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@ -164,7 +164,25 @@ enum DecompTypes { |
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\f[norm = \forkthree{\frac{\|\texttt{src1}-\texttt{src2}\|_{L_{\infty}} }{\|\texttt{src2}\|_{L_{\infty}} }}{if \(\texttt{normType} = \texttt{NORM_RELATIVE_INF}\) } |
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{ \frac{\|\texttt{src1}-\texttt{src2}\|_{L_1} }{\|\texttt{src2}\|_{L_1}} }{if \(\texttt{normType} = \texttt{NORM_RELATIVE_L1}\) } |
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{ \frac{\|\texttt{src1}-\texttt{src2}\|_{L_2} }{\|\texttt{src2}\|_{L_2}} }{if \(\texttt{normType} = \texttt{NORM_RELATIVE_L2}\) }\f] |
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*/ |
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As example for one array consider the function \f$r(x)= \begin{pmatrix} x \\ 1-x \end{pmatrix}, x \in [-1;1]\f$. |
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The \f$ L_{1}, L_{2} \f$ and \f$ L_{\infty} \f$ norm for the sample value \f$r(-1) = \begin{pmatrix} -1 \\ 2 \end{pmatrix}\f$ |
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is calculated as follows |
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\f{align*} |
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\| r(-1) \|_{L_1} &= |-1| + |2| = 3 \\
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\| r(-1) \|_{L_2} &= \sqrt{(-1)^{2} + (2)^{2}} = \sqrt{5} \\
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\| r(-1) \|_{L_\infty} &= \max(|-1|,|2|) = 2 |
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\f} |
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and for \f$r(0.5) = \begin{pmatrix} 0.5 \\ 0.5 \end{pmatrix}\f$ the calculation is |
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\f{align*} |
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\| r(0.5) \|_{L_1} &= |0.5| + |0.5| = 1 \\
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\| r(0.5) \|_{L_2} &= \sqrt{(0.5)^{2} + (0.5)^{2}} = \sqrt{0.5} \\
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\| r(0.5) \|_{L_\infty} &= \max(|0.5|,|0.5|) = 0.5. |
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\f} |
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The following graphic shows all values for the three norm functions \f$\| r(x) \|_{L_1}, \| r(x) \|_{L_2}\f$ and \f$\| r(x) \|_{L_\infty}\f$. |
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It is notable that the \f$ L_{1} \f$ norm forms the upper and the \f$ L_{\infty} \f$ norm forms the lower border for the example function \f$ r(x) \f$. |
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 |
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*/ |
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enum NormTypes { NORM_INF = 1, |
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NORM_L1 = 2, |
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NORM_L2 = 4, |
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