Vamos ter uma função $f$ que pega uma string e remove todos os pares de caracteres idênticos adjacentes. Por exemplo

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

Observe que quando dois pares se sobrepõem, removemos apenas um deles.

Chamaremos uma string perfeitamente emparelhada se um aplicativo repetido eventualmente produzir a string vazia. Por exemplo, a string acima de $abbbacc$ não está perfeitamente emparelhada porque, se aplicarmos $f$ novamente, ainda obteremos $aba$ . No entanto, uma string como $eabbccadde$ está perfeitamente emparelhada porque, se aplicarmos $f$ três vezes, obteremos a string vazia

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

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

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

Sua tarefa é escrever um código de computador perfeitamente emparelhado que use uma string (de ASCII imprimível) e decida se ele está perfeitamente emparelhado. A bytestring de sua fonte deve ser uma sequência perfeitamente emparelhada , embora seu código não precise necessariamente ser restrito ao ASCII imprimível.

Você pode gerar dois valores distintos: um para o caso em que a entrada está perfeitamente emparelhada e outro para os casos em que não está.

Esta é uma questão de código-golfe, para que as respostas sejam pontuadas pelo tamanho em bytes de sua origem, com menos bytes sendo melhores.

Casos de teste

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

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

Sem comentários. Retorna Trueou False.

Experimente online!

Edit: -22 bytes graças ao @Cat Wizard

Python 2 , 94 bytes

ss=''

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

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

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A etapa inteira da atualização é ss=[cc+ss,ss[1:]][cc==ss[:1]]cancelada para apenas =[+,[.

05AB1E , 26 24 22 20 18 bytes

-2 bytes graças a ovs . Emite 0 se a sequência estiver perfeitamente emparelhada, 1 caso contrário.

ΔγʒgÉ}JJ}ĀqqĀÉgʒγΔ

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ΔγʒgÉ} JJ} ĀqqĀÉgʒγΔ - Programa completo.
Δ} - Até o resultado não mudar mais:
 γʒ} - Divida a sequência em partes de caracteres iguais e filtre por:
   gÉ - O comprimento é estranho?
      JJ - E após a filtragem, junte as peças novamente, mas faça isso
                     duas vezes para salvar 2 bytes, como nas versões anteriores.
         Ā - Verifique se o resultado está vazio
          q - Terminar (sair) da execução. O restante do código é ignorado.
           qĀÉgʒγΔ - Espelhe a peça não correspondida para ajudar no layout de origem.

Versões prévias

Este baseia-se puramente em comportamento indefinido (portanto, não há "código morto") e gera [['0']] para cadeias perfeitamente emparelhadas e [['1']] para cadeias não perfeitamente correspondentes:

ΔγεDgÉ£}JJ}ĀĀ£ÉgDεγΔ 

E a versão de 22 bytes, explicada, que é exatamente a UB acima, mas sem abusar, e produzindo valores sensatos .

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.

Cubix , 54 bytes

U#;[email protected]>??>i..??O.@1^^...u--u.u!ww;..#..U..;..;!^^!

Não produz nada se a sequência estiver perfeitamente emparelhada e 1não.
Experimente aqui

Cubificado

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

Explicação

A maioria dos caracteres é necessária para preencher perfeitamente o código. Substituindo aqueles com .(no-op), obtemos

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

Isso pode ser dividido em três etapas:

• Verifique com a sequência vazia (a esquerda ie a ?).
• Faça um loop, jogando caracteres na pilha e exibindo duplicatas (tudo na parte inferior e direita).
• Verifique se a pilha está vazia (o material na parte superior).

V , 20 , 18 bytes

òóˆ±òø‚

::‚øò±ˆóò

Experimente online!

Hexdump:

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

Saídas 0 para verdade, 1 para falsidade. Obrigado ao nmjcman101 por salvar indiretamente 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

R , 142 126 bytes

Lógica mais rigorosa e alguns bytes de comentário publicados por @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

Experimente 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:

Experimente online!

Função detector recursiva seguida de comentário com todos os caracteres da função na ordem inversa.

APL (Dyalog) , 38 bytes

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

Experimente online!

Retina , 28 26 bytes

+`(.)\1

C`^$

$^`C1\).(`+

Experimente online!

Saídas `C1\).(`+0`C1\).(`+para casos falsos e verdadeiros `C1\).(`+1`C1\).(`+.

Flak cerebral , 228 200 bytes

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

Experimente online!

Esta é uma prova de conceito. Provavelmente poderia ser mais curto. No entanto, ele não usa nenhum comentário.

Saída 0,0se a entrada estiver perfeitamente emparelhada e 0,1se a entrada não estiver.

sed 4.2.2 , 34 bytes

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

Experimente online!

Seqüências de caracteres emparelhadas dão saída vazia, outras não emparelhadas ct:

A versão palindrômica trivial é 32 :;ss(.)\1ss;t;/./cc/./;t;1\).(;:. A solução antiga foi :;ss((..??\??))\1ss1;t;;/./cc/./t:(alterada porque a atual abusava cmenos, edit: yay agora só há 1 caractere depois c: D)

(observe que ;é o separador de instruções)

: declara um rótulo vazio

:t declara o rótulo t

ss((..??\??))\1ss1é uma substituição, no sed, você pode alterar o delimitador para uma substituição, e foi isso que fiz ao alterá-lo s; portanto, o que isso faz é substituir o primeiro (como indicado 1no final)

• partida de ((..??\??))\1

  • . qualquer personagem
  • .?? seguido por um caractere opcional opcional
  • \?? e um opcional ?
  • seguido pela mesma coisa ao lado

• com nada

Agora, essa substituição está emparelhada com ela mesma, então o ;s antes e depois também são cancelados

t e volte ao rótulo até que não haja mais substituições bem-sucedidas

/..?/se .(curinga) seguido por .?um caractere opcional for correspondido

• cc altere o buffer para c

Brain-Flak , 112 110 108 bytes

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

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

Experimente online!

Isso se baseia na minha resposta de Os colchetes são compatíveis? .

Tentei não usar comentários, mas fiquei preso tentando fazer o pop nilads ( {}) par. O problema está na maneira mais fácil de emparelhar um par de colchetes: envolvê-lo em outro par do mesmo tipo. Embora isso seja fácil para outras nilads, a {...}mônada cria loops. Para sair do loop, você precisa pressionar um 0, mas depois de sair do loop, é necessário exibir o 0, o que agrava o problema.

A solução pré-emparelhada de 66 bytes é:

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

Experimente online!

Saídas 1ou 1,0se a entrada for um emparelhamento perfeito, 0,0se não.

Versão sem comentários, 156 bytes

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

Experimente online!

Como o Assistente de Gato apontou, a primeira resposta não funciona para todos os intérpretes, pois nem todos lidam com #comentários. Esta versão não contém comentários.

Japonês, 24 22 bytes

Saídas falsepara verdade e truepara falsey.

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

Tente

Brain-Flak, 96 bytes

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

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

Non-perfectly paired (original) version:

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

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Haskell, 66 bytes

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

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Cat Wizard saved 6 bytes.

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`

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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 $[ccc]$:

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:

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.

JavaScript (ES6), 76 bytes

Returns a boolean.

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

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

Wolfram Language (Mathematica), 70 64 bytes

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

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Without comments, 92 bytes

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

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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

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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)

Gol><>, 30 bytes

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

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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

Cubix, 30 bytes

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

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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.

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:|

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@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.

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

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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!

Jelly, 26 24 22 bytes

ẠƬµF€ḂLḣgŒŒgḣLḂ$$€FµƬẠ

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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.

Attache, 82 bytes

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

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

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';'//'