問題
入力がない場合、次の文字列を出力または返すプログラムまたは関数を作成します。
(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)
ルール
- 最短のプログラムが勝ちます。
- 末尾の空白が許可されます。
- 末尾の改行が許可されます。
- 許可されている関数の未使用のパラメーター。
入力がない場合、次の文字列を出力または返すプログラムまたは関数を作成します。
(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)
回答:
pΙ2○3V?hG]‘Γ
説明:
pΙ2○3V?hG]‘ Push compressed string of the first half of the string (this is the kind of thing this compressor was made for)
Γ mirror, swap chars & palindromize with one character overlap
"(<>.".∞ηJÀ24£.∞
説明
"(<>." # push this string
# STACK: "(<>."
.∞ # intersected mirror
# STACK: "(<>.<>)"
η # compute prefixes
J # join to string
# STACK: "((<(<>(<>.(<>.<(<>.<>(<>.<>)"
À # rotate left
24£ # take the first 24 chars
# STACK: "(<(<>(<>.(<>.<(<>.<>(<>."
.∞ # intersected mirror
# STACK: "(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"
ウェブサイトにアクセスできなかったため、これをすべて手動で把握する必要がありました-.-
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ほとんどの文字は比較的簡単です:括弧の場合
([]["fill"]+[])[13] // '('
([]["fill"]+[])[14] // ')'
完全に停止するには、jsに1.1e + 21のような科学的記数を作成させ、文字列に変換した後にドット文字を取得します。
角かっこはより難しいので、斜体のhtmlオブジェクト文字列を作成して山かっこを盗むために関数を実行する必要があります。これの主なトリックは、イタリックという単語の「c」を取得することです。これには、「function」からcを取得するために別の関数を作成する必要があります
ここには多くの改善の余地がありますが、大部分は角括弧を取得するためにどの関数が使用されますか。また、イタリック体の方が良い場合に文字「c」を取得する簡単な方法を思い出しますが、それを見つけるには古いファイルを調べなければなりません
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s=>"(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"
-2バイト、Kevin Cruijssenのおかげ
RIP C#
;
は、ラムダ応答ではカウントされません。また()
、単一の文字(未使用null
オブジェクト)に置き換えることができます。2番目の関連メタ投稿。
s=>
、まったく問題ありません。
3?4?@KritixiLithosのおかげでバイト
i(<>.<>)òÙxlHÄ$xGòxÍî
i(<>.<>) 'insert (<>.<>)
ò ò 'recursively
Ù 'duplicate this (bottom line) down
x 'delete the first character
l 'and break when there's only one character left
H 'go to the top of the buffer
Ä 'duplicate that line up
$x 'delete the last character
G 'and go back to the bottom of the buffer
x 'delete the extra ) left by the loop
Íî 'and remove all newlines
hx
だけで置き換えることができX
ます。
dkHxògJ
よりも1バイト短くなりdkVHgJ0x
ます。
Íî
事はあまりにも動作します
i(<>.<>)òÙxlHÄ$xGòddÍî
は23 を取得
lX
ために切り替えたxl
:D
cat(substring('(<>.<>)',-4:6,2:12),sep='')
Rで部分文字列が機能する方法を利用substring('(<>.<>)',-4:6,2:12)
する非常に単純なプログラム。次のベクトルを生成します。
> substring('(<>.<>)',-4:6,2:12)
[1] "(<" "(<>" "(<>." "(<>.<" "(<>.<>" "(<>.<>)" "<>.<>)" ">.<>)"
[9] ".<>)" "<>)" ">)"
cat
空の区切り文字を使用すると、必要な形式でSTDOUTに出力されます。
2単純なputs()
ソリューションよりも4バイト少ない。いくつかのエレガントな再帰的ソリューションがあるはずですが、これまでのところ、すべての試行でオーバーヘッドが大きくなりすぎています。
f(i){for(i=1;i++<12;)printf("%.*s",i,"(<>.<>)"+i/7*i%7);}
私はコードゴルフをやったことがないので、これが初めてです。フィードバックは大歓迎です。
++++[>+++++<-]>[>++>+++>+++>++>++[<]>-]>.>.<.>.>++.<<.>.>.>>++++++.<<<<.>.>.>>.<<<.<.>.>.>>.<<<.>.<<.>.>.>>.<<<.>.>+.<<.>.>>.<<<.>.>.<.>>.<<<.>.>.>.<<<.>.>.<<.>.>.<.>.
私の意見では、最も明白な解決策を探しました。最初に、セル1〜5を文字「()。<>」のいずれかに設定します。次に、正しいセルに移動してキャラクターを出力します。
更新:文字を「テープ」に表示する順序を変更して、プログラムをより短く、より効率的にしました。
更新2:しばらくして私の投稿を再訪して、セルを設定するための短いシーケンスを使用すると、いくつかのバイトを節約できることに気付きました。
g=(i=2)=>i>12?'':'(<>.<>)'.slice(i-14,i)+g(i+1)
f=(i=11)=>i?f(i-1)+'(<>.<>)'.slice(i-13,i+1):''
// Both of the above work and are the same size
document.write('<pre>Actual g(): ' + g() + '\nActual f(): ' + f() + '\nExpected: (<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)</pre>');
RodのPythonの回答と同じ方法を使用します。
A(<>ι(<ιι.ι.<ι.ιι.‖B
パターンに気づいたDestructible Lemonに感謝します(-4)。
AST:
Program
├A: Assign
│├'(<>': String '(<>'
│└ι: Identifier ι
├Print
│└'(<': String '(<'
├Print
│└ι: Identifier ι
├Print
│└ι: Identifier ι
├Print
│└'.': String '.'
├Print
│└ι: Identifier ι
├Print
│└'.<': String '.<'
├Print
│└ι: Identifier ι
├Print
│└'.': String '.'
├Print
│└ι: Identifier ι
├Print
│└ι: Identifier ι
├Print
│└'.': String '.'
└‖B: Reflect butterfly
└Multidirectional
どのような‖B
行いは、基本的に視覚的にpalindromizeキャンバスです。
-a
。
Print
sがあります。あなたが何を意味するのかわかりません。
私はこれが足りないことを知っていますが、他のソリューションよりもはるかに短いです...
f(){puts("(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)");}
これは私がややトリッキーにしようとしているバージョンです...
char*e="(<>.<>)<>.<>)>.<>).<>)<>)>)";f(i,n){for(n=2;n<8;n++)for(i=0;i<n;i++)putchar(e[i]);puts(&e[i]);}
完全にトリッキーな再帰バージョン...
char*e="(<>.<>)";main(i,j){if(i<6){for(++i,j=0;j<i;putchar(e[j++]));main(i);}else if(i<12)printf(e+i++-6),main(i);}
アダム・ウェストに敬意を表してバットマンを眺めている空っぽの群衆...
s=>{string b="NanananaNanaBATMAN!NanaBATMAN!NANANanaBATMAN!NANAnanaNanaBATMAN!NANABATMAN!NanaBATMAN!NANABATMAN!naNABATMAN!NANABATMAN!naNANAnaNANABATMAN!naNANANABATMAN!naNABATMAN!naNANAnanaNA",r="",t;for(int i=0;i<176;)if(b[i]=='B'){r+="<>";i+=7;}else{t=b.Substring(i,4);if(t=="Nana")r+="(";else if(t=="nana")r+="<";else if(t=="NAna")r+=">";else if(t=="NANA")r+=".";else r+=")";i+=4;}return r;}
raze"(<>.<>)"{(y-5)_x}/:(!)11
ストリートスターのおかげで-2バイト
説明の編集:
言語は右から左に解釈されます。
使い方
raze"(<>.<>)"{(y-5)_x}/:(!)11
(!)11 /yields indices [0-10], right parameter of function (y)
/: /each right: loop right parameter of a dyadic function
{ } /function
"(<>.<>)" /left parameter of function (x)
(y-5) /subtract 5 from y (indices)
_ /remove first y chars from x, or last y chars if negative
raze /flatten string output to produce the final string
_x
1バイト前のスペースを削除し、さらに1バイト節約til 11
する(!)11
ために変更する= 2バイトを保存する:)
map take[2..7]++map drop[1..5]>>=($"(<>.<>)")
ハードコーディングされたソリューションに勝つための最初のHaskellの回答!
この答えが機能する方法は、文字列に適用される関数のリストを作成すること(<>.<>)
です。まず、左と中央を構築します
map take[2..7]
これにより、サイズ2から7までのすべてのプレフィックスが得られます。次に、
map drop[1..5]
サイズ6から2までのすべての接尾辞が与えられます。
関数のリストを取得したら、モナドバインド(>>=
)を使用しますconcatMap
。concatmapで使用する関数($"(<>.<>)")
は、入力を文字列に適用します(<>.<>)
。
これにより、文字列が作成されます。
µ
x->"(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"
ボーリングが、Javaでこれを短くする方法はありません..です単なる文字列がすでに11バイトである一時を初期化...( String t="something";
)、および使用substring
時間のカップルは確かにあまりにも多くのバイトがかかります。..
リテラルリターンの最短の選択肢は、おそらくこれ(58バイト)です。
x->"(<(x(x.(x.<(x.x(x.x)x.x)>.x).x)x)>)".replace("x","<>")
=>
)を使用して追加します。
進行中のゴルフ。
>++++++++[<+++++>-]>++++++[<++++++++++>-]>++++++[<++++++++++>-]<++>>++++++++[<+++++>-]<++++++<<<.>.<.>.>.<<.>.>.>.<<<.>.>.>.<<.<.>.>.>.<<.>.<<.>.>.>.<<.>.<<+.>.>.>.<<.>.<<.>>.>.<<.>.<<.>>>.<<.>.<<.>.>.<<.>>.<<.
import Data.List
import Control.Arrow
concat$tail$init$uncurry(++)$inits&&&tail.tails$"(<>.<>)"
これは、出力を含む文字列リテラルに必要な49バイトよりもはるかに長くなりますが、構造体を利用するためにできる最善の方法です。いつものように、私は矢印が大好きinits &&& tails
で、入力の先行部分文字列のリストと後続部分文字列のリストのタプルを作成します。次に、これらの2つのタプル要素は、タプルを++
に渡すことで1つのリストにまとめられ、そのリストはconcat
1つの大きな文字列にエネーレートされます。tail
そしてinit
呼び出しは重複を避ける(<>.<>)
中央(によって生成された一つにinits
、によって他のtails
)アカウントに長さ2以上のサブストリングのみを取って、開始と終了から不要な括弧をドロップ。
“(<>.<>)”ḣJœ|ṫJ$ḊṖ
“(<>.<>)”ḣJœ|ṫJ$ḊṖ Main link. No arguments.
“(<>.<>)” Set the argument and return value to s := "(<>.<>)".
J Yield all indices of s, i.e., [1, 2, 3, 4, 5, 6, 7].
ḣ Dyadic head; yield s's prefixes of lengths 1 to 7.
$ Combine the two links to the left into a chain.
J Indices; yield [1, 2, 3, 4, 5, 6, 7].
ṫ Dyadic tail; yield s's postfixes of lengths 1 to 7.
œ| Multiset union; concatenate the results to both sides,
discarding the copy of s. (s is both a prefix and a postfix.)
Ḋ Dequeue; remove the first prefix, i.e., "(".
Ṗ Pop; remove the last postfix, i.e., ")".
main(n){while(write(++n<13,n/7*(n-7)+"(<>.<>)",n/7?14-n:n));}
この終了は、標準への書き込みので、TIOにタイムアウトした後、入力が失敗したが、それはしますターミナルで終了します。プログラムは評価の特定の順序(未定義の動作)に依存しており、gccなどでは機能しません。
@Steadyboxに感謝して、バイトを節約しました。
$ cat crowd.c
main(n){while(write(++n<12,n/7*(n-7)+"(<>.<>)",n/7?14-n:n));}
$ clang -o crowd crowd.c 2> /dev/null
$ ./crowd 0> /dev/null
(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)
main(n){while(++n<13)write(1,n/7*(n-7)+"(<>.<>)",n/7?14-n:n);}
さらに1バイトのコストで、ソリューションの移植性が大幅に向上します。
(<
そこに行方不明です。
現在のpython 2の答えに勝つことに抵抗できませんでした
print "(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"
(<>.<>)
:"(<" + "(<>" + "(<>." + ... + "(<>.<>)" + "<>.<>)" + ">.<>)" + ... + ">)"