25
\$\begingroup\$

Let's have a function \$f\$ that takes a string and removes all pairs of adjacent identical characters. For example

\$f(a\color{red}{bb}ba\color{red}{cc}) = aba\$

Note that when two pairs overlap we only remove one of them.

We will call a string perfectly paired if repeated application eventually yields the empty string. For example the string above \$abbbacc\$ is not perfectly paired because if we apply \$f\$ again we still get \$aba\$. However a string like \$eabbccadde\$ is perfectly paired because if we apply \$f\$ three times we get the empty string

\$f(ea\color{red}{bbcc}a\color{red}{dd}e) = eaae\$

\$f(e\color{red}{aa}e) = ee\$

\$f(\color{red}{ee}) =\$


Your task is to write perfectly paired computer code that takes a string (of printable ASCII) and decides if it is perfectly paired. The bytestring of your source must be itself a perfectly paired string, although your code doesn't necessarily have to be restricted to printable ASCII.

You may output two distinct values: one for the case where the input is perfectly paired, and another one for cases where it isn't.

This is a question so answers will be scored by the size in bytes of their source with fewer bytes being better.


Test Cases

\$abbbacc \rightarrow \mathrm{False}\\ abcba \rightarrow \mathrm{False}\\ abab \rightarrow \mathrm{False}\\ abbbaabacc \rightarrow \mathrm{True}\\ eabbccadde \rightarrow \mathrm{True}\\ bbbb \rightarrow \mathrm{True}\$

\$\endgroup\$
  • 1
    \$\begingroup\$ While it may be too late to change it now, I feel like the "twist" part of the challenge is rendered almost pointless if you allow comments or similar "dead" code. \$\endgroup\$ – Geobits Jun 25 '18 at 17:26
  • 11
    \$\begingroup\$ @Geobits I disagree. For one I think that disallowing dead code is just a mire of vague definitions and never turns out fun anyway. For two I think allowing comments lowers the bar of entry. For three I believe that the comment-less code will inevitably be better scoring than the comment-full code. Maybe the twist isn't fun but it would definitely be less fun if I added unenfoceable restrictions to make answers do it in a particular way. \$\endgroup\$ – Sriotchilism O'Zaic Jun 25 '18 at 17:32
  • 4
    \$\begingroup\$ Unary doesn't give a damn about your source restriction rule, mwahahahaha (that is ... as long as the answer has an even number of bytes). \$\endgroup\$ – Arnauld Jun 25 '18 at 17:41
  • 2
    \$\begingroup\$ @Geobits One thing that might have encouraged more creative answers is to factor the number of steps to get to the empty string into the scoring. Using comments tends to cause this number to be quite high because the comments naturally nest where as a lower score requires you to interlace the pairs quite a bit. Obviously it is too late to make that change though. \$\endgroup\$ – Sriotchilism O'Zaic Jun 25 '18 at 18:12
  • 1
    \$\begingroup\$ @dylnan The empty string can be, looping forever however is not valid output. \$\endgroup\$ – Sriotchilism O'Zaic Jun 25 '18 at 18:18

25 Answers 25

10
\$\begingroup\$

Haskell, 146 124 bytes

((""##))
a===bb=bb==a
((aa:bb))##((cc:dd))|aa===cc=bb##dd|1==1=((cc:aa:bb))##dd
a##""=""===a
""##cc=((cc!!00:cc!!00:""))##cc

No comments. Returns either True or False.

Try it online!

Edit: -22 bytes thanks to @Cat Wizard

\$\endgroup\$
  • 2
    \$\begingroup\$ This is the least haskell like haskell I've ever seen \$\endgroup\$ – Cubic Jun 26 '18 at 15:14
5
\$\begingroup\$

Python 2, 94 bytes

ss=''

for cc in input():ss=[cc+ss,ss[1:]][cc==ss[:1]]

print''==ss##tnirp[,+[=:)(tupni nirof=

Try it online!

The whole update step ss=[cc+ss,ss[1:]][cc==ss[:1]] cancels to just =[+,[.

\$\endgroup\$
5
\$\begingroup\$

05AB1E, 26 24 22 20 18 bytes

-2 bytes thanks to ovs. Outputs 0 if the string is perfectly paired, 1 otherwise.

ΔγʒgÉ}JJ}ĀqqĀÉgʒγΔ

Try it online!

ΔγʒgÉ}JJ}ĀqqĀÉgʒγΔ – Full program.
Δ       }          – Until the result no longer changes:
 γʒ  }               – Split the string into chunks of equal characters and filter by:
   gÉ                – Is the length odd?
      JJ           – And after filtering, join the parts back together, but do this
                     twice to save 2 bytes, as in the previous versions.
         Ā         – Check whether the result is empty
          q        – Terminate (quit) execution. The rest of the code is ignored.
           qĀÉgʒγΔ – Mirror the non-matched part to help with the source layout.

Previous versions

This one purely relies on undefined behaviour (so there's no "dead code"), and outputs [['0']] for perfectly paired strings and [['1']] for non-perfectly matched strings:

ΔγεDgÉ£}JJ}ĀĀ£ÉgDεγΔ 

And the 22-byte version, explained, which is just the above but not abusing UB, and yielding sane values.

ΔγεDgÉ£}JJ}ĀqqĀ£ÉgDεγΔ – Full program.
Δ         }            – Until fixed point is reached (starting from the input value):
 γε    }                 – Group equal adjacent values, and for each chunk,
   DgÉ                     – Duplicate, get its length mod by 2.
      £                    – And get the first ^ characters of it. This yields the
                             first char of the chunk or "" respectively for odd-length
                             and even-length chunks respectively.
         JJ                – Join the result to a string, but do this twice to help
                             us with the source layout, saving 2 bytes.
            Ā           – Check if the result is an empty string.
             q          – Terminate the execution. Any other commands are ignored.
              qĀ£ÉgDεγΔ – Mirror the part of the program that isn't otherwise removed
                          anyways. This part forgoes }JJ} because that substring will
                          always be trimmed by the algorithm anyway.
\$\endgroup\$
5
\$\begingroup\$

Cubix, 54 bytes

U#;!u1@.Oi>??>i..??O.@1^^...u--u.u!ww;..#..U..;..;!^^!

Outputs nothing if the string is perfectly paired and 1 otherwise.
Try it here

Cubified

      U # ;
      ! u 1
      @ . O
i > ? ? > i . . ? ? O .
@ 1 ^ ^ . . . u - - u .
u ! w w ; . . # . . U .
      . ; .
      . ; !
      ^ ^ !

Explanation

Most of the characters are filler needed to perfectly pair the code. Replacing those with . (no-op), we get

      U # ;
      ! u 1
      @ . O
i . ? . > i . . ? . . .
. . ^ . . . . u - . . .
. . . w ; . . . . . . .
      . ; .
      . ; !
      ^ ^ !

This can be broken into three steps:

  • Check against the empty string (the left i and the ?).
  • Loop, throwing characters onto the stack and popping duplicates (everything on the bottom and right).
  • Check if the stack is empty (the stuff at the top).
\$\endgroup\$
4
\$\begingroup\$

V, 20, 18 bytes

òóˆ±òø‚

::‚øò±ˆóò

Try it online!

Hexdump:

00000000: f2f3 88b1 f2f8 820a 0a3a 3a82 f8f2 b188  .........::.....
00000010: f3f2                                     ..                   ....

Outputs 0 for truthy, 1 for falsy. Thanks to nmjcman101 for indirectly saving 2 bytes.

ò        ò        " Recursively...
 ó                "   Remove...
  <0x88>          "     Any printable ASCII character
        ±         "     Followed by itself
          ø       " Count...
           <0x82> "   The number of non-empty strings

::<0x82>øò±<0x88>óò      " NOP to ensure that the code is paired
\$\endgroup\$
  • \$\begingroup\$ You could replace ^$ with . and return 0 for truthy, anything else for falsy? I'm a bit foggy on the rules after not doing this for a while. \$\endgroup\$ – nmjcman101 Jun 25 '18 at 19:05
  • \$\begingroup\$ I think that would work except that the rules said You may output two distinct values one for the case where the input is perfectly paired and one for otherwise.. That does give me an idea though... \$\endgroup\$ – DJMcMayhem Jun 25 '18 at 19:06
3
\$\begingroup\$

R, 142 126 bytes

Tighter logic and some comment bytes golfed by @Giuseppe

f=function(x,p="(.)\\1")"if"(grepl(p,x),f(sub(p,"",x)),!nchar(x))##x(rahcn!,x,,p(bus(f,)x,p(lperg("fi")"1\\).("=p,x(noitcnuf=f

Try it online!

f=function(x,p="(.)\\1")"if"(nchar(x),"if"(grepl(p,x),f(sub(p,"",x)),0),1)##)1,)0,xp(bus(f,)x,p(lperg("fi",)x(rahcn("fi")"1).("=p,x(noitcnuf=f

Original:

Try it online!

Recursive detector function followed by comment with all the characters in the function in reverse order.

\$\endgroup\$
  • \$\begingroup\$ Your code currently throws an error. Here is a working version at 142 bytes. \$\endgroup\$ – ovs Jun 25 '18 at 20:09
  • \$\begingroup\$ Thank you. Must have been a cut-and-paste mishap. \$\endgroup\$ – ngm Jun 25 '18 at 20:18
  • \$\begingroup\$ 126 bytes -- you might be able to compress the comment some more as well... \$\endgroup\$ – Giuseppe Jun 26 '18 at 14:35
  • \$\begingroup\$ I am wondering if `\\ˋ will simplify or needs to be duplicated in the comments. \$\endgroup\$ – JayCe Jun 28 '18 at 2:03
  • \$\begingroup\$ @JayCe you'd think it wouldn't need to be in the comments, but try this and it doesn't seem to work. I don't know why. \$\endgroup\$ – ngm Jun 28 '18 at 12:52
2
\$\begingroup\$

APL (Dyalog), 38 bytes

''≡{'(.)\1'⎕R''⊢⍵}⍣≡⍝⍝≡⍣}⍵⊢R⎕'1\).('{≡

Try it online!

\$\endgroup\$
2
\$\begingroup\$

Retina, 28 26 bytes

+`(.)\1

C`^$

$^`C1\).(`+

Try it online!

Outputs `C1\).(`+0`C1\).(`+ for falsy and `C1\).(`+1`C1\).(`+ for truthy cases.

\$\endgroup\$
2
\$\begingroup\$

Brain-Flak, 228 200 bytes

(()){{{}([]<<{{(({}<<>>)<<>>[({})]){{{{}(<<<<>>()>>)((<<>>))}}}{}{}<<>>{}<<>>}}{}<<>>>>[[]])}}{}(<<({{(())(<<()>>)}}<<>>)>>){{{{}{}}((){{}{}{}{}{}}(()())())[[]((){{}{}}())[]]((){{}{}}[[][]]()){{}{}}}}

Try it online!

This is a bit of a proof of concept. It could probably be shorter. It doesn't use any comments however.

Outputs 0,0 if the input is perfectly paired and 0,1 if the input is not.

\$\endgroup\$
2
\$\begingroup\$

sed 4.2.2, 34 bytes

:;:t;ss((..??\??))\1ss1;t;/..??/cc

Try it online!

Paired strings give empty output, unpaired ones give ct:

The trivial palindromic version is at 32 :;ss(.)\1ss;t;/./cc/./;t;1\).(;:. Old solution was :;ss((..??\??))\1ss1;t;;/./cc/./t: (changed because current one abused c less, edit: yay now there's only 1 character after c :D)

(note that ; is the statement separator)

: declares an empty label

:t declares the label t

ss((..??\??))\1ss1 is a substitution, in sed you can change the delimiter to a substitution, and this is what I did by changing it to s, so what this does is substitute the first (as is denoted by the 1 at the end)

  • match of ((..??\??))\1

    • . any character
    • .?? followed by an optional optional character
    • \?? and an optional ?
    • followed by the same thing right beside it
  • with nothing

Now this substitution is paired with itself, so the ;s before and after it get cancelled out too

t and loop back to the label until there are no more successful substitutions

/..?/ if . (wildcard) followed by .? an optional character is matched

  • cc change the buffer to c
\$\endgroup\$
2
\$\begingroup\$

Brain-Flak, 112 110 108 bytes

(()){{}({<<(({}<<>>)<<>>[({})]){{((<<>>)<<>>)}}{}{##{

}<<>>>>{}<<>>}<<>>)}{}((){{<<>>[[]]}})##}{}])}{([)}(}

Try it online!

This is based on my answer from Are the brackets matched?.

Tried not to use comments, but got stuck trying to make the pop nilads ({}) pair up. The problem lies in the easiest way to pair up a pair of brackets is to surround it in another pair of the same kind. While this is easy for other nilads, the {...} monad creates loops. In order to exit the loop you have to push a 0, but once you've exited the loop, you then have to pop the 0, which compounds the problem.

The 66 byte pre-paired solution is:

(()){{}({<(({}<>)<>[({})]){((<>)<>)}{}{}<>>{}<>}<>)}{}((){<>[[]]})

Try it online!

Outputs 1 or 1,0 if the input is a perfect pairing, 0,0 if not.

No comment version, 156 bytes

(()){{{}({<<(({}<<>>)<<>>[({{}((<<[[]]>>)){}}{}(<<[]>>){{}{}}{})]){{((<<>>)<<>>)}}{{}{}{}}{}{}<<>>>>{}<<>>}<<>>)}}{{}{}}{}((){<<>>[[]]})(<<()()>>){{}{}{}}{}

Try it online!

As Cat Wizard pointed out, the first answer does not work for all interpreters, as not all handle # comments. This version contains no comments.

\$\endgroup\$
  • \$\begingroup\$ Note that this only works in the ruby brainflak interpreter and thus is not a pure brainflak answer \$\endgroup\$ – Sriotchilism O'Zaic Jun 26 '18 at 13:02
  • \$\begingroup\$ @CatWizard Is there a canon Brain-Flak interpreter? As far as I know, Rain-Flak (Ruby) is the original interpreter. (Also, I'm working on a solution without comments) \$\endgroup\$ – Jo King Jun 26 '18 at 13:39
  • \$\begingroup\$ Not really. Rain-Flak is the original interpreter but it's comments syntax is unique to it. We wrote a Brain-Flak standard a while back, I don't remember where it ended up though. \$\endgroup\$ – Sriotchilism O'Zaic Jun 26 '18 at 13:43
  • \$\begingroup\$ @CatWizard Finished the no comment version \$\endgroup\$ – Jo King Jun 26 '18 at 14:14
2
\$\begingroup\$

Japt, 24 22 bytes

Outputs false for truthy and true for falsey.

&&!!e"(.)%1"PP"1%).("e

Try it

\$\endgroup\$
  • \$\begingroup\$ Would «e"(.)%1 work? \$\endgroup\$ – Oliver Jun 28 '18 at 16:22
  • \$\begingroup\$ @Oliver, that's what I originally had before the source restrictions were brought to my attention. Still trying to figure out a way to get it working with «, though. \$\endgroup\$ – Shaggy Jun 28 '18 at 16:35
  • \$\begingroup\$ @Oliver, it doesn't work, sadly. \$\endgroup\$ – Shaggy Jun 28 '18 at 17:11
  • \$\begingroup\$ I suspect you may have missed the restricted-source/source-layout part of the challenge, @Oliver. \$\endgroup\$ – Shaggy Jun 28 '18 at 18:12
  • \$\begingroup\$ I did...my bad. \$\endgroup\$ – Oliver Jun 28 '18 at 19:04
2
\$\begingroup\$

Brain-Flak, 96 bytes

{<<>>(({})<<>>[(([{}<<>>]))]){((<<[[]]>>))}{}{}{}{}<<>>}<<>>{{{}{}{}{}{}}((<<>>){{}{}}){{{}}}}{}

Try it online!

Outputs nothing if the input is perfectly paired, and 0 otherwise.

Non-perfectly paired (original) version:

{<>(({})<>[({}<>)]){((<()>))}{}{}{}<>}<>{((<>))}{}

Try it online!

\$\endgroup\$
2
\$\begingroup\$

Cubix, 30 bytes

1O@;??;@ii??O;>>;;;..1Wcc1??1W

Try it online!

Outputs 1 if the string is perfectly paired and nothing otherwise.

Cubified

      1 O @
      ; ? ?
      ; @ i
i ? ? O ; > > ; ; ; . .
1 W c c 1 ? ? 1 W . . .
. . . . . . . . . . . .
      . . .
      . . .
      . . .

Simplified

      1 O @
      ; ? .
      . @ .
i ? . . . . > ; ; ; . .
. W c . . . ? 1 W . . .
. . . . . . . . . . . .
      . . .
      . . .
      . . .

The logic and general structure are the same as in Mnemonic's answer, but without an explicit check for the empty string.

\$\endgroup\$
2
\$\begingroup\$

Haskell, 66 bytes

null.foldr((%))"";;cc%((r:t))|cc==r=t;;t%s=t:s--s:t=s%=r|t:dlof.un

Try it online!

Cat Wizard saved 6 bytes.

\$\endgroup\$
2
\$\begingroup\$

Add++, 146 bytes

D,g,@~~,L2_|*;;*|_2L,@,g,D
D,ff,@^^,BG€gBF;;FBg€GB,@D1:?:

xx:?

aa:1
`bb
Bxx;;B
Waa*bb,`yy,$ff>xx,`aa,xx|yy,`bb,Byy,xx:yy

O;;O:,B,`,|,`,>$,`,*W`

Try it online!

Fun fact: This was 272 bytes long before the explanation was started, now it beats Java.

Outputs True for perfectly balanced strings, and False otherwise

To my great satisfaction, this beats the boring palindromize version by 2 bytes, to prevent the result being printed twice. I have also aimed to have as little dead code as possible, nevertheless there are still some commented-out sections, and the code exits with an error code of 1, after printing the correct value.

NB : A bug with the BF commands was fixed while this answer was in development.

How it works

The code starts by defining the two key functions, \$\mathit{ff}\$ and \$g\$. These two functions are used to calculate the next step in the process of removing pairs, and work entirely from \$\mathit{ff}\$ i.e. only \$\mathit{ff}\$ is called from the main program, never \$g\$. If we define the input string as \$S\$, \$\mathit{ff}(S)\$ modifies \$S\$ in the following way:

First, identical adjacent characters in \$S\$ are grouped together. For an example of \$abbbaabacc\$, this yields the array \$[[a], [bbb], [aa], [b], [a], [cc]]\$. Over each of the sublists (i.e. the identical groups), we run the function \$g\$, and replace the sublists with the result of the function.

\$g\$ starts by unpacking the group, splatting the characters onto the stack. It then pushes the number of characters on the stack and takes the absolute difference with \$2\$ and that number. We'll call this difference \$x\$. Lets see how this transforms the respective inputs of \$[a]\$, \$[bb]\$ and \$[ccc]\$:

$$[a] \Rightarrow [a, 1]$$ $$[bb] \Rightarrow [b, b, 0]$$ $$[ccc] \Rightarrow [c, c, c, 1]$$

As you can see \$x\$ indicates how many of the next character we wish to keep. For simple pairs, we remove them entirely (yielding 0 of the next character), for lone characters we leave them untouched, or yield 1 of them, and for groups where \$x > 2\$, we want \$x - 2\$ of the character. In order to generate \$x\$ of the character, we repeat the character with *, and the function naturally returns the top element of the stack: the repeated string.

After \$g(s)\$ has been mapped over each group \$s\$, we splat the array to the stack to get each individual result with BF. Finally, the ^ flag at the function definition (D,ff,@^^,) tells the return function to concatenate the strings in the stack and return them as a single string. For pairs, which yielded the empty string from \$g\$, this essentially removes them, as the empty string concatenated with any string \$r\$ results in \$r\$. Anything after the two ;; is a comment, and is thus ignored.

The first two lines define the two functions, \$\mathit{ff}\$ and \$g\$, but don't execute \$\mathit{ff}\$ just yet. We then take input and store it in the first of our 4 variables. Those variables are:

  • \$\mathit{xx}\$ : The initial input and previous result of applying \$\mathit{ff}\$
  • \$\mathit{yy}\$ : The current result of applying \$\mathit{ff}\$
  • \$\mathit{aa}\$ : The loop condition
  • \$\mathit{bb}\$ : Whether \$\mathit{yy}\$ is truthy

As you can see, all variables and functions (aside from \$g\$) have two letter names, which allows them to be removed from the source code fairly quickly, rather than having a comment with a significant amount of \$\mathit{xyab}\$. \$g\$ doesn't do this for one main reason:

If an operator, such as , is run over a user defined function \$\mathit{abc}\$, the function name needs to be enclosed in {...}, so that the entire name is taken by the operator. If however, the name is a single character, such as \$g\$, the {...} can be omitted. In this case, if the function name was \$\mathit{gg}\$, the code for \$\mathit{ff}\$ and \$g\$ would have to change to

D,gg,@~~,L2_|*;;*|_2L,@D             (NB: -2 bytes)
D,ff,@^^,BG€{gg}BF;;FB}gg{€GB,@D?:   (NB: +6 bytes)

which is 4 bytes longer.

An important term to introduce now is the active variable. All commands except assignment assign their new value to the active variable and if the active variable is being operated on, it can be omitted from function arguments. For example, if the active variable is \$x = 5\$, then we can set \$x = 15\$ by

x+10 ; Explicit argument
+10  ; Implicit argument, as x is active

The active variable is \$x\$ by default, but that can be changed with the ` command. When changing the active variable, it is important to note that the new active variable doesn't have to exist beforehand, and is automatically assigned as 0.

So, after defining \$\mathit{ff}\$ and \$g\$, we assign the input to \$\mathit{xx}\$ with xx:?. We then need to manipulate our loop conditions ever so slightly. First, we want to make sure that we enter the while loop, unless \$\mathit{xx}\$ is empty. Therefore, we assign a truthy value to \$\mathit{aa}\$ with aa:1, the shortest such value being \$1\$. We then assign the truthiness of \$\mathit{xx}\$ to \$\mathit{bb}\$ with the two lines

`bb
Bxx

Which first makes \$\mathit{bb}\$ the active variable, then runs the boolean command on \$\mathit{xx}\$. The respective choices of \$\mathit{aa} := 1\$ and \$\mathit{bb} := \neg \neg \mathit{xx}\$ matter, as will be shown later on.

Then we enter our while loop:

Waa*bb,`yy,$ff>xx,`aa,xx|yy,`bb,Byy,xx:yy

A while loop is a construct in Add++: it operates directly on code, rather than variables. Constructs take a series of code statements, separated with , which they operate on. While and if statements also take a condition directly before the first , which consist of a single valid statement, such as an infix command with variables. One thing to note: the active variable cannot be omitted from the condition.

The while loop here consists of the condition aa*bb. This means to loop while both \$\mathit{aa}\$ and \$\mathit{bb}\$ are truthy. The body of the code first makes \$\mathit{yy}\$ the active variable, in order to store the result of \$\mathit{ff}(x)\$. This is done with

`yy,$ff>xx

We then activate our loop condition \$\mathit{aa}\$. We have two conditions for continued looping:

  • 1) The new value doesn't equal the old value (loop while unique)
  • 2) The new value isn't the empty string

One of Add++'s biggest drawbacks is the lack of compound statements, which necessitates having a second loop variable. We assign our two variables:

$$\mathit{aa} := \mathit{xx} \neq \mathit{yy}$$ $$\mathit{bb} := \neg \neg (\mathit{yy})$$

With the code

`aa,xx|yy,`bb,Byy

Where | is the inequality operator, and B converts to boolean. We then update the \$\mathit{xx}\$ variable to be the \$\mathit{yy}\$ variable with xx:yy, in preperation for the next loop.

This while loop eventually reduces the input into one of two states: the empty string or a constant string, even when applied to \$\mathit{ff}\$. When this happens, either \$\mathit{aa}\$ or \$\mathit{bb}\$ result in False, breaking out of the loop.

After the loop is broken, it can break for one of two reasons, as stated above. We then output the value of \$\mathit{aa}\$. If the loop was broken due to \$\mathit{x} = \mathit{y}\$, then both the output and \$\mathit{aa}\$ are False. If the loop was broken because \$\mathit{yy}\$ was equal to the empty string, then \$\mathit{bb}\$ is falsy and \$\mathit{aa}\$ and the output are truthy.

We then reach our final statement:

O

The program can now be in one of three states, in all of which the active variable is \$\mathit{bb}\$:

  • 1) The input was empty. In this case, the loop didn't run, \$\mathit{aa} = 1\$ and \$\mathit{bb} = \mathrm{False}\$. The correct output is \$\mathrm{False}\$.
  • 2) The input was perfectly balanced. If so, the loop ran, \$\mathit{aa} = \mathrm{True}\$ and \$\mathit{bb} = \mathrm{False}\$. The correct output is \$\mathrm{False}\$
  • 3) The input was not perfectly balanced. If so, the loop ran, \$\mathit{aa} = \mathrm{False}\$ and \$\mathit{bb} = \mathrm{True}\$. The correct output is \$\mathrm{True}\$

As you can see, \$\mathit{bb}\$ is equal to the expected output (albeit reversed from the logical answer), so we simply output it. The final bytes that help us beat Java come from the fact that \$\mathit{bb}\$ is the active variable, so can be omitted from the argument, leaving us to output either \$\mathrm{True}\$ or \$\mathrm{False}\$, depending on whether the input is perfectly balanced or not.

\$\endgroup\$
1
\$\begingroup\$

JavaScript (ES6), 76 bytes

Returns a boolean.

ff=ss=>ss==(ss=ss.replace(/(.)\1/,''))?!ss:ff(ss)//)(:!?,/1\).(/(ecalper.=(>

Try it online!

Suggested by @Shaggy: 58 bytes by returning an empty string for perfectly paired or throwing an error otherwise.

\$\endgroup\$
  • 1
    \$\begingroup\$ If one of the "return values" can be an error (waiting for confirmation on that) then this could be 66 bytes. \$\endgroup\$ – Shaggy Jun 25 '18 at 17:31
  • \$\begingroup\$ Programs can by default output via exit code. In this answer's particular case, the possible outputs would be exit code 0 for perfectly paired strings and exit code 1 for non-perfectly paired strings, which are two distinct values therefore fulfilling the criteria; so the 58 byter must be perfectly valid. \$\endgroup\$ – Don't be a x-triple dot Jun 25 '18 at 18:50
1
\$\begingroup\$

Wolfram Language (Mathematica), 70 64 bytes

{#//.{aa___,bb__,bb_,cc___}:>{aa,cc}}=={{}}(**(,{>:}_,{.#{&)**)&

Try it online!

Without comments, 92 bytes

((#//.bb_:>StringReplace[00[ecalpeRgnirtS>aa__:_.#&];bb,{{,}};00>00;00;aa:_~~aa_:>""]))==""&

Try it online!

\$\endgroup\$
1
\$\begingroup\$

Lua, 178 bytes

p=...S={}for a in p:gmatch"."do E=S[#S]~=a;S[E and#S+1 or#S]=E and a or X end;print(#S==0)--)0S#(tnirp;dne X ro a dna E=]S#ro 1+S#dna E[S;a=~]S#[S=E od"."hctamg:p ni a rof}{=S.=p

Try it online!

While it is a terribly long solution, this does make quite a bit of use of Lua-specific quirks. This is actually a minified brute force stack algorithm. The program is made complicated by the fact that Lua's patterns don't allow replacing pairs and regex is not built in.

Explanation:

p=... -- command-line argument
S={} -- the stack
for c in p:gmatch"." do -- shorter than "for i=1,#p do ..."
    E=S[#S]~=c -- check whether we have the right letter on top of stack
    -- could've saved some bytes by doing == instead of ~=
    -- but the double negation is necessary for ternary operator
    -- to work with nil values
    S[E and #S+1 or #S]=E and c or X -- Lua's awesome "ternary operator"
end
-- i'm sure there is a better way to output this (table indexing?)
print(#S==0)
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1
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Gol><>, 30 bytes

1ll1**F:}}:{=Q{~~||lzBBzl{Q={F

Try it online!

Everything after the first B is excess code and is not executed. A function that returns the top of stack as 1 if the input is a perfect pairing, 0 otherwise.

Explanation:

1       Push 1 as the end string marker
 ll1**  Push n, where n (len+1)*(len+2), 
        This is larger than the amount of steps needed to determine pairing
      F           |  Repeat that many times
       :}}:{=        Compare the first two characters of the string
             Q   |   If they are equal
              {~~    Pop both of them
        String is also rotated by 1
        If the string becomes empty, the 1 is compared to itself and removed.
                   lzB   Return whether the length of the stack is 0
                      Bzl{Q={F  Excess code to match unpaired symbols
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1
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Haskell, 92 bytes

gg""

gg((aa:cc:bb))dd|aa==cc=gg bb dd

gg  aa""=""==aa

gg  aa((bb:c))=gg((bb:aa))c--c:=c:|

Try it online!

@nimi's answer is pretty cool, it doesn't use any comments. This one is shorter but does use a comment.

@xnor's answer is also pretty cool, it does use comments and is shorter than this one.

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0
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Python 2, 114 bytes

import re

e=lambda i,nn=1:e(*re.subn('(.)\\1','',i))if nn else''==i##ieslef'1).('(nbus.er*(e:1=,i adbmal=r tropmi

Try it online!

Returns True for perfectly-paired strings, False otherwise.

(Actually fails to verify itself, because (.) won't match the newlines in the code! But @Cat Wizard said this is okay, because newlines aren't printable ASCII characters, so my program needn't handle them.)


This is a perfectly-paired version of:

import re;p=lambda s,n=1:p(*re.subn('(.)\\1','',s))if n else''==i

for which a “lazier” perfectization of code + '##' + f(code[::-1]) would give 120 bytes. (That is, renaming the variables etc. to introduce more collapsed pairs inside the comment half of the code saved 6 bytes.)


re.subn is a little-known variant of re.sub that returns a tuple (new_string, number_of_substitutions_made). It's pretty good for finding regex substitution fixpoints!

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0
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Jelly, 26 24 22 bytes

ẠƬµF€ḂLḣgŒŒgḣLḂ$$€FµƬẠ

Try it online!

Weirdly seems to work without moving the backwards code to an unused link.

Returns 0 if the input is perfectly paired, 1 otherwise.

Active code:

ŒgḣLḂ$$€FµƬẠ
Œg            Group runs 'abbbcc'->['a','bbb','cc']
       €      For each of these strings:
      $       Monad{
     $            Monad{
   L                  Find the length...
    Ḃ                 ...mod 2. 
                      } -> [1, 1, 0] in this example.
  ḣ               Take this many characters from the string.
                  } -> [['a'], ['b'], []]
        F     Flatten -> ['a', 'b']
          Ƭ   Repeat...
         µ    The last monadic chain until a fixed point is reached.
           Ạ  All. If it is not a perfectly paired string, all elements in the 
              result of Ƭ will be nonempty and 1 is returned.
              If it is perfectly paired, the last element is [] which is falsy
              and 0 is returned.
\$\endgroup\$
0
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Attache, 82 bytes

{0=#Fixpoint[{Replace[_,/"(.)\\1",""]},_]}??}]_,}],"1).("/,_[ecalpeR{[tniopxiF#=0{

Try it online!

Nothing incredible here. Fixpoints a function which removes consecutive pairs.

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0
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Java 8, 158 156 154 bytes

n->{for(;n.matches(".*(.)\\1.*");n=n.replaceAll("(.)\\1",""));return  n.isEmpty();}//};)(ytpmEsi.ruter;,"1).("(Aecalper.n=n;)"*.1).(*."(sehctam.n;(rof{>-n

Returns a boolean (true/false).

-2 bytes thanks to @raznagul.

Try it online.

Explanation:

n->{                              // Method with String parameter and boolean return-type
  for(;n.matches(".*(.)\\1.*");   //  Loop as long as the String still contains pairs
    n=n.replaceAll("(.)\\1","")); //   Remove all pairs
  return  n.isEmpty();}           //  Return whether the String is empty now
//};)(ytpmEsi.ruter;,"1).("(Aecalper.n=n;)"*.1).(*."(sehctam.n;(rof{>-n
                                  // Comment reversed of the source code,
                                  // minus the pairs: '\\';'ll';'\\';'""))';'n  n';'//'
\$\endgroup\$
  • 1
    \$\begingroup\$ By renaming s to n and adding a second space to return s.isEmpty you can remove s n from the comment, saving 2 bytes in total. \$\endgroup\$ – raznagul Jun 26 '18 at 15:36

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