[ImageJ-devel] Bug in creation of CompositeXYProjectors in DefaultDatasetView
Aivar Grislis
grislis at wisc.edu
Mon Jul 15 15:46:32 CDT 2013
I think CompositeXYProjector is meant to handle the following cases:
1) Rendering LUT images, a single converter is used. Grayscale images
are included here.
2) Rendering RGB images, three converters are used. These use red-only,
green-only, and blue-only LUTs.
3) I believe it's also intended to work with images with > 3 channels,
using C, M, and Y for the excess channels.
The existing code works well for cases 1 & 2. Case 3 adds the
possibility of overflow, if your red converter gives you a value of 255
for the red component but your magenta converter adds another 255.
Currently the code just limits the value to 255 in that case. Some sort
of blending might work better here, but the bigger issue is RGBCMY is
not an additive color system. If you see a cyan blotch you don't know
if its in both the G & B channels or just the C channel.
Aivar
On 7/15/13 2:40 PM, Lee Kamentsky wrote:
> Thanks for answering Aivar,
>
> I think what your reply did for me is to have me take a step back and
> consider what we're modeling. If you look at my replies below, I think
> that the best solution is to use a model where the background is white
> and each successive layer filters out some of that background, like a
> gel. A layer attenuates the underlying layer by a fraction of (1 -
> alpha/255 * (1 - red/255)), resulting in no attenuation for 255 and
> attenuation of alpha/255 for zero. We can then use a red converter
> that returns a value of 255 for the blue and green channels and the
> model and math work correctly.
>
> On Mon, Jul 15, 2013 at 1:59 PM, Aivar Grislis <grislis at wisc.edu
> <mailto:grislis at wisc.edu>> wrote:
>
>> I have an ImgPlus backed by an RGB PlanarImg of UnsignedByteType
>> and ARGBType.alpha(value) is 255 for all of them, so aSum is 765.
>> It would appear that the correct solution would be to divide aSum
>> by 3.
> Isn't it unusual to define an alpha for each color component,
> generally you have a single A associated with a combined RGB? So
> averaging the three alphas might make sense here, because I think
> they should all be the same value.
>
> I think you're right, the model always is that each pixel has an alpha
> value that applies to R, G and B. The image I was using was the Clown
> example image. DefaultDatasetView.initializeView constructs three
> RealLUTConverters for the projector, one for red, one for green and
> one for blue which sends you down this rabbit hole.
>
>> In addition, there's no scaling of the individual red, green and
>> blue values by their channel's alpha. If the input were two
>> index-color images, each of which had different alphas, the code
>> should multiply the r, g and b values by the alphas before
>> summing and then divide by the total alpha in the end. The alpha
>> in this case *should* be the sum of alphas divided by the number
>> of channels.
> I think alpha processing is more cumulative, done layer by layer
> in some defined layer order. For a given pixel say the current
> output pixel value is ARGB1 and you are compositing a second image
> with value ARGB2 on top of it: For the red channel the output
> color should be ((255 - alpha(ARGB2)) * red(ARGB1) + alpha(ARGB2)
> * red(ARGB2)) / 255. The alpha of ARGB1 is not involved.
>
> I think that's a valid interpretation. I've always used (alpha(ARGB1)
> * red(ARGB1) + alpha(ARGB2) * red(ARGB2)) / (alpha(ARGB1) +
> alpha(ARGB2)) because I assumed the alpha indicated the
> strength of the blending of each source. In any case, the code as it
> stands doesn't do either of these.
>
>
> In other words, if you add a layer that is completely opaque you
> no longer have to consider any of the colors or alpha values
> underneath it.
>
>
> I think the bigger issue here is this code is specifically
> designed to composite red, green and blue image layers. It's a
> special case since for a given pixel the red comes from the red
> layer, blue from blue layer, and green from green layer. These
> layers shouldn't be completely opaque, since the colors wouldn't
> combine at all then or completely transparent since then they
> wouldn't contribute any color. I don't think transparency is
> useful here.
>
> So this is an argument for blending instead of layering - transparency
> would be useful if the images were blended and treated as if on a par
> with each other, allowing the user to emphasize one channel or the other.
>
>
> It's also possible that a multichannel image with > 3 channels is
> being displayed with more color channels, namely cyan, magenta,
> and yellow. The code here is designed to stop overflow, but I'm
> not convinced those extended color channels would combine
> meaningfully.
>
> Aivar
>
>> In addition, there's no scaling of the individual red, green and
>> blue values by their channel's alpha. If the input were two
>> index-color images, each of which had different alphas, the code
>> should multiply the r, g and b values by the alphas before
>> summing and then divide by the total alpha in the end. The alpha
>> in this case *should* be the sum of alphas divided by the number
>> of channels.
> I think alpha processing is cumulative layer by layer.
>
> This brings up some interesting questions:
>
> 1) If the first, bottom-most layer is transparent, what color
> should show through? Black, white? Or perhaps it's best to
> ignore this base layer transparency.
>
> Maybe the model should be that the background is white and successive
> layers are like gel filters on top. In that case, you'd have:
> red = (255 - alpha(ARGB2) *(255 - red(ARGB2))/255) * red(ARGB1)
>
> And maybe that points to what the true solution is. For the default,
> we could change things so that red channel would have blue = 255 and
> green = 255 and the first composition would change only the red channel.
>
>
> 2) If you wanted to composite several transparent images, how do
> you calculate the transparency of the composite? I'm not sure
> this is something we need to do.
>
> Aivar
>
>
> On 7/15/13 10:31 AM, Lee Kamentsky wrote:
>> Hi all,
>> I'm looking at the code for
>> net.imglib2.display.CompositeXYProjector and as I step through
>> it, it's clear that the alpha calculation isn't being handled
>> correctly. Here's the code as it stands now, line 190 roughly:
>>
>> for ( int i = 0; i < size; i++ )
>> {
>> sourceRandomAccess.setPosition( currentPositions[ i ], dimIndex );
>> currentConverters[ i ].convert( sourceRandomAccess.get(), bi );
>> // accumulate converted result
>> final int value = bi.get();
>> final int a = ARGBType.alpha( value );
>> final int r = ARGBType.red( value );
>> final int g = ARGBType.green( value );
>> final int b = ARGBType.blue( value );
>> aSum += a;
>> rSum += r;
>> gSum += g;
>> bSum += b;
>> }
>> if ( aSum > 255 )
>> aSum = 255;
>> if ( rSum > 255 )
>> rSum = 255;
>> if ( gSum > 255 )
>> gSum = 255;
>> if ( bSum > 255 )
>> bSum = 255;
>> targetCursor.get().set( ARGBType.rgba( rSum, gSum, bSum, aSum ) );
>>
>> I have an ImgPlus backed by an RGB PlanarImg of UnsignedByteType
>> and ARGBType.alpha(value) is 255 for all of them, so aSum is 765.
>> It would appear that the correct solution would be to divide aSum
>> by 3. In addition, there's no scaling of the individual red,
>> green and blue values by their channel's alpha. If the input were
>> two index-color images, each of which had different alphas, the
>> code should multiply the r, g and b values by the alphas before
>> summing and then divide by the total alpha in the end. The alpha
>> in this case *should* be the sum of alphas divided by the number
>> of channels.
>>
>> However, I think the problem is deeper than that. For an RGB
>> ImgPlus, there are three LUTs and each of them has an alpha of
>> 255, but that alpha only applies to one of the colors in the LUT.
>> When you're compositing images and weighing them equally, if two
>> are black and one is white, then the result is 1/3 of the white
>> intensity - if you translate that to red, green and blue images,
>> the resulting intensity will be 1/3 of that desired. This might
>> sound weird, but the only solution that works out mathematically
>> is for the defaultLUTs in the DefaultDatasetView to use color
>> tables that return values that are 3x those of ColorTables.RED,
>> GREEN and BLUE. Thinking about it, I'm afraid this *is* the
>> correct model and each channel really is 3x brighter than possible.
>>
>> It took me quite a bit of back and forth to come up with the
>> above... I hope you all understand what I'm saying and understand
>> the problem and counter-intuitive solution and have the patience
>> to follow it. Dscho, if you made it this far - you're the
>> mathematician, what's your take?
>>
>> --Lee
>>
>>
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>
>
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