215 lines
6.3 KiB
C++
215 lines
6.3 KiB
C++
/* -*- mode: C++; c-basic-offset: 2; indent-tabs-mode: nil -*- */
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/*
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* Main authors:
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* Patrick Pekczynski <pekczynski@ps.uni-sb.de>
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* Mikael Lagerkvist <lagerkvist@gecode.org>
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* Christian Schulte <schulte@gecode.org>
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*
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* Copyright:
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* Patrick Pekczynski, 2004
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* Mikael Lagerkvist, 2006
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* Christian Schulte, 2007
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*
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* This file is part of Gecode, the generic constraint
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* development environment:
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* http://www.gecode.org
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*
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* Permission is hereby granted, free of charge, to any person obtaining
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* a copy of this software and associated documentation files (the
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* "Software"), to deal in the Software without restriction, including
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* without limitation the rights to use, copy, modify, merge, publish,
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* distribute, sublicense, and/or sell copies of the Software, and to
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* permit persons to whom the Software is furnished to do so, subject to
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* the following conditions:
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*
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* The above copyright notice and this permission notice shall be
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* included in all copies or substantial portions of the Software.
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*
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* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
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* EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
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* MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND
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* NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE
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* LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION
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* OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION
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* WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
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*
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*/
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#include <gecode/driver.hh>
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#include <gecode/int.hh>
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#include <gecode/minimodel.hh>
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using namespace Gecode;
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/**
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* \brief Options taking two additional parameters
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*
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* \relates LangfordNumber
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*/
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class LangfordNumberOptions : public Options {
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public:
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int k, n; /// Parameters to be given on the command line
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/// Initialize options for example with name \a s
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LangfordNumberOptions(const char* s, int k0, int n0)
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: Options(s), k(k0), n(n0) {}
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/// Parse options from arguments \a argv (number is \a argc)
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void parse(int& argc, char* argv[]) {
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Options::parse(argc,argv);
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if (argc < 3)
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return;
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n = atoi(argv[1]);
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k = atoi(argv[2]);
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}
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/// Print help message
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virtual void help(void) {
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Options::help();
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std::cerr << "\t(unsigned int) default: " << n << std::endl
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<< "\t\tparameter n" << std::endl
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<< "\t(unsigned int) default: " << k << std::endl
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<< "\t\tparameter k" << std::endl;
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}
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};
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/**
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* \brief %Example: Langford's number problem
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*
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* See problem 24 at http://www.csplib.org/.
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*
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* \ingroup Example
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*/
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class LangfordNumber : public Script {
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protected:
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int k, n; ///< Problem parameters
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IntVarArray y; ///< Sequence variables
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public:
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/// Propagation to use for model
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enum {
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PROP_REIFIED, ///< Use reified constraints
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PROP_EXTENSIONAL, ///< Use extensional constraints
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PROP_EXTENSIONAL_CHANNEL ///< Use extensional and channel constraints
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};
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/// Construct model
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LangfordNumber(const LangfordNumberOptions& opt)
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: Script(opt), k(opt.k), n(opt.n), y(*this,k*n,1,n) {
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switch (opt.propagation()) {
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case PROP_REIFIED:
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{
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// Position of values in sequence
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IntVarArgs pv(*this,k*n,0,k*n-1);
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Matrix<IntVarArgs> p(pv,n,k);
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/*
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* The occurences of v in the Langford sequence are v numbers apart.
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*
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* Let \#(i, v) denote the position of the i-th occurence of
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* value v in the Langford Sequence. Then
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*
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* \f$ \forall i, j \in \{1, \dots, k\}, i \neq j:
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* \forall v \in \{1, \dots, n\}: \#(i, v) + (v + 1) = \#(j, v)\f$
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*
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*/
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for (int i=0; i<n; i++)
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for (int j=0; j<k-1; j++)
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rel(*this, p(i,j)+i+2 == p(i,j+1));
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distinct(*this, pv, opt.ipl());
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// Channel positions <-> values
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for (int i=0; i<n; i++)
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for (int j=0; j<k; j++)
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element(*this, y, p(i,j), i+1);
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}
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break;
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case PROP_EXTENSIONAL:
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{
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IntArgs a(n-1);
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for (int v=2; v<=n; v++)
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a[v-2]=v;
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for (int v=1; v<=n; v++) {
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// Construct regular expression for all symbols but v
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if (v > 1)
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a[v-2]=v-1;
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REG ra(a), rv(v);
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extensional(*this, y, *ra+rv+(ra(v,v)+rv)(k-1,k-1)+*ra);
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}
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}
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break;
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case PROP_EXTENSIONAL_CHANNEL:
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{
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// Boolean variables for channeling
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BoolVarArgs bv(*this,k*n*n,0,1);
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Matrix<BoolVarArgs> b(bv,k*n,n);
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// Post channel constraints
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for (int i=0; i<n*k; i++)
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channel(*this, b.col(i), y[i], 1);
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// For placing two numbers three steps apart, we construct the
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// regular expression 0*100010*, and apply it to the projection of
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// the sequence on the value.
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REG r0(0), r1(1);
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for (int v=1; v<=n; v++)
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extensional(*this, b.row(v-1),
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*r0 + r1 + (r0(v,v) + r1)(k-1,k-1) + *r0);
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}
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break;
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}
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// Symmetry breaking
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rel(*this, y[0], IRT_LE, y[n*k-1]);
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// Branching
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branch(*this, y, INT_VAR_SIZE_MIN(), INT_VAL_MAX());
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}
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/// Print solution
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virtual void print(std::ostream& os) const {
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os << "\t" << y << std::endl;
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}
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/// Constructor for cloning \a l
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LangfordNumber(LangfordNumber& l)
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: Script(l), k(l.k), n(l.n) {
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y.update(*this, l.y);
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}
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/// Copy during cloning
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virtual Space*
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copy(void) {
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return new LangfordNumber(*this);
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}
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};
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/** \brief Main-function
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* \relates LangfordNumber
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*/
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int
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main(int argc, char* argv[]) {
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LangfordNumberOptions opt("Langford Numbers",3,9);
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opt.ipl(IPL_DOM);
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opt.propagation(LangfordNumber::PROP_EXTENSIONAL_CHANNEL);
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opt.propagation(LangfordNumber::PROP_REIFIED,
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"reified");
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opt.propagation(LangfordNumber::PROP_EXTENSIONAL,
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"extensional");
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opt.propagation(LangfordNumber::PROP_EXTENSIONAL_CHANNEL,
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"extensional-channel");
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opt.parse(argc, argv);
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if (opt.k < 1) {
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std::cerr << "k must be at least 1!" << std::endl;
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return 1;
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}
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if (opt.k > opt.n) {
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std::cerr << "n must be at least k!" << std::endl;
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return 1;
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}
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Script::run<LangfordNumber,DFS,LangfordNumberOptions>(opt);
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return 0;
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}
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// STATISTICS: example-any
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