### 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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