107
\$\begingroup\$

I always used a Mandelbrot image as the 'graphical' version of Hello World in any graphical application I got my hands on. Now it's your guys' turn.

  • Language must be capable of graphical output or drawing charts (saving image files disallowed)
  • Render a square image or graph. The size at least 128 and at most 640 across*
  • The fractal coordinates range from approximately -2-2i to 2+2i
  • The pixels outside of the Mandelbrot set should be colored according to the number of iterations before the magnitude exceeds 2 (excluding* black & white)
  • Each iteration count must have a unique color*, and neighboring colors should preferably be easily distinguishable by the eye
  • The other pixels (presumably inside the Mandelbrot set) must be colored either black or white
  • At least 99 iterations
  • ASCII art not allowed

* unless limited by the platform, e.g. graphical calculator

Allowed:
Allowed
Disallowed:
Disallowed
(shrunken images)

Winning conditions:

Shortest version (size in bytes) for each language will get a mention in this post, ordered by size.
No answer will ever be 'accepted' with the button.

Leaderboard:

/* Configuration */

var QUESTION_ID = 23423; // Obtain this from the url
// It will be like https://XYZ.stackexchange.com/questions/QUESTION_ID/... on any question page
var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";
var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk";
var OVERRIDE_USER = 17419; // This should be the user ID of the challenge author.

/* App */

var answers = [], answers_hash, answer_ids, answer_page = 1, more_answers = true, comment_page;

function answersUrl(index) {
  return "https://api.stackexchange.com/2.2/questions/" +  QUESTION_ID + "/answers?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + ANSWER_FILTER;
}

function commentUrl(index, answers) {
  return "https://api.stackexchange.com/2.2/answers/" + answers.join(';') + "/comments?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + COMMENT_FILTER;
}

function getAnswers() {
  jQuery.ajax({
    url: answersUrl(answer_page++),
    method: "get",
    dataType: "jsonp",
    crossDomain: true,
    success: function (data) {
      answers.push.apply(answers, data.items);
      answers_hash = [];
      answer_ids = [];
      data.items.forEach(function(a) {
        a.comments = [];
        var id = +a.share_link.match(/\d+/);
        answer_ids.push(id);
        answers_hash[id] = a;
      });
      if (!data.has_more) more_answers = false;
      comment_page = 1;
      getComments();
    }
  });
}

function getComments() {
  jQuery.ajax({
    url: commentUrl(comment_page++, answer_ids),
    method: "get",
    dataType: "jsonp",
    crossDomain: true,
    success: function (data) {
      data.items.forEach(function(c) {
        if (c.owner.user_id === OVERRIDE_USER)
          answers_hash[c.post_id].comments.push(c);
      });
      if (data.has_more) getComments();
      else if (more_answers) getAnswers();
      else process();
    }
  });  
}

getAnswers();

var SCORE_REG = /<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/;

var OVERRIDE_REG = /^Override\s*header:\s*/i;

function getAuthorName(a) {
  return a.owner.display_name;
}

function process() {
  var valid = [];
  
  answers.forEach(function(a) {
    var body = a.body;
    a.comments.forEach(function(c) {
      if(OVERRIDE_REG.test(c.body))
        body = '<h1>' + c.body.replace(OVERRIDE_REG, '') + '</h1>';
    });
    
    var match = body.match(SCORE_REG);
    if (match)
      valid.push({
        user: getAuthorName(a),
        size: +match[2],
        language: match[1],
        link: a.share_link,
      });
    else console.log(body);
  });
  
  valid.sort(function (a, b) {
    var aB = a.size,
        bB = b.size;
    return aB - bB
  });

  var languages = {};
  var place = 1;
  var lastSize = null;
  var lastPlace = 1;
  valid.forEach(function (a) {
    if (a.size != lastSize)
      lastPlace = place;
    lastSize = a.size;
    ++place;
    
    var answer = jQuery("#answer-template").html();
    answer = answer.replace("{{PLACE}}", lastPlace + ".")
                   .replace("{{NAME}}", a.user)
                   .replace("{{LANGUAGE}}", a.language)
                   .replace("{{SIZE}}", a.size)
                   .replace("{{LINK}}", a.link);
    answer = jQuery(answer);
    jQuery("#answers").append(answer);

    var lang = a.language;
    lang = jQuery('<a>'+lang+'</a>').text();
    
    languages[lang] = languages[lang] || {lang: a.language, lang_raw: lang, user: a.user, size: a.size, link: a.link};
  });

  var langs = [];
  for (var lang in languages)
    if (languages.hasOwnProperty(lang))
      langs.push(languages[lang]);

  langs.sort(function (a, b) {
    if (a.lang_raw.toLowerCase() > b.lang_raw.toLowerCase()) return 1;
    if (a.lang_raw.toLowerCase() < b.lang_raw.toLowerCase()) return -1;
    return 0;
  });

  for (var i = 0; i < langs.length; ++i)
  {
    var language = jQuery("#language-template").html();
    var lang = langs[i];
    language = language.replace("{{LANGUAGE}}", lang.lang)
                       .replace("{{NAME}}", lang.user)
                       .replace("{{SIZE}}", lang.size)
                       .replace("{{LINK}}", lang.link);
    language = jQuery(language);
    jQuery("#languages").append(language);
  }

}
body {
  text-align: left !important;
  display: block !important;
}

#answer-list {
  padding: 10px;
  width: 290px;
  float: left;
}

#language-list {
  padding: 10px;
  width: 290px;
  float: left;
}

table thead {
  font-weight: bold;
}

table td {
  padding: 5px;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<link rel="stylesheet" type="text/css" href="https://cdn.sstatic.net/Sites/codegolf/all.css?v=ffb5d0584c5f">
<div id="language-list">
  <h2>Shortest Solution by Language</h2>
  <table class="language-list">
    <thead>
      <tr><td>Language</td><td>User</td><td>Score</td></tr>
    </thead>
    <tbody id="languages">

    </tbody>
  </table>
</div>
<div id="answer-list">
  <h2>Leaderboard</h2>
  <table class="answer-list">
    <thead>
      <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr>
    </thead>
    <tbody id="answers">

    </tbody>
  </table>
</div>
<table style="display: none">
  <tbody id="answer-template">
    <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr>
  </tbody>
</table>
<table style="display: none">
  <tbody id="language-template">
    <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr>
  </tbody>
</table>

\$\endgroup\$
14
  • 9
    \$\begingroup\$ "Easily distinguished by the eye" is hard to make objective. ...Also, apart from your personal association of the two, the Mandelbrot set has nothing to do with Hello World, so it's best to omit that from the title unless you are deliberately trolling the search engines. \$\endgroup\$ Commented Mar 7, 2014 at 21:50
  • 4
    \$\begingroup\$ I've seen a few people now mentioning that they render the Mandelbrot as a "Hello World". I've done that too, for something like 30 years. The Mandelbrot is the perfect "Hello World" because it shows both that you have pixel access to the display and gives a good feel for compute bound performance on the new platform. \$\endgroup\$
    – Roger Dahl
    Commented Mar 9, 2014 at 2:00
  • 7
    \$\begingroup\$ Great idea to ask a question which requires a combination of mathematical and aesthetic sensitivities, then impose all the the design decisions in advance. \$\endgroup\$
    – jwg
    Commented Mar 10, 2014 at 8:25
  • 3
    \$\begingroup\$ Anyone manages to make one in brainfuck WINS, I'd say :D \$\endgroup\$
    – MadTux
    Commented Mar 10, 2014 at 18:45
  • 3
    \$\begingroup\$ Not mine and not golfed, but this code is a marvel of obfuscated python producing a mandelbrot fractal \$\endgroup\$
    – Aaron
    Commented May 26, 2017 at 8:37

41 Answers 41

112
\$\begingroup\$

Sharp EL-9300 Graphics Calculator, 296 bytes

This was my secondary school graphing calculator in the early '90s. I remember writing a mandelbrot generator for it way back then. And sure enough, its still sitting there in the NV memory:

ClrG
DispG
Range -2.35,2.35,.5,-1.55,1.55,0.5
y=-1.55
Label ly
x=-2.35
Label lx
n=1
zx=0
zy=0
Label ln
tzx=zx²-zy²+x
zy=(2*zx*zy)+y
zx=tzx
If zx²+zy²>4Goto esc
n=n+1
If n<20Goto ln
Label esc
If fpart (n/2)=0Goto npl
Plot x,y
Label npl
x=x+.05
If x<=2.35Goto lx
y=y+.05
If y<=1.55Goto ly
Wait

It took about 90 minutes to render. Only 20 iterations, but at this resolution it probably doesn't make much difference.

This is totally ungolfed. I'm sure I could save a bit of space, but I just wanted to share this historical curiosity.

I love that the only control statements available are ifs and gotos.

Here's a photo. I don't have any other means to get the graphical output out: enter image description here

\$\endgroup\$
14
  • 2
    \$\begingroup\$ Me too, but my NV memory went blank after years of shelf-time. \$\endgroup\$ Commented Mar 8, 2014 at 9:10
  • 2
    \$\begingroup\$ zx²+zy²>4 couldn't that be Abs(x)>2? \$\endgroup\$ Commented Mar 8, 2014 at 9:31
  • 32
    \$\begingroup\$ Interesting. So you've been a nerd for quite a while. \$\endgroup\$
    – devnull
    Commented Mar 9, 2014 at 3:34
  • 4
    \$\begingroup\$ @devnull Touché! Guilty as charged. \$\endgroup\$ Commented Mar 9, 2014 at 3:58
  • 6
    \$\begingroup\$ Nice "Screenshot" \$\endgroup\$
    – meawoppl
    Commented Mar 9, 2014 at 10:01
94
\$\begingroup\$

I came across this the other day. I don't take credit for it, but damn, is it awesome:

Python 2:

_                                      =   (
                                        255,
                                      lambda
                               V       ,B,c
                             :c   and Y(V*V+B,B,  c
                               -1)if(abs(V)<6)else
               (              2+c-4*abs(V)**-0.4)/i
                 )  ;v,      x=1500,1000;C=range(v*x
                  );import  struct;P=struct.pack;M,\
            j  ='<QIIHHHH',open('M.bmp','wb').write
for X in j('BM'+P(M,v*x*3+26,26,12,v,x,1,24))or C:
            i  ,Y=_;j(P('BBB',*(lambda T:(T*80+T**9
                  *i-950*T  **99,T*70-880*T**18+701*
                 T  **9     ,T*i**(1-T**45*2)))(sum(
               [              Y(0,(A%3/3.+X%v+(X/v+
                               A/3/3.-x/2)/1j)*2.5
                             /x   -2.7,i)**2 for  \
                               A       in C
                                      [:9]])
                                        /9)
                                       )   )

enter image description here http://preshing.com/20110926/high-resolution-mandelbrot-in-obfuscated-python/

\$\endgroup\$
9
  • 13
    \$\begingroup\$ Seems to be disallowed: the regions are not easily distinguishable, or even at all. \$\endgroup\$
    – primo
    Commented Mar 8, 2014 at 13:32
  • 5
    \$\begingroup\$ Also, this writes to a file. \$\endgroup\$
    – Lie Ryan
    Commented Mar 8, 2014 at 21:49
  • 43
    \$\begingroup\$ disallowed or not, this is pretty awesome :D \$\endgroup\$
    – Navin
    Commented Mar 9, 2014 at 2:28
  • 18
    \$\begingroup\$ @DigitalTrauma, heck, +1 for most beautiful input! \$\endgroup\$
    – Brian S
    Commented Mar 10, 2014 at 15:26
  • 23
    \$\begingroup\$ Does this count as a quine? ;-) \$\endgroup\$ Commented Mar 10, 2014 at 21:04
55
\$\begingroup\$

LaTeX, 673 bytes

\countdef\!1\!129\documentclass{article}\usepackage[margin=0pt,papersize=\!bp]{geometry}\usepackage{xcolor,pgf}\topskip0pt\offinterlineskip\def~{99}\let\rangeHsb~\countdef\c2\countdef\d3\countdef\e4\begin{document}\let\a\advance\let\p\pgfmathsetmacro\makeatletter\def\x#1#2#3{#10
\@whilenum#1<#2\do{#3\a#11}}\d0\x\c{\numexpr~+1}{\expandafter\edef\csname\the\c\endcsname{\hbox{\noexpand\color[Hsb]{\the\d,1,1}\/}}\a\d23
\ifnum\d>~\a\d-~\fi}\def\/{\rule{1bp}{1bp}}\x\c\!{\hbox{\x\d\!{\p\k{4*\d/(\!-1)-2}\p\K{2-4*\c/(\!-1)}\def\z{0}\def\Z{0}\x\e~{\p\:{\z*\z-\Z*\Z+\k}\p\Z{2*\z*\Z+\K}\let\z\:\p\:{\z*\z+\Z*\Z}\ifdim\:pt>4pt\csname\the\e\endcsname\e~\fi}\ifnum\e=~\/\fi}}}\stop

Result 129x129 (129 × 129)

The PDF image consists of colored square units with size 1bp × 1bp.

Ungolfed

% count register \size contains the width and height of the square
\countdef\size=1
\size=31
\documentclass{article}
\usepackage[margin=0pt,papersize=\size bp]{geometry}
\usepackage{xcolor,pgf}
\topskip0pt
\offinterlineskip
\def\iterations{99}
\let\rangeHsb\iterations
\countdef\c2
\countdef\d3
\countdef\e4
\begin{document}
\let\p\pgfmathsetmacro
\makeatletter
% \Loop: for (#1 = 0; #1 < #2; #1++) {#3}
\def\Loop#1#2#3{%
  #1=0
  \@whilenum#1<#2\do{#3\advance#11}%
}
\d0%
\Loop\c{\numexpr\iterations+1\relax}{%
  \expandafter\edef\csname\the\c\endcsname{%
    \hbox{\noexpand\color[Hsb]{\the\d,1,1}\noexpand\pixel}%
  }%
  \advance\d23 \ifnum\d>\iterations\advance\d-\iterations\fi
}
\def\pixel{\rule{1bp}{1bp}}
% \c: row
% \d: column
% \e: iteration
\Loop\c\size{%
  \typeout{c: \the\c}%
  \hbox{%
    \Loop\d\size{%
      \pgfmathsetmacro\k@re{4*\d/(\size-1)-2}%
      \pgfmathsetmacro\K@im{2-4*\c/(\size-1)}%
      \def\z@re{0}%
      \def\Z@im{0}%
      \Loop\e\iterations{%
         % calculate z(n+1) = z^2(n) + k
         \pgfmathsetmacro\temp{\z@re*\z@re-\Z@im*\Z@im+\k@re}%
         \pgfmathsetmacro\Z@im{2*\z@re*\Z@im+\K@im}%
         \let\z@re\temp
         % calculate abs(z)^2
         \pgfmathsetmacro\temp{\z@re*\z@re+\Z@im*\Z@im}%
         \ifdim\temp pt>4pt\csname\the\e\endcsname\e\iterations\fi
      }%   
      \ifnum\e=\iterations\pixel\fi
    }%
  }%
}
\stop
\$\endgroup\$
1
  • \$\begingroup\$ I wonder if functional shading is shorter / meets the rule (not loading the pre-defined shading but define it yourself). Also including tikz includes both pgf and xcolor. \$\endgroup\$
    – Symbol 1
    Commented Mar 26, 2022 at 3:33
42
\$\begingroup\$

x86 DOS Assembly, 208 177 173 bytes

The full binary, in HEX, that I created by hand, is:

DBE3BE00A0B81300CD1056BA640007BF87F9FDBDC7008BCDE81A008AC3AA4979F7B9C70083EF784D79EE33C0CD16B80300CD10CD208BC12BC289441CDF441CDF06A701DEF9D95C088BC52BC289441CDF441CDF06A701DEF9D95C0CD9EED914D95404D95410D95C14B301D904D84C04DE0EA901D8440CD95404D94410D86414D84408D914D80CD95C10D84C04D95414D84410DF06AB01DED99BDFE09B9E7207433ADA72C632DBC3320002000400

The sample image is:

Mandlebrot screen shot with black cropped

The full source in readable ASM is fairly long (I used this to figure out how I was coding this sucker):

.286
CODE SEGMENT
ASSUME CS:code, DS:code
ORG 0100h
  
; *****************************************************************************
start:
  ; Mandlebrot coordinates
  zr   = DWORD PTR [SI+0]
  zi   = DWORD PTR [SI+4]
  cr   = DWORD PTR [SI+8]
  ci   = DWORD PTR [SI+12]
  zrsq = DWORD PTR [SI+16]
  zisq = DWORD PTR [SI+20]

  ; Temp int
  Temp = WORD PTR  [SI+28]

  ; ===========================================================================
  ; Initialize

  ; Initialize the FPU
  FNINIT

  ; SI points to our memory
  mov si, 0A000h ; So we can push it

  ; Shave off some bytes by reusing 100
  mov dx, 100

  ; Switch to MCGA
  mov ax, 013h
  int 010h

  ; ES:DI is the end of our drawing area
  push si
  pop es
  mov di, 63879
  std ; We're using stosb backwards

  ; Initialize our X and Y
  mov bp, 199
  mov cx, bp


  ; ===========================================================================
  ; Main draw loop
  
MainLoop:
  ; Get our next mandelbrot value
  call GMV

  ; Store it
  mov al, bl
  stosb

  ; Decrement our X
  dec cx
  jns MainLoop

  ; Decrement our Y
  mov cx, 199
  sub di, 120
  dec bp
  jns MainLoop


  ; ===========================================================================
  ; Done

  ; Wait for a key press
  xor ax, ax
  int 016h

  ; Change back to text mode
  mov ax, 3
  int 010h

  ; Exit to DOS
  int 020h



; *****************************************************************************
; GMV: Get Mandelbrot Value
; Gets the value for the next Mandelbrot pixel
; Returns:
;   BL - The color to use
GMV:
  ; ===========================================================================
  ; Initialize

  ; cr = (x - 100) / 50;
  mov ax, cx
  sub ax, dx                  ; \
  mov Temp, ax                ;  > ST0 = Current X - 100
  FILD Temp                   ; /
  FILD Divisor                ; ST0 = 50, ST1 = Current X - 100
  FDIVP                       ; ST0 = (Current X - 100) / 50
  FSTP cr                     ; Store the result in cr

  ; ci = (y - 100) / 50;
  mov ax, bp
  sub ax, dx                  ; \
  mov Temp, ax                ;  > ST0 = Current Y - 100
  FILD Temp                   ; /
  FILD Divisor                ; ST0 = 50, ST1 = Current Y - 100
  FDIVP                       ; ST0 = (Current Y - 100) / 50
  FSTP ci                     ; Store the result in ci

  ; zr = zi = zrsq = zisq = 0;
  FLDZ
  FST zr
  FST zi
  FST zrsq
  FSTP zisq

  ; numiteration = 1;
  mov bl, 1

  ; ===========================================================================
  ; Our main loop

  ; do {
GMVLoop:

  ; zi = 2 * zr * zi + ci;
  FLD zr
  FMUL zi
  FIMUL TwoValue
  FADD ci
  FST zi ; Reusing this later

  ; zr = zrsq - zisq + cr;
  FLD zrsq
  FSUB zisq
  FADD cr
  FST zr ; Reusing this since it already is zr

  ; zrsq = zr * zr;
  ;FLD zr ; Reused from above
  FMUL zr
  FSTP zrsq

  ; zisq = zi * zi;
  ;FLD zi ; Reused from above
  FMUL zi
  FST zisq ; Reusing this for our comparison

  ; if ((zrsq + zisq) < 4)
  ;   return numiteration;
  FADD zrsq
  FILD FourValue
  FCOMPP
  FSTSW ax
  FWAIT
  sahf
  jb GMVDone

  ;} while (numiteration++ < 200);
  inc bx
  cmp bl, dl
  jb GMVLoop

  ;return 0;
  xor bl, bl

GMVDone:  
  ret
;GMV



; *****************************************************************************
; Data

; Divisor
Divisor DW 50
; Two Value
TwoValue DW 2
; 4 Value
FourValue DW 4

CODE ENDS
END start

This is designed for compiling with TASM, runs in MCGA, and waits for a keypress before ending the program. The colors are just the default MCGA palette.

EDIT: Optimized it, now it draws backwards (same image though), and saved 31 bytes!

EDIT 2: To assuage the OP, I have recreated the binary by hand. By doing so, I also shaved another 4 bytes off. I documented every single step of the process, showing all of my work so anybody can follow along if they really want to, here (warning, it's boring and very long): https://web.archive.org/web/20160308132501/http://lightning.memso.com/media/perm/mandelbrot2.txt

I used a couple regex's in EditPadPro, to find all the ; Final: ... entries in the file and dump them as hex binary to a .com file. The resulting binary is what you see at the top of this post.

\$\endgroup\$
10
  • 6
    \$\begingroup\$ I could hand code the entire thing in binary if it makes it easier for you, but that would be like asking anybody using a high level language to avoid using automatic constructs, macros, etc. That's ALL that assembly really is, just a bunch of macros. The resulting binary to run a full JavaScript, Perl, etc. includes the binary of the library. With ASM, the final hex value is everything, libraries, ALL CODE, included. \$\endgroup\$ Commented Mar 10, 2014 at 18:50
  • 7
    \$\begingroup\$ No. I can hand convert ASM to binary if really necessary. It will come out with the exact same 177 bytes that my assembler helped with. The resulting code can be pasted by anybody with a binary editor into a new file, saved out, 177 bytes, and it will work as expected. Apparently SO is divided on ASM submissions, so maybe you should clarify if you feel it does not count: meta.codegolf.stackexchange.com/questions/260/… \$\endgroup\$ Commented Mar 10, 2014 at 19:14
  • 10
    \$\begingroup\$ Alright, so, to prove this a valid entry, I spent the time it took to hand translate this to binary. I have updated my answer accordingly. \$\endgroup\$ Commented Mar 12, 2014 at 5:34
  • 2
    \$\begingroup\$ Machine code should count if the author actually codes in it directly. I used to code directly in Z80 machine code on my ZX Spectrum in 1983. I didn't even know what an assembler was. I never transferred this skill to any other processor or platform though. That it might be hard to decide who really wrote directly in machine code and who wrote in assembly then lied about it is an entirely separate thing. \$\endgroup\$ Commented Mar 12, 2014 at 12:23
  • 10
    \$\begingroup\$ The thing is, there is no compiler with assembly. You just use macros. Saying that it doesn't count is like saying you can't use any predefined #define statements in C. It's just time consuming to manually replace it all. \$\endgroup\$ Commented Mar 12, 2014 at 16:33
27
\$\begingroup\$

Java, 505 405 324 bytes

Just a standard calculation, with golfitude now with extra golfitude.

enter image description here

Golfed:

import java.awt.*;class M{public static void main(String[]v){new Frame(){public void paint(Graphics g){for(int t,s,n=640,i=n*n;--i>0;g.setColor(new Color(s*820)),g.drawLine(i/n,i%n+28,i/n,i%n),setSize(n,668)){float c=4f/n,a=c*i/n-2,b=i%n*c-2,r=a,e=b,p;for(s=t=99;t-->0&&r*r+e*e<4;s=t,p=r*r-e*e+a,e=r*e*2+b,r=p);}}}.show();}}

With line breaks:

import java.awt.*;
class M{
    public static void main(String[]v){
        new Frame(){
            public void paint(Graphics g){
                for(int t,s,n=640,i=n*n;--i>0;g.setColor(new Color(s*820)),g.drawLine(i/n,i%n+28,i/n,i%n),setSize(n,668)){
                    float c=4f/n,a=c*i/n-2,b=i%n*c-2,r=a,e=b,p;
                    for(s=t=99;t-->0&&r*r+e*e<4;s=t,p=r*r-e*e+a,e=r*e*2+b,r=p);
                }
            }
        }.show();
    }
}
\$\endgroup\$
13
  • \$\begingroup\$ f.setSize(n,668); - depends heavily on the used Theme, but I'll accept it. \$\endgroup\$ Commented Mar 8, 2014 at 9:11
  • \$\begingroup\$ You can drop the imports in Java because they're auto-generated anyway. \$\endgroup\$ Commented Mar 8, 2014 at 9:23
  • \$\begingroup\$ I also see double where float could be used if you tried \$\endgroup\$ Commented Mar 8, 2014 at 9:29
  • \$\begingroup\$ JFrame => Frame shaves off 2 chars. Although you can't close the window anymore. ;) \$\endgroup\$
    – EthanB
    Commented Mar 8, 2014 at 15:01
  • 2
    \$\begingroup\$ Your class do not need to be public. Further, use Java 8 to get rid of the final modifier. And you must not omit the imports in order to be a complete submission. \$\endgroup\$ Commented Mar 10, 2014 at 4:53
21
\$\begingroup\$

Javascript (ECMAScript 6) - 315 308 Characters

document.body.appendChild(e=document.createElement("canvas"));v=e.getContext("2d");i=v.createImageData(e.width=e.height=n=600,n);j=0;k=i.data;f=r=>k[j++]=(n-c)*r%256;for(y=n;y--;)for(x=0;x++<n;){c=s=a=b=0;while(c++<n&&a*a+b*b<5){t=a*a-b*b;b=2*a*b+y*4/n-2;a=t+x*4/n-2}f(87);f(0);f(0);k[j++]=255}v.putImageData(i,0,0)

Default Output

(d=document).body.appendChild(e=d.createElement`canvas`);v=e.getContext`2d`;i=v.createImageData(e.width=e.height=n=600,n);j=0;k=i.data;f=r=>k[j++]=(n-c)*r%256;for(y=n;y--;)for(x=0;x++<n;){c=s=a=b=0;while(c++<n&&a*a+b*b<5){t=a*a-b*b;b=2*a*b+y*4/n-2;a=t+x*4/n-2}f(87);f(0);f(0);k[j++]=255}v.putImageData(i,0,0)

  • Change n to vary the image size (and number of iterations).
  • Change the values passed in the f(87);f(0);f(0); calls (near the end) to change the RGB colour values. (f(8);f(8);f(8); is greyscale.)

With f(8);f(23);f(87);:

enter image description here

(d=document).body.appendChild(e=d.createElement`canvas`);v=e.getContext`2d`;i=v.createImageData(e.width=e.height=n=600,n);j=0;k=i.data;f=r=>k[j++]=(n-c)*r%256;for(y=n;y--;)for(x=0;x++<n;){c=s=a=b=0;while(c++<n&&a*a+b*b<5){t=a*a-b*b;b=2*a*b+y*4/n-2;a=t+x*4/n-2}f(8);f(23);f(87);k[j++]=255}v.putImageData(i,0,0)

\$\endgroup\$
5
  • 2
    \$\begingroup\$ Nice. d=document would save you a few more. (Also, Is there a reason for creating the canvas? Does codegolf assume a certain level of HTML available?) \$\endgroup\$ Commented Mar 10, 2014 at 15:47
  • 1
    \$\begingroup\$ You can write document.createElement`canvas` and save 2 bytes. Same as the getContext`2d` . \$\endgroup\$ Commented Oct 7, 2019 at 10:17
  • \$\begingroup\$ I didn't assume a HTML Canvas as this is a pure JavaScript solution. \$\endgroup\$
    – MT0
    Commented Oct 7, 2019 at 10:42
  • \$\begingroup\$ or you can 'drop' canvas at all like here \$\endgroup\$ Commented Oct 7, 2019 at 19:59
  • \$\begingroup\$ The first image looks like a Virtual Boy game. \$\endgroup\$
    – Joao-3
    Commented Jan 7, 2023 at 19:36
20
\$\begingroup\$

J, 73 bytes

load'viewmat'
(0,?$~99 3)viewmat+/2<|(j./~i:2j479)(+*:) ::(3:)"0^:(i.99)0

mandelbrot set

Edit, some explaining:

x (+*:) y           NB. is x + (y^2)
x (+*:) ::(3:) y    NB. returns 3 when (+*:) fails (NaNs)
j./~i:2j479         NB. a 480x480 table of complex numbers in required range
v =: (j./~i:2j479)(+*:) ::(3:)"0 ]     NB. (rewrite the above as one verb)
v z0                NB. one iteration of the mandelbrot operation (z0 = 0)
v v z0              NB. one iteration on top of the other
(v^:n) z0           NB. the result of the mandelbrot operation, after n iterations
i.99                NB. 0 1 2 3 4 ... 98
(v^:(i.99))0        NB. returns 99 tables, one for each number of iterations
2<| y               NB. returns 1 if 2 < norm(y), 0 otherwise
2<| (v^:(i.99))0    NB. 99 tables of 1s and 0s
+/...               NB. add the tables together, element by element.
NB. we now have one 480x480 table, representing how many times each element exceeded norm-2.
colors viewmat M    NB. draw table 'M' using 'colors'; 'colors' are rgb triplets for each level of 'M'.
$~99 3              NB. 99 triplets of the numbers 99,3
?$~99 3             NB. 99 random triplets in the range 0 - 98 and 0 - 2
0,?$~99 3           NB. prepend the triplet (0,0,0): black
\$\endgroup\$
3
  • 1
    \$\begingroup\$ +1 but would it be possible for you to explain a little how that code work? In particular i m curious to know how (where in the code) does it pick the colors? \$\endgroup\$
    – plannapus
    Commented Mar 10, 2014 at 8:40
  • 1
    \$\begingroup\$ @MarkJeronimus, I can make it 70 but I kept some things for clarity. I, thus, took the liberty to ignore the LF when counting. \$\endgroup\$
    – Eelvex
    Commented Mar 10, 2014 at 12:39
  • \$\begingroup\$ @plannapus, OK, added some comments. The color picking is done with (0,?$~99 3) which produces 100 rgb triplets, one for each level. Because of the randomness, you might get less than 100 triplets so some levels will have a smoother transition (but still have different colors). \$\endgroup\$
    – Eelvex
    Commented Mar 10, 2014 at 13:07
20
\$\begingroup\$

Mathematica, 214 191 215 19 30

Since version 10.0 there is a built-in: (19 bytes)

MandelbrotSetPlot[]

mandelbrot


To conform to the coordinate range requirements, 11 additional bytes are required. (30 bytes)

MandelbrotSetPlot@{-2-2I,2+2I}

m2


A hand-rolled case:

m=Compile[{{c,_Complex}},Length[FixedPointList[#^2+c&,0,99,SameTest→(Abs@#>=2&)]]];
ArrayPlot[Table[m[a+I b],{b,-2,2,.01},{a,-2,2,.01}],DataRange→{{-2,2},{-2,2}},
ColorRules→{100→Black},ColorFunction→(Hue[Log[34,#]]&)]

green

\$\endgroup\$
6
  • \$\begingroup\$ {b, -2, 2, .01}, {a, -2, 2, .01} is shorter and closer to the rules \$\endgroup\$ Commented Mar 7, 2014 at 23:50
  • \$\begingroup\$ @MarkJeronimus Thanks. I used the suggested range for the iterating picture. \$\endgroup\$
    – DavidC
    Commented Mar 8, 2014 at 3:20
  • \$\begingroup\$ You had it almost right, then you made the inside non-black. The last frame in the GIF is black inside and an allowed answer. EDIT: and I count 195 bytes. \$\endgroup\$ Commented Mar 8, 2014 at 9:20
  • \$\begingroup\$ I missed the point about being black. The count increased because some single characters became two characters in the cut-and-paste to SE. \$\endgroup\$
    – DavidC
    Commented Mar 8, 2014 at 13:34
  • \$\begingroup\$ Your built-in solution uses a very loose interpretation of The fractal coordinates range from approximately -2-2i to 2+2i. \$\endgroup\$ Commented Sep 25, 2017 at 12:58
17
\$\begingroup\$

Python with Pylab+Numpy, 151 bytes

I couldn't bear to see a non-DQ'ed Python entry, but I think I really outdid myself on this one, and I made it down to 153 characters!

import numpy as n
from pylab import*
i=99
x,y=n.mgrid[-2:2:999j,-2:2:999j]
c=r=x*1j+y
x-=x
while i:x[(abs(r)>2)&(x==0)]=i;r=r*r+c;i-=1
show(imshow(x))

Also, notably, the second to last line raises 4 distinct runtime warnings, a new personal record!

enter image description here

\$\endgroup\$
7
  • \$\begingroup\$ I count 152. No space is required between import and *, and not defining f at all should be shorter, unless i've misunderstood something, which is possible. You should also change it around such that 0 iterations and 1 iterations are distiguished (they're currently both grey). \$\endgroup\$
    – primo
    Commented Mar 9, 2014 at 14:02
  • \$\begingroup\$ Weird. Does wc include the eof? Fixed and slightly smaller. Just a moment. \$\endgroup\$
    – meawoppl
    Commented Mar 9, 2014 at 17:46
  • \$\begingroup\$ I get 151 with wc. First golf, so not sure how to score it. \$\endgroup\$
    – meawoppl
    Commented Mar 9, 2014 at 18:01
  • \$\begingroup\$ I count 150, without trailing newline. Some interpreters/compilers demand one, but the python interpreter does fine without. Not sure about wc, but maybe try stat -c %s instead. Are the black upper and lower borders part of the image? \$\endgroup\$
    – primo
    Commented Mar 9, 2014 at 18:09
  • 1
    \$\begingroup\$ You can save 1 character by using from numpy import* instead of import numpy as n and mgrid instead of n.mgrid. \$\endgroup\$
    – user344
    Commented Jun 15, 2014 at 15:15
16
\$\begingroup\$

C + Allegro 4.2.2 - 248 bytes

#include<allegro.h>
x=-1,y,K=400;float a,h,c,d,k;main(i){set_gfx_mode('SAFE',K,K,allegro_init(),0);while(x++<K)
for(y=0;y<K;y++){for(a=h=i=0;a*a+h*h<4&&++i<256;k=a,a=a*a-h*h+x*0.01-2,h=2*k*h+y*0.01-2);
putpixel(screen,x,y,i);}while(1);}END_OF_MAIN()

Output:

MSet 1

\$\endgroup\$
8
  • \$\begingroup\$ You should mention that this is Allegro 4 (which is quite different from Allegro 5). Which exact version is this? \$\endgroup\$ Commented Mar 10, 2014 at 4:57
  • \$\begingroup\$ it's either 246 or 249 long \$\endgroup\$ Commented Mar 10, 2014 at 6:42
  • \$\begingroup\$ @Victor Allegro 4.2.2. \$\endgroup\$
    – Oberon
    Commented Mar 10, 2014 at 14:34
  • 1
    \$\begingroup\$ @MarkJeronimus Isn't there supposed to be a newline between ... allegro.h> and x=-1, ...? I suppose Notepad++ counts it as \r\n = 0D 0A. \$\endgroup\$
    – Oberon
    Commented Mar 10, 2014 at 21:05
  • 2
    \$\begingroup\$ I think 0.01 can be typed as .01. \$\endgroup\$
    – Yytsi
    Commented Jun 21, 2016 at 9:53
15
\$\begingroup\$

Windows PowerShell (v4), 299 bytes

Mandelbrot fractal picture

# Linewrapped here for show:

$M='System.Windows.Forms';nal n New-Object;Add-Type -A System.Drawing,$M;(
$a=n "$M.Form").backgroundimage=($b=n Drawing.Bitmap 300,300);0..299|%{
$r=$_;0..299|%{$i=99;$k=$C=n numerics.complex($_/75-2),($r/75-2);while(((
$k=$k*$k).Magnitude-lt4)-and$i--){$k+=$C}$b.SetPixel($_,$r,-5e6*++$i)}};$a.Show()


# The single line 299 char entry version:

$M='System.Windows.Forms';nal n New-Object;Add-Type -A System.Drawing,$M;($a=n "$M.Form").backgroundimage=($b=n Drawing.Bitmap 300,300);0..299|%{$r=$_;0..299|%{$i=99;$k=$C=n numerics.complex($_/75-2),($r/75-2);while((($k=$k*$k).Magnitude-lt4)-and$i--){$k+=$C}$b.SetPixel($_,$r,-5e6*++$i)}};$a.Show()

Instructions

  • Run a normal PowerShell console (ISE might not work)
  • Copy/paste code in, press Enter
  • Wait - it takes a minute or more to run
  • The only way to quit is to close the console

Comment

  • There's a tiny bit of rule-testing going on with the colours inside the set; the rules say "The other pixels (presumably inside the Mandelbrot set) must be colored either black or white'"; the code is colouring the pixels completely black RGB(0,0,0) ... it just happens to be a transparent black RGBA(0,0,0,0). So what shows up is the form background colour of the current Windows theme, a slightly off-white RGB(240,240,240) in this case.
\$\endgroup\$
3
  • \$\begingroup\$ I'd change lt2 to lt4 to make it a "mandelbrot set" instead of the image you have now, Many points of the set are swallowed up by the color bands. \$\endgroup\$ Commented Sep 30, 2015 at 11:05
  • \$\begingroup\$ Apparently Magnitude is a*a+b*b not sqrt(a*a+b*b) \$\endgroup\$ Commented Sep 30, 2015 at 13:41
  • \$\begingroup\$ I thought I tested that earlier, but I went in search of an answer to "where's the horizontal line on the left gone?", and after a bit, I found it exactly where you said, -lt4. Which is good - thank you. I've updated my answer with corrected code, and image. (Will have to rethink my understanding of what it's doing, since I'm missing something). \$\endgroup\$ Commented Sep 30, 2015 at 18:02
15
\$\begingroup\$

Python + PIL, 166 bytes

import Image
d=600;i=Image.new('RGB',(d,d))
for x in range(d*d):
 z=o=x/9e4-2-x%d/150.j-2j;c=99
 while(abs(z)<2)*c:z=z*z+o;c-=1
 i.putpixel((x/d,x%d),5**8*c)
i.show()

Output (will open in the default *.bmp viewer):

\$\endgroup\$
7
  • 1
    \$\begingroup\$ You can shave off 3 if you get rid of the y loop. r=range(d*d), use x/d and x%d for x and y. \$\endgroup\$
    – Geobits
    Commented Mar 8, 2014 at 16:15
  • \$\begingroup\$ @Geobits the idea actually saved 10, thanks! \$\endgroup\$
    – primo
    Commented Mar 8, 2014 at 16:28
  • 1
    \$\begingroup\$ Complex types can be initialized like: c=1+2j, which I think would save you a couple characters with: z=o=x/9e4-2+(x%d/150.-2)*1j;c=99 \$\endgroup\$
    – meawoppl
    Commented Mar 9, 2014 at 3:10
  • \$\begingroup\$ @meawoppl another 7 :D \$\endgroup\$
    – primo
    Commented Mar 9, 2014 at 5:07
  • \$\begingroup\$ Technically disallowed: this doesn't any graphical output feature of Python itself (and Image.show() implicitly saves a temporary file). \$\endgroup\$
    – nneonneo
    Commented Mar 9, 2014 at 5:51
13
\$\begingroup\$

BBC Basic (228 bytes)

What about languages that nobody ever heard of in code golf? Most likely could be optimized, but I'm not quite where - improvements possible. Based of http://rosettacode.org/wiki/Mandelbrot_set#BBC_BASIC, but I tried to code golf it as much as possible.

VDU23,22,300;300;8,8,8,8
ORIGIN0,300
GCOL1
FORX=0TO600STEP2
i=X/200-2
FORY=0TO300STEP2
j=Y/200
x=0
y=0
FORI=1TO128
IFx*x+y*y>4EXIT FOR
t=i+x*x-y*y
y=j+2*x*y
x=t
NEXT
COLOUR1,I*8,I*4,0
PLOTX,Y:PLOTX,-Y
NEXT
NEXT

The generated Mandelbrot set

The > symbol on image is prompt, and it's automatically generated after running the program.

\$\endgroup\$
1
  • \$\begingroup\$ No need to plot twice, just go with a more inefficient version. Doesn;t it support NEXT Y,X? \$\endgroup\$ Commented Mar 9, 2014 at 11:59
12
\$\begingroup\$

APL, 194 chars/bytes*

m←{1{⍺=99:0⋄2<|⍵:⍺⋄(⍺+1)∇c+⍵*2}c←⍵}¨⍉v∘.+0j1×v←¯2+4÷s÷⍳s←640
'F'⎕WC'Form'('Coord' 'Pixel')('Size'(s s))
'B'⎕WC'Bitmap'('CMap'(0,,⍨⍪0,15+10×⍳24))('Bits'(24⌊m))
'F.I'⎕WC'Image'(0 0)('Picture' 'B')

This is for Dyalog APL with ⎕IO ⎕ML←1 3

Most of the space is taken by API calls to show a bitmap in a window (lines 2, 3, 4)
If there was a shortcut to do it, the code would be down to 60 chars (line 1)

Ungolfed version (only line 1)

s←640            ⍝ size of the bitmap
v←(4×(⍳s)÷s)-2   ⍝ vector of s reals, uniform between ¯2 and 2
m←(0j1×v)∘.+v    ⍝ square matrix of complex numbers from ¯2j¯2 to 2j2
m←{              ⍝ transform each number in matrix m according to the following
  1{             ⍝   function that takes iteration counter as ⍺ and current value as ⍵
    ⍺=99: 0      ⍝     if we have done 99 iterations, return 0
    2<|⍵: ⍺      ⍝     if |⍵| > 2 return the number of iterations done
    (⍺+1)∇c+⍵*2  ⍝     otherwise, increment the iterations and recurse with the new value
  }c←⍵           ⍝   save the initial value as c
}¨m    

Screenshot:

screenshot

*: Dyalog has its own single byte charset, with the APL symbols mapped to the upper 128 byte values, so the entire code can be stored in 194 bytes.

\$\endgroup\$
10
\$\begingroup\$

R, 199 211 characters

Old solution at 199 characters:

r=seq(-2,2,l=500);c=t(sapply(r,function(x)x+1i*r));d=z=array(0,dim(c));a=1:25e4;for(i in 1:99){z[a]=c[a]+z[a]^2;s=abs(z[a])<=2;d[a[!s]]=i;a=a[s]};image(d,b=0:99,c=c(1,sample(rainbow(98))),ax=F,asp=1)

With indentation:

r=seq(-2,2,l=500)
c=t(sapply(r,function(x)x+1i*r)) #Produces the initial imaginary number matrix
d=z=array(0,dim(c)) #empty matrices of same size as c 
a=1:25e4            #(z will store the magnitude, d the number of iterations before it reaches 2)
for(i in 1:99){     #99 iterations
    z[a]=c[a]+z[a]^2
    s=abs(z[a])<=2
    d[a[!s]]=i
    a=a[s]
    }
image(d,b=0:99,c=c(1,sample(rainbow(98))),ax=F,asp=1) #Colors are randomly ordered (except for value 0)

enter image description here

Edit: Solution at 211 characters that colors the inside of the set and the outside of the first layer differently:

r=seq(-2,2,l=500);c=t(sapply(r,function(x)x+1i*r));d=z=array(0,dim(c));a=1:25e4;for(i in 1:99){z[a]=c[a]+z[a]^2;s=abs(z[a])<=2;d[a[!s]]=i;a=a[s]};d[a[s]]=-1;image(d,b=-1:99,c=c(1:0,sample(rainbow(98))),ax=F,asp=1)

With indentation:

r=seq(-2,2,l=500)
c=t(sapply(r,function(x)x+1i*r))
d=z=array(0,dim(c))
a=1:25e4
for(i in 1:99){
    z[a]=c[a]+z[a]^2
    s=abs(z[a])<=2
    d[a[!s]]=i
    a=a[s]
    }
d[a[s]]=-1 #Gives the inside of the set the value -1 to differenciate it from value 0.
image(d,b=-1:99,c=c(1,sample(rainbow(99))),ax=F,asp=1)

enter image description here

\$\endgroup\$
5
  • \$\begingroup\$ technically black on the outside is disallowed. Did you miss that or is it difficult to implement? \$\endgroup\$ Commented Mar 8, 2014 at 9:19
  • \$\begingroup\$ @MarkJeronimus both actually :) I'll try to have a look of how to do that but i m not 100% confident that I ll find a way to do that cleanly. \$\endgroup\$
    – plannapus
    Commented Mar 8, 2014 at 9:20
  • \$\begingroup\$ @MarkJeronimus Done! \$\endgroup\$
    – plannapus
    Commented Mar 8, 2014 at 9:40
  • 5
    \$\begingroup\$ Second place in the hideous colors division. \$\endgroup\$
    – meawoppl
    Commented Mar 9, 2014 at 10:02
  • 1
    \$\begingroup\$ @meawoppl blame rainbow() :) \$\endgroup\$
    – plannapus
    Commented Mar 10, 2014 at 7:23
10
\$\begingroup\$

TI-80 BASIC, 125 106 bytes

ZDECIMAL
FOR(Y,-2,2,.1
FOR(X,-2,2,.1
0->S
0->T
1->N
LBL N
N+1->N
IF S²+T²≥4
GOTO B
S²-T²+X->I
2ST+Y->T
I->S
IF N<20
GOTO N
LBL B
IF FPART (N/2
PT-ON(X,Y
END
END

Based on Digital Trauma's answer. enter image description here

\$\endgroup\$
10
\$\begingroup\$

Mathematica 10.0, 19 chars

MandelbrotSetPlot[]

MandelbrotSetPlot is a new function in Mathematica 10.0.

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ How convenient, that this built in function just happens to satisfy all my requirements (except location, which can be set with 13 more characters). Except this is a standard loophole. \$\endgroup\$ Commented Mar 30, 2014 at 10:03
  • 24
    \$\begingroup\$ Code golf is generally won by specialist languages with single-character tokens, or by systems like Mathematica that have a massive number of special functions built-in. To use them is not cheating, any more than using single-character commands would be in APL. \$\endgroup\$ Commented Apr 17, 2014 at 15:08
10
\$\begingroup\$

Java - Processing (271 bytes)

void setup(){int h=100,e=5*h,i;float d,v,w,a,b,c;size(e,e);colorMode(HSB,h);loadPixels();d=4./e;v=2;for(int x=1;x<=e;x++){v-=d;w=2;for(int y=0;y<e;){w-=d;a=b=c=0;i=-1;while(a*a+b*b<4&&++i<h){c=a*a-b*b+v;b=2*a*b+w;a=c;}pixels[e*++y-x]=color(i*9%h,h,h-i);}}updatePixels();}

Expanded:

void setup(){
  int h=100, e=5*h, i; //init of size "e", max hue "h", iterator "i"
  float d,v,w,a,b,c; //init of stepwidth "d", y-coord "v", x-coord "w", Re(z) "a", Im(z) "b", temp_a "c"
  size(e,e);
  colorMode(HSB,h);
  loadPixels();
  d = 4./e;
  v = 2;
  for(int x = 1; x <= e; x++){
    v -= d;
    w = 2;
    for(int y = 0; y < e;){
      w -= d;
      a = b = c = 0;
      i = -1;
      while(a*a + b*b < 4 && ++i < h){
        c = a*a - b*b + v;
        b = 2*a*b + w;
        a = c;
      }
      pixels[e * ++y - x] = color(i*9 % h, h, h-i);
    }
  }
  updatePixels();
}

\$\endgroup\$
1
  • \$\begingroup\$ Aw, maaaaan, I wanted to do this. +1. \$\endgroup\$ Commented Sep 28, 2017 at 12:57
8
\$\begingroup\$

Applesoft BASIC, 302 286 280 bytes

This picks random points to draw, so it will run forever and may never fill in the full plane.

1HGR:POKE49234,0:DIMco(10):FORc=0TO10:READd:co(c)=d:NEXT:DATA1,2,3,5,6,1,2,3,5,6,0
2x=INT(RND(1)*280):y=INT(RND(1)*96):x1=x/280*3-2:y1=y/191*2-1:i=0:s=x1:t=y1
3s1=s*s-t*t+x1:t=2*s*t+y1:s=s1:i=i+1:IFs*s+t*t<4ANDi<20THENGOTO3
4c=co(i/2):IFc THENHCOLOR=c:HPLOTx,y:HPLOTx,191-y
5GOTO2

Turns out Applesoft BASIC is really forgiving about lack of spaces. Only one space is necessary in the entire program.

Output after 14 hours:

        output

GIF:

        gif

Before golfing:

10 HGR : POKE 49234,0
20 DIM co(10) : FOR c = 0 TO 10 : READ d : co(c) = d : NEXT
30 DATA 1, 2, 3, 5, 6, 1, 2, 3, 5, 6, 0
100 x = INT(RND(1) * 280) : y = INT(RND(1) * 96)
110 x1 = x / 280 * 3 - 2 : y1 = y / 191 * 2 - 1
120 i = 0:s = x1:t = y1
130 s1 = s * s - t * t + x1
140 t = 2 * s * t + y1:s = s1: i = i + 1
150 IF s * s + t * t < 4 AND i < 20 THEN GOTO 130
160 c = co(i/2) : IF c THEN HCOLOR= c : HPLOT x,y : HPLOT x,191 - y
170 GOTO 100

Note: POKE 49234,0 (in Applesoft BASIC) puts the machine into full graphics mode.

A version optimized for B&W displays:

110 HGR:POKE 49234,0:HCOLOR=3
120 FOR x = 0 TO 279:FOR y = 0 TO 95
130 x1 = x / 280 * 3 - 2:y1 = y / 191 * 2 - 1
140 i = 0:s = x1:t = y1:c = 0
150 s1 = s * s - t * t + x1
160 t = 2 * s * t + y1:s = s1:c = 1 - c:i = i + 1
170 IF s * s + t * t < 4 AND i < 117 THEN GOTO 150
180 IF c = 0 THEN HPLOT x,y:HPLOT x,191 - y
190 NEXT:NEXT

Output after 12 hours:

        b&w

A version that will work in GW-BASIC (DOS):

5 CLS
6 SCREEN 1
20 DIM co(10) : FOR c = 0 TO 10 : READ d : co(c) = d : NEXT
30 DATA 1, 2, 3, 5, 6, 1, 2, 3, 5, 6, 0
100 x = INT(RND(1) * 280) : y = INT(RND(1) * 96)
110 x1 = x / 280 * 3 - 2 : y1 = y / 191 * 2 - 1
120 i = 0 : s = x1 : t = y1
130 s1 = s * s - t * t + x1
140 t = 2 * s * t + y1 : s = s1 : i = i + 1
150 IF s * s + t * t < 4 AND i < 20 THEN GOTO 130
160 c = co(i/2) : PSET (x,y),C : PSET (x,191 - y),C
170 GOTO 100
\$\endgroup\$
2
  • \$\begingroup\$ Would be smaller (and slower) if you didn't plot two pixels at once but choose a random pixel on the entire screen. \$\endgroup\$ Commented May 31, 2017 at 6:09
  • 1
    \$\begingroup\$ @MarkJeronimus It's already so slow that the color version has not yet finished, after posting this 5 days ago. I don't think I can afford it to be any slower :P \$\endgroup\$
    – MD XF
    Commented May 31, 2017 at 20:11
8
\$\begingroup\$

R, 140 136 128 124 123 110 109 bytes*

image(outer(j<-1:396/99-2,j,Vectorize(function(x,y,n=99){while((n=n-1)&abs(F)<2)F=F*F+x+1i*y;n})),c=colors())

Try it at rdrr.io

6y after the Q was asked, but I love Mandelbrot sets, and the earlier R solution was 211 characters...

enter image description here

*Or just 62 bytes as a pixel-shader function in R≥4.1:
\(x,y,n=99){while((n=n-1)&abs(F)<2)F=F*F+x+1i*y;colors()[n+1]}

\$\endgroup\$
1
  • 2
    \$\begingroup\$ No problem, submissions are always welcome. En welkom. \$\endgroup\$ Commented May 19, 2020 at 21:03
7
\$\begingroup\$

JavaScript, 284 283 ...(12 improvements)... 191 190 188 199 bytes

currenty (2019-10-05) this is shortest js solution (the shortest one before this has 285 bytes)

enter image description here

Expanded

// Fractal calculations
c=512;

// p has pixels, p+= allways join 4 digit number implicit casted to string
// 4 char string interpred as base64 gives 3bytes = 1 RGB pixel
for( p=i=''; j=x=y=0,++i<=c*c; p+= j<c ? 9*c+9*j : 'AAAA' ) 
  while( x*x+y*y<4 && ++j-c )
    [x,y] = [ x*x-y*y+i%c/128-2, 2*x*y+i/c/128-2 ];
  
// draw pixels in 512x512 BMP base64 image
document.write("<img src=data:;base64,Qk0bAAwAAAAAABsAAAAMAAAAAAIAAgEAGAAA"+p)

After small constans changes

c=512;for(p=i='';j=x=y=0,++i<=c*c;p+=j<c?2*c+9*j:'AAAA')while(x*x+y*y<4&&++j-c)[x,y]=[x*x-y*y+i%c/128-2,2*x*y+i/c/128-2]
document.write("<img src=data:;base64,Qk0bAAwAAAAAABsAAAAMAAAAAAIAAgEAGAAA"+p)

\$\endgroup\$
3
  • \$\begingroup\$ This is very cool, but the pixels inside the set should be black or white. I'm not sure what either of those are in base 64 \$\endgroup\$ Commented Apr 22, 2022 at 0:11
  • \$\begingroup\$ @MatthewJensen you are right - AAAA - fixed \$\endgroup\$ Commented May 5, 2022 at 15:48
  • 1
    \$\begingroup\$ A few minor changes gets 194 bytes for(c=i=512,p='<img src=data:;base64,Qk0bAAwAAAAAABsAAAAMAAAAAAIAAgEAGAAA';i++<c*c;p+=j<c?9*c+9*j:'AAAA')for(j=x=y=0;x*x+y*y<4&&++j-c;)[x,y]=[x*x-y*y+i%c/128-2,2*x*y+i/c/128-2];document.write(p) \$\endgroup\$ Commented Jun 16, 2022 at 1:01
6
\$\begingroup\$

GLSL - 225 bytes:

void main(){vec2 c=gl_FragCoord.xy/iResolution.y*4.-2.,z=c,v;for(int i=0;i<99;i++){z=vec2(z.x*z.x-z.y*z.y,2.*z.x*z.y)+c;if(length(z)>2.&&v.y<1.)v=vec2(float(i)/99.,1.);}gl_FragColor=(v.y<1.)?vec4(v,v):texture2D(iChannel0,v);}

Defining variables in the code (242 bytes):

uniform vec3 r;uniform sampler2D t;void main(){vec2 c=gl_FragCoord.xy/r.y*4.-2.,z=c,v;for(int i=0;i<99;i++){z=vec2(z.x*z.x-z.y*z.y,2.*z.x*z.y)+c;if(length(z)>2.&&v.y<1.)v=vec2(float(i)/99.,1.);}gl_FragColor=(v.y<1.)?vec4(v,v):texture2D(t,v);}

See it in ShaderToy

Mandelbrot Golf

This requires a suitable palette texture be loaded as iChannel0. (The colouring here is from the "random pixel" texture on ShaderToy).

\$\endgroup\$
3
  • \$\begingroup\$ Variable declarations should be counted too, unless they can be auto-generated from the code. (color scheme is fine if it's only available as an external setting) \$\endgroup\$ Commented Mar 10, 2014 at 6:40
  • \$\begingroup\$ @MarkJeronimus: For the ShaderToy environment, these variables are fixed. Otherwise, for standard shaders, I would have picked shorter variable names. \$\endgroup\$
    – nneonneo
    Commented Mar 10, 2014 at 7:25
  • \$\begingroup\$ Is this the fastest of them all? \$\endgroup\$
    – Demi
    Commented Aug 3, 2015 at 7:15
6
\$\begingroup\$

Octave (212 136 bytes)

(Now including some ideas due to @ChrisTaylor.)

[y,x]=ndgrid(-2:.01:2);z=c=x+i*y;m=c-c;for n=0:99;m+=abs(z)<2;z=z.^2+c;end;imagesc(m);colormap([hsv(128)(1+mod(0:79:7890,128),:);0,0,0])

With whitespace:

[y,x] = ndgrid(-2:.01:2);
z = c = x + i*y;
m = c-c;
for n=0:99
    m += abs(z)<2;
    z = z.^2 + c;
end
imagesc(m)
colormap([hsv(128)(1+mod(0:79:7900,128),:);
          0,0,0])

Output:

Mandelbrot steps to abs(z)>2

To convert to Matlab, change "m+=abs(z)<2" to "m=m+(abs(z)<2)". [+3 bytes]

To make the aspect ratio 1:1, add ";axis image". [+11 bytes]

My first answer (212 bytes):

[x,y]=meshgrid(-2:.01:2);z=c=x+i*y;m=0*e(401);for n=0:99;m+=abs(z)<2;z=z.^2+c;endfor;t=[0*e(1,7);2.^[6:-1:0]];[s{1:7}]=ndgrid(num2cell(t,1){:});t=1+sum(cat(8,s{:}),8);imagesc(m);colormap([hsv(128)(t(:),:);0,0,0])
\$\endgroup\$
6
  • \$\begingroup\$ There's probably a shorter way to get a discontinuous colormap.... \$\endgroup\$
    – aschepler
    Commented Mar 8, 2014 at 20:40
  • \$\begingroup\$ Yes, much better now. \$\endgroup\$
    – aschepler
    Commented Mar 9, 2014 at 1:13
  • \$\begingroup\$ +1 nice and concise solution. But your aspect ratio is not 1:1 (cf. rule n°2: output should be square). \$\endgroup\$
    – plannapus
    Commented Mar 10, 2014 at 7:39
  • \$\begingroup\$ Fixing the aspect ratio will take 11 more bytes: append ";axis image". Is that required to qualify? \$\endgroup\$
    – aschepler
    Commented Mar 10, 2014 at 14:39
  • \$\begingroup\$ i think it was just me nitpicking :) , the OP doesn't seem to have a problem with it since he didn't say anything. \$\endgroup\$
    – plannapus
    Commented Mar 11, 2014 at 7:46
6
\$\begingroup\$

Excel VBA, 251 246 224 223 221 bytes

Saved 5 bytes thanks to ceilingcat Saved 23 bytes thanks to Taylor Scott

Sub m
D=99
For x=1To 4*D
For y=1To 4*D
p=0
q=0
For j=1To 98
c=2*p*q
p=p^2-q^2-2+(x-1)/D
q=c+2+(1-y)/D
If p^2+q^2>=4Then Exit For
Next
j=-j*(j<D)
Cells(y,x).Interior.Color=Rnd(-j)*1E6*j/D
Next y,x
Cells.RowHeight=48
End Sub

Output:

Output with D = 99

I made a version that did this a long time ago but it had a lot of extras like letting the user pick the basic color and easy-to-follow math. Golfing it way down was an interesting challenge. The Color method uses 1E6 as a means to get a wide range of colors since the valid colors are 0 to 2^24. Setting it to 10^6 gave nice contrast areas.

Explanation / Auto-Formatting:

Sub m()

    'D determines the number of pixels and is factored in a few times throughout
    D = 99
    For x = 1 To 4 * D
    For y = 1 To 4 * D
        'Test to see if it escapes
        'Use p for the real part and q for the imaginary
        p = 0
        q = 0
        For j = 1 To 98
            'This is a golfed down version of complex number math that started as separate generic functions for add, multiple, and modulus
            c = 2 * p * q
            p = p ^ 2 - q ^ 2 - 2 + (x - 1) / D
            q = c + 2 + (1 - y) / D
            If p ^ 2 + q ^ 2 >= 4 Then Exit For
        Next

        'Correct for no escape
        j = -j * (j < D)

        'Store the results
        'Rnd() with a negative input is deterministic
        'This is what gives us the distinct color bands
        Cells(y, x).Interior.Color = Rnd(-j) * 1000000# * j / D

    Next x, y

    'Resize for pixel art
    Cells.RowHeight = 48

End Sub

I also played around with D=999 and j=1 to 998 to get a much larger and more precise image. The results are irrelevant to the challenge because they're way too large but they are neat.

D=999

\$\endgroup\$
5
  • \$\begingroup\$ @ceilingcat Thanks. That was a carryover from my original which had specialized functions for complex number math. \$\endgroup\$ Commented Sep 27, 2017 at 12:07
  • \$\begingroup\$ Does it really need to be >=4 or can you get away with >4? Also, can replace j<99 with j<D. \$\endgroup\$
    – ceilingcat
    Commented Sep 27, 2017 at 22:38
  • \$\begingroup\$ @EngineerToast you can drop the () from the sub name,you should change (j<99) to (j<d) and you can, for the purpose of making square cells use only Cells.RowHeight=48 in place of the Cells.RowHeight=9,Cells.ColumnWidth=1 - this does making messing around with your output more difficult but has been accepted as valid by the community - \$\endgroup\$ Commented Sep 28, 2017 at 5:03
  • 1
    \$\begingroup\$ @TaylorScott I remember the RowHeight trick from the VBA tips post and had meant to integrate it after I got all my pretty pictures. That was a nice chunk'o'bytes, thanks. \$\endgroup\$ Commented Sep 28, 2017 at 12:31
  • \$\begingroup\$ I believe that you can get away with removing another byte by swapping 2^20 with 1E6 \$\endgroup\$ Commented Oct 8, 2017 at 17:55
6
\$\begingroup\$

PostScript, 290 285 282 271 261 bytes

Screenshot:

screeshot

Golfed code:

-2 .01 2{/a exch def -2 .01 2{/b exch def/x 0 def/y 0 def 0 1 99{/i exch def x x mul y y mul add 4 ge{i log 2 div 1 1 sethsbcolor a 2 add 100 mul b 2 add 100 mul 1 1 rectfill exit}if/x x x mul y y mul sub a add/y 2 x mul y mul b add def def}for}for}for showpage

Ungolfed code:

-2 .01 2 {
    /a exch def
    -2 .01 2 {
        /b exch def
        /x 0 def
        /y 0 def
        0 1 99 {
           /i exch def                       % iteration count
           x x mul y y mul add 4 ge {
               i log 2 div 1 1 sethsbcolor   % rainbow colors outside
               a 2 add 100 mul
               b 2 add 100 mul 1 1 rectfill
               exit
           } if
           /x x x mul y y mul sub a add
           /y 2 x mul y mul b add
           def def
       } for
    } for
} for
showpage
\$\endgroup\$
1
  • \$\begingroup\$ Golfed some more for 198 bytes: [/d{-2 .01 2}(]){exch def}([){mul}/p{add}/s{2 p 100[}/q{x x[y y[}>>begin d{/a]d{/b]0/x 0/y]]0 1 99{q p 4 ge{log .5[1 1 sethsbcolor a s b s 1 1 rectfill exit}if q sub a p 2 x[y[b p/y]/x]=}for}for}for \$\endgroup\$ Commented May 5, 2022 at 16:51
6
\$\begingroup\$

Mindustry, 492 bytes

set x
set y
op div v x 44
op sub v v 2
op div w y 44
op sub w w 2
set a
set b
set n
op pow c a 2
op pow d b 2
op sub A c d
op mul B 2 a
op add a A v
op mul B B
op add b B w
op pow c a 2
op pow d b 2
op add s c d
op sqrt r s
op add n n 1
jump 23 greaterThan r 2
jump 11 lessThan n 100
op mul r n 42
op mod r r 256
op mul g n 83
op mod g g 256
op mul b n 160
op mod b b 256
draw color r g b
draw rect x y 1 1
drawflush
op add x x 1
jump 2 lessThan x 176
op add y y 1
set x
jump 2 lessThan y 176

How the result looks: Setup of the blocks

What this is

Mindustry contains, among many other things, a few blocks related to running computations (processors, memory cells, switches, screens, messages). The above code is the "assembly" language of the processor (bottom left).

The resolution of the image drawn is 176x176, because this is the largest possible screen block (the other is 80x80). The processor can only process up to 1500 operations per second, so the program takes at least an hour to run.

How to run it

The code should be copied into the processor block, which must be connected with the display. Ideally, this should be constructed on a sandbox map.

Scoring

In the in-game program editor, this would be shown as 37 assembly blocks, but since it is stored as text internally, the byte count of the textual representation is used to score it.

Golfing

Since this language is assembly-like and its text representation is not designed to take few bytes, a lot of things are quite verbose. The main way in which this program is golfed is by removing the last argument to many of these calls if it is the default value for this call, since it will be inserted automatically. For example, set x 0 can be set x and when copying it into the processor, the 0 is added.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Very cool "processor"! Looking at the game and source code, the program is stored in plain text, that is to say the score is the sum of the characters that make it up, including newlines. Also, no matter what size display is in the game, at least one display is always allowed by the rule exception. \$\endgroup\$ Commented Oct 11 at 21:21
  • \$\begingroup\$ I've updated my answer to incorporate your feedback. \$\endgroup\$
    – Dornteufel
    Commented Oct 12 at 20:30
5
\$\begingroup\$

gnuplot 110 (105 without newlines)

Obligatory gnuplot entry. It's been done countless times but this one is from scratch (not that it's difficult). I like how gnuplot golfs its commands intrinsically :)

f(z,w,n)=abs(z)>2||!n?n:f(z*z+w,w,n-1)
se vi map
se si sq
se isos 256
sp [-2:2] [-2:2] f(0,x+y*{0,1},99) w pm

ungolfed:

f(z,w,n)=abs(z)>2||n==0?n:f(z*z+w,w,n-1)
set view map
set size square
set isosamples 256
splot [-2:2] [-2:2] f(0,x*{1,0}+y*{0,1},99) with pm3d

However, I'm DEEPLY disappointed at the entry of complex numbers. x*{1,0}+y*{0,1} must be the saddest existing way of constructing a complex number.

Oops, the image: gnuplot mandelbrot

Set isosamples higher for better resolution. We could also say unset tics and unset colorbox for a pure image, but I think this version qualifies just fine.

\$\endgroup\$
2
  • \$\begingroup\$ Bet it's copy/pasta from the first google hit "gnuplot mandel". For starters, *{1,0} is unity and is more like a code-bowling way of saying *1, and probably can be dropped. (untested) \$\endgroup\$ Commented Mar 10, 2014 at 6:26
  • 1
    \$\begingroup\$ No it's not a copy-paste. It's a very straight forward formula and it wasn't even necessary to search for it. However, I did find the pages you get with that search when I was looking for a better way of initializing complex numbers (their implementation is different, well, as different as it can be in this case). Thanks for the tip about the real part, it works. Fixing. \$\endgroup\$
    – orion
    Commented Mar 10, 2014 at 7:53
5
\$\begingroup\$

Matlab (89 bytes)

[X,Y]=ndgrid(-2:.01:2);C=X+i*Y;Z=C-C;K=Z;
for j=1:99,Z=Z.*Z+C;K=K+(abs(Z)<2);end,imagesc(K)

Output -

enter image description here

Doesn't satisfy the requirement that the inner cells must be black or white, but that can be satisfied by either (1) using imshow(K) instead of imagesc(K) (requires 1 fewer byte but needs the image processing toolbox) or (2) appending colormap hot (requires 12 more bytes).

Ungolfed version -

Z = zeros(N);
K = Z;

[X,Y]=ndgrid(-2:.01:2);
C = X+1i*Y;

for j = 1:99
  Z = Z.*Z + C;
  K(K==0 & abs(Z) > 2) = j;
end

imagesc(K)
\$\endgroup\$
7
  • \$\begingroup\$ Using a library is fine if it's packaged in Matlab by default and any user can guess it's being used from the code or the error messages. \$\endgroup\$ Commented Mar 10, 2014 at 19:19
  • \$\begingroup\$ Nice, you beat me. I like the C-C in place of my 0*e(401). Plus, you're not using N. And we can get a little shorter using my m+=abs(z)<2 idea in place of your K(~K&abs(Z)>2)=j. \$\endgroup\$
    – aschepler
    Commented Mar 11, 2014 at 12:33
  • \$\begingroup\$ The default colormap jet and colormap hot are both incorrect though - they only have 64 distinct colors. colormap(hot(101)) doesn't look visually distinguishable to me. colormap([0,0,0;jet(100)]) is maybe acceptable but iffy. \$\endgroup\$
    – aschepler
    Commented Mar 11, 2014 at 12:39
  • \$\begingroup\$ Does that work? On Octave, K=K+abs(Z)<2 means K=((K+abs(Z))<2). (So I was wrong about the one byte estimate to eliminate +=.) \$\endgroup\$
    – aschepler
    Commented Mar 11, 2014 at 13:06
  • 2
    \$\begingroup\$ The Mandelbrot set rotated by 90 degrees is still the Mandelbrot set. \$\endgroup\$ Commented Mar 5, 2015 at 23:19
5
\$\begingroup\$

Floater, 620 pixels

A language I made up when I got inspired by my own challenge, as well as from the esoteric language Piet.

enter image description here

\$\endgroup\$
1
  • 3
    \$\begingroup\$ Link to a language and description of the code? Or actually, what is the code? \$\endgroup\$
    – MD XF
    Commented Sep 26, 2017 at 3:20
4
\$\begingroup\$

JavaScript + HTML5 (356B)

(Note: lines ending with '//' are added here for some readability)

Performant version (375B):

<body onload='var
w,h=w=C.width=C.height=500,X=C.getContext("2d"),I=X.createImageData(w,h),D=I.data, //
y=0,f=255,T=setInterval(function(x,i,j,k,l,c,o){for(x=0;x<w;){                     //
for(i=x*4/w-2,j=y*4/h-2,k=l=0,c=f;--c&&k*k+l*l<4;)t=k*k-l*l+i,l=2*k*l+j,k=t
D[o=(y*w+x++)*4]=(c*=0xc0ffeeee)&f
D[++o]=c>>8&f
D[++o]=c>>16&f
D[++o]=f}X.putImageData(I,0,0)
++y-h||clearInterval(T)},0)'><canvas id=C>

Slow version (356B): remove the 'var' and parameters in the inner function so that the global scope is used.

Try it out: http://jsfiddle.net/neuroburn/Bc8Rh/

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ Forgive me if I don't understand your instructions on making the short version. \$\endgroup\$ Commented Mar 8, 2014 at 21:33
  • \$\begingroup\$ No problem. Remove var w, at the beginning, and change function(x,i,j,k,l,c,o) to function(). \$\endgroup\$ Commented Mar 8, 2014 at 21:35

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