# Draw the Pentaflake

First of all ... I would like to wish everyone a Merry Christmas (sorry if I am a day late for your timezone).

To celebrate the occasion, we are going to draw a snowflake. Because the year is 2015 and Christmas is on the 25th (for a large portion of persons), we will draw a Pentaflake. The Pentaflake is a simple fractal composed of pentagons. Here are a few examples (taken from here): Each Pentaflake has an order n. The Pentaflake of order 0 is simply a pentagon. For all other orders n, a Pentaflake is composed of 5 Pentaflakes of the previous order arranged around a 6th Pentaflake of the previous order. For example, a Pentaflake of order 1 is composed of 5 pentagons arranged around a central pentagon.

# Input

The order n. This may be given in any way except that of a predefined variable.

# Output

An image of the order n Pentaflake. Must be at least 100px wide and 100px long. It may be saved to a file, displayed to the user, or outputted to STDOUT. Any other form of output is not allowed. All image formats existing before this challenge are allowed.

# Winning

As codegolf, the person with the least number of bytes wins.

• -1 because snowflakes only have 6 fold symmetry! =D Dec 25, 2015 at 16:30
• @flawr According to this article only about .1% of snowflakes actually have 6-fold symmetry ... or any symmetry at all. However, those snowflakes that do have symmetry can have 3-fold symmetry in addition to 6-fold symmetry :P Dec 26, 2015 at 14:27
• Well this article only studied way less than .1% of all snowflakes, and it is meaningless anyway, as they only studied american snowflakes. I bet metric snowflakes are way more symmetrical! (PS: Beautiful images! Snowflake #167 is especially interesting!) (I just noticed that metric snowflakes must have 10-fold symmetry.) Dec 26, 2015 at 14:30
• It will be okay as long as it outputs using one of the above methods. However, n cannot be predefined in your script file. You can read n from STDIN, prompt it from the user, take it as a function / commad line argument ... basically anything you want except for directly embedding it in your code. Dec 27, 2015 at 0:20
• Don't want to +1 this because it has 25 :( Jan 6, 2016 at 3:23

# Matlab, 226

function P(M);function c(L,X,Y,O);hold on;F=.5+5^.5/2;a=2*pi*(1:5)/5;b=a(1)/2;C=F^(2*L);x=cos(a+O*b)/C;y=sin(a+O*b)/C;if L<M;c(L+1,X,Y,~O);for k=1:5;c(L+1,X+x(k),Y+y(k),O);end;else;fill(X+x*F, Y+y*F,'k');end;end;c(0,0,0,0);end


Ungolfed:

function P(M);
function c(L,X,Y,O);          %recursive function
hold on;
F=.5+5^.5/2;                  %golden ratio
a=2*pi*(1:5)/5;               %full circle divided in 5 parts (angles)
b=a(1)/2;
C=F^(2*L);
x=cos(a+O*b)/C;               %calculate the relative position ofnext iteration
y=sin(a+O*b)/C;
if L<M;                       %current recursion (L) < Maximum (M)? recurse
c(L+1,X,Y,~O);            %call recursion for inner pentagon
for k=1:5;
c(L+1,X+x(k),Y+y(k),O)%call recursion for the outer pentagons
end;
else;                         %draw
fill(X+x*F, Y+y*F,'k');
end;
end;
c(0,0,0,0);
end


Fifth iteration (already took quite a while to render). A slight alteration of the code (unfortunately more bytes) results in this beauty=) Oh, and another one: • Thanks for pointing me to this challenge, I went and added another solution, hope you don't mind;) I'm safely away from your byte-count, anyway, I just found it too interesting to miss. Dec 26, 2015 at 19:26

## Mathematica, 200 bytes

a=RotationTransform
b=Range
r@k_:={Re[t=I^(4k/5)],Im@t}
R@k_:=a[Pi,(r@k+r[k+1])/2]
Graphics@Nest[GeometricTransformation[#,ScalingTransform[{1,1}(Sqrt@5-3)/2]@*#&/@Append[R/@b@5,a@0]]&,Polygon[r/@b@5],#]&


The last line is a function which can be applied to an integer n.

Mathematica function names are long. Somebody should entropy-encode them and make a new language from it. :)

When applied to 1: When applied to 2: ## MATLAB, 235233 217 bytes

Update: a bunch of suggestions from @flawr helped me lose 16 bytes. Since only this allowed me to beat flawr's solution, and that I wouldn't have found the challenge without flawr's help in the first place, consider this a joint submission by us:)

N=input('');f=2*pi/5;c=1.5+5^.5/2;g=0:f:6;p=[cos(g);sin(g)];R=[p(:,2),[-p(2,2);p(1,2)]];for n=1:N,t=p;q=[];for l=0:4,q=[q R^l*[c-1+t(1,:);t(2,:)]/c];end,p=[q -t/c];end,p=reshape(p',5,[],2);fill(p(:,:,1),p(:,:,2),'k');


This is another MATLAB solution, this one based on a philosophy of iterated function systems. I was mostly interested in developing the algorithm itself, and I haven't golfed too much on the solution. There's surely room for improvement. (I contemplated using a hard-coded fixed-point approximation for c, but that wouldn't be nice.)

Ungolfed version:

N=input('');                                % read order from stdin

f=2*pi/5;                                   % angle of 5-fold rotation
c=1.5+5^.5/2;                               % scaling factor for contraction

g=0:f:6;
p=[cos(g);sin(g)];                          % starting pentagon, outer radius 1
R=[p(:,2),[-p(2,2);p(1,2)]];                % 2d rotation matrix with angle f

for n=1:N,                                  % iterate the points
t=p;
q=[];
for l=0:4,
q=[q R^l*[c-1+t(1,:);t(2,:)]/c];     % add contracted-rotated points
end,
p=[q -t/c];                             % add contracted middle block
end,

p=reshape(p',5,[],2);                 % reshape to 5x[]x2 matrix to separate pentagons
fill(p(:,:,1),p(:,:,2),'k');          % plot pentagons


Result for N=5 (with a subsequent axis equal off for prettiness, but I hope that doesn't count byte-wise): • I think you could save a few bytes by using R=[p(:,2),[-p(2,2);p(1,2)]]; (and eliminating the previous R,C,S) and you can use q=[q R^l*[c-1+t(1,:);t(2,:)]/c] and I think c=1.5+5^.5/2; Jan 1, 2016 at 11:13
• @flawr obviously you're right:) 1. thanks for the rotation matrix, 2. thanks for the new q, I even had a needless pair of parentheses in there... 3. thanks, but what is this magic??:D 4. since the solution is now shorter than your original, I consider this to be partly your submission as well. Jan 3, 2016 at 21:45

# Mathematica, 124 bytes

Mathematica supports new syntax for Table since version 10: Table[expr, n], which saves another byte. Table[expr, n] is equivalent to Table[expr, {n}].

f@n_:=(p=E^Array[π.4I#&,5];Graphics@Map[Polygon,ReIm@Fold[{g,s}~Function~Join[.62(.62g#+#&/@s),{-.39g}],p,p~Table~n],{-3}])


The core of this function is using complex numbers to do tranformations and then convert them to points by ReIm.

Test case:

f • π takes up two bytes in UTF-8, so you come out to 125 bytes total. Jan 4, 2016 at 4:58
• OMFG what is this Jan 5, 2016 at 22:04

## Mathematica, 199 196 bytes

Edging out Peter Richter's answer by a hair, here's one of my own. It leans heavily on graphics functionality, and less on math and FP. The CirclePoints builtin is new in 10.1.

c=CirclePoints;g=GeometricTransformation;
p@0=Polygon@c[{1,0},5];
p@n_:=GraphicsGroup@{
p[n-1],
g[
p[n-1]~g~RotationTransform[Pi/5],
TranslationTransform/@{GoldenRatio^(2n-1),n*Pi/5}~c~5
]
};
f=Graphics@*p


Edit: Thanks to DumpsterDoofus for GoldenRatio

• You can save 3 bytes by replacing ((1+Sqrt@5)/2) with GoldenRatio. Also in the second line I think it should be p@0=Polygon@c[{1,0},5]; instead of p@0=Polygon@cp[{1,0},5];. (BTW I'm actually Peter, I've got two profiles lol). Dec 27, 2015 at 3:42
• Yes! Good call. I spotted the typo, too, but forgot to fix it. D'oh, Dec 27, 2015 at 3:43

# Mathematica, 130 bytes

r=Exp[Pi.4I Range@5]
p=1/GoldenRatio
f@0={r}
f@n_:=Join@@Outer[1##&,r,p(f[n-1]p+1),1]~Join~{-f[n-1]p^2}
Graphics@*Polygon@*ReIm@*f


I use a similar technique to njpipeorgan's answer (in fact I stole his 2Pi I/5 == Pi.4I trick), but implemented as a recursive function.

Example usage (using % to access the anonymous function that was output on the last line):

 % ## Wolfram Language (Mathematica), 9690 84bytes

Region@Polygon@ReIm@Nest[Join[Tr/@Tuples@{.62p,t=.39#},-t]&,{p=(-1)^(.4Range@5)},#]& 