1 /* |
|
2 * LDForge: LDraw parts authoring CAD |
|
3 * Copyright (C) 2013, 2014 Santeri Piippo |
|
4 * |
|
5 * This program is free software: you can redistribute it and/or modify |
|
6 * it under the terms of the GNU General Public License as published by |
|
7 * the Free Software Foundation, either version 3 of the License, or |
|
8 * (at your option) any later version. |
|
9 * |
|
10 * This program is distributed in the hope that it will be useful, |
|
11 * but WITHOUT ANY WARRANTY; without even the implied warranty of |
|
12 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the |
|
13 * GNU General Public License for more details. |
|
14 * |
|
15 * You should have received a copy of the GNU General Public License |
|
16 * along with this program. If not, see <http://www.gnu.org/licenses/>. |
|
17 */ |
|
18 |
|
19 #include "RingFinder.h" |
|
20 #include "../Misc.h" |
|
21 |
|
22 RingFinder g_RingFinder; |
|
23 |
|
24 // ============================================================================= |
|
25 // |
|
26 bool RingFinder::findRingsRecursor (double r0, double r1, Solution& currentSolution) |
|
27 { |
|
28 // Don't recurse too deep. |
|
29 if (m_stack >= 5) |
|
30 return false; |
|
31 |
|
32 // Find the scale and number of a ring between r1 and r0. |
|
33 assert (r1 >= r0); |
|
34 double scale = r1 - r0; |
|
35 double num = r0 / scale; |
|
36 |
|
37 // If the ring number is integral, we have found a fitting ring to r0 -> r1! |
|
38 if (isInteger (num)) |
|
39 { |
|
40 Component cmp; |
|
41 cmp.scale = scale; |
|
42 cmp.num = (int) round (num); |
|
43 currentSolution.addComponent (cmp); |
|
44 |
|
45 // If we're still at the first recursion, this is the only |
|
46 // ring and there's nothing left to do. Guess we found the winner. |
|
47 if (m_stack == 0) |
|
48 { |
|
49 m_solutions.push_back (currentSolution); |
|
50 return true; |
|
51 } |
|
52 } |
|
53 else |
|
54 { |
|
55 // Try find solutions by splitting the ring in various positions. |
|
56 if (isZero (r1 - r0)) |
|
57 return false; |
|
58 |
|
59 double interval; |
|
60 |
|
61 // Determine interval. The smaller delta between radii, the more precise |
|
62 // interval should be used. We can't really use a 0.5 increment when |
|
63 // calculating rings to 10 -> 105... that would take ages to process! |
|
64 if (r1 - r0 < 0.5) |
|
65 interval = 0.1; |
|
66 else if (r1 - r0 < 10) |
|
67 interval = 0.5; |
|
68 else if (r1 - r0 < 50) |
|
69 interval = 1; |
|
70 else |
|
71 interval = 5; |
|
72 |
|
73 // Now go through possible splits and try find rings for both segments. |
|
74 for (double r = r0 + interval; r < r1; r += interval) |
|
75 { |
|
76 Solution sol = currentSolution; |
|
77 |
|
78 m_stack++; |
|
79 bool res = findRingsRecursor (r0, r, sol) && findRingsRecursor (r, r1, sol); |
|
80 m_stack--; |
|
81 |
|
82 if (res) |
|
83 { |
|
84 // We succeeded in finding radii for this segment. If the stack is 0, this |
|
85 // is the first recursion to this function. Thus there are no more ring segments |
|
86 // to process and we can add the solution. |
|
87 // |
|
88 // If not, when this function ends, it will be called again with more arguments. |
|
89 // Accept the solution to this segment by setting currentSolution to sol, and |
|
90 // return true to continue processing. |
|
91 if (m_stack == 0) |
|
92 m_solutions.push_back (sol); |
|
93 else |
|
94 { |
|
95 currentSolution = sol; |
|
96 return true; |
|
97 } |
|
98 } |
|
99 } |
|
100 |
|
101 return false; |
|
102 } |
|
103 |
|
104 return true; |
|
105 } |
|
106 |
|
107 // ============================================================================= |
|
108 // Main function. Call this with r0 and r1. If this returns true, use bestSolution |
|
109 // for the solution that was presented. |
|
110 // |
|
111 bool RingFinder::findRings (double r0, double r1) |
|
112 { |
|
113 m_solutions.clear(); |
|
114 Solution sol; |
|
115 |
|
116 // Recurse in and try find solutions. |
|
117 findRingsRecursor (r0, r1, sol); |
|
118 |
|
119 // Compare the solutions and find the best one. The solution class has an operator> |
|
120 // overload to compare two solutions. |
|
121 m_bestSolution = null; |
|
122 |
|
123 for (QVector<Solution>::iterator solp = m_solutions.begin(); solp != m_solutions.end(); ++solp) |
|
124 { |
|
125 const Solution& sol = *solp; |
|
126 |
|
127 if (m_bestSolution == null || sol.isSuperiorTo (m_bestSolution)) |
|
128 m_bestSolution = / |
|
129 } |
|
130 |
|
131 return (m_bestSolution != null); |
|
132 } |
|
133 |
|
134 // ============================================================================= |
|
135 // |
|
136 bool RingFinder::Solution::isSuperiorTo (const Solution* other) const |
|
137 { |
|
138 // If this solution has less components than the other one, this one |
|
139 // is definitely better. |
|
140 if (getComponents().size() < other->getComponents().size()) |
|
141 return true; |
|
142 |
|
143 // vice versa |
|
144 if (other->getComponents().size() < getComponents().size()) |
|
145 return false; |
|
146 |
|
147 // Calculate the maximum ring number. Since the solutions have equal |
|
148 // ring counts, the solutions with lesser maximum rings should result |
|
149 // in cleaner code and less new primitives, right? |
|
150 int maxA = 0, |
|
151 maxB = 0; |
|
152 |
|
153 for (int i = 0; i < getComponents().size(); ++i) |
|
154 { |
|
155 maxA = max (getComponents()[i].num, maxA); |
|
156 maxB = max (other->getComponents()[i].num, maxB); |
|
157 } |
|
158 |
|
159 if (maxA < maxB) |
|
160 return true; |
|
161 |
|
162 if (maxB < maxA) |
|
163 return false; |
|
164 |
|
165 // Solutions have equal rings and equal maximum ring numbers. Let's |
|
166 // just say this one is better, at this point it does not matter which |
|
167 // one is chosen. |
|
168 return true; |
|
169 } |
|