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Author Topic: Darkening, brightening and information loss  (Read 9231 times)
dreed
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« on: July 10, 2011, 03:50:37 PM »
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With the "Fill Light" slider in LR, it is tempting to push it to the left (darken when over exposed to the right) and then pull it out again to the right using "Fill Light".

To my way of thinking, this would result in the lightened areas at the end having less colour resolution than they started with because the darkening transformation takes away data that cannot be replaced - if a lossy transformation such as multiple (by a fraction) or divide are used with integers.

Is this correct?
Or does LR use floating point math internally for all of those pixels (the 97.9% red comes to mind) with originally integer values, allowing them to be scaled up and down without any loss of information?
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Jonathan Ratzlaff
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« Reply #1 on: July 10, 2011, 08:20:26 PM »
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Lightroom by definition is non destructive so anything you do has no permanent effect on the original.  That isone of the main reasons most people use it myself included
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Scott Hargis
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« Reply #2 on: July 10, 2011, 11:34:14 PM »
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The Fill Light slider's default position is zero, so there's no moving it to the left unless you've already made an adjustment to the right. And since LR is non-destructive, the back-and-forth isn't going to change the image at all. You'll end up where you started.
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kikashi
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« Reply #3 on: July 11, 2011, 02:30:29 AM »
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The Fill Light slider's default position is zero, so there's no moving it to the left unless you've already made an adjustment to the right. And since LR is non-destructive, the back-and-forth isn't going to change the image at all. You'll end up where you started.
I expect he meant pull the Exposure slider to the left, then restore the dark areas with Fill Light.

Jeremy
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stamper
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« Reply #4 on: July 11, 2011, 03:52:59 AM »
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I know that Raw is described as non destructive but in reality that can't be correct? If so then there wouldn't be a need for HDR or programmes to extend dynamic range?
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dreed
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« Reply #5 on: July 11, 2011, 04:24:26 AM »
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Let me make the question clearer...

A pixel in your pixel has an RGB value, of say, (10,10,10).

If I then darken it by 66%, I might end up with (6,6,6) or (7,7,7) - depending if the integer math rounds up or truncates.

If I then brighten it by 33% I've either got (9,9,9) or (10,10,10).

Thus I'm curious to know if LR uses floating point math for all of its internal operations on pixels before they are presented or integer. This is important because the use of integer means that all brightening operations should be done first because there's a potential loss of information from darkening.

The problem is arguably less relevant for brighter colours.
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Hans Kruse
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« Reply #6 on: July 11, 2011, 05:21:22 AM »
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From what I know Lightroom (and ACR) will not do the calculations in the order you have made the adjustments. As long as all you adjustments are done on top of a RAW file (and we are not talking about round trips to TIFF or anything like this) then Lightroom can reorder the calculations to get the best and most correct result. I don't know what precision Lightroom uses internally for the calculations but would expect no precision loss (that would be too silly Wink ). And be sure there will no integer representation between the various operations as this would cause severe imprecision.

As long as your pictures looks good to you, you can safely use fill light Smiley But I would advise to not only use fill light to adjust. If you have an overexposed picture you can use a combination of reduction in exposure and recovery or even the tone curve to drag down the light level in the over exposed area as far as you can recover the details. Then the picture typically will become too dark and you can then use the brightness slider to compensate and to some degree the fill light slider. You can also use the fill light and the blacks slider to iteratively come to the look you like for what remains after you have recovered for the over exposure. Alternatively use the graduated filter if it is a sky that is overexposed or too bright.

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Costas
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« Reply #7 on: July 11, 2011, 05:35:12 AM »
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Deleted - answered above
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sandymc
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« Reply #8 on: July 11, 2011, 07:13:25 AM »
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Let me make the question clearer...

A pixel in your pixel has an RGB value, of say, (10,10,10).

If I then darken it by 66%, I might end up with (6,6,6) or (7,7,7) - depending if the integer math rounds up or truncates.

If I then brighten it by 33% I've either got (9,9,9) or (10,10,10).

Thus I'm curious to know if LR uses floating point math for all of its internal operations on pixels before they are presented or integer. This is important because the use of integer means that all brightening operations should be done first because there's a potential loss of information from darkening.

The problem is arguably less relevant for brighter colours.

Short answer is that LR always keeps the the original 10,10,10 pixel stored, regardless of what adjustments you make. That's what makes it "non-destructive". To display, print or whatever, it starts with that original pixel value, and does all the adjustments on that to get to an output value. So the kind of math doesn't matter, and you can go back and forth on adjustments as many times as you want.

Sandy
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dreed
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« Reply #9 on: July 11, 2011, 07:34:38 AM »
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Short answer is that LR always keeps the the original 10,10,10 pixel stored, regardless of what adjustments you make. That's what makes it "non-destructive".

That the (10,10,10) is kept is unrelated to my question.

Quote
To display, print or whatever, it starts with that original pixel value, and does all the adjustments on that to get to an output value. So the kind of math doesn't matter, and you can go back and forth on adjustments as many times as you want.

It does matter because the order of operations and what type of operations are performed impact the end result. For example, if you do integer math (no fractions) with 20/6*6 then you end up with 18. If you do floating point, then you end up with 20 again (or a very close approximation thereof.)

Now someone else has eluded to the possibility that whilst LR presents a sequence of operations in your develop list, the actual transformation performed on the picture is a compressed one where LR works out what the final transformation matrix should be and applies that to the picture rather than apply a sequence of individual transformations. That's actually quite an important detail.
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elied
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« Reply #10 on: July 11, 2011, 08:04:47 AM »
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Surely in 14 bit, or even 12, quantization errors are trivial?

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Now someone else has eluded to the possibility that whilst LR presents a sequence of operations in your develop list, the actual transformation performed on the picture is a compressed one where LR works out what the final transformation matrix should be and applies that to the picture rather than apply a sequence of individual transformations. That's actually quite an important detail.
Either way, internal workflow cannot be altered by the user.
« Last Edit: July 11, 2011, 08:09:50 AM by elied » Logged

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digitaldog
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« Reply #11 on: July 11, 2011, 08:20:25 AM »
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Thus I'm curious to know if LR uses floating point math for all of its internal operations on pixels before they are presented or integer. This is important because the use of integer means that all brightening operations should be done first because there's a potential loss of information from darkening.

I could be wrong but I don’t believe it uses floating point math for this (Schewe?) but I do know the order of edits is not in the order applied by the user but rather one optimized for image processing.

The ‘non destructive’ discussion based on the idea that the original isn’t touched doesn’t wash with me. If that’s the definition, we’ve had this kind of editing from the day “Save As” was an option in any application.
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Andrew Rodney
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dreed
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« Reply #12 on: July 11, 2011, 09:08:50 AM »
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The ‘non destructive’ discussion based on the idea that the original isn’t touched doesn’t wash with me. If that’s the definition, we’ve had this kind of editing from the day “Save As” was an option in any application.

Indeed we have, but if you chose Save instead then that action was destructive. In LR, there is no concept of a Save, you have to use Export to get the image with its deltas applied to it in a file.

The closest LR has to Save is saving the metadata out to XMP files.
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digitaldog
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« Reply #13 on: July 11, 2011, 09:37:35 AM »
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In LR, there is no concept of a Save, you have to use Export to get the image with its deltas applied to it in a file.

Yup, and if the original data is a raw, the new data is built from scratch so to speak from the raw data, this process I’d consider non destructive (its pixel creation).

Take a rendered image, apply edits in LR and export out a new iteration. No damage?
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Andrew Rodney
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Hans Kruse
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« Reply #14 on: July 11, 2011, 10:02:24 AM »
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The closest LR has to Save is saving the metadata out to XMP files.

That's not a save at all. Lightroom stores all edits you do in the Lightroom catalog (or database) and you can choose have this data replicated in an xmp file but only of the file is the original RAW file. xmp save cannont be done for virtual copies.

I understand your question about precision, but think of it: The Adobe engineers would be rather stupid if the lost precision on the way, wouldn't they? As mentioned Lightroom reorders the operations to achieve the best result and you can be assured that the internal algorithms will work with the needed precision to not loose quality. And I have no doubt that the question about quality is a lot more than just the precision of the number representations along the way.
« Last Edit: July 11, 2011, 10:52:18 AM by Hans Kruse » Logged

NikoJorj
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« Reply #15 on: July 11, 2011, 10:37:15 AM »
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Surely in 14 bit, or even 12, quantization errors are trivial?
Yes, and from what I've understood the math is done in 16-bits integer, ie more than the 14 or 12-bits data used in DSLRs.
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Nicolas from Grenoble
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stamper
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« Reply #16 on: July 11, 2011, 10:50:04 AM »
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Short answer is that LR always keeps the the original 10,10,10 pixel stored, regardless of what adjustments you make. That's what makes it "non-destructive". To display, print or whatever, it starts with that original pixel value, and does all the adjustments on that to get to an output value. So the kind of math doesn't matter, and you can go back and forth on adjustments as many times as you want.

Sandy

A lot of posters state that LR is non-destructive. That implies the raw data can be destroyed, so the question is what program does destroy it?
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kikashi
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« Reply #17 on: July 11, 2011, 12:03:48 PM »
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Take a rendered image, apply edits in LR and export out a new iteration. No damage?
Correct - no damage. Even after the export, you can backtrack to any stage you like in the history of your LR editing changes, which LR will preserve for you permanently.

Of course, if you overwrite the initial rendered image with the newly exported one, you're opening a whole can of worms.

A lot of posters state that LR is non-destructive. That implies the raw data can be destroyed, so the question is what program does destroy it?
rm, especially rm -P  Wink

Jeremy
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digitaldog
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« Reply #18 on: July 11, 2011, 12:28:39 PM »
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Correct - no damage. Even after the export, you can backtrack to any stage you like in the history of your LR editing changes, which LR will preserve for you permanently.

I’m not referring to the original, that’s not important (again, the Save As analogy). The question is about the new iterations FROM the rendered original. No damage?
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Andrew Rodney
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kikashi
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« Reply #19 on: July 11, 2011, 02:24:58 PM »
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I’m not referring to the original, that’s not important (again, the Save As analogy). The question is about the new iterations FROM the rendered original. No damage?
Improvements, I'd rather hope!

It will depend on the saved format, won't it? If you use JPEG, there'll be loads of damage. TIFF, probably none. But we were talking about non-destructive editing.

Perhaps we're at cross-purposes.

Jeremy
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