对于此挑战，ASCII艺术棉被将是一块24个字符宽，18行高的文本块，其中包含=-<>/\水平和垂直对称的类似棉被样式的字符。

被子示例：

========================
------------------------
//\\//\\\//\/\\///\\//\\
<<><<>>>>><<>><<<<<>><>>
/\\/\\\\/\/\/\/\////\//\
------------------------
/\/////\\///\\\//\\\\\/\
\///\/\/\\\\////\/\/\\\/
\///\/\/\\\\////\/\/\\\/
/\\\/\/\////\\\\/\/\///\
/\\\/\/\////\\\\/\/\///\
\/\\\\\//\\\///\\/////\/
------------------------
\//\////\/\/\/\/\\\\/\\/
<<><<>>>>><<>><<<<<>><>>
\\//\\///\\/\//\\\//\\//
------------------------
========================

所有被子都具有相同的形式：

• 他们总是24乘18。
• 顶线（第1行）和底线（第18行）=一直贯穿。
• 第2、6、13和17 -行一直贯穿。
• 线4和15是相同的随机水平对称的模式<和>。
• 所有其他线条（3、5、7、8、9、10、11、12、14、16）充满/且\完全随机，从而使整个被子保持水平和垂直对称。

请注意，将被子完全垂直或水平对折时，字符的形状完全匹配。不要将此与字符本身匹配起来混淆。例如线3和线16是不相同的，它们是垂直镜像。

挑战

编写程序或函数，以打印或返回随机的ASCII艺术被子。

由于许多硬编码的行和对称性，唯一真正的随机性来自第3、4、5、7、8、9行的前12个字符：

• 在第4行的前12个字符应能是任何字符的长度12的字符串<和>。
• 第3、5、7、8、9行的前12个字符应该可以是任意长度的12个字符串，/并且\（彼此独立）。
• 然后将这些随机的字符串进行相应镜像以制成整个被子。

以字节为单位的最短答案将获胜。Tiebreaker是较早的帖子。

您可以使用伪随机数生成器。（不，你不需要证明所有的12字符的字符串<>或/\可以与你的语言的PRNG产生。）

输出可以选择包含尾随换行符，但是除了被子必需外，不包含尾随空格或其他字符。

CJam，61 60 58 55 54 52 51字节

在Sp3000和Optimizer的帮助下有所缩短。

"=-/</-///"{C*1{"<\/>"%1$W%\_W%er}:F~+mrC<1FN}%s3F(

在这里测试。

说明

与这些对称ASCII艺术挑战一样，我通常会生成一个象限，然后通过两个适当的镜像操作将其扩展为完整的象限。

对于这个解释，我应该从函数开始，在函数的F某个地方定义它，因为它在三个地方用于三种不同的事物：

{"<\/>"%1$W%\_W%er}:F

这期望在堆栈的顶部有一个整数，在堆栈的下面有一个字符串。其目的是反转字符串并交换一些字符，以获得正确的镜像。整数是1或，3并指示是否1应同时交换（3）方括号和斜杠或（）仅交换方括号。下面是它的工作原理：

"<\/>"            "Push a string with all four relevant characters.";
      %           "% applied to a string and an integer N (in any order) selects every
                   Nth character, starting from the first. So with N = 1 this just
                   leaves the string unchanged, but with N = 3 it returns a string
                   containing only < and >.";
       1$         "Copy the string we want to mirror.";
         W%       "% also takes negative arguments. Giving it -1 reverses the string.";
           \_     "Swap the two strings and duplicate the <\/> or <> string.";
             W%   "Reverse that one. Due to the symmetry of this string, we'll now
                   have the characters to be swapped at corresponding indices.";
               er "Perform element-wise transliteration on the reversed input string
                   to complete the mirroring operation.";

现在，剩下的代码：

"=-/</-///"                            "This string encodes the 9 different line types.
                                        Note that for the /\ and <> lines we only use
                                        one of the characters. This idea is due to
                                        Sp3000. Thanks! :)";
           {                   }%      "Map this block onto the characters.";
            C*                         "Repeat the character 12 times, turning it into
                                        a string.";
              1{...}:F~                "Define and call F on the resulting string. The
                                        reversal doesn't do anything, but the character
                                        swapping creates strings containing both \/ and
                                        <>.";
                       +mr             "Add the two halves together and shuffle them.";
                          C<           "Truncate to 12 characters. We've now got our
                                        random half-lines.";
                            1F         "Call F again to mirror the half-line.";
                              N        "Push a newline.";
                                 s     "Join all those separate strings together by
                                        converting the array to a string.";
                                  3F   "Perform one more mirroring operation on the
                                        half-quilt, but this time only swap < and >.
                                        This yields the correct full quilt, except
                                        there are two newlines in the centre.";
                                    (  "This slices the leading newline off the second
                                        half and pushes it onto the stack.";

然后，在程序末尾自动打印出两半和该换行符。

Python 3中，257个 229 192 185 176 149 143字节

from random import*
k,*L=80703,
while k:s=eval("''"+".join(sample('--==<>\/'[k%4*2:][:2],2))"*12);L=[s]+L+[s[::(-1)**k]];k//=4
*_,=map(print,L)

在@xnor的帮助下，我们终于赶上了JS！

样本输出：

========================
------------------------
///////////\/\\\\\\\\\\\
>><<<>><<<><><>>><<>>><<
/\/\\/\/\\/\/\//\/\//\/\
------------------------
//\\////\\/\/\//\\\\//\\
/////\\\/\/\/\/\///\\\\\
/\\//\\/////\\\\\//\\//\
\//\\//\\\\\/////\\//\\/
\\\\\///\/\/\/\/\\\/////
\\//\\\\//\/\/\\////\\//
------------------------
\/\//\/\//\/\/\\/\/\\/\/
>><<<>><<<><><>>><<>>><<
\\\\\\\\\\\/\///////////
------------------------
========================

说明

（稍有过时，将在以后更新）

"444046402"对行进行编码，每个数字均引用的相关2个字符的子字符串的起始索引'--==\/<>'。通过对两个字符进行重复改组（不幸的是使用sample(...,2)，因为使用了random.shuffle）和字符串连接，从而由内而外构建了每一行。

第四行扩展的简化示例如下：

''.join(['<','>']).join(['>','<']).join(['>','<']).join(['<','>']).join(['>','<'])

这将产生><>><><<><：

               ''
    <>         .join(['<','>'])
   >  <        .join(['>','<'])
  >    <       .join(['>','<'])
 <      >      .join(['<','>'])
>        <     .join(['>','<'])

整个被子也由内而外建造，因为施工从第9排/第10排开始，向外进行。为此，我们先从一个空列表开始L，然后在通过时将行添加到行的前后

L=[s]+L+[[s,s[::-1]][n<"5"]]

的n<"5"条件是检查我们是否有一排包括><，在这种情况下，我们附加相同的行到后面，否则它的反面。

最后，*_,=是要强制评估是否map发生打印，这只是一种较短的方法print("\n".join(L))。

很长一段时间我有功能

g=lambda s:s.translate({60:62,62:60,92:47,47:92})

这需要一个字符串，并转换/\><到\/<>分别，但我终于成功地摆脱它:)

Python 2，300字节

该程序使用join, lambda, replace, sample, import和其他详细功能，因此不会赢得任何高尔夫奖项。

from random import*
f=lambda a,b,t:t.replace(a,'*').replace(b,a).replace('*',b)
k=lambda a:''.join(sample(a*12,12))
c='-'*24
e=k('<>')
h=e+f('<','>',e[::-1])
j=[d+f('/','\\',d[::-1])for d in[k('\\/')for i in'quilt']]
g=['='*24,c,j[0],h,j[1],c]+j[2:]
print'\n'.join(g+[f('/','\\',d)for d in g[::-1]])

自动高尔夫球手掌握的代码：

from random import *

change = lambda a,b,t: t.replace(a,'*').replace(b,a).replace('*',b)
pick = lambda a: ''.join(sample(a*12, 12))

innerline = '-' * 24
line4h = pick('<>')
line4 = line4h + change('<', '>', line4h[::-1])
diag = [d + change('/', '\\', d[::-1]) for d in [pick('\\/') for i in 'quilt']]

quilt = ['='*24, innerline, diag[0], line4, diag[1], innerline] + diag[2:]
print '\n'.join(quilt + [change('/', '\\', d) for d in quilt[::-1]])

示例输出：

========================
------------------------
\\\\/\////\\//\\\\/\////
<><<>>>><><><><><<<<>><>
/\\\\////\\\///\\\\////\
------------------------
\\\\//\///\\//\\\/\\////
//\//\\\\/\/\/\////\\/\\
\/\\\\/\//\/\/\\/\////\/
/\////\/\\/\/\//\/\\\\/\
\\/\\////\/\/\/\\\\//\//
////\\/\\\//\\///\//\\\\
------------------------
\////\\\\///\\\////\\\\/
<><<>>>><><><><><<<<>><>
////\/\\\\//\\////\/\\\\
------------------------
========================

杀伤人员地雷（53 58）

这不是很为对称的，因为我认为这是，很遗憾。修复程序花了我5个字符，现在我已经无法运行了。

L←+,3-⌽⋄'==--/\<><'[↑(732451451260688⊤⍨18/8)+L{L?12⍴2}¨⍳9]

说明：

• L←+,3-⌽：L是一个返回其参数的函数，后跟3-其参数的反函数
• L{L?12⍴2}¨⍳9：从[1,2]生成9行12个随机值加上它们的反向，然后对这9行进行反向
• 732451451260688⊤⍨18/8：生成列表0 2 4 6 4 2 4 4 4 4 4 4 2 4 _7_ 4 2 0（那是该死的不对称之处）
• +：对于每行，将相应的数字添加到每个值
• ↑：格式为矩阵
• '==--/\<><'[... ]：对于矩阵中的每个数字，请从该位置的字符串中选择字符

输出：

========================
------------------------
///////\\///\\\//\\\\\\\
<<><<><><<<<>>>><><>><>>
\\\\\//\/\\\///\/\\/////
------------------------
/\///\\/\/\/\/\/\//\\\/\
\////////\//\\/\\\\\\\\/
\\/\\\//\///\\\/\\///\//
//\///\\/\\\///\//\\\/\\
/\\\\\\\\/\\//\////////\
\/\\\//\/\/\/\/\/\\///\/
------------------------
/////\\/\///\\\/\//\\\\\
<<><<><><<<<>>>><><>><>>
\\\\\\\//\\\///\\///////
------------------------
========================

PHP，408，407，402，387，379个字节

我不是一个好高尔夫球手，但是这个问题听起来很有趣，所以我尝试了一下。

<?$a=str_replace;$b=str_repeat;function a($l,$a,$b){while(strlen($s)<$l){$s.=rand(0,1)?$a:$b;}return$s;}$c=[$b('=',12),$b('-',12),a(12,'/','\\'),a(12,'<','>'),a(12,'/','\\'),$b('-',12)];while(count($c)<9){$c[]=a(12,'/','\\');}for($d=9;$d--;){$c[$d].=strrev($a(['/','<','\\','>',1,2],[1,2,'/','<','\\','>'],$c[$d]));$c[]=$a(['/','\\',1],[1,'/','\\'],$c[$d]);}echo implode("
",$c);

非高尔夫代码

<?php

    function randomString($length, $a, $b) {
        $string = '';
        while(strlen($string) < $length) {
            $string .= rand(0, 1) ? $a : $b;
        }
        return $string;
    }

    if(isset($argv[1])) {
        srand(intval($argv[1]));
    }

    $lines = [
        str_repeat('=', 12),
        str_repeat('-', 12),
        randomString(12, '/', '\\'),
        randomString(12, '<', '>'),
        randomString(12, '/', '\\'),
        str_repeat('-', 12)
    ];
    while(count($lines) < 9) {
        $lines[] = randomString(12, '/', '\\');
    }

    for($index = count($lines); $index--;) {
        $lines[$index] .= strrev(str_replace(['/', '<', '\\', '>', 1, 2], [1, 2, '/', '<', '\\', '>'], $lines[$index]));
        $lines[] = str_replace(['/', '\\', 1], [1, '/', '\\'], $lines[$index]);
    }

    echo implode("\n", $lines) . "\n";

?>

非高尔夫版本有一些好处：您可以将其传递给整数以进行播种，rand()并每次获得相同的被子作为种子：

php quilt.php 48937

例如，这将导致这种美丽的手工编织被子：

========================
------------------------
/\\\////\\\/\///\\\\///\
><>><>><<<><><>>><<><<><
/\\\///\//\/\/\\/\\\///\
------------------------
/\\/\\\/\\/\/\//\///\//\
/\\\\/\//\\/\//\\/\////\
\/\\/\/\////\\\\/\/\//\/
/\//\/\/\\\\////\/\/\\/\
\////\/\\//\/\\//\/\\\\/
\//\///\//\/\/\\/\\\/\\/
------------------------
\///\\\/\\/\/\//\///\\\/
><>><>><<<><><>>><<><<><
\///\\\\///\/\\\////\\\/
------------------------
========================

编辑：原来我的第一个版本没有返回正确的被子。所以我修好了。有趣的是，解决方法甚至更短。

的JavaScript（ES6）169 195 201

编辑 6个字节保存的@nderscore。当心，反引号内的换行符很重要且很重要。

Edit2简化了行的建立，不需要reverse和concat

F=(o='')=>[...z='/\\/-/<\\-='].map((c,i,_,y=[z,'\\/\\-\\>/-='],q=[for(_ of-z+z)Math.random(Q=k=>q.map(v=>r=y[v^!k][i]+r+y[v^k][i],r='')&&r+`
`)*2])=>o=Q()+o+Q(i!=5))&&o

运行代码片段进行测试（在Firefox中）

F=(o='')=>[...z='/\\/-/<\\-='].map((c,i,_,y=[z,'\\/\\-\\>/-='],q=[for(_ of-z+z)Math.random(Q=k=>q.map(v=>r=y[v^!k][i]+r+y[v^k][i],r='')&&r+`
`)*2])=>o=Q()+o+Q(i!=5))&&o

Q.innerHTML=F()

<pre id=Q style='font-size:9px;line-height:9px'>

红宝石， 162 155

我喜欢这一行，因为它使我学会了在字符串文字和中使用反斜杠String#tr。否则，代码并不是非常聪明，只是紧凑。

a='/\\'
b='\\\/'
t=Array.new(9){x=''
12.times{x+=a[rand(2)]}
x+x.reverse.tr(a,b)}
t[0]=?=*24
t[1]=t[5]=?-*24
t[3].tr!a,'<>'
puts t,((t.reverse*'
').tr a,b)

佩斯，57 59 61

J"\<>/"K"\/"L+b_mXdK_Kbjbym+d_XdJ_JmsmOk12[\=\-K-JKK\-KKK

J"\<>/"K"\/"jbu++HGXHK_Km+d_XdJ_JmsmOk12[KKK\-K-JKK\-\=)Y

非常感谢@Jakube提出了这57个字节的版本。

算法与马丁算法非常相似。（修订）解释。

在线尝试

说明：

=G"\<>/"                            : Set G to be the string "\<>/"
K"\/"                               : Set K to be the string "\/"
Jm+d_XdG_GmsmOk12[\=\-K"<>"K\-KKK;  : Set J to be the top half of the carpet
                 [\=\-K"<>"K\-KKK;  : Make a list of possible chars for each row
          msmOk12                   : for each element of that list,
                                    : make a list of 12 randomly chosen characters
                                    : from it, then join them
Jm+d_XdG_G                          : for each element of that list,
                                    : make a new list with the old element,
                                    : and its horizontal reflection
jb+J_mXdK_KJ                        : Print the whole carpet
     mXdK_KJ                        : For each element of J make its vertical reflection

J，56 54字节

'=/\<>--></\'{~(-,|.)0,(3(2})8$5,3#1)+(,1-|.)"1?8 12$2

用法：

   '=/\<>--></\'{~(-,|.)0,(3(2})8$5,3#1)+(,1-|.)"1?8 12$2
========================
------------------------
///\\\\/\///\\\/\////\\\
><<><><>><>><<><<><><>><
\\/\//\\/\//\\/\//\\/\//
------------------------
\/\/\//////\/\\\\\\/\/\/
/\/\\//\//\\//\\/\\//\/\
//\\\\/////\/\\\\\////\\
\\////\\\\\/\/////\\\\//
\/\//\\/\\//\\//\//\\/\/
/\/\/\\\\\\/\//////\/\/\
------------------------
//\/\\//\/\\//\/\\//\/\\
><<><><>><>><<><<><><>><
\\\////\/\\\///\/\\\\///
------------------------
========================

1个字节，感谢@FUZxxl。

解释即将推出。

在这里在线尝试。

Python 295 287 227 bytes

Not that great but I'll post it anyway:

from random import*
m,d=[],dict(zip("\/<>=-","/\><=-"))
v=lambda w:[d[x]for x in w]
for e in '=-/>/-///':f=[choice([e,d[e]])for x in[0]*12];t=''.join(f+v(f[::-1]));print t;m+=''.join(e=='/'and v(t)or t),
print'\n'.join(m[::-1])

If you want an explanation just ask me.

Javascript (ES7 Draft) 174 168 146

Some inspiration taken from @edc65

Edit: Thanks to edc65 for some ideas to optimize building of rows.

F=(o='')=>[for(i of'555357531')(z=[for(_ of c='==--/\\<>golf')Math.random(r=x=>z.reduce((s,y)=>c[w=i^y^x]+s+c[w^1],'')+`
`)*2],o=r()+o+r(i<7))]&&o

Demonstration: (Firefox only)

F=(o='')=>[for(i of'555357531')(z=[for(_ of c='==--/\\<>golf')Math.random(r=x=>z.reduce((s,y)=>c[w=i^y^x]+s+c[w^1],'')+`
`)*2],o=r()+o+r(i<7))]&&o

document.write('<pre>' + F() + '</pre>');

Commented:

// define function, initialize output to ''
F = (o = '') =>
    // loop through character indexes of first 9 lines
    [
        for (i of '555357531')(
            // build array of 12 random 0/1's, initialize character list
            z = [
                for (_ of c = '==--/\\<>golf')
                    Math.random(
                        // define function for generating lines
                        // if x is true, opposite line is generated
                        r = x =>
                            z.reduce(
                                (s, y) => 
                                    c[w = i ^ y ^ x] + s + c[w ^ 1],
                                ''
                            ) + `\n`
                    ) * 2
            ],
            // build new lines and wrap output in them
            // x true in the second line for indexes < 7 (not character '>')
            o = r() + o + r(i < 7)
        )
    ]
    // return output
    && o

Pharo 4, 236

|s i f v h|s:='====----/\/\/<><<>'.f:=[:c|s at:(s indexOf:c)+i].v:=#(),'=-/</-///'collect:[:c|h:=(String new:12)collect:[:x|i:=2atRandom.f value:c].i:=1.h,(h reverse collect:f)].i:=3.String cr join:v,(v reverse collect:[:k|k collect:f])

or formatted normally:

|s i f v h|
s:='====----/\/\/<><<>'.
f:=[:c|s at:(s indexOf:c)+i].
v:=#() , '=-/</-///'
  collect:[:c|
    h:=(String new:12)collect:[:x|i:=2atRandom.f value:c].
    i:=1.
    h,(h reverse collect:f)].
i:=3.
String cr join:v,(v reverse collect:[:k|k collect:f])

Explanation:

The string s:='====----/\/\/<><<>' together with the block f:=[:c|s at:(s indexOf:c)+i] are here both for tossing the characters and for reversing the patterns...

• For i=1, it performs horizontal reversion (/ <-> \ , < <-> > ).

• For i=3, it performs vertical reversion (/ <-> \)

• For i=1 or 2 atRandom, it toss among / or \, < or >

'=-/</-///' encodes the character type c that will feed the block f for the 9 first lines.

#() , '=-/</-///' is a concatenation trick for transforming the String into an Array and thus collecting into an Array.

The rest is simple concatenation after applying the horizontal/vertical symetry.

========================
------------------------
\\/\/\\\\/\/\/\////\/\//
>>>><><><>><><<><><><<<<
\/\/\//\///\/\\\/\\/\/\/
------------------------
/\//\/\/////\\\\\/\/\\/\
\\//\//\\\/\/\///\\/\\//
////\\/\/\//\\/\/\//\\\\
\\\\//\/\/\\//\/\/\\////
//\\/\\///\/\/\\\//\//\\
\/\\/\/\\\\\/////\/\//\/
------------------------
/\/\/\\/\\\/\///\//\/\/\
>>>><><><>><><<><><><<<<
//\/\////\/\/\/\\\\/\/\\
------------------------
========================

Squeak 4.X, 247

|r s h m n p|s:='==--/\<>'.r:=(1<<108)atRandom.h:=''.m:=0.(16to:0by:-2),(0to:16by:2)do:[:i|n:=3bitAnd:28266>>i.p:=0.(11to:0by:-1),(0to:11)do:[:j|h:=h,(s at:n*2+1+(r-1>>(6*i+j)+(101>>(3-n*4+m+p))bitAnd:1)).j=0and:[p:=1]].i=0and:[m:=2].h:=h,#[13]].h

Or formatted:

|r s h m n p|
s:='==--/\<>'.
r:=(1<<108)atRandom.
h:=''.
m:=0.
(16to:0by:-2),(0to:16by:2) do:[:i|
  n:=3 bitAnd: 28266>>i.
  p:=0.
  (11to:0 by:-1),(0to:11) do:[:j|
    "h:=h,(s at:n*2+1+((r-1bitAt:6*i+j+1)bitXor:(101bitAt:3-n*4+m+p))). <- originally"
    h:=h,(s at:n*2+1+(r-1>>(6*i+j)+(101>>(3-n*4+m+p))bitAnd:1)).
    j=0and:[p:=1]].
  i=0and:[m:=2].
  h:=h,#[13]].
h

Explanations:

s:='==--/\<>'. obviously encodes the four posible pairs.

r:=(1<<108)atRandom. toss 108 bits (in a LargeInteger) for 9 rows*12 columns (we toss the == and -- unecessarily but performance is not our problem).

h:=''is the string where we will concatenate (Schlemiel painter because a Stream would be too costly in characters).

(16to:0by:-2),(0to:16by:2)do:[:i| is iterating on the rows (*2)

(11to:0by:-1),(0to:11) do:[:j| is iterating on the columns

28266 is a magic number encoding the pair to be used on first 9 lines.
It is the bit pattern 00 01 10 11 10 01 10 10 10, where 00 encodes '==', 01 '--', 10 '/\' and 11 '<>'.

101 is a magic number encoding the horizontal and vertical reversion.
It is the bit pattern 0000 0000 0110 1010, encoding when to reverse (1) or not (0) the first (0) or second (1) character of each pair '==' '--' '/\' and '<>', for the vertical symetry and horizontal symetry.

n:=3 bitAnd: 28266>>i gives the encoding of character pair for row i/2 (0 for '==', 1 for '--', 2 for '/\' and 3 for '<>').

(r-1 bitAt: 6*i+j+1) pick the random bit for row i/2 column j (1 is the rank of lowest bit thus we have a +1, k atRandom toss in interval [1,k] thus we have a -1).

(101 bitAt: 3-n*4+m+p) pick the reversal bit: (3-n)*4 is the offset for the group of 4 bits corresponding to the pair code n, m is the vertical reversion offset (0 for first 9, 2 for last 9 rows), p is the horizontal reversion offset (0 for first 12, 1 for last 12 columns)+1 because low bit rank is 1.

bitXor: performs the reversion (it answer an offset 0 or 1), and s at:2*n+1+bitXor_offset pick the right character in s.

But (A>>a)+(B>>b) bitAnd: 1 costs less bytes than (A bitAt:a+1)bitXor:(B bitAt:b+1)thus the bitXor was rewritten and offset +1 on p is gone...

h,#[13] is an ugly squeakism, we can concatenate a String with a ByteArray (containing code for carriage return).

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