Skip to main content

performance tuning - How to implement Bresenham’s algorithm as a system of equations?


Bresenham's line drawing algorithm is usually implemented via loops. But in Mathematica we can take advantage of its ability to solve Diophantine equations. From educational viewpoint it is quite interesting to write Bresenham in a form suitable for Solve or other superfunction. Such implementation also potentially can be very terse and sufficiently efficient for most practical applications.


I have implemented Bresenham for the first octant and obtained general solution via reflections (code for version 10):


Clear[bresenhamSolve];
bresenhamSolve[p1_, p2_] /; 
   GreaterEqual @@ Abs[p2 - p1] && MatchQ[Sign[p2 - p1], {1, 1} | {1, 0}] := 
  Block[{ab = First@Solve[{p1, p2}.{a, 1} == b], x, y}, {x, y} /. 
    Solve[{a x + y + err == b /. ab, -1/2 < err <= 1/2, {x, y} \[Element] Integers, 
      p1 <= {x, y} <= p2}, {x, y, err}]

   ];
bresenhamSolve[p1_, p2_] /; 
   Less @@ Abs[p2 - p1] && MatchQ[Sign[p2 - p1], {1, 1} | {0, 1}] := 
  Reverse /@ bresenhamSolve[Reverse[p1], Reverse[p2]];
bresenhamSolve[p1_, p2_] := 
  With[{s = 2 UnitStep[p2 - p1] - 1}, 
   Replace[bresenhamSolve[p1 s, p2 s], {x_, y_} :> s {x, y}, {1}]];

Of course this code cannot be called terse.


This implementation is identical to halirutan's implementation:



lines = DeleteCases[Partition[#, 2] & /@ Tuples[Range[-30, 30, 6], {4}], {p_, p_}];
Length[lines]
bresenhamSolve @@@ lines == bresenham @@@ lines


14520

True

Visualization:



p1 = {2, 3}; p2 = {20, 13};
Graphics[{EdgeForm[{Thick, RGBColor[203/255, 5/17, 22/255]}], 
  FaceForm[RGBColor[131/255, 148/255, 10/17]], 
  Rectangle /@ (bresenhamSolve[p1, p2] - .5), {RGBColor[0, 43/255, 18/85], Thick, 
   Line[{p1, p2}]}}, GridLines -> (Range[#1, #2 + 1] & @@@ Transpose[{p1, p2}] - .5), 
 Frame -> True]


plot




Manipulate[Row[{Graphics[{EdgeForm[{Thick, RGBColor[203/255, 5/17, 22/255]}], 
     FaceForm[RGBColor[131/255, 148/255, 10/17]], 
     Rectangle /@ (bresenhamSolve @@ Round[pts] - .5), {RGBColor[0, 43/255, 18/85], Thick,
       Arrow@pts}}, GridLines -> ({Range[-50, 50], Range[-50, 50]} - .5), Frame -> True, 
    PlotRange -> {{-20, 20}, {-20, 20}}, ImageSize -> 500], 
   Column[Round@pts]}], {{pts, {{-11, -13}, {8, 15}}}, Locator}]


manipulate




Checking symmetry:


n = 20; center = {0, 0};
perimeterOfSquare = {x, y} /. 
   Solve[{x, y} \[Element] 
     RegionBoundary[Rectangle[{-n, -n} + center, {n, n} + center]], {x, y}, Integers];
ArrayPlot[SparseArray[
  Rule @@@ Tally[# - center & /@ 
     Flatten[bresenhamSolve[center, #] & /@ perimeterOfSquare + n + 1, 1]], {2 n + 1, 
   2 n + 1}], Mesh -> True, PlotRange -> {All, All, {1, 9}}, ClippingStyle -> Red, 
 PixelConstrained -> True]



arrayplot



Timing comparison with halirutan's implementation:


p1 = {2, 3}; p2 = {2001, 1300};
Timing[bresenhamSolve[p1, p2];]
Timing[bresenham[p1, p2];]



{0.171601, Null}

{0.0312002, Null}

My question:


Is it possible to write general implementation without splitting it into special cases and with good performance? Instead of Solve one can use other superfunction(s).




Comments

Post a Comment

Popular posts from this blog

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]
Post a Comment
Read more

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....
Post a Comment
Read more