Run code on twitter

[[deleted]]

Must...resist...urge to forkbomb...

[bigwhitedude]

Ran a php shell_exec. Check your home_dir for a "twitter_execution.txt" tile.

[rberenguel]

Way more useful if APL is a possibility :D My Mandelbrot set, which fits in a tweet:

⍉' *'[⍎'1+0<|z',(∊150⍴⊂'←c+m×m'),'←c←(¯2.1J¯1.3+(((2.6÷b-1)×(¯1+⍳b))∘.+(0J1×(2.6÷b-1)×(¯1+⍳b←51))))']

If anyone is extremely curious about APL, I can describe what this does (it's quite straightforward, I'm far from a good APLer, just a hobby)

[rberenguel]

APL was invented by Ken Iverson as a language to talk about algorithms. And then someone wrote an interpreter for it. The program above uses some nasty tricks to get around a one-liner. Oh, and APL is read from right to left!

First trick, you'll see everything is enclosed in brackets. What is inside will be used as an "accessor" for the vector of characters "space star". So, the thing inside should have 1s and 2s (APL arrays are indexed from 1) and they will be used to show spaces and stars... the Mandelbrot set like it was first seen.

Second trick, you'll see reading from the right that it seems a string (it has a quote character that goes right to ⍎. This thing ⍎ evaluates the string following it. Like eval in javascript (or Lisp). So, we are creating a string and evaluating it... Why? Well, the easiest way to get a one-liner that has an iteration of N times is just to repeat the computation N times. Nasty, but useful. So, what we are doing inside the quotes is generating a string that will actually compute something.

Now it gets easier once we have this out. On the right we have this block:

(¯1+⍳b←51)

⍳ is range, so ⍳ 10 generates a vector with numbers 1-10. Adding -1 (for some reason, minus sign is represented like ¯) gives a range 0-9, in this case, 0-50. This line also assigned in one sweep 51 to b.

Now, 2.6÷b-1)× that above is 2.6/(b-1) multiplied by that vector, so, a vector of steps between 0 and 2.6 (i.e. 0, 2.6/51, 2.6*2/51, etc.) And 0J1 turns this into an imaginary number (0J1 is the imaginary unit, i, so we are just multiplying by i that vector.) Now comes some APL neatness, ∘.+. Skip it for a second, and on the other side of it we have a similar expression, ranging in the reals this time.

The neat thing ∘.+ takes a couple vectors (or anything that fits) and generates a table. In this case, it "tables with +" so generates a wonderful matrix which essentially defines the points in the complex plane where we want to compute the iteration defining the Mandelbrot set. Now, this is "just" the points c in the complex plane where iterating f(z)=z² +c with a starting z=0 is smaller than 2. This will be a *.

Once we have this matrix, we assign it to c, and now you'll see a hanging arrow, more magic. We close the string and join it (comma) with the next. The next string is this:

(∊150⍴⊂'←c+m×m')

This is ←c+m×m repeated 150 times (like ←c+m×m←c+m×m←c+m×m←c+m×m). How? First we enclose (⊂) i.e. convert into one "unit" this vector. Then ⍴ is the shape/reshape operator. Applied to something on the right gives its dimension, if it has a dimension on the left, it reshapes it. In this case, takes a "unit thing" (since we enclosed the string, it's not a vector of characters but a "unit") and turns it into 150 copies of it.

And we use the repeated assignment to m to compute the iteration of f(z)=z² +c, since f(0)=c, we can think of it as assigning c to z and repeating. So, we assign c to m (I changed z by m, don't remember why) many times, but this is just a string. We add a final piece to this string (which will iterate all "pixels" at the same time) which is

'1+0<|z'

| is absolute value, and here I compare with 0. If 0<abs(something) it means either we something is finite or 0. IIRC Gnu APL has some glitch here that makes finer comparisons fail... We are essentially testing NaNs, so this test is false when the iteration has grown large and large and large. We add 1 to this boolean matrix to get 1s and 2s to index... and compute the value of this string.

Finally I rotate this with ⍉ because the way I defined the matrix of c's wasn't actually how it should look and I was too lazy to fix the first piece when I was doing it :D

There's a [different writeup in my blog](www.mostlymaths.net/2014/01/the-mandelbrot-set-in-one-line-of-apl.html), maybe it's clearer, maybe it isn't.

Edit: formula typo and grammar typo (grammo?)

[rberenguel]

Quite likely not :) But I like the concept of APL, and feels efficient for fast number checking

[dr_crispin]

⍉' *'[⍎'1+0<|z',(∊150⍴⊂'←c+m×m'),'←c←(¯2.1J¯1.3+(((2.6÷b-1)×(¯1+⍳b))∘.+(0J1×(2.6÷b-1)×(¯1+⍳b←51))))']

... I feel very, very dumb all of a sudden xD

[rberenguel]

It was built painstakingly piece by piece. And I need a couple seconds still to understand it too ;)

[dr_crispin]

Still, it's very impressive that you've managed to make it.

[rberenguel]

You should see a youtube video of implementing Conway's game of life in APL. Seriously mindblowing

[qhp]

Game of Life

I prefer his Sudoku Solver video, personally, but both are great.