src/misc/ringFinder.cc

changeset 629
b75c6cce02e2
parent 628
6b13e4c2e97b
child 630
42ec68fcad9e
child 675
450827da2376
--- a/src/misc/ringFinder.cc	Mon Jan 20 23:44:22 2014 +0200
+++ /dev/null	Thu Jan 01 00:00:00 1970 +0000
@@ -1,178 +0,0 @@
-/*
- *  LDForge: LDraw parts authoring CAD
- *  Copyright (C) 2013, 2014 Santeri Piippo
- *
- *  This program is free software: you can redistribute it and/or modify
- *  it under the terms of the GNU General Public License as published by
- *  the Free Software Foundation, either version 3 of the License, or
- *  (at your option) any later version.
- *
- *  This program is distributed in the hope that it will be useful,
- *  but WITHOUT ANY WARRANTY; without even the implied warranty of
- *  MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
- *  GNU General Public License for more details.
- *
- *  You should have received a copy of the GNU General Public License
- *  along with this program.  If not, see <http://www.gnu.org/licenses/>.
- */
-
-#include "ringFinder.h"
-#include "../misc.h"
-
-RingFinder g_RingFinder;
-
-// =============================================================================
-// This is the main algorithm of the ring finder. It tries to use math to find
-// the one ring between r0 and r1. If it fails (the ring number is non-integral),
-// it finds an intermediate radius (ceil of the ring number times scale) and
-// splits the radius at this point, calling this function again to try find the
-// rings between r0 - r and r - r1.
-//
-// This does not always yield into usable results. If at some point r == r0 or
-// r == r1, there is no hope of finding the rings, at least with this algorithm,
-// as it would fall into an infinite recursion.
-// -----------------------------------------------------------------------------
-bool RingFinder::findRingsRecursor (double r0, double r1, Solution& currentSolution)
-{
-	// Don't recurse too deep.
-	if (m_stack >= 5)
-		return false;
-
-	// Find the scale and number of a ring between r1 and r0.
-	assert (r1 >= r0);
-	double scale = r1 - r0;
-	double num = r0 / scale;
-
-	// If the ring number is integral, we have found a fitting ring to r0 -> r1!
-	if (isInteger (num))
-	{
-		Component cmp;
-		cmp.scale = scale;
-		cmp.num = (int) round (num);
-		currentSolution.addComponent (cmp);
-
-		// If we're still at the first recursion, this is the only
-		// ring and there's nothing left to do. Guess we found the winner.
-		if (m_stack == 0)
-		{
-			m_solutions.push_back (currentSolution);
-			return true;
-		}
-	}
-	else
-	{
-		// Try find solutions by splitting the ring in various positions.
-		if (isZero (r1 - r0))
-			return false;
-
-		double interval;
-
-		// Determine interval. The smaller delta between radii, the more precise
-		// interval should be used. We can't really use a 0.5 increment when
-		// calculating rings to 10 -> 105... that would take ages to process!
-		if (r1 - r0 < 0.5)
-			interval = 0.1;
-		else if (r1 - r0 < 10)
-			interval = 0.5;
-		else if (r1 - r0 < 50)
-			interval = 1;
-		else
-			interval = 5;
-
-		// Now go through possible splits and try find rings for both segments.
-		for (double r = r0 + interval; r < r1; r += interval)
-		{
-			Solution sol = currentSolution;
-
-			m_stack++;
-			bool res = findRingsRecursor (r0, r, sol) && findRingsRecursor (r, r1, sol);
-			m_stack--;
-
-			if (res)
-			{
-				// We succeeded in finding radii for this segment. If the stack is 0, this
-				// is the first recursion to this function. Thus there are no more ring segments
-				// to process and we can add the solution.
-				//
-				// If not, when this function ends, it will be called again with more arguments.
-				// Accept the solution to this segment by setting currentSolution to sol, and
-				// return true to continue processing.
-				if (m_stack == 0)
-					m_solutions.push_back (sol);
-				else
-				{
-					currentSolution = sol;
-					return true;
-				}
-			}
-		}
-
-		return false;
-	}
-
-	return true;
-}
-
-// =============================================================================
-// Main function. Call this with r0 and r1. If this returns true, use bestSolution
-// for the solution that was presented.
-// -----------------------------------------------------------------------------
-bool RingFinder::findRings (double r0, double r1)
-{
-	m_solutions.clear();
-	Solution sol;
-
-	// Recurse in and try find solutions.
-	findRingsRecursor (r0, r1, sol);
-
-	// Compare the solutions and find the best one. The solution class has an operator>
-	// overload to compare two solutions.
-	m_bestSolution = null;
-
-	for (QVector<Solution>::iterator solp = m_solutions.begin(); solp != m_solutions.end(); ++solp)
-	{
-		const Solution& sol = *solp;
-
-		if (m_bestSolution == null || sol.isBetterThan (m_bestSolution))
-			m_bestSolution = &sol;
-	}
-
-	return (m_bestSolution != null);
-}
-
-// =============================================================================
-// -----------------------------------------------------------------------------
-bool RingFinder::Solution::isBetterThan (const Solution* other) const
-{
-	// If this solution has less components than the other one, this one
-	// is definitely better.
-	if (getComponents().size() < other->getComponents().size())
-		return true;
-
-	// vice versa
-	if (other->getComponents().size() < getComponents().size())
-		return false;
-
-	// Calculate the maximum ring number. Since the solutions have equal
-	// ring counts, the solutions with lesser maximum rings should result
-	// in cleaner code and less new primitives, right?
-	int maxA = 0,
-		maxB = 0;
-
-	for (int i = 0; i < getComponents().size(); ++i)
-	{
-		maxA = max (getComponents()[i].num, maxA);
-		maxB = max (other->getComponents()[i].num, maxB);
-	}
-
-	if (maxA < maxB)
-		return true;
-
-	if (maxB < maxA)
-		return false;
-
-	// Solutions have equal rings and equal maximum ring numbers. Let's
-	// just say this one is better, at this point it does not matter which
-	// one is chosen.
-	return true;
-}
\ No newline at end of file

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