stack

So all match = len 1, remaining bracket = len 2

(basically so I could do S[-1] without worries)

Martin Büttner

this problem sounds like a job for a .net regex

Sp3000

Ahaha can you use regexes in a short way in .NET?

FryAmTheEggman

:o

Sp3000

Test cases from CheckIO: pastebin.com/vRbMLAjm

Martin Büttner

15:41

@Sp3000 for the purposes of PPCG regex-only answers would be valid

Sp3000

Well I mean, how do you invoke regexes in .NET? I'm assuming it's not import re :P

(never used .NET so I don't know what a full program looks like)

Martin Büttner

yes, of course, if you actually do it in .NET, it's quite a lot of boilerplate characters... but if this was a PPCG challenge, I'd answer with only the regex.

you'd need a class, and a function inside

Sp3000

Ah k :o

Didn't think of regexes, maybe that's even shorter in Python

... maybe PCRE (I don't think Python can do recursion...)

Martin Büttner

either that or balancing groups

but recursion is probably shorter

Sp3000

Hmm I'm not sure how to collapse the three types of brackets together though

FryAmTheEggman

15:54

Warg formatting

def c(s,S=[],b='{[(}])'):
 for C in s:
  i=b.find(C)
  S+=C*(i>-1)
  if i>3:
    if b[b.find(S[-1])%3]!=S[-2]:return 0
    S.pop();S.pop()
 return not S

Sp3000

Seems to fail on {[(3+1)+2]+}

FryAmTheEggman

Rats, and I had just got it to 131 ;p

Sp3000

:P

My regex is at 95 chars and it's half broken :(

FryAmTheEggman

I feel incredibly dumb.

i> 2

def c(s,S=[],b='{[(}])'):
 for C in s:
  i=b.find(C)
  print i
  if i>2:
    if b[i%3]!=S[-1]:return 0
    S.pop()
  else:S+=C*(i>-1)
 return not S

131 bytes I think

Sp3000

You left a print i in there :P

But it seems to work (huzzah)

FryAmTheEggman

16:08

XD

Hrm, it might not work for "})]" or something...

Yeah

Sp3000

Ah list index

I'll add that to the cases :P

FryAmTheEggman

Ok, just add an and S to the i>2 line

I think that should fix it..?

Sp3000

Each run isn't independent I'm finding :/

(Funny thing is })] sometimes passes due to a bracket left in by the last run)

FryAmTheEggman

XD

I pass all of these (on the first run ;) ):

h="((5+3)*2+1)"
i="(3+{1-1)}"
j=']]()'
k='{[(3+1)+2]+}'

Sp3000

Hmm well... this is awkward. I can't use anchors for a recursive regex

FryAmTheEggman

16:13

(By pass I mean get correct answer...)

Sp3000

(same way I used the word pass :P)

@MartinBüttner Is there a way in PCRE to match something iff there's something already in a specific group?

Oh (?(1)yes|no), same as Python it seems

Sp3000

16:31

regex101.com/r/yN3yJ0/2 how to anchor :/

So it's about 4am over here so I should probably head off to sleep :) Will see if I can get some progress tomorrow

Also @FryAmTheEggman I'll do the function thing later for square roots, my brain hasn't been working well today and I'm still not quite convinced xnor's suggestion's correct ( though I'm pretty sure it is)

Martin Büttner

21:18

@Sp3000 yes.

@Sp3000 wrap everything in ^(...)$. replace (?R) with (?1)

FryAmTheEggman

Wow, that get 55.

Nice :)

Martin Büttner

it can also be shortened at the cost of efficiency

unrolling the loop is nice for that, but it requires you to double the negated character class

Martin Büttner

21:52

40 bytes:

^(([^][(){}]|$(?1)$|\[(?1)]|{(?1)})*)$

might be possible but hard to beat with CJam

FryAmTheEggman

22:07

That's tiny! Nice :)

I have to go now, hopefully this room doesn't totally die, it was fun :)

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