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So for example I have this code:

points = {}; k = 0;

x = .1; y = .3; K = .9;
  {x, y} = {FractionalPart[x + K y], FractionalPart[x]};
  AppendTo[points, {x, y}];

The problem is that the list of points is growing and it slows down with time. What I want is to be able to add a point (or another primitive) directly on the dynamic graph without storing unnecessary chunks of data.

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AppendTo[] is notoriously slow... – J. M. Oct 1 '12 at 9:08
@J.M. And increasing the number of points to plot ad infinitum doesn't help – belisarius has settled Oct 1 '12 at 9:18
For systems like Matlab this works because it uses a more PostScript-like approach to rendering (just adding on top)... – Yves Klett Oct 1 '12 at 12:55

1 Answer 1

up vote 9 down vote accepted

As it was mentioned in the comments: your loop is infinite, which eventually will cause the slowdown of any computation that accumulates data in the memory. The important rule of thumb for Dynamic updating is: only update when necessary and only update what is necessary. Accordingly, you can speed up the performance of the dynamic drawing by wrapping only points and k in Dynamic. By this way, only the list of points is updated (and the label) and Mathematica does not have to redraw the whole Graphics object again and again (which involves a lot of extra computation).

points = {};
Graphics[Point@Dynamic@points, PlotLabel -> Dynamic@k]

x = .1; y = .3; K = .9;
 {x, y} = {FractionalPart[x + K y], FractionalPart[x]};
 points = Append[points, {x, y}];
 , {k, 20000}]

Starting from @belisarius' comment, I came up with a more economic version (time scales linearly with k). If one does not have to keep all the points we can apply a reasonable resolution to bin the ranges and saving new datapoints in a matrix, overwriting previous data.

resolution = 256;  (* divide the (0,1) range into 256 bins *)
array = Array[0 &, {resolution, resolution}];

Dynamic@ArrayPlot[array, PlotLabel -> Dynamic@k]

x = .1; y = .3; K = .9;
 {x, y} = {FractionalPart[x + K y], FractionalPart[x]};
 array = ReplacePart[array, (Min[#, resolution] & /@ (Round[{x, y}*resolution] + 1)) -> 1],
 {k, 1000000}]

enter image description here

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Yes it's much faster but execution still slows down with time. The goal is to get rid off list and just add primitives directly to the existing graph. – swish Oct 1 '12 at 10:14
@swish: I think that in any case a list must be maintained to store points if you want to build up your graphics by successively introducing points. – István Zachar Oct 1 '12 at 10:32
How about keeping a rasterized pre-image, and adding a point at a time. It's slow, but O(1) – belisarius has settled Oct 1 '12 at 12:33
Thanks for the tip @belisarius. See edit. – István Zachar Oct 1 '12 at 13:39
Thank you all, it is really fast now. I was playing with ArrayPlot too. – swish Oct 1 '12 at 14:25

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