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caryclark@google.com07393ca2013-04-08 11:47:37 +00001/*
2http://stackoverflow.com/questions/2009160/how-do-i-convert-the-2-control-points-of-a-cubic-curve-to-the-single-control-poi
3*/
4
5/*
6Let's call the control points of the cubic Q0..Q3 and the control points of the quadratic P0..P2.
7Then for degree elevation, the equations are:
8
9Q0 = P0
10Q1 = 1/3 P0 + 2/3 P1
11Q2 = 2/3 P1 + 1/3 P2
12Q3 = P2
13In your case you have Q0..Q3 and you're solving for P0..P2. There are two ways to compute P1 from
14 the equations above:
15
16P1 = 3/2 Q1 - 1/2 Q0
17P1 = 3/2 Q2 - 1/2 Q3
18If this is a degree-elevated cubic, then both equations will give the same answer for P1. Since
19 it's likely not, your best bet is to average them. So,
20
21P1 = -1/4 Q0 + 3/4 Q1 + 3/4 Q2 - 1/4 Q3
22
23
24SkDCubic defined by: P1/2 - anchor points, C1/C2 control points
25|x| is the euclidean norm of x
26mid-point approx of cubic: a quad that shares the same anchors with the cubic and has the
27 control point at C = (3·C2 - P2 + 3·C1 - P1)/4
28
29Algorithm
30
31pick an absolute precision (prec)
32Compute the Tdiv as the root of (cubic) equation
33sqrt(3)/18 · |P2 - 3·C2 + 3·C1 - P1|/2 · Tdiv ^ 3 = prec
34if Tdiv < 0.5 divide the cubic at Tdiv. First segment [0..Tdiv] can be approximated with by a
35 quadratic, with a defect less than prec, by the mid-point approximation.
36 Repeat from step 2 with the second resulted segment (corresponding to 1-Tdiv)
370.5<=Tdiv<1 - simply divide the cubic in two. The two halves can be approximated by the mid-point
38 approximation
39Tdiv>=1 - the entire cubic can be approximated by the mid-point approximation
40
41confirmed by (maybe stolen from)
42http://www.caffeineowl.com/graphics/2d/vectorial/cubic2quad01.html
43// maybe in turn derived from http://www.cccg.ca/proceedings/2004/36.pdf
44// also stored at http://www.cis.usouthal.edu/~hain/general/Publications/Bezier/bezier%20cccg04%20paper.pdf
45
46*/
47
48#include "SkPathOpsCubic.h"
49#include "SkPathOpsLine.h"
50#include "SkPathOpsQuad.h"
51#include "SkReduceOrder.h"
52#include "SkTDArray.h"
commit-bot@chromium.orgb76d3b62013-04-22 19:55:19 +000053#include "SkTSort.h"
caryclark@google.com07393ca2013-04-08 11:47:37 +000054
55#define USE_CUBIC_END_POINTS 1
56
57static double calc_t_div(const SkDCubic& cubic, double precision, double start) {
58 const double adjust = sqrt(3.) / 36;
59 SkDCubic sub;
60 const SkDCubic* cPtr;
61 if (start == 0) {
62 cPtr = &cubic;
63 } else {
64 // OPTIMIZE: special-case half-split ?
65 sub = cubic.subDivide(start, 1);
66 cPtr = &sub;
67 }
68 const SkDCubic& c = *cPtr;
69 double dx = c[3].fX - 3 * (c[2].fX - c[1].fX) - c[0].fX;
70 double dy = c[3].fY - 3 * (c[2].fY - c[1].fY) - c[0].fY;
71 double dist = sqrt(dx * dx + dy * dy);
72 double tDiv3 = precision / (adjust * dist);
73 double t = SkDCubeRoot(tDiv3);
74 if (start > 0) {
75 t = start + (1 - start) * t;
76 }
77 return t;
78}
79
80SkDQuad SkDCubic::toQuad() const {
81 SkDQuad quad;
82 quad[0] = fPts[0];
83 const SkDPoint fromC1 = {(3 * fPts[1].fX - fPts[0].fX) / 2, (3 * fPts[1].fY - fPts[0].fY) / 2};
84 const SkDPoint fromC2 = {(3 * fPts[2].fX - fPts[3].fX) / 2, (3 * fPts[2].fY - fPts[3].fY) / 2};
85 quad[1].fX = (fromC1.fX + fromC2.fX) / 2;
86 quad[1].fY = (fromC1.fY + fromC2.fY) / 2;
87 quad[2] = fPts[3];
88 return quad;
89}
90
91static bool add_simple_ts(const SkDCubic& cubic, double precision, SkTDArray<double>* ts) {
92 double tDiv = calc_t_div(cubic, precision, 0);
93 if (tDiv >= 1) {
94 return true;
95 }
96 if (tDiv >= 0.5) {
97 *ts->append() = 0.5;
98 return true;
99 }
100 return false;
101}
102
103static void addTs(const SkDCubic& cubic, double precision, double start, double end,
104 SkTDArray<double>* ts) {
105 double tDiv = calc_t_div(cubic, precision, 0);
106 double parts = ceil(1.0 / tDiv);
107 for (double index = 0; index < parts; ++index) {
108 double newT = start + (index / parts) * (end - start);
109 if (newT > 0 && newT < 1) {
110 *ts->append() = newT;
111 }
112 }
113}
114
115// flavor that returns T values only, deferring computing the quads until they are needed
116// FIXME: when called from recursive intersect 2, this could take the original cubic
117// and do a more precise job when calling chop at and sub divide by computing the fractional ts.
118// it would still take the prechopped cubic for reduce order and find cubic inflections
119void SkDCubic::toQuadraticTs(double precision, SkTDArray<double>* ts) const {
120 SkReduceOrder reducer;
121 int order = reducer.reduce(*this, SkReduceOrder::kAllow_Quadratics, SkReduceOrder::kFill_Style);
122 if (order < 3) {
123 return;
124 }
125 double inflectT[5];
126 int inflections = findInflections(inflectT);
127 SkASSERT(inflections <= 2);
128 if (!endsAreExtremaInXOrY()) {
129 inflections += findMaxCurvature(&inflectT[inflections]);
130 SkASSERT(inflections <= 5);
131 }
commit-bot@chromium.orgb76d3b62013-04-22 19:55:19 +0000132 SkTQSort<double>(inflectT, &inflectT[inflections - 1]);
caryclark@google.com07393ca2013-04-08 11:47:37 +0000133 // OPTIMIZATION: is this filtering common enough that it needs to be pulled out into its
134 // own subroutine?
135 while (inflections && approximately_less_than_zero(inflectT[0])) {
136 memmove(inflectT, &inflectT[1], sizeof(inflectT[0]) * --inflections);
137 }
138 int start = 0;
139 do {
140 int next = start + 1;
141 if (next >= inflections) {
142 break;
143 }
144 if (!approximately_equal(inflectT[start], inflectT[next])) {
145 ++start;
146 continue;
147 }
148 memmove(&inflectT[start], &inflectT[next], sizeof(inflectT[0]) * (--inflections - start));
149 } while (true);
150 while (inflections && approximately_greater_than_one(inflectT[inflections - 1])) {
151 --inflections;
152 }
153 SkDCubicPair pair;
154 if (inflections == 1) {
155 pair = chopAt(inflectT[0]);
156 int orderP1 = reducer.reduce(pair.first(), SkReduceOrder::kNo_Quadratics,
157 SkReduceOrder::kFill_Style);
158 if (orderP1 < 2) {
159 --inflections;
160 } else {
161 int orderP2 = reducer.reduce(pair.second(), SkReduceOrder::kNo_Quadratics,
162 SkReduceOrder::kFill_Style);
163 if (orderP2 < 2) {
164 --inflections;
165 }
166 }
167 }
168 if (inflections == 0 && add_simple_ts(*this, precision, ts)) {
169 return;
170 }
171 if (inflections == 1) {
172 pair = chopAt(inflectT[0]);
173 addTs(pair.first(), precision, 0, inflectT[0], ts);
174 addTs(pair.second(), precision, inflectT[0], 1, ts);
175 return;
176 }
177 if (inflections > 1) {
178 SkDCubic part = subDivide(0, inflectT[0]);
179 addTs(part, precision, 0, inflectT[0], ts);
180 int last = inflections - 1;
181 for (int idx = 0; idx < last; ++idx) {
182 part = subDivide(inflectT[idx], inflectT[idx + 1]);
183 addTs(part, precision, inflectT[idx], inflectT[idx + 1], ts);
184 }
185 part = subDivide(inflectT[last], 1);
186 addTs(part, precision, inflectT[last], 1, ts);
187 return;
188 }
189 addTs(*this, precision, 0, 1, ts);
190}