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<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>&nbsp;</TD>
<TD>&nbsp;</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">&#160;</A><A NAME="14645">&#160;</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>&nbsp;</TD>
<TD>&nbsp;</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>&nbsp;</TD>
<TD>&nbsp;</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">&#160;</A><A NAME="14661">&#160;</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>&nbsp;</TD>
<TD>&nbsp;</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>&nbsp;</TD>
<TD>&nbsp;</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">&#160;</A><A NAME="14679">&#160;</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>
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<ADDRESS>
<I>Petra Nass</I>
<BR><I>1999-06-15</I>
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