Create a fractal tree

Question

What I would like to see is a fractal tree being drawn where the you can input an integer, and the output will be a fractal tree with the entered amount of branch steps.

Rules:

The fractal should be a line fractal tree: $line fractal tree$
the shape of the fractal should also be the same as this picture.
Each branch should decrease by 25% in width for each time it splits
The final layer of branches should be an 1px wide line
Shortest code wins

Tip: this site might be useful.

This question seems different enough from "Sorry, young man..." and I'm voting to reopen. For one, this doesn't require an L-system, and it is more precise in its description of the output. Answers to the other aren't necessarily the best answers here. — luser droog, Commented Jan 18, 2014 at 22:09
If there's no new answers in the next day, I'll mark the shortest code as accepted. — joeytje50, Commented Jan 23, 2014 at 14:20
@w4nteD The challenge of course does not specify any leaves, but I daresay they would be fairly easy to add to most of these programs. After the recursion decides to return, draw a leaf before returning. The current branch angle will be available for orientation. You could draw a leaf shape in the same direction, or maybe two at slight angles similar to the branching. — luser droog, Commented Sep 19, 2015 at 17:30

MtnViewMark · Accepted Answer · 2014-01-23 04:12:58Z

14

Context Free, 82 65 57 characters

rule T{9*[y 1]SQUARE[]2*{f 90}T{s.75y 8r 25}}startshape T

A Tree

See: http://www.contextfreeart.org/gallery/view.php?id=3384

Golf'd further with help from Context Free Art users minimaleye, MtnViewJohn, and kipling.

edited Jan 23, 2014 at 4:12

answered Jan 21, 2014 at 7:03

MtnViewMark

5,01918 silver badges26 bronze badges

\$\begingroup\$ According to a comment on that link you posted, it can be reduced to 64 characters. Nice one! \$\endgroup\$
– joeytje50
Commented Jan 21, 2014 at 15:19
1

\$\begingroup\$ Upvoting not only because it's a good answer with pretty output, but because I have never even heard of this language before. \$\endgroup\$
– Michael Stern
Commented Jan 21, 2014 at 16:46

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friol · Accepted Answer · 2014-01-19 19:11:17Z

9

Logo, 88, 86

to t:d if:d=0[stop]setpensize:d*.75 fd:d*7 rt 25 t:d-1 lt 50 t:d-1 rt 25 bk:d*7end t [numsteps]

Tree - size 11

Edit: Made the branches decrease as pointed out in the comments.

edited Jan 19, 2014 at 19:11

answered Jan 19, 2014 at 17:47

friol

1,4688 silver badges14 bronze badges

\$\begingroup\$ Your branches don't get shorter the further up the tree we go. This was one of the requirements, since "the shape of the fractal should also be the same as this picture". Also, you should include the code to create the tree in the character count. Other than that, well done on the short code! \$\endgroup\$
– joeytje50
Commented Jan 19, 2014 at 18:14
\$\begingroup\$ Also, your branch width isn't decreasing geometrically at 25% each time - it's linear based on the iteration number. I set my pen size to (4/3)^(:d-1) to accomplish this. The lowest I got in my Logo attempt was 108. Nice work! \$\endgroup\$
– GuitarPicker
Commented Aug 13, 2016 at 6:44

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

9

Mathematica 127

k=12;

r=#2/.{x__Real}:>.1{{7,-#},{#,7}}.{x}+y&;
f@n_:={f@1=N@Polygon@{y={0,.7^k},0y,x={.002,0},x+y},r[-4,p=f[n-1]],4~r~p}
Graphics@f@k

enter image description here

Value .002 adjusted to produce 1 pixel width of the final branches.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jan 21, 2014 at 18:50

ybeltukov

1,87111 silver badges11 bronze badges

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Jim Bauwens · Accepted Answer · 2015-04-03 14:07:18Z

HTML+CSS (no JavaScript) 14791 14630 (multiplatform), 294 (webkit only)

You can't specify how many levels you go.. And yes it is pretty big. But it is pure HTML/CSS, without any JavaScript.

<style>q{background-color:#000}q::before{content:""}q.start{width:15px;height:100px;position:absolute;top:500px;left:500px}q q{position:absolute;width:75%;height:75%;top:-55%}q q:nth-child(1){left:-90%;transform:rotate(-27deg)}q q:nth-child(2){left:110%;transform:rotate(27deg)}</style><q 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Webkit short version:

<style>b{background:#000;-webkit-box-reflect: left;}b.i{width:7px;height:100px;position:absolute;top:500px;left:500px}b b{position:absolute;width:75%;height:75%;top:-70%}b b{right:300%;transform:rotate(-27deg)}</style><b class="i"><b><b><b><b><b><b><b><b></b><b></b></b></b></b></b></b></b></b>

Fractaltree
Edit: removed the cross browser prefixes, firefox and google chrome render it properly without it. Edit 2: Added a much shorter version that uses -webkit-box-reflect. Only works in webkit based browsers though.

<style>q{background-color:#000}q::before{content:""}q.start{width:15px;height:100px;position:absolute;top:500px;left:500px}q q{position:absolute;width:75%;height:75%;top:-55%}q q:nth-child(1){left:-90%;transform:rotate(-27deg)}q q:nth-child(2){left:110%;transform:rotate(27deg)}</style><q class="start"><q></q><q></q></q> With only three branches and removing non needed CSS components for cross-browser compatibility the code is 321 characters long. Adding a level of branches is done by replacing <q></q> with <q><q></q><q></q></q>. — Jim Bauwens, Commented Sep 26, 2014 at 18:01
Ha, this got an "excessively long" automatic flag. :D It would be interesting if you posted the code to generate this (or explain how you did it). (Assuming you didn't do it by hand!) — Doorknob, Commented Sep 26, 2014 at 21:11
Doorknob: The core CSS code that does the real 'magic' is in the beginning, it 'recursivly' positions sub HTML in the correct way using transforms, such that they form the tree. The HTML itself I did manually, but isn't hard. I start with <q></q> and use my code editor to replace that with <q><q></q><q></q></q>. Every time I repeat this replacement operation I add a level of branches to my fractal. The CSS automatically wil render the added levels correctly. My previous comment demonstrates this. — Jim Bauwens, Commented Sep 27, 2014 at 10:24

mousetail 'he-him' · Accepted Answer · 2023-05-12 07:42:22Z

9

Python - 215

import turtle
t=turtle.Turtle()
def f(w):
    if w>=1:
        s=t.pensize()
        t.pensize(w)
        t.fd(w*9)
        t.rt(20)
        f(w*.75)
        t.lt(40)
        f(w*.75)
        t.right(20)
        t.bk(w*9)
        t.pensize(w)
t.speed(9)
f(9)
t.ht()
raw_input()

Close enough?

edited May 12, 2023 at 7:42

mousetail 'he-him'

13.5k1 gold badge39 silver badges85 bronze badges

answered Jan 18, 2014 at 0:12

mniip

9,4924 gold badges31 silver badges57 bronze badges

\$\begingroup\$ I'll accept it, because I just said that the "shape" should be the same. I didn't specify any orientation rules. \$\endgroup\$
– joeytje50
Commented Jan 18, 2014 at 0:15
\$\begingroup\$ a little more documentation (or at least meaningful names) could be very nice \$\endgroup\$
– Elisha
Commented Jan 19, 2014 at 18:08
2

\$\begingroup\$ Documentation? This is a complete program. Run as-is, no arguments, no nothing. Meaningful names? You cannot fit a lot of meaning into 1 character, and last time I checked this is code-golf \$\endgroup\$
– mniip
Commented Jan 19, 2014 at 18:49
\$\begingroup\$ I think "commentary" might be the word. I know that to a Python hacker, it must be very clear, and maybe to English-speakers who can recognize the abbreviated names, and turtle-graphics fans (like me) it is, but for others who don't know the language but want to vote you up, give them some help, you know? Tell a little story, even if it's just "the problem domain maps naturally to this feature set". Does that help?? I think that's what's going on here. ... Or at least, that's a practical way to interpret such comments. :) \$\endgroup\$
– luser droog
Commented Jan 22, 2014 at 4:38
\$\begingroup\$ I think that n>=1 can be reduced to just n \$\endgroup\$
– xenia
Commented Sep 19, 2015 at 19:19

| Show 1 more comment

Community · Accepted Answer · 2020-06-17 09:04:33Z

7

Mathematica 199

f[p : {_, _}, r_, s_, d_] := Module[{q}, If[d == 0, Return[]];
  q = p + {Cos@r, Sin@r}*d;
  DeleteCases[
   Flatten@{{Thickness[.002*1.25^d], Line@{p, q}}, 
     f[q, r - s, s, d - 1], f[q, r + s, s, d - 1]}, Null]]
g@d_ := Graphics[f[{0, 0}, Pi/2, Pi/9, d]]

Example

8 is the depth of the tree.

g[8]

tree

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jan 18, 2014 at 0:54

DavidC

25.4k2 gold badges52 silver badges105 bronze badges

Add a comment |

luser droog · Accepted Answer · 2014-01-21 02:52:50Z

Postscript 216 214 209

Uses the criteria "linewidth==1 pixel" as the recursion bound. Edit: juggled some constants. Edit: tweak, tweak.

/b{gsave dup scale rotate
0 0 0 12 0 0 moveto translate lineto stroke
currentlinewidth 0 dtransform dup mul exch dup mul add sqrt 1 ge
{28 .75 b -28 .75 b}if grestore}def
2 setlinewidth
200 20 translate
0 6 b

binary-token workfile:

0 6 200 20 2
/.{<920>dup 1 4 3 roll put cvx exec}def/${//. 73 .}def
%/b{gsave dup scale rotate  % gsave=78(N)<4E> dup=56(8)<38> scale=139<8B> rotate=136<88>
/b{
%(N8)$<8B88>$
<4E388B88>$
0 0 0 12 0 0
%moveto translate lineto stroke
% moveto=107(k)<6B> lineto=99(c)<63> translate=173<AD> stroke=167<A7>
%(k)$<AD>$(c)$<A7>$
%<6B63ADA7>$ % <--this typo makes a pot leaf. :-)
<6BAD63A7>$
%currentlinewidth 0 dtransform % clw=38(&) dxfm=55(7)
0(&7)$
%dup mul exch dup mul add sqrt 1 ge % mul=108 sqrt=355 ge=74 add=1 exch=62
(8l>8l)$ add sqrt 1 ge
{28 .75 b -28 .75 b}
%if grestore}def % grestore=77(M) if=84(T) def=51(3)<33>
(TM)$
}
%(3)$
%setlinewidth % slw=155<9B>
%translate % trsl8=173<AD>
<339BAD>$ b

Postscript 172 169

Same program using binary token strings.

/.{<920>dup 1 4 3 roll put cvx exec}def/${//. 73 .}def
0 6 200 20 2/b{<4E388B88>$ 0 0 0 12 0 0<6BAD63A7>$
0(&78l>8l)$ add sqrt 1 ge{28 .75 b -28 .75 b}(TM)$}<339BAD>$
b

output2

robert · Accepted Answer · 2016-05-23 07:15:57Z

CoffeeScript (using Canvas) (289 bytes)

The following creates branches until the pixel width is less than 1px. The width and height of the image created is 500px. To see the virtual image that the program draws, add document.body.appendChild(c) to the last line of the code.

c=document.createElement 'canvas'
c.width=c.height=500
t=c.getContext '2d'
m=Math
b=m.PI/8
f=(w,l,x,y,a)->
    t.beginPath()
    t.moveTo x,y
    X=x+m.cos(a)*l
    Y=y+m.sin(a)*l
    t.lineWidth=w
    t.lineTo X,Y
    t.stroke()
    if w >= 1
        f w*3/4,l*3/4,X,Y,a+b
        f w*3/4,l*3/4,X,Y,a-b
f 10,90,250,0,m.PI/2

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Create a fractal tree

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Context Free, 82 65 57 characters

Logo, 88, 86

Mathematica 127

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Python - 215

Mathematica 199

Postscript 216 214 209

Postscript 172 169

CoffeeScript (using Canvas) (289 bytes)

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Create a fractal tree

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Context Free, 82 65 57 characters

Logo, 88, 86

Mathematica 127

HTML+CSS (no JavaScript) 14791 14630 (multiplatform), 294 (webkit only)

Python - 215

Mathematica 199

Postscript 216 214 209

Postscript 172 169

CoffeeScript (using Canvas) (289 bytes)

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