# Mandelbrot image in every language

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:

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.

/* 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 COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk";
var OVERRIDE_USER = 17419; // This should be the user ID of the challenge author.

/* App */

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

}

jQuery.ajax({
method: "get",
dataType: "jsonp",
crossDomain: true,
success: function (data) {
data.items.forEach(function(a) {
});
comment_page = 1;
}
});
}

jQuery.ajax({
method: "get",
dataType: "jsonp",
crossDomain: true,
success: function (data) {
data.items.forEach(function(c) {
if (c.owner.user_id === OVERRIDE_USER)
});
else process();
}
});
}

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

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

function process() {
var valid = [];

var body = a.body;
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],
});
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;

.replace("{{NAME}}", a.user)
.replace("{{LANGUAGE}}", a.language)
.replace("{{SIZE}}", a.size)

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)
language = jQuery(language);
jQuery("#languages").append(language);
}

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

width: 290px;
float: left;
}

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

font-weight: bold;
}

table td {
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="language-list">
<h2>Shortest Solution by Language</h2>
<table class="language-list">
<tr><td>Language</td><td>User</td><td>Score</td></tr>
<tbody id="languages">

</tbody>
</table>
</div>
<tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr>

</tbody>
</table>
</div>
<table style="display: none">
</tbody>
</table>
<table style="display: none">
<tbody id="language-template">
</tbody>
</table>

• "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. Mar 7, 2014 at 21:50
• 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. Mar 9, 2014 at 2:00
• Great idea to ask a question which requires a combination of mathematical and aesthetic sensitivities, then impose all the the design decisions in advance.
– jwg
Mar 10, 2014 at 8:25
• Anyone manages to make one in brainfuck WINS, I'd say :D Mar 10, 2014 at 18:45
• Not mine and not golfed, but this code is a marvel of obfuscated python producing a mandelbrot fractal May 26, 2017 at 8:37

# 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:

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

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)
)   )

• Seems to be disallowed: the regions are not easily distinguishable, or even at all. Mar 8, 2014 at 13:32
• Also, this writes to a file. Mar 8, 2014 at 21:49
• disallowed or not, this is pretty awesome :D Mar 9, 2014 at 2:28
• @DigitalTrauma, heck, +1 for most beautiful input! Mar 10, 2014 at 15:26
• Does this count as a quine? ;-) Mar 10, 2014 at 21:04

## 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
}
\d0%
\Loop\c{\numexpr\iterations+1\relax}{%
\expandafter\edef\csname\the\c\endcsname{%
\hbox{\noexpand\color[Hsb]{\the\d,1,1}\noexpand\pixel}%
}%
}
\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

• I wonder if functional shading is shorter / meets the rule (not loading the pre-defined shading but define it yourself). Also including tikz includes both pgf and xcolor. Mar 26 at 3:33

## x86 DOS Assembly, 208177 173 bytes

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

DBE3BE00A0B81300CD1056BA640007BF87F9FDBDC7008BCDE81A008AC3AA4979F7B9C70083EF784D79EE33C0CD16B80300CD10CD208BC12BC289441CDF441CDF06A701DEF9D95C088BC52BC289441CDF441CDF06A701DEF9D95C0CD9EED914D95404D95410D95C14B301D904D84C04DE0EA901D8440CD95404D94410D86414D84408D914D80CD95C10D84C04D95414D84410DF06AB01DED99BDFE09B9E7207433ADA72C632DBC3320002000400


The sample image is:

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
FST zi ; Reusing this later

; zr = zrsq - zisq + cr;
FLD zrsq
FSUB zisq
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;
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.

• 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. Mar 10, 2014 at 18:50
• 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/… Mar 10, 2014 at 19:14
• 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. Mar 12, 2014 at 5:34
• 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. Mar 12, 2014 at 12:23
• 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. Mar 12, 2014 at 16:33

## Java, 505405 324 bytes

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

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();
}
}

• f.setSize(n,668); - depends heavily on the used Theme, but I'll accept it. Mar 8, 2014 at 9:11
• You can drop the imports in Java because they're auto-generated anyway. Mar 8, 2014 at 9:23
• I also see double where float could be used if you tried Mar 8, 2014 at 9:29
• JFrame => Frame shaves off 2 chars. Although you can't close the window anymore. ;) Mar 8, 2014 at 15:01
• Your class do not need to be public. Further, use Java 8 to get rid of the final modifier. And you must not omit the imports in order to be a complete submission. Mar 10, 2014 at 4:53

# 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)


(d=document).body.appendChild(e=d.createElementcanvas);v=e.getContext2d;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);:

(d=document).body.appendChild(e=d.createElementcanvas);v=e.getContext2d;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)

• 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?) Mar 10, 2014 at 15:47
• You can write document.createElementcanvas  and save 2 bytes. Same as the getContext2d . Oct 7, 2019 at 10:17
• I didn't assume a HTML Canvas as this is a pure JavaScript solution.
– MT0
Oct 7, 2019 at 10:42
• or you can 'drop' canvas at all like here Oct 7, 2019 at 19:59

# J, 73 bytes

load'viewmat'
(0,?$~99 3)viewmat+/2<|(j./~i:2j479)(+*:) ::(3:)"0^:(i.99)0  Edit, some explaining: x (+*:) y NB. is x + (y^2) x (+*:) ::(3:) y NB. returns 3 when (+*:) fails (NaNs) j./~i:2j479 NB. a 480x480 table of complex numbers in required range v =: (j./~i:2j479)(+*:) ::(3:)"0 ] NB. (rewrite the above as one verb) v z0 NB. one iteration of the mandelbrot operation (z0 = 0) v v z0 NB. one iteration on top of the other (v^:n) z0 NB. the result of the mandelbrot operation, after n iterations i.99 NB. 0 1 2 3 4 ... 98 (v^:(i.99))0 NB. returns 99 tables, one for each number of iterations 2<| y NB. returns 1 if 2 < norm(y), 0 otherwise 2<| (v^:(i.99))0 NB. 99 tables of 1s and 0s +/... NB. add the tables together, element by element. NB. we now have one 480x480 table, representing how many times each element exceeded norm-2. colors viewmat M NB. draw table 'M' using 'colors'; 'colors' are rgb triplets for each level of 'M'.$~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

• +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? Mar 10, 2014 at 8:40
• @MarkJeronimus, I can make it 70 but I kept some things for clarity. I, thus, took the liberty to ignore the LF when counting. Mar 10, 2014 at 12:39
• @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). Mar 10, 2014 at 13:07 # 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,#]]&)]  • {b, -2, 2, .01}, {a, -2, 2, .01} is shorter and closer to the rules Mar 7, 2014 at 23:50 • @MarkJeronimus Thanks. I used the suggested range for the iterating picture. Mar 8, 2014 at 3:20 • 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. Mar 8, 2014 at 9:20 • I missed the point about being black. The count increased because some single characters became two characters in the cut-and-paste to SE. Mar 8, 2014 at 13:34 • Your built-in solution uses a very loose interpretation of The fractal coordinates range from approximately -2-2i to 2+2i. Sep 25, 2017 at 12:58 # 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! • 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). Mar 9, 2014 at 14:02 • Weird. Does wc include the eof? Fixed and slightly smaller. Just a moment. Mar 9, 2014 at 17:46 • I get 151 with wc. First golf, so not sure how to score it. Mar 9, 2014 at 18:01 • 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? Mar 9, 2014 at 18:09 • You can save 1 character by using from numpy import* instead of import numpy as n and mgrid instead of n.mgrid. Jun 15, 2014 at 15:15 # 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: • You should mention that this is Allegro 4 (which is quite different from Allegro 5). Which exact version is this? Mar 10, 2014 at 4:57 • it's either 246 or 249 long Mar 10, 2014 at 6:42 • @Victor Allegro 4.2.2. Mar 10, 2014 at 14:34 • @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. Mar 10, 2014 at 21:05 • I think 0.01 can be typed as .01. Jun 21, 2016 at 9:53 ### Windows PowerShell (v4), 299 bytes # 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.
• 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. Sep 30, 2015 at 11:05
• Apparently Magnitude is a*a+b*b not sqrt(a*a+b*b) Sep 30, 2015 at 13:41
• 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). Sep 30, 2015 at 18:02

## 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):

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

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

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

# APL, 194 chars/bytes*

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


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

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

Ungolfed version (only line 1)

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


Screenshot:

*: 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.

### 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)


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)


• technically black on the outside is disallowed. Did you miss that or is it difficult to implement? Mar 8, 2014 at 9:19
• @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. Mar 8, 2014 at 9:20
• @MarkJeronimus Done! Mar 8, 2014 at 9:40
• Second place in the hideous colors division. Mar 9, 2014 at 10:02
• @meawoppl blame rainbow() :) Mar 10, 2014 at 7:23

# 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);
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();
}


• Aw, maaaaan, I wanted to do this. +1. Sep 28, 2017 at 12:57

# 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


# Mathematica 10.0, 19 chars

MandelbrotSetPlot[]


MandelbrotSetPlot is a new function in Mathematica 10.0.

• 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. Mar 30, 2014 at 10:03
• Code golf is generally won by specialist languages with single-character tokens, or by systems like Mathematica that have a massive number of special functions built-in. To use them is not cheating, any more than using single-character commands would be in APL. Apr 17, 2014 at 15:08

# R, 140136128124123 110 bytes*

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


Try it at rdrr.io

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

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

• No problem, submissions are always welcome. En welkom. May 19, 2020 at 21:03

# Applesoft BASIC, 302286 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:

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:

### 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

• Would be smaller (and slower) if you didn't plot two pixels at once but choose a random pixel on the entire screen. May 31, 2017 at 6:09
• @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 May 31, 2017 at 20:11

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

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

Expanded

// Fractal calculations
c=512;

// p has pixels, p+= allways join 4 digit number implicit casted to string
// 4 char string interpred as base64 gives 3bytes = 1 RGB pixel
for( p=i=''; j=x=y=0,++i<=c*c; p+= j<c ? 9*c+9*j : 'AAAA' )
while( x*x+y*y<4 && ++j-c )
[x,y] = [ x*x-y*y+i%c/128-2, 2*x*y+i/c/128-2 ];

// draw pixels in 512x512 BMP base64 image
document.write("<img src=data:;base64,Qk0bAAwAAAAAABsAAAAMAAAAAAIAAgEAGAAA"+p)

After small constans changes

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

• This is very cool, but the pixels inside the set should be black or white. I'm not sure what either of those are in base 64 Apr 22 at 0:11
• @MatthewJensen you are right - AAAA - fixed May 5 at 15:48

# 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);}


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

• 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) Mar 10, 2014 at 6:40
• @MarkJeronimus: For the ShaderToy environment, these variables are fixed. Otherwise, for standard shaders, I would have picked shorter variable names. Mar 10, 2014 at 7:25
• Is this the fastest of them all?
– Demi
Aug 3, 2015 at 7:15

# 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:

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]

[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])

• There's probably a shorter way to get a discontinuous colormap.... Mar 8, 2014 at 20:40
• Yes, much better now. Mar 9, 2014 at 1:13
• +1 nice and concise solution. But your aspect ratio is not 1:1 (cf. rule n°2: output should be square). Mar 10, 2014 at 7:39
• Fixing the aspect ratio will take 11 more bytes: append ";axis image". Is that required to qualify? Mar 10, 2014 at 14:39
• i think it was just me nitpicking :) , the OP doesn't seem to have a problem with it since he didn't say anything. Mar 11, 2014 at 7:46

## PostScript, 290285282271 261 bytes

#### Golfed code:

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


#### Ungolfed code:

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

• Golfed some more for 198 bytes: [/d{-2 .01 2}(]){exch def}([){mul}/p{add}/s{2 p 100[}/q{x x[y y[}>>begin d{/a]d{/b]0/x 0/y]]0 1 99{q p 4 ge{log .5[1 1 sethsbcolor a s b s 1 1 rectfill exit}if q sub a p 2 x[y[b p/y]/x]=}for}for}for May 5 at 16:51

# 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:

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.

• 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) Mar 10, 2014 at 6:26
• No it's not a copy-paste. It's a very straight forward formula and it wasn't even necessary to search for it. However, I did find the pages you get with that search when I was looking for a better way of initializing complex numbers (their implementation is different, well, as different as it can be in this case). Thanks for the tip about the real part, it works. Fixing. Mar 10, 2014 at 7:53

## 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 -

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)

• 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. Mar 10, 2014 at 19:19
• 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. Mar 11, 2014 at 12:33
• 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. Mar 11, 2014 at 12:39
• 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 +=.) Mar 11, 2014 at 13:06
• The Mandelbrot set rotated by 90 degrees is still the Mandelbrot set. Mar 5, 2015 at 23:19

# Excel VBA, 251246224223 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:

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.

• @ceilingcat Thanks. That was a carryover from my original which had specialized functions for complex number math. Sep 27, 2017 at 12:07
• Does it really need to be >=4 or can you get away with >4? Also, can replace j<99 with j<D. Sep 27, 2017 at 22:38
• @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 - Sep 28, 2017 at 5:03
• @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. Sep 28, 2017 at 12:31
• I believe that you can get away with removing another byte by swapping 2^20 with 1E6 Oct 8, 2017 at 17:55

# JavaScript + HTML5 (356B)

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/

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

# 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/

• OP asks for color unless your platform doesn't support color. Looks great, though, and nice concise code. Welcome to PPCG! Mar 13, 2014 at 16:45
• I start from this size 285B and improve it more in this answer Oct 17, 2019 at 10:04

## 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: