Generate a mandelbrot fractal [closed]

Your task is to draw the mandelbrot set in ascii. It should look something like

The complex number c lies in the mandelbrot set, when the sequence z(n+1) = z(n)^2 + c, z(0) = 0 remains bounded. For the purpose of this challenge you can consider the sequence bounded for c if |z(32)| < 2.

Plot the mandelbrot set on the complex plane from (-2 - i) to (1 + i), and a minimum resolution of 40x30, using ascii characters in your favourite language, using as few characters as possible.

• Since the specification is a little bit unclear with respect to that point I can propose the following low-res ;-) 1-char php solution: * Jul 6, 2011 at 17:32
• Well I was hoping I could just say "don't be a jerk" :D. We'll go with at least 40x30 then. Jul 6, 2011 at 18:40
• I have a personal preference for tasks that solve a class of problems rather than one instance. Making the region to use an input would makes this questions qualify. In any case, the specification is a bit light. In the future you can get help with these kinds of issues on the meta sandbox or the puzzle lab chat before you task goes live Jul 6, 2011 at 19:16
• @Hannesh, I agree with dmckee. If you change the question, I'll be happy to add arguments. Jul 7, 2011 at 0:01
• Here's a source code in the shape of the mandelbrot set, which then generates high resolution mandelbrot set (preshing.com/20110926/…) - OT for this question but I thought people here might like it. Oct 1, 2011 at 11:54

TI-BASIC, 256

PROGRAM:M
:Input "ITER. ",D
:For(A,Xmin,Xmax,ΔX)
:For(B,Ymin,Ymax,ΔY)
:0→X
:0→Y
:0→I
:D→M
:While X^2+Y^2≤4 and I<M
:X^2-Y^2+A→R
:2XY+B→Y
:R→X
:I+1→I
:End
:If I≠M
:Then
:I→C
:Else
:0→C
:End
:If C<1
:Pt-On(A,B)
:End
:End
:End

• Source. May 27, 2017 at 0:18
• Since you've taken the code from elsewhere you should probably make this answer a Community Wiki. May 27, 2017 at 0:37
• @WheatWizard I remember Dennis saying on TNB last night that doing so is unnecessary. May 27, 2017 at 0:38
• It is not necessary. There will not probably not be any repercussions for not doing it. I'm just suggesting it because I think its the right thing to do. Its ultimately your choice. May 27, 2017 at 0:40
• @WheatWizard I guess it is a bit douchey to possibly get rep for something I didn't write. But there's a meta post by DJ talking about how CW isn't a rep waiver. May 27, 2017 at 0:41