Mandelbrot image in every language

91

34

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 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>

Mark Jeronimus

Posted 2014-03-07T21:39:15.050

Reputation: 6 451

@CharlesStaats: for me it is harder to make neighboring colors easily distinguishable. – sergiol – 2017-03-19T16:55:54.120

I've noticed a curious pattern. At the 0-indexed iteration 3 the shape invariably seems like a bike saddle. – sergiol – 2017-03-21T20:34:44.733

Not mine and not golfed, but this code is a marvel of obfuscated python producing a mandelbrot fractal

– Aaron – 2017-05-26T08:37:16.310

Does my HP42s have enough pixels to do this within the rules?

– SIGSTACKFAULT – 2017-09-28T13:00:46.383

Read the asterisk – Mark Jeronimus – 2017-09-28T14:19:54.883

8"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. – Jonathan Van Matre – 2014-03-07T21:50:47.857

I'm just trying to discourage using sequential rgb indices. Thx about the title. I forgot to double-check that after finishing writing. – Mark Jeronimus – 2014-03-07T22:13:07.043

1

Related: ASCII Mandelbrot (although some of the answers posted there aren't ASCII and could probably fit better as answers to this question).

– Peter Taylor – 2014-03-07T22:44:41.793

Why do you require that neighboring colors be easily distinguishable? Making them vary continuously (but still be distinct) is actually much harder. – Charles Staats – 2014-03-08T18:54:13.297

3I'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. – Roger Dahl – 2014-03-09T02:00:10.367

6Great idea to ask a question which requires a combination of mathematical and aesthetic sensitivities, then impose all the the design decisions in advance. – jwg – 2014-03-10T08:25:08.743

3Anyone manages to make one in brainfuck WINS, I'd say :D – MadTux – 2014-03-10T18:45:12.470

@Charles Staats: because the brot would become too dark or too plain (see my example). Also noone will try the 'harder' things because they will always increase the code size. – Mark Jeronimus – 2014-03-10T19:17:40.557

1@jwg that's because it's a rosetta-stone and not an art contest. – Mark Jeronimus – 2014-03-10T19:18:01.647

I get what it is. I am pointing out what it should be. – jwg – 2014-03-11T12:29:54.453

1Instead of refusing to "accept" an answer, the way I usually like to run a golfing contest is to accept the shortest answer at the time, and commit to updating the tick should a shorter answer come in later. That way, people actually have a (15 rep) motivation to post the shortest answer. – Chris Jester-Young – 2014-05-30T10:07:38.457

Answers

94

Sharp EL-9300 Graphics Calculator, 296 bytes

This was my secondary school graphing calculator, getting on for 20 years ago! 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.

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 gotos.

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

Digital Trauma

Posted 2014-03-07T21:39:15.050

Reputation: 64 644

1Me too, but my NV memory went blank after years of shelf-time. – Mark Jeronimus – 2014-03-08T09:10:09.747

2zx²+zy²>4 couldn't that be Abs(x)>2? – Mark Jeronimus – 2014-03-08T09:31:12.637

1Maybe you should get a new battery... – NothingsImpossible – 2014-03-08T17:48:49.247

I had a program just like that on my old TI-95, it took about 2 hours to run. Unfortunately, I no longer have that old calculator... – Mark Ormston – 2014-03-08T22:32:56.937

26Interesting. So you've been a nerd for quite a while. – devnull – 2014-03-09T03:34:56.297

1@devnull Touché! Guilty as charged. – Digital Trauma – 2014-03-09T03:58:29.827

@NothingsImpossible I'd like to say that this is battery drain due to such a computationally-intensive task. But really I think its mostly because I put in rechargeable AAAs which have a slightly lower voltage than regular alkaline – Digital Trauma – 2014-03-09T04:03:57.023

4Nice "Screenshot" – meawoppl – 2014-03-09T10:01:15.967

Didn't the OP specifically specify that graphing calculators cannot be used? – alexyorke – 2014-03-09T17:05:13.267

1@alexy13 OP made an exception for languages that had monochrome output and mentioned calculators as an example. – Sylwester – 2014-03-09T18:08:36.877

I did something similar on my Casio-7200G back in the day, that at least had subroutines of some kind :-) – yatima2975 – 2014-03-09T22:10:34.613

@MarkJeronimus The Abs() version would be more golfy for sure. In fact the calculator can do |x|>2. I think when I wrote this, I was trying to optimise performance and there is an implicit sqrt() in the abs(), so the comparison I have may be faster. Also this calculator has a Complex mode, but the plot command is not supported in that mode, so I had do do my own complex emulation in Real mode, and I don't think Abs() would be useful in that case - the zx and zy variables are supposed to be real and imaginary parts of z. – Digital Trauma – 2014-03-10T02:58:09.437

Weird about the abs function. I'm used to TI which has ABS( as a single token and runs fine when you don't use closing brackets, e.g. y←Abs(x. My program just plotted pixels in one of the Image variables, and then went to the image menu and opened it. – Mark Jeronimus – 2014-03-10T19:06:24.343

83

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/

ɐɔıʇǝɥʇuʎs

Posted 2014-03-07T21:39:15.050

Reputation: 4 449

@o0'. Yeah, I'm surprised indentation didn't cause any errors here because Python is usually really picky about that. – HyperNeutrino – 2017-03-19T19:20:21.513

12Seems to be disallowed: the regions are not easily distinguishable, or even at all. – primo – 2014-03-08T13:32:11.277

5Also, this writes to a file. – Lie Ryan – 2014-03-08T21:49:26.210

40disallowed or not, this is pretty awesome :D – Navin – 2014-03-09T02:28:29.327

2@DigitalTrauma I guess we'd see more of this output if it was allowed. – Eelvex – 2014-03-09T20:14:49.347

18@DigitalTrauma, heck, +1 for most beautiful input! – Brian S – 2014-03-10T15:26:08.903

20Does this count as a quine? ;-) – Blazemonger – 2014-03-10T21:04:01.810

3I'm shocked you can trick python into running something so oddly packed... – o0'. – 2014-03-11T11:49:41.347

47

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

Heiko Oberdiek

Posted 2014-03-07T21:39:15.050

Reputation: 3 841

36

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): 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.

Mark Ormston

Posted 2014-03-07T21:39:15.050

Reputation: 799

Someone had edited my response to say you can use Python to create the file (I guess if you don't have a hex editor). I don't use Python, but their method was: python.exe -c "open('mandel.exe', 'wb').write('[ Paste Hex Here ]'.decode('hex_codec'))" – Mark Ormston – 2014-03-09T08:17:17.263

1Machine-code doesn't count. If that counts, then any language producing byte-code or machine-code should be shorter. I count 820 after changing everything to 1-character long names. – Mark Jeronimus – 2014-03-10T06:32:09.533

@MarkJeronimus I would guess that for most of the popular languages, small programs will get significantly larger when compiled to machine code. Always counting the full binary, of course. – Christopher Creutzig – 2014-03-10T07:55:42.907

3I 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. – Mark Ormston – 2014-03-10T18:50:44.723

Any computer language is a translation/mapping of a human-readable form to computer-readable form. When asking for a language, one always implies the human-readable form, in this case ASM. Have you ever seen people optimizing C, C# and Java code for smallest binary code (on this page)? – Mark Jeronimus – 2014-03-10T19:13:08.117

5

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: http://meta.codegolf.stackexchange.com/questions/260/should-executable-binaries-be-considered-a-reasonable-solution-for-code-golf

– Mark Ormston – 2014-03-10T19:14:49.610

1I can translate Basic code to assembly code if I take the time. That doesn't make it a valid submission. Read the comment on the first post of the page you linked. I'm with @AShelly. I quote: "You should be scored on the language you write in, not your compiler's output. (Unless you really are writing in raw binary...)" – Mark Jeronimus – 2014-03-11T18:42:33.853

6Alright, 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. – Mark Ormston – 2014-03-12T05:34:06.743

1Machine 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. – hippietrail – 2014-03-12T12:23:12.953

7The 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. – Mark Ormston – 2014-03-12T16:33:21.987

28

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();
    }
}

Geobits

Posted 2014-03-07T21:39:15.050

Reputation: 19 061

You don't need Java 8 to remove the public modifier. – mbomb007 – 2015-03-05T20:00:16.980

@mbomb007 Which one? I removed the one from the class already. Removing the one on main allows it to compile, but not run. Removing the other won't compile. – Geobits – 2015-03-05T20:04:14.990

Oh. Never mind. – mbomb007 – 2015-03-05T20:05:12.373

Besides the golf of changing class M{public static void main to interface M{static void main when converting from Java 7 to 8, you can golf 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); to for(s=t=99;t-->0&r*r+e*e<4;p=r*r-e*e+a,e=r*e*2+b,r=p)s=t; to save 2 bytes. – Kevin Cruijssen – 2017-09-25T10:33:18.957

f.setSize(n,668); - depends heavily on the used Theme, but I'll accept it. – Mark Jeronimus – 2014-03-08T09:11:15.227

You can drop the imports in Java because they're auto-generated anyway. – Mark Jeronimus – 2014-03-08T09:23:45.583

I also see double where float could be used if you tried – Mark Jeronimus – 2014-03-08T09:29:44.990

JFrame => Frame shaves off 2 chars. Although you can't close the window anymore. ;) – EthanB – 2014-03-08T15:01:18.820

@MarkJeronimus Thanks for the advice, especially about imports. New to golf, so it helps. I'll be coming back to this later today, had to get some sleep ;) – Geobits – 2014-03-08T15:22:46.113

2Your 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. – Victor Stafusa – 2014-03-10T04:53:26.683

@Victor Ah, edited the imports back in; was wondering about that. Not installing Java 8 right now, but thanks for the tip.

– Geobits – 2014-03-11T01:26:12.540

@Geobits After working with your other answer, I saw this answer and noticed you can make similar reductions by including setSize(..) in the frame's paint method's for-loop, which would also allow you to place n=640 in the for-loop, saving you additional bytes on the final modifier and extra int declaration. – FThompson – 2014-04-17T03:20:20.337

@Vulcan Yea, I was going to get back to this one, but it's sleepy-time for now. Will hit it up in the morning. Thanks again :) – Geobits – 2014-04-17T03:22:25.007

21

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)

MT0

Posted 2014-03-07T21:39:15.050

Reputation: 3 373

1You can write document.createElement`canvas` and save 2 bytes. Same as the getContext`2d`. – Ismael Miguel – 2019-10-07T10:17:13.097

I didn't assume a HTML Canvas as this is a pure JavaScript solution. – MT0 – 2019-10-07T10:42:44.247

or you can 'drop' canvas at all like here

– Kamil Kiełczewski – 2019-10-07T19:59:25.413

2Nice. 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?) – Matthew Wilcoxson – 2014-03-10T15:47:05.180

19

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

Eelvex

Posted 2014-03-07T21:39:15.050

Reputation: 5 204

1+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? – plannapus – 2014-03-10T08:40:00.520

1@MarkJeronimus, I can make it 70 but I kept some things for clarity. I, thus, took the liberty to ignore the LF when counting. – Eelvex – 2014-03-10T12:39:40.420

@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). – Eelvex – 2014-03-10T13:07:40.590

17

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

DavidC

Posted 2014-03-07T21:39:15.050

Reputation: 24 524

Your built-in solution uses a very loose interpretation of The fractal coordinates range from approximately -2-2i to 2+2i. – Jonathan Frech – 2017-09-25T12:58:51.713

@JonathanFrech, Yes, thanks. – DavidC – 2017-09-25T13:06:41.727

{b, -2, 2, .01}, {a, -2, 2, .01} is shorter and closer to the rules – Mark Jeronimus – 2014-03-07T23:50:19.707

@MarkJeronimus Thanks. I used the suggested range for the iterating picture. – DavidC – 2014-03-08T03:20:30.333

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. – Mark Jeronimus – 2014-03-08T09:20:39.533

I missed the point about being black. The count increased because some single characters became two characters in the cut-and-paste to SE. – DavidC – 2014-03-08T13:34:20.193

16

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

meawoppl

Posted 2014-03-07T21:39:15.050

Reputation: 261

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). – primo – 2014-03-09T14:02:55.817

Weird. Does wc include the eof? Fixed and slightly smaller. Just a moment. – meawoppl – 2014-03-09T17:46:21.607

I get 151 with wc. First golf, so not sure how to score it. – meawoppl – 2014-03-09T18:01:39.583

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? – primo – 2014-03-09T18:09:37.787

150 here too. Besides, you sure it's 640 or less wide, and not 999? The image looks antialiased – Mark Jeronimus – 2014-03-10T06:35:28.803

I cropped out of the figure matplotlib draws using my desktop in gimp, so the image size might be derped somehow. It is doing 999 now, but changing that to anything 640x640 will look basically the same, and have the same char count. I will fix when I get home tonight and recrop w/ pic count in the borders. – meawoppl – 2014-03-10T17:27:22.353

1You can save 1 character by using from numpy import* instead of import numpy as n and mgrid instead of n.mgrid. – nyuszika7h – 2014-06-15T15:15:17.737

15

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

Oberon

Posted 2014-03-07T21:39:15.050

Reputation: 2 881

2I think 0.01 can be typed as .01. – Yytsi – 2016-06-21T09:53:44.427

You should mention that this is Allegro 4 (which is quite different from Allegro 5). Which exact version is this? – Victor Stafusa – 2014-03-10T04:57:06.463

it's either 246 or 249 long – Mark Jeronimus – 2014-03-10T06:42:27.930

@Victor Allegro 4.2.2. – Oberon – 2014-03-10T14:34:01.287

@MarkJeronimus Notepad++ tells me it's 248 characters long. – Oberon – 2014-03-10T15:33:14.150

Notepad++ is wrong. http://i.imgur.com/vnRMzOj.png

– Mark Jeronimus – 2014-03-10T19:27:11.417

1@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. – Oberon – 2014-03-10T21:05:39.040

247 characters for me if I remove newlines except for the one after #include <allegro.h>. http://bpaste.net/show/z45cZWcn5qTKSBZ8YN9e/

– nyuszika7h – 2014-06-15T15:11:24.720

14

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):

primo

Posted 2014-03-07T21:39:15.050

Reputation: 30 891

Works in Python 3 at the cost of one additional byte: x//d instead of x/d in the putpixel line. – rosuav – 2019-11-15T11:55:06.497

1You 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. – Geobits – 2014-03-08T16:15:59.257

@Geobits the idea actually saved 10, thanks! – primo – 2014-03-08T16:28:32.610

1Complex 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 – meawoppl – 2014-03-09T03:10:47.340

@meawoppl another 7 :D – primo – 2014-03-09T05:07:04.723

Technically disallowed: this doesn't any graphical output feature of Python itself (and Image.show() implicitly saves a temporary file). – nneonneo – 2014-03-09T05:51:31.777

This post totally got me started on the 153 character version I posted below :) – meawoppl – 2014-03-09T08:27:43.687

14

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.

TessellatingHeckler

Posted 2014-03-07T21:39:15.050

Reputation: 2 412

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. – Mark Jeronimus – 2015-09-30T11:05:20.697

Apparently Magnitude is a*a+b*b not sqrt(a*a+b*b) – Mark Jeronimus – 2015-09-30T13:41:38.053

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). – TessellatingHeckler – 2015-09-30T18:02:24.077

12

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.

Konrad Borowski

Posted 2014-03-07T21:39:15.050

Reputation: 11 185

No need to plot twice, just go with a more inefficient version. Doesn;t it support NEXT Y,X? – Mark Jeronimus – 2014-03-09T11:59:07.740

10

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)

PLZ HELP FIND SHORTCUT KTHX

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:

(Freeware version run in OS X under Wine. Yes I'm cheap like that.)

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. Every statement in this footnote is possibly true. Keep calm and carry on golfing.

Tobia

Posted 2014-03-07T21:39:15.050

Reputation: 5 455

10

Mathematica 10.0, 19 chars

MandelbrotSetPlot[]

MandelbrotSetPlot is a new function in Mathematica 10.0.

enter image description here

alephalpha

Posted 2014-03-07T21:39:15.050

Reputation: 23 988

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. – Mark Jeronimus – 2014-03-30T10:03:08.903

20Code 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. – Michael Stern – 2014-04-17T15:08:36.200

9

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

plannapus

Posted 2014-03-07T21:39:15.050

Reputation: 8 610

technically black on the outside is disallowed. Did you miss that or is it difficult to implement? – Mark Jeronimus – 2014-03-08T09:19:45.103

@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. – plannapus – 2014-03-08T09:20:42.553

@MarkJeronimus Done! – plannapus – 2014-03-08T09:40:34.230

5Second place in the hideous colors division. – meawoppl – 2014-03-09T10:02:34.927

1@meawoppl blame rainbow() :) – plannapus – 2014-03-10T07:23:12.813

9

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();
}

Jan K

Posted 2014-03-07T21:39:15.050

Reputation: 91

Aw, maaaaan, I wanted to do this. +1. – SIGSTACKFAULT – 2017-09-28T12:57:58.960

8

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

12Me21

Posted 2014-03-07T21:39:15.050

Reputation: 6 110

6

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

MD XF

Posted 2014-03-07T21:39:15.050

Reputation: 11 605

Would be smaller (and slower) if you didn't plot two pixels at once but choose a random pixel on the entire screen. – Mark Jeronimus – 2017-05-31T06:09:58.903

1@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 – MD XF – 2017-05-31T20:11:11.050

6

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])

aschepler

Posted 2014-03-07T21:39:15.050

Reputation: 717

Comment from user Martin Novy: "m=c-c; can be shortened to m=0; ... it works the same in this program, I have just tested it in Octave"

– Martin Ender – 2014-12-26T15:09:22.650

There's probably a shorter way to get a discontinuous colormap.... – aschepler – 2014-03-08T20:40:58.287

Yes, much better now. – aschepler – 2014-03-09T01:13:24.770

+1 nice and concise solution. But your aspect ratio is not 1:1 (cf. rule n°2: output should be square). – plannapus – 2014-03-10T07:39:27.907

Fixing the aspect ratio will take 11 more bytes: append ";axis image". Is that required to qualify? – aschepler – 2014-03-10T14:39:05.637

i think it was just me nitpicking :) , the OP doesn't seem to have a problem with it since he didn't say anything. – plannapus – 2014-03-11T07:46:39.867

6

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).

nneonneo

Posted 2014-03-07T21:39:15.050

Reputation: 11 445

Is this the fastest of them all? – Demi – 2015-08-03T07:15:51.980

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) – Mark Jeronimus – 2014-03-10T06:40:26.700

@MarkJeronimus: For the ShaderToy environment, these variables are fixed. Otherwise, for standard shaders, I would have picked shorter variable names. – nneonneo – 2014-03-10T07:25:14.133

5

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.

orion

Posted 2014-03-07T21:39:15.050

Reputation: 3 095

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) – Mark Jeronimus – 2014-03-10T06:26:57.750

1No 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. – orion – 2014-03-10T07:53:03.583

5

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)

Chris Taylor

Posted 2014-03-07T21:39:15.050

Reputation: 167

Shouldn't this image be rotated -90 degrees? – mbomb007 – 2015-03-05T19:50:16.120

2The Mandelbrot set rotated by 90 degrees is still the Mandelbrot set. – Chris Taylor – 2015-03-05T23:19:23.417

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. – Mark Jeronimus – 2014-03-10T19:19:52.950

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. – aschepler – 2014-03-11T12:33:45.990

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. – aschepler – 2014-03-11T12:39:52.980

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 +=.) – aschepler – 2014-03-11T13:06:41.600

Ah, good point! Edited. – Chris Taylor – 2014-03-11T13:28:16.827

4

Tcl/Tk, 316

322 324 336 348 349 351 352 353 354 355

Now a shorter version using 3 letter #RGB shorthands style color triplets (instead of #RRGGBB triplets), which results in different colors.

And some more golfing.

rename set s
grid [canvas .c -w 640 -he 640]
.c cr i 320 320 -i [s p [image c photo -w 640 -h 640]]
time {incr x
s y 0
time {incr y
s a 0
s b 0
s n 0
while \$n<99 {s A [expr $a*$a-$b*$b+$x[s f *4/639.-2]]
if [s b [expr 2*$a*$b+$y$f]]*$b+[s a $A]*$a>4 break
incr n}
$p p [format #%03x [expr $n*41]] -t $x $y} 640} 640

fractal


Tcl/Tk, 325

331 333 345 357 358 360 361 362 364 365

I think I would win if the criterium was beauty!

rename set s
grid [canvas .c -w 640 -he 640]
.c cr i 320 320 -i [s p [image c photo -w 640 -h 640]]
time {incr x
s y 0
time {incr y
s a 0
s b 0
s n 0
while \$n<99 {s A [expr $a*$a-$b*$b+$x[s f *4/639.-2]]
if [s b [expr 2*$a*$b+$y$f]]*$b+[s a $A]*$a>4 break
incr n}
$p p [format #%06x [expr $n*16777215/99]] -t $x $y} 640} 640

Presentation:

fractal

sergiol

Posted 2014-03-07T21:39:15.050

Reputation: 3 055

1Nice. You can reduce a few characters (down to 380, I think) by adding rename set s at the top and then replacing all the set by s – Rolazaro Azeveires – 2017-04-04T00:00:37.367

4

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

Engineer Toast

Posted 2014-03-07T21:39:15.050

Reputation: 5 769

@ceilingcat Thanks. That was a carryover from my original which had specialized functions for complex number math. – Engineer Toast – 2017-09-27T12:07:18.113

Does it really need to be >=4 or can you get away with >4? Also, can replace j<99 with j<D. – ceilingcat – 2017-09-27T22:38:47.023

@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 - – Taylor Scott – 2017-09-28T05:03:20.810

1@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. – Engineer Toast – 2017-09-28T12:31:15.553

I believe that you can get away with removing another byte by swapping 2^20 with 1E6 – Taylor Scott – 2017-10-08T17:55:28.107

4

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

ɲeuroburɳ

Posted 2014-03-07T21:39:15.050

Reputation: 401

Forgive me if I don't understand your instructions on making the short version. – Mark Jeronimus – 2014-03-08T21:33:02.293

No problem. Remove var w, at the beginning, and change function(x,i,j,k,l,c,o) to function(). – ɲeuroburɳ – 2014-03-08T21:35:28.510

4

Javascript, 285B

Based off my code and some improvements on MT0's code, I've got this down to 285B in colour:

document.body.appendChild(V=document.createElement('Canvas'));j=(D=(X=V.getContext('2d')).createImageData(Z=V.width=V.height=255,Z)).data;for(x=Z*Z;x--;){k=a=b=c=0;while(a*a+b*b<4&&Z>k++){c=a*a-b*b+4*(x%Z)/Z-3;b=2*a*b+4*x/(Z*Z)-2;a=c;}j[4*x]=99*k%256;j[4*x+3]=Z;}X.putImageData(D,0,0);

in action: http://jsfiddle.net/acLhe/7/

was: Coffeescript, 342B

document.body.appendChild V=document.createElement 'Canvas'
N=99
Z=V.width=V.height=400
P=[]
P.push "rgba(0,0,0,"+Math.random()*i/N+')' for i in [N..0]
X=V.getContext '2d'
for x in [0..Z]
 for y in [0..Z]
  k=a=b=0
  [a,b]=[a*a-b*b+4*x/Z-3,2*a*b+4*y/Z-2] while a*a+b*b<4 and N>k++
  X.fillStyle=P[k-1]
  X.fillRect x,y,1,1

Coffeescript is supposed to be readable :-/ see it in action: http://jsfiddle.net/acLhe/6/

Mandelbrot Coffeescript

alexander-brett

Posted 2014-03-07T21:39:15.050

Reputation: 1 485

I start from this size 285B and improve it more in this answer

– Kamil Kiełczewski – 2019-10-17T10:04:30.233

OP asks for color unless your platform doesn't support color. Looks great, though, and nice concise code. Welcome to PPCG! – Jonathan Van Matre – 2014-03-13T16:45:31.223

4

QBasic, QuickBasic, QB64 - 156 153

SCREEN 13
FOR J=0TO 191
B=J/48-2
FOR I=0TO 191
A=I/48-2
X=A
Y=B
C=0
DO
U=X*X
V=Y*Y
Y=2*X*Y+B
X=U-V+A
C=C+1
LOOP UNTIL C>247OR U+V>4
PSET(I,J),C
NEXT
NEXT

Standard DOS palette:

enter image description here

Mark Jeronimus

Posted 2014-03-07T21:39:15.050

Reputation: 6 451

3

Perl + GD, 264

$I=new GD::Image $s=499,$s;Z(0,0,0);Z(map rand 256,1..3)for
0..99;for$x(0..$s){for$y(0..$s){for($H=$K=$c=$t=0;$c++<99&&$H*$H+$K*$K<4;){sub
Z{$I->colorAllocate(@_)}($H,$K)=($H*$H-$K*$K+4*$x/$s-2,2*$H*$K+4*$y/$s-2)}use
GD;$I->setPixel($x,$y,$c<99&&$c)}}print $I->png

Mandelbrot fractal from Perl+GD

Golfed from this code

mob

Posted 2014-03-07T21:39:15.050

Reputation: 2 506

2Nominated: most ugly color scheme. – meawoppl – 2014-03-09T09:41:34.793

3

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

Mark Jeronimus

Posted 2014-03-07T21:39:15.050

Reputation: 6 451

2Link to a language and description of the code? Or actually, what is the code? – MD XF – 2017-09-26T03:20:49.530

2

SmileBASIC, 126 bytes

X=RNDF()*4-2Y=RNDF()*4-2@N
N=N+16I=X+S*S-T*T
T=Y+S*T*2S=I
IF N<#L&&S*S+T*T<4GOTO@N
GPSET X*50+99,Y*50+99,RGB(99XOR N,N,N)EXEC.

enter image description here

12Me21

Posted 2014-03-07T21:39:15.050

Reputation: 6 110

Nice use of randoms to save on loops – Mark Jeronimus – 2017-02-19T09:07:20.280

Should make it 240x240, but I guess that would be longer. – snail_ – 2017-03-20T12:28:34.137

Code golf is about what you can get away with, not what you should do. – 12Me21 – 2017-06-17T15:28:03.887

2

QBasic, 140 bytes

SCREEN 13
FOR i=0TO 199
FOR j=0TO 199
x=0
y=0
FOR c=0TO 255
t=y*y
c=c-255*(x*x+t>4)
y=2*x*y+j/50-2
x=x*x-t+i/50-2
NEXT
PSET(i,j),c
NEXT
NEXT

Output:


FreeBASIC, 140 bytes

Using FreeBASIC's SCREEN 16, we can generate a more detailed image at the same code length:

SCREEN 16
FOR i=0TO 383
FOR j=0TO 383
x=0
y=0
FOR c=0TO 255
t=y*y
c=c-255*(x*x+t>4)
y=2*x*y+j/96-2
x=x*x-t+i/96-2
NEXT
PSET(i,j),c
NEXT
NEXT

Compile as fbc -lang qb mandel.bas.

Higher resolutions are also available.

primo

Posted 2014-03-07T21:39:15.050

Reputation: 30 891

Please count newlines as a byte too – Mark Jeronimus – 2017-09-27T14:32:05.380

@MarkJeronimus my mistake, I though I had. – primo – 2017-09-28T04:31:09.653

2

JavaScript, 284 283 ...(11 improvements)... 208 191 190 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 contains 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+=9*c+9*j) 
  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+=2*c+9*j)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}`)

Kamil Kiełczewski

Posted 2014-03-07T21:39:15.050

Reputation: 531

1

Python 3, 290 Bytes

from PIL import Image
import numpy as n
s=640;d=n.zeros((s,s,3),dtype=n.uint8)
for k in range(s*s):
    r=4/s*(k%s)-2;i=2-4/s*(k//s);z=c=complex(r,i)
    for i in range(50,9950,100):
        if abs(z)>2:
            d[k//s,k%s]=[i*5,i*9,i*7];break
        z=z*z+c
Image.fromarray(d).show()

mandelbrot

r3dapple

Posted 2014-03-07T21:39:15.050

Reputation: 11

1

Julia

Well, better late than never:

function mandelbrot(x0,y0,side,N=800,L=55,R=3.)
    m = [0 for i=1:N,j=1:N]
    delta = side/N
    for i=1:N, j=1:N
        c = x0+delta*i+(y0+delta*j)*im
        z, h = 0+0*im, 0
        while (h<L) && (abs(z)<R)
            z = z^2+c
            h+=1
        end
        m[j,i]=h
    end
    return m
end
n=2.6
m = mandelbrot(-n/1.3,-n/2, n)
using Winston, Color
imagesc(m)
title("Mandelbrot Set")

Colormap Mod. :

function RGB_cm()
    colormap = [RGB(0,0,0) for t=1:255*5]
    rgb = [255,0,0]
    for t in 0:(255*5-1)
    c = [0, 0, 0]
    i = ifloor(t/255)
    c[(i+3)%3!=0?(i+3)%3:3] = (-1)^i
    rgb+=c
    colormap[t+1] = RGB(rgb[1],rgb[2],rgb[3])
    end
    colormap[(end-25):end] = RGB(0,0,0)
    return colormap
end
c = RGB_cm()
Winston.colormap(c)

Output:

Mandelbrot Set

CCP

Posted 2014-03-07T21:39:15.050

Reputation: 632

1I'm sure you can save much on whitespace. – Erik the Outgolfer – 2017-09-25T17:15:15.217

Should I byte-count only the first program or both? – Mark Jeronimus – 2014-03-29T10:10:17.793

I think both, because without the color map mod the image would not satisfy the rules. – CCP – 2014-03-29T15:35:36.087