# 完全なペアリングの検出

25

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

2つのペアが重複する場合、そのうちの1つのみを削除することに注意してください。

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

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

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

あなたの仕事は、（印刷可能なASCIIの）文字列を受け取り、それが完全にペアリングされているかどうかを判断する、完全にペアリングされたコンピューターコードを記述することです。 ソースのバイト文字列それ自体が完全にペアの文字列である必要がありますが、コードは必ずしも印刷可能なASCIIに制限する必要はありません。

2つの異なる値を出力できます。1つは入力が完全にペアリングされている場合、もう1つはペアリングされていない場合です。

これは質問であるため、回答はソースのバイト数でスコア付けされ、バイト数は少ない方が良いでしょう。

## テストケース

$abbbacc\to \mathrm{F}\mathrm{a}\mathrm{l}\mathrm{s}\mathrm{e}\phantom{\rule{0ex}{0ex}}abcba\to \mathrm{F}\mathrm{a}\mathrm{l}\mathrm{s}\mathrm{e}\phantom{\rule{0ex}{0ex}}abab\to \mathrm{F}\mathrm{a}\mathrm{l}\mathrm{s}\mathrm{e}\phantom{\rule{0ex}{0ex}}abbbaabacc\to \mathrm{T}\mathrm{r}\mathrm{u}\mathrm{e}\phantom{\rule{0ex}{0ex}}eabbccadde\to \mathrm{T}\mathrm{r}\mathrm{u}\mathrm{e}\phantom{\rule{0ex}{0ex}}bbbb\to \mathrm{T}\mathrm{r}\mathrm{u}\mathrm{e}$$abbbacc \rightarrow \mathrm{False}\\ abcba \rightarrow \mathrm{False}\\ abab \rightarrow \mathrm{False}\\ abbbaabacc \rightarrow \mathrm{True}\\ eabbccadde \rightarrow \mathrm{True}\\ bbbb \rightarrow \mathrm{True}$

1

ジオビット

11
@Geobits私は同意しません。デッドコードを拒否することは漠然とした定義の単なる泥沼であり、とにかく決して面白くないと思う人がいます。2人の場合、コメントを許可することでエントリーの水準が下がると思います。3人にとって、コメントのないコードは、コメントのないコードよりも必然的に得点が高くなると考えています。ひねりは面白くないかもしれませんが、特定の方法で答えを出すために強化できない制限を追加すると、間違いなく面白くないでしょう。

4

アーナウルド

2
@Geobitsより創造的な答えを促したかもしれないことの1つは、スコアリングに空の文字列を取得するためのステップ数を考慮することです。コメントを使用すると、このスコアが非常に高くなる傾向があります。これは、スコアが低いとペアをかなりインターレースする必要があるため、コメントが自然にネストするためです。しかし、明らかにその変更を行うには遅すぎます。

1
@dylnan空の文字列は可能ですが、永久にループすることは有効な出力ではありません。

10

((""##))
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


コメントはありません。Trueまたはを返しますFalse

オンラインでお試しください！

2
これは私が今まで見たハスケルのような最小のハスケルです
キュービック

5

# Python 2、94バイト

ss=''

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

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

オンラインでお試しください！

5

# 05AB1E、26 24 22 20 18バイト

ovsのおかげで-2バイト。文字列が完全にペアになっている場合は0、それ以外の場合は1を出力します。

ΔγʒgÉ}JJ}ĀqqĀÉgʒγΔ


オンラインでお試しください！

ΔγʒgÉ} JJ}ĀqqĀÉgʒγΔ–完全なプログラム。
Δ} –結果が変わらなくなるまで：
γʒ} –文字列を等しい文字のチャンクに分割し、以下でフィルタリングします。
ジェ-長さは奇数ですか？
JJ –フィルタリング後、パーツを結合し直しますが、これを行います
以前のバージョンと同様に、2回保存して2バイトを節約します。
Ā–結果が空かどうかを確認します
q –実行を終了（終了）します。残りのコードは無視されます。
qĀÉgʒγΔ–一致しない部分をミラーリングして、ソースレイアウトを支援します。


### 以前のバージョン

これは純粋に未定義の動作に依存するため（「デッドコード」はありません）、完全にペア化された文字列には[['0']]を、完全に一致しない文字列には[['1']]を出力します：

ΔγεDgÉ£}JJ}ĀĀ£ÉgDεγΔ


また、説明した22バイトバージョンは、上記のとおりですが、UBを悪用せず、適切値を生成します。

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.


5

# Cubix、54バイト

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


ここで試してみてください

### 立方体

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


### 説明

ほとんどの文字は、コードを完全にペアにするために必要なフィラーです。それらを.（no-op）に置き換えると、

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


これは3つのステップに分けることができます。

• 空の文字列（左i?）を確認します。
• ループ、スタックにキャラクターを投げて、重複（下と右のすべて）をポップします。
• スタックが空かどうかを確認します（上部のもの）。

4

# V、20、18のバイト

òóˆ±òø‚

::‚øò±ˆóò


オンラインでお試しください！

Hexdump：

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


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


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{f}\mathit{f}$$\mathit{ff}$ and $g$$g$. These two functions are used to calculate the next step in the process of removing pairs, and work entirely from $\mathit{f}\mathit{f}$$\mathit{ff}$ i.e. only $\mathit{f}\mathit{f}$$\mathit{ff}$ is called from the main program, never $g$$g$. If we define the input string as $S$$S$, $\mathit{f}\mathit{f}\left(S\right)$$\mathit{ff}(S)$ modifies $S$$S$ in the following way: First, identical adjacent characters in $S$$S$ are grouped together. For an example of $abbbaabacc$$abbbaabacc$, this yields the array $\left[\left[a\right],\left[bbb\right],\left[aa\right],\left[b\right],\left[a\right],\left[cc\right]\right]$$[[a], [bbb], [aa], [b], [a], [cc]]$. Over each of the sublists (i.e. the identical groups), we run the function $g$$g$, and replace the sublists with the result of the function. $g$$g$ starts by unpacking the group, splatting the characters onto the stack. It then pushes the number of characters on the stack and takes the absolute difference with $2$$2$ and that number. We'll call this difference $x$$x$. Lets see how this transforms the respective inputs of $\left[a\right]$$[a]$, $\left[bb\right]$$[bb]$ and $\left[ccc\right]$$[ccc]$: $\left[a\right]⇒\left[a,1\right]$ $\left[bb\right]⇒\left[b,b,0\right]$ $\left[ccc\right]⇒\left[c,c,c,1\right]$ As you can see $x$$x$ indicates how many of the next character we wish to keep. For simple pairs, we remove them entirely (yielding 0 of the next character), for lone characters we leave them untouched, or yield 1 of them, and for groups where $x>2$$x > 2$, we want $x-2$$x - 2$ of the character. In order to generate $x$$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\left(s\right)$$g(s)$ has been mapped over each group $s$$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$$g$, this essentially removes them, as the empty string concatenated with any string $r$$r$ results in $r$$r$. Anything after the two ;; is a comment, and is thus ignored. The first two lines define the two functions, $\mathit{f}\mathit{f}$$\mathit{ff}$ and $g$$g$, but don't execute $\mathit{f}\mathit{f}$$\mathit{ff}$ just yet. We then take input and store it in the first of our 4 variables. Those variables are: • $\mathit{x}\mathit{x}$$\mathit{xx}$ : The initial input and previous result of applying $\mathit{f}\mathit{f}$$\mathit{ff}$ • $\mathit{y}\mathit{y}$$\mathit{yy}$ : The current result of applying $\mathit{f}\mathit{f}$$\mathit{ff}$ • $\mathit{a}\mathit{a}$$\mathit{aa}$ : The loop condition • $\mathit{b}\mathit{b}$$\mathit{bb}$ : Whether $\mathit{y}\mathit{y}$$\mathit{yy}$ is truthy As you can see, all variables and functions (aside from $g$$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{x}\mathit{y}\mathit{a}\mathit{b}$$\mathit{xyab}$. $g$$g$ doesn't do this for one main reason: If an operator, such as €, is run over a user defined function $\mathit{a}\mathit{b}\mathit{c}$$\mathit{abc}$, the function name needs to be enclosed in {...}, so that the entire name is taken by the operator. If however, the name is a single character, such as $g$$g$, the {...} can be omitted. In this case, if the function name was $\mathit{g}\mathit{g}$$\mathit{gg}$, the code for $\mathit{f}\mathit{f}$$\mathit{ff}$ and $g$$g$ would have to change to D,gg,@~~,L2_|*;;*|_2L,@D (NB: -2 bytes) D,ff,@^^,BG€{gg}BF;;FB}gg{€GB,@D?: (NB: +6 bytes)  which is 4 bytes longer. An important term to introduce now is the active variable. All commands except assignment assign their new value to the active variable and if the active variable is being operated on, it can be omitted from function arguments. For example, if the active variable is $x=5$$x = 5$, then we can set $x=15$$x = 15$ by x+10 ; Explicit argument +10 ; Implicit argument, as x is active  The active variable is $x$$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{f}\mathit{f}$$\mathit{ff}$ and $g$$g$, we assign the input to $\mathit{x}\mathit{x}$$\mathit{xx}$ with xx:?. We then need to manipulate our loop conditions ever so slightly. First, we want to make sure that we enter the while loop, unless $\mathit{x}\mathit{x}$$\mathit{xx}$ is empty. Therefore, we assign a truthy value to $\mathit{a}\mathit{a}$$\mathit{aa}$ with aa:1, the shortest such value being $1$$1$. We then assign the truthiness of $\mathit{x}\mathit{x}$$\mathit{xx}$ to $\mathit{b}\mathit{b}$$\mathit{bb}$ with the two lines bb Bxx  Which first makes $\mathit{b}\mathit{b}$$\mathit{bb}$ the active variable, then runs the boolean command on $\mathit{x}\mathit{x}$$\mathit{xx}$. The respective choices of $\mathit{a}\mathit{a}:=1$$\mathit{aa} := 1$ and $\mathit{b}\mathit{b}:=\mathrm{¬}\mathrm{¬}\mathit{x}\mathit{x}$$\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{a}\mathit{a}$$\mathit{aa}$ and $\mathit{b}\mathit{b}$$\mathit{bb}$ are truthy. The body of the code first makes $\mathit{y}\mathit{y}$$\mathit{yy}$ the active variable, in order to store the result of $\mathit{f}\mathit{f}\left(x\right)$$\mathit{ff}(x)$. This is done with

yy,$ff>xx  We then activate our loop condition $\mathit{a}\mathit{a}$$\mathit{aa}$. We have two conditions for continued looping: • 1) The new value doesn't equal the old value (loop while unique) • 2) The new value isn't the empty string One of Add++'s biggest drawbacks is the lack of compound statements, which necessitates having a second loop variable. We assign our two variables: $\mathit{a}\mathit{a}:=\mathit{x}\mathit{x}\ne \mathit{y}\mathit{y}$ $\mathit{b}\mathit{b}:=\mathrm{¬}\mathrm{¬}\left(\mathit{y}\mathit{y}\right)$ With the code aa,xx|yy,bb,Byy  Where | is the inequality operator, and B converts to boolean. We then update the $\mathit{x}\mathit{x}$$\mathit{xx}$ variable to be the $\mathit{y}\mathit{y}$$\mathit{yy}$ variable with xx:yy, in preperation for the next loop. This while loop eventually reduces the input into one of two states: the empty string or a constant string, even when applied to $\mathit{f}\mathit{f}$$\mathit{ff}$. When this happens, either $\mathit{a}\mathit{a}$$\mathit{aa}$ or $\mathit{b}\mathit{b}$$\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{a}\mathit{a}$$\mathit{aa}$. If the loop was broken due to $\mathit{x}=\mathit{y}$$\mathit{x} = \mathit{y}$, then both the output and $\mathit{a}\mathit{a}$$\mathit{aa}$ are False. If the loop was broken because $\mathit{y}\mathit{y}$$\mathit{yy}$ was equal to the empty string, then $\mathit{b}\mathit{b}$$\mathit{bb}$ is falsy and $\mathit{a}\mathit{a}$$\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{b}\mathit{b}$$\mathit{bb}$: • 1) The input was empty. In this case, the loop didn't run, $\mathit{a}\mathit{a}=1$$\mathit{aa} = 1$ and $\mathit{b}\mathit{b}=\mathrm{F}\mathrm{a}\mathrm{l}\mathrm{s}\mathrm{e}$$\mathit{bb} = \mathrm{False}$. The correct output is $\mathrm{F}\mathrm{a}\mathrm{l}\mathrm{s}\mathrm{e}$$\mathrm{False}$. • 2) The input was perfectly balanced. If so, the loop ran, $\mathit{a}\mathit{a}=\mathrm{T}\mathrm{r}\mathrm{u}\mathrm{e}$$\mathit{aa} = \mathrm{True}$ and $\mathit{b}\mathit{b}=\mathrm{F}\mathrm{a}\mathrm{l}\mathrm{s}\mathrm{e}$$\mathit{bb} = \mathrm{False}$. The correct output is $\mathrm{F}\mathrm{a}\mathrm{l}\mathrm{s}\mathrm{e}$$\mathrm{False}$ • 3) The input was not perfectly balanced. If so, the loop ran, $\mathit{a}\mathit{a}=\mathrm{F}\mathrm{a}\mathrm{l}\mathrm{s}\mathrm{e}$$\mathit{aa} = \mathrm{False}$ and $\mathit{b}\mathit{b}=\mathrm{T}\mathrm{r}\mathrm{u}\mathrm{e}$$\mathit{bb} = \mathrm{True}$. The correct output is $\mathrm{T}\mathrm{r}\mathrm{u}\mathrm{e}$$\mathrm{True}$ As you can see, $\mathit{b}\mathit{b}$$\mathit{bb}$ is equal to the expected output (albeit reversed from the logical answer), so we simply output it. The final bytes that help us beat Java come from the fact that $\mathit{b}\mathit{b}$$\mathit{bb}$ is the active variable, so can be omitted from the argument, leaving us to output either $\mathrm{T}\mathrm{r}\mathrm{u}\mathrm{e}$$\mathrm{True}$ or $\mathrm{F}\mathrm{a}\mathrm{l}\mathrm{s}\mathrm{e}$$\mathrm{False}$, depending on whether the input is perfectly balanced or not. 1 # 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. 1 If one of the "return values" can be an error (waiting for confirmation on that) then this could be 66 bytes. Shaggy 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. Mr. Xcoder 1 # 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! 1 # 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) 1 # 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  1 # 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. 1 # 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. 0 # 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! 0 # 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{
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.


0

# Attache, 82 bytes

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


Try it online!

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

0

# Java 8, 158156 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';'//'

1
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.
raznagul