<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2 Final//EN"> <!--Converted with LaTeX2HTML 98.1p1 release (March 2nd, 1998) originally by Nikos Drakos (nikos@cbl.leeds.ac.uk), CBLU, University of Leeds * revised and updated by: Marcus Hennecke, Ross Moore, Herb Swan * with significant contributions from: Jens Lippmann, Marek Rouchal, Martin Wilck and others --> <HTML> <HEAD> <TITLE>The Laplacian Pyramid</TITLE> <META NAME="description" CONTENT="The Laplacian Pyramid"> <META NAME="keywords" CONTENT="vol2"> <META NAME="resource-type" CONTENT="document"> <META NAME="distribution" CONTENT="global"> <META HTTP-EQUIV="Content-Type" CONTENT="text/html; charset=iso-8859-1"> <LINK REL="STYLESHEET" HREF="vol2.css"> <LINK REL="next" HREF="node320.html"> <LINK REL="previous" HREF="node318.html"> <LINK REL="up" HREF="node318.html"> <LINK REL="next" HREF="node320.html"> </HEAD> <BODY > <!--Navigation Panel--> <A NAME="tex2html5444" HREF="node320.html"> <IMG WIDTH="37" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="next" SRC="icons.gif/next_motif.gif"></A> <A NAME="tex2html5441" HREF="node318.html"> <IMG WIDTH="26" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="up" SRC="icons.gif/up_motif.gif"></A> <A NAME="tex2html5435" HREF="node318.html"> <IMG WIDTH="63" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="previous" SRC="icons.gif/previous_motif.gif"></A> <A NAME="tex2html5443" HREF="node1.html"> <IMG WIDTH="65" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="contents" SRC="icons.gif/contents_motif.gif"></A> <BR> <B> Next:</B> <A NAME="tex2html5445" HREF="node320.html">Pyramidal Algorithm with one</A> <B> Up:</B> <A NAME="tex2html5442" HREF="node318.html">Pyramidal Algorithm</A> <B> Previous:</B> <A NAME="tex2html5436" HREF="node318.html">Pyramidal Algorithm</A> <BR> <BR> <!--End of Navigation Panel--> <H3><A NAME="SECTION002044100000000000000"> The Laplacian Pyramid</A> </H3> The Laplacian Pyramid has been developed by Burt and Adelson in 1981 [<A HREF="node370.html#burt">4</A>] in order to compress images. After the filtering, only one sample out of two is kept. The number of pixels decreases by a factor two at each scale. <P> The convolution is done with the filter <I>h</I> by keeping one sample out of two (see figure <A HREF="node319.html#fig_shema_lap1">14.7</A>): <BR> <DIV ALIGN="CENTER"> <!-- MATH: \begin{eqnarray} c_{j+1}(k) = \sum_l h(l-2k) c_j(l) \end{eqnarray} --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG WIDTH="265" HEIGHT="65" ALIGN="MIDDLE" BORDER="0" SRC="img676.gif" ALT="$\displaystyle c_{j+1}(k) = \sum_l h(l-2k) c_j(l)$"></TD> <TD> </TD> <TD> </TD> <TD WIDTH=10 ALIGN="RIGHT"> (14.38)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> <P> <BR> <DIV ALIGN="CENTER"><A NAME="fig_shema_lap1"> </A><A NAME="14645"> </A> <TABLE WIDTH="50%"> <CAPTION><STRONG>Figure 14.7:</STRONG> Passage from <I>c</I><SUB>0</SUB> to <I>c</I><SUB>1</SUB>, and from <I>c</I><SUB>1</SUB> to <I>c</I><SUB>2</SUB>.</CAPTION> <TR><TD><IMG WIDTH="683" HEIGHT="341" SRC="img677.gif" ALT="\begin{figure} \centerline{ \hbox{ \psfig{figure=fig_shema_lap1.ps,bbllx=3.5cm,bblly=14.5cm,bburx=18cm,bbury=22cm,height=6cm,width=10cm,clip=} }} \end{figure}"></TD></TR> </TABLE> </DIV> <BR> <P> To reconstruct <I>c</I><SUB><I>j</I></SUB> from <I>c</I><SUB><I>j</I>+1</SUB>, we need to calculate the difference signal <I>w</I><SUB><I>j</I>+1</SUB>. <BR> <DIV ALIGN="CENTER"> <!-- MATH: \begin{eqnarray} w_{j+1}(k) = c_j(k) - \tilde{c}_j(k) \end{eqnarray} --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG WIDTH="232" HEIGHT="44" ALIGN="MIDDLE" BORDER="0" SRC="img678.gif" ALT="$\displaystyle w_{j+1}(k) = c_j(k) - \tilde{c}_j(k)$"></TD> <TD> </TD> <TD> </TD> <TD WIDTH=10 ALIGN="RIGHT"> (14.39)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> where <!-- MATH: $\tilde{c}_j$ --> <IMG WIDTH="23" HEIGHT="40" ALIGN="MIDDLE" BORDER="0" SRC="img679.gif" ALT="$\tilde{c}_j$"> is the signal reconstructed by the following operation (see figure <A HREF="node319.html#fig_shema_lap2">14.8</A>): <BR> <DIV ALIGN="CENTER"> <!-- MATH: \begin{eqnarray} \tilde{c}_j(k) = 2 \sum_l h(k-2l) c_j(k) \end{eqnarray} --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG WIDTH="263" HEIGHT="65" ALIGN="MIDDLE" BORDER="0" SRC="img680.gif" ALT="$\displaystyle \tilde{c}_j(k) = 2 \sum_l h(k-2l) c_j(k)$"></TD> <TD> </TD> <TD> </TD> <TD WIDTH=10 ALIGN="RIGHT"> (14.40)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> <P> <BR> <DIV ALIGN="CENTER"><A NAME="fig_shema_lap2"> </A><A NAME="14661"> </A> <TABLE WIDTH="50%"> <CAPTION><STRONG>Figure 14.8:</STRONG> Passage from <I>C</I><SUB>1</SUB> to <I>C</I><SUB>0</SUB>.</CAPTION> <TR><TD><IMG WIDTH="680" HEIGHT="224" SRC="img681.gif" ALT="\begin{figure} \centerline{ \hbox{ \psfig{figure=fig_shema_lap2.ps,bbllx=3.5cm,bblly=7cm,bburx=18cm,bbury=12cm,height=4cm,width=10cm,clip=} }} \end{figure}"></TD></TR> </TABLE> </DIV> <BR> <P> In two dimensions, the method is similar. The convolution is done by keeping one sample out of two in the two directions. We have: <BR> <DIV ALIGN="CENTER"> <!-- MATH: \begin{eqnarray} c_{j+1}(n,m) = \sum_{k,l} h(k-2n,l-2m) c_j(k,l) \end{eqnarray} --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG WIDTH="396" HEIGHT="71" ALIGN="MIDDLE" BORDER="0" SRC="img682.gif" ALT="$\displaystyle c_{j+1}(n,m) = \sum_{k,l} h(k-2n,l-2m) c_j(k,l)$"></TD> <TD> </TD> <TD> </TD> <TD WIDTH=10 ALIGN="RIGHT"> (14.41)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> and <!-- MATH: $\tilde{c}_j$ --> <IMG WIDTH="23" HEIGHT="40" ALIGN="MIDDLE" BORDER="0" SRC="img683.gif" ALT="$\tilde{c}_j$"> is: <BR> <DIV ALIGN="CENTER"> <!-- MATH: \begin{eqnarray} \tilde{c}_j(n,m) = 2 \sum_{k,l} h(n-2l,m-2l) c_{j+1}(k,l) \end{eqnarray} --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG WIDTH="405" HEIGHT="71" ALIGN="MIDDLE" BORDER="0" SRC="img684.gif" ALT="$\displaystyle \tilde{c}_j(n,m) = 2 \sum_{k,l} h(n-2l,m-2l) c_{j+1}(k,l)$"></TD> <TD> </TD> <TD> </TD> <TD WIDTH=10 ALIGN="RIGHT"> (14.42)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> <P> The number of samples is divided by four. If the image size is <IMG WIDTH="72" HEIGHT="41" ALIGN="MIDDLE" BORDER="0" SRC="img685.gif" ALT="$N\times N$">, then the pyramid size is <!-- MATH: $\frac{4}{3}N^2$ --> <IMG WIDTH="49" HEIGHT="49" ALIGN="MIDDLE" BORDER="0" SRC="img686.gif" ALT="$\frac{4}{3}N^2$">. We get a pyramidal structure (see figure <A HREF="node319.html#fig_pyr">14.9</A>). <P> <BR> <DIV ALIGN="CENTER"><A NAME="fig_pyr"> </A><A NAME="14679"> </A> <TABLE WIDTH="50%"> <CAPTION><STRONG>Figure 14.9:</STRONG> Pyramidal Structure</CAPTION> <TR><TD><IMG WIDTH="445" HEIGHT="559" SRC="img687.gif" ALT="\begin{figure} \centerline{ \hbox{ \psfig{figure=fig_struc_pyr.ps,bbllx=4cm,bblly=7.5cm,bburx=17cm,bbury=23cm,height=8cm,width=6.5cm,clip=} }} \end{figure}"></TD></TR> </TABLE> </DIV> <BR> <P> The laplacian pyramid leads to an analysis with four wavelets [<A HREF="node370.html#bijaoui">3</A>] and there is no invariance to translation. <P> <HR> <!--Navigation Panel--> <A NAME="tex2html5444" HREF="node320.html"> <IMG WIDTH="37" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="next" SRC="icons.gif/next_motif.gif"></A> <A NAME="tex2html5441" HREF="node318.html"> <IMG WIDTH="26" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="up" SRC="icons.gif/up_motif.gif"></A> <A NAME="tex2html5435" HREF="node318.html"> <IMG WIDTH="63" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="previous" SRC="icons.gif/previous_motif.gif"></A> <A NAME="tex2html5443" HREF="node1.html"> <IMG WIDTH="65" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="contents" SRC="icons.gif/contents_motif.gif"></A> <BR> <B> Next:</B> <A NAME="tex2html5445" HREF="node320.html">Pyramidal Algorithm with one</A> <B> Up:</B> <A NAME="tex2html5442" HREF="node318.html">Pyramidal Algorithm</A> <B> Previous:</B> <A NAME="tex2html5436" HREF="node318.html">Pyramidal Algorithm</A> <!--End of Navigation Panel--> <ADDRESS> <I>Petra Nass</I> <BR><I>1999-06-15</I> </ADDRESS> </BODY> </HTML>