<html>
<head>
<meta content="text/html; charset=ISO-8859-1"
http-equiv="Content-Type">
</head>
<body bgcolor="#FFFFFF" text="#000000">
<div class="moz-cite-prefix">I think CompositeXYProjector is meant
to handle the following cases:<br>
<br>
1) Rendering LUT images, a single converter is used. Grayscale
images are included here.<br>
<br>
2) Rendering RGB images, three converters are used. These use
red-only, green-only, and blue-only LUTs.<br>
<br>
3) I believe it's also intended to work with images with > 3
channels, using C, M, and Y for the excess channels.<br>
<br>
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.<br>
<br>
Aivar<br>
<br>
<br>
On 7/15/13 2:40 PM, Lee Kamentsky wrote:<br>
</div>
<blockquote
cite="mid:CAHLFyjdWxW_736_czjTbDJgYRJ215gw7bBkUY5nYB_phCzjPjg@mail.gmail.com"
type="cite">
<meta http-equiv="Context-Type" content="text/html;
charset=ISO-8859-1">
<div dir="ltr">Thanks for answering Aivar,
<div>
<div class="gmail_extra"><br>
</div>
<div class="gmail_extra">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.</div>
<div class="gmail_extra"><br>
<div class="gmail_quote">On Mon, Jul 15, 2013 at 1:59 PM,
Aivar Grislis <span dir="ltr"><<a
moz-do-not-send="true" href="mailto:grislis@wisc.edu"
target="_blank">grislis@wisc.edu</a>></span> wrote:<br>
<blockquote class="gmail_quote">
<div>
<div>
<blockquote type="cite">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.</blockquote>
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.<br>
</div>
</div>
</blockquote>
<div>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.</div>
<blockquote class="gmail_quote">
<div>
<div>
<blockquote type="cite">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.</blockquote>
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.<br>
</div>
</div>
</blockquote>
<div>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</div>
<div>strength of the blending of each source. In any case,
the code as it stands doesn't do either of these.</div>
<blockquote class="gmail_quote">
<div>
<div> <br>
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. <br>
</div>
</div>
</blockquote>
<blockquote class="gmail_quote">
<div>
<div> <br>
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.<br>
</div>
</div>
</blockquote>
<div>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. </div>
<blockquote class="gmail_quote">
<div>
<div> <br>
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.<br>
<br>
Aivar<br>
<br>
<blockquote type="cite">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.</blockquote>
I think alpha processing is cumulative layer by
layer. <br>
<br>
This brings up some interesting questions:<br>
<br>
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.<br>
</div>
</div>
</blockquote>
<div>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:</div>
<div>
red = (255 - alpha(ARGB2) *(255 - red(ARGB2))/255) *
red(ARGB1) </div>
<div><br>
</div>
<div>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.</div>
<blockquote class="gmail_quote">
<div>
<div> <br>
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.<br>
<br>
Aivar<br>
<br>
<br>
On 7/15/13 10:31 AM, Lee Kamentsky wrote:<br>
</div>
<blockquote type="cite">
<div dir="ltr">
<div>Hi all, </div>
<div>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:</div>
<div><br>
</div>
<div><span> </span>for ( int i = 0; i < size;
i++ )</div>
<div><span> </span>{</div>
<div><span> </span>sourceRandomAccess.setPosition(
currentPositions[ i ], dimIndex );</div>
<div><span> </span>currentConverters[ i
].convert( sourceRandomAccess.get(), bi );</div>
<div><span> </span>// accumulate converted result</div>
<div><span> </span>final int value = bi.get();</div>
<div><span> </span>final int a = ARGBType.alpha(
value );</div>
<div><span> </span>final int r = ARGBType.red(
value );</div>
<div><span> </span>final int g = ARGBType.green(
value );</div>
<div><span> </span>final int b = ARGBType.blue(
value );</div>
<div><span> </span>aSum += a;</div>
<div><span> </span>rSum += r;</div>
<div><span> </span>gSum += g;</div>
<div><span> </span>bSum += b;</div>
<div><span> </span>}</div>
<div><span> </span>if ( aSum > 255 )</div>
<div><span> </span>aSum = 255;</div>
<div><span> </span>if ( rSum > 255 )</div>
<div><span> </span>rSum = 255;</div>
<div> <span> </span>if ( gSum > 255 )</div>
<div><span> </span>gSum = 255;</div>
<div><span> </span>if ( bSum > 255 )</div>
<div><span> </span>bSum = 255;</div>
<div><span> </span>targetCursor.get().set(
ARGBType.rgba( rSum, gSum, bSum, aSum ) );</div>
<div><br>
</div>
<div>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.</div>
<div><br>
</div>
<div>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.</div>
<div><br>
</div>
<div>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?</div>
<div><br>
</div>
<div>--Lee</div>
</div>
<br>
<fieldset></fieldset>
<br>
<pre>_______________________________________________
ImageJ-devel mailing list
<a moz-do-not-send="true" href="mailto:ImageJ-devel@imagej.net" target="_blank">ImageJ-devel@imagej.net</a>
<a moz-do-not-send="true" href="http://imagej.net/mailman/listinfo/imagej-devel" target="_blank">http://imagej.net/mailman/listinfo/imagej-devel</a>
</pre>
</blockquote>
<br>
</div>
<br>
_______________________________________________<br>
ImageJ-devel mailing list<br>
<a moz-do-not-send="true"
href="mailto:ImageJ-devel@imagej.net">ImageJ-devel@imagej.net</a><br>
<a moz-do-not-send="true"
href="http://imagej.net/mailman/listinfo/imagej-devel"
target="_blank">http://imagej.net/mailman/listinfo/imagej-devel</a><br>
<br>
</blockquote>
</div>
<br>
</div>
</div>
</div>
</blockquote>
<br>
</body>
</html>