Morgantao
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Hi everybody.
Does anyone know how Photoshop's "Poster Edges" works? I'm trying to recreate the filter, but every time I think I got it I see something else that's not right... The picture attached shows the result of the "Poster Edges" filter with these settings: Edge Thikness = 0 Edge Intensity = 0 Posterization = 1 ![]() |
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Posted: January 10, 2012 1:20 pm | ||||
Morgantao
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Posted: January 10, 2012 1:37 pm | ||||
inujima
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Posted: January 10, 2012 2:19 pm | ||||
Morgantao
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Posted: January 10, 2012 2:28 pm | ||||
inujima
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Although I do not know an algorithm of Cutout filter well, it is sure that have statistical processing to many pixels.
I think it is impossible to reproduce completely Cutout filter by FF3. |
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Posted: January 10, 2012 3:15 pm | ||||
Morgantao
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WOW Inujima, your Poster Edges is pretty much spot on!
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NOOOOOOOOO! I refuse to believe that! EVEREYTHING is possible with FF! ![]() I'll take the next best thing, if anyone can get close to Cutout... |
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Posted: January 10, 2012 3:27 pm | ||||
inujima
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Perhaps, Cutout filter algorithm is doing like this.
step1: Analyzes an image histogram and limits colors according to Number of Level. step2: Replaces each pixel colors by nearest color in the color list of step1. step3: Vectorization FilterForge cannot do step1 and step3. |
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Posted: January 10, 2012 4:33 pm | ||||
Morgantao
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Any Script Head wants to program a verctorization process?
![]() OK, A challange: Anyone, from your experiance, out of the almost 8600 filters, can you point me to something that will give me a result somewhat close to the Cutout picture above? Whoever points me to the closest filter, or posts one here will get a prize: A bubble gum, with barely any teeth marks on it! ![]() |
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Posted: January 10, 2012 5:18 pm | ||||
inujima
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Having written above about the Cutout algorithm was a perfect mistake.
I corrected it as a result of experiments. step1:Converts source image to grayscale. step2:Posterization. step3:Traces boundary of connected domain and creates polygon. step4:Picks up source image colors in the polygon of step3. step5:Average colors of step4 and fill the polygon with averaged color. step6:Repeats step3~5 for every connected domain. I think this is fairly near the answer. ![]() |
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Posted: January 12, 2012 2:57 am | ||||
Morgantao
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Other than the poligon part, it sounds like FF can do it...
Is there a way to modify step 3 into something that FF can do? Maybe separate each of the posterized steps usig threshold, then using these as alphas? Edit: Looks like step 5 is also over my head... ![]() I guess I can forget about the filter I had in mind. Too many Photoshop filters that need to be duplicate in FF (cutout, smart blur, diffuse, dust and scratches, photocopy) |
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Posted: January 12, 2012 8:18 am | ||||
inujima
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Even if making polygon part is avoided, at present, there is no way to pick out connected domain or manipulate image by each connected domains.
To do this, we have to wait until appearing of Bitmap Script Component. My thoughts about other Photoshp filters. Smart Blur - cannot duplicate Diffuse - cannot duplicate but can make similar one Dust and Scratches - can duplicate Photocopy - can duplicate |
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Posted: January 13, 2012 5:55 am | ||||
Morgantao
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For argument's sake, let's say I don't mind that each connected domain will be it's own avarage color. I went with steps 1 and 2 of your cutout algorithm, then I used 9 Impulse Tonemaps (10 wdth each) and made each one an alpha (inverted) of the original image.
How would I go about making an avarage color of everything in those new images? When I try to get the avarage color I get mostly nothing, because most of each step is 0 alpha... |
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Posted: January 13, 2012 4:20 pm | ||||
inujima
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There is a method of averaging at areas using Blur component. however this method is thousands of times slower than taking average value ordinarily like doing by photoshop etc.
Area Average.ffxml |
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Posted: January 13, 2012 5:36 pm | ||||
Morgantao
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I'm not gonna pretend to undersand what you just did there, but it seems to do the trick
![]() Too bad it's slow, not to mention my filter needs 9 repeats of this... |
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Posted: January 13, 2012 6:36 pm |
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