# Tag Info

## New answers tagged image

1

You have a hard problem. The following - just an extended comment - offers one approach to attack it. Maybe you or others can build on it. Start with your first two lines of code: image00 = Import[FileNameJoin[{NotebookDirectory[], "image_00.png"}]] imageAdj01 = ImageAdjust[image00, {0.5, 0., 0.3} ] Then apply a filter: HarmonicMeanFilter[...

4

rp = RegionPlot3D[RegionDifference[Cuboid[], CantorMesh[1, 3]], Lighting -> "Neutral", Boxed -> False, PlotTheme -> "Monochrome", ImageSize -> Large] coords = DeleteCases[_List?(FreeQ[1/3 | 2/3])] @ Tuples[Subdivide[3], 3]; nng = Show[NearestNeighborGraph[coords, VertexSize -> 0, VertexCoordinates -> coords, ...

8

Update Mesh Clear; all = Tuples[{0, 1, 2}, 3]; erase = Tuples[{0, 2}, 3]; rest = Complement[all, erase]; Graphics3D[{Lighting -> "Accent", EdgeForm[Blue], FaceForm[], Cuboid[#] & /@ rest}, Boxed -> False] Solid Clear; all = Tuples[{0, 1, 2}, 3]; erase = Tuples[{0, 2}, 3]; rest = Complement[all, erase]; Graphics3D[Cuboid[#] & /@ ...

1

As promised above, I played a bit more. This time using Nestlist, what make things much easier. Please play with it. tr = TranslationTransform[{2, 0}].RotationTransform[Pi/9].ScalingTransform[0.99{1,1}]; Graphics[{tx,Polygon[#, VertexTextureCoordinates -> Automatic] & /@ NestList[tr, pol[[1]], 400]}] tr = TranslationTransform[{1,0}]....

2

Perhaps ImageCompose? Fold[ImageCompose[#, #2[[1]], #2[[2]], #2[[2]]] &, back, {{img1, {Left, Bottom}}, {img2, {Right, Top}}}] If you wish to have a Graphics object, just wrap with Show: Show[%, ImageSize -> 600, Frame -> True, PlotRangePadding -> Scaled[.02]]

2

Need to specify positions for Inset in a specific way. Graphics[{Inset[back, {Center, Center}, {Center, Center}, ImageDimensions[back]], Inset[img1, Scaled[{0, 0}]], Inset[img2, Scaled[{1, 1}]]}]

1

Clear["Global`*"] region1 = RegionProduct[ DiscretizeGraphics[ Text[Style["Fally", Bold, FontFamily -> "Times"]], _Text, MaxCellMeasure -> 0.1], MeshRegion[{{0}, {2}}, Line[{1, 2}]]]; region2 = RegionProduct[ DiscretizeGraphics[ Text[Style["Mol", Bold, FontFamily -> "Times"]], ...

3

I am now much closer to an answer. It is a two-step process. First the original image must be transformed. Second, the transformed image is used to get the pixel color in a modified Julia code where the position of the pixel is not the original z0, but the last position before escaping the radius of 2. Part 1 - Transforming the original image The ...

4

I have tested code from the paper Through the Looking-Glass, and What the Quadratic Camera Found There to understand how they been able to take Figure 1-5. Let use this code to produce Julia set points from the picture inputs[[1]] inputs = Import /@ {CloudObject[ "https://www.wolframcloud.com/objects/09bf1cae-1109-4490-a35a-\ 960ff9199863"], ...

0

nd = 10; centernumber = 9; PieChart[{{Labeled[1, Style[centernumber, FontSize -> Scaled[.1], White], {{0, 0}, {.5, .5}}]}, Labeled[1, Style[#, FontSize -> Scaled[.05]]] & /@ Range[nd], Labeled[1, Style[#, FontSize -> Scaled[.05], White]] & /@ (centernumber Range[nd])}, ChartStyle -> { "Rainbow", None}, ...

2

This is not an answer, but I want to show an image. This is the left reflection of the Julia set with c= -0.79 + I 0.15. No colors shown. Black represents points not escaping after 100 iterations. The four gray levels represents points escaping at the first, second and third iteration, and at higher iterations. The shape of the three darker bands are visible ...

5

You do not need excotic function, the built in tools will do the job. Due to bad weather, I did some experimenting. What I did not yet try is to use Nest, that I think may come in handy. For the time being, I did everything "by hand". Maybe I will add more, if I have the time. Have fun: inputs = Import /@ { CloudObject["https://www....

2

Update 2 This is a better implementation of the parts function that does not assume that the spacings are the same. It finds groups of rows in the image that are not composed of entirely white pixels and splits by those groups. ClearAll@parts parts[image_] := Module[{width, height, imageData}, {width, height} = ImageDimensions@image; imageData = ...

5

thickness = 10; img = Import["https://i.stack.imgur.com/PvtRV.png"]; 1. ContourDetect + EdgeDetect + Dilation ImageMultiply[img, ColorNegate @ Dilation[EdgeDetect @ ImagePad[ColorNegate @ ContourDetect[img], 20], thickness]] 2. ComponentMeasurements img0 = ImagePad[img, 10]; Show[img0, Graphics[{AbsoluteThickness[thickness], Red, ...

3

Using ImageMesh and textures is a little more awkward than I expected it to be, but it does the job without needing image manipulation like EdgeDetect: img = ImageCrop[Import["https://i.stack.imgur.com/PvtRV.png"]]; mesh = ImageMesh[AlphaChannel[img]]; coords = MeshCoordinates[mesh]; mmx = MinMax[coords[[All, 1]]]; mmy = MinMax[coords[[All, 2]]]; ...

2

When you ask for a big image you also need to specify a high raster resolution. Like so: eq = HoldForm[μ = (Subscript[μ, max] s^n)/(Subscript[k, s] + s^n)]; With[{size = 600}, Rasterize[TraditionalForm[eq], RasterSize -> size, ImageSize -> size]]

3

img = Import["https://i.stack.imgur.com/i0HUj.png"]; imgmesh = ImageMesh[img]; tworects = imgmesh // MeshCoordinates // Sort // Subsets[#, {2}] & // Select[VectorLess] // Rectangle @@@ # & // Select[RegionWithin[imgmesh, #] &] // Subsets[#, {2}] & // Select[RegionEqual[RegionUnion @@ #, imgmesh] &] // ...

5

A simple idea is to identify the corners and then create all possible rectangles that can be created from those corners: img = ColorConvert[ Import["https://i.stack.imgur.com/i0HUj.png"], "Grayscale" ]; corners = ImageCorners[img]; HighlightImage[img, corners] A rectangle can be created using three points, so we simply create ...

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