Skip to main content

Is there a way to convert an image into a Graph?


I'm trying to convert an image with several overlapping dots into a Graph. The goal is to be able to derive the Kirchhoff matrix for the randomly created "network of resistors" a.k.a dots with the function KirchhoffMatrix[]. Any thoughts or ideas?


Here's the image I'm trying to extract the graph from: enter image description here


Haha, sorry I suppose I have been pretty vague with this question. I am currently researching percolation simulations. The dots in the image represent metallic particles. The dots the overlap represent metallic particles connected via a resistor. We randomize the dots on the screen, group the ones that are touching with MorphologicalComponents[] // Colorize , and then we Isolate the 'BackBone Cluster' a.k.a the largest component with DeleteSmallComponents[]. We check to see if the cluster spans all the way from the top to the bottom and if it does, the sample percolated. We are then trying to find the conductance of the sample from the BackBone cluster. We have been using the Finite Element Method for this, which requires the KirchhoffMatrix[].


Your answer, Vitaliy Kaurov, looks like it will very thoroughly answer my question. I'm going to test things before I accept it but thank you for putting so much time into it!!



Answer



==== Method 1 ===


Here is a way to get a graph from an image. MorphologicalGraph can get you started.


img = Import["http://i.stack.imgur.com/9HXZ5.png"];

g = MorphologicalGraph[img]

enter image description here


And here is your KirchhoffMatrix of the graph. Please note that MatrixPlot averages values for the best visual representation, - actual plot would be too detailed to be a good representation.


m = KirchhoffMatrix[g]; MatrixPlot[m]

enter image description here


A good guess is that your graph, due to its random nature, not all entirely connected. This maybe important in your case, because with no connection you have no conductance. Here is the way to find all connected components, highlight them and compute their own KirchhoffMatrix.


cc = ConnectedComponents[g];
index = Length /@ ConnectedComponents[g];

ctrl = Sort[MapThread[Rule, {cc, index}], #1[[2]] > #2[[2]] &];
Manipulate[
 GraphicsRow[{HighlightGraph[g, Subgraph[g, n], 
    GraphStyle -> "LargeNetwork"], 
   MatrixPlot[KirchhoffMatrix[Subgraph[g, n]], 
    ColorFunction -> "DarkBands"]}, 
  ImageSize -> 600], {{n, ctrl[[1, 1]], "Vertex number"}, ctrl}, 
 FrameMargins -> 0]

enter image description here



Let's do some math now. Here is a famous property:



The number of times 0 appears as an eigenvalue in the KirchhoffMatrix is the number of connected components in the graph.



The number of connected comments (separate sub-graphs) is found as


 ConnectedComponents[g] // Length


159




And as you can see it is equal to the number of zero eigenvalues :


 ev = Chop[Eigenvalues[N@m]]; Count[ev, 0]


159



Isn't it great to see programming and math matching up perfectly?


Here is the way to list all connected sub-graphs separately. Mouse-over will give KirchhoffMatrix.


Tooltip[#, 
   MatrixPlot[KirchhoffMatrix[Subgraph[g, #]], 

    ColorFunction -> "DarkBands", 
    ImageSize -> 200]] & /@ (Subgraph[g, #, ImageSize -> 80, 
     VertexSize -> 0] & /@ ctrl[[All, 1]])

enter image description here


Note some extra image processing before MorphologicalGraph would result in different graphs. For example finding first SkeletonTransform or DistanceTransform would change the game:


GraphicsRow[{DistanceTransform[Dilation[img, 3]] // ImageAdjust,
  SkeletonTransform[img]}, ImageSize -> 600]

enter image description here



So if you now apply MorphologicalGraph function to these images you of course will get different graphs as result. This maybe useful because there is no exact guarantee that MorphologicalGraph will give you always what you need in all cases, especially with low-res images. Below is an example of possible issue - with real question being where exactly is the boundary between these two cases?


Grid@{{r = .6; 
   img = ColorNegate@
     Graphics[{Disk[{0, 0}, r], Disk[{.5, .9}, r], Disk[{1, 0}, r]}],
   MorphologicalGraph[img, VertexSize -> .1]},
  {r = 1.2; 
   img = ColorNegate@
     Graphics[{Disk[{0, 0}, r], Disk[{.5, .9}, r], Disk[{1, 0}, r]}],
   MorphologicalGraph[img, VertexSize -> .1]}}


enter image description here


So this is just the word of caution - you may need some image pre-processing before you get your MorphologicalGraph. Or you may use something else - see method below.


==== Method 2 ===


This is just an outline, not complete solution:


1) This question and answers by @Verbeia and @Szabolcs and Fred Simons on MathGroup are a starting point


2) They work only if you know center and radius of every point in your image - which you do NOT


3) You can get center and radius of every point by using for example ComponentMeasurements function. Then you can apply methods by @Verbeia and @Szabolcs and Fred Simons on MathGroup.



  • Example 1: This Wolfram Blog gives code for finding what you need even in the case of overlapping particles, like this (image - courtesy of Wolfram Research):



enter image description here



enter image description here


Comments

Post a Comment

Popular posts from this blog

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 - 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 - 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