Commit fa3eb035 authored by Turnhout, M.C. van's avatar Turnhout, M.C. van
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2/3 paints discussion

parent 4318a2f1
......@@ -24,9 +24,9 @@ The flag is a 8\,bits RGB image, so
I_\mathrm{max} = 2^8 - 1 = 255
\end{equation}
\section{Three colour flag analysis}
\section{Three dye flag analysis}
We will first see whether we can deconvolve and reconstruct this simple three colour flag with the algorithm in chapter \ref{algebra}.
We will first see whether we can deconvolve and reconstruct this simple three colour flag with the algorithm in chapter \ref{algebra} and the actual (pure) `colours' in the flag (yellow, green, black).
We will also write out (many) numerical examples in order to help the reader get some feeling and insight in what is going on.
......@@ -154,11 +154,11 @@ We therefore did not deconvolve with the actual colours (absorptions) in the fla
\end{figure}
\section{Two colour flag analysis}
\section{Two dye flag analysis}
In this section, we are only interested in deconvolving the yellow and green from the flag (figure \ref{flagJA}) and we will not treat `black' as an input colour, or as a colour to be deconvolved (in yellow and green).
In this section, we are only interested in deconvolving the yellow and green from the flag (figure \ref{flagJA}) and we will not treat `black' as an input dye, or as a colour to be deconvolved (in yellow and green).
However, since colour deconvolution in RGB \textsl{has} to be done with \textsl{three} `input colours' (matrix $\mat{K}$ needs to be 3 by 3), we'll have come up with an alternative third `colour'.
However, since colour deconvolution in RGB \textsl{has} to be done with \textsl{three} `input dyes' (`colours', matrix $\mat{K}$ needs to be 3 by 3), we'll have come up with an alternative `complementary' dye (`colour').
\subsection{Adding a third `perpendicular' absorption column}
......@@ -197,7 +197,7 @@ Note the negative absorption coefficients: this column represents `a dye' that p
\begin{equation}
\col{\hat{R}}_c = 256\cdot \mathrm{e}^{\begin{bmatrix*}[r] 0.0796 \\ -0.9962\\ 0.0355\end{bmatrix*} \cdot 1} - 1 = \begin{bmatrix*}[r] 277\\ 95 \\ 265 \end{bmatrix*} \label{flagJA2pRc}
\end{equation}
In other words: this third `dye colour' is not a real `colour' and cannot be (properly) represented in RGB space. It is a column that (precisely) covers the part of the 3D absorption space that is not covered by the other two columns, it is `the (absorption) remainder' for use in deconvolution.
In other words: this complementary `dye colour' is not a real `colour' and cannot be (properly) represented in RGB space. It is a column that (precisely) covers the part of the 3D absorption space that is not covered by the other two columns, it is `the (absorption) remainder' for use in deconvolution.
\subsection{Deconvolution}
......@@ -249,7 +249,7 @@ The black patches in the flag has 75\,\% of the yellow dye of the stripes (figur
The black in the flag patches are `as green' as the green patches (figures \ref{flagJA2pag}, \ref{flagJA2pg}), and they contain a certain amount of our complementary `perpendicular' dye (figures \ref{flagJA2pac}, \ref{flagJA2pc}). Note that these RGB values have been scaled by $\frac{255}{300}$ to make them representable in RGB, the actual RGB colour for this image does not exist.
Also note, finally, that the combination of 75\,\% yellow, 100\,\% green and 5 units of our third dye, \textsl{does} result in `proper black'. And it should be obvious from figure \ref{flagJA2precon} that to add the effects of (absorbing) dyes, you cannot `add RGB values'.
Also note, finally, that the combination of 75\,\% yellow, 100\,\% green and 5 units of our third dye, \textsl{does} result in `proper black'. And it should be obvious from figure \ref{flagJA2precon} that to add the effects of (absorbing) dyes, you cannot work with, or `think in', RGB values.
\subsection{Adding a third `Ruifrok' absorption column}
......@@ -307,7 +307,25 @@ But this combination also, results in `proper black' in the reconstructed flag (
\subsection{Two or three, colours or dyes?}
One may be tempted to conclude that this deconvolution of the flag with two dyes did not `work': we find yellow dye and green dye where there isn't any yellow or green in the flag (figures \ref{flagJA2pas}--\ref{flagJA2rrecon})!
But that reasoning may be too hasty.\\
\noindent In the 3 dye deconvolution of the flag, we asked `how much of these (figure \ref{flagJAreconn}) three paints does an artist use to paint this flag?' And we got the correct answer.\\
\noindent In the last example we still asked `how much of these three paints does an artist use to paint this flag?' but we switched the actual black (colour) in the flag for a purple-blue-ish paint (figure \ref{flagJA2rc}). And given that the artist had to produce a black colour with yellow, green and purple-blue-ish paint\dots\ we still got the correct answer.
The \textsl{colour} `yellow' may not be present in the flag, that doesn't mean that the artist did not use yellow \textsl{paint} for (all) non-`yellow' colours in the flag\footnote{The fact that the `paint' in the second perpendicular example `does not exist' changes the numbers, but does not change the argument.}.\\
\noindent So `where is the \textsl{colour} yellow?' may be an ambiguous question, in colour deconvolution you actually ask: `where is the \textsl{dye} (paint) yellow?'\footnote{More precisely, you ask: `\textsl{how much} of dye \dots\ is there?'}.
If there are `colours' in your image that are \textsl{not} the result of the combination of only two dyes, and you \textsl{force} the painter (algorithm) to use a third paint (dye) that was not used by the `original artist' (sample) to produce those `colours', then you \textsl{will} force the painter (algorithm) to also use the two paints (dyes). \\
\noindent So if the colours in your image are actually the result of \textsl{three} dyes, you \textsl{have to} include all three actual dyes in your deconvolution, or the answer to `where is the dye (paint) \dots?' will be wrong.
If the colours in your image are truly the result of only \textsl{two} dyes, the answer to `where is the dye (paint) \dots?' will be correct with deconvolution with those two dyes, save for image/colour artefacts.\\
\noindent Well. As long as you are deconvolving pure (not blended) paints, that is. Although the results for `black' are promising, it is time to move on and to investigate (known) blends of colours.
\section{Two blending flag colours analysis}
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