text:
Display random colored pixels
I've always liked screens full of randomly colored pixels. They're interesting to look at and the programs that draw them are fun to watch.
The challenge
Fill your screen, or a graphical window, with colored pixels.
The rules

Your program must have an even chance of picking all colors (i.e in the range #000000 to #FFFFFF), or all colors that can be displayed on your system.
Your program must continue to display random pixels until manually stopped (it cannot terminate on its own).
Pixels can be any size, as long as your output has at least 40x40 "pixels".
Your program must run at such a speed that it can replace every pixel on the screen/window at least once after running for three minutes.
Your program must choose truly random colors and points to replace, i.e. random with all points/colors equally likely. It cannot just look random. It must use a pRNG or better, and the output cannot be the same every time.
Your program must have an equal chance of picking all colors each iteration.
Your program must replace only one pixel at once.
Your program cannot use the internet nor your filesystem (/dev/random and /dev/urandom excepted).

Example
Your output could look like this if stopped at a random time:

The winner
The shortest answer in each language wins. Have fun!

By "truly random", I assume pseudorandom is OK (like Math.random() or similar)

@OldBunny2800 It must use a pRNG or better

Since the program can't use the internet, I'll just leave this in a comment: https://babelia.libraryofbabel.info/slideshow.html

So does the "3 minute limit" apply to setup time, too? Or just once the "program" begins? Asking because I have a Minecraft redstone solution, which runs very quickly once it is going, but takes a while to set up initially (it has to "place" each "pixel" in the 40x40 grid) before it can start changing colors.

Truly random by definition exclude pseudo-random, yet doesn't define a distribution. I assume you mean a uniform PRNG, where all events are equiprobable and independent of each other.

What if I the display of my system is too small for 40x40?

Related: Random Pixel Poking

Geometry Dash Leaked 2.2 Editor - 16778817 objects (theoretically). Sadly, this does not meet the 3 minute requirement because it would crash a supercomputer and is also not possible to store in your computer (look at the object count) :(

Minecraft 1.12 Redstone Command Blocks, 4,355 2,872 bytes

(Size determined by saved structure block file size.)
Here is a full YouTube overview, but I'll try to outline the code below.
Setup Routine:

This sets up the 40x40 grid of Minecraft armor stands. Armor stands are necessary because Minecraft has no way to substitute variables into world coordinates. So the workaround is to refer to the location of these armor stand entities.
(impulse) summon armor_stand 2 ~ 1 {CustomName:"A"} /create named armor stand
(chain) fill -2 ~ -2 43 ~ 43 stone                  /create big stone square
(chain) fill -1 ~ -1 42 ~ 42 air                    /leave just a ring of stone
(chain) setblock -4 ~ -12 redstone_block            /kicks off next sequence

This named armor stand is basically our "cursor" to place all the armor stands that we will need. The redstone block in the last step "powers" nearby blocks (including our command blocks), so kicks off the next loop:
(repeat) execute @e[name=A] ~ ~ ~ summon armor_stand ~-1 ~ ~   /create new armor stand
(chain) tp @e[name=A] ~1 ~ ~                                   /move "cursor" one block
(chain) execute @e[name=A] ~ ~ ~ testforblock ~1 ~ ~ stone     /if at end of row,
(conditional) tp @e[name=A] ~-40 ~ ~1                          /go to start of next row
(chain) execute @e[name=A] ~ ~ ~ testforblock ~ ~ ~2 stone     /If at last row
(conditional) setblock ~6 ~ ~ air                              /stop looping
(conditional) kill @e[name=A]                                  /kill cursor

At this point our grid is complete:

Random Color Selector

The purple repeaters in the center of this picture choose a random color via the following command:
(repeat) execute @r[type=armor_stand,r=9] ~ ~ ~ setblock ~ ~-2 ~ redstone_block

That "@r[]" is the magic sauce, it selects a random entity in the world that matches the given conditions. In this case, it finds an armor stand inside a radius of 9 blocks, and we've set up 16 armor stands, one for each wool color. Under the selected color, it places a redstone block (which powers the two command blocks on either side).
Random Pixel Selector
Placing the redstone block under the selected wool color triggers two more command blocks:
(impulse) execute @r[type=armor_stand] ~ ~ ~ setblock ~ ~3 ~ wool X
(impulse) setblock ~ ~ ~1 air

This first line uses our same magic @r command to choose any armor stand on the entire map (no radius restriction, so that includes the 40x40 grid), and places a wool of the selected color above its head. The X determines the color, and ranges from 0 to 15. The second command removes the redstone block so it is ready to go again.
I have 5 purple repeater blocks, and redstone works in "ticks" 20 times a second, so I'm placing 100 pixels per second (minus some color overlaps). I've timed it, and I usually get the entire grid covered in about 3 minutes.
This was fun, I'll try to look for other challenges that might also work in Minecraft. Huge thanks to lorgon111 for his YouTube Command Block tutorial series.
EDIT: Made some serious reductions in the size of the saved structure, now at 2,872 saved bytes:

Scooted things in a bit (in all 3 dimensions) so I could reduce the overall size of the saved area.
Changed the different colored wools to stone, they were just decorative anyway.
Removed the glowstone lamp.
Changed all air blocks to void blocks (the red squares).

Tested by pulling the saved structure into a new world, everything still works as designed.
EDIT 2: Read-only Dropbox link to the NBT structure file
Walk through is in my YouTube video, but here are the steps:

In Minecraft 1.12, create a new creative superflat world using the "Redstone Ready" preset. Make it peaceful mode.
Once the world exists, copy the NBT file into a new \structures folder you create under the current world save.
Back in the game, do /give @p structure_block, and /tp @p -12, 56, -22 to jump to the right spot to get started.
Dig a hole and place the structure block at -12, 55, -22.
Right-click the structure block, click the mode button to switch it to "Load".
Type in "random_pixels", turn "include entities" ON, and click "Load"
If it finds the structure file, it will preview the outline. Right-click again and click "load" to bring the structure into the world.
Press the button to run the setup routine.
When it completes, flip the switch to run the wool randomization.

Holy crap... you actually did it. And it's your first answer, wow! Welcome to the site, and this is an amazing way to start!

@MDXF In addition to your +10, I'm giving +60

@NoOneIsHere I was thinking about that too. I might do it. When will you award?

@MDXF 7 days, then you can

@NoOneIsHere All right :P But if you have an answer, it'll have to be +100...

Thanks, everyone. It won't ever be competitive with more traditional languages, but I'm going to work on reducing the byte count as much as possible. I already dropped quite a bit by just excluding those two written signs from the structure block. I'll have to figure out whether a "void block" takes up less saved space than an air block.

I am going to try and golf this if you give the the world file (I will not post another answer but just give you the file)

Well, shortest answer in each language wins and this is the shortest answer in Minecraft :P have an extra +15

Thanks, @MDXF. Post edited, bytes reduced by 35%. So I've got the (very small) NBT file, can I share a dropbox link here? Or is that against policy?

That... is very impressive. Dropbox and Google Drive links are fine AFAIK, we generally prefer GitHub or GitHub Gist but however you want to do it isn't a problem.

@Christopher Dropbox link to the structure file added to the end of the post, along with instructions for use. Much smaller than the whole world download. Let me know if you have any trouble.

+1 for using something bonkers to do this. It's not often we see answers utilizing Minecraft.

Hm... I bet you could golf this by experimenting with removing random characters from the NBT file. Probably not a good idea but it might do the job.

sh + ffmpeg, 52 bytes
ffplay -f rawvideo -s cif -pix_fmt rgb24 /dev/random

Does ffmpeg count as an esolang? :D
Sadly the pix_fmt is required, as ffmpeg defaults to yuv420p. That fails the "must have equal likelihood of every possible pixel color" requirement. Conveniently, cif is a shortcut for a fairly large video size that uses less space than "40x40".

Unsurprisingly, optimizing this gif with gifsicle did absolutely nothing. It's 4MiB.

Does this "replace only one pixel at once"?

Technically? ffmpeg will wait for the entire frame to be filled with pixels before displaying it, though. I misread "must only replace 1 pixel at a time" as "may". :/

And of course gifsicle couldn't save any bytes in the gif: there is no way to compress arbitrary data and the random pixels are all chaotically arranged, meaning that the most efficient way of encoding them is one pixel at a time. Numberphile (I think?) had a video on "what is information" and did random noise to intentionally mess with YouTube's compression.  VSauce had a video on video compression at one point too, but I forget what it was called.

@Draco18s, I know, the note about it being incompressible was making fun of myself for trying to run gifsicle on it anyway.

Oh sure, its just that the reason its uncompressable is sometimes worth noting. :)

Gifsicle is the wrong tool. Precomp gets it down from 4 MB to 3 MB :) That's because the GIF algorithm expands the randomness instead of compressing it and Precomp reverses this.

The rules: Your program cannot use the internet nor your filesystem (/dev/random and /dev/urandom excepted).

And this doesn't break that rule? This answer is almost a year old, either way. It breaks a more important rule in any case and is here as a novelty.

@Draco18s An extremely powerful compressing tool could figure out the seed and compress the gif to a few bytes.

@JonathanFrech It would need to encode the seed and the algorithm used. Using a Mersenne Twister instead of whatever /dev/random uses would give a different result for the same seed. Wikipedia lists 26 different PRGNs, 4 hardware (true) RNGs, 3 cryptographic, and 4 utilizing "external entropy" (including /dev/random) which could be any of the aforementioned previous or something unlisted.

@Draco18s I am not saying that it is likely that such a software exists. I am just saying that [...] there is no way to compress arbitrary data [...] to me does not seem appropriate under the circumstances of this post.

C on POSIX, 98 96 95 92 bytes
-3 thanks to Tas
#define r rand()
f(){for(srand(time(0));printf("\e[%d;%dH\e[%d;4%dm ",r%40,r%40,r%2,r%8););}

This chooses between 16 colors (dark grey, red, green, blue, orange, cyan, purple, light grey, black, pink, light blue, yellow, light cyan, magenta, white) and prints them directly to the terminal.
Note that if your GPU is too slow, this may seem like it's updating the entire screen at once. It's actually going pixel by pixel, but C is fast.

Alternate solution that makes the colors more distinct:
f(){for(srand(time(0));printf("\e[%d;%dH\e[%d;3%dm█",rand()%40,rand()%40,rand()%2,rand()%8););}

Proof that it goes pixel by pixel (screenshot from alternate program):

Wow, that looks almost 3-dimensional...

It's missing a ; to compile correctly but seems to work great other than that! B-)

@cleblanc  Oh, you're right! I forgot to add that after I switched from while to for.

You could probably shave a few bytes by #define r rand() and then using r%40, r%40, r%2, r%8

By not using 32-bit color this does not meet the requirements.

@wberry What do you mean?

Oops I meant 24 bit color.  "all colors in the range #000000 to #FFFFFF".  The number of possibilities being 16777216 not 16.

@wberry "Your program must have an even chance of picking all colors (i.e in the range #000000 to #FFFFFF), or all colors that can be displayed on your system." These are all the colors that the POSIX terminal can display.

JS+HTML 162+32 (194) 124+13 (137) bytes
Thanks to Luke and other commenters for saving me lots of bytes.

r=n=>n*Math.random()|0
setInterval("b=c.getContext`2d`;b.fillStyle='#'+r(2**24).toString(16);b.fillRect(r(99),r(99),1,1)",0)
<canvas id=c>

Actually <canvas id=c> should be sufficient as I think the default size for the canvas-element is 300 x 150 px and so exceeds your "viewport size" of 99 x 99 px. Anyway, nice solution.

@insertusernamehere, yes, thanks

Doesn't work in Firefox now though. It does work in Chrome

Nicely done :) Similar to the solution I was going to try to work up when I got to a computer in the morning. Think you've squeezed most all you can out of this; the only thing I can spot right now is that you can omit the closing > on the canvas tag (although you will need it for it to work within a Snippet).

OK, OK, I've got it.  Save 2 bytes by doing b=c.getContext2d.  (Not sure how to format this, but if "2d" is a template literal, it doesn't need the parentheses.)  (Deleting my previous suggestions.)

@Rick, lol, fair enough

@Shaggy, I tried, it doesn't work in a stack snippet, maybe its specific to canvas and the way it gets a context, not sure.

Save 3 bytes by passing a string instead of a function to setInterval.  See  https://jsfiddle.net/5f1ww0g5/1/

Along with Rick's suggestion, use a with statement to get rid of all instances of b.: r=n=>n*Math.random()|0 setInterval("with(c.getContext\2d`)fillStyle='#'+r(2**24).toString(16),fillRect(r(99),r(99),1,1)")`

@darrylyeo, that would replace 6 bytes with 6 bytes

@darrylyeo 's suggestion should save you a byte.  See https://jsfiddle.net/5f1ww0g5/2/

Does this really have an even chance of picking all colors in the RRGGBB range? The r(2**24).toString(16) doesn't appear to be zero padded, so e.g., #3FA becomes #33FFAA instead of #0003FA right?

@apsillers, sure, you are right but does Math.random() even produce uniformly distributed values. The fact is it doesn't. Most of the random algorithms on this page are going to be rather crappy at doing that.

MATL, 28 bytes
40tI3$l`3l2$r,40Yr]4$Y(t3YGT

Try it at MATL Online. I have added a half-second pause (.5Y.) to this version.

Explanation
40       % Push the number literal 40 to the stack
t        % Duplicate
I        % Push the number 3 to the stack
3$l      % Create a 40 x 40 x 3 matrix of 1's (40 x 40 RGB image)
`        % Do...while loop
  3l1$r  % Generate 3 random numbers (RGB)
  ,      % Do twice loop
    40Yr % Generate two integers between 1 and 40. These will be the
  ]      % row and column of the pixel to replace
  4$Y(   % Replace the pixel with the random RGB value
  t      % Make a copy of the RGB image
  3YG    % Display the image
  T      % Push a literal TRUE to create an infinite loop

I love it! ----

out of curiosity, how did you giffify your code output?

@Octopus I just used LICEcap. One of these days, I'm going to incorporate animated GIF output into the online compiler.

Seeing as how this answer uses less commands/methods to achieve a similar result, couldn't you convert it to MATL to possibly achieve a lower score?

I saw an already replaced pixel get replaced again ;).

TI-BASIC (84+C(S)E only), 37 35 bytes
:For(A,1,5!
:For(B,1,5!
:Pxl-On(A,B,randInt(10,24
:End
:End
:prgmC //"C" is the name of this program

Due to hardware limitations, this will eventually crash, since every time a program is nested within a program in TI-BASIC, 15 KB of RAM are allocated to "keep a bookmark" in the parent program. This would run fine on a "theoretical" calculator with infinite RAM, but if we want it to run indefinitely on a real calculator, we can just wrap it in a While 1 loop for an extra 2 bytes:
:While 1
:...
:End

The TI-83 family calculators with color screens (TI 84+CE and CSE) support 15 colors. They have color codes 10 through 24. This cycles through all the pixels in a 120 by 120 (5!) square and assigns each a random color.
Result:

Yay, another BASIC answer! Just to be sure, does this run forever?

@MDXF Now it does! ;-). Missed that part the first time. +5 bytes.

You can make the for loops start at 0 if you want to cover 100x100 pixels.

@kamoroso94 Ah, true. I could also do 5! if I wanted 120 pixels.

As the actual bytecount for the code is a lot larger than 35 bytes, I'm guessing that isn't the golfed code. I suggest you include the actual golfed version at the top of the answer instead of the ungolfed one.

@MDXF TI-BASIC is token-based. I.e. For( is 1 byte, Pxl-On( is 1 byte, randInt( is 2 bytes, etc.

@ScottMilner Ah, I didn't know that! Thanks!

Bash + coreutils, 59  56 bytes
for((;;)){ printf "\e[48;5;`shuf -i 0-16777215 -n1`m ";}

\e[48;5;COLORm is the escape secuence to background color.
Each "pixel" has the chance to be in the [0..16777215] range every time.

Wow, that's a really good idea! +1

You can shove off 3 bytes, if you use jot: for((;;)){ printf "\e[48;5;\jot -r 1 0 16777215`m ";}`

I stole this idea and golfed it down to 45 bytes (\e is 1 escape char, and shuf | xargs makes it shorter (no need for -n 1, and no for). https://codegolf.stackexchange.com/a/158142/7017 . I can delete it if you want (I'm not really sure if I should just propose this as a comment instead of posting it. My goal is to get a bit of points to one day bounty the amazing answers to the amazing "tetris in game of life" challenge & answer! see https://codegolf.stackexchange.com/q/11880/7017

MATLAB, 56 bytes
x=rand(40,40,3);while imagesc(x),x(randi(4800))=rand;end

Output looks like the image below. One "pixel" changes at a time, and only one of the RGB-colors changes.
Why? The colors in MATLAB are represented as a 3D-matrix, one layer for R,G and B. The code above changes only one of the layers per iteration. All pixels and all layers can be changed, so if you wait a bit all colors are equally possible in all positions.
Add pause(t) inside the loop to pause t seconds between each image.
You must stop it with Ctrl+C.

Clever use of the handle to the graphics object as the conditional in the where

A rule has been clarified - Your program must have an equal chance of picking all colors/points each iteration. Does your program fulfill this?

Javascript+HTML 118 + 13 (131 bytes)

r=_=>255*Math.random()|0;setInterval('x=c.getContext`2d`;x.fillRect(r(),r(),1,1,x.fillStyle=`rgb(${[r(),r(),r()]})`)')
<canvas id=c>

This will produce evenly distributed RGB colors. You cannot use Hex colors without proper padding as numbers like #7 is not a valid color, or #777 and #777777 are the same color (2x the odds)
The canvas element is 300x150 by default, but I'm actually drawing on a  255x255 square, so there are off canvas pixels, so the effective area is 255x150.
Works only on Google Chrome.

Great first post! Welcome to the site!

I don't believe this has any chance of displaying any colors with a channel set at 255, such as red, green, blue, or white.  I think you have to multiply by 256 instead.

Processing, 90 bytes
void draw(){float n=noise(millis());int i=(int)(n*9999);set(i%99,i/99,(int)(n*(-1<<24)));}

expanded and commented:
void draw(){
  float n=noise(millis());//compute PRNG value
  int i=(int)(n*9999);    //compute 99x99 pixel index
  set(i%99,i/99,          //convert index to x,y
  (int)(n*(-1<<24)));     //PRNG ARGB color = PRNG value * 0xFFFFFFFF
}

Ideally I could use a pixel index instead of x,y location, but Processing's pixels[] access requires loadPixels() pre and updatePixels() post, hence the use of set(). point() would work too, but has more chars and requires stroke(). The random area is actually 99x99 to save a few bytes(instead of 100x100), but that should cover 40x40 with each pixel in such an area to be replaced.
Perlin noise() is is used instead of random() to keep it more pseudo-random and a byte shorter. The value is computed once, but used twice: once for the random position, then again for the colour.
The colour is actually ARGB(00000000 to FFFFFFFF) (not RGB) (bonus points ? :D).

Awesome, and good to see you on this site!

Like-wise ! and thanks! :)

+1 set() is a very neat trick and can be very useful for Tips for golfing in Processing

ooooooh! I didn't know about that. I'll add it there, thanks :)

Excel VBA, 131 102 85 Bytes
Anonymous VBE immediate window function that uses a helper function (see below) to output an array of randomly colored cells to the range A1:AN40 of the activesheet object.
Note: This solution is restricted to 32-Bit installs of MS Excel (and therefore of Office as a whole) as 8^8 will not compile on 64-Bit versions of VBA
Randomize:Cells.RowHeight=48:For Each c In[A1:AN40]:c.Interior.Color=(8^8-1)*Rnd:Next

Sample Output

Previous Version
Randomize:Cells.ColumnWidth=2:For Each c In Range("A1:AN40"):c.Interior.Color=RGB(n,n,n):Next

Helper Function
Outputs a random int in the range [0,255]
Function n
n=Int(255*Rnd)
End Function

x86 machine language (real mode) for IBM PC, 20 19 bytes
 0:       b8 12 00                mov    $0x12,%ax
 3:       31 db                   xor    %bx,%bx
 5:       cd 10                   int    $0x10
 7:       0f c7 f0                rdrand %ax
 a:       88 e1                   mov    %ah,%cl
 c:       0f c7 f2                rdrand %dx
 f:       b4 0c                   mov    $0xc,%ah
11:       eb f2                   jmp    0x5

This requires a processor with the rdrand instruction and a VGA adapter (real or emulated). The above can be copied into a boot block or MS-DOS *.COM file.
To try this out, compile the following and save the output to a file like floppy.img and boot the image on a virtual machine.
#include<stdio.h>
#include<string.h>
#include<unistd.h>
int main(){
  char buffer[ 1440*1024 ];
  memcpy( buffer, "\xb8\x12\x00\x31\xdb\xcd\x10\x0f\xc7\xf0\x88\xe1\x0f\xc7\xf2\xb4\x0c\xeb\xf2", 20 );
  memcpy( buffer + 510, "\x55\xaa", 2 );
  write( 1, buffer, sizeof buffer );
}

C++ on Windows, 125 bytes
#include<Windows.h>
#include<ctime>
#define r rand()%256
int main(){for(srand(time(0));;)SetPixel(GetDC(0),r,r,RGB(r,r,r));}

Newlines necessary and included in byte count.
Loops forever, randomly picks a position between 0 and 255 (inclusive) for row and column values, assigns random R,G,B values between 0-255 (inclusive)

I realize you're golfing this, but man that leak of a device context each time through the loop really gets my spidey-senses tingling!

Save 15 bytes by making it C on Windows instead of C++ on Windows, as C compilers will let you leave off the #includes. (And yes, there are non-Visual Studio C compilers for Windows.)

C#, 369 288 287 bytes
namespace System.Drawing{class P{static void Main(){var g=Graphics.FromHwnd((IntPtr)0);var w=Windows.Forms.Screen.GetBounds(Point.Empty);for(var r=new Random();;)g.FillRectangle(new SolidBrush(Color.FromArgb(r.Next(256),r.Next(256),r.Next(256))),r.Next(w.Width),r.Next(w.Height),1,1);}}}

Saved 88 bytes thanks to @CodyGray.
A full program that get's the handle to the screen and it's size and then starts randomly drawing pixels on it. Note that this might grind your graphics card to a halt when ran. Also if the screen or any control decides to repaint at any time the pixels will be lost and have to be redrawn.
Note: When running keep focus on the window as to kill it you have to either Alt+F4 or press the close button and doing so when you can't see the screen is a bit hard.
I couldn't record this working, with ScreenToGif, as that kept forcing a repaint so the pixels would get removed. However, here is a screenshot of it running after about 10-15 seconds, any longer and I think I may have ground my PC to a halt! The gap in the top right corner is where the screen forced a repaint just as I took the screenshot.

Full/Formatted version:
namespace System.Drawing
{
    class P
    {
        static void Main()
        {
            var g = Graphics.FromHdc((IntPtr)0);
            var w = Windows.Forms.Screen.GetBounds(Point.Empty);

            for (var r = new Random();;)
                g.FillRectangle(new SolidBrush(Color.FromArgb(r.Next(256), r.Next(256), r.Next(256))),
                                r.Next(w.Width), r.Next(w.Height), 1, 1);
        }
    }
}

A version for 308 227 226 bytes that only draws on the region 0-40:
namespace System.Drawing{class P{static void Main(){var g=Graphics.FromHdc((IntPtr)0);for(var r=new Random();;)g.FillRectangle(new SolidBrush(Color.FromArgb(r.Next(256),r.Next(256),r.Next(256))),r.Next(40),r.Next(40),1,1);}}}

Example output for this one:

Graphics.FromHwnd((IntPtr)0) would also work, and be much shorter than P/Invoking GetDC. Of course, your code is still like Tas's, and leaks resources like crazy, which is why it grinds your machine to a halt after only a short time of running.

@CodyGray Oh I know it leaks, should still work within the challenges specs though. And GetDC does not return IntPtr.Zero so sending that to FromHdc will not work.

Read the suggestion more carefully. :-) I'm not suggesting that you call FromHdc, I'm suggesting that you call FromHwnd. Notice that when you call GetDC to obtain the DC for a window handle, you pass in a NULL pointer (0) as the window handle.

@CodyGray Ah yes that works, thanks! They're named to similar, should have read your comment more carefully...

C# Console, 233 220 189 188 bytes
namespace System{using static Console;class P{static void Main(){for(var r=new Random();;){BackgroundColor=(ConsoleColor)r.Next(16);SetCursorPosition(r.Next(40),r.Next(40));Write(" ");}}}}

Uses "all" (windows) 16 console colors.
Thanks for the "feature" to be able to alias classes in C# via the using directive.
Edit #1

Removed some spaces.
Removed zeros from Random.Next()
Went to namespace system{...}

Edit #2

Minimum size of grid is 40x40
One byte by declaring Random in for loop header
Removed public from Main method

Edit #3
Turns out using C=Console; is not the best there is. using static Console is much like the VB.Net way to "Import" classes
I give up: TheLethalCoder made this happen

Original code for adapting window size at 207 bytes:
namespace System{using static Console;class P{static void Main(){for(var r=new Random();;){BackgroundColor=(ConsoleColor)r.Next(16);SetCursorPosition(r.Next(WindowWidth),r.Next(WindowHeight));Write(" ");}}}}

Original Image:

namespace System should save bytes, r.Next can remove the zero I believe, Remove irrelevant white space,

Also just titling C# is fine

Thanks! I leave Console in the title since I feel for this challenge it provides valuable context. I don't see how namespace System{} is shorter than using System;. Care to explain?

Because you can then drop the System from System.Console

Hittin my head on the table right now.

Pixels can be any size, as long as your output has at least 40x40 "pixels". So r.Next(C.WindowWidth),r.Next(C.WindowHeight) can change to r.Next(40),r.Next(40) (or 41?) although it doesn't look as cool...

Move the random declaration into the for loop to save a byte: for(var r=...

Oh and one last one I've just seen, you don't need the access modifier on Main, i.e. remove public.

188: namespace System{using static Console;class P{static void Main(){for(var r=new Random();;){BackgroundColor=(ConsoleColor)r.Next(16);SetCursorPosition(r.Next(40),r.Next(40));Write(" ");}}}} I'm excited first use I've seen for using static in code golf.

I had this strong but unfounded memory that public is nescessary for the Main. Unfounded memories are the best!

Oh please, I've been waiting for that feature so long... and it's just there?!

Don't give up, I've spent a lot of time golfing in C# recently and use it daily for work. It does take a while to learn most of the language quirks you can abuse for code golf. My answers still get a lot of improvements suggested on them :)

Thanks. Will not give up! :) Will use what I learned on the next challenge!

@KevinCruijssen what makes you think that?

Wow! That looks amazing!

6502 Assembly, 92 bytes
loo: lda $fe
sta $00
lda $fe
and #$3
clc
adc #$2
sta $01
lda $fe
ldy #$0
sta ($00),y
jmp loo

Output:

Explanation:
loop: lda $fe       ; accumulator = random
      sta $00       ; store accumulator
      lda $fe       ; accumulator = random
      and #$3       ; accumulator &= 3
      clc           ; clear carry
      adc #$2       ; accumulator += 2
      sta $01       ; store accumulator
      lda $fe       ; accumulator = random
      ldy #$0       ; register Y = 0
      sta ($00),y   ; store register Y
      jmp loop      ; loop

What is yoiur target 6502 platform here? BBC Micro? Apple ][? Commodore VIC-20 etc...?

Do label names have to be at least three characters long?

6502 Assembly, 582 bytes
Whew, this was fun. Does more or less the same thing as my Applesoft BASIC solution.
start:
lda #15
sta $0
sta $1
loo:
lda $fe
and #3
cmp #0
beq g_l
cmp #1
beq g_r
cmp #2
beq g_d
dec $1
d_w:
lda $1
and #$1f
asl
tax
lda ypo,x
sta $2
inx
lda ypo,x
sta $3
lda $0
and #$1f
tay
lda ($2),y
tax
inx
txa
sta ($2),y
jmp loo
g_d:
inc $1
jmp d_w
g_l:
dec $0
jmp d_w
g_r:
inc $0
jmp d_w
ypo:
dcb $00,$02,$20,$02,$40,$02,$60,$02
dcb $80,$02,$a0,$02,$c0,$02,$e0,$02
dcb $00,$03,$20,$03,$40,$03,$60,$03
dcb $80,$03,$a0,$03,$c0,$03,$e0,$03
dcb $00,$04,$20,$04,$40,$04,$60,$04
dcb $80,$04,$a0,$04,$c0,$04,$e0,$04
dcb $00,$05,$20,$05,$40,$05,$60,$05
dcb $80,$05,$a0,$05,$c0,$05,$e0,$05

Is this the ungolfed version of your other 6502 assembly answer? :-)

@CodyGray Nope, it does things a bit differently. It's more like my Applesoft BASIC answer (uses a random walk), and the other 6502 assembly answer is more like all the other ones.

I disagree with the byte counts: this is way shorter than 583 bytes of assembly?

@OlivierDulac https://tio.run/##Zc/bioMwEAbg@3mKgLnoggs5LsWXKTGxe8CtXQ1Un97G5IcG9mqYPzOfY@@Wr31/ePbu932Jbo4djcGxRlpKLeOiFEnjNJUnfh3I3QJrNPnfO2sE9cMf@7yMpZVo59IqtIHC4A8nXB5wZGG4vJJbRopuzfl2n9q1fFXR9@1fqMu2eG1Ht@XsxNVbu2Xo2IuryxuIf9I56Sco3dKl93zMkaWDUjZ25UBRZTPmXlk6I835PmWi5UK1XKEa1I9cy8wZmUP1qEM1kx0NR8PRcHTlaDgajoajK8fAMXAMHFM5Bo6BY@CYyrFwLBwLx1aOhWPhWDj2CQ ?

Python 3.6 + Tkinter, 281 bytes
from tkinter import*
from random import*
from threading import*
a=randrange
x=40
d={"width":x,"height":x}
w=Tk()
c=Canvas(w,**d)
c.pack()
i=PhotoImage(**d)
c.create_image((20,20),image=i)
def r():
 while 1:i.put(f"{a(0,0xffffff):0>6f}",(a(0,x),a(0,x)))
Thread(r).start()
mainloop()

As tkinter is a standard library, you don't need to include it in the header

I get the same error -- Windows 10, Python 3.6.0 I can guess why init is seeing four passed parameters -- the method is being passed a "self" parameter implicitly.  But I have no idea why that is generating an error, since the documentation shows the Canvas constructor accepts three parameters (master, x, y)  as shown in the code.

Should be fixed now.

also @Ilikemydog Tkinter is not always in stdlib. On windows it's optional when installing, and I don't think my arch partition has Tkinter installed either. Might as well include it.

ther is typo - "img" instead of "i". And still it does not work, this time I get c.create_image((20,20),i)   File "C:\Python36\lib\tkinter\__init__.py", line 2483, in create_image     return self._create('image', args, kw)   File "C:\Python36\lib\tkinter\__init__.py", line 2474, in _create     *(args + self._options(cnf, kw)))) _tkinter.TclError: unknown option "pyimage1"

Can't you reduce ffffff to fff? And can't you put Tk() directly  in the argument?  And can't you do image=i=PhotoImage(**d) directly in the argument?

JavaScript using Canvas 340 316 324 bytes

function r(t,e){return Math.floor(e*Math.random()+t)}function f(){x.fillStyle="rgba("+r(1,255)+","+r(1,255)+","+r(1,255)+", 1)",x.fillRect(r(0,40),r(0,40),1,1)}c=document.createElement("canvas"),c.width=40,c.height=40,x=c.getContext("2d"),document.getElementsByTagName("body")[0].appendChild(c),c.interval=setInterval(f,1);

full version

Welcome to the site! :)

Does this honor the rule, "Your program must continue to display random pixels until manually stopped (it cannot terminate on its own)."?

This appears to simply generate a bunch of random pixels when run. It does not continue to display pixels one by one until stopped.

Seems that I missed that rule.  I mistakenly thought the goal was to create an image not an animation. I will update my answer when I return to my computer.

@TimPenner for future reference, var is usually optional (but very good to include) in JavaScript, so you can usually ignore it. You can also always use one letter variable names to save some bytes.

@TimPenner I think you might be able to redefine document to also save a byte or two, if you use it more than once in your next version also

@MDXF it's animated now

@Octopus it does now

Nice first effort! I golfed your entry down to 228 and gave you a list of pointers for your next JavaScript golf: https://jsfiddle.net/pg4aLw97/1 Feel free to ask if anything is unclear :)

HTML+SVG+PHP, 245 Bytes
<?$u=$_GET;$u[rand()%40][rand()%40]=sprintf("%06x",rand()%16777216);echo'<meta http-equiv="refresh" content="0.1; url=?'.http_build_query($u).'" /><svg>';foreach($u as$x=>$a)foreach($a as$y=>$c)echo"<rect x=$x y=$y width=1 height=1 fill=#$c />";

Expanded
$u=$_GET; # Get the Url
$u[rand()%40][rand()%40]=sprintf("%06x",rand()%16777216); # Set One Value in a 2 D Array
echo'<meta http-equiv="refresh" content="0.1; url=?'.http_build_query($u).'" /><svg>'; # refresh the site after 0.1 second follow the new Get parameter
foreach($u as$x=>$a) #loop through x Coordinates as Key
  foreach($a as$y=>$c) #loop through y Coordinates as Key value is the color
    echo"<rect x=$x y=$y width=1 height=1 fill=#$c />"; #print the rects for the SVG

Example for Output without meta tag and in a greater version

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Python, 133 bytes
I'm not quite sure if this fits the specs, because it's on a canvas in a 40x40 area.
from turtle import*
from random import*
ht()
up()
speed(0)
R=randint
while 1:goto(R(0,39),R(0,39));dot(1,eval('('+'R(0,255),'*3+')'))

Try it online - version without eval, which doesn't work in Trinket

Note quite sure why, maybe your turtle dist is different from mine, but my default canvas is 300 by 400 while the default colormode is 1.0 not 255.

Trinket.io is pretty unique.

Yeah - I just tried changing colormode and it wouldn't have any of it :/

Why does dot(1,R(0,255),R(0,255),R(0,255)) not work either? dot is meant to accept (width,*color). It's not Python :p

Because Trinket.io uses skulpt.js, which is an incomplete implementation of Python.

shortC, 66 56 bytes
Dr rand()
AOZR"\e[%d;%dH\e[%d;4%dm ",r%40,r%40,r%2,r%8);

-10 bytes thanks to Dr. Rand. :P
A                                                                  main function
 O                                                                 for
  Z                                                                seed rand()
   R                                                               print
    "\e[%d;%dH                                                     coordinate placeholder string
              \e[%d;4%dm "                                         color placeholder string
                          ,rand()%40,rand()%40,rand()%2,rand()%8   fill placeholders
                                                                ); interpreter hole

No TIO link because you obviously can't print to the kind of terminal this requires online.

Processing, 112 bytes
void setup(){size(40,40);}void draw(){stroke(random(255),random(255),random(255));point(random(40),random(40));}

I can't guarantee that every pixel is replaced every 3 minutes, but looking at it it appears to be doing so.  At least the odds of it missing a pixel, out of 1600 total, updating a random one 30x per second, totaling 5400 updates per 3 minutes, makes it unlikely that one would be missed.
Ungolfed:
The program is really straightforward.  Open a window at 40x40 pixels, and every frame (default 30 per second) get a random color, and draw a point at a random coordinate between 0 and the parameter.  40 for pixel coordinates, 255 for colors.
void setup()
{
  size(40,40);
}

void draw()
{
  stroke(random(255),random(255),random(255));
  point(random(40),random(40));
}

Can you remove the size(40,40); and change point to point(random(99),random(99));? (or 100 instead if 99 isn't working)

@KritixiLithos in Processing, the size function MUST be the first line of code in the setup function (reasons why are complicated), so I can't get around that.  I can change all occurrences of 40 to 99, I just didn't since it didn't save me any space.  I mainly went with the minimum size to increase the odds that every pixel will be refreshed in a 3-minute period

Heh, that was what I first thought as well, until someone told me about it and it somehow worked out, as you can see here in a submission of mine.

Removing the whole setup() function on the latest version of Processing on my machine works fine without errors.

I guess my version of Processing is quite old, and I don't use it enough to bother upgrading right now.

Yep, you can remove the setup() function completely. Also the default is 60 fps, not 30. You can probably also save a few bytes by generating a random int and using that as the color instead of individual RGB values.

I need the minimum size to be more confident that every pixel is replaced in the time frame given in the OP.  Also, if I used the same random int for each pass of the draw function, I would only get shades of gray.

@Cody No, colors are stored as int values behind the scenes. You can generate a random one and use that as the color. Actually I just noticed George's Processing answer does this, so I'll go upvote him. :p

JavaScript (ES7) + HTML using SVG, 129 + 10 = 139 bytes
SVG version, heavily inspired by @Octopus's <canvas> approach.
JS
r=n=>n*Math.random()|0
setInterval('s.innerHTML+=`<rect x=${r(40)} y=${r(40)} fill=#${r(2**24).toString(16)} width=1 height=1>`')

HTML
<svg id=s>

Since the Stack Snippet likes to break the script by parsing the <rect> tag, here's a CodePen.

LOGO, 71 bytes
The only language that I know of, being not esoteric/specially designed for codegolfing, and have forever function. Any idea on the map part?
pu forever[setxy random 100 random 100 setpixel map[random 256][1 1 1]]

Perl (on *nix), 69 bytes
The \x1bs are literal escape characters.
Relies on stty command and works well on OS X.
{printf"\x1b[%i;%iH\x1b[48;5;%im ",map{rand$_}`stty size`=~/\d+/g,256;redo}

Similar to other approaches, but I liked the combination of all params into one call to printf so thought I'd share. Kills my terminal.

Twice as many pixels, 83 bytes
{printf"\x1b[%i;%iH\x1b[48;5;%i;38;5;%im▄",map{rand$_}`stty size`=~/\d+/g,256,256;redo}

This approach uses a unicode block ▄ and a random foreground and background colour which gives a more square pixel. Kills my terminal too, but looks cooler.

Don't forget you can use the literal escape byte, ASCII 0x1B (27) instead of \e. And on StackExchange this can be represented by ␛.

@MDXF Yeah... I need to test that as a script, because that wasn't working via -e! I also need to up the byte count of the one with unicode. Oops! Thanks for the reminder!

Bash, 104 bytes
The \es are literal escape characters.
These are pretty much translations of my Perl submission, but using bash fork bomb style syntax! Not as clever as the other bash entry that writes directly to the display I'm afraid.
:(){ printf "\e[%i;%iH\e[48;5;%im " $((RANDOM%`tput lines`)) $((RANDOM%`tput cols`)) $((RANDOM%256));:;};:

More pixels, 130 bytes
Same as my Perl answer, this uses a unicode character for the foreground and colours the background of each pixel too.
:(){ printf "\e[%i;%iH\e[48;5;%i;38;5;%im▄" $((RANDOM%`tput lines`)) $((RANDOM%`tput cols`)) $((RANDOM%256)) $((RANDOM%256));:;};:

Not bad. However, I still don't understand the "\e[48;5;%im". Why are the 48 and 5 necessary?

That access the full 256 colours, rather than just the standard 16 colour palette. In terminals that support it, it's possible to use \e[48;2;RRR;GGG;BBBm but very few terminal emulators support it... :(

Holy mother of code.... I had no idea that was possible. My life is a lie. Thanks a million :P

@MDXF ^^ and also 38;5 denotes foreground and 48;5 deonates background. http://misc.flogisoft.com/bash/tip_colors_and_formatting

@MDXF No worries! Glad to help! Shame I missed this when it was brand new, but that Minecraft answer... unreal!

Super Chip (48)?, 12 bytes
0x00FF 'enter high resolution mode (64x128 pixels)
0xA209 'set I to 0x209 (second byte of draw instruction)
0xC03F 'set register 0 to a random number from 0 to 63
0xC13F 'set register 1 to a random number from 0 to 63
0xD101 'draw a sprite. x=register 1, y=register 0, height=1
0x1204 'jump to third instruction

I'm not sure of the exact name, but I had to use this instead of normal Chip-8 because of the 40x40 pixel limitation.

Link to this (very interesting) language?

You can use the Octo emulator: https://johnearnest.github.io/Octo/

QBIC, 34 bytes
screen 12{pset(_r640|,_r480|),_r16

Unfortunately, QBIC doesn't set a SCREEN mode by default, so that eats some bytes. PSET is an illegal command in QBIC's default graphical context.
Explanation
screen 12           Set the screen to a mode supporting (colored) graphics
{                   DO until the compiler is killed by ctrl-scroll lock
pset                PSET is a QBasic command to set one pixel
(_r640|,_r480|)     it takes a set of coords (x, y) which QBIC chooses at random with _r
,_r16               and a color (1,16)

Sample output
@Anonymous found an error in QBIC's rnd() function. Fixing that gives the output below. Thanks!

As should be obvious from the screen shot, your solution doesn't satisfy the randomness conditions. Given a coordinate, not all colours are equally likely and given a colour, not all coordinates are equally likely; they're highly correlated and that's putting it mildly. The problem is the random number generator you're using: FUNCTION getRandomNumber! (Bottom!, Top!) ↵ RANDOMIZE TIMER 'this sets the random number generator ↵ IF Bottom > Top THEN ... END IF ↵ getRandomNumber = INT((Top - Bottom + 1) * RND + Bottom) ↵ END FUNCTION Your random numbers are just hashes of the current time.

@Anonymous It's fixed now, thanks!

GW-Basic, 47 bytes (tokenised)
0 SCREEN 7:RANDOMIZE TIMER:WHILE 1:PSET(RND*40 MOD 40,RND*40 MOD 40),RND*16 AND 15:WEND

To get it down to 47 bytes, you'll have to strip spaces and trailing zeroes and end of file character manually.

Minecraft 1.12.2, 747 bytes(compressed), 836+60(contain+filename)
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To prove whole world is not always bigger (bin is .rar file)
level.dat
    Data
        GameRules
            gameLoopFunction: a
        GameType: 3 # so you don't fall and mob won't spawn at night
        generatorName: flat
        generatorOptions: 3;256*1;1 # full of stone
        SpawnX: 0
        SpawnZ: 0
        version: 19133 # avaliable
data
    functions
        minecraft
            a.mcfunction
                summon area_effect_cloud
                # won't show something when die
                spreadplayers 0 0 0 20 0 @e[type=!player]
                execute @r[type=!cow] ~ ~ ~ tp @s ~ ~-1 ~
                execute @r[type=!cow] ~ ~ ~ tp @s ~ ~-2 ~
                execute @r[type=!cow] ~ ~ ~ tp @s ~ ~-4 ~
                execute @r[type=!cow] ~ ~ ~ tp @s ~ ~-8 ~
                # minecraft bug, tp @r[type=!cow] ~ ~-1 ~
                # always tp player
                execute @e[type=!player] ~ ~ ~ function b
                tp @a ~ 275 ~
            b.mcfunction
                setblock ~ ~-1 ~ wool 15
                setblock ~ ~ ~ wool 14
                setblock ~ ~1 ~ wool 13
                setblock ~ ~2 ~ wool 12
                setblock ~ ~3 ~ wool 11
                setblock ~ ~4 ~ wool 10
                setblock ~ ~5 ~ wool 9
                setblock ~ ~6 ~ wool 8
                setblock ~ ~7 ~ wool 7
                setblock ~ ~8 ~ wool 6
                setblock ~ ~9 ~ wool 5
                setblock ~ ~10 ~ wool 4
                setblock ~ ~11 ~ wool 3
                setblock ~ ~12 ~ wool 2
                setblock ~ ~13 ~ wool 1
                setblock ~ ~14 ~ wool

Minecraft 1.12.2, 806 bytes(compressed), 1352+60(contain+filename)
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01e0h: 96 D5 25 02 03 0B 8F 01 04 8F 01 20 83 7C 6D FD
01f0h: 80 00 00 09 6C 65 76 65 6C 2E 64 61 74 0A 03 02
0200h: 10 70 3C 51 27 6D D4 01 1F 8B 08 00 C9 78 D3 5B
0210h: 00 FF 35 8D 41 0A C2 30 14 44 47 D2 6A 1A 28 78
0220h: 14 C5 65 B7 62 41 A4 05 75 21 EE 3E F2 5B 84 9A
0230h: 84 24 55 3C 9C 77 B3 1F 74 37 8F 79 C3 18 C0 20
0240h: DB 52 22 8D B2 67 CB 81 92 0B 0D 3D 18 59 37 50
0250h: 52 D0 F5 04 E7 B7 67 00 4A 61 F1 E4 10 EF CE 02
0260h: FB 8F 41 21 E5 71 1C 38 6A 2C FB 29 1F 9C F3 BB
0270h: D1 DE 92 28 33 82 C2 FC E4 E9 65 AF 90 F9 0F 2E
0280h: 02 32 F8 FF B5 5E FC 88 7C 53 AD AA 35 F0 05 F9
0290h: 41 3F 00 96 00 00 00 1B 7A 52 1B 32 02 03 0B 00
02a0h: 05 00 10 00 00 00 00 80 00 00 18 64 61 74 61 2F
02b0h: 66 75 6E 63 74 69 6F 6E 73 2F 6D 69 6E 65 63 72
02c0h: 61 66 74 0A 03 02 BB 8F 67 2F 25 6D D4 01 C4 B1
02d0h: B7 E8 28 02 03 0B 00 05 00 10 00 00 00 00 80 00
02e0h: 00 0E 64 61 74 61 2F 66 75 6E 63 74 69 6F 6E 73
02f0h: 0A 03 02 D1 6A 8E 26 25 6D D4 01 F6 87 4C DB 1E
0300h: 02 03 0B 00 05 00 10 00 00 00 00 80 00 00 04 64
0310h: 61 74 61 0A 03 02 EE 28 D1 1F 25 6D D4 01 1D 77
0320h: 56 51 03 05 04 00

level.dat
    Data
        GameRules
            gameLoopFunction: a
        GameType: 3 # so you don't fall and mob won't spawn at night
        generatorName: flat
        generatorOptions: 3;0;1 # just air
        SpawnX: 0
        SpawnZ: 0
        version: 19133 # avaliable
data
    functions
        minecraft
            a.mcfunction
                summon area_effect_cloud
                # won't show something when die
                fill -20 3 -20 20 3 20 wool
                # init. A .mca is at least 12KB too expensive
                # (though flat generation may work fine)
                scoreboard objectives add a dummy
                execute @r[type=!cow] ~ ~ ~ scoreboard players add @s a 1
                execute @r[type=!cow] ~ ~ ~ scoreboard players add @s a 2
                execute @r[type=!cow] ~ ~ ~ scoreboard players add @s a 4
                execute @r[type=!cow] ~ ~ ~ scoreboard players add @s a 8
                # minecraft bug, scoreboard players add @r[type=!cow] a 1
                # always add score to player
                execute @e[type=!player] ~ ~ ~ function b
            b.mcfunction
                spreadplayers 0 0 0 20 0 @s
                execute @s ~ 4 ~ setblock ~ ~ ~ wool
                # setblock ~ 4 ~ wool uses the old location
                execute @s[score_a=15] ~ 4 ~ setblock ~ ~ ~ wool 15
                execute @s[score_a=14] ~ 4 ~ setblock ~ ~ ~ wool 14
                execute @s[score_a=13] ~ 4 ~ setblock ~ ~ ~ wool 13
                execute @s[score_a=12] ~ 4 ~ setblock ~ ~ ~ wool 12
                execute @s[score_a=11] ~ 4 ~ setblock ~ ~ ~ wool 11
                execute @s[score_a=10] ~ 4 ~ setblock ~ ~ ~ wool 10
                execute @s[score_a=9] ~ 4 ~ setblock ~ ~ ~ wool 9
                execute @s[score_a=8] ~ 4 ~ setblock ~ ~ ~ wool 8
                execute @s[score_a=7] ~ 4 ~ setblock ~ ~ ~ wool 7
                execute @s[score_a=6] ~ 4 ~ setblock ~ ~ ~ wool 6
                execute @s[score_a=5] ~ 4 ~ setblock ~ ~ ~ wool 5
                execute @s[score_a=4] ~ 4 ~ setblock ~ ~ ~ wool 4
                execute @s[score_a=3] ~ 4 ~ setblock ~ ~ ~ wool 3
                execute @s[score_a=2] ~ 4 ~ setblock ~ ~ ~ wool 2
                execute @s[score_a=1] ~ 4 ~ setblock ~ ~ ~ wool 1

So, what is this, exactly? Is this a compressed version of the structure block from my answer? Or your own program/save file?

@BradC the whole map with level.dat etc.

Also, no structure block, no reading your solution

Cool, a new independent Minecraft solution, then. I'd love to see more about how yours works, its a little hard to tell from a block of raw binary :)

@BradC added explanation

Powershell, 128 126 124 88 bytes
-2 bytes thanks @Veskah
1..($c=[Console])::WindowWidth*$c::WindowHeight|%{Write-Host ' '-b(random 16)-n}
for(){}

Output:

VS code output:

Don't need the semi-colon and there's an extra space in the write-host

not sure this satisfies the condition of overwriting every cell at least once within 3 minutes

Thanks. Fixed. ¯\(ツ)/¯

Your latest edit is not functional for the terms of the challenge.

A golf of your last working answer that brings it down to 105 bytes and ensures all cells can be rewritten within 3 minutes as required: mode 40,40;for(){[Console]::SetCursorPosition((random(40)),(random(40)));Write-Host ' '-b(random 16)-n}

Nice. You can create a new post with your PS code. And i don't understand why the latest edit is not functional for the terms of the challenge.

Commodore 64 (old ROM I think), ~28 tokenized BASIC Bytes used
 0poke646,rnd(.)*16:?"{reverse on} ";:goto

The memory location 646 is used as a colour out in the BASIC kernal (pretty sure this is the old kernal, not the C64c kernal), which is one of the 16 colours available in its palette. Then a reverse space is printed after. goto is used as an infinite loop until the screen is filled.
This is the cheapest in terms of bytes. There is a nicer way to do this, as follows:
Commodore 64 (new and old ROM), ~46 tokenized BASIC bytes
 0 fori=0to999:poke1024+i,160:poke55296+i,rnd(.)*16:next

This will poke firstly a reversed (or inversed) space to the screen RAM (at 0x0400 - 0x07e7), and then alter the colour attribute for each character position (mapped at 0xd800 - 0xdbe7) between 0 and 15 inclusive.
As soon as the for/next loop ends, the ready. prompt is shown. To stop this, add the following line, costing another 7 tokenized BASIC bytes:
 1 go to 1

TI-Basic (83/84(+(SE)) only), 33 bytes
For(I,0,62
For(J,0,94
If int(2rand
Pxl-Change(I,J
End
End
prgmC

where C is the name of the program.
Without calling itself (+2 bytes):
While 1
For(I,0,62
For(J,0,94
If int(2rand
Pxl-Change(I,J
End
End
End

The TI-83 family calculators with monochrome screens (TI-83, TI-83+, TI-83+SE, TI-84+, and TI-84+SE) support only black and white as colors. This program goes through all the pixels on the screen (except for the rightmost column and bottommost row) and has a 50% chance of inverting it. This may take more than 3 minutes depending on the calculator; if so, replacing 62 and 94 with 39 each should work.

R, 143 138 bytes
-4 bytes thanks to user2390246
r=runif
s=sample
g=1600
x=1:40
k=hsv(r(g),r(g),r(g))
while(1){image(x,x,matrix(1:g,40),c=k)
Sys.sleep(99/g)
k[s(g,1)]=hsv(r(1),r(1),r(1))}

image makes the required 40x40 grid of rectangles with a matrix m of values (fixed to be 1:1600 in a 40x40 square), with a randomly selected color for each rectangle, k. It plots the image, waits long enough to actually render the image, then randomly changes one of the colors. Uses hsv encoding since it is a value between 0 and 1, which allows us to use runif to generate random colors.
This generates a new plot each time.

It won't affect the score, but shouldn't image(x,x,m,c=k) be outside the while loop?

I don't think you need to define m, you only use it once.

@user2390246 for the image question, nah, it has to be re-called every time since it's really drawing a new image each time after changing one of the colors. Good call on the m, though. I originally thought I had to change that matrix and never went back and changed it.

PHP, 194 bytes
My previous answer did not animate; this one does.
<?session_register("i");$i||imagecreatetruecolor(64,64);imagesetpixel($i,rand(64)%,rand()%64,rand()%2**24);header("Content-Type: image/png");header("Location: ".basename(__FILE__));imagepng($i);

save to file, call in browser. It´s as fast as your browser can refresh.
Probably not able to set 4096 pixels in 180 seconds though.
breakdown
<?                                                  # start PHP
session_register("i");                              # register image in session
$i||imagecreatetruecolor(64,64);                    # if no image exists, create one
imagesetpixel($i,rand(64)%,rand()%64,rand()%2**24); # set a random pixel to random color
header("Content-Type: image/png");                  # send HTTP header for image
header("Location: ".basename(__FILE__));            # send HTTP header to reload
imagepng($i);                                       # send image data

Java 7 with JavaFX, 686 674 bytes
import javafx.scene.*;import javafx.scene.canvas.*;import javafx.scene.layout.*;public class M extends javafx.application.Application{public static void main(String[]a){launch(a);}GraphicsContext g;public M(){}public void start(javafx.stage.Stage x){BorderPane r=new BorderPane();Canvas c=new Canvas(40,40);g=c.getGraphicsContext2D();Pane p=new Pane();p.getChildren().addAll(c);r.setCenter(p);x.setScene(new Scene(r,40,40));x.show();new javafx.animation.AnimationTimer(){public void handle(long n){M.this.g.setFill(javafx.scene.paint.Color.color(Math.random(),Math.random(),Math.random()));M.this.g.fillRect((int)(Math.random()*40),(int)(Math.random()*40),1,1);}}.start();}}

Explanation:
import javafx.scene.*;import javafx.scene.canvas.*;import javafx.scene.layout.*;
                                            // Required imports

public class M extends javafx.application.Application{
                                            // The class (NOTE: The public is mandatory for JavaFX)
  public static void main(String[]a){       //  Mandatory main method
    launch(a);                              //   Call the launch of the Application
  }                                         //  End of main-method

  GraphicsContext g;                        //  GraphicsContext on class-level which we'll use in the animation

  public M(){}                              //  Empty constructor (again, mandatory for JavaFX)

  public void start(javafx.stage.Stage x){  //  Overwritten start-method of Application
    BorderPane r=new BorderPane();          //   Create a BorderPane
    Canvas c=new Canvas(40,40);             //   Create a Canvas with the minimum size
    g=c.getGraphicsContext2D();             //   Retrieve the GraphicsContext and set it to the class-level variable
    Pane p=new Pane();                      //   Create a Pane
    p.getChildren().addAll(c);              //   Add the Canvas to the Pane
    r.setCenter(p);                         //   Put it in the center (otherwise it won't show in the window)
    x.setScene(new Scene(r,40,40));         //   Create a new Scene
    x.show();                               //   And show the screen
    new javafx.animation.AnimationTimer(){  //   Create a new animation
      public void handle(long n){           //    Overwritten handle-method of this animation
        M.this.g.setFill(javafx.scene.paint.Color.color(Math.random(),Math.random(),Math.random()));
                                            //     Set a random color
        M.this.g.fillRect((int)(Math.random()*40),(int)(Math.random()*40),1,1);
                                            //     And add this color as a 1x1 pixel in a random position
      }                                     //    End of handle-method
    }.start();                              //   And start the animation
  }                                         //  End of start-method
}                                           // End of class

Output:

Mathematica 112 Bytes
r:=RandomReal[];b:=⌈40r⌉;a=r&~Array~{40,40,3};Dynamic[a[[b,b]]={r,r,r};ArrayPlot[a,ColorFunction->RGBColor]]

Very straightforward, uses Dynamic around an ArrayPlot.  Not terribly fast; limited by the maximum speed dynamic can update, but should fulfill the goal on most computers.
Starts out with a random 40x40 grid and recolors a random pixel one at a time.
Ceiling characters are 3 bytes each, Ouch!

Ruby + ruby2d: 101+9 = 110 bytes
Uses the -rruby2d flag. 80x60 "pixel" image. Timed the module to run 100 update ticks in 2 seconds, so it theoretically replaces all pixels on the screen in under 2 minutes.
s=[]
80.times{|x|60.times{|y|s<<Square.new(x:x*8,y:y*8,size:8)}}
update{s.sample.color='random'}
show

Processing.js, 142 bytes
for(var i=0;i<99;i++){for(var j=0;j<99;j++){fill(random(0,255),random(0,255),random(0,255),random(0,255));rect((i-1)*40,(j-1)*40,i*40,j*40);}}

TryItOnline

SmileBASIC, 43 41 40 bytes
GPSET RND(40),RND(40),-1-RND(1<<24)EXEC.

I use -1-RND(1<<24) to generate the random color. RND(1<<24) generates a number from 0x000000 to 0xFFFFFF, and -1 - n (equivalent to bitwise NOT) will set the alpha to 0xFF (and flip the bits of the color data but that won't affect the randomness).

The console will still show over the text by default; I'd add a CLS in there to make the answer more friendly.

? would also work, because it'll scroll all the text offscreen after 29 iterations. I think I could add it for 1 or 2 characters but I don't think it's necessary.

Ruby, 78 Bytes
loop{print"\e[#{rand(41)};#{rand(41)}H\e[#{rand(30..37)};#{rand(1..3)}m█\e[0m"}

If you count the bytes you'll see that it's more than 78 bytes. That's because I had to replace the ASCII escape character (decimal value 27) with "\e" as stackexchange sanitizes the post. Both can be used in ruby code so the above code will work but for the purpose of golfing you can replace them with the literal escape character.
The code prints a box into the terminal at a random position from 0;0 to 40;40 with a random color.
The result will look something like this:

(result may vary depending on terminal and color scheme)
Explanation
loop {                      # Loop forever
  print "                   # Prints one long string to the terminal
    \e[                     # Start of an ANSI escape sequence. In this case it
                            # will move the cursor
    #{rand(41)};            # Random y position
      #{rand(41)}H          # Random x position
        \e[                 # Specify the color of the following character
          #{rand(30..37)};  # Set random hue
            #{rand(1..3)}m  # Set random brightness
                    █       # The only thing that will be visible in the terminal
                    \e[0m"} # Reset the color code to keep the blocks from
                            # interfering with each other

Java, 401 383 373 bytes
Golfed version:
import java.awt.image.*;import java.util.*;class R{public static void main(String[]a){new javax.swing.JFrame(){int w=200;BufferedImage i=new BufferedImage(w,w,1);Random r=new Random();{setSize(w,w);show();new Thread(()->{for(;;){i.setRGB(r.nextInt(w),r.nextInt(w),r.nextInt(1<<24));repaint();}}).start();}public void paint(java.awt.Graphics g){g.drawImage(i,0,0,null);}};}}

Ungolfed version:
import java.awt.image.*;
import java.util.*;

class R {
    public static void main(String[] a) {
        new javax.swing.JFrame() {
            int w = 200;                           // a constant for width and height
            BufferedImage i = new BufferedImage(
                                        w, w,      // image width and height
                                        1);        // image type BufferedImage.TYPE_INT_RGB
            Random r = new Random();
            {
                setSize(w, w);                     // frame big enough for image
                show();                            // make the frame visible
                new Thread(() -> {                 // create a forever running thread
                    for(;;) {
                        i.setRGB(
                             r.nextInt(w),         // random x position in range 0 .. w-1
                             r.nextInt(w),         // random y position in range 0 .. w-1
                             r.nextInt(1 << 24));  // random RGB value in range 0x000000 .. 0xFFFFFF
                        repaint();                 // trigger a call to the paint method
                    }
                }).start();                        // start the thread
            }

            public void paint(java.awt.Graphics g) {
                g.drawImage(i, 0, 0, null);        // draw the whole image at x=0, y=0
            }
        };
    }
}

Result:

Racket (ISL w/ lambda + universe.rkt), 168 bytes
(define(r x)(random x))(big-bang(empty-scene 40 40)[to-draw(lambda(x)x)][on-tick(lambda(x)(place-image(square 1'solid(make-color(r 255)(r 255)(r 255)))(r 40)(r 40)x))])

Ungolfed:
(define (r x) (random x))
(big-bang (empty-scene 40 40)
  [to-draw (lambda (x) x)]
  [on-tick (lambda (x) (place-image (square 1 "solid" (make-color (r 255) (r 255) (r 255))) (r 40) (r 40) x))])

Pretty simple, just places a randomly-colored 1-pixel square at a random location from 0 to 40.
Output after ~3 minutes:

Can anyone guess what book I've been reading, given my language choice?

Batch 165 156 bytes

\E must be replaced with escape character OR download original from pastebin
requires windows 10 with virtual terminal sequences enabled
note: code contains trailing whitespace

@mode 40,40&@Setlocal EnableDelayedExpansion&@For /l %%i in ()Do @Set/AY=!random!%%40+1,X=!random!%%40+1,c=!random!%%256+1&<nul Set/p=\E[!Y!;!X!H\E[48;5;!c!m

Python, 162 bytes
Uses OpenCV library to show the RGB array. Pair of bytes can be saved if define smaller array or a square array.
 Note: imshow and waitkey methods are from cv2 module; zeros and random.randint are from numpy module.
from cv2 import *
from numpy import *
h=200
w=200
c=zeros((h,w,3),"u1")
R=random.randint
while 1:
    c[R(h),R(w)]=R(256,size=3)
    imshow("",c)
    waitKey(1)

Python 3 with Pygame, 171 Bytes
import pygame as p,random as r
r=r.randint
a=p.PixelArray(p.display.set_mode((40,40)))
while(1):
    a[r(0,39),r(0,39)]=(r(0,255),r(0,255),r(0,255))
    p.display.update()

Tcl/Tk, 139
Must be run in the interactive shell
gri [can .c -w 40 -he 40]
set x 0
wh 1 {.c cr o $x [set y [exp int(rand()*40)]] $x [set x $y] -f #[form %06x [exp int(rand()*255**3)]]
upd}

Unfortunately, converting expr int(rand()* into a proc makes the script have one byte more!

To stop, one just needs to click the ineffable "X" button.

I've reused the code of present answer on https://codegolf.stackexchange.com/a/148124/29325

bash and utils, 45 bytes
Inspired (well, stolen from?) @marcosm answer, here is a quite golfed down version (from his 56 bytes down to 45, shaving the -n 1, and xarg-ing into a printf %s):
shuf -i 0-16777215|xargs printf "\e[48;5;%sm "  # \e is replaced by 1 character, Escape

PostScript, 151 bytes
realtime srand[/PageSize[640 640]>>setpagedevice
10 10 scale([){rand 16#7fffffff div}def(]){rand 64 mod}def{[[[setrgbcolor]]1 1 rectfill flushpage}loop

Needs ghostscript for flushpage; not very fast! Maybe 100 pixels/second on my machine.


text:
Let's get the gang back together!
A narcissist program is a program that only accepts its own source code. A bigotous program is a program that only accepts the source code of programs semantically equivalent to itself. A gang member is somewhere in between.
In particular, we will define a gang as set of programs G, which may or may not be in the same programming language, such that each program, when executed, will only accept members of its gang, and reject everything else (since they all do the same thing, this means that are semantically equivalent). A gang member is a program that is part of a gang, and the set of programs that it accepts is its gang.
So, a narcissist program is a gang of size 1, and a class of bigots is a maximally sized gang (every equivalent program gets to be part of the gang).
Your job is to create a gang of programs. It must be of size at least 2, and can have programs in different languages (you can even count the same program multiple times if it's a polyglot).
Your score is (average length of gang member)/(# of unique programming languages), which you are trying to minimize (so you are rewarded for using different languages). It is considered cheating to use built in language features that allow a program to access its own code.
Note: We'll define acceptance v.s. rejection via https://codegolf.meta.stackexchange.com/a/12685/16842 or https://codegolf.meta.stackexchange.com/a/12684/16842, your choice.

What does "accept" mean? Can acceptance and rejection be the output of two different values?

Full programs only, or are functions allowed?

@L3viathan I think using this criteria would be best.

@StephenS functions are fine

IMHO, you should explicitly define exactly what "accept"/"reject" mean in the question; it's unclear as it currently stands.

Questions should be self-contained in case the linked ones are deleted or something in the future. Please copy your definitions of accept/reject into your question description.

0 points (average length 0 bytes, 2 unique language)
I'm assuming we can define acceptance and rejection as the output of two different values.
The gang size is 2 (the languages are 05AB1E and 2sable), and the program is the empty program.
For some reason, both an empty 05AB1E program and an empty 2sable program appear to outputs 2 (plus a newline in the case of 2sable) when the input is not empty (i.e. a different program; rejection). They output nothing if there is no input (i.e. the same program; acceptance).
Since I "can even count the same program multiple times if its a polygot", these are two programs.
(I know neither 05AB1E nor 2sable at all and just tried golfing languages in TIO until I found some that seemed to work, so this answer might be wrong if they output something else for some specific inputs.)

I'm assuming OP will edit in a restriction of "no empty programs" - it is [tag:quine] after all

@totallyhuman It's not clear to me whether that refers to the gang size or the program size.

I added another language, so the minimum size requirement is met (assuming it refers to gang size).

@L3viathan "Your job is to create a gang of programs. It must be of size at least 2"

@StephenS The gang is of size 2 now.

Clever abuse of the rules...

@StephenS I think its already implied due to https://codegolf.meta.stackexchange.com/a/5337/16842.

@StephenS "if you can answer a question with an empty program and that question is scored by length ... and is tagged as quine ...  just notify the OP."

The empty Charcoal program outputs Parsing failed whether or not the input is empty—it does not get as far as reading the input.

@AndersKaseorg on TIO it didn't seem that way

@L3viathan When I go to https://tio.run/#charcoal with empty code and empty input and click Execute, I get Parsing failed. (You do have to remember to click Execute.)

@AndersKaseorg Could have sworn it worked yesterday. I replaced Charcoal with 2sable.

∞ points (Average length ∞, 1 unique language)
I might not win the challenge, but I sure have the biggest gang. (Watch out!)
main=interact$ \x->if length(lines x)==1&&take 279x==quine q++"--"then"Yarr!"else"It's one of them PyRulez guys!";quine s=s++show s;q="main=interact$ \\x->if length(lines x)==1&&take 279x==quine q++\"--\"then\"Yarr!\"else\"It's one of them PyRulez guys!\";quine s=s++show s;q="--

The language is Haskell. You can keep adding any letter to the end of the program except newlines. Outputs Yarr! for gang members and It's one of them PyRulez guys! for everything else.

141.9 points (Average length 141.9 bytes, 1 unique language, 255 gang members)
f x=elem '\n'x||take 141x/=k q++"--"||length x>142;k s=s++show s;q="f x=elem '\\n'x||take 141x/=k q++\"--\"||length x>142;k s=s++show s;q="--

+1 for having the worst possible entry. (I was wondering whether I needed to put in a rule about infinitely large gangs, but then I realized I didn't need to.)

Also, you can decrease your score substantially if you remove some of the longer gang members.

140.75 points (average length 281.5 bytes, 2 unique languages)
These programs use cryptographic hashing to identify the gang members, and the first line of each program is identical and encodes the hash of the remainder of each program. Each program then compares the first line to the given list and the hash of the remainder, and returns true/True if it's accepted and false/False if it's rejected.
JavaScript (Node.js), 338 bytes
x=["69fa7652b24a3aa2a56122db2f3ce27395a3be0a6e6eed00b01592b9","604e0bb526a9af96f90a964c3849be96862bb3d31b9fcfbc83d6c7aa"]
c=[]
process.stdin.on('data',d=>c.push(d)).on('end',_=>console.log((d=Buffer.concat(c)).slice(0,121)=='x='+JSON.stringify(x)&&x.some(y=>y==require('crypto').createHash('sha224').update(d.slice(121)).digest('hex'))));

Python 2, 225 bytes
x=["69fa7652b24a3aa2a56122db2f3ce27395a3be0a6e6eed00b01592b9","604e0bb526a9af96f90a964c3849be96862bb3d31b9fcfbc83d6c7aa"]
import hashlib,sys,json
I=sys.stdin
print I.read(121)=='x='+json.dumps(x,separators=',:')and hashlib.sha224(I.read()).hexdigest()in x

What about hash collisions?

0 points (Average length 24 bytes, ∞ unique languages)
This is noncompeting since the MetaGolfScript family is banned, but I thought it would be fun to post it anyways.
The languages are MetaGolfScript-N, for every natural number N. There is one program in each language, which all have the following source code:
{s ":p`:s;p" + =}:p`:s;p

The real question is which infinitely large gang would win in a gang fight, mine or the @BlackCap gang.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes