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