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->
       /
      /
     /
    /
   /
  /
 /
/
\
 \
  \
   \
    \
     \
      \
       \
       /
      /
     /
    /
   /
  /
 /
/
\
 \
  \
   \
    \
     \
      \
       \
       /
      /
     /
    /
   /
  /
 /
/
\
 \
  \
   \
    \
     \
      \
       \
       /
      /
     /
    /
   /
  /
 /
/
\
 \
  \
   \
    \
     \
      \
       \

Answer 1

Charcoal, 16 10 9 bytes

ＦＮ«↖Ｉθ→‖Ｔ

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

Answer 2

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);}

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Answer 3

MATL, 17 bytes

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

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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

Answer 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++));}

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Answer 5

SOGL V0.12, 13 12 9 bytes

╝F{±↔}P}ø

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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

Answer 6

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)]

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Answer 7

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 /.

Answer 8

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!

Answer 9

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

Answer 10

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

Answer 11

Pyth, 20 bytes

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

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Answer 12

PowerShell, 81 bytes

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

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

Answer 13

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

Answer 14

05AB1E, 17 16 bytes

F<„/\Nèú.sNƒR}»,

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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

Answer 15

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

Answer 16

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

Answer 17

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)));}

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Answer 18

Dyalog APL, 39 36 35 34 bytes

{↑((,⍵ ⍵⍴(⌽,⊢)⍳⍵)/¨' '),¨⍵/⍵⍴'/\'}

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1 byte saved thanks to Zacharý

Answer 19

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

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

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Answer 20

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

Answer 21

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))}}

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Answer 22

Jelly, 15 bytes

ḶṚ⁶ẋm0ż⁾/\x$ṁ²Y

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Full program.

Answer 23

Haskell, 86 85 bytes

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

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Saved one byte thanks to Laikoni

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

Answer 24

J, 39 35 33 32 25 bytes

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

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Answer 25

Dyalog APL, 41 40 bytes

⎕IO must be 0.

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

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Answer 26

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);}

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Lifted from HatsuPointerKun's C++ answer.