11
$\begingroup$

I want to convert images of GoL configurations into binary matrices to run in Mathematica's CellularAutomaton. I know that there's a collection of GoL patterns in http://conwaylife.com and other sites, but I want to do this from images, as sometimes the same pattern is named differently by different people.

Other than using Binarize or MorphologicalBinarize, I don't know which image processing functions of Mathematica would be helpful.

enter image description here

enter image description here

enter image description here

enter image description here

So far I've got:

Manipulate[
 bin = MorphologicalBinarize[imag, {t1, t2}], {t1, 0, 1, 0.1}, {t2, 
  0, 1, 0.1}] (* find the appropriate thresholds for the image *)

MorphologicalComponents @ bin // ArrayPlot

and I'm thinking about doing some clustering and deletion of duplicates.

$\endgroup$
  • $\begingroup$ This is done for the more complicated case of the Wireworld CA here. $\endgroup$ – C. E. Oct 19 '16 at 20:09
  • $\begingroup$ @C.E. Thanks, that's interesting. The image processing case is quite different though, the image there has no mesh, and even though it has colors it's easier to process spatially. Quote form the link: "It seems like every cell of WireWorld structure corresponds to exactly one pixel in that frame". $\endgroup$ – andandandand Oct 19 '16 at 20:32
  • $\begingroup$ You're right, the Wireworld CA is more complicated, but it's easier to extract the information from that image. $\endgroup$ – C. E. Oct 20 '16 at 0:10
7
$\begingroup$

First import the image and remove the white edge around it (if there is no edge then omit this step):

img = Binarize@Import["https://i.stack.imgur.com/BiZpI.png"];
img = ImagePad[img, -BorderDimensions[img]];

Mathematica graphics

Count the number of cells in each direction:

data = ImageData[img];

rows[m_] := Unitize@Total[m]
columns[m_] := rows[Transpose@m]

{nx, ny} = {
   SequenceCount[rows[data], {1 ..}],
   SequenceCount[columns[data], {1 ..}]
   };

Create an empty grid:

grid = With[{r = rows[data], c = columns[data]}, Image[
   ConstantArray[r, Length[c]] Transpose@ConstantArray[c, Length[r]]
   ]]

Mathematica graphics

MorphologicalComponents can find each cell in this grid:

MorphologicalComponents[grid] // Colorize

Mathematica graphics

This result can be used as the label matrix for ComponentMeasurements, which allows us to get information from the real image while using components that were found using the empty grid:

cells = ComponentMeasurements[
   {img, MorphologicalComponents[grid]},
   {"Centroid", "Median"}
   ];

The problem now is that we have all the data, but we don't know what the matrix index should be for each cell. We solve that by sorting the cells based on their centroids, first by their y value and then by their x values. I'm giving the y value extra weight to achieve this:

rank[_ -> {{x_, y_}, _}] := x + 10^15 y

ArrayPlot[
 Reverse@Partition[SortBy[cells, rank][[All, 2, 2]], nx],
 ColorRules -> {0 -> Black, 1 -> White}
 ]

Mathematica graphics

$\endgroup$
5
$\begingroup$
img = Import["https://i.stack.imgur.com/LCY7B.png"];

{dim1, dim2} = ImageDimensions@img/30 // Round

{7, 5}

r = Length @ ImageCorners @ ImageTake[img, {1, dim1}] - 1
c = Length @ ImageCorners @ ImageTake[img, All, {1, dim2}] - 1

17

13

p = ImagePartition[img, Scaled[{1/r, 1/c}]];
m = Map[DominantColors[#, 1, "LABColor"][[1]] &, p, {-1}]

enter image description here

Position[Table[
  m[[i, j]] == LABColor[0, 0, 0], {i, 1, c}, {j, 1, r}],
    True]

{{2, 6}, {2, 12}, {3, 5}, {3, 6}, {3, 7}, {3, 11}, {3, 12}, {3, 13}, {4, 4}, {4, 7}, {4, 11}, {4, 14}, {5, 3}, {5, 4}, {5, 5}, {5, 13}, {5, 14}, {5, 15}, {6, 4}, {6, 6}, {6, 12}, {6, 14}, {7, 6}, {7, 7}, {7, 11}, {7, 12}, {8, 2}, {8, 7}, {8, 11}, {8, 16}, {9, 7}, {9, 11}, {10, 2}, {10, 3}, {10, 7}, {10, 11}, {10, 15}, {10, 16}, {11, 4}, {11, 7}, {11, 11}, {11, 14}, {12, 6}, {12, 12}}


NOTE: Because m[[2, 6]] == LABColor[0, 0, 0] yields True, I thought that Position[m, LABColor[0, 0, 0]] will work; unfortunately it returns {}.

$\endgroup$
  • $\begingroup$ @core979: thanks, any idea why it fails with img = Import["https://i.stack.imgur.com/BiZpI.png"]? $\endgroup$ – andandandand Oct 19 '16 at 21:51
  • $\begingroup$ I guess it's due to the quality of the images - run ImageHistogram @ img on both to see. Doing img = ImageCrop@Binarize@Import["https://i.stack.imgur.com/BiZpI.png"] helps a bit (setting manually {r, c} = {29, 18} is easier in this case), but right now I don't know how to ovecome the imperfect LABColors... $\endgroup$ – corey979 Oct 19 '16 at 22:15
5
$\begingroup$

Here's an approach that uses correlation with the boxes. Each box is 12 by 12 (hence the "divide by 12" in the PixelValuePositions). The first two lines remove the extra alpha channel and fake-color data from the image. The output is a list of the box positions.

img = Import["https://i.stack.imgur.com/LCY7B.png"];
imgNeg = ColorNegate[ColorConvert[RemoveAlphaChannel[img], "Grayscale"]];
pos = Ceiling[PixelValuePositions[
      ImageConvolve[imgNeg, BoxMatrix[5]]//ImageAdjust, 1.]/12.];
{img, Rotate[SparseArray[pos -> 1] // ArrayPlot, Pi/2]}

enter image description here

pos

{{6, 12}, {12, 12}, {5, 11}, {6, 11}, {7, 11}, {11, 11}, {12, 11}, {13, 11}, 
{4, 10}, {7, 10}, {11, 10}, {14, 10}, {3, 9}, {4, 9}, {5, 9}, {13, 9}, {14, 9}, 
{15, 9}, {4, 8}, {6, 8}, {12, 8}, {14, 8}, {6, 7}, {7, 7}, {11, 7}, {12, 7}, {2, 6}, 
{7, 6}, {11, 6}, {16, 6}, {7, 5}, {11, 5}, {2, 4}, {3, 4}, {7, 4}, {11, 4}, {15, 4},
{16,4}, {4, 3}, {7, 3}, {11, 3}, {14, 3}, {6, 2}, {12, 2}}

It's missing the top row and right-hand column because there is nothing there (in the SparseArray).

$\endgroup$
  • $\begingroup$ Thanks, that's nice! However, take a look at this case: img = Import["https://i.stack.imgur.com/BiZpI.png"]; imgNeg = ColorNegate[ ColorConvert[RemoveAlphaChannel[img], "Grayscale"]]; imgData = ImageData[ImageConvolve[imgNeg, BoxMatrix[5]] // ImageAdjust]; SparseArray[Ceiling[Position[imgData, 1.]/12.] -> 1] // ArrayPlot $\endgroup$ – andandandand Oct 19 '16 at 19:44
  • $\begingroup$ I've tried adjusting the divisor, but it's still off. img = Import["https://i.stack.imgur.com/BiZpI.png"]; imgNeg = ColorNegate[ ColorConvert[RemoveAlphaChannel[img], "Grayscale"]]; imgData = ImageData[ImageConvolve[imgNeg, BoxMatrix[5]] // ImageAdjust]; Manipulate[ positions = Ceiling[PixelValuePositions[ ImageConvolve[imgNeg, BoxMatrix[5]] // ImageAdjust, 1.]/divisor]; Row[{img, ArrayPlot[SparseArray[positions -> 1], Mesh -> True]}] , {divisor, 1., 12., 1.}] $\endgroup$ – andandandand Oct 19 '16 at 20:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.