Make a zigzag pattern

Question

Your task is to take in one integer input and print a zigzag pattern using slashes and backslashes.

The integer input determines the length of each zig and zag, as well as the number of zigs and zags
The pattern always starts from right to left

Test Cases

4->
   /
  /
 /
/
\
 \
  \
   \
   /
  /
 /
/
\
 \
  \
   \
2->
 /
/
\
 \
0->
1->
/
8->
       /
      /
     /
    /
   /
  /
 /
/
\
 \
  \
   \
    \
     \
      \
       \
       /
      /
     /
    /
   /
  /
 /
/
\
 \
  \
   \
    \
     \
      \
       \
       /
      /
     /
    /
   /
  /
 /
/
\
 \
  \
   \
    \
     \
      \
       \
       /
      /
     /
    /
   /
  /
 /
/
\
 \
  \
   \
    \
     \
      \
       \

Can we output an array/list of strings for each line? Are training or leading newlines or spaces allowed? — Shaggy, Sep 21, 2017 at 23:50
Is leading whitespace okay as long as the pattern is unaffected? — Emigna, Sep 22, 2017 at 8:10

Jonathan Frech · Accepted Answer · 2017-09-22 13:38:55Z

10

C (gcc), 108 102 101 98 80 76 72 bytes

Saved six bytes thanks to Kevin Cruijssen; removing parentheses and golfing N-n-1 to N+~n
Saved a byte by moving Z's incrementation into the loop condition
Saved three bytes by using printf("%c\n",...) instead of putchar(...) and ,puts("")
Saved eighteen (!) bytes thanks to HatsuPointerKun; using printf("%*s",n,""); to print n spaces instead of using a loop j;for(j=n;j--;)putchar(32); and combining both printf(...); calls
Saved four bytes by using printf("%*c",-~n,...); instead of printf("%*s%c",n,"",...);
Saved four bytes thanks to nwellnhof; moving everything inside one loop instead of two

j;f(k){for(j=0;j<k*k;j++)printf("%*c\n",j/k%2?j%k+1:k-j%k,j/k%2?92:47);}

Try it online!

edited Sep 22, 2017 at 13:38

answered Sep 22, 2017 at 8:07

Jonathan Frech

7,3111 gold badge20 silver badges43 bronze badges

\$\begingroup\$ Z,n,j;f(N){for(Z=0;Z<N;Z++)for(n=N;n--;putchar(Z%2?92:47),puts(""))for(j=Z%2?N+~n:n;j--;)putchar(32);} 102 bytes. Removed the curly brackets by putting everything inside the loops; and changed N-n-1 to N+~n. \$\endgroup\$
– Kevin Cruijssen
Sep 22, 2017 at 8:27
1

\$\begingroup\$ @KevinCruijssen Thanks. Saved another byte by swapping both Z%2?...:... and replacing Z<N;Z++ with Z++<N;. \$\endgroup\$
– Jonathan Frech
Sep 22, 2017 at 8:37
1

\$\begingroup\$ You can save several bytes by using the printf magic like i did in my answer. That way you will get rid of the for loop used to print spaces. For more details, see this stack overflow answer about left padding spaces with printf \$\endgroup\$
– HatsuPointerKun
Sep 22, 2017 at 9:46
\$\begingroup\$ @HatsuPointerKun Thanks; that is a really short way to repeat spaces in C. \$\endgroup\$
– Jonathan Frech
Sep 22, 2017 at 10:01
\$\begingroup\$ 4 bytes shorter: i;f(N){for(i=0;i<N*N;i++)printf("%*c\n",i/N%2?i%N+1:N-i%N,i/N%2?92:47);}. Try it online! \$\endgroup\$
– nwellnhof
Sep 22, 2017 at 13:25

| Show 1 more comment

Neil · Accepted Answer · 2017-10-04 12:49:11Z

10

Charcoal, 16 10 9 bytes

ＦＮ«↖Ｉθ→‖Ｔ

Try it online! Link is to verbose version of code.

edited Oct 4, 2017 at 12:49

answered Sep 21, 2017 at 20:00

Neil

156k12 gold badges68 silver badges263 bronze badges

\$\begingroup\$ this works (Was InputNumber broken in golfy mode too?) \$\endgroup\$
– ASCII-only
Oct 4, 2017 at 11:29
\$\begingroup\$ @ASCII-only No, thus the separate links to the given succinct version and the approximate verbose version. \$\endgroup\$
– Neil
Oct 4, 2017 at 12:05
\$\begingroup\$ Oh >_> didn't look to see which link I was opening \$\endgroup\$
– ASCII-only
Oct 4, 2017 at 12:06
\$\begingroup\$ @ASCII-only Well there's only one link now ;-) \$\endgroup\$
– Neil
Oct 4, 2017 at 12:49

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

MATL, 17 bytes

:"GXy@o?P47}92]*c

Try it online!

Explanation

:         % Implicit input, n. Push range [1 2 ... n]
"         % For each k in that range
  G       %   Push n again
  Xy      %   Identity matrix of that size
  @       %   Push k
  o?      %   If it's odd
    P     %     Flip the matrix upside down
    47    %     Push 47 (ASCII for '/')
  }       %   Else
    92    %     Push 92 (ASCII for '\')
  ]       %   End
  *       %   Multiply each entry of the matrix by that number
  c       %   Convert to char. Char 0 is shown as space
          % Implicit end. Implicit display

the default. · Accepted Answer · 2017-09-22 08:44:15Z

4

C# (.NET Core), 117 103 101 bytes

a=>{for(int z=a+1,e=0;e<a*a;)System.Console.WriteLine(e++/a%2<1?"/".PadLeft(--z):@"\".PadLeft(z++));}

Try it online!

edited Sep 22, 2017 at 8:44

answered Sep 22, 2017 at 3:08

the default.

8,3951 gold badge34 silver badges61 bronze badges

\$\begingroup\$ You can save 14 bytes like this: a=>{var o="";for(int z=a+1,e=0;e<a*a;)o+=(e++/a%2<1?"/".PadLeft(--z):@"\".PadLeft(z++))+"\n";return o;} 103 bytes You don't need all those parenthesis; you can combine the int; and only add +"\n" once. \$\endgroup\$
– Kevin Cruijssen
Sep 22, 2017 at 6:56
\$\begingroup\$ Hmm, you can save 2 more bytes by printing directly, instead of returning a string: a=>{for(int z=a+1,e=0;e<a*a;)System.Console.WriteLine(e++/a%2<1?"/".PadLeft(--z):@"\".PadLeft(z++));} 101 bytes \$\endgroup\$
– Kevin Cruijssen
Sep 22, 2017 at 7:37

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

3

SOGL V0.12, 13 12 9 bytes

╝F{±↔}P}ø

Try it Here!

could be 8 bytes ╝F{±↔}P} if the 0 test-case wasn't required

Explanation:

       }   implicitly started loop repeated input times
╝            create a down-right diagonal of the input
 F           get the current looping index, 1-indexed
  {  }       that many times
   ±↔          reverse the diagonal horizontally
      P      print that
        ø  push an empty string - something to implicitly print if the loop wasn't executed

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 21, 2017 at 19:58

dzaima

20.2k2 gold badges39 silver badges75 bronze badges

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Rod · Accepted Answer · 2017-09-22 11:15:46Z

3

Python 2, 69 68 62 bytes

^{-1 byte thanks to Jonathan Frech}

lambda n:[[~i,i][i/n%2]%n*' '+'/\\'[i/n%2]for i in range(n*n)]

Try it online!

edited Sep 22, 2017 at 11:15

answered Sep 21, 2017 at 20:00

Rod

18.3k4 gold badges29 silver badges89 bronze badges

\$\begingroup\$ 68 bytes. \$\endgroup\$
– Jonathan Frech
Sep 21, 2017 at 20:10

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Keyu Gan · Accepted Answer · 2017-09-22 21:52:58Z

3

Mathematica, 84 90 bytes

(n=#;Grid@Array[If[Abs[n-(s=Mod[#-1,2n])-.5]==#2-.5,If[s<n,"‌/","\\"],""]&,{n^2,n‌}])&

Thank Jenny_mathy for -6 bytes.

I have no idea why \ is obviously darker than /.

edited Sep 22, 2017 at 21:52

answered Sep 21, 2017 at 22:55

Keyu Gan

2,1339 silver badges25 bronze badges

2

\$\begingroup\$ 84 bytes (n=#;Grid@Array[If[Abs[n-(s=Mod[#-1,2n])-.5]==#2-.5,If[s<n,"/","\\"],""]&,{n^2,n}])& \$\endgroup\$
– ZaMoC
Sep 22, 2017 at 10:52

Add a comment |

jq170727 · Accepted Answer · 2017-09-24 00:47:37Z

Jq 1.5, 94 89 bytes

["/","\\"][range($n)%2]as$s|range($n)|[(range(if$s=="/"then$n-.-1 else. end)|" "),$s]|add

Explanation

  ["/","\\"][range($n)%2] as $s                         # for $s= / \ / \ $n times 
| range($n)                                             # for .=0 to $n-1
| [(range(if $s=="/" then $n-.-1 else . end)|" "), $s]  # form list of spaces ending with $s
| add                                                   # concatenate

Sample Run

$ jq -Mnr --argjson n 5 '["/","\\"][range($n)%2]as$s|range($n)|[(range(if$s=="/"then$n-.-1 else. end)|" "),$s]|add'
    /
   /
  /
 /
/
\
 \
  \
   \
    \
    /
   /
  /
 /
/
\
 \
  \
   \
    \
    /
   /
  /
 /
/

Try it online!

Community · Accepted Answer · 2020-06-17 09:04:33Z

Java 8, 140 134 116 bytes

n->{String r="";for(int a=0,b,c;a++<n;)for(b=n;b-->0;r+=a%2>0?"/\n":"\\\n")for(c=b-n+b|-a%2;++c<b;r+=" ");return r;}

-24 bytes thanks to @Nevay.

Explanation:

Try it here.

n->{                // Method with integer parameter and String return-type
  String r="";      //  Result-String
  for(int a=0,b,c;  //  Index integers
      a++<n;)       //  Loop (1) from 0 to the input (exclusive)
    for(b=n;        //   Reset `b` to the input
        b-->0;      //   Inner loop (2) from the input to 0 (exclusive)
                    //     After every iteration: 
        r+=a%2>0?"/\n":"\\\n") 
                    //      Append either of the slashes + a new-line
      for(c=b-n+b|-a%2;++c<b;r+=" ");
                    //    Append the correct amount of spaces
                    //   End of inner loop (2) (implicit / single-line body)
                    //  End of loop (1) (implicit / single-line body)
  return r;         //  Return the result-String
}                   // End of method

The condition of the innermost loop can be written as c-->f*(b-n-~b) (-6 bytes). — Nevay, Sep 22, 2017 at 14:31
116 bytes: n->{String r="";for(int a=0,b,c;a++<n;)for(b=n;b-->0;r+=a%2>0?"/\n":"\\\n")for(c=b-n+b|-a%2;++c<b;r+=" ");return r;} — Nevay, Sep 24, 2017 at 14:09

Community · Accepted Answer · 2020-06-17 09:04:33Z

3

Javascript ES8, 83 79 78 76 75 74 71 bytes

*reduced 1 byte with ES8 thanks to Shaggy

A=(m,i=0)=>i<m*m?`/\\`[x=i/m&1].padStart(x?i%m+1:m-i%m)+`
`+A(m,++i):""

Test here

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 22, 2017 at 11:29

DanielIndie

1,2305 silver badges5 bronze badges

\$\begingroup\$ whoever downvoted my solution, could you explain why? Am i missing something? \$\endgroup\$
– DanielIndie
Sep 22, 2017 at 12:19
2

\$\begingroup\$ I'm not the one who downvoted, but I assume that it is because functions need to be repeatable to be valid. Yours could be fixed pretty easily by making i a default parameter. The byte count seem off as well. \$\endgroup\$
– Emigna
Sep 22, 2017 at 12:21
1

\$\begingroup\$ Adding a TIO link is always appreciated as well, so that people can easily test your solution. \$\endgroup\$
– Emigna
Sep 22, 2017 at 12:23
1

\$\begingroup\$ @Emigna fixed it (char wise and Link wise) :) \$\endgroup\$
– DanielIndie
Sep 22, 2017 at 13:19
1

\$\begingroup\$ 74 bytes with some ES8. Also, for JS you can just use a Stack Snippet rather than TIO. \$\endgroup\$
– Shaggy
Sep 22, 2017 at 15:20

| Show 1 more comment

Leaky Nun · Accepted Answer · 2017-09-21 19:57:26Z

2

Pyth, 20 bytes

Lm+*;dbQjs<*,_y\/y\\

Try it online!

answered Sep 21, 2017 at 19:57

Leaky Nun

49k5 gold badges104 silver badges280 bronze badges

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AdmBorkBork · Accepted Answer · 2017-09-21 20:21:50Z

2

PowerShell, 81 bytes

param($a)if($a){1..$a|%{((1..$a|%{" "*--$_+'\'}),($a..1|%{" "*--$_+'/'}))[$_%2]}}

Try it online!

Ugh, this is ugly. So much repeated code, plus 7 bytes required to account for 0 special case. Golfing suggestions welcome.

answered Sep 21, 2017 at 20:21

AdmBorkBork

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Jakube · Accepted Answer · 2017-09-21 23:15:21Z

Pyth, 17 bytes

js<*_+RV"\/"_B*L;

Try it online: Demonstration

Explanation:

js<*_+RV"\/"_B*L;QQQ   implicit Qs at the end
              *L;Q     list with ["", " ", "  ", ..., " "*(input-1)]
            _B         bifurcate with reverse: [["" to "   "], ["   " to ""]]
     +RV"\/"           append to each one either "\" or "/": 
                       [["\", to "   \"], ["   /" to "/"]]
    _                  reverse
   *              Q    repeat input times
  <                Q   but only take the first input many
 s                     flatten the list of lists
j                      print on each line

Bassintag · Accepted Answer · 2017-09-22 09:21:53Z

2

Python 3: 90 Bytes 82 Bytes

lambda n:"\n".join(" "*(abs(i%(n*2)-n+i//n%2)-1)+"/\\"[i//n%2]for i in range(n*n))

Thanks to @Jonathan Frech for pointing out that print wasn't needed and that the first zig was the wrong way

edited Sep 22, 2017 at 9:21

answered Sep 22, 2017 at 8:50

Bassintag

312 bronze badges

\$\begingroup\$ ] for -> ]for. \$\endgroup\$
– Jonathan Frech
Sep 22, 2017 at 9:04
\$\begingroup\$ You do not need the print(...), a function returning a string would be valid. Also, I think your initial zig is oriented the wrong way (\ rather than /). \$\endgroup\$
– Jonathan Frech
Sep 22, 2017 at 9:08
\$\begingroup\$ @JonathanFrech Thanks! I changed it \$\endgroup\$
– Bassintag
Sep 22, 2017 at 9:23
1

\$\begingroup\$ (abs(...)-1) -> ~-abs(...). \$\endgroup\$
– Jonathan Frech
Sep 22, 2017 at 9:26

Add a comment |

Emigna · Accepted Answer · 2017-09-22 10:59:41Z

2

05AB1E, 17 16 bytes

F<„/\Nèú.sNƒR}»,

Try it online!

Explanation

F                  # for N in [0 ... input-1] do
  „/\              # push the string "/\"
     Nè            # cyclically index into this string with N
 <     ú           # prepend input-1 spaces to this string
        .s         # get suffixes
          NƒR}     # reverse the list of suffixes input+1 times
              »,   # join on newline and print

Current best attempt using canvas:

F„/\Nè©53NèΛ2®ð«4Λ

edited Sep 22, 2017 at 10:59

answered Sep 22, 2017 at 8:43

Emigna

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HatsuPointerKun · Accepted Answer · 2017-09-22 11:15:05Z

2

C++, 92 91 bytes

-1 bytes thanks to Kevin Cruijssen

void m(int n){for(int i=0,j;i<n;++i)for(j=0;j<n;++j)printf("%*c\n",i%2?j+1:n-j,i%2?92:47);}

Thanks to the power of the magic printf

edited Sep 22, 2017 at 11:15

answered Sep 22, 2017 at 9:37

HatsuPointerKun

2,1811 gold badge6 silver badges13 bronze badges

\$\begingroup\$ You can put the int i=0,j in the for-loop for(int i=0,j;i<n;++i) to save a byte. \$\endgroup\$
– Kevin Cruijssen
Sep 22, 2017 at 11:10
\$\begingroup\$ 88 bytes \$\endgroup\$
– ceilingcat
Jan 10, 2019 at 21:33

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Roberto Graham · Accepted Answer · 2017-09-22 15:29:47Z

2

Java (OpenJDK 8), 131 106 98 96 94 91 bytes

i->{for(int j=0;j<i*i;System.out.printf("%"+(j/i%2<1?i-j%i:j%i+1)+"c\n",47+45*(j++/i%2)));}

Try it online!

edited Sep 22, 2017 at 15:29

answered Sep 22, 2017 at 13:50

Roberto Graham

1,3356 silver badges10 bronze badges

1

\$\begingroup\$ You can remove quite a few parentheses: i->{for(int j=0;j<i*i;System.out.printf("%"+(j/i%2<1?i-j%i+1:j%i+2)+"s",j++/i%2<1?"/\n":"\\\n"));} (98 bytes). \$\endgroup\$
– Nevay
Sep 22, 2017 at 14:39

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dzaima · Accepted Answer · 2017-09-26 19:04:27Z

2

Dyalog APL, 39 36 35 34 bytes

{↑((,⍵ ⍵⍴(⌽,⊢)⍳⍵)/¨' '),¨⍵/⍵⍴'/\'}

Try it online!

1 byte saved thanks to Zacharý

edited Sep 26, 2017 at 19:04

answered Sep 24, 2017 at 13:35

dzaima

20.2k2 gold badges39 silver badges75 bronze badges

\$\begingroup\$ Dang it, beat me by one byte. You can make ⎕IO be 0, and then remove ¯1+. \$\endgroup\$
– Adalynn
Sep 24, 2017 at 13:41
\$\begingroup\$ @Zacharý I was just about to do that :p \$\endgroup\$
– dzaima
Sep 24, 2017 at 13:42
\$\begingroup\$ Oh, one more thing: (⌽,⊢)⍳⍵ instead of (⌽⍳⍵),⍳⍵ \$\endgroup\$
– Adalynn
Sep 24, 2017 at 13:42
\$\begingroup\$ @Zacharý Yeah, I'm yet to understand the tacks, tacitness and stuff that comes with that :/ \$\endgroup\$
– dzaima
Sep 24, 2017 at 13:46
\$\begingroup\$ Don't worry, I don't fully understand how trains/forks/whatever-they're-called work either. \$\endgroup\$
– Adalynn
Sep 24, 2017 at 13:48

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Xcali · Accepted Answer · 2017-09-22 02:49:59Z

1

Perl 5, 70 + 1 (`-n`) = 71 bytes

$n=$_**2;$_=$"x--$_.'/';say&&s% /%/ %||s%\\ % \\%||y%\\/%/\\%while$n--

Try it online!

answered Sep 22, 2017 at 2:49

Xcali

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jrtapsell · Accepted Answer · 2017-09-24 21:18:59Z

1

Kotlin, 102 bytes

(0..q-1).map{r->if(r%2<1)q-1 downTo 0 else{0..q-1}.map{(1..it).map{print(' ')};println('/'+45*(r%2))}}

Try it online!

answered Sep 24, 2017 at 21:18

jrtapsell

9736 silver badges15 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Excel VBA, ₈₄ 83 Bytes

Anonymous VBE immediate window function that takes input from range [A1] and outputs to the VBE immediate window

For i=1To[A1]:For j=1To[A1]:?IIf(i mod 2,Space([A1]-j)&"/",Space(j-1)&"\"):Next j,i

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 24, 2017 at 20:41

Taylor Alex Raine

8,2002 gold badges24 silver badges48 bronze badges

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Erik the Outgolfer · Accepted Answer · 2017-09-22 15:10:19Z

0

Jelly, 15 bytes

ḶṚ⁶ẋm0ż⁾/\x$ṁ²Y

Try it online!

Full program.

answered Sep 22, 2017 at 15:10

Erik the Outgolfer

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jferard · Accepted Answer · 2017-09-23 07:31:39Z

0

Haskell, 86 85 bytes

f n=take(n*n)$cycle$[(' '<$[x..n-1])++"/"|x<-[1..n]]++[(' '<$[2..x])++"\\"|x<-[1..n]]

Try it online!

Saved one byte thanks to Laikoni

Repeat a zig ++ a zag, and take the first n*n lines.

edited Sep 23, 2017 at 7:31

answered Sep 22, 2017 at 15:17

jferard

1,7945 silver badges7 bronze badges

\$\begingroup\$ cycle$ ... instead of cycle( ... ) saves a byte. \$\endgroup\$
– Laikoni
Sep 22, 2017 at 19:15

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FrownyFrog · Accepted Answer · 2017-09-23 19:19:50Z

0

J, 39 35 33 32 25 bytes

' /\'{~[,/@$(|.,:+:)@=@i.

Try it online!

edited Sep 23, 2017 at 19:19

answered Sep 23, 2017 at 13:38

FrownyFrog

3,6621 gold badge12 silver badges20 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Dyalog APL, 41 40 bytes

⎕IO must be 0.

{⍪/((⌽⍵ ⍵⍴S↑'/')(⍵ ⍵⍴'\'↑⍨S←⍵+1))[2|⍳⍵]}

Try it online!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 24, 2017 at 13:21

Adalynn

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Adalynn · Accepted Answer · 2018-12-12 18:35:37Z

0

D, 105 bytes

import std.stdio;void m(T)(T n){for(T i,j;i<n;++i)for(j=0;j<n;++j)printf("%*c\n",i%2?j+1:n-j,i%2?92:47);}

Try it online!

Lifted from HatsuPointerKun's C++ answer.

answered Dec 12, 2018 at 18:35

Adalynn

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Stack Exchange Network

Make a zigzag pattern

Test Cases

26 Answers 26

C (gcc), 108 102 101 98 80 76 72 bytes

Charcoal, 16 10 9 bytes

MATL, 17 bytes

Explanation

C# (.NET Core), 117 103 101 bytes

SOGL V0.12, 13 12 9 bytes

Python 2, 69 68 62 bytes

Mathematica, 84 90 bytes

Jq 1.5, 94 89 bytes

Java 8, 140 134 116 bytes

Javascript ES8, 83 79 78 76 75 74 71 bytes

Pyth, 20 bytes

PowerShell, 81 bytes

Pyth, 17 bytes

Explanation:

Python 3: 90 Bytes 82 Bytes

05AB1E, 17 16 bytes

C++, 92 91 bytes

Java (OpenJDK 8), 131 106 98 96 94 91 bytes

Dyalog APL, 39 36 35 34 bytes

Perl 5, 70 + 1 (`-n`) = 71 bytes

Kotlin, 102 bytes

Excel VBA, ₈₄ 83 Bytes

Jelly, 15 bytes

Haskell, 86 85 bytes

J, 39 35 33 32 25 bytes

Dyalog APL, 41 40 bytes

D, 105 bytes

Your Answer

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code-golf
ascii-art
or ask your own question.

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Hot Network Questions

Make a zigzag pattern

Test Cases

26 Answers 26

C (gcc), 108 102 101 98 80 76 72 bytes

Charcoal, 16 10 9 bytes

MATL, 17 bytes

Explanation

C# (.NET Core), 117 103 101 bytes

SOGL V0.12, 13 12 9 bytes

Python 2, 69 68 62 bytes

Mathematica, 84 90 bytes

Jq 1.5, 94 89 bytes

Java 8, 140 134 116 bytes

Javascript ES8, 83 79 78 76 75 74 71 bytes

Pyth, 20 bytes

PowerShell, 81 bytes

Pyth, 17 bytes

Explanation:

Python 3: 90 Bytes 82 Bytes

05AB1E, 17 16 bytes

C++, 92 91 bytes

Java (OpenJDK 8), 131 106 98 96 94 91 bytes

Dyalog APL, 39 36 35 34 bytes

Perl 5, 70 + 1 (-n) = 71 bytes

Kotlin, 102 bytes

Excel VBA, 84 83 Bytes

Jelly, 15 bytes

Haskell, 86 85 bytes

J, 39 35 33 32 25 bytes

Dyalog APL, 41 40 bytes

D, 105 bytes

Your Answer

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Hot Network Questions

Perl 5, 70 + 1 (`-n`) = 71 bytes

Excel VBA, ₈₄ 83 Bytes

Not the answer you're looking for? Browse other questions tagged
code-golf
ascii-art
or ask your own question.