The Squiggly Sequence

Question

All credits to Adnan for coming up with this challenge.

My last challenge, before I go on break.

Task

Given positive integer n, if n is odd, repeat / that many times; if n is even, repeat \ that many times.

(Seriously, the testcases would be much clearer than this description, so just look at the testcases.)

Specs

• Any reasonable input/output format.
• Standard loopholes apply.

Testcases

n output
1 /
2 \\
3 ///
4 \\\\
5 /////
6 \\\\\\

Answer 1

Actually, 10 bytes

Golfing suggestions welcome. Try it online!

;'\'/2(%I*

How it works

       Implicit input of n
;      Duplicate n
'\'/   The strings "/" and "\"
2(%    n mod 2
I      If n mod 2, then "/", else "\"
*      Multiply by n. Implicit print at the end.

Answer 2

Python 3, 43 Bytes

n=int(input())
a='\\'*n
if n%2:a='/'*n
print(a)

In python the \ symbol is also used as an operator when inside a string (e.g: \n means newline) so \\ has to be used to get the backslash symbol.

This code works by assuming the input (n) is odd, and then checks if it is even. If so, then it redefines the variable a (the output) before displaying it.

This is shorter than using an if-else statement by 5 bytes as it does the checking after defining the variable, meaning that the else command is not needed.

Answer 3

><>, 15 + 3 = 18 bytes

\/:l2-),:2%0go:

+3 bytes for the -v flag to initialise the stack with the input. Try it online!

><> is a toroidal 2D language, so the following runs in a loop:

\/          Mirrors that reflect the instruction pointer. Since the code is
            one row high, the reflection wraps and the IP ends up moving in the
            same direction anyway, so these are no-ops.

:l2-)       Push (input > length of stack - 1) - note that the actual subtraction
            is by 2 to account for the fact that we duplicated an element with :

,           Divide the input by the above. If (input == length of stack - 1),
            then the program errors out due to division by 0. Otherwise, the
            input is unchanged.

:2%0g       Push the char at (input % 2, 0), which will be one of the two slashes
            at the beginning of the code

o           Output this char

:           Duplicate the input, increasing the length of the stack by 1

Answer 4

PHP, 63 52 bytes

Version 1 - 63 bytes

$i=$argv[1];$x=($i%2?"/":"\\");echo sprintf("%'".$x.$i."s",$x);

Version 2 - 52 bytes

^{Thanks to manatwork for saving 11 bytes}

$i=$argv[1];$x=$i%2?"/":"\\";printf("%'$x{$i}s",$x);

Run with php -f squiggle.php <number>

https://repl.it/DFdi/2 (remember to change the input on repl)

Answer 5

Bash + coreutils, 29 28 bytes

yes $1_45r2%*92+P|sed $1q|dc

Previous answer, easier to understand:

yes $[92-$1%2*45]P|sed $1q|dc

In the new script I just transferred the calculation of the ASCII code from bash to dc entirely.

Answer 6

Husk, 6 bytes

R¹!"/¦

Try it online!

-4 bytes from Zgarb.

¦ is an alias for \ in a string.

Husk, 10 bytes

*⁰;!⁰"/\\"

Try it online!

Explanation

*⁰;!⁰"/\\"
   !⁰"/\\" get element at index n in string "/\" (wraps around)
  ;        create singleton list from it
*⁰         repeat list n times

Answer 7

Japt, 7 bytes

çUg"\\/

Try it

Answer 8

Forth (gforth), 55 bytes

: b 1- dup for dup 2 mod if 92 else 47 then emit next ;

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called as <number> b.

Answer 9

05AB1E, 7 5 bytes

-2 bytes thanks to @Neil.

„\/è×

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     è   # 0-based index...
         # implicit input...
     è   # implicit mod 2 in...
 „\/     # "\/"
      ×  # repeat implicit input times
         # implicit output

Answer 10

Vyxal, 8 bytes

?£₂k/i¥ẋ

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Answer 11

Vyxal, 31 bits¹, 3.875 bytes

‛\/iẋ

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'\/    # string literal "\\/"
   i   # modular index
    ẋ  # repeat

Answer 12

TypeScript's Type System, No Ignore, 541 311 155 148 140 118 bytes

Based off of a previous even/odd answer

type E<N>=N extends`11${infer R}`?R extends""?"/":E<R>:"\\"
type X<A,B=A>=B extends`1${infer R}`?`${E<A>}${X<A,R>}`:""

X takes in a repeated 1 N times to represent the number N, for example X<"11111"> represents 5 passed to the X function.

TypeScript Playground && Tests

• Drastic improvements by Jacob via even/odd, input simplification, and joining together types

Answer 13

Uiua, 11 bytes

▽∶?@/@\\◿2.

Try it!

▽∶?@/@\\◿2.
          .  # duplicate
        ◿2   # modulo two
  ?@/@\\     # choose slash depending on result
▽∶           # repeat it input times

Answer 14

S.I.L.O.S, 55 bytes

readIO
c=i
c%2
c*45
c-92
c|
lbla
printChar c
i-1
if i a

Try it online!

Port of this answer in C.

Answer 15

Haskell, 32 bytes

f n=replicate n$"\\/"!!(mod n 2)

Answer 16

RETURN, 21 bytes

[$2÷%'/'\2øø4ø2␂{␋},]

Try it here.

Anonymous lambda. Usage:

5[$2÷%'/'\2øø4ø2␂{␋},]!

Explanation

[                    ]  [lambda]%
 $2÷%                   [copy input and mod 2]%
     '/'\2øø            [choose either / or \ depending on input mod 2]%
            4ø          [get input again]%
              2␂{      [copy the /\ char and the input into a new stack]%
                  ␋},   [repeat the stack by the input and output the stack]%

Answer 17

Bash, 63 bytes

q(){ seq -s0 0 $1|cut -c-$1|tr 0-9 $(($1%2))|sed 'y/10/\/\\/';}

Answer 18

Rust, 210 190 Bytes

fn main(){let x:usize=std::env::args().nth(1).unwrap().parse::<usize>().unwrap();let l:&str=match x%2{0=>"\\",1=>"/",_=>"",};println!("{}",std::iter::repeat(l).take(x).collect::<String>());}

Ungolfed:

fn main() {
    let x : usize = std::env::args().nth(1)
                                    .unwrap()
                                    .parse::<usize>()
                                    .unwrap();
    let l : &str = match x % 2 {
        0 => "\\",
        1 => "/",
        _ => "",
    };
    println!("{}", std::iter::repeat(l).take(x)
                                       .collect::<String>());
}

I don't usually golf (so I bet this isn't a great score, at least for Rust), but I figured this would be good Rust practice. This compiles into an executable run as squiggly.exe x where x is the number of times to repeat the character.

Answer 19

awk, 26 bytes

NF=$1,$1=OFS=$1%2?"/":"\\"

$1 is the input and then part of the output

NF is the number of fields

OFS is the output field separator

It prints if the expression before the comma is truthy.

Answer 20

Forth, 58 54 bytes

Leaves n%2 as junk on the stack. Output is sent to stdout. 47 is /, and 92 is \. Computes y=92-45*x, where x is n%2 and emits (outputs) the character with that codepoint.

: f dup 2 mod swap 0 DO dup 92 swap 45 * - emit LOOP ;

Try it online

Answer 21

Lua (33 44 Bytes)

Never tried golfing with Lua before.

Shortest program I could come up with:

n=io.read()print(({'\\','/'})[n%2+1]:rep(n))

Answer 22

PL-SQL (Oracle), 530 characters

DECLARE
-- Enter your input here, here example is taken as 3 ---
  n number := 3 ;
  k varchar2(20);
BEGIN 
     -- Checking the input number for odd --
     if mod(n,2)=1 then      
     for j in 1..n loop 
      k := k||'/';           
      end loop; 
      DBMS_OUTPUT.PUT_line (n||' '||k); 
     end if;

     -- Checking the input number for even --
         if mod(n,2)=0 then
         for j in 1..n loop
         k := k||'\';         
          end loop;
          DBMS_OUTPUT.PUT_line (n||' '||k); 
         end if; 
END;

Answer 23

Acc!!, 90 bytes

N
Count i while _%60-10 {
_*10+_%60*50-2880+N
}
_/60
Count j while _-j {
Write 92-_%2*45
}

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With comments

# Read a line of stdin, convert it to a decimal number, and store it in the accumulator
N
Count i while _%60-10 {
    _*10+_%60*50-2880+N
}
_/60
# Loop that many times
Count j while _-j {
    # Output \ if accumulator mod 2 is 0, or / if it is 1
    Write 92-_%2*45
}

Answer 24

MAWP, 38 bytes

%@!!2P2WA{~[67W5M;1A].}~[99W25W1MM;1A]

Made using Dion's modulus function from here.

Try it!

Answer 25

x86-16 machine code, PC DOS, 18 bytes

Binary:

00000000: a082 002c 30a8 0191 b85c 0a74 02b0 2fcd  ...,0....\.t../.
00000010: 10c3                                     ..

Listing:

A0 0082     MOV  AL, BYTE PTR[82H]  ; first char of command line into AL
2C 30       SUB  AL, '0'            ; ASCII convert input char
A8 01       TEST AL, 1              ; set ZF if even
91          XCHG AX, CX             ; CL = number of times to repeat
B8 0A5C     MOV  AX, 0A5CH          ; AL = '\', AH = 0AH (BIOS write char)
74 02       JZ   WRITE              ; if even, write with a '\'
B0 2F       MOV  AL, '/'            ; otherwise change to a '/'
        WRITE:
CD 10       INT  10H                ; write AL to console CX number of times
C3          RET                     ; return to DOS

Standalone PC DOS executable (COM) program. Input n on command line (note: this is limited to single digit values).

Answer 26

Pyth, 5 bytes

*@"\/

Try it online!

*@"\/
@("\/", Q) * Q  // After implicit input is read and re-structured to appear more readable.
@("\/", Q)      // Get element at index Q(input)
@("\/", Q) * Q  // Repeat Q times.

Answer 27

Arturo, 31 bytes

$->n->repeat(0=n%2)?->{\}->"/"n

or

$->n[s:[{\}"/"]repeat s\[n%2]n]

Try it

Answer 28

Scala, 56 bytes

Golfed version. Try it online!

def c(i:Int)=for(_<-1 to i){print(if(i%2<1)"\\"else"/")}

Ungolfed version. Try it online!

object M {
  implicit class IntExtensions(val i: Int) extends AnyVal {
    def times[T](f: => T): Unit = {
      for (_ <- 1 to i) {
        f
      }
    }
  }
  //def c(i:Int)=print(Iterator.fill(i)(if(i%2<1)"\\"else"/").mkString)
  //def c(i:Int)=Iterator.fill(i)(if(i%2<1)"\\"else"/").foreach(print)
  def c(i: Int) = i.times { print(if (i % 2 < 1) "\\" else "/") }
  def main(args: Array[String]): Unit = {
    for (i <- 0 until 10) {
      c(i)
      println()
    }
  }
}

Answer 29

Rust, 28 bytes

|n|[r"\","/"][n%2].repeat(n)

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Answer 30

R, 32 bytes

\(n)intToUtf8(rep(92-45*n%%2,n))

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