118 lines
3.6 KiB
C++
118 lines
3.6 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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* Vincent Barichard <Vincent.Barichard@univ-angers.fr>
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*
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* Copyright:
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* Vincent Barichard, 2012
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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/minimodel.hh>
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#include <gecode/float.hh>
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using namespace Gecode;
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/**
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* \brief %Example: Cartesian Heart
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*
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* There are many mathematical curves that produce heart shapes.
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* With a good solving effort, coordinates of a filled heart shape
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* can be computed by solving the cartesian equation:
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*
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* \f[
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* x^2+2\left(y-p\times\operatorname{abs}(x)^{\frac{1}{q}}\right)^2 = 1
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* \f]
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*
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* By setting \f$p=0.5\f$ and \f$q=2\f$, it yields to the equation:
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*
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* \f[
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* x^2+2\left(y-\frac{\operatorname{abs}(x)^{\frac{1}{2}}}{2}\right)^2 = 1
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* \f]
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*
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* To get reasonable interval starting sizes, \f$x\f$ and \f$y\f$
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* are restricted to \f$[-20;20]\f$.
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*
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* \ingroup Example
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*/
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class CartesianHeart : public Script {
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protected:
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/// The numbers
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FloatVarArray f;
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/// Minimum distance between two solutions
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FloatNum step;
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public:
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/// Actual model
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CartesianHeart(const Options& opt)
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: Script(opt), f(*this,2,-20,20), step(opt.step()) {
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int q = 2;
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FloatNum p = 0.5;
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// Post equation
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rel(*this, sqr(f[0]) + 2*sqr(f[1]-p*nroot(abs(f[0]),q)) == 1);
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branch(*this, f[0], FLOAT_VAL_SPLIT_MIN());
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branch(*this, f[1], FLOAT_VAL_SPLIT_MIN());
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}
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/// Constructor for cloning \a p
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CartesianHeart(CartesianHeart& p)
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: Script(p), step(p.step) {
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f.update(*this, p.f);
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}
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/// Copy during cloning
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virtual Space* copy(void) {
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return new CartesianHeart(*this);
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}
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/// Add constraints to current model to get next solution (not too close)
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virtual void constrain(const Space& _b) {
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const CartesianHeart& b = static_cast<const CartesianHeart&>(_b);
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rel(*this,
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(f[0] >= (b.f[0].max()+step)) ||
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(f[1] >= (b.f[1].max()+step)) ||
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(f[1] <= (b.f[1].min()-step)));
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}
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/// Print solution coordinates
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virtual void print(std::ostream& os) const {
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os << "XY " << f[0].med() << " " << f[1].med()
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<< std::endl;
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}
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};
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/** \brief Main-function
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* \relates CartesianHeart
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*/
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int main(int argc, char* argv[]) {
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Options opt("CartesianHeart");
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opt.solutions(0);
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opt.step(0.01);
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opt.parse(argc,argv);
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Script::run<CartesianHeart,BAB,Options>(opt);
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return 0;
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}
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// STATISTICS: example-any
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