Title: Compute the perimeter density matrix

Body:
Introduction

The perimeter density matrix is an infinite binary matrix M defined as follows.
Consider a (1-based) index (x, y), and denote by M[x, y] the rectangular sub-matrix spanned by the corner (1, 1) and (x, y).
Suppose that all values of M[x, y] except Mx, y, the value at index (x, y), have already been determined.
Then the value Mx, y is whichever of 0 or 1 that puts the average value of M[x, y] closer to 1 / (x + y).
In case of a tie, choose Mx, y = 1.

This is the sub-matrix M[20, 20] with zeros replaced by dots for clarity:

```<code>1 . . . . . . . . . . . . . . . . . . .
. . . . . 1 . . . . . . . . . . . . . .
. . 1 . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . 1 . . . . . . . . . . . . . . .
. 1 . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . 1 . .
. . . . . . . . . . . . . . 1 . . . . .
. . . . . . . . . . . . 1 . . . . . . .
. . . . . . . . . . 1 . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . . 1 . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . 1 . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . 1 . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
</code>```

For example, we have M1, 1 = 1 at the upper left corner, since 1 / (1 + 1) = ½, and the average of the 1 × 1 sub-matrix M[1, 1] is either 0 or 1; that's a tie, so we choose 1.

Consider then the position (3, 4).
We have 1 / (3 + 4) = 1/7, and the average of the sub-matrix M[3, 4] is 1/6 if we choose 0, and 3/12 if we choose 1.
The former is closer to 1/7, so we choose M3, 4 = 0.

Here is the sub-matrix M[800, 800] as an image, showing some of its intricate structure.

The task

Given a positive integer N < 1000, output the N × N sub-matrix M[N, N], in any reasonable format.
The lowest byte count wins.

Body:
Racket 294 bytes

(define(g x y)(if(= 1 x y)1(let*((s(for*/sum((i(range 1(add1 x)))(j(range 1(add1 y)))#:unless(and(= i x)(= j y)))
(g i j)))(a(/ s(* x y)))(b(/(add1 s)(* x y)))(c(/ 1(+ x y))))(if(<(abs(- a c))(abs(- b c)))0 1))))
(for((i(range 1(add1 a))))(for((j(range 1(add1 b))))(print(g i j)))(displayln""))

Ungolfed:

(define(f a b)
  (define (g x y)
    (if (= 1 x y) 1
        (let* ((s (for*/sum ((i (range 1 (add1 x)))
                             (j (range 1 (add1 y)))
                             #:unless (and (= i x) (= j y)))
                    (g i j)))
               (a (/ s (* x y)))
               (b (/ (add1 s) (* x y)))
               (c (/ 1 (+ x y))))
          (if (< (abs(- a c))
                 (abs(- b c)))
              0 1))))
  (for ((i (range 1 (add1 a))))
    (for ((j (range 1 (add1 b))))
      (print (g i j)))
    (displayln ""))
  )

Testing:

(f 8 8)

Output:

10000000
00000100
00100000
00000000
00001000
01000000
00000000
00000000


问:
Polyglot Anagrams Cops' Thread
This challenge has two threads.  This is the cops' thread. The robbers' thread is located here.

Your challenge is to choose an OEIS sequence and write two full programs in two different languages that produces that nth item in the sequence when given an n via STDIN, or an other forms of standard input, where n is any positive number.  However your two programs must be anagrams, meaning each can be rearranged from the other's letters.

Programs must output the decimal of the number followed by optional whitespace to STDOUT.  Programs may output to STDERR however it should be ignored and if the hidden case does so it must be clearly stated that this is the case.

If you wish you may also output by character code.  However if you do so in your hidden solution you must state such in the body of your submission.

You will then present the OEIS number, the source code for and the name of one language it is in.

Robbers will crack your submission if they find an anagram of the original  submission that runs in a language other than the one you already presented.  To crack an answer they must only find any language and program which produces the sequence and is an anagram of the original, not necessarily the answer the you were thinking about.

Thus you are incentivized to make it as hard as possible to find any language that does the task using their list of symbols.

Scoring

This is code-golf so the shortest un-cracked program is the winner.

Languages

Languages will be considered different if the two proposed solutions do not complete the task in both languages.  This will include different versions of the same language as long as neither the cop's solution or the robber's solution produce the correct output in the other's language.

i.e. If the there are two solutions 1 and 2 that are in language A and B respectively solution 2 must not produce the correct output in language A and solution 1 must not produce the correct output in language B.

Safety

Once your submission has been uncracked for a week you may post your solution and declare your post safe.  If after a week you choose not to post a solution your answer may still be cracked.

答:
<s>2sable, 15 bytes, A000290</s>, Cracked!

Hopping on the same n2 train :p.

```<code>*?"!#$&<=@\^{|}
</code>```

Try it online!

ContentLicense: CC BY-SA 3.0

Title: Polyglot Anagrams Cops' Thread

Body:
<s>2sable, 15 bytes, A000290</s>, Cracked!

Hopping on the same n2 train :p.

*?"!#$&<=@\^{|}

Try it online!


答:
<s>CJam, 7 bytes, A005843</s> Cracked!

```<code>ri2*e#^
</code>```

This is a basic 2*n sequence.

Explanation:

```<code>r          e# read input
 i         e# convert to integer
  2*       e# multiply it by 2
    e#^    e# this is a comment that is ignored by the interpreter
</code>```

Try it online!

Original Solution, Carrot

```<code>#^i*2er
</code>```

Carrot is an esolang created by me. I have stopped developing it a long time ago. The reason I chose this is because I hoped that it would be hard for other languages to comment out the unnecessary parts of the code.

Explanation:

```<code>#^            This pushes the input to the stack (anything before the ^ is the stack)
  i           Convert stack to integer
   *2         Multiply it by 2
     er       These are ignored because they are not Carrot commands
              Implicit output
</code>```

Try it online!

Title: Polyglot Anagrams Cops' Thread

Body:
This challenge has two threads.  This is the cops' thread. The robbers' thread is located here.

Your challenge is to choose an OEIS sequence and write two full programs in two different languages that produces that nth item in the sequence when given an n via STDIN, or an other forms of standard input, where n is any positive number.  However your two programs must be anagrams, meaning each can be rearranged from the other's letters.

Programs must output the decimal of the number followed by optional whitespace to STDOUT.  Programs may output to STDERR however it should be ignored and if the hidden case does so it must be clearly stated that this is the case.

If you wish you may also output by character code.  However if you do so in your hidden solution you must state such in the body of your submission.

You will then present the OEIS number, the source code for and the name of one language it is in.

Robbers will crack your submission if they find an anagram of the original  submission that runs in a language other than the one you already presented.  To crack an answer they must only find any language and program which produces the sequence and is an anagram of the original, not necessarily the answer the you were thinking about.

Thus you are incentivized to make it as hard as possible to find any language that does the task using their list of symbols.

Scoring

This is code-golf so the shortest un-cracked program is the winner.

Languages

Languages will be considered different if the two proposed solutions do not complete the task in both languages.  This will include different versions of the same language as long as neither the cop's solution or the robber's solution produce the correct output in the other's language.

i.e. If the there are two solutions 1 and 2 that are in language A and B respectively solution 2 must not produce the correct output in language A and solution 1 must not produce the correct output in language B.

Safety

Once your submission has been uncracked for a week you may post your solution and declare your post safe.  If after a week you choose not to post a solution your answer may still be cracked.

Body:
<s>CJam, 7 bytes, A005843</s> Cracked!

ri2*e#^

This is a basic 2*n sequence.

Explanation:

r          e# read input
 i         e# convert to integer
  2*       e# multiply it by 2
    e#^    e# this is a comment that is ignored by the interpreter

Try it online!

Original Solution, Carrot

#^i*2er

Carrot is an esolang created by me. I have stopped developing it a long time ago. The reason I chose this is because I hoped that it would be hard for other languages to comment out the unnecessary parts of the code.

Explanation:

#^            This pushes the input to the stack (anything before the ^ is the stack)
  i           Convert stack to integer
   *2         Multiply it by 2
     er       These are ignored because they are not Carrot commands
              Implicit output

Try it online!


问:
Alternating Fibonacci
In the alternating Fibonacci sequence, you first start with 1 and 1 as usual.

However, instead of always adding the last two values to get the next number, you alternate starting with adding, and every other time you subtract instead.

The sequence starts like this:

```<code>1
1
2    # 1 + 1
-1   # 1 - 2
1    # 2 + -1
-2   # -1 - 1
-1   # 1 + -2
-1   # -2 - -1
-2   # -1 + -1
1    # -1 - -2
-1   # -2 + 1
2    # 1 - -1
1    # -1 + 2
1    # 2 - 1
</code>```

etc.

Notice that after it starts over once it gets to 1 and 1 again.

Given a number N, print the Nth term of the alternating fibonacci sequence.

Remember, this is code-golf, so the code with the smallest number of bytes wins.

答:
Batch, 49 bytes

```<code>@cmd/cset/a"n=%1%%12,~!(n%%3)*(1|-!(n%%5*(n/4)))"
</code>```

Takes input as a command-line parameter. Explanation: Closed form uses the following observations:

Sequence is cyclic with period 12

Every third term is ±2 while other terms are ±1

Terms after the third are negative except multiples of 5 (after reducing modulo 12)

We therefore start by reducing modulo 12 (to save 2 bytes). We then reduce modulo three and invert the result, which is 1 for multiples of 3 or 0 otherwise. We then bitwise not that value, giving us -2 for multiples of 3 or -1 otherwise. We then reduce modulo 5 and separately divide by 4, giving zero for terms 1, 2, 3, 5, 10 and 12 (0). Inverting and negating gives us -1 for those values and zero for other values. We then bitwise or that with 1 and multiply with the earlier calculation.

Title: Alternating Fibonacci

Body:
Batch, 49 bytes

@cmd/cset/a"n=%1%%12,~!(n%%3)*(1|-!(n%%5*(n/4)))"

Takes input as a command-line parameter. Explanation: Closed form uses the following observations:

Sequence is cyclic with period 12

Every third term is ±2 while other terms are ±1

Terms after the third are negative except multiples of 5 (after reducing modulo 12)

We therefore start by reducing modulo 12 (to save 2 bytes). We then reduce modulo three and invert the result, which is 1 for multiples of 3 or 0 otherwise. We then bitwise not that value, giving us -2 for multiples of 3 or -1 otherwise. We then reduce modulo 5 and separately divide by 4, giving zero for terms 1, 2, 3, 5, 10 and 12 (0). Inverting and negating gives us -1 for those values and zero for other values. We then bitwise or that with 1 and multiply with the earlier calculation.


答:
<s>MATL, 13 bytes, A002275 </s> Cracked!

```<code>i:"@ax'1']
v!
</code>```

Try it online!

Explanation:

```<code>i           % Grab input
 :          % Push (range(1,input))
  "         % For each element in this range:
   @        %   Push it
    a       %   Is is truthy?
     x      %   Delete it
      '1'   %   Push '1'
         ]  %   End loop
v           % Join all of these '1's together
 !          % Transpose and display
</code>```

Title: Polyglot Anagrams Cops' Thread

Body:
<s>MATL, 13 bytes, A002275 </s> Cracked!

i:"@ax'1']
v!

Try it online!

Explanation:

i           % Grab input
 :          % Push (range(1,input))
  "         % For each element in this range:
   @        %   Push it
    a       %   Is is truthy?
     x      %   Delete it
      '1'   %   Push '1'
         ]  %   End loop
v           % Join all of these '1's together
 !          % Transpose and display


问:
Compute the perimeter density matrix

Introduction

The perimeter density matrix is an infinite binary matrix M defined as follows.
Consider a (1-based) index (x, y), and denote by M[x, y] the rectangular sub-matrix spanned by the corner (1, 1) and (x, y).
Suppose that all values of M[x, y] except Mx, y, the value at index (x, y), have already been determined.
Then the value Mx, y is whichever of 0 or 1 that puts the average value of M[x, y] closer to 1 / (x + y).
In case of a tie, choose Mx, y = 1.

This is the sub-matrix M[20, 20] with zeros replaced by dots for clarity:

```<code>1 . . . . . . . . . . . . . . . . . . .
. . . . . 1 . . . . . . . . . . . . . .
. . 1 . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . 1 . . . . . . . . . . . . . . .
. 1 . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . 1 . .
. . . . . . . . . . . . . . 1 . . . . .
. . . . . . . . . . . . 1 . . . . . . .
. . . . . . . . . . 1 . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . . 1 . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . 1 . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . 1 . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
</code>```

For example, we have M1, 1 = 1 at the upper left corner, since 1 / (1 + 1) = ½, and the average of the 1 × 1 sub-matrix M[1, 1] is either 0 or 1; that's a tie, so we choose 1.

Consider then the position (3, 4).
We have 1 / (3 + 4) = 1/7, and the average of the sub-matrix M[3, 4] is 1/6 if we choose 0, and 3/12 if we choose 1.
The former is closer to 1/7, so we choose M3, 4 = 0.

Here is the sub-matrix M[800, 800] as an image, showing some of its intricate structure.

The task

Given a positive integer N < 1000, output the N × N sub-matrix M[N, N], in any reasonable format.
The lowest byte count wins.

ContentLicense: CC BY-SA 4.0

Title: Compute the perimeter density matrix

Body:
Perl, 151 + 1 = 152 bytes

Run with the -n flag.  The code will only work correctly every other iteration within the same instance of the program.  To get it to work correctly every time, add 5 bytes by prepending my%m; to the code.

for$b(1..$_){for$c(1..$_){$f=0;for$d(1..$b){$f+=$m{"$d,$_"}/($b*$c)for 1..$c}$g=1/($b+$c);print($m{"$b,$c"}=abs$f-$g>=abs$f+1/($b*$c)-$g?1:_).$"}say""}''

Readable:

for$b(1..$_){
    for$c(1..$_){
        $f=0;
        for$d(1..$b){
            $f+=$m{"$d,$_"}/($b*$c)for 1..$c
        }
        $g=1/($b+$c);
        print($m{"$b,$c"}=abs$f-$g>=abs$f+1/($b*$c)-$g?1:_).$"
    }
    say""
}

Output for input of 100:

[encoded data, 10000 characters]


问:
Test a number for narcissism
A Narcissistic Number is a number which is the sum of its own digits, each raised to the power of the number of digits.

For example, take \$153\$ (3 digits):

\$1^3 + 5^3 + 3^3 = 153\$

\$1634\$:

\$1^4 + 6^4 + 3^4 + 4^4 = 1634 = 1 + 1296 + 81 + 256 = 1634\$

The Challenge:

Your code must take input from the user and output True or False depending upon whether the given number is a Narcissistic Number.

Error checking for text strings or other invalid inputs is not required. 1 or 0 for the output is acceptable. Code that simply generates a list of Narcissistic Numbers, or checks the user input against a hardcoded list, does not qualify.

OEIS A005188

答:
C#, 103 Bytes

Golfed:

```<code>bool N(int n){double a=0;foreach(var d in n+""){a+=Math.Pow(int.Parse(d+""),n+"".Length);}return a==n;}
</code>```

Ungolfed:

```<code>public bool N(int n)
{
  double a = 0;

  foreach (var digit in n.ToString())
  {
    a += Math.Pow(int.Parse(digit + ""), n.ToString().Length);
  }

  return a == n;
}
</code>```

Testing:

```<code>Console.WriteLine(new NarcissisticNumber().N(153));
True

Console.WriteLine(new NarcissisticNumber().N(1634));
True
</code>```

Title: Test a number for narcissism

Body:
C#, 103 Bytes

Golfed:

bool N(int n){double a=0;foreach(var d in n+""){a+=Math.Pow(int.Parse(d+""),n+"".Length);}return a==n;}

Ungolfed:

public bool N(int n)
{
  double a = 0;

  foreach (var digit in n.ToString())
  {
    a += Math.Pow(int.Parse(digit + ""), n.ToString().Length);
  }

  return a == n;
}

Testing:

Console.WriteLine(new NarcissisticNumber().N(153));
True

Console.WriteLine(new NarcissisticNumber().N(1634));
True


答:
Brain-Flak, 24 bytes, A000290, Safe

Yet another square solution. This time there is nothing but parentheses

```<code>({(({}[()])()){}[()]}{})
</code>```

The intended solution was in Brain-Flueue, a version of brain-flak that uses queues instead of stacks.  The program was:

```<code>({(({})[()]){}}{})[()()]
</code>```

The languages are considered distinct because neither of the two programs halt when run in the other language.

Title: Polyglot Anagrams Cops' Thread

Body:
Brain-Flak, 24 bytes, A000290, Safe

Yet another square solution. This time there is nothing but parentheses

({(({}[()])()){}[()]}{})

The intended solution was in Brain-Flueue, a version of brain-flak that uses queues instead of stacks.  The program was:

({(({})[()]){}}{})[()()]

The languages are considered distinct because neither of the two programs halt when run in the other language.


答:
Python 2, <s>171</s> <s>170</s> 168 bytes

a,b,c=r"\\","/"*4," "
f,g=c*2,c+a+b+c
d=(a*2+b)*6,g*6,(f+b+f)*6,g[::-1]*6,(b+a*2)*6,('\\//'+f+a+"\\")*6,(a+f*2+a)*6,(a+"\\"+f+'//\\')*6
for e in 0,1,2:print'\n'.join(d)
```

Not pretty and not clever. Just sets variables for the most often used groups of strings then combines them and prints the result 3 times. May try and golf it more later if I don't find a better approach.

1 byte saved by using raw input on the a assignment. Thanks @nedla2004

-2 by assigning a couple of variables but still not a serious competitor

Title: Make me a blanket!

Body:
I want a blanket that looks like this. Each strip goes over, under, over, under. Can you print it?

```<code>\\\\////\\\\////\\\\////\\\\////\\\\////\\\\////
 \\////  \\////  \\////  \\////  \\////  \\////
  ////    ////    ////    ////    ////    ////
 ////\\  ////\\  ////\\  ////\\  ////\\  ////\\
////\\\\////\\\\////\\\\////\\\\////\\\\////\\\\
\//  \\\\//  \\\\//  \\\\//  \\\\//  \\\\//  \\\
\\    \\\\    \\\\    \\\\    \\\\    \\\\    \\
\\\  //\\\\  //\\\\  //\\\\  //\\\\  //\\\\  //\
\\\\////\\\\////\\\\////\\\\////\\\\////\\\\////
 \\////  \\////  \\////  \\////  \\////  \\////
  ////    ////    ////    ////    ////    ////
 ////\\  ////\\  ////\\  ////\\  ////\\  ////\\
////\\\\////\\\\////\\\\////\\\\////\\\\////\\\\
\//  \\\\//  \\\\//  \\\\//  \\\\//  \\\\//  \\\
\\    \\\\    \\\\    \\\\    \\\\    \\\\    \\
\\\  //\\\\  //\\\\  //\\\\  //\\\\  //\\\\  //\
\\\\////\\\\////\\\\////\\\\////\\\\////\\\\////
 \\////  \\////  \\////  \\////  \\////  \\////
  ////    ////    ////    ////    ////    ////
 ////\\  ////\\  ////\\  ////\\  ////\\  ////\\
////\\\\////\\\\////\\\\////\\\\////\\\\////\\\\
\//  \\\\//  \\\\//  \\\\//  \\\\//  \\\\//  \\\
\\    \\\\    \\\\    \\\\    \\\\    \\\\    \\
\\\  //\\\\  //\\\\  //\\\\  //\\\\  //\\\\  //\
</code>```

Trailing spaces at the end of each line and trailing newlines are acceptable.

Remember, this is code-golf, so the code with the fewest bytes wins.

Leaderboards

Here is a Stack Snippet to generate both a regular leaderboard and an overview of winners by language.

To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:

```<code># Language Name, N bytes
</code>```

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

```<code># Ruby, <s>104</s> <s>101</s> 96 bytes
</code>```

If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:

```<code># Perl, 43 + 2 (-p flag) = 45 bytes
</code>```

You can also make the language name a link which will then show up in the leaderboard snippet:

```<code># [><>](http://esolangs.org/wiki/Fish), 121 bytes
</code>```

var QUESTION_ID=99023,OVERRIDE_USER=12537;function answersUrl(e){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"https://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="
"+e.body.replace(OVERRIDE_REG,"")+"
")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery("#answer-template").html();t=t.replace("{{PLACE}}",n+".").replace("{{NAME}}",e.user).replace("{{LANGUAGE}}",e.language).replace("{{SIZE}}",e.size).replace("{{LINK}}",e.link),t=jQuery(t),jQuery("#answers").append(t);var o=e.language;/s.lang?1:e.lang<s.lang?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;```
<pre class="snippet-code-css lang-css prettyprint-override"><code>body{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;float:left}table thead{font-weight:700}table td{padding:5px}</code>```

 <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b">

Leaderboard

	Author	Language	Size

Winners by Language

Language	User	Score

{{PLACE}}	{{NAME}}	{{LANGUAGE}}	{{SIZE}}	Link

{{LANGUAGE}}	{{NAME}}	{{SIZE}}	Link

```

Body:
Python 2, <s>171</s> <s>170</s> 168 bytes

a,b,c=r"\\","/"*4," "
f,g=c*2,c+a+b+c
d=(a*2+b)*6,g*6,(f+b+f)*6,g[::-1]*6,(b+a*2)*6,('\\//'+f+a+"\\")*6,(a+f*2+a)*6,(a+"\\"+f+'//\\')*6
for e in 0,1,2:print'\n'.join(d)

Not pretty and not clever. Just sets variables for the most often used groups of strings then combines them and prints the result 3 times. May try and golf it more later if I don't find a better approach.

1 byte saved by using raw input on the a assignment. Thanks @nedla2004

-2 by assigning a couple of variables but still not a serious competitor


问:
Product catalog
This problem is about separating a string representing a product identifier into three components.

The first part consists of upper and lower letters of arbitrary length which represents the warehouse.

The second part is digits which represents the product number. This part is also of arbitrary length.

The last part is qualifiers as size and colours, and this part continues to the end of the string. The qualifiers are guaranteed to start with a capital letter and consist of alphanumeric characters.

Each part should be printed clearly separated. It is guaranteed that each part is non-empty.

The winner is the one who uses least bytes to solve this problem.

Example:

Input: UK7898S14

Output:

UK

7898

S14

Here UK is United Kingdom, 7898 is the product code, and S14 is size 14.

Example 2:

Input: cphDK1234CYELLOWS14QGOOD

Output:

cphDK

1234

CYELLOWS14QGOOD

Here cphDK is Copenhagen, Denmark, 1234 is the product code, CYELLOWS14QGOOD represents yellow colour, size 14, and good quality.

答:
C#, 74 bytes

```<code>v=>new System.Text.RegularExpressions.Regex("\\d+").Replace(v,"\n$&\n",1);
</code>```

Replace 1st set of digits with carriage return, set of digits, and another carriage return, as Johan Karlsson did for JavaScript.

Title: Product catalog

Body:
C#, 74 bytes

v=>new System.Text.RegularExpressions.Regex("\\d+").Replace(v,"\n$&\n",1);

Replace 1st set of digits with carriage return, set of digits, and another carriage return, as Johan Karlsson did for JavaScript.


答:
JavaScript, 56 characters

n=prompt();n.split('').reduce((a,i)=>a+i**n.length,0)==n

This makes use of the exponentiation operator, so you have to be running a modern browser for this to work.

Title: Test a number for narcissism

Body:
A Narcissistic Number is a number which is the sum of its own digits, each raised to the power of the number of digits.

For example, take \$153\$ (3 digits):

\$1^3 + 5^3 + 3^3 = 153\$

\$1634\$:

\$1^4 + 6^4 + 3^4 + 4^4 = 1634 = 1 + 1296 + 81 + 256 = 1634\$

The Challenge:

Your code must take input from the user and output True or False depending upon whether the given number is a Narcissistic Number.

Error checking for text strings or other invalid inputs is not required. 1 or 0 for the output is acceptable. Code that simply generates a list of Narcissistic Numbers, or checks the user input against a hardcoded list, does not qualify.

OEIS A005188


答:
SOML, 106 bytes

```<code>3{"\\\\////”6*p" \\//// ”6*p"  ////  ”6*p" ////\\ ”6*p"////\\\\”6*p"\//  \\\”6*p"\\    \\”6*p"\\\  //\”6*p
</code>```

a non-competing version using a function that I've only recently added: (<s>83</s> <s>67</s> 66 bytes)

explanation:

```<code>  →$\\→#////→@”6*p"→~3{"##@~ #@ ~$@$~ @# ~@##~\//$#\~#$$#~#\$//\~”
  →$                                                                in the further code replace "$" with "  "
    \\→#                                                            replace "#" with "\\"
        ////→@                                                      replace "@" with "////"
              ”6*p"→~3{"##@~ #@ ~$@$~ @# ~@##~\//$#\~#$$#~#\$//\~” the code to exchange stuff in

so that results in:
”6*p"→~3{"\\\\////~ \\//// ~  ////  ~ ////\\ ~////\\\\~\//  \\\~\\    \\~\\\  //\~”
”6*p"→~3{"A~B~C~D~E~F~G~H~”  modified version of the program (removing "/","\" and " " and replaced with A,B,C,ect.)
”6*p"→~                      replace in the further program "~" with ”6*p" which is:
”                            end string
 6*                          repeat the last thing in stack 6 times
   p                         output the result
    "                        start a string
resulting program:
3{"A”6*p"B”6*p"C”6*p"D”6*p"E”6*p"F”6*p"G”6*p"H”6*p"”
shortened example:
3{"A”6*p"B”6*p"H”6*p"”
3{                      repeat 3 times
  "A”                   push "A" (original: "\\\\////")
     6*p                output it multiplied by 6
        "B”             push "B" (original: " \\//// ")
           6*p          output it multiplied by 6
              "H”       push "H" (original: "\\\  //\")
                 6*p    output it multiplied by 6
                    "”  push an empty string (shorter to do ~” than ”6*p)
</code>```

Title: Make me a blanket!

Body:
SOML, 106 bytes

3{"\\\\////”6*p" \\//// ”6*p"  ////  ”6*p" ////\\ ”6*p"////\\\\”6*p"\//  \\\”6*p"\\    \\”6*p"\\\  //\”6*p

a non-competing version using a function that I've only recently added: (<s>83</s> <s>67</s> 66 bytes)

explanation:

  →$\\→#////→@”6*p"→~3{"##@~ #@ ~$@$~ @# ~@##~\//$#\~#$$#~#\$//\~”
  →$                                                                in the further code replace "$" with "  "
    \\→#                                                            replace "#" with "\\"
        ////→@                                                      replace "@" with "////"
              ”6*p"→~3{"##@~ #@ ~$@$~ @# ~@##~\//$#\~#$$#~#\$//\~” the code to exchange stuff in

so that results in:
”6*p"→~3{"\\\\////~ \\//// ~  ////  ~ ////\\ ~////\\\\~\//  \\\~\\    \\~\\\  //\~”
”6*p"→~3{"A~B~C~D~E~F~G~H~”  modified version of the program (removing "/","\" and " " and replaced with A,B,C,ect.)
”6*p"→~                      replace in the further program "~" with ”6*p" which is:
”                            end string
 6*                          repeat the last thing in stack 6 times
   p                         output the result
    "                        start a string
resulting program:
3{"A”6*p"B”6*p"C”6*p"D”6*p"E”6*p"F”6*p"G”6*p"H”6*p"”
shortened example:
3{"A”6*p"B”6*p"H”6*p"”
3{                      repeat 3 times
  "A”                   push "A" (original: "\\\\////")
     6*p                output it multiplied by 6
        "B”             push "B" (original: " \\//// ")
           6*p          output it multiplied by 6
              "H”       push "H" (original: "\\\  //\")
                 6*p    output it multiplied by 6
                    "”  push an empty string (shorter to do ~” than ”6*p)


问:
Creating a Crossed Square

Creating a Crossed Square

You are to take input of an integer of one or more and output a square made of any printable character of your choice with a diagonal cross through the centre.

The general idea is for the output to be a hollow square that has a diagonal cross through it.:

```<code>Input: 7
Output:
*******
*#   #*
* # # *
*  #  *
* # # *
*#   #*
*******
</code>```

In the above example the '*'s represent the outer box and the '#'s represent the diagonal cross.

Note that the above example uses two different characters so that it is easier to see what the output looks like, your program should use one character only.

Input

An integer of 1 or more, it is guaranteed to be odd.

Output

A square that is made up of a character of your choice with a cross through the middle.

The cross must be diagonal

The square may be output via the function or written to output

Trailing newlines are okay

Can output as a graphic, diagram or image if you wish too

Examples

```<code>Input: 1
Output:
*

Input: 3
Output:
***
***
***

Input: 5
Output:
*****
** **
* * *
** **
*****

Input: 7
Output:
*******
**   **
* * * *
*  *  *
* * * *
**   **
*******
</code>```

Specs

Functions or full programs are allowed

You can get input by your preferred means

Standard loopholes are disallowed

Programs must work without any additional statements i.e. usings in C#, they must be included in the entry

You can output from a function or print the result

This is code golf so the shortest solution wins.

答:
Perl, 83 +1 = 84 bytes

Run with the -n flag.

```<code>$\="*
*";print$c="*"x($_+1);for$b(1..$_){@a=($")x$_;@a[$b-1,-$b]=(a,a);print@a}say$c
</code>```

The literal newline saves 1 byte over \n or $/.

Readable:

```<code>$\="*\n*";
print$c="*"x($_+1);
for$b(1..$_){
    @a=($")x$_;
    @a[$b-1,-$b]=(a,a);
    print@a
}
say$c
</code>```

The code prints the top line and saves it in $c, then prints a bunch of spaces with the appropriate slots replaced with as, then prints the top line again.

The assignment to the $\ variable tells the interpreter to print the contents (an asterisk, a newline, and another asterisk) after every print, but this does NOT occur after a say.

Title: Creating a Crossed Square

Body:
Perl, 83 +1 = 84 bytes

Run with the -n flag.

$\="*
*";print$c="*"x($_+1);for$b(1..$_){@a=($")x$_;@a[$b-1,-$b]=(a,a);print@a}say$c

The literal newline saves 1 byte over \n or $/.

Readable:

$\="*\n*";
print$c="*"x($_+1);
for$b(1..$_){
    @a=($")x$_;
    @a[$b-1,-$b]=(a,a);
    print@a
}
say$c

The code prints the top line and saves it in $c, then prints a bunch of spaces with the appropriate slots replaced with as, then prints the top line again.

The assignment to the $\ variable tells the interpreter to print the contents (an asterisk, a newline, and another asterisk) after every print, but this does NOT occur after a say.


Title: Build me a brick wall!

Body:
Challenge

I need help building a brick wall! Throw together some code for me
 using no input and produce the following output wall shown below:

```<code>_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__
_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__
_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__
_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__
_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__
</code>```

Now, this wall is exactly 10 characters high and 70 characters
 wide.

As stated above, no inputs, only code. Least amount of bytes, we have
 to efficiently build this brick wall. Obviously, this program has to
 function to get the wall built..? or it obviously won't build itself!
 Good luck!

Winner

The winner goes to Marinus using the APL language, requiring a whole 12 bytes!
https://codegolf.stackexchange.com/a/99028/61857

Thank you everyone for participating!

Body:
Ruby, 30 bytes

puts ('_|__'*175).scan /.{70}/

Output:

_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__
_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__
_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__
_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__
_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__


Title: This is my pillow

Body:
...will you help me immortalize it?

I've had this pillow a few years now, and apparently it's time to get rid of it. Can you please write a function or program, that I can bring with me and use to recreate this pillow whenever I want to reminisce a bit.

It must work with no input arguments.

The output should look exactly like this (trailing newlines and spaces are OK).

```<code>/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
</code>```

This is code golf, so the shortest code in bytes win!

Leaderboard

var QUESTION_ID=98701,OVERRIDE_USER=31516;function answersUrl(e){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"https://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="
"+e.body.replace(OVERRIDE_REG,"")+"
")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery("#answer-template").html();t=t.replace("{{PLACE}}",n+".").replace("{{NAME}}",e.user).replace("{{LANGUAGE}}",e.language).replace("{{SIZE}}",e.size).replace("{{LINK}}",e.link),t=jQuery(t),jQuery("#answers").append(t);var o=e.language;/s.lang?1:e.lang<s.lang?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;```
<pre class="snippet-code-css lang-css prettyprint-override"><code>body{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;float:left}table thead{font-weight:700}table td{padding:5px}</code>```

 <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b">

Leaderboard

	Author	Language	Size

Winners by Language

Language	User	Score

{{PLACE}}	{{NAME}}	{{LANGUAGE}}	{{SIZE}}	Link

{{LANGUAGE}}	{{NAME}}	{{SIZE}}	Link

```

Body:
PHP, <s>142</s> 124 bytes

<?=str_replace(['#','-'],['/////','\\\\\\\\\\'],wordwrap(str_repeat(str_repeat("#-",24).str_repeat("-#",24), 4),12,"\n",1));

Previous version:

<?for($i=1;$i<4;$i++)echo wordwrap(str_repeat("/////\\\\\\\\\\",24),60,"\n",1)."\n".wordwrap(str_repeat("\\\\\\\\\\/////",24),60,"\n",1)."\n";

Output:

C:\PHP\>php this-is-my-pillow.php
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////
\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////\\\\\/////


答:
<s>Python 2, 70 Bytes, A000217</s> Cracked!

I have a feeling this won't be cracked in the language I used for the other version, we will see :)

o=input()
v=0
i=1
while o:
 v+=i
 i+=1
print v

#|  d00->1@@@++-^,,[
```

<hr>

<p>I realized afterwards I had incorrectly obfuscated the code (it doesn't change the posted answer's validity). Here's the code I started with in Haystack:</p>

```v
0
v
0
i
1
-
>      d0[v
^-1@+@d+1@?,,o|
```

Title: Polyglot Anagrams Cops' Thread

Body:
This challenge has two threads.  This is the cops' thread. The robbers' thread is located here.

Your challenge is to choose an OEIS sequence and write two full programs in two different languages that produces that nth item in the sequence when given an n via STDIN, or an other forms of standard input, where n is any positive number.  However your two programs must be anagrams, meaning each can be rearranged from the other's letters.

Programs must output the decimal of the number followed by optional whitespace to STDOUT.  Programs may output to STDERR however it should be ignored and if the hidden case does so it must be clearly stated that this is the case.

If you wish you may also output by character code.  However if you do so in your hidden solution you must state such in the body of your submission.

You will then present the OEIS number, the source code for and the name of one language it is in.

Robbers will crack your submission if they find an anagram of the original  submission that runs in a language other than the one you already presented.  To crack an answer they must only find any language and program which produces the sequence and is an anagram of the original, not necessarily the answer the you were thinking about.

Thus you are incentivized to make it as hard as possible to find any language that does the task using their list of symbols.

Scoring

This is code-golf so the shortest un-cracked program is the winner.

Languages

Languages will be considered different if the two proposed solutions do not complete the task in both languages.  This will include different versions of the same language as long as neither the cop's solution or the robber's solution produce the correct output in the other's language.

i.e. If the there are two solutions 1 and 2 that are in language A and B respectively solution 2 must not produce the correct output in language A and solution 1 must not produce the correct output in language B.

Safety

Once your submission has been uncracked for a week you may post your solution and declare your post safe.  If after a week you choose not to post a solution your answer may still be cracked.

Body:
<s>Python 2, 70 Bytes, A000217</s> Cracked!

I have a feeling this won't be cracked in the language I used for the other version, we will see :)

o=input()
v=0
i=1
while o:
 v+=i
 i+=1
print v

#|  d00->1@@@++-^,,[

I realized afterwards I had incorrectly obfuscated the code (it doesn't change the posted answer's validity). Here's the code I started with in Haystack:

v
0
v
0
i
1
-
>      d0[v
^-1@+@d+1@?,,o|


答:
Vim, <s>26</s> 22 strokes

Assume the q register to be empty (if not type qqq)

```<code>175i_|__<esc>0qq70li<enter><esc>q8@q
</code>```

Title: Build me a brick wall!

Body:
Vim, <s>26</s> 22 strokes

Assume the q register to be empty (if not type qqq)

175i_|__<esc>0qq70li<enter><esc>q8@q


问:
The meta-polyglot quine
In my time on PPCG, I've noticed that quine problems and polyglot problems are quite popular. Also, meta-solutions to problems, that is, scripts which generate a program which is the solution to a problem, tend to get a lot of positive feedback from the community. Therefore, I have created this challenge, which implements these three ideas.

Your task then, reader and code-golf enthusiast, is to create as short as possible a script which can run in two languages A and B to generate quines for A and B. When your program is run in language A, it should generate a program that is a quine in language B but not in language A and vice versa. The languages A and B can be different version of the same language, provided you keep in mind that the generated quines should only work in one of the versions.

Keep in mind that the standard loopholes should be considered closed and only proper quines are allowed.

Good luck, fewest characters wins!

答:
JavaScript / C 278 bytes

At a staggering 278 bytes:

```<code>//\
console.log('int main(){char*A="int main(){char*A=%c%s%c;printf(A,34,A,34);}";printf(A,34,A,34);}');/*
int main(){puts("A='A=;B=String.fromCharCode(39);console.log(A.slice(0,2)+B+A+B+A.slice(2));';B=String.fromCharCode(39);console.log(A.slice(0,2)+B+A+B+A.slice(2));");}//*/
</code>```

The C quine:

int main(){char*A="int main(){char*A=%c%s%c;printf(A,34,A,34);}";printf(A,34,A,34);}

The JavaScript quine:

A='A=;B=String.fromCharCode(39);console.log(A.slice(0,2)+B+A+B+A.slice(2));';B=String.fromCharCode(39);console.log(A.slice(0,2)+B+A+B+A.slice(2));

Title: The meta-polyglot quine

Body:
JavaScript / C 278 bytes

At a staggering 278 bytes:

//\
console.log('int main(){char*A="int main(){char*A=%c%s%c;printf(A,34,A,34);}";printf(A,34,A,34);}');/*
int main(){puts("A='A=;B=String.fromCharCode(39);console.log(A.slice(0,2)+B+A+B+A.slice(2));';B=String.fromCharCode(39);console.log(A.slice(0,2)+B+A+B+A.slice(2));");}//*/

The C quine:

int main(){char*A="int main(){char*A=%c%s%c;printf(A,34,A,34);}";printf(A,34,A,34);}

The JavaScript quine:

A='A=;B=String.fromCharCode(39);console.log(A.slice(0,2)+B+A+B+A.slice(2));';B=String.fromCharCode(39);console.log(A.slice(0,2)+B+A+B+A.slice(2));


问:
GO OUT AND VOTE
Today is November 8th, 2016, Election Day in the United States of America.

If you are a U.S. citizen eligible to vote, then go out and vote if you haven't already before answering this challenge. Do not discuss who you voted for. It only matters that you voted.

If you are not a U.S. citizen or not eligible to vote, then, before answering this challenge, do the U.S. a favor by telling anyone you know who is an eligible citizen to go out and vote if they haven't already.

Challenge

Write a program that indicates that you voted, like a digital "I Voted" sticker.

It should take no input and must output in a reasonable way the phrase I Voted where the I, o, and e are red (#FF0000) and the V, t, and d are blue (#0000FF). The background must be white (#FFFFFF).

For example:

These colors are of course representative of the American flag (though not the official colors). Red comes first simply because it comes first in the common idiom "red white and blue".

To be valid, an answer must:

Use the colors specified in the arrangement specified.

Use a single legible font and font size. The example uses 72pt Times New Roman bold but any common font above 6pt is probably fine.

Have just the phrase I Voted on a single line, capitalized correctly, with a clear space between the two words. It shouldn't look like IVoted.

Not indicate who the answerer voted for or supports for president or any down-ballot races. Let's not start any internet debates. This is about celebrating voting, not candidates.

Any reasonable way of displaying or producing the output is valid, such as:

Drawing the text to an image that is then displayed, saved, or output raw.

Writing the text to a console using color formatting. In this case you may approximate pure red and blue if necessary, and it's ok if only the area directly behind the text can be made white.

Displaying the text on a WPF/Windows form.

Outputting an HTML/RTF/PDF file with the text.

Please post an image of your output.

The shortest answer in bytes wins.

答:
Python 2 (or 3), (75 bytes)

```<code>a="\033[3%sm"
r,b=a%"1;47",a%4
print(r+"I "+b+"V"+r+"o"+b+"t"+r+"e"+b+"d")
</code>```

Resized my terminal font size to 72 for a bigger picture, outputs in colour in most terminals.

Title: GO OUT AND VOTE

Body:
Python 2 (or 3), (75 bytes)

a="\033[3%sm"
r,b=a%"1;47",a%4
print(r+"I "+b+"V"+r+"o"+b+"t"+r+"e"+b+"d")

Resized my terminal font size to 72 for a bigger picture, outputs in colour in most terminals.


问:
Polyglot Anagrams Robbers' Thread
This is the robbers' thread of this challenge

The cops will choose an OEIS sequence and write two full programs in two different languages that produces that nth item in the sequence when given an n via STDIN.  The two programs must be anagrams of each other, meaning each can be rearranged from the other's letters.

They will present the OEIS number, the source code for and the name of one language it is in.

You must find an anagram of the original cop's submission that runs in a language other than the one used by the cop.  To crack an answer you must only find any language and program which produces the sequence and is an anagram of the original, not necessarily the answer the cop was thinking about.

You may output by character code or include STDERR in your solution, but only if the cop has indicated their hidden solution does so.  Decimal output to STDOUT is always a valid crack.

Thus cops are incentivized to make it as hard as possible to find any language that does the task using their list of symbols.

Scoring

The person with the most cracks will be the winner of this section.  Only the first crack for each solution will count towards one's score.

Leader Board

Big thanks to every one that took place in this challenge.

Here is the leader board as it stands

```<code>Place  User             Score
-----------------------------
1      Adnan            3
1      Kritixi Lithos   3
1      Emigna           3
1      milk             3
5      Steven H.        2
5      Wheat Wizard     2
5      jimmy23013       2
8      ETH Productions  1
8      Dennis           1
8      alleks           1
8      Martin Ender     1
8      Oliver           1
8      Conor O'Brien    1
</code>```

答:
2sable, 7 bytes, Kritixi Lithos, A005843

Code:

```<code>r^#ei2*
</code>```

Explanation:

```<code>r         # Reverse the stack, which is a no-op
 ^        # XOR the input with itself, always leading to 0
  #       # If 1, break (which stops the program)
   e      # Compute input nPr input, which always leads to 1
    i     # If equal to one, do the following..
     2*   #   Multiply the input by 2.
</code>```

Try it online!

Title: Polyglot Anagrams Robbers' Thread

Body:
2sable, 7 bytes, Kritixi Lithos, A005843

Code:

r^#ei2*

Explanation:

r         # Reverse the stack, which is a no-op
 ^        # XOR the input with itself, always leading to 0
  #       # If 1, break (which stops the program)
   e      # Compute input nPr input, which always leads to 1
    i     # If equal to one, do the following..
     2*   #   Multiply the input by 2.

Try it online!


答:
MATL, 7 bytes, A000217, cracked

```<code>:sp{1}x
</code>```

The sequence is n(n+1)/2 (triangular numbers), starting at input n=1 as specified by the challenge: 1, 3, 6, 10, ... (Output for input 0 is not guaranteed to be the same in the two programs).

The program in the other language exits with an error (after producing the correct output in STDOUT).

Try it online!

```<code>:            % Push [1 2 ... n], where n is implicit input
 s           % Sum of that array. Gives the desired result
   p         % Product of that. Gives the same number
    {1}      % Push a cell array containing number 1
       x     % Delete it
</code>```

Title: Polyglot Anagrams Cops' Thread

Body:
MATL, 7 bytes, A000217, cracked

:sp{1}x

The sequence is n(n+1)/2 (triangular numbers), starting at input n=1 as specified by the challenge: 1, 3, 6, 10, ... (Output for input 0 is not guaranteed to be the same in the two programs).

The program in the other language exits with an error (after producing the correct output in STDOUT).

Try it online!

:            % Push [1 2 ... n], where n is implicit input
 s           % Sum of that array. Gives the desired result
   p         % Product of that. Gives the same number
    {1}      % Push a cell array containing number 1
       x     % Delete it


问:
Koch Snowflake - codegolf
The Koch snowflake (also known as the Koch star and Koch island) is a mathematical curve and one of the earliest fractal curves to have been described. It is based on the Koch curve, which appeared in a 1904 paper titled "On a continuous curve without tangents, constructible from elementary geometry" (original French title: "Sur une courbe continue sans tangente, obtenue par une construction géométrique élémentaire") by the Swedish mathematician Helge von Koch.

Here are some ascii representations of various iterations:

```<code>n=1
__
\/

n=2
__/\__
\    /
/_  _\
  \/

n=3
      __/\__
      \    /
__/\__/    \__/\__
\                /
/_              _\
  \            /
__/            \__
\                /
/_  __      __  _\
  \/  \    /  \/
      /_  _\
        \/
</code>```

Since there is obviously a limit to the resolution of the ascii representation, we must enlarge the size of the snowflake by a factor of 3 for each iteration to show the extra detail.

Write the shortest code to output the snowflake in the same style for n=4

Your program should not take any input.

Your program should write the snowflake to the console.

答:
Python, 650 612 594 574 characters

```<code>n='\n'
S='_a/G\F I\n'
A=dict(zip(S,('III','   ','__/','  G','\  ','F__','   ','III','')))
B=dict(zip(S,('III','   ','\  ',' aF','/a ','  G','   ','III','')))
C=dict(zip(S,('___','aaa','/  ','GII','II\\','  F','   ','III','')))
def T(s):
 a=b=c=d=r=u''
 for k in s:
    a+=A[k];b+=B[k];c+=C[k]
    if k=='I':a=a[:-3]+('II\\'if'a '==d[1:3]else'GII'if' a'==d[:2]else 3*k)
    d=d[3:]
    if k==n:d=c.replace('____','__/F').replace('aaaa','aa  ').replace('/  a','/a  ').replace('a  F','  aF');r+=a+n+b+n+d+n;a=b=c=''
 return r
print T(T(T('__\n\G\n'))).translate({97:95,71:47,73:32,70:92})
</code>```

This works by expanding the triangle by a factor of 3 each time.  To do that, we need to keep track of whether each symbol is a left or right boundary (e.g. how / is expanded depends on which side of the / is the inside).  We use different symbols for the two possible cases, as follows:

```<code>_: _, outside on the top
a: _, outside on the bottom
/: /, outside on the left
G: /, outside on the right
\: \, outside on the left
F: \, outside on the right
<space>: inside
I: outside
</code>```

The d variable handles the special case where the expansion of an a needs to extend into the 3x3 in the next row.

ContentLicense: CC BY-SA 2.5

Title: Koch Snowflake - codegolf

Body:
Python, 650 612 594 574 characters

n='\n'
S='_a/G\F I\n'
A=dict(zip(S,('III','   ','__/','  G','\  ','F__','   ','III','')))
B=dict(zip(S,('III','   ','\  ',' aF','/a ','  G','   ','III','')))
C=dict(zip(S,('___','aaa','/  ','GII','II\\','  F','   ','III','')))
def T(s):
 a=b=c=d=r=u''
 for k in s:
    a+=A[k];b+=B[k];c+=C[k]
    if k=='I':a=a[:-3]+('II\\'if'a '==d[1:3]else'GII'if' a'==d[:2]else 3*k)
    d=d[3:]
    if k==n:d=c.replace('____','__/F').replace('aaaa','aa  ').replace('/  a','/a  ').replace('a  F','  aF');r+=a+n+b+n+d+n;a=b=c=''
 return r
print T(T(T('__\n\G\n'))).translate({97:95,71:47,73:32,70:92})

This works by expanding the triangle by a factor of 3 each time.  To do that, we need to keep track of whether each symbol is a left or right boundary (e.g. how / is expanded depends on which side of the / is the inside).  We use different symbols for the two possible cases, as follows:

_: _, outside on the top
a: _, outside on the bottom
/: /, outside on the left
G: /, outside on the right
\: \, outside on the left
F: \, outside on the right
<space>: inside
I: outside

The d variable handles the special case where the expansion of an a needs to extend into the 3x3 in the next row.


问:
Find the Emirps!

An emirp is a non-palindromic prime which, when reversed, is also prime.

The list of base 10 emirps can be found on OEIS. The first six are:

```<code>13, 17, 31, 37, 71, 73
</code>```

However, due to the reversal rule, emirps are different in each base. For example, the first six binary emirps are:

```<code>Bin  | 1011, 1101, 10111, 11101, 101001, 100101
Dec  | (11 , 13  , 23   , 29   , 37    , 41   )
</code>```

...and in hexadecimal, they are:

```<code>Hex |  17, 1F, 35, 3B, 3D, 53
Dec | (23, 31, 53, 59, 61, 83)
</code>```

Fun Fact: there are no emirps in unary as every number is a palindrome.

The Challenge

Your task is to create a function (or full program) which takes two parameters,
\$ n \$ and \$ b \$, and generates a list of the first \$ n \$ emirps in base \$ b \$.

Rules/Details:

\$ n \$ and \$ b \$ are both positive integers larger than \$ 0 \$.

You can assume \$ 2 ≤ b ≤ 16 \$: that is to say, the base will be between binary and hexidecimal.

You should be able to compute for values of \$ n \$ up to \$ ~100 \$.

The generated list can be in base \$ b \$, or your language's standard integer base, as long as you specify this in your answer.

Builtin emirp checks are not permitted (builtin primality tests are fine)

You cannot hard-code the emirps, or read from any external files.

Standard loopholes are banned, as always.

This is code-golf, so the shortest answer (in bytes) wins.

Test Cases

For each test case, I've included the list in base b and its base 10 equivalents.

```<code>B = 2, N = 10

BIN: [1011, 1101, 10111, 11101, 100101, 101001, 101011, 101111, 110101, 111101]
DEC: [11, 13, 23, 29, 37, 41, 43, 47, 53, 61]

B = 3, N = 5

BASE3: [12, 21, 102, 201, 1011]
DEC:   [5, 7, 11, 19, 31]

B = 12, N = 7

BASE12: [15, 51, 57, 5B, 75, B5, 107]
DEC: [17, 61, 67, 71, 89, 137, 151]

B = 16, N = 4

HEX: [17, 1F, 35, 3B]
DEC: [23, 31, 53, 59]
</code>```

You can test your program further against my (ungolfed) Python example on repl.it

答:
05AB1E, 17 bytes

Uses CP-1252 encoding.

Input order is n, b

Output is in base-10.

```<code>µN²BÂD²öpŠÊNpPD–½
</code>```

Try it online!

Explanation

```<code>                    # implicit input a,b
µ                   # loop until counter is a
 N²B                # convert current iteration number to base b
    ÂD              # create 2 reversed copies
      ²ö            # convert one reversed copy to base 10
        p           # check for primality
         ŠÊ         # compare the normal and reversed number in base b for inequality
           Np       # check current iteration number for primality
             P      # product of all
              D     # duplicate
               –    # if 1, print current iteration number
                ½   # if 1, increase counter
</code>```

Title: Find the Emirps!

Body:
05AB1E, 17 bytes

Uses CP-1252 encoding.

Input order is n, b

Output is in base-10.

µN²BÂD²öpŠÊNpPD–½

Try it online!

Explanation

                    # implicit input a,b
µ                   # loop until counter is a
 N²B                # convert current iteration number to base b
    ÂD              # create 2 reversed copies
      ²ö            # convert one reversed copy to base 10
        p           # check for primality
         ŠÊ         # compare the normal and reversed number in base b for inequality
           Np       # check current iteration number for primality
             P      # product of all
              D     # duplicate
               –    # if 1, print current iteration number
                ½   # if 1, increase counter


答:
<s>WinDbg, 39 bytes, A000007</s> Cracked by jimmy23013

```<code>~e.block{j(0>=@$t0)?@$t0+(1<7);??0}t":"
</code>```

The difficult sequence of 0**n.

Input is done by passing a value in the pseudo-register $t0.

My original solution was C#:

```<code>@t=>(object)(0<@t?0:1)??"$$700~lk.{}+";
</code>```

Title: Polyglot Anagrams Cops' Thread

Body:
<s>WinDbg, 39 bytes, A000007</s> Cracked by jimmy23013

~e.block{j(0>=@$t0)?@$t0+(1<7);??0}t":"

The difficult sequence of 0**n.

Input is done by passing a value in the pseudo-register $t0.

My original solution was C#:

@t=>(object)(0<@t?0:1)??"$$700~lk.{}+";


答:
<s>Excel, 12 bytes, A000012</s> Cracked

```<code>=IF(1=1,1,1)
</code>```

Maybe not the toughest, but a fun one to crack.

Title: Polyglot Anagrams Cops' Thread

Body:
<s>Excel, 12 bytes, A000012</s> Cracked

=IF(1=1,1,1)

Maybe not the toughest, but a fun one to crack.


答:
Encapsulated Postscript, 103 bytes

Copy-paste and save as .eps to display it:

```<code>%!PS
/run{100 500 moveto setrgbcolor show}def/Courier 60 selectfont(I  o e)1 0 0 run(  V t d)0 0 1 run
</code>```

Explanation:

```<code>/run{100 500 moveto setrgbcolor show}def
</code>```

Declares a procedure "run" that expects to have in the stack a string and 3 rgb values. The procedure moves the cursor to 100 500, takes the 3 rgb values from the stack to set the color, then takes the string from the stack and displays it,

```<code>/Courier 60 selectfont
</code>```

Sets the Courier font with size 60

```<code>(I  o e)1 0 0 run
</code>```

Puts the string and the 3 rgb componets in the stack and calls "run"

Title: GO OUT AND VOTE

Body:
Today is November 8th, 2016, Election Day in the United States of America.

If you are a U.S. citizen eligible to vote, then go out and vote if you haven't already before answering this challenge. Do not discuss who you voted for. It only matters that you voted.

If you are not a U.S. citizen or not eligible to vote, then, before answering this challenge, do the U.S. a favor by telling anyone you know who is an eligible citizen to go out and vote if they haven't already.

Challenge

Write a program that indicates that you voted, like a digital "I Voted" sticker.

It should take no input and must output in a reasonable way the phrase I Voted where the I, o, and e are red (#FF0000) and the V, t, and d are blue (#0000FF). The background must be white (#FFFFFF).

For example:

These colors are of course representative of the American flag (though not the official colors). Red comes first simply because it comes first in the common idiom "red white and blue".

To be valid, an answer must:

Use the colors specified in the arrangement specified.

Use a single legible font and font size. The example uses 72pt Times New Roman bold but any common font above 6pt is probably fine.

Have just the phrase I Voted on a single line, capitalized correctly, with a clear space between the two words. It shouldn't look like IVoted.

Not indicate who the answerer voted for or supports for president or any down-ballot races. Let's not start any internet debates. This is about celebrating voting, not candidates.

Any reasonable way of displaying or producing the output is valid, such as:

Drawing the text to an image that is then displayed, saved, or output raw.

Writing the text to a console using color formatting. In this case you may approximate pure red and blue if necessary, and it's ok if only the area directly behind the text can be made white.

Displaying the text on a WPF/Windows form.

Outputting an HTML/RTF/PDF file with the text.

Please post an image of your output.

The shortest answer in bytes wins.

Body:
Encapsulated Postscript, 103 bytes

Copy-paste and save as .eps to display it:

%!PS
/run{100 500 moveto setrgbcolor show}def/Courier 60 selectfont(I  o e)1 0 0 run(  V t d)0 0 1 run

Explanation:

/run{100 500 moveto setrgbcolor show}def

Declares a procedure "run" that expects to have in the stack a string and 3 rgb values. The procedure moves the cursor to 100 500, takes the 3 rgb values from the stack to set the color, then takes the string from the stack and displays it,

/Courier 60 selectfont

Sets the Courier font with size 60

(I  o e)1 0 0 run

Puts the string and the 3 rgb componets in the stack and calls "run"


答:
05AB1E, 12 bytes, tuskiomi, A000012

```<code>1,(FI1=1=1,)
</code>```

Try it online!

Explanation

```<code>1,            # print 1
  (           # negate input
   F          # that many times do (i.e. no times)
    I1=1=1,)  # the rest of the code
</code>```

Title: Polyglot Anagrams Robbers' Thread

Body:
This is the robbers' thread of this challenge

The cops will choose an OEIS sequence and write two full programs in two different languages that produces that nth item in the sequence when given an n via STDIN.  The two programs must be anagrams of each other, meaning each can be rearranged from the other's letters.

They will present the OEIS number, the source code for and the name of one language it is in.

You must find an anagram of the original cop's submission that runs in a language other than the one used by the cop.  To crack an answer you must only find any language and program which produces the sequence and is an anagram of the original, not necessarily the answer the cop was thinking about.

You may output by character code or include STDERR in your solution, but only if the cop has indicated their hidden solution does so.  Decimal output to STDOUT is always a valid crack.

Thus cops are incentivized to make it as hard as possible to find any language that does the task using their list of symbols.

Scoring

The person with the most cracks will be the winner of this section.  Only the first crack for each solution will count towards one's score.

Leader Board

Big thanks to every one that took place in this challenge.

Here is the leader board as it stands

```<code>Place  User             Score
-----------------------------
1      Adnan            3
1      Kritixi Lithos   3
1      Emigna           3
1      milk             3
5      Steven H.        2
5      Wheat Wizard     2
5      jimmy23013       2
8      ETH Productions  1
8      Dennis           1
8      alleks           1
8      Martin Ender     1
8      Oliver           1
8      Conor O'Brien    1
</code>```

Body:
05AB1E, 12 bytes, tuskiomi, A000012

1,(FI1=1=1,)

Try it online!

Explanation

1,            # print 1
  (           # negate input
   F          # that many times do (i.e. no times)
    I1=1=1,)  # the rest of the code


Title: Make me a blanket!

Body:
I want a blanket that looks like this. Each strip goes over, under, over, under. Can you print it?

```<code>\\\\////\\\\////\\\\////\\\\////\\\\////\\\\////
 \\////  \\////  \\////  \\////  \\////  \\////
  ////    ////    ////    ////    ////    ////
 ////\\  ////\\  ////\\  ////\\  ////\\  ////\\
////\\\\////\\\\////\\\\////\\\\////\\\\////\\\\
\//  \\\\//  \\\\//  \\\\//  \\\\//  \\\\//  \\\
\\    \\\\    \\\\    \\\\    \\\\    \\\\    \\
\\\  //\\\\  //\\\\  //\\\\  //\\\\  //\\\\  //\
\\\\////\\\\////\\\\////\\\\////\\\\////\\\\////
 \\////  \\////  \\////  \\////  \\////  \\////
  ////    ////    ////    ////    ////    ////
 ////\\  ////\\  ////\\  ////\\  ////\\  ////\\
////\\\\////\\\\////\\\\////\\\\////\\\\////\\\\
\//  \\\\//  \\\\//  \\\\//  \\\\//  \\\\//  \\\
\\    \\\\    \\\\    \\\\    \\\\    \\\\    \\
\\\  //\\\\  //\\\\  //\\\\  //\\\\  //\\\\  //\
\\\\////\\\\////\\\\////\\\\////\\\\////\\\\////
 \\////  \\////  \\////  \\////  \\////  \\////
  ////    ////    ////    ////    ////    ////
 ////\\  ////\\  ////\\  ////\\  ////\\  ////\\
////\\\\////\\\\////\\\\////\\\\////\\\\////\\\\
\//  \\\\//  \\\\//  \\\\//  \\\\//  \\\\//  \\\
\\    \\\\    \\\\    \\\\    \\\\    \\\\    \\
\\\  //\\\\  //\\\\  //\\\\  //\\\\  //\\\\  //\
</code>```

Trailing spaces at the end of each line and trailing newlines are acceptable.

Remember, this is code-golf, so the code with the fewest bytes wins.

Leaderboards

Here is a Stack Snippet to generate both a regular leaderboard and an overview of winners by language.

To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:

```<code># Language Name, N bytes
</code>```

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

```<code># Ruby, <s>104</s> <s>101</s> 96 bytes
</code>```

If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:

```<code># Perl, 43 + 2 (-p flag) = 45 bytes
</code>```

You can also make the language name a link which will then show up in the leaderboard snippet:

```<code># [><>](http://esolangs.org/wiki/Fish), 121 bytes
</code>```

var QUESTION_ID=99023,OVERRIDE_USER=12537;function answersUrl(e){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"https://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="
"+e.body.replace(OVERRIDE_REG,"")+"
")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery("#answer-template").html();t=t.replace("{{PLACE}}",n+".").replace("{{NAME}}",e.user).replace("{{LANGUAGE}}",e.language).replace("{{SIZE}}",e.size).replace("{{LINK}}",e.link),t=jQuery(t),jQuery("#answers").append(t);var o=e.language;/s.lang?1:e.lang<s.lang?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;```
<pre class="snippet-code-css lang-css prettyprint-override"><code>body{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;float:left}table thead{font-weight:700}table td{padding:5px}</code>```

 <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b">

Leaderboard

	Author	Language	Size

Winners by Language

Language	User	Score

{{PLACE}}	{{NAME}}	{{LANGUAGE}}	{{SIZE}}	Link

{{LANGUAGE}}	{{NAME}}	{{SIZE}}	Link

```

Body:
Perl, <s>132</s> <s>131</s> 113 bytes

@a=((0)x4,1..4);map{say+(map$a[7-$_]?$a[$_]*$r?'/':'\\':$a[$_]?'/':$",0..7)x6;push@a,shift@a;$_%4||($r=!$r)}0..23

Ungolfed:

use strict;
use warnings;
use feature 'say';

my @a = ((1) x 4, (0) x 4);  # print '\' if true
my @b = ((0) x 4, (1) x 4);  # print '/' if true
my $r = 0;                   # print '\' over '/' if true

for (0 .. 23) {
    say((map { $a[$_] ? ($b[$_] * $r ? '/' : '\\') : ($b[$_] ? '/' : ' ') } 0 .. 7) x 6);
    unshift(@a, pop(@a));    # circular shift to left
    push(@b, shift(@b));     # circular shift to right
    $r = !$r if !($_ % 4);   # change print priority
}


答:
Pyth, 36 bytes

```<code>V24sm@" \// \/\\"im<3%k8++BNdt-NdT48
</code>```

Try it online: Demonstration

Explanation:

We can determine the symbol by checking 3 conditions:

```<code>A := row % 8 > 3
B := (row + column) % 8 > 3
C := (row - column - 1) % 8 > 3
</code>```

If we interpret [A,B,C] as a binary number, we get the following mapping:

```<code>01234567
 \// \/\
</code>```

We can also interpret [A,B,C] as decimal number and perform a modular indexed lookup in the string. This doesn't make a difference because 10 mod 8 = 2.

Now to the code:

V24 iterates N (row-idx) over [0, 1, ..., 23].

sm...48 maps the numbers d (column-idx) in [0, 1, ..., 47] to chars and print the combined string.

++BNd generates the list [N, N+d], +...t-Nd appends N-d-1. So we have get the list [N, N+d, N-d-1]. m<3%k8 checks for each computed number k, if 3 < k % 8, so this gives the list with the conditions [A, B, C].

i...T convents this to a decimal number and then @" \// \/\\" does the lookup in the string.

More or less the same code in Python2: 98 bytes:

```<code>R=range(8)
for r in R*3:print''.join(" \// \/\\"[4*(r>3)+2*((r+c)%8>3)+((r-c-1)%8>3)]for c in R*6)
</code>```

Title: Make me a blanket!

Body:
Pyth, 36 bytes

V24sm@" \// \/\\"im<3%k8++BNdt-NdT48

Try it online: Demonstration

Explanation:

We can determine the symbol by checking 3 conditions:

A := row % 8 > 3
B := (row + column) % 8 > 3
C := (row - column - 1) % 8 > 3

If we interpret [A,B,C] as a binary number, we get the following mapping:

01234567
 \// \/\

We can also interpret [A,B,C] as decimal number and perform a modular indexed lookup in the string. This doesn't make a difference because 10 mod 8 = 2.

Now to the code:

V24 iterates N (row-idx) over [0, 1, ..., 23].

sm...48 maps the numbers d (column-idx) in [0, 1, ..., 47] to chars and print the combined string.

++BNd generates the list [N, N+d], +...t-Nd appends N-d-1. So we have get the list [N, N+d, N-d-1]. m<3%k8 checks for each computed number k, if 3 < k % 8, so this gives the list with the conditions [A, B, C].

i...T convents this to a decimal number and then @" \// \/\\" does the lookup in the string.

More or less the same code in Python2: 98 bytes:

R=range(8)
for r in R*3:print''.join(" \// \/\\"[4*(r>3)+2*((r+c)%8>3)+((r-c-1)%8>3)]for c in R*6)


答:
CJam, 70 bytes, Shebang, A000217

```<code>ri),:+

e#|  d00->1@@@++-^,[o=input(v=0i=1whil o v=i i+=1pnt v
</code>```

Try it here.

Title: Polyglot Anagrams Robbers' Thread

Body:
CJam, 70 bytes, Shebang, A000217

ri),:+

e#|  d00->1@@@++-^,[o=input(v=0i=1whil o v=i i+=1pnt v

Try it here.


答:
Python, <s>245</s> <s>236</s> <s>234</s> <s>233</s> <s>230</s> <s>216</s> <s>212</s> <s>198</s> 195 bytes

OK, longer than my last (and any other) answer but would be interested in feedback on the approach.

for a in(40,4496,6200,5456,3240,1188,720,228)*3:print((`a%6561/2178`+`a%2178/729`+`a%729/243`+`a%243/81`+`a%81/27`+`a%27/9`+`a%9/3`+`a%3/1`)*6).replace('0','\\').replace('1','/').replace('2',' ')
```

Edit

-9 due to @nedla2004 being more on the ball than me

-2 by taking the lambda outside of the loop and so losing 2 indent spaces

-1 by using in' '*3 instead of in 0,1,2 since I don't use h anyway. it's just a counter.

-3 Why, why, why did I leave a newline and 2 indents between the second for and the print??? It's late. Will revisit tomorrow.

-14 Can actually lose the lambda completely and just include the base 3 decoder directly after the print statement. Looks messy but after all, this is code golf :)

-4 No point setting a variable for the integer list. Just use it directly in the second for loop.

-14 and no point using the outer loop. Just multiply the integer tuple by 3 (shamelessly stolen from @nedla2004 to get under 200 :))

-3 Saved 3 by making \=0, /=1 and space=2. This makes the integer list shorter as three of the base 3 numbers now have leading 0's

How it works (and it does)

Since only 3 characters are used:

l is a list of the 8 repeating patterns as integer equivalents of their base 3 representation assuming that " "=0, "\"=1 and "/"=2

<s>The lambda</s> The first code after the print statement is a lightweight converter from integer to a base 3 string

The first loop loops 3 times and the second prints each line with the base 3 characters multiplied by 6 and replaced with /,\ or space.

I'm sure I could use a regex instead of the nested replace() but I'm too tired to try right now. This was just an experiment and longer than my previous Python effort but have posted just for any comments on the approach (and also because I have never worked in base 3 before and I quite enjoyed working out the converter).

Title: Make me a blanket!

Body:
Python, <s>245</s> <s>236</s> <s>234</s> <s>233</s> <s>230</s> <s>216</s> <s>212</s> <s>198</s> 195 bytes

OK, longer than my last (and any other) answer but would be interested in feedback on the approach.

for a in(40,4496,6200,5456,3240,1188,720,228)*3:print((`a%6561/2178`+`a%2178/729`+`a%729/243`+`a%243/81`+`a%81/27`+`a%27/9`+`a%9/3`+`a%3/1`)*6).replace('0','\\').replace('1','/').replace('2',' ')

Edit

-9 due to @nedla2004 being more on the ball than me

-2 by taking the lambda outside of the loop and so losing 2 indent spaces

-1 by using in' '*3 instead of in 0,1,2 since I don't use h anyway. it's just a counter.

-3 Why, why, why did I leave a newline and 2 indents between the second for and the print??? It's late. Will revisit tomorrow.

-14 Can actually lose the lambda completely and just include the base 3 decoder directly after the print statement. Looks messy but after all, this is code golf :)

-4 No point setting a variable for the integer list. Just use it directly in the second for loop.

-14 and no point using the outer loop. Just multiply the integer tuple by 3 (shamelessly stolen from @nedla2004 to get under 200 :))

-3 Saved 3 by making \=0, /=1 and space=2. This makes the integer list shorter as three of the base 3 numbers now have leading 0's

How it works (and it does)

Since only 3 characters are used:

l is a list of the 8 repeating patterns as integer equivalents of their base 3 representation assuming that " "=0, "\"=1 and "/"=2

<s>The lambda</s> The first code after the print statement is a lightweight converter from integer to a base 3 string

The first loop loops 3 times and the second prints each line with the base 3 characters multiplied by 6 and replaced with /,\ or space.

I'm sure I could use a regex instead of the nested replace() but I'm too tired to try right now. This was just an experiment and longer than my previous Python effort but have posted just for any comments on the approach (and also because I have never worked in base 3 before and I quite enjoyed working out the converter).


答:
Perl, 262 bytes

```<code>($b,$n)=@ARGV;$,=',';sub c{my$z;for($_=pop;$_;$z=(0..9,a..z)[$_%$b].$z,$_=($_-$_%$b)/$b){};$z}sub d{my$z;for(;c(++$z)ne@_[0];){}$z}for($p=2;@a<$n;$p++){$r=qr/^1?$|^(11+?)\1+$/;(c($p)eq reverse c$p)||((1x$p)=~$r)||(1x d($x=reverse c($p)))=~$r?1:push@a,c($p);}say@a
</code>```

Readable:

```<code>($b,$n)=@ARGV;
$,=',';
sub c{
    my$z;
    for($_=pop;$_;$z=(0..9,a..z)[$_%$b].$z,$_=($_-$_%$b)/$b){};
    $z
}
sub d{
    my$z;
    for(;c(++$z)ne@_[0];){}
    $z
}
for($p=2;@a<$n;$p++){
    $r=qr/^1?$|^(11+?)\1+$/;
    (c($p)eq reverse c$p)||((1x$p)=~$r)||(1x d($x=reverse c($p)))=~$r?1:push@a,c($p)
}
say@a
</code>```

c converts a given number into base $b, and d converts a given number from base $b back into decimal by finding the first number which returns said base-$b number when passed to c.  The for loop then checks if it's a palindrome and if both of the numbers are prime using the composite regex.

Title: Find the Emirps!

Body:
An emirp is a non-palindromic prime which, when reversed, is also prime.

The list of base 10 emirps can be found on OEIS. The first six are:

```<code>13, 17, 31, 37, 71, 73
</code>```

However, due to the reversal rule, emirps are different in each base. For example, the first six binary emirps are:

```<code>Bin  | 1011, 1101, 10111, 11101, 101001, 100101
Dec  | (11 , 13  , 23   , 29   , 37    , 41   )
</code>```

...and in hexadecimal, they are:

```<code>Hex |  17, 1F, 35, 3B, 3D, 53
Dec | (23, 31, 53, 59, 61, 83)
</code>```

Fun Fact: there are no emirps in unary as every number is a palindrome.

The Challenge

Your task is to create a function (or full program) which takes two parameters,
\$ n \$ and \$ b \$, and generates a list of the first \$ n \$ emirps in base \$ b \$.

Rules/Details:

\$ n \$ and \$ b \$ are both positive integers larger than \$ 0 \$.

You can assume \$ 2 ≤ b ≤ 16 \$: that is to say, the base will be between binary and hexidecimal.

You should be able to compute for values of \$ n \$ up to \$ ~100 \$.

The generated list can be in base \$ b \$, or your language's standard integer base, as long as you specify this in your answer.

Builtin emirp checks are not permitted (builtin primality tests are fine)

You cannot hard-code the emirps, or read from any external files.

Standard loopholes are banned, as always.

This is code-golf, so the shortest answer (in bytes) wins.

Test Cases

For each test case, I've included the list in base b and its base 10 equivalents.

```<code>B = 2, N = 10

BIN: [1011, 1101, 10111, 11101, 100101, 101001, 101011, 101111, 110101, 111101]
DEC: [11, 13, 23, 29, 37, 41, 43, 47, 53, 61]

B = 3, N = 5

BASE3: [12, 21, 102, 201, 1011]
DEC:   [5, 7, 11, 19, 31]

B = 12, N = 7

BASE12: [15, 51, 57, 5B, 75, B5, 107]
DEC: [17, 61, 67, 71, 89, 137, 151]

B = 16, N = 4

HEX: [17, 1F, 35, 3B]
DEC: [23, 31, 53, 59]
</code>```

You can test your program further against my (ungolfed) Python example on repl.it

Body:
Perl, 262 bytes

($b,$n)=@ARGV;$,=',';sub c{my$z;for($_=pop;$_;$z=(0..9,a..z)[$_%$b].$z,$_=($_-$_%$b)/$b){};$z}sub d{my$z;for(;c(++$z)ne@_[0];){}$z}for($p=2;@a<$n;$p++){$r=qr/^1?$|^(11+?)\1+$/;(c($p)eq reverse c$p)||((1x$p)=~$r)||(1x d($x=reverse c($p)))=~$r?1:push@a,c($p);}say@a

Readable:

($b,$n)=@ARGV;
$,=',';
sub c{
    my$z;
    for($_=pop;$_;$z=(0..9,a..z)[$_%$b].$z,$_=($_-$_%$b)/$b){};
    $z
}
sub d{
    my$z;
    for(;c(++$z)ne@_[0];){}
    $z
}
for($p=2;@a<$n;$p++){
    $r=qr/^1?$|^(11+?)\1+$/;
    (c($p)eq reverse c$p)||((1x$p)=~$r)||(1x d($x=reverse c($p)))=~$r?1:push@a,c($p)
}
say@a

c converts a given number into base $b, and d converts a given number from base $b back into decimal by finding the first number which returns said base-$b number when passed to c.  The for loop then checks if it's a palindrome and if both of the numbers are prime using the composite regex.


Title: Polyglot Anagrams Robbers' Thread

Body:
CJam, 39 bytes, milk, A000007

l~0{"e.bockj(>=@$t0)@$t0+(<7)??0t:"};1?

Try it here.


问:
Determine the optimal cruise control options
A cruise control has 3 different options to move the handle to set the speed you want to drive with.

Towards you: Adds 1 speed.

Upwards: Increases speed to the next multiple of 10 (e.g. 20-->30, 32-->40)

Downwards: Decreases speed to the next multiple of 10 (e.g. 20-->10, 32-->30)

Input

2 integers: the first is the starting speed and the second is your desired speed, both non-negative and in any form you like (array, two arguments, etc.)

Task

Determine the optimal way of using the handle to reach the desired speed and print out the moves in the correct order.

Rules

If you have the choice between pulling towards you and going upwards (like from 39 to 40) you can choose either option, but stay with whatever you choose for similar cases

You may use any 3 different (preferably visible) symbols to distinguish between the moves in the output (T, U and D for example).

The symbols can be seperated by new lines, spaces, etc. but don't have to be

Here are some test cases:

```<code>start speed, desired speed  -->  output
30, 40  -->  U
30, 43  -->  UTTT
43, 30  -->  DD
51, 39  -->  DDDTTTTTTTTT
29, 30  -->  T or U
29, 50  -->  TUU or UUU
12, 12  -->
</code>```

This is code-golf so the shortest answer in bytes wins.

Body:
JavaScript (ES6), <s>91</s> <s>84</s> 75 bytes

Saved 4 bytes thanks to @Neil

f=(s,d)=>s-d?(q=s+10-s%10)>d?s>d?0+f(s-(s%10||10),d):1+f(s+1,d):2+f(q,d):""

Uses 0 for D, 1 for T, and 2 for U.


答:
Batch, 175 bytes

```<code>@set/as=%1,d=%2,e=d/10*10
:d
@if %s% gtr %d% echo d&set/as=~-s/10*10&goto d
:u
@if %s% lss %e% echo u&set/as=s/10*10+10&goto u
:t
@if %s% neq %d% echo t&set/as+=1&goto t
</code>```

Fairly straightforward this time. Takes input as command-line parameters, which it saves into s and d. e is d rounded down to the previous multiple of 10. If s is greater than d, then we obviously need to invoke d until s becomes lower than d. Otherwise, we need to check whether s is lower than e; if so, we can invoke u until s equals e. At this point s is now between e and d and we can simply invoke t until we reach d. I looked into for loops but they use inclusive endpoints so would have become too verbose.

Title: Determine the optimal cruise control options

Body:
A cruise control has 3 different options to move the handle to set the speed you want to drive with.

Towards you: Adds 1 speed.

Upwards: Increases speed to the next multiple of 10 (e.g. 20-->30, 32-->40)

Downwards: Decreases speed to the next multiple of 10 (e.g. 20-->10, 32-->30)

Input

2 integers: the first is the starting speed and the second is your desired speed, both non-negative and in any form you like (array, two arguments, etc.)

Task

Determine the optimal way of using the handle to reach the desired speed and print out the moves in the correct order.

Rules

If you have the choice between pulling towards you and going upwards (like from 39 to 40) you can choose either option, but stay with whatever you choose for similar cases

You may use any 3 different (preferably visible) symbols to distinguish between the moves in the output (T, U and D for example).

The symbols can be seperated by new lines, spaces, etc. but don't have to be

Here are some test cases:

```<code>start speed, desired speed  -->  output
30, 40  -->  U
30, 43  -->  UTTT
43, 30  -->  DD
51, 39  -->  DDDTTTTTTTTT
29, 30  -->  T or U
29, 50  -->  TUU or UUU
12, 12  -->
</code>```

This is code-golf so the shortest answer in bytes wins.

Body:
Batch, 175 bytes

@set/as=%1,d=%2,e=d/10*10
:d
@if %s% gtr %d% echo d&set/as=~-s/10*10&goto d
:u
@if %s% lss %e% echo u&set/as=s/10*10+10&goto u
:t
@if %s% neq %d% echo t&set/as+=1&goto t

Fairly straightforward this time. Takes input as command-line parameters, which it saves into s and d. e is d rounded down to the previous multiple of 10. If s is greater than d, then we obviously need to invoke d until s becomes lower than d. Otherwise, we need to check whether s is lower than e; if so, we can invoke u until s equals e. At this point s is now between e and d and we can simply invoke t until we reach d. I looked into for loops but they use inclusive endpoints so would have become too verbose.


答:
Mathematica 112 bytes

```<code>Cases[Table[Prime@n~IntegerDigits~#2,{n,500}],x_/;x!=(z=Reverse@x)&&PrimeQ[z~(f=FromDigits)~#2]:>x~f~#2]~Take~#&
</code>```

Example

Find the first 10 Emips in hex; return them in decimal.

```<code>Cases[Table[Prime@n~IntegerDigits~#2, {n, 500}],
x_ /; x != (z = Reverse@x) && PrimeQ[z~(f = FromDigits)~#2] :> x~f~#2]~Take~# &[10, 16]

{23, 31, 53, 59, 61, 83, 89, 113, 149, 179}
</code>```

Ungolfed

```<code>Take[Cases[                                             (* take #1 cases; #1 is the first input argument *)
   Table[IntegerDigits[Prime[n], #2], {n, 500}],        (* from a list of the first 500 primes, each displayed as a list of digits in base #2 [second argument] *)
   x_ /;                                                (* x, a list of digits, such that *)
   x != (z = Reverse[x]) && PrimeQ[FromDigits[z, #2]]   (* the reverse of the digits is not the same as the list of digits; and the reverse list, when composed, also constitutes a prime *)
   :> FromDigits[x, #2]],                               (* and return the prime *)
   #1] &                                                (* [this is where #1 goes, stating how many cases to Take] *)
</code>```

Title: Find the Emirps!

Body:
Mathematica 112 bytes

Cases[Table[Prime@n~IntegerDigits~#2,{n,500}],x_/;x!=(z=Reverse@x)&&PrimeQ[z~(f=FromDigits)~#2]:>x~f~#2]~Take~#&

Example

Find the first 10 Emips in hex; return them in decimal.

Cases[Table[Prime@n~IntegerDigits~#2, {n, 500}],
x_ /; x != (z = Reverse@x) && PrimeQ[z~(f = FromDigits)~#2] :> x~f~#2]~Take~# &[10, 16]

{23, 31, 53, 59, 61, 83, 89, 113, 149, 179}

Ungolfed

Take[Cases[                                             (* take #1 cases; #1 is the first input argument *)
   Table[IntegerDigits[Prime[n], #2], {n, 500}],        (* from a list of the first 500 primes, each displayed as a list of digits in base #2 [second argument] *)
   x_ /;                                                (* x, a list of digits, such that *)
   x != (z = Reverse[x]) && PrimeQ[FromDigits[z, #2]]   (* the reverse of the digits is not the same as the list of digits; and the reverse list, when composed, also constitutes a prime *)
   :> FromDigits[x, #2]],                               (* and return the prime *)
   #1] &                                                (* [this is where #1 goes, stating how many cases to Take] *)


问:
Solve the Halting Problem for Befinge
Let's define a simple 2D language, which we'll give the incredibly original name befinge. Befinge has 5 instructions:

<>^v, as in most 2D esolangs, redirect the instruction pointer in their respective directions.

. is  a no-op.

The instruction pointer starts out at the top-left corner going right. If the instruction pointer gets to an edge, the program halts. Every Befinge program will obviously either halt or go into an infinite loop which does nothing. Here are two examples:

Halting:

```<code>>.v
..<
</code>```

Non-Halting:

```<code>>....v
..v..<
..>v..
^..<..
</code>```

The halting problem is not solvable for a Turing-complete language, but it is for this one. Your task is to write a program (or function) that takes as input a string representing the befinge program and returns a truthy or falsey value depending on whether it halts or not.

You can assume that the input will consist only of these characters and will be padded with spaces to form a rectangle.

You can use any set of five characters for the instructions (e.g. adws ).

Test Cases

Halting:

```<code>.
</code>```

```<code>v>
>^
</code>```

```<code>....v....
....>...v
.^..<....
.......v<
.......v.
....^..<.
</code>```

```<code>v<>v>v^
>v^>^>v
<>>^v<v
v^<>v^<
</code>```

Non-Halting:

```<code>>..v
^..<
</code>```

```<code>>v<
v<.
>v.
v<.
>.^
</code>```

```<code>>.>.>.v
.><.<.<
</code>```

This is code-golf, so the shortest program (in bytes) wins.

答:
JavaScript (ES6), <s>158</s> 127 bytes

```<code>f=(a,x=0,y=0,d=1,e=0,c=a[y]&&a[y][x])=>c<'~'?(c>'.'&&(a[y][x]='~',d=(c=='>')-(c=='<'),e=(c=='v')-(c=='^')),f(a,x+d,y+e,d,e)):!c
</code>```

Takes input as a two-dimensional character array and returns true for halting and false for an infinite loop. Works by setting visited direction characters to ~s as it recursively traverses them. Edit: Saved 31 bytes by updating my direction vector before recursing.

Abusing the instruction characters (1=^ 4=< 5=. 6=> 9=v) takes me down to 101 bytes:

```<code>f=(a,x=0,y=0,d=1,e=0,c=a[y]&&a[y][x])=>+c?(c-5&&(a[y][x]='0',d=~-c%4,e=~-(c>>2)),f(a,x+d,y+e,d,e)):!c
</code>```

Title: Solve the Halting Problem for Befinge

Body:
JavaScript (ES6), <s>158</s> 127 bytes

f=(a,x=0,y=0,d=1,e=0,c=a[y]&&a[y][x])=>c<'~'?(c>'.'&&(a[y][x]='~',d=(c=='>')-(c=='<'),e=(c=='v')-(c=='^')),f(a,x+d,y+e,d,e)):!c

Takes input as a two-dimensional character array and returns true for halting and false for an infinite loop. Works by setting visited direction characters to ~s as it recursively traverses them. Edit: Saved 31 bytes by updating my direction vector before recursing.

Abusing the instruction characters (1=^ 4=< 5=. 6=> 9=v) takes me down to 101 bytes:

f=(a,x=0,y=0,d=1,e=0,c=a[y]&&a[y][x])=>+c?(c-5&&(a[y][x]='0',d=~-c%4,e=~-(c>>2)),f(a,x+d,y+e,d,e)):!c


答:
Java 8, 128

This code directly calculates each character and has no imports or dependencies.

Golfed:

()->{String s="";for(int x,y=0;y<10;++y){for(x=0;x<70;++x)s+=((y%2<1&(x+3)%4<1)|(y%2>0&(x+1)%4<1))?"|":"_";s+="\n";}return s;}
```

<p>Ungolfed:</p>

<pre class="lang-java prettyprint-override"><code>() -> {
  String s = "";
  for (int x, y = 0; y < 10; ++y) {
    for (x = 0; x < 70; ++x)
      s += ((y % 2 < 1 & (x + 3) % 4 < 1) | (y % 2 > 0 & (x + 1) % 4 < 1)) ? "|" : "_";
    s += "\n";
  }
  return s;
}
</code>```

Title: Build me a brick wall!

Body:
Challenge

I need help building a brick wall! Throw together some code for me
 using no input and produce the following output wall shown below:

```<code>_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__
_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__
_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__
_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__
_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__
</code>```

Now, this wall is exactly 10 characters high and 70 characters
 wide.

As stated above, no inputs, only code. Least amount of bytes, we have
 to efficiently build this brick wall. Obviously, this program has to
 function to get the wall built..? or it obviously won't build itself!
 Good luck!

Winner

The winner goes to Marinus using the APL language, requiring a whole 12 bytes!
https://codegolf.stackexchange.com/a/99028/61857

Thank you everyone for participating!

Body:
Java 8, 128

This code directly calculates each character and has no imports or dependencies.

Golfed:

()->{String s="";for(int x,y=0;y<10;++y){for(x=0;x<70;++x)s+=((y%2<1&(x+3)%4<1)|(y%2>0&(x+1)%4<1))?"|":"_";s+="\n";}return s;}

Ungolfed:

() -> {
  String s = "";
  for (int x, y = 0; y < 10; ++y) {
    for (x = 0; x < 70; ++x)
      s += ((y % 2 < 1 & (x + 3) % 4 < 1) | (y % 2 > 0 & (x + 1) % 4 < 1)) ? "|" : "_";
    s += "\n";
  }
  return s;
}


Title: Build me a brick wall!

Body:
Swift 3, 69 bytes

(3...702).map{print($0%4<1 ?"|":"_",terminator:($0-2)%70<1 ?"\n":"")}

Output after compiling and running:

_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__
_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__
_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__
_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__
_|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|__

Try online here (In this tool it run as script so it will produce a warning due the map result is unused.