Draw the Sierpinski Arrowhead Curve

Question

Introduction

The Sierpinski Arrowhead Curve is a curve that's limit is Sierpinski's Triangle.

It first starts like this:

 _
/ \

Then, each line is replaced with a rotated version of the first one:

  _
 / \
 \ /
_/ \_

Next:

     _
    / \
    \ /
   _/ \_
  /     \
  \_   _/
 _  \ /  _
/ \_/ \_/ \

Your task

Given a number n, output the n-th iteration of the Sierpinski Arrowhead Curve.

You may choose to 0- or 1-index, but please specify in your answer.

You may generate an image, or use Ascii Art in the format I have above.

You may not use built-ins to generate this curve.

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

Answer 1

Octave, 240 236 221 bytes

This is made with the same idea used here but I had to change it to fit the sierpinsky arrowhead curve.

m=input(0);g=2*pi/6;u=cos(g);v=sin(g);A=[1,0];B=[u,v];C=[-u,v];D=-A;E=-B;F=-C;for k=1:m;f=[E;F;A];b=[A;B;C];A=[B;A;F];d=[C;D;E];C=[D;C;B];E=[F;E;D];B=b;D=d;F=f;end;A=[0,0;cumsum(A)];plot(A(:,1),A(:,2));axis off;axis equal

Answer 2

Haskell + diagrams, 176 bytes

import Diagrams.Prelude
import Diagrams.Backend.SVG
g n=renderSVG"a"(mkWidth 99).strokeT.a n
a 0=hrule 1
a n|b<-a(n-1)=b%6<>b<>b%(-6);a%n=rotateBy(1/n).reflectY$a::Trail V2 Double

Makes a svg file with transparent background called "a".

g 0 outputs a horizontal line, g 1 is /¯\.

Answer 3

MSWLogo (Version 6.5b), 102 bytes

Takes the two functions shapeL, shapeR given here and merges them by adding an extra argument :a, which calls the opposite function when negated.

to s :n :a :l
if :n=0[fd :l stop]
rt :a
s :n-1(-:a):l
lt :a
s :n-1 :a :l
lt :a
s :n-1(-:a):l
rt :a
end

A function s is defined, which takes number of iterations :n (1-based), angle :a, length :l. It is recursive, calling a lower iteration of itself with the angle :a negated in two instances to get the orientation correct.

• rt :a, lt :a rotate the turtle (triangle thingy whose path is traced) right, left by :a degrees.
• fd :l moves the turtle forward by :l steps.

The function is to be called with :a equal to 60.

Here, repeat is essentially a FOR loop, with built-in counter repcount. pu and pd mean "pen up" and "pen down", which stop the turtle from drawing while its position is being set using setxy.

The drawings of each iteration have been called with length :l equal to power 2 (7-repcount), which decreases exponentially; this is because the definition uses the same :l in the recursive step, so with fixed :l the overall size of the output will increase exponentially with :n.

Answer 4

Python 2, 124 bytes

Based off of the code in the Wikipedia article.

from turtle import*
def c(o,a):
 if o:o-=1;c(o,-a);lt(a);c(o,a);lt(a);c(o,-a)
 else:fd(9)
n=input()
if n%2==0:lt(60)
c(n,60)

Order 0 is a straight line.

Answer 5

Mathematica / Wolfram Language 73 bytes

s=1;Region@Line@AnglePath[Nest[Join@@({#,s=-s,s}&/@#)&,{#~Mod~2},#]Pi/3]&

Simple explanation：
AnglePath[{θ1,θ2,θ3,…}] gives the list of 2D coordinates corresponding to a path that starts at {0,0}, then takes a series of steps of unit length at successive relative angles θi.

n=1

Graphics@Line@AnglePath[60°{1,-1,-1}]

n=2

Graphics@Line@AnglePath[60°{0,1,1, -1,-1,-1, -1,1,1}]

n=3

Graphics@Line@AnglePath[60°{1,-1,-1, 1,1,1, 1,-1,-1, -1,1,1, -1,-1,-1, -1,1,1, -1,-1,-1, 1,1,1, 1,-1,-1}]

Answer 6

Mathematica, 62 bytes

Graphics@Line@AnglePath[Pi/3Nest[Flatten@{-#,1,#,1,-#}&,0,#]]&

Answer 7

JavaScript (ES6), 180 bytes

f=(n,d=0,r=n=>` `.repeat(n))=>n?f(--n,d=3-d%3).map(s=>r([l=s.length/2,0,1,~n&1][d]+l)+s+r([,1,0,~n&1][d]+l)).concat(f(n,d+1).map(s=>s+r(!(d%3))+a.shift(),a=f(n,d+2))):[`_/\\`[d%3]]

<input type=number min=1 oninput=o.textContent=f(this.value).join`\n`><pre id=o>

Returns an array of strings. Getting the spacing right was the hardest part! Pure string version for 205 bytes:

f=(n,d=0,r=n=>` `.repeat(n))=>n?f(--n,d=3-d%3).replace(/.+/g,s=>r([l=s.length/2,0,1,~n&1][d]+l)+s+r([,1,0,~n&1][d]+l))+`\n`+f(n,d+1).replace(/.+/g,s=>s+r(!(d%3))+a.shift(),a=f(n,d+2).split`\n`):`_/\\`[d%3]