85a326c5e4
interpretation... (Have not compiled/checked/changed paradisEO.) That is, the current code compiles with gcc-3.4 and the checks (besides t-MGE1bit) all pass.
71 lines
2.3 KiB
C++
71 lines
2.3 KiB
C++
/** -*- mode: c++; c-indent-level: 4; c++-member-init-indent: 8; comment-column: 35; -*-
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-----------------------------------------------------------------------------
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eoParetoRanking.h
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(c) Maarten Keijzer, Marc Schoenauer, 2001
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This library is free software; you can redistribute it and/or
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modify it under the terms of the GNU Lesser General Public
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License as published by the Free Software Foundation; either
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version 2 of the License, or (at your option) any later version.
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This library is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
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Lesser General Public License for more details.
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You should have received a copy of the GNU Lesser General Public
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License along with this library; if not, write to the Free Software
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Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA
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Contact: todos@geneura.ugr.es, http://geneura.ugr.es
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Marc.Schoenauer@polytechnique.fr
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mkeijzer@dhi.dk
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*/
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//-----------------------------------------------------------------------------
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#ifndef eoParetoRanking_h
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#define eoParetoRanking_h
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#include <eoPerf2Worth.h>
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#include <eoDominanceMap.h>
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/**
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Straightforward pareto ranking. Every individual gets a rank according to the number
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of elements it dominates. Note that without niching, this technique will usually not
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find the whole front of non-dominated solutions, but will quite likely converge
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on a single spot on the front.
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*/
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template <class EOT>
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class eoParetoRanking : public eoPerf2WorthCached<EOT, double>
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{
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public:
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using eoParetoRanking< EOT>::value;
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eoParetoRanking(eoDominanceMap<EOT>& _dominanceMap)
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: eoPerf2WorthCached<EOT, double>(), dominanceMap(_dominanceMap)
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{}
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void calculate_worths(const eoPop<EOT>& _pop)
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{
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dominanceMap(_pop);
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value() = dominanceMap.sum_dominators(); // get rank: 0 means part of current front
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// calculate maximum
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double maxim = *std::max_element(value().begin(), value().end());
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// higher is better, so invert the value
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for (unsigned i = 0; i < value().size(); ++i)
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{
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value()[i] = maxim - value()[i];
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}
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}
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private :
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eoDominanceMap<EOT>& dominanceMap;
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};
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#endif
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