Difference between revisions of "Talk:Presentation"

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m (Combinatorial APS?)
m (Combinatorial APS?)
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== Combinatorial APS? ==
 
== Combinatorial APS? ==
For some problems that are at least partly discrete, the basic APS applies a \emph{quantisation} operator. If the quantum is $q_{d}$ for
+
For some problems that are at least partly discrete, the basic APS applies a ''quantisation'' operator. If the quantum is $q_{d}$ for
 
the dimension $d$, the operator is defined by <br />
 
the dimension $d$, the operator is defined by <br />
  

Revision as of 22:55, 13 June 2013

Combinatorial APS?

For some problems that are at least partly discrete, the basic APS applies a quantisation operator. If the quantum is $q_{d}$ for the dimension $d$, the operator is defined by

$$ x_{i,d}\leftarrow q_{d}\left\lfloor \left(0.5+x_{i,d}/q_{d}\right)\right.\tag{1} $$

where $\left\lfloor \right.$ is the floor operator.

Now, for combinatorial problems, it is certainly better to define specific operators for Expansion, Contraction, and Local search, similarly to what has been done, for example, for PSO.