Draw the "Cool S"

Question

Introduction

We all know the cool S (also known as Superman S, Stüssy S, Super S, Skater S, Pointy S, Graffiti S etc. etc.): billions of schoolchildren around the world drew this S and immediately felt proud of themselves. In case you've forgotten or had a completely uncool childhood, here is an image of said cool S:

Given a scale factor n as input (where \$1\leq n\leq 20\$), output the Cool S in ASCII art.

How to Draw It

From the Wikipedia page on the Cool S:

Output

The Cool S when n = 1 is:

   ^
  / \
 /   \
/     \
|  |  |
|  |  |
\  \  /
 \  \/
 /\  \
/  \  \
|  |  |
|  |  |
\     /
 \   /
  \ /
   v

And for different values of n, you simply make the output n times bigger. For example, n=2:

     ^  
    / \
   /   \
  /     \
 /       \
/         \
|    |    |
|    |    |
|    |    |
|    |    |
\    \    /
 \    \  /
  \    \/
  /\    \
 /  \    \
/    \    \
|    |    |
|    |    |
|    |    |
|    |    |
\         /
 \       /
  \     /
   \   /
    \ /
     v

Note that the vertical sections are two times longer and the spacing between the vertical lines is two times wider.

And when n=3:

       ^
      / \
     /   \
    /     \
   /       \
  /         \
 /           \
/             \
|      |      |
|      |      |
|      |      |
|      |      |
|      |      |
|      |      |
\      \      /
 \      \    /
  \      \  /
   \      \/
   /\      \
  /  \      \
 /    \      \
/      \      \
|      |      |
|      |      |
|      |      |
|      |      |
|      |      |
|      |      |
\             /
 \           /
  \         /
   \       /
    \     /
     \   /
      \ /
       v

Note: Although not required, your code may also be able to support n=0:

 ^
/ \
\\/
/\\
\ /
 v

Winning

The shortest program in bytes wins.

Answer 1

Charcoal, 58 53 47 43 41 bytes

Ｎθ≔⊕⊗θδ↗θ/⊗θ↘δ^‖Ｂ↓‖Ｍ← vＭδ⁰⊗θ↗⊕θＭ⁰δ↗θ/⊗θ⟲Ｔ

Try it online!

I just wanted to try another approach, this draws the outside via reflections (thanks to Neil for expanding the idea) and then draws the inside part. As Charcoal has :Left as default direction to draw lines, I make use of that direction as much as possible to save some bytes by drawing the S horizontally, like this:

     /----\    /----\     
    /      \  /      \    
   /        \/        \   
  /         /          \  
 /         /            \ 
v     ----/    /----     ^
 \            /         / 
  \          /         /  
   \        /\        /   
    \      /  \      /    
     \----/    \----/     

And then I just need to rotate the canvas 90 degrees counterclockwise.

Answer 2

Charcoal, 47 42 41 bytes

Ｆv^«↓⊗θ↘⊗⊕θ←↓⊗θ↙⊕⊗θ↖ι↖⊕⊗θ→↑⊗θ↗⊕θＭθ⁺⊗θ⊕θ⟲⁴

Try it online! Link is to verbose version of code. Explanation: Draws the following lines in order:

   ^
  / \
 /   \
/     \
|  1  |
|  1  |
\  2  /
 \  2/
 8\  2
8  \  2
7  |  3
7  9  3
6     4
 6   4
  6 4
   5

Where 5 is the current character of the string v^. At the end of the first loop the cursor is then positioned at point 9. The entire canvas is then rotated so that the other half of the Cool S can be drawn. (The canvas actually gets rotated twice, but this is just an implementation detail.)

Charcoal doesn't support RotateCopy(:Up, 4) but if it did then this would work for 33 bytes:

↖^↖⊕⊗θ→↑⊗θ↗⊕θ‖ＢＭ↓↙⊗θ→↓⊗θ⟲Ｃ↑⁴Ｊ⁰¦⁰v

Answer 3

Python 3, 255 249 248 209 bytes

-6 bytes thanks to Kevin Cruijssen

-1 byte thanks to Kevin Cruijssen

-39 bytes thanks to Rod and Jo King

n=int(input())
m=2*n
a,b,q,c,l='\ \n/|'
f=m*b
s=q+q.join([f[d:]+c+b*2*d+b+a+f[d:]for d in range(m+1)]+[l+f+l+f+l]*m+[d*b+a+f+a+f[d*2:]+c+d*b for d in range(n)]+[n*b+a+f+a+c+n*b])
print(f,'^'+s+q+s[::-1]+f,'v')

Try it online!

It now handles n=0.

Answer 4

Canvas, 36 32 29 bytes

«|＊‼ｌ├／Ｌ１^╋；╶╵＼∔∔│α╶«├：╵╋：↔↕∔

Try it here!

A whole lot of stack manipulation. (outdated) explanation:

«|*                                an array of input*2 "|"s
   ‼                               cast to a 2D object (needed because bug)
    :                              duplicate that (saved for the center line)
     l├                            height+2
       /                           create a diagonal that long
        L1^╋                       and in it, at (width; 1) insert "^"
            ;∔                     append the vertical bars
                               ^
                              /
          so far done:       / 
                            /  
                            |  
                            |  
              ⁸╵                   input+1
                \                  antidiagonal with that size
                 ∔                 appended to the above
                  │                mirror horizontally
                              ^
                             / \
                            /   \
                           /     \
                current:   |     |
                           |     |
                           \     /
                            \   /                                                       |
                   α               get the 2nd to last popped thing - the antidiagonal  |
                    └∔             append it to the vertical line copied way before:    \
                      ⁸«├          input/2 + 2                                            \
                         :╵        duplicate + 1
                           ╋       at (input/2 + 2; input/2 + 3) in the big part insert  ^
                            :↔↕∔   mirror a copy vertically & horizontally and append that to the original

Answer 5

C (gcc), 260 254 bytes

-6 bytes thanks to ceilingcat.

f(n){int s=2*n++,t=s+1,I[]={1,t,s,n,n,s,t,1},A[]={s,1,1,1,2*t,1,t,t,1,t,1,n,t,t,1,t,t,1,1,1,t,s,1,1},x;for(s=8;s--;)for(n=0;n<I[s];n++,puts(""))for(t=3;t--;)x=s*3+t,printf("%*c",n*("AAAA?BAAAAC@?ABAAACA@AAA"[x]-65)+A[x],"w!!!0]}}}]]00]]}}}]!0_!!"[x]-1);}

Try it online!

Rundown

We can divide the shape into parts:

 ^           Top cap
/ \          Top slope
|||          Sides
\\/          Twist, part 1
/\\          Twist, part 2
|||          Sides
\ /          Bottom slope
 v           Bottom cap

Each part could be described by a number of lines, three chars, and three relationships to certain values that decides the field-width at each line.

A first iteration came to be:

#define g(x,s,A,B,C)for(i=0;i<x;i++)printf("%*c%*c%*c\n",A,*s,B,s[1],C,s[2]);
f(n)
{
    int s=2*n++,t=s+1,i;
    
    g(1,  "  ^",  1,      1,  t-1)
    g(t, "/ \\",t-i,      1,2*i+1)
    g(s,  "|||",  1,      t,    t)
    g(n,"\\\\/",i+1,      t,t-2*i)
    g(n,"/\\\\",n-i,  2*i+1,    t)
    g(s,  "|||",  1,      t,    t)
    g(t, "\\/ ",i+1,2*t-2*i,    1)
    g(1,  "  v",  1,      1,  t-1)
}

The calls to the g() macro looks very much like a table could be constructed and looped over. Field-widths are sometimes related to the index counter, and sometimes not. We can generalise the field-width to be F * i + A, where F is some factor to multiply i with, and A is some value to add to the width. So the last width of the fourth call above would be -2 * i + t, for example.

Thus we get:

f(n){int s=2*n++,t=s+1,         s = size of "side" parts, t = size of top and bottom slopes
I[]={1,t,s,n,n,s,t,1},          The number of lines per part.
A[]={...},x;                    A[] holds the values to add to each field-width.
for(s=8;s--;)                   Loop through the parts.
for(n=0;n<I[s];n++,puts(""))    I[s] decides how many lines to the part. Ends with newline.
for(t=3;t--;)                   Go through the three chars of each line.
x=s*3+t,                        Calculate offset.
printf("%*c",                   Print the char.
n*("..."[x]-65)+A[x],           Build field-width. The string holds the index factor, A[]
                                holds the offset part.
"..."[x]-1);}                   The char itself is grabbed from the string.
                                Shifted by 1 to eliminated double backspaces.

In the end it was not much shorter than a tightened version of the g() calling one, but shorter is shorter.

Answer 6

Python 2, 227 208 207 202 196 181 bytes

I=n=2*input()
R,L,S,P='/\ |'
k=n*[2*(P+S*n)+P]
exec"k=[R+S+2*S*I+L]+k+-~I%2*[L+S*n+L+S*I+R];I-=1;"*-~n
print'\n'.join(t.center(2*n+3)for t in['^']+k+[a[::-1]for a in k[::-1]]+['v'])

Try it online!

Thks to Jo King for 1 byte; and then another 5 bytes total (via n => 2*n).

Works for n=0 as well.

Answer 7

C (gcc), 379 353 344 334 bytes

I used a couple of #defines for subexpression elimination and several globals to communicate between the internal functions. The main loop goes {0,1,2,3,3,2,1,0} to construct the S.

Thanks to Jonathan Frech for the suggestions.

#define z(a,b...)printf("%*c%*c%*c\n"+a,b);}
#define y(a){for(i=~-a*t;v*i<v*a*!t+t;i+=v)
i,n,p,r,t,u,v;a(){z(6,r+2,94+t*24)b()y(-~r)z(3,-i-~r,47+u,i*2+2,92-u)c()y(r)z(0,~r,124,~r,124,~r,124)d()y(-~n)z(0,i+1,92-u,2*(n-t*i)+1,92,2*(n-!t*i)+1,47+u)(*x[])()={a,b,c,d};f(s){r=2*s;for(p=0;p<8;x[7*t-p++*(2*t-1)](n=s))t=p>3,v=2*!t-1,u=t*45;}

Try it online!

Answer 8

PHP, 378 374 378 377 376 335 331 328 bytes

-3 bytes, thanks to manatwork

-4 bytes, used str_pad instead of str_repeat

-41 bytes, thanks to manatworks' suggestions

-1 byte, merged two increments into a +=2

-1 byte, removed superfluous \

-4 bytes by echoing once. Forgot I needed to pass the string into the function so this is more bytes

Works for n = 0 as well.

function s($b){return str_pad($w,$b);}echo s($i=1+$a=2*$argv[1]).'^
';for(;$i;$j++,$y=$z.$y)echo$z=s(--$i).'/'.s(++$j).'\
';for(;$k<$a;$k++)$x.='|'.s($a).'|'.s($a).'|
';echo$x;for(;$l<=$a/2;)echo s($m++).$c='\\',s($a).$c.s($a-$l++*2).'/
';for(;$m;$n+=2)echo s(--$m).'/'.s($n).$c.s($a).'\
';echo$x.strtr($y,'/\\','\/').s($a+1).v;

Try it online!

Answer 9

Java, 435 bytes

The function itself takes 435 bytes. There is certainly room for improvement, "high level" by analyzing the rules about where to place which character (in the end the S is point-symmetric), and "low-level", by classical golfing (maybe pulling out another variable or combining two of the for-loops). But it's a first shot with this rather ungolfy language:

import static java.util.Arrays.*;
import static java.lang.System.*;

public class CoolS
{
    public static void main(String[] args)
    {
        print(1);
        print(2);
        print(3);
    }
    static void print(int n){int i,r,d=3+6*n,w=3+n*4,h=6+n*10,m=n+n,v=w/2,k=h-1,j=w-1;char t[],S='/',B='\\',P='|',s[][]=new char[h][w];for(char x[]:s)fill(x,' ');s[0][v]='^';s[k][v]='v';for(i=0;i<1+m;i++){r=i+1;t=s[r];t[v-r]=S;t[v+r]=B;t=s[k-r];t[v-r]=B;t[v+r]=S;}for(i=0;i<m;i++){r=2+m+i;t=s[r];t[0]=t[v]=t[j]=P;t=s[k-r];t[0]=t[v]=t[j]=P;}for(i=0;i<1+n;i++){r=2+m+m+i;t=s[r];t[i]=t[i+1+m]=B;t[j-i]=S;t=s[d-i];t[i]=S;t[v-i]=t[j-i]=B;}for(char x[]:s)out.println(x);}
}

Answer 10

><>, 1024 bytes

i'0'-i1+:?!v1-'0'-$a*+1v
v          <           <
>~     :1-2*3+ >1-' 'v       v                                             oa&0~~~<
 v&1:oa*2:~o'^'^?:o  <                                              >~'/'o$:v    <^        <
>>' 'o1-: ?v   v                  >@ '\'o$v v   -1o<           >          v >:?vv^ -1<     ^     <
:^         <   ~                  ~>      \:>:?v       v  >'/'v^o' '^!?:-1<:<  > ' 'o^ >'\'o:1-?v^
- vo' '<:&o'/' < <                ^^?:o' '\  \ >' '/'\'\~$^ > o:?v~1>$  :?v$^   ~!   oa  $+2$-1 < 
^1\1-:?^~'\'oao v                /        \-1<:@@~o/o1-    v^a<  >1+^v:$-2<vo'\'<      ^@~<v             <
 v^?:+1v!?:-1&+2/            /   /        \        \' '^!?:<         <     >$@@:>1-' 'o:?v^      >~'\'o$v^    oa<
       >         ^     /-1   /-1 /        \        \       \:0:<                ^        <   v      -1< :v o' '<^ +1$-2  <      v    -1<
 >~&~1&         >:2*:1->'|'o1>@@:>' 'o1-:?^~@'|'o:?^~ao:  ?^~&?^                 1+2*0     >:>:?!^' 'o^ >>1-: ?^~'/'o:2-?^@@2*ao>' 'o:?^'v'o ;

Takes one number 1-99 as input and draws the Cool S.
The program uses only one stack to store the number of spaces.
It is almost not golfed, coded just for practice and fun :D
Try it online!

Answer 11

Python 3, 321 307 bytes

Thanks to @EsolangingFruit for saving 14 bytes

n=int(input())
b,f='\/'
c,l=4*n+3,10*n+6
r=range
h=c//2
L=[c*[' ']for _ in r(l)]
L[0][h],L[-1][h]='^v'
for i in r(h):a=L[h-i];a[i],a[c+~i]=f,b
for i in r(2*n):L[h-~i][0::h]='|'*3
for i in r(n+1):a=L[h+h+i];a[c+~i],a[i:c-1:h]=f,b*2
for i in r(1,l//2):L[l+~i]=L[i][::-1]
print('\n'.join(''.join(i)for i in L))

Try it online!

Python 2, 303 bytes

n=int(input())
b,f='\/'
c,l,r=4*n+3,10*n+6,range
h=c/2
L=[c*[' ']for _ in r(l)]
L[0][h],L[-1][h]='^v'
for i in r(h):a=L[h-i];a[i],a[c+~i]=f,b
for i in r(2*n):L[h-~i][0::h]='|'*3
for i in r(n+1):a=L[h+h+i];a[c+~i],a[i:c-1:h]=f,b*2
for i in r(1,l/2):L[l+~1]=L[i][::-1]
print'\n'.join(''.join(i)for i in L)

Try it online!

Answer 12

Regenerate, 180 bytes

( {2*$~1}) ^\n(( {#3-1}|$1)/(  $4| )\\\n){#1+1}((\|$1\|$1\|\n){#1})(( $8|)(\\$1\\)( {#10-2}|$1)/\n){$~1+1}(( {#12-1}|$8)/(  $13|)$9\n){$~1+1}$5(( $15|)\\( {#16-2}|$4)/\n){#1+1}$1 v

Takes input as a command-line argument. Try it here!

Explanation

Hoo boy, this is long. It's not actually super-complicated, though; once you get a few basic principles of golfing in Regenerate, this just uses the same techniques over and over.

( {2*$~1})

Group 1: Take the command-line argument ($~1), multiply it by 2, and add that many spaces.

 ^\n

Complete the first line with one more space, ^, and a newline.

(...){#1+1}

Group 2 generates one line of the / \ section. Repeat it (length of group 1, plus 1) times. It contains groups 3 and 4.

( {#3-1}|$1)/

Group 3: Add spaces, one fewer than the last time we matched group 3; or, if group 3 hasn't been matched before, start with the contents of group 1 (arg * 2 spaces) instead. After group 3, add a forward slash.

(  $4| )\\\n

Group 4: Add spaces, two more than the last time we matched group 4; or, if group 4 hasn't been matched before, start with one space instead. After group 4, add a backslash and a newline.

((\|$1\|$1\|\n){#1})

Group 5 is the whole vertical lines section, which we're capturing because we can reuse it verbatim later. Specifically, this is (length of group 1) repetitions of group 6, which consists of: pipe, group 1, pipe, group 1, pipe, newline.

(...){$~1+1}

Group 7 generates one line of the \ \/ section. Repeat it (argument plus 1) times. It contains groups 8, 9, and 10.

( $8|)

Group 8: Add spaces, one more than the last time we matched group 8; or, if group 8 hasn't been matched before, start with empty string instead.

(\\$1\\)

Group 9: backslash, group 1, backslash (capture for reuse later).

( {#10-2}|$1)/\n

Group 10: Add spaces, two fewer than the last time we matched group 10; or, if group 10 hasn't been matched before, start with the contents of group 1 instead. After group 10, add a forward slash and a newline.

(...){$~1+1}

Group 11 generates one line of the /\ \ section. Repeat it (argument plus 1) times. It contains groups 12 and 13.

( {#12-1}|$8)/

Group 12: Add spaces, one fewer than the last time we matched group 12; or, if group 12 hasn't been matched before, start with the most recent contents of group 8 instead. After group 12, add a forward slash.

(  $13|)$9\n

Group 13: Add spaces, two more than the last time we matched group 13; or, if group 13 hasn't been matched before, start with empty string instead. After group 13, add the contents of group 9 (two backslashes with some spaces in between) and a newline.

$5

Add the contents of group 5 again (vertical pipes section).

(...){#1+1}

Group 14 generates one line of the \ / section. Repeat it (length of group 1, plus 1) times. It contains groups 15 and 16.

( $15|)\\

Group 15: Add spaces, one more than the last time we matched group 15; or, if group 15 hasn't been matched before, start with empty string instead. After group 15, add a backslash.

( {#16-2}|$4)/\n

Group 16: Add spaces, two fewer than the last time we matched group 16; or, if group 16 hasn't been matched before, start with the most recent contents of group 4 instead. After group 16, add a forward slash and a newline.

$1 v

For the last line, add the contents of group 1, a space, and v.

Answer 13

C (gcc), 271 261 253 230 bytes

-10 bytes by moving 2* inside S macro

-8 bytes by placing u,d in the global scope and replacing (n-u) with d

-23 bytes by abusing the C preprocessor, thanks to @ceilingcat

#define S(i,c)printf("%*s"#c,2*i,"");
#define R(m)}for(u=0,d=n*m+1;d--;u++){S(0+
#define V(x,y)S(x,\\\n)R(2-1)0,|)S(n,|)S(n,|\n)R(y)u,\\)S(
u,d;f(n){{S(n+1,^\n)R(2)d,/)V(u+1,1)n,\\)S(d,/\n)R(1)d,/)S(u,\\)V(n,2)d+1,/\n)}S(n+1,v\n)}

Try it online!

Answer 14

Vyxal C, 49 bytes

\^?d›(\/n꘍øM)\|?꘍m∞?dḊ?›(\\?꘍m?›n-d꘍\/J)W:ṘRJṪ\vJ

Try it Online!

Will add explanation tomorrow.