Graphical Representation of Koch Snowflake

Question

Generate a Koch Snowflake

A Koch snowflake is a triangle that for each n, another equilateral point is added in the middle of each side: http://en.wikipedia.org/wiki/Koch_snowflake#Properties

We already had a kolmogrov-complexity Koch Snowflake challenge for n=4. The new challenge is to draw a Koch snowflake with any n between 1 and 10.

Rules

The snowflakes may not be hardcoded in the program or in files - they must be generated by your program.

Your program must support all sized n between 1 and 10.

The number of sides must be input by the user through std-in.

You must print a graphical representation of the snowflake to the screen.

Sample Koch Snowflakes with n equaling 1, 2, 3, and 4 (green lines for clarity only, do not reproduce them):

In the event of a tie-breaker, the program with the highest number of upvotes wins (pop-contest).

Answer 1

MATLAB, 119 115

In an unusual turn of events, I found that this program actually worked better as I golfed it. First, it became much faster due to vectorization. Now, it displays a helpful prompt ~n:~ reminding the user of which quantity to enter!

Newlines are not part of the program.

x=exp(i*pi/3);
o='~n:~';
P=x.^o;
for l=2:input(o);
P=[kron(P(1:end-1),~~o)+kron(diff(P)/3,[0 1 1+1/x 2]) 1];
end;
plot(P)

n = 9:

o is an arbitrary string which is equal to [0 2 4 0] modulo 6. e^iπ/3 raised to these powers gives the vertices of an equilateral triangle in the complex plane. The first kron is used to make a copy of the list of points with each one duplicated 4 times. ~~o is the convenient way to get a vector of 4 ones. Secondly diff(P) finds the vector between each pair of consecutive points. Multiples of this vector (0, 1/3, (1 + e^-iπ/3)/3, and 2/3) are added to each of the old points.

Answer 2

T-SQL: 686 (excluding formatting)

For SQL Server 2012+.

Even though this will never be a contender, I had to see if I could get it done in T-SQL. Gone for the approach of starting with the three initial edges, then recursing through each edge and replacing them with 4 edges for each level. Finally unioning it all up into a single geometry for the level specified for @i

DECLARE @i INT=8,@ FLOAT=0,@l FLOAT=9;
WITH R AS(
    SELECT sX,sY,eX,eY,@l l,B,1i
    FROM(VALUES(@,@,@l,@,0),(@l,@,@l/2,SQRT(@l*@l-(@l/2)*(@l/2)),-120),(@l/2,SQRT(@l*@l-(@l/2)*(@l/2)),@,@,-240))a(sX,sY,eX,eY,B)
    UNION ALL
    SELECT a.sX,a.sY,a.eX,a.eY,l/3,a.B,i+1
    FROM R 
        CROSS APPLY(VALUES(sX,sY,sX+(eX-sX)/3,sY+(eY-sY)/3,sX+((eX-sX)/3)*2,sY+((eY-sY)/3)*2))x(x1,y1,x2,y2,x3,y3)
        CROSS APPLY(VALUES(x2+((l/3)*SIN(RADIANS(B-210.))),y2+((l/3)*COS(RADIANS(B-210.)))))n(x4,y4)
        CROSS APPLY(VALUES(x1,y1,x2,y2,B),
            (x3,y3,eX,eY,B),
            (x2,y2,x4,y4,B+60),
            (x4,y4,x3,y3,B-60)
        )a(sX,sY,eX,eY,B)
WHERE @i>i)
SELECT Geometry::UnionAggregate(Geometry::Parse(CONCAT('LINESTRING(',sX,' ',sY,',',eX,' ',eY,')')))
FROM R 
WHERE i=@i

Answer 3

LOGO: 95

to w:c ifelse:c=1[fd 2 lt 60][w:c-1 w:c-1 lt 180 w:c-1 w:c-1]end
to k:c repeat 3[w:c rt 180]end

Defines function k with a single level parameter.

Edit

In the this online editor http://www.calormen.com/jslogo/ you can add k readword to use prompt for input, but for some reason this command does not support the standard abbreviation rw.

The 102 characters solution below works in USBLogo with standard input as specified in the question. However the code needed slight changes as UCBLogo has some weird parser. It requires to and end to be in separate lines and space before : is required but on the other hand : are optional.

to w c
ifelse c=1[fd 2 lt 60][w c-1 w c-1 lt 180 w c-1 w c-1]
end
to k c
repeat 3[w c rt 180]
end
k rw

Answer 4

Mathematica - 177

r[x_, y_] := Sequence[x, RotationTransform[π/3, x]@y, y]
Graphics@Polygon@Nest[
 ReplaceList[#, {___, x_, y_, ___}  Sequence[x,r[2 x/3 + y/3, 2 y/3 + x/3], y]
  ] &, {{0, 0}, {.5, √3/2}, {1, 0}, {0, 0}}, Input[]]

Bonus clip of varying the angle of middle piece

Answer 5

BBC BASIC, 179

REV 1

INPUTn
z=FNt(500,470,0,0,3^n/9)END
DEFFNt(x,y,i,j,a)
LINEx-36*a,y-21*a,x+36*a,y-a*21PLOT85,x,y+a*42IFa FORj=-1TO1FORi=-1TO1z=FNt(x+24*a*i,y+j*14*a*(i*i*3-2),i,j,-j*a/3)NEXTNEXT
=0

As before, but in black and white, in ungolfed (but streamlined) and golfed versions. Not a winner, despite the fact that doing it ths way avoids the need for a special treament for n=1.

  INPUTn
        REM call function and throw return value away to z
  z=FNt(500,470,0,0,3^n/9)
  END

        REM x,y=centre of triangle. a=scale.
        REM loop variables i,j are only passed in order to make them local, not global.
  DEFFNt(x,y,i,j,a)

        REM first draw a line at the bottom of the triangle. PLOT85 then draws a filled triangle,
        REM using the specified point (top of the triangle) and the last two points visited (those used to draw the line.)
  LINEx-36*a,y-21*a,x+36*a,y-21*a
  PLOT85,x,y+42*a

        REM if the absolute magnitude of a is sufficient to be truthy, recurse to the next level.
        REM j determines if the triangle will be upright, inverted or (if j=0) infinitely small.
        REM i loops through the three triangles of each orientation, from left to right.
  IFa FORj=-1TO1 FORi=-1TO1:z=FNt(x+24*a*i,y+j*14*a*(i*i*3-2),i,j,-j*a/3):NEXT:NEXT

        REM return 0
  =0

REV 0

According to the OP's answer to @xnor, filled in snowflakes are OK. This answer was inspired by xnor's comment. The colours are just for fun and to show the way it is constructed. Take a triangle (magenta in this case) and overplot with 6 triangles 1/3 of the base.

  INPUTn
  z=FNt(500,470,n,1,0,0)
  END

  DEFFNt(x,y,n,o,i,a)
  a=3^n/22
  GCOLn
  MOVEx-36*a,y-o*21*a
  MOVEx+36*a,y-o*21*a
  PLOT85,x,y+o*42*a
  IFn>1FORi=-1TO1:z=FNt(x+24*a*i,y+14*a*(i*i*3-2),n-1,-1,i,a):z=FNt(x+24*a*i,y-14*a*(i*i*3-2),n-1,1,i,a)NEXT
  =0

Answer 6

Mathematica 72

Region@Line@AnglePath[Nest[Join@@({#,1,4,1}&/@#)&,{4,4,4},Input[]-1]π/3]

n=3

Thanks for alephalpha.

Answer 7

Python 3 - 139

Uses the turtle graphics library.

from turtle import*
b=int(input())
a=eval(("FR"*3)+".replace('F','FLFRFLF')"*~-b)
for j in a:fd(9/b*("G">j));rt(60*(2-3*("Q">j))*("K"<j))])

Answer 8

Python 3, 117 bytes

from turtle import*
ht();n=int(input())-1
for c in eval("'101'.join("*n+"'0'*4"+")"*n):rt(60+180*(c<'1'));fd(99/3**n)

Method:

• The solution uses Python's turtle graphics.
• n is input - 1
• Starting from the string 0000 we join its every character with 101 n times iteratively with the eval trick (thanks to @xnor for that).
• For every character in the final string we turn 60 degrees right or 120 degrees left based on the character value (1 or 0) and then move forward a length (99/3^n) which guarantees a similar size for all n's.
• The last 0 in the string will be useless but it just redraws the same line the first 0 draws.

Example output for input = 3:

Answer 9

R: 240 175

Because I trying to get my head around R, here's another version. There's likely to be lot better ways to do this and I'm happy to receive pointers. What I've done seems very convoluted.

n=readline();d=c(0,-120,-240);c=1;while(c<n){c=c+1;e=c();for(i in 1:length(d)){e=c(e,d[i],d[i]+60,d[i]-60,d[i])};d=e};f=pi/180;plot(cumsum(sin(d*f)),cumsum(cos(d*f)),type="l")

Answer 10

Wise fwom youw gwave...

I knew I'd want to try implementing this in Befunge-98 using TURT, but I couldn't figure out how to go about it and I sat on it for several months. Now, only recently, I figured out a way to do it without using self-modification! And so...

Befunge-98 with the TURT fingerprint, 103

;v;"TRUT"4(02-&:0\:0\1P>:3+:4`!*;
>j$;space makes me cry;^
^>1-:01-\:0\:01-\:0
0>I@
^>6a*L
^>ca*R
^>fF

Let's get some implementation details out of the way first:

• I tested this using CCBI 2.1, whose implementation of TURT (the turtle graphics fingerprint) makes I "print" the image to a SVG file. If you run this in CCBI without the command-argument --turt-line=PATH, it will come out as a file named CCBI_TURT.svg by default. This is the closest I could come to "print a graphical representation of the snowflake to the screen" with available Funge interpreters I could find. Maybe someday there will be a better interpreter out there that has a graphical display for the turtle, but for now...
• There has to be a newline at the end for CCBI to run it without hanging (this is included with the character count). I know, right?

Basically, this works by using the stack as a sort of makeshift L-system and expanding it on the fly. On each pass, if the top number on the stack is:

• -2, then it prints and stops;
• -1, the turtle turns counterclockwise 60 degrees;
• 0, it turns clockwise 120 degrees;
• 1, it moves forward (by 15 pixels here, but you can change it by modifying the f on the last line);
• some number n that's 2 or greater, then it is expanded on the stack to n-1, -1, n-1, 0, n-1, -1, n-1.

For n = 10, this process takes a very long time (a couple of minutes on my system), and the resulting SVG is ~10MB in size and invisible when viewed in the browser because you can't adjust the size of the brush using TURT. IrfanView seems to work decently if you have the right plugins. I'm not super familiar with SVG, so I don't know what the preferred method is for viewing those files (especially when they're really big).

Hey, at least it works - which, considering it's Befunge, is something to be thankful for on its own.

Answer 11

APL(Dyalog Extended), 57 bytes ^SBCS

'A=↑1 +=⌽¯60 -=⌽60'⌂turtle'A' 'A-A++A-A'⌂lsys⍣⎕⊢'A++A++A'

APL has builtins for generation of fractal L-systems and drawing them using turtle graphics.

Requires Dyalog for Windows.

Answer 12

Python 2, 127 bytes

from turtle import *
def L(l,d=input()-1):
 for x in[1,-2,1,0]:[L(l,d-1),rt(60*x)] if d>0 else fd(l)
for i in'lol':lt(120);L(9)