一堆空白的凝视

36

问题

在没有输入的情况下，编写输出或返回以下字符串的程序或函数：

(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)

规则

  • 最短的程序获胜。
  • 允许尾随空格。
  • 允许尾随换行符。
  • 允许使用的功能未使用的参数。
code-golf  ascii-art  kolmogorov-complexity 
— LiefdeWen
source
5
注意：此字符串是以下子字符串的串联(<>.<>)："(<" + "(<>" + "(<>." + ... + "(<>.<>)" + "<>.<>)" + ">.<>)" + ... + ">)"
— JungHwan Min
2
这比“对所有文本进行硬编码！”要少一些。如果是更大的人群...
— –humanhuman
@totallyhuman下次会考虑
— LiefdeWen

Answers:

13

SOGL V0.12，12个字节

pΙ2○3V?hG]‘Γ

说明：

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

在这里尝试！

— 宰马
source
很高兴看到您现在有了在线翻译：）
— Emigna
你怎么做到这一点？伟大的工作！
— abhiagNitk
18

05AB1E，16个字节

"(<>.".∞ηJÀ24£.∞

在线尝试！

说明

"(<>."             # 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: "(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"
— 艾米尼亚
source
15

JS（Jsfuck），10,614字节

没有访问该网站的权限，必须手动解决所有问题-.- 

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说明：

大多数字符都相对容易：对于括号，我正在执行 

   ([]["fill"]+[])[13] // '('

   ([]["fill"]+[])[14] // ')'

通过使js创建一个像1.1e + 21这样的科学记数法并转换为字符串后取点字符来获得句号

方括号更难，我们必须执行该函数以创建斜体html对象字符串并窃取尖括号。这样做的主要技巧是为斜体字获取“ c”，这需要建立另一个函数以从“ function”中获取c

这里有很多改进的余地，主要是使用哪些函数来获取方括号。我还记得，如果斜体字更好，则更容易获得字符“ c”，但是我必须查看旧文件才能找到它

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展开摘要

— 鲁涅尔
source
3
即使对于JSFuck来说，这看起来也太长了，但是对于手动完成的工作量却为+1。
— F1Krazy
我打电话给BS手动做。您必须做一些事情来自动生成它。
— Patrick Roberts
实际上是10660字节。
— CalculatorFeline
6
@PatrickRoberts我敢打赌，他手动找到了答案，并使用了ctrlc和v的功能
— bleh
1
@PatrickRoberts这不是一个超级困难的程序，它只是了解如何构建每个角色的一个案例。作为一个基本示例，获取't'字符的方法是创建一个空数组'[]'，将其强制转换为数字以使数字为0：'+ []'，然后将not运算符应用于其进行强制转换为布尔值true'！+ []'。然后转换为字符串，方法是再次添加一个空数组并将其放在方括号中，以得到字符串“ true”'（！+ [] + []）'，最后使用'（！+ [] + []）[+ []]'
— 鲁格尼尔
12

C＃（单声道），54 52字节

s=>"(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"

在线尝试！

-2个字节，感谢Kevin Cruijssen

RIP C＃

— Skidsdev
source
我知道这种感觉（Java）。但是，您可以节省两个字节。;lambda-answer不计算尾随。并且()可以替换为单个字符（未使用的null对象）。第二个相关的元发布。
— Kevin Cruijssen
1
@KevinCruijssen是的，我确实指定了允许使用的未使用函数参数，因此s=>非常好。
— LiefdeWen
-1：没有努力以任何方式使答案正确。
— MechMK1
重点是@ MechMK1，实际上在C＃中尽可能地打高尔夫球。字符串的字面输出是用此特定语言输出此特定字符串的最简便的方法。参见Erlantz的C＃答案
— Skidsdev，
@Skidsdev我仍然认为这与Code Golf的精神背道而驰，无聊的写作和无聊的阅读。我的意思是，我的投票不算在内，因为我没有积极参与，所以不必担心。可以这么说，这只是我的2美分。
— MechMK1
7

V，28 27 25 22字节

3？4？字节感谢@KritixiLithos

i(<>.<>)òÙxlHÄ$xGòxÍî

在线尝试！

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
               $x        'delete the last character
                 G       'and go back to the bottom of the buffer
                   x     'delete the extra ) left by the loop
                    Íî   'and remove all newlines
— nmjcman101
source
您可以替换hx为just X。
— Kritixi Lithos
并且dkHxògJ比短1个字节dkVHgJ0x。
— Kritixi Lithos
是的，DJ的Íî东西也可以工作
— nmjcman101
我得到i(<>.<>)òÙxlHÄ$xGòddÍî23
— Kritixi Lithos
@KritixiLithos我真的只是切换lX为xl过：d
— nmjcman101
7

R，42个字节

cat(substring('(<>.<>)',-4:6,2:12),sep='')

相当简单的程序，它利用R中子串的工作方式。因此substring('(<>.<>)',-4:6,2:12)产生以下向量

> substring('(<>.<>)',-4:6,2:12)
 [1] "(<"      "(<>"     "(<>."    "(<>.<"   "(<>.<>"  "(<>.<>)" "<>.<>)"  ">.<>)"  
 [9] ".<>)"    "<>)"     ">)"

cat 空分隔符会将其以所需格式输出到STDOUT。

在线尝试！

— 米奇
source
5

TrumpScript，70字节

我知道，这是一个无聊的解决方案。

say "(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"
America is great
— 灰尘
source
尽管如此，做得好。
— 亚当
3

C（gcc），59 57字节

2 4个字节少于一个简单的puts()解决方案。肯定会有一些优雅的递归解决方案，但是到目前为止，每次尝试的开销都变得太大。

f(i){for(i=1;i++<12;)printf("%.*s",i,"(<>.<>)"+i/7*i%7);}

在线尝试！

— 腹肌
source
3

Brainfuck，198个 175个 167字节

我从来没有做过代码高尔夫球，所以这是我的第一个。反馈受到高度赞赏。

++++[>+++++<-]>[>++>+++>+++>++>++[<]>-]>.>.<.>.>++.<<.>.>.>>++++++.<<<<.>.>.>>.<<<.<.>.>.>>.<<<.>.<<.>.>.>>.<<<.>.>+.<<.>.>>.<<<.>.>.<.>>.<<<.>.>.>.<<<.>.>.<<.>.>.<.>.

在线尝试！

我认为，这是最明显的解决方案。首先，将单元格1-5设置为字母“（）。<>”之一。然后，我只是转到正确的单元格并输出角色。

更新：我更改了字符在“磁带”上出现的顺序，使程序更短，更有效。

更新2：不久后重新访问我的帖子并意识到，使用较短的顺序来设置单元格可以节省一些字节。

— 灰尘
source
欢迎来到PPCG！:)
— Shaggy
不错的第一答案，通常是无法立即看出其工作原理的答案，需要解释一下，但是这一挑战非常微不足道，因此并不是必需的。但是，如果您将来进行1200字节的头脑操练，那么一个简单的解释就可以走很长的路要走。
— LiefdeWen
我将来一定会这样做。
— 扬尘
好的，我为我的代码添加了一个简短的描述（有点不好）。
— 扬尘
2

JavaScript，47个字节

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>');
展开摘要

使用与Rod的Python答案相同的方法。

— 阿特耶
source
2

木炭，20字节

Ａ(<>ι(<ιι.ι.<ι.ιι.‖Ｂ

在线尝试！

感谢Destructible Lemon注意到了模式（-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

是什么‖Ｂ做的是基本上视觉palindromize画布。

— 暴民埃里克
source
请解释。
— CalculatorFeline
@CalculatorFeline添加了AST，由所返回-a。
— 外围大佬埃里克（Erik the Outgolfer）'17年
嗯 那比我想的要有用。
— CalculatorFeline
我认为，即使您只是添加了对象隐含的打印内容，这也会更加清楚
— Destructible Lemon 2015年
@DestructibleLemon但是AST已经Print在其中了，不确定您的意思。
— 暴民埃里克（Erik the Outgolfer）'17年
2

C（gcc），61字节

我知道这很la脚，但比我的其他解决方案要短得多...

f(){puts("(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)");}

C（gcc），103字节

这是我要变得有些棘手的版本...

char*e="(<>.<>)<>.<>)>.<>).<>)<>)>)";f(i,n){for(n=2;n<8;n++)for(i=0;i<n;i++)putchar(e[i]);puts(&e[i]);}

C（gcc），117115字节

完全复杂的递归版本...

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);}

在线尝试！

— 克莱布朗
source
1
也请分享您的非硬编码解决方案。尽管更长，但我敢肯定它比这个琐碎的有趣得多。
— Laikoni
2

一群空白的目光注视着蝙蝠侠，以纪念亚当·韦斯特。

C＃（.NET Core），393字节

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;}

在线尝试！

— 查理
source
2

q，29个字节

raze"(<>.<>)"{(y-5)_x}/:(!)11

-2字节归功于streetster

编辑说明：

语言是从右到左解释的。

怎么运行的

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
— 王
source
4
这里的解释会很好。
— 狮ry-恢复莫妮卡
1
您可以前去除空间_x为1个字节，变化til 11到(!)11另1个字节=节省2个字节保存:)
— streetster
@streetster好东西:)
— B.Wong
2

Haskell，55 45字节

map take[2..7]++map drop[1..5]>>=($"(<>.<>)")

在线尝试！

Haskell的第一个答案胜过硬编码的解决方案！

说明

该答案的工作方式是通过构建要应用于字符串的函数列表(<>.<>)。首先，我们用

map take[2..7]

这为我们提供了从2到7的所有前缀。然后我们用

map drop[1..5]

这给了我们所有从6号到2号的后缀。

获得函数列表后，我们将使用单子绑定（>>=），concatMap但该绑定较短。我们concatmap所使用的函数是($"(<>.<>)")将输入应用于字符串(<>.<>)。

这使字符串。

— 小麦巫师
source
1

Java 8，52字节

x->"(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"

无聊，但是在Java中没有任何方法可以使它更短。仅初始化一个临时字符串就已经是11个字节了。（String t="something";），并且使用substring几次肯定会花费太多字节。

文字返回的最短替代方法可能是以下内容（58个字节）：

x->"(<(x(x.(x.<(x.x(x.x)x.x)>.x).x)x)>)".replace("x","<>")

在这里尝试。

— 凯文·克鲁伊森
source
=>如果要添加，也可以在JavaScript中使用“胖箭头”功能（）使用。
— 毛茸茸的
@Shaggy Nah，如果需要，您可以保留JavaScript答案。同样适用于C＃，它也已经有一个单独的答案。
— Kevin Cruijssen
1

Python 2，57字节

print''.join("(<>.<>)"[max(0,n-5):n+2]for n in range(11))

我说这看起来像一群猫头鹰在他们的领导者后面偷看。只是说。

— 丹尼尔
source
1

疯子，245210字节

打高尔夫球。

>++++++++[<+++++>-]>++++++[<++++++++++>-]>++++++[<++++++++++>-]<++>>++++++++[<+++++>-]<++++++<<<.>.<.>.>.<<.>.>.>.<<<.>.>.>.<<.<.>.>.>.<<.>.<<.>.>.>.<<.>.<<+.>.>.>.<<.>.<<.>>.>.<<.>.<<.>>>.<<.>.<<.>.>.<<.>>.<<.

在线尝试！

— 全人类
source
1

Haskell，95个字节

import Data.List
import Control.Arrow
concat$tail$init$uncurry(++)$inits&&&tail.tails$"(<>.<>)"

这远比带有输出的字符串文字所需的49个字节长，但是我可以尽最大努力利用该结构。像往常一样，我喜欢箭头，并且inits &&& tails确实生成输入的前导子串列表和尾随子串列表的元组。然后，通过将元组传递到++，将这两个元组元素放到一个列表中，并将该列表concat赋给一个大字符串。的tail和init呼叫避免重复(<>.<>)在中间（一个由产生inits，另外通过tails），并从开始和结束删除不需要的括号，只服用长度为2或更多的考虑的子串。

— 贝吉
source
可以请您解释一下吗？
— CalculatorFeline
@CalculatorFeline完成
— Bergi
1

果冻，18字节

“(<>.<>)”ḣJœ|ṫJ$ḊṖ

在线尝试！

怎么运行的

“(<>.<>)”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., ")".
— 丹尼斯
source
1

C（clang），62 61字节

main(n){while(write(++n<13,n/7*(n-7)+"(<>.<>)",n/7?14-n:n));}

这终止仅在TIO超时由于写入到标准后输入失败，但它会在终端终止。该程序依赖于特定的评估顺序（未定义的行为），因此无法与gcc一起使用。

感谢@Steadybox提供了节省一个字节的想法！

在线尝试！

验证

$ 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
(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)

备用版本，62字节

main(n){while(++n<13)write(1,n/7*(n-7)+"(<>.<>)",n/7?14-n:n);}

以一个字节为代价，该解决方案变得更加可移植。

在线尝试！

— 丹尼斯
source
1
60个字节作为函数，但可能不可靠。
— Steadybox
— 丹尼斯
@Dennis是的，(<那里不见了。
— 暴民埃里克（Erik the Outgolfer）'17年
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