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1 /* | 1 /* |
2 http://stackoverflow.com/questions/2009160/how-do-i-convert-the-2-control-points
-of-a-cubic-curve-to-the-single-control-poi | 2 http://stackoverflow.com/questions/2009160/how-do-i-convert-the-2-control-points
-of-a-cubic-curve-to-the-single-control-poi |
3 */ | 3 */ |
4 | 4 |
5 /* | 5 /* |
6 Let's call the control points of the cubic Q0..Q3 and the control points of the
quadratic P0..P2. | 6 Let's call the control points of the cubic Q0..Q3 and the control points of the
quadratic P0..P2. |
7 Then for degree elevation, the equations are: | 7 Then for degree elevation, the equations are: |
8 | 8 |
9 Q0 = P0 | 9 Q0 = P0 |
10 Q1 = 1/3 P0 + 2/3 P1 | 10 Q1 = 1/3 P0 + 2/3 P1 |
11 Q2 = 2/3 P1 + 1/3 P2 | 11 Q2 = 2/3 P1 + 1/3 P2 |
12 Q3 = P2 | 12 Q3 = P2 |
13 In your case you have Q0..Q3 and you're solving for P0..P2. There are two ways t
o compute P1 from | 13 In your case you have Q0..Q3 and you're solving for P0..P2. There are two ways t
o compute P1 from |
14 the equations above: | 14 the equations above: |
15 | 15 |
16 P1 = 3/2 Q1 - 1/2 Q0 | 16 P1 = 3/2 Q1 - 1/2 Q0 |
17 P1 = 3/2 Q2 - 1/2 Q3 | 17 P1 = 3/2 Q2 - 1/2 Q3 |
18 If this is a degree-elevated cubic, then both equations will give the same answe
r for P1. Since | 18 If this is a degree-elevated cubic, then both equations will give the same answe
r for P1. Since |
19 it's likely not, your best bet is to average them. So, | 19 it's likely not, your best bet is to average them. So, |
20 | 20 |
21 P1 = -1/4 Q0 + 3/4 Q1 + 3/4 Q2 - 1/4 Q3 | 21 P1 = -1/4 Q0 + 3/4 Q1 + 3/4 Q2 - 1/4 Q3 |
22 | |
23 SkDCubic defined by: P1/2 - anchor points, C1/C2 control points | |
24 |x| is the euclidean norm of x | |
25 mid-point approx of cubic: a quad that shares the same anchors with the cubic an
d has the | |
26 control point at C = (3·C2 - P2 + 3·C1 - P1)/4 | |
27 | |
28 Algorithm | |
29 | |
30 pick an absolute precision (prec) | |
31 Compute the Tdiv as the root of (cubic) equation | |
32 sqrt(3)/18 · |P2 - 3·C2 + 3·C1 - P1|/2 · Tdiv ^ 3 = prec | |
33 if Tdiv < 0.5 divide the cubic at Tdiv. First segment [0..Tdiv] can be approxima
ted with by a | |
34 quadratic, with a defect less than prec, by the mid-point approximation. | |
35 Repeat from step 2 with the second resulted segment (corresponding to 1-Tdiv) | |
36 0.5<=Tdiv<1 - simply divide the cubic in two. The two halves can be approximated
by the mid-point | |
37 approximation | |
38 Tdiv>=1 - the entire cubic can be approximated by the mid-point approximation | |
39 | |
40 confirmed by (maybe stolen from) | |
41 http://www.caffeineowl.com/graphics/2d/vectorial/cubic2quad01.html | |
42 // maybe in turn derived from http://www.cccg.ca/proceedings/2004/36.pdf | |
43 // also stored at http://www.cis.usouthal.edu/~hain/general/Publications/Bezier/
bezier%20cccg04%20paper.pdf | |
44 | |
45 */ | 22 */ |
46 | 23 |
47 #include "SkPathOpsCubic.h" | 24 #include "SkPathOpsCubic.h" |
48 #include "SkPathOpsLine.h" | |
49 #include "SkPathOpsQuad.h" | 25 #include "SkPathOpsQuad.h" |
50 #include "SkReduceOrder.h" | |
51 #include "SkTArray.h" | |
52 #include "SkTSort.h" | |
53 | |
54 #define USE_CUBIC_END_POINTS 1 | |
55 | |
56 static double calc_t_div(const SkDCubic& cubic, double precision, double start)
{ | |
57 const double adjust = sqrt(3.) / 36; | |
58 SkDCubic sub; | |
59 const SkDCubic* cPtr; | |
60 if (start == 0) { | |
61 cPtr = &cubic; | |
62 } else { | |
63 // OPTIMIZE: special-case half-split ? | |
64 sub = cubic.subDivide(start, 1); | |
65 cPtr = ⊂ | |
66 } | |
67 const SkDCubic& c = *cPtr; | |
68 double dx = c[3].fX - 3 * (c[2].fX - c[1].fX) - c[0].fX; | |
69 double dy = c[3].fY - 3 * (c[2].fY - c[1].fY) - c[0].fY; | |
70 double dist = sqrt(dx * dx + dy * dy); | |
71 double tDiv3 = precision / (adjust * dist); | |
72 double t = SkDCubeRoot(tDiv3); | |
73 if (start > 0) { | |
74 t = start + (1 - start) * t; | |
75 } | |
76 return t; | |
77 } | |
78 | 26 |
79 SkDQuad SkDCubic::toQuad() const { | 27 SkDQuad SkDCubic::toQuad() const { |
80 SkDQuad quad; | 28 SkDQuad quad; |
81 quad[0] = fPts[0]; | 29 quad[0] = fPts[0]; |
82 const SkDPoint fromC1 = {(3 * fPts[1].fX - fPts[0].fX) / 2, (3 * fPts[1].fY
- fPts[0].fY) / 2}; | 30 const SkDPoint fromC1 = {(3 * fPts[1].fX - fPts[0].fX) / 2, (3 * fPts[1].fY
- fPts[0].fY) / 2}; |
83 const SkDPoint fromC2 = {(3 * fPts[2].fX - fPts[3].fX) / 2, (3 * fPts[2].fY
- fPts[3].fY) / 2}; | 31 const SkDPoint fromC2 = {(3 * fPts[2].fX - fPts[3].fX) / 2, (3 * fPts[2].fY
- fPts[3].fY) / 2}; |
84 quad[1].fX = (fromC1.fX + fromC2.fX) / 2; | 32 quad[1].fX = (fromC1.fX + fromC2.fX) / 2; |
85 quad[1].fY = (fromC1.fY + fromC2.fY) / 2; | 33 quad[1].fY = (fromC1.fY + fromC2.fY) / 2; |
86 quad[2] = fPts[3]; | 34 quad[2] = fPts[3]; |
87 return quad; | 35 return quad; |
88 } | 36 } |
89 | |
90 static bool add_simple_ts(const SkDCubic& cubic, double precision, SkTArray<doub
le, true>* ts) { | |
91 double tDiv = calc_t_div(cubic, precision, 0); | |
92 if (tDiv >= 1) { | |
93 return true; | |
94 } | |
95 if (tDiv >= 0.5) { | |
96 ts->push_back(0.5); | |
97 return true; | |
98 } | |
99 return false; | |
100 } | |
101 | |
102 static void addTs(const SkDCubic& cubic, double precision, double start, double
end, | |
103 SkTArray<double, true>* ts) { | |
104 double tDiv = calc_t_div(cubic, precision, 0); | |
105 double parts = ceil(1.0 / tDiv); | |
106 for (double index = 0; index < parts; ++index) { | |
107 double newT = start + (index / parts) * (end - start); | |
108 if (newT > 0 && newT < 1) { | |
109 ts->push_back(newT); | |
110 } | |
111 } | |
112 } | |
113 | |
114 // flavor that returns T values only, deferring computing the quads until they a
re needed | |
115 // FIXME: when called from recursive intersect 2, this could take the original c
ubic | |
116 // and do a more precise job when calling chop at and sub divide by computing th
e fractional ts. | |
117 // it would still take the prechopped cubic for reduce order and find cubic infl
ections | |
118 void SkDCubic::toQuadraticTs(double precision, SkTArray<double, true>* ts) const
{ | |
119 SkReduceOrder reducer; | |
120 int order = reducer.reduce(*this, SkReduceOrder::kAllow_Quadratics); | |
121 if (order < 3) { | |
122 return; | |
123 } | |
124 double inflectT[5]; | |
125 int inflections = findInflections(inflectT); | |
126 SkASSERT(inflections <= 2); | |
127 if (!endsAreExtremaInXOrY()) { | |
128 inflections += findMaxCurvature(&inflectT[inflections]); | |
129 SkASSERT(inflections <= 5); | |
130 } | |
131 SkTQSort<double>(inflectT, &inflectT[inflections - 1]); | |
132 // OPTIMIZATION: is this filtering common enough that it needs to be pulled
out into its | |
133 // own subroutine? | |
134 while (inflections && approximately_less_than_zero(inflectT[0])) { | |
135 memmove(inflectT, &inflectT[1], sizeof(inflectT[0]) * --inflections); | |
136 } | |
137 int start = 0; | |
138 int next = 1; | |
139 while (next < inflections) { | |
140 if (!approximately_equal(inflectT[start], inflectT[next])) { | |
141 ++start; | |
142 ++next; | |
143 continue; | |
144 } | |
145 memmove(&inflectT[start], &inflectT[next], sizeof(inflectT[0]) * (--infl
ections - start)); | |
146 } | |
147 | |
148 while (inflections && approximately_greater_than_one(inflectT[inflections -
1])) { | |
149 --inflections; | |
150 } | |
151 SkDCubicPair pair; | |
152 if (inflections == 1) { | |
153 pair = chopAt(inflectT[0]); | |
154 int orderP1 = reducer.reduce(pair.first(), SkReduceOrder::kNo_Quadratics
); | |
155 if (orderP1 < 2) { | |
156 --inflections; | |
157 } else { | |
158 int orderP2 = reducer.reduce(pair.second(), SkReduceOrder::kNo_Quadr
atics); | |
159 if (orderP2 < 2) { | |
160 --inflections; | |
161 } | |
162 } | |
163 } | |
164 if (inflections == 0 && add_simple_ts(*this, precision, ts)) { | |
165 return; | |
166 } | |
167 if (inflections == 1) { | |
168 pair = chopAt(inflectT[0]); | |
169 addTs(pair.first(), precision, 0, inflectT[0], ts); | |
170 addTs(pair.second(), precision, inflectT[0], 1, ts); | |
171 return; | |
172 } | |
173 if (inflections > 1) { | |
174 SkDCubic part = subDivide(0, inflectT[0]); | |
175 addTs(part, precision, 0, inflectT[0], ts); | |
176 int last = inflections - 1; | |
177 for (int idx = 0; idx < last; ++idx) { | |
178 part = subDivide(inflectT[idx], inflectT[idx + 1]); | |
179 addTs(part, precision, inflectT[idx], inflectT[idx + 1], ts); | |
180 } | |
181 part = subDivide(inflectT[last], 1); | |
182 addTs(part, precision, inflectT[last], 1, ts); | |
183 return; | |
184 } | |
185 addTs(*this, precision, 0, 1, ts); | |
186 } | |
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