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author wolffd
date Tue, 10 Feb 2015 15:05:51 +0000
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<h2>Example of more complex parameter tieing</h2>

We now give a more complex pattern of parameter tieing.
(This example is due to Rainer Deventer.)
The structure is as follows:
<p>
<img src="Figures/rainer_tied.gif" height="400">
<!--<img src="rainer_dbn.jpg" height="600">-->
<p>
Since nodes 2 and 3 in slice 2 (N7 and N8)
have different parents than their counterparts in slice 1 (N2 and N3),
they must be put into different equivalence classes.
Hence we define
<pre>
eclass1 = [1 2 3 4 5];
eclass2 = [1 6 7 4 5];
</pre>
The dotted bubbles represent the equivalence classes.
Node 7 is the representative node for equivalence class
6, and node 8 is the rep. for class 7, so we need to write
<pre>
bnet.CPD{6} = xxx_CPD(bnet, 7, xxx);
bnet.CPD{7} = xxx_CPD(bnet, 8, xxx);
</pre>
In general, you can use the following code fragment:
<pre>
eclass = bnet.equiv_class(:);
for e=1:max(eclass)
  i = bnet.rep_of_eclass(e);
  bnet.CPD{e} = xxx_CPD(bnet,i);
end
</pre>
<!--
which is equivalent to
<pre>
E = max(eclass);
rep = zeros(1,E);
for e=1:E
  mems = find(eclass==e);
  rep(e) = mems(1);
end
for e=1:E
  bnet.CPD{e} = xxx_CPD(bnet, rep(e));
end
</pre>
-->