Mandelbrot image in every language

Question 1

I always used aMandelbrot 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:

(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 pagevar 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>  <h2>Shortest Solution by Language</h2>  <table>    <thead>      <tr><td>Language</td><td>User</td><td>Score</td></tr>    </thead>    <tbody>    </tbody>  </table></div><div>  <h2>Leaderboard</h2>  <table>    <thead>      <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr>    </thead>    <tbody>    </tbody>  </table></div><table>  <tbody>    <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr>  </tbody></table><table>  <tbody>    <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr>  </tbody></table>

Question 2

"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.

Question 3

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.

Question 4

Great idea to ask a question which requires a combination of mathematical and aesthetic sensitivities, then impose all the the design decisions in advance.

Question 5

Anyone manages to make one in brainfuck WINS, I'd say :D

Question 6

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

Question 7

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:

ClrGDispGRange -2.35,2.35,.5,-1.55,1.55,0.5y=-1.55Label lyx=-2.35Label lxn=1zx=0zy=0Label lntzx=zx²-zy²+xzy=(2*zx*zy)+yzx=tzxIf zx²+zy²>4Goto escn=n+1If n<20Goto lnLabel escIf fpart (n/2)=0Goto nplPlot x,yLabel nplx=x+.05If x<=2.35Goto lxy=y+.05If y<=1.55Goto lyWait

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 areifs andgotos.

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

Question 8

Me too, but my NV memory went blank after years of shelf-time.

Question 9

zx²+zy²>4 couldn't that beAbs(x)>2?

Question 10

Interesting. So you've been anerd for quite a while.

Question 11

@devnull Touché! Guilty as charged.

Question 12

Nice "Screenshot"

Question 13

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').writefor 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/

Question 14

Seems to be disallowed: the regions are not easily distinguishable, or even at all.

Question 15

Also, this writes to a file.

Question 16

disallowed or not, this is pretty awesome :D

Question 17

@DigitalTrauma, heck, +1 for most beautiful input!

Question 18

Does this count as a quine? ;-)

Question 19

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

(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

Question 20

I wonder if functional shading is shorter / meets the rule (not loading the pre-defined shading but define it yourself). Also includingtikz includes bothpgf andxcolor.

Question 21

x86 DOS Assembly,208177 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):

.286CODE SEGMENTASSUME CS:code, DS:codeORG 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 useGMV:  ; ===========================================================================  ; 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, blGMVDone:    ret;GMV; *****************************************************************************; Data; DivisorDivisor DW 50; Two ValueTwoValue DW 2; 4 ValueFourValue DW 4CODE ENDSEND 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.

Question 22

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.

Question 23

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/…

Question 24

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.

Question 25

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.

Question 26

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.

Question 27

Java,505405 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();    }}

Question 28

f.setSize(n,668); - depends heavily on the used Theme, but I'll accept it.

Question 29

You can drop the imports in Java because they're auto-generated anyway.

Question 30

I also seedouble wherefloat could be used if you tried

Question 31

JFrame =>Frame shaves off 2 chars. Although you can't close the window anymore. ;)

Question 32

Your class do not need to be public. Further, use Java 8 to get rid of thefinal modifier. And you must not omit the imports in order to be a complete submission.

Question 33

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)

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

Withf(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)

Question 34

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?)

Question 35

You can writedocument.createElement`canvas` and save 2 bytes. Same as thegetContext`2d`.

Question 36

I didn't assume a HTML Canvas as this is a pure JavaScript solution.

Question 37

or you can 'drop' canvas at all likehere

Question 38

The first image looks like a Virtual Boy game.

Question 39

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 rangev =: (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 iterationsi.99                NB. 0 1 2 3 4 ... 98(v^:(i.99))0        NB. returns 99 tables, one for each number of iterations2<| y               NB. returns 1 if 2 < norm(y), 0 otherwise2<| (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 - 20,?$~99 3           NB. prepend the triplet (0,0,0): black

Question 40

+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?

Question 41

@MarkJeronimus, I can make it 70 but I kept some things for clarity. I, thus, took the liberty to ignore the LF when counting.

Question 42

@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).

Question 43

Mathematica,21419121519 30

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

MandelbrotSetPlot[]

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

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

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

Question 44

{b, -2, 2, .01}, {a, -2, 2, .01} is shorter and closer to the rules

Question 45

@MarkJeronimus Thanks. I used the suggested range for the iterating picture.

Question 46

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.

Question 47

I missed the point about being black. The count increased because some single characters became two characters in the cut-and-paste to SE.

Question 48

Your built-in solution uses a very loose interpretation ofThe fractal coordinates range from approximately -2-2i to 2+2i.

Question 49

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

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

enter image description here

Question 50

I count 152. No space is required betweenimport and*, and not definingf 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).

Question 51

Weird. Does wc include the eof? Fixed and slightly smaller. Just a moment.

Question 52

I get 151 with wc. First golf, so not sure how to score it.

Question 53

I count 150, without trailing newline. Some interpreters/compilers demand one, but the python interpreter does fine without. Not sure aboutwc, but maybe trystat -c %s instead. Are the black upper and lower borders part of the image?

Question 54

You can save 1 character by usingfrom numpy import* instead ofimport numpy as n andmgrid instead ofn.mgrid.

Question 55

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

Question 56

You should mention that this is Allegro 4 (which is quite different from Allegro 5). Which exact version is this?

Question 57

it's either 246 or 249 long

Question 58

@Victor Allegro 4.2.2.

Question 59

@MarkJeronimus Isn't there supposed to be a newline between... allegro.h> andx=-1, ...? I suppose Notepad++ counts it as\r\n =0D 0A.

Question 60

I think0.01 can be typed as.01.

Question 61

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.

Question 62

I'd changelt2 tolt4 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.

Question 63

Apparently Magnitude isa*a+b*b notsqrt(a*a+b*b)

Question 64

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).

Digital Trauma 73.8k10 gold badges117 silver badges268 bronze badges · Accepted Answer · 2024-09-14 18:15:59Z

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:

ClrGDispGRange -2.35,2.35,.5,-1.55,1.55,0.5y=-1.55Label lyx=-2.35Label lxn=1zx=0zy=0Label lntzx=zx²-zy²+xzy=(2*zx*zy)+yzx=tzxIf zx²+zy²>4Goto escn=n+1If n<20Goto lnLabel escIf fpart (n/2)=0Goto nplPlot x,yLabel nplx=x+.05If x<=2.35Goto lxy=y+.05If y<=1.55Goto lyWait

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 areifs andgotos.

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