How do I make an algorithm to zig-zag fill a grid at any size as shown in the image below?
Here is my algorithm which doesn't work. (Starting bottom left to top right corner instead):
x1 = 0;
y1 = grid_h-1;
var a = 0;
put(x1,y1);
while(!((x1 = grid_w-1) and (y1 = 0))) { //If it isn't at the top right corner
if a = 2 {
x1 += 1;
put(x1,y1);
while(x1 != grid_w-1) { //While x1 isn't at the right
//Go diagonally down
x1 += 1;
y1 += 1;
put(x1,y1);
}
y1 -= 1;
put(x1,y1);
while(y1 != 0) { //While y1 isn't at the top
//Go diagonally up
x1 -= 1;
y1 -= 1;
put(x1,y1);
}
} else if a = 1 {
while(x1 != grid_w-1) { //While x1 isn't at the right
//Go diagonally down
x1 += 1;
y1 += 1;
put(x1,y1);
}
y1 -= 1;
put(x1,y1);
while(y1 != 0) { //While y1 isn't at the top
//Go diagonally up
x1 -= 1;
y1 -= 1;
put(x1,y1);
}
x1 += 1;
put(x1,y1);
} else {
y1 -= 1;
if (y1 = 0) { a = 1; } //At top?
put(x1,y1);
while(y1 != grid_h-1) { //While y1 isn't at the bottom
//Go diagonally down
x1 += 1;
y1 += 1;
put(x1,y1);
}
x1 += 1;
put(x1,y1);
while(x1 != 0) { //While x1 isn't at the left
//Go diagonally up
x1 -= 1;
y1 -= 1;
put(x1,y1);
if (y1 = 0) { a = 2; } //At top?
}
}
}
Any simpler way to do this?
The key observation here is that you go northeast when the Manhattan distance to the top left square is odd and southwest otherwise.
Of course, you must consider hitting one of the edges. For example, when you walk southwest and hit the bottom or south edge, you move east instead; when you hit the left or west edge, you move south. You can either catch the three cases (south edge, west edge, unrestrained movement) or you can move and correct your position when you have walked out of bounds.
After hitting an adge, your new position should leave you moving the other way. That is, each correction involves an odd number of steps. (Steps here is the Manhattan distance between the point you'd have gone to normally and the point you ended up in.)
If your zigzagging algorithm works correctly, you will end up visiting each cell once. That is you make h × w moves, where h and w are the height and width. You can use this as termination criterion instead of checking whether you are in the last square.
Here's example code for this solution. The additional boolean parameter down specifies whether the first step is down or left.
function zigzag(width, height, down) {
var x = 0;
var y = 0;
var n = width * height;
if (down === undefined) down = false;
while (n--) {
var even = ((x + y) % 2 == 0);
put(x, y);
if (even == down) { // walk southwest
x--;
y++;
if (y == height) {
y--; x += 2;
}
if (x < 0) x = 0;
} else { // walk northeast
x++;
y--;
if (x == width) {
x--; y += 2;
}
if (y < 0) y = 0;
}
}
return res;
}
Here's the solution abusing if-statements.
x1 = 0;
y1 = 0;
put(x1,y1);
var a = 0;
while(!((x1 = grid_w-1) and (y1 = grid_h-1))) {
switch(a) { //Down, Right-Up, Right, Left-Down
case 0: y1++; break;
case 1: x1++;y1--; break;
case 2: x1++; break;
case 3: x1--;y1++; break;
}
put(x1,y1);
if (a = 2) { //If moved right.
if (x1 = grid_w-1) or (y1 = 0) { //If at the right or top edge. Go left-down.
a = 3
} else if (y1 = grid_h-1) { //At bottom edge. Go right-up.
a = 1
}
} else if (y1 = 0) { ///At top edge.
if (x1 = grid_w-1) { //If at the right corner. Go down.
a = 0;
} else { //Go right.
a = 2;
}
} else if (a = 3) { ///If moved left-down.
if (y1 = grid_h-1) { //At bottom n edge. Go right.
a = 2
} else if (x1 = 0) { //At left edge and not bottom. Go down.
a = 0
}
} else if (a = 0) { //If moved down.
if (x1 = 0) { //If at the left corner. Go right-up.
a = 1
} else if (x1 = grid_w-1) { //If at the right corner. Go left-down.
a = 3
} else { //Go right
a = 2
}
} else if (a = 1) { //If right-up.
if (x1 = grid_w-1) { //If at the right corner.
if (a = 2) { //If moved right. Go left-down.
a = 3
} else { //Go down.
a = 0
}
}
}
}
Doesn't work well if one of the size is 1.
Basically we can use state diagram along with recursion to solve this.
permitted_directions = {
"start":["down", "side"],
"down":["north_east", "south_west"],
"north_east":["north_east", "side","down"],
"side":["north_east", "south_west"],
"south_west":["south_west","down", "side"]
}
def is_possible(x, y, pos):
if pos == "down":
if x+1 < row and y >=0 and y < col:
return (True, x+1, y)
if pos == "side":
if x >= 0 and x < row and y+1 >=0 and y+1 < col:
return (True, x, y+1)
if pos == "north_east":
if x-1 >= 0 and x-1 < row and y+1 >= 0 and y+1 < col:
return (True, x-1, y+1)
if pos == "south_west":
if x+1 >= 0 and x+1 < row and y-1 >= 0 and y-1 < col:
return (True, x+1, y-1)
return (False, 0, 0)
def fill_the_grid(grid, x, y, position, prev):
grid[x][y] = prev
prev = (x, y)
for pos in permitted_directions[position]:
possible, p, q = is_possible(x, y, pos)
if possible:
return fill_the_grid(grid, p, q, pos, prev)
return grid
I'm trying to detect when a line intersects a circle in javascript. I found a function that works almost perfectly but I recently noticed that it does not work when the intersecting line is perfectly horizontal or vertical. Since I don't have a great understanding of how this function actually works, I'm not sure how to edit it to get the results I'd like.
function lineCircleCollision(circleX,circleY,radius,lineX1,lineY1,lineX2,lineY2) {
var d1 = pDist(lineX1,lineY1,circleX,circleY);
var d2 = pDist(lineX2,lineY2,circleX,circleY);
if (d1<=radius || d2<=radius) {
return true;
}
var k1 = ((lineY2-lineY1)/(lineX2-lineX1));
var k2 = lineY1;
var k3 = -1/k1;
var k4 = circleY;
var xx = (k1*lineX1-k2-k3*circleX+k4)/(k1-k3);
var yy = k1*(xx-lineX1)+lineY1;
var allow = true;
if (lineX2>lineX1) {
if (xx>=lineX1 && xx<=lineX2) {}
else {allow = false;}
} else {
if (xx>=lineX2 && xx<=lineX1) {}
else {allow = false;}
}
if (lineY2>lineY1) {
if (yy>=lineY1 && yy<=lineY2) {}
else {allow = false;}
} else {
if (yy>=lineY2 && yy<=lineY1) {}
else {allow = false;}
}
if (allow) {
if (pDist(circleX,circleY,xx,yy)<radius) {
return true;
}
else {
return false;
}
} else {
return false;
}
}
function pDist(x1,y1,x2,y2) {
var xd = x2-x1;
var yd = y2-y1;
return Math.sqrt(xd*xd+yd*yd);
}
You can express the line as two relations:
x = x1 + k * (x2 - x1) = x1 + k * dx
y = y1 + k * (y2 - y1) = y1 + k * dy
with 0 < k < 1. A point on the circle satisfies the equation:
(x - Cx)² + (y - Cy)² = r²
Replace x and y by the line equations and you'll get a quadratic equation:
a*k² + b*k + c = 0
a = dx² + dy²
b = 2*dx*(x1 - Cx) + s*dy*(y1 - Cy)
c = (x1 - Cx)² + (y1 - Cy)² - r²
Solve that and if any of the two possible solutions for k lies in the range between 0 and 1, you have a hit. This method checks real intersections and misses the case where the line is entirely contained in the circle, so an additional check whether the line's end points lie within the circle is necessary.
Here's the code:
function collision_circle_line(Cx, Cy, r, x1, y1, x2, y2) {
var dx = x2 - x1;
var dy = y2 - y1;
var sx = x1 - Cx;
var sy = y1 - Cy;
var tx = x2 - Cx;
var ty = y2 - Cy;
if (tx*tx + ty*ty < r*r) return true;
var c = sx*sx + sy*sy - r*r;
if (c < 0) return true;
var b = 2 * (dx * sx + dy * sy);
var a = dx*dx + dy*dy;
if (Math.abs(a) < 1.0e-12) return false;
var discr = b*b - 4*a*c;
if (discr < 0) return false;
discr = Math.sqrt(discr);
var k1 = (-b - discr) / (2 * a);
if (k1 >= 0 && k1 <= 1) return true;
var k2 = (-b + discr) / (2 * a);
if (k2 >= 0 && k2 <= 1) return true;
return false;
}
Another way to view the intersection check is that we're finding the point on the line segment closest to the circle center and then determining whether it's close enough. Since distance to the circle center is a convex function, there are three possibilities: the two endpoints of the segment, and the closest point on the line, assuming that it's on the segment.
To find the closest point on the line, we have an overdetermined linear system
(1 - t) lineX1 + t lineX2 = circleX
(1 - t) lineY1 + t lineY2 = circleY,
expressed as a matrix:
[lineX2 - lineX1] [t] = [circleX - lineX1]
[lineY2 - lineY1] [circleY - lineY1].
The closest point can be found by solving the normal equation
[(lineX2 - lineX1) (lineY2 - lineY1)] [lineX2 - lineX1] [t] =
[lineY2 - lineY1]
[(lineX2 - lineX1) (lineY2 - lineY1)] [circleX - lineX1]
[circleY - lineY1],
expressed alternatively as
((lineX2 - lineX1)^2 + (lineY2 - lineY1)^2) t =
(lineX2 - lineX1) (circleX - lineX1) + (lineY2 - lineY1) (circleY - lineY1),
and solved for t:
(lineX2 - lineX1) (circleX - lineX1) + (lineY2 - lineY1) (circleY - lineY1)
t = ---------------------------------------------------------------------------.
((lineX2 - lineX1)^2 + (lineY2 - lineY1)^2)
Assuming that t is between 0 and 1, we can plug it in and check the distance. When t is out of range, we can clamp it and check only that endpoint.
Untested code:
function lineCircleCollision(circleX, circleY, radius, lineX1, lineY1, lineX2, lineY2) {
circleX -= lineX1;
circleY -= lineY1;
lineX2 -= lineX1;
lineY2 -= lineY1;
var t = (lineX2 * circleX + lineY2 * circleY) / (lineX2 * lineX2 + lineY2 * lineY2);
if (t < 0) t = 0;
else if (t > 1) t = 1;
var deltaX = lineX2 * t - circleX;
var deltaY = lineY2 * t - circleY;
return deltaX * deltaX + deltaY * deltaY <= radius * radius;
}
If you do not need the point just want to know if line intersects then:
compute distance from circle center P0(x0,y0) and line endpoints P1(x1,y1),P2(x2,y2)
double d1=|P1-P0|=sqrt((x1-x0)*(x1-x0)+(y1-x0)*(y1-x0));
double d2=|P2-P0|=sqrt((x2-x0)*(x2-x0)+(y2-x0)*(y2-x0));
order d1,d2 ascending
if (d1>d2) { double d=d1; d1=d2; d2=d; }
check intersection
if ((d1<=r)&&(d2>=r)) return true; else return false;
r is circle radius
[notes]
you do not need sqrt distances
if you leave them un-sqrted then just compare them to r*r instead of r
I want to split a bezier curve into a polygonal chain with n straight lines. The number of lines being dependent on a maximum allowed angle between 2 connecting lines.
I'm looking for an algorithm to find the most optimal solution (ie to reduce as much as possible the number of straight lines).
I know how to split a bezier curve using Casteljau or Bernstein polynomals. I tried dividing the bezier into half calculate the angle between the straight lines, and split again if the angle between the connecting lines is within a certain threshold range, but i may run into shortcuts.
Is there a known algorithm or pseudo code available to do this conversion?
Use de Casteljau algorithm recursively until the control points are approximately collinear. See for instance http://www.antigrain.com/research/adaptive_bezier/index.html.
This was a fascinating topic. The only thing I'm adding is tested C# code, to perhaps save somebody the trouble. And I tried to write for clarity as opposed to speed, so it mostly follows the AGG web site's PDF doc (see above) on the Casteljau algorithm. The Notation follows the diagram in that PDF.
public class Bezier
{
public PointF P1; // Begin Point
public PointF P2; // Control Point
public PointF P3; // Control Point
public PointF P4; // End Point
// Made these global so I could diagram the top solution
public Line L12;
public Line L23;
public Line L34;
public PointF P12;
public PointF P23;
public PointF P34;
public Line L1223;
public Line L2334;
public PointF P123;
public PointF P234;
public Line L123234;
public PointF P1234;
public Bezier(PointF p1, PointF p2, PointF p3, PointF p4)
{
P1 = p1; P2 = p2; P3 = p3; P4 = p4;
}
/// <summary>
/// Consider the classic Casteljau diagram
/// with the bezier points p1, p2, p3, p4 and lines l12, l23, l34
/// and their midpoint of line l12 being p12 ...
/// and the line between p12 p23 being L1223
/// and the midpoint of line L1223 being P1223 ...
/// </summary>
/// <param name="lines"></param>
public void SplitBezier( List<Line> lines)
{
L12 = new Line(this.P1, this.P2);
L23 = new Line(this.P2, this.P3);
L34 = new Line(this.P3, this.P4);
P12 = L12.MidPoint();
P23 = L23.MidPoint();
P34 = L34.MidPoint();
L1223 = new Line(P12, P23);
L2334 = new Line(P23, P34);
P123 = L1223.MidPoint();
P234 = L2334.MidPoint();
L123234 = new Line(P123, P234);
P1234 = L123234.MidPoint();
if (CurveIsFlat())
{
lines.Add(new Line(this.P1, this.P4));
return;
}
else
{
Bezier bz1 = new Bezier(this.P1, P12, P123, P1234);
bz1.SplitBezier(lines);
Bezier bz2 = new Bezier(P1234, P234, P34, this.P4);
bz2.SplitBezier(lines);
}
return;
}
/// <summary>
/// Check if points P1, P1234 and P2 are colinear (enough).
/// This is very simple-minded algo... there are better...
/// </summary>
/// <returns></returns>
public bool CurveIsFlat()
{
float t1 = (P2.Y - P1.Y) * (P3.X - P2.X);
float t2 = (P3.Y - P2.Y) * (P2.X - P1.X);
float delta = Math.Abs(t1 - t2);
return delta < 0.1; // Hard-coded constant
}
The PointF is from System.Drawing, and the Line class follows:
public class Line
{
PointF P1; PointF P2;
public Line(PointF pt1, PointF pt2)
{
P1 = pt1; P2 = pt2;
}
public PointF MidPoint()
{
return new PointF((P1.X + P2.X) / 2f, (P1.Y + P2.Y) / 2f);
}
}
A sample call creates the Bezier object with 4 points (begin, 2 control, and end), and returns a list of lines that approximate the Bezier:
TopBezier = new Bezier(Point1, Point2, Point3, Point4 );
List<Line> lines = new List<Line>();
TopBezier.SplitBezier(lines);
Thanks to Dr Jerry, AGG, and all the other contributors.
There are some alternatives for RSA flattening that are reported to be faster:
RSA vs PAA:
http://www.cis.usouthal.edu/~hain/general/Theses/Ahmad_thesis.pdf
RSA vs CAA vs PAA:
http://www.cis.usouthal.edu/~hain/general/Theses/Racherla_thesis.pdf
RSA = Recursive Subdivision Algorithm
PAA = Parabolic Approximation Algorithm
CAA = Circular Approximation Algorithm
According to Rachela, CAA is slower than the PAA by a factor of 1.5–2. CAA is as slow as RSA, but achieves required flatness better in offset curves.
It seems that PAA is best choice for actual curve and CAA is best for offset's of curve (when stroking curves).
I have tested PAA of both thesis, but they fail in some cases. Ahmad's PAA fails in collinear cases (all points on same line) and Rachela's PAA fails in collinear cases and in cases where both control points are equal. With some fixes, it may be possible to get them work as expected.
A visual example on my website -> DXF -> polybezier.
it is basically a recursive split with casteljau.
Bezier2Poly.prototype.convert = function(array,init) {
if (init) {
this.vertices = [];
}
if (!init && (Math.abs(this.controlPointsDiff(array[0], array[2])) < this.threshold
|| Math.abs(this.controlPointsDiff({x:array[2].x-array[1].x, y:array[2]-array[1].y}, array[2])) < this.threshold)) {
this.vertices.push(array[2]);
} else {
var split = this.splitBezier(array);
this.convert(split.b1);
this.convert(split.b2);
}
return this.vertices;
}
And judgement by: calculating the angle between the controlpoints and the line through the endpoint.
Bezier2Poly.prototype.controlPointsDiff = function (vector1, vector2) {
var angleCp1 = Math.atan2(vector1.y, vector1.x);
var angleCp2 = Math.atan2(vector2.y, vector2.x);
return angleCp1 - angleCp2;
}
i solve it with qt for any svg path including bezier curve , i found in svg module a static function in qsvghandler.cpp which parsePathDataFast from your svg path to QPainterPath and the cherry on the cake!! QPainterPath have three native functions to convert your path to polygon (the big one toFillPolygon and the others which split in a list of polygon toSubpathPolygons or toFillPolygons) along with nice stuff like bounding box, intersected, translate ... ready to use with Boost::Geometry now, not so bad!
the header parsepathdatafast.h
#ifndef PARSEPATHDATAFAST_H
#define PARSEPATHDATAFAST_H
#include <QPainterPath>
#include <QString>
bool parsePathDataFast(const QStringRef &dataStr, QPainterPath &path);
#endif // PARSEPATHDATAFAST_H
the code parsepathdatafast.cpp
#include <QtCore/qmath.h>
#include <QtMath>
#include <QChar>
#include <QByteArray>
#include <QMatrix>
#include <parsepathdatafast.h>
Q_CORE_EXPORT double qstrtod(const char *s00, char const **se, bool *ok);
// '0' is 0x30 and '9' is 0x39
static inline bool isDigit(ushort ch)
{
static quint16 magic = 0x3ff;
return ((ch >> 4) == 3) && (magic >> (ch & 15));
}
static qreal toDouble(const QChar *&str)
{
const int maxLen = 255;//technically doubles can go til 308+ but whatever
char temp[maxLen+1];
int pos = 0;
if (*str == QLatin1Char('-')) {
temp[pos++] = '-';
++str;
} else if (*str == QLatin1Char('+')) {
++str;
}
while (isDigit(str->unicode()) && pos < maxLen) {
temp[pos++] = str->toLatin1();
++str;
}
if (*str == QLatin1Char('.') && pos < maxLen) {
temp[pos++] = '.';
++str;
}
while (isDigit(str->unicode()) && pos < maxLen) {
temp[pos++] = str->toLatin1();
++str;
}
bool exponent = false;
if ((*str == QLatin1Char('e') || *str == QLatin1Char('E')) && pos < maxLen) {
exponent = true;
temp[pos++] = 'e';
++str;
if ((*str == QLatin1Char('-') || *str == QLatin1Char('+')) && pos < maxLen) {
temp[pos++] = str->toLatin1();
++str;
}
while (isDigit(str->unicode()) && pos < maxLen) {
temp[pos++] = str->toLatin1();
++str;
}
}
temp[pos] = '\0';
qreal val;
if (!exponent && pos < 10) {
int ival = 0;
const char *t = temp;
bool neg = false;
if(*t == '-') {
neg = true;
++t;
}
while(*t && *t != '.') {
ival *= 10;
ival += (*t) - '0';
++t;
}
if(*t == '.') {
++t;
int div = 1;
while(*t) {
ival *= 10;
ival += (*t) - '0';
div *= 10;
++t;
}
val = ((qreal)ival)/((qreal)div);
} else {
val = ival;
}
if (neg)
val = -val;
} else {
bool ok = false;
val = qstrtod(temp, 0, &ok);
}
return val;
}
static inline void parseNumbersArray(const QChar *&str, QVarLengthArray<qreal, 8> &points)
{
while (str->isSpace())
++str;
while (isDigit(str->unicode()) ||
*str == QLatin1Char('-') || *str == QLatin1Char('+') ||
*str == QLatin1Char('.')) {
points.append(toDouble(str));
while (str->isSpace())
++str;
if (*str == QLatin1Char(','))
++str;
//eat the rest of space
while (str->isSpace())
++str;
}
}
/**
static QVector<qreal> parsePercentageList(const QChar *&str)
{
QVector<qreal> points;
if (!str)
return points;
while (str->isSpace())
++str;
while ((*str >= QLatin1Char('0') && *str <= QLatin1Char('9')) ||
*str == QLatin1Char('-') || *str == QLatin1Char('+') ||
*str == QLatin1Char('.')) {
points.append(toDouble(str));
while (str->isSpace())
++str;
if (*str == QLatin1Char('%'))
++str;
while (str->isSpace())
++str;
if (*str == QLatin1Char(','))
++str;
//eat the rest of space
while (str->isSpace())
++str;
}
return points;
}
**/
static void pathArcSegment(QPainterPath &path,
qreal xc, qreal yc,
qreal th0, qreal th1,
qreal rx, qreal ry, qreal xAxisRotation)
{
qreal sinTh, cosTh;
qreal a00, a01, a10, a11;
qreal x1, y1, x2, y2, x3, y3;
qreal t;
qreal thHalf;
sinTh = qSin(xAxisRotation * (M_PI / 180.0));
cosTh = qCos(xAxisRotation * (M_PI / 180.0));
a00 = cosTh * rx;
a01 = -sinTh * ry;
a10 = sinTh * rx;
a11 = cosTh * ry;
thHalf = 0.5 * (th1 - th0);
t = (8.0 / 3.0) * qSin(thHalf * 0.5) * qSin(thHalf * 0.5) / qSin(thHalf);
x1 = xc + qCos(th0) - t * qSin(th0);
y1 = yc + qSin(th0) + t * qCos(th0);
x3 = xc + qCos(th1);
y3 = yc + qSin(th1);
x2 = x3 + t * qSin(th1);
y2 = y3 - t * qCos(th1);
path.cubicTo(a00 * x1 + a01 * y1, a10 * x1 + a11 * y1,
a00 * x2 + a01 * y2, a10 * x2 + a11 * y2,
a00 * x3 + a01 * y3, a10 * x3 + a11 * y3);
}
// the arc handling code underneath is from XSVG (BSD license)
/*
* Copyright 2002 USC/Information Sciences Institute
*
* Permission to use, copy, modify, distribute, and sell this software
* and its documentation for any purpose is hereby granted without
* fee, provided that the above copyright notice appear in all copies
* and that both that copyright notice and this permission notice
* appear in supporting documentation, and that the name of
* Information Sciences Institute not be used in advertising or
* publicity pertaining to distribution of the software without
* specific, written prior permission. Information Sciences Institute
* makes no representations about the suitability of this software for
* any purpose. It is provided "as is" without express or implied
* warranty.
*
* INFORMATION SCIENCES INSTITUTE DISCLAIMS ALL WARRANTIES WITH REGARD
* TO THIS SOFTWARE, INCLUDING ALL IMPLIED WARRANTIES OF
* MERCHANTABILITY AND FITNESS, IN NO EVENT SHALL INFORMATION SCIENCES
* INSTITUTE BE LIABLE FOR ANY SPECIAL, INDIRECT OR CONSEQUENTIAL
* DAMAGES OR ANY DAMAGES WHATSOEVER RESULTING FROM LOSS OF USE, DATA
* OR PROFITS, WHETHER IN AN ACTION OF CONTRACT, NEGLIGENCE OR OTHER
* TORTIOUS ACTION, ARISING OUT OF OR IN CONNECTION WITH THE USE OR
* PERFORMANCE OF THIS SOFTWARE.
*
*/
static void pathArc(QPainterPath &path,
qreal rx,
qreal ry,
qreal x_axis_rotation,
int large_arc_flag,
int sweep_flag,
qreal x,
qreal y,
qreal curx, qreal cury)
{
qreal sin_th, cos_th;
qreal a00, a01, a10, a11;
qreal x0, y0, x1, y1, xc, yc;
qreal d, sfactor, sfactor_sq;
qreal th0, th1, th_arc;
int i, n_segs;
qreal dx, dy, dx1, dy1, Pr1, Pr2, Px, Py, check;
rx = qAbs(rx);
ry = qAbs(ry);
sin_th = qSin(x_axis_rotation * (M_PI / 180.0));
cos_th = qCos(x_axis_rotation * (M_PI / 180.0));
dx = (curx - x) / 2.0;
dy = (cury - y) / 2.0;
dx1 = cos_th * dx + sin_th * dy;
dy1 = -sin_th * dx + cos_th * dy;
Pr1 = rx * rx;
Pr2 = ry * ry;
Px = dx1 * dx1;
Py = dy1 * dy1;
/* Spec : check if radii are large enough */
check = Px / Pr1 + Py / Pr2;
if (check > 1) {
rx = rx * qSqrt(check);
ry = ry * qSqrt(check);
}
a00 = cos_th / rx;
a01 = sin_th / rx;
a10 = -sin_th / ry;
a11 = cos_th / ry;
x0 = a00 * curx + a01 * cury;
y0 = a10 * curx + a11 * cury;
x1 = a00 * x + a01 * y;
y1 = a10 * x + a11 * y;
/* (x0, y0) is current point in transformed coordinate space.
(x1, y1) is new point in transformed coordinate space.
The arc fits a unit-radius circle in this space.
*/
d = (x1 - x0) * (x1 - x0) + (y1 - y0) * (y1 - y0);
sfactor_sq = 1.0 / d - 0.25;
if (sfactor_sq < 0) sfactor_sq = 0;
sfactor = qSqrt(sfactor_sq);
if (sweep_flag == large_arc_flag) sfactor = -sfactor;
xc = 0.5 * (x0 + x1) - sfactor * (y1 - y0);
yc = 0.5 * (y0 + y1) + sfactor * (x1 - x0);
/* (xc, yc) is center of the circle. */
th0 = qAtan2(y0 - yc, x0 - xc);
th1 = qAtan2(y1 - yc, x1 - xc);
th_arc = th1 - th0;
if (th_arc < 0 && sweep_flag)
th_arc += 2 * M_PI;
else if (th_arc > 0 && !sweep_flag)
th_arc -= 2 * M_PI;
n_segs = qCeil(qAbs(th_arc / (M_PI * 0.5 + 0.001)));
for (i = 0; i < n_segs; i++) {
pathArcSegment(path, xc, yc,
th0 + i * th_arc / n_segs,
th0 + (i + 1) * th_arc / n_segs,
rx, ry, x_axis_rotation);
}
}
bool parsePathDataFast(const QStringRef &dataStr, QPainterPath &path)
{
qreal x0 = 0, y0 = 0; // starting point
qreal x = 0, y = 0; // current point
char lastMode = 0;
QPointF ctrlPt;
const QChar *str = dataStr.constData();
const QChar *end = str + dataStr.size();
while (str != end) {
while (str->isSpace())
++str;
QChar pathElem = *str;
++str;
QChar endc = *end;
*const_cast<QChar *>(end) = 0; // parseNumbersArray requires 0-termination that QStringRef cannot guarantee
QVarLengthArray<qreal, 8> arg;
parseNumbersArray(str, arg);
*const_cast<QChar *>(end) = endc;
if (pathElem == QLatin1Char('z') || pathElem == QLatin1Char('Z'))
arg.append(0);//dummy
const qreal *num = arg.constData();
int count = arg.count();
while (count > 0) {
qreal offsetX = x; // correction offsets
qreal offsetY = y; // for relative commands
switch (pathElem.unicode()) {
case 'm': {
if (count < 2) {
num++;
count--;
break;
}
x = x0 = num[0] + offsetX;
y = y0 = num[1] + offsetY;
num += 2;
count -= 2;
path.moveTo(x0, y0);
// As per 1.2 spec 8.3.2 The "moveto" commands
// If a 'moveto' is followed by multiple pairs of coordinates without explicit commands,
// the subsequent pairs shall be treated as implicit 'lineto' commands.
pathElem = QLatin1Char('l');
}
break;
case 'M': {
if (count < 2) {
num++;
count--;
break;
}
x = x0 = num[0];
y = y0 = num[1];
num += 2;
count -= 2;
path.moveTo(x0, y0);
// As per 1.2 spec 8.3.2 The "moveto" commands
// If a 'moveto' is followed by multiple pairs of coordinates without explicit commands,
// the subsequent pairs shall be treated as implicit 'lineto' commands.
pathElem = QLatin1Char('L');
}
break;
case 'z':
case 'Z': {
x = x0;
y = y0;
count--; // skip dummy
num++;
path.closeSubpath();
}
break;
case 'l': {
if (count < 2) {
num++;
count--;
break;
}
x = num[0] + offsetX;
y = num[1] + offsetY;
num += 2;
count -= 2;
path.lineTo(x, y);
}
break;
case 'L': {
if (count < 2) {
num++;
count--;
break;
}
x = num[0];
y = num[1];
num += 2;
count -= 2;
path.lineTo(x, y);
}
break;
case 'h': {
x = num[0] + offsetX;
num++;
count--;
path.lineTo(x, y);
}
break;
case 'H': {
x = num[0];
num++;
count--;
path.lineTo(x, y);
}
break;
case 'v': {
y = num[0] + offsetY;
num++;
count--;
path.lineTo(x, y);
}
break;
case 'V': {
y = num[0];
num++;
count--;
path.lineTo(x, y);
}
break;
case 'c': {
if (count < 6) {
num += count;
count = 0;
break;
}
QPointF c1(num[0] + offsetX, num[1] + offsetY);
QPointF c2(num[2] + offsetX, num[3] + offsetY);
QPointF e(num[4] + offsetX, num[5] + offsetY);
num += 6;
count -= 6;
path.cubicTo(c1, c2, e);
ctrlPt = c2;
x = e.x();
y = e.y();
break;
}
case 'C': {
if (count < 6) {
num += count;
count = 0;
break;
}
QPointF c1(num[0], num[1]);
QPointF c2(num[2], num[3]);
QPointF e(num[4], num[5]);
num += 6;
count -= 6;
path.cubicTo(c1, c2, e);
ctrlPt = c2;
x = e.x();
y = e.y();
break;
}
case 's': {
if (count < 4) {
num += count;
count = 0;
break;
}
QPointF c1;
if (lastMode == 'c' || lastMode == 'C' ||
lastMode == 's' || lastMode == 'S')
c1 = QPointF(2*x-ctrlPt.x(), 2*y-ctrlPt.y());
else
c1 = QPointF(x, y);
QPointF c2(num[0] + offsetX, num[1] + offsetY);
QPointF e(num[2] + offsetX, num[3] + offsetY);
num += 4;
count -= 4;
path.cubicTo(c1, c2, e);
ctrlPt = c2;
x = e.x();
y = e.y();
break;
}
case 'S': {
if (count < 4) {
num += count;
count = 0;
break;
}
QPointF c1;
if (lastMode == 'c' || lastMode == 'C' ||
lastMode == 's' || lastMode == 'S')
c1 = QPointF(2*x-ctrlPt.x(), 2*y-ctrlPt.y());
else
c1 = QPointF(x, y);
QPointF c2(num[0], num[1]);
QPointF e(num[2], num[3]);
num += 4;
count -= 4;
path.cubicTo(c1, c2, e);
ctrlPt = c2;
x = e.x();
y = e.y();
break;
}
case 'q': {
if (count < 4) {
num += count;
count = 0;
break;
}
QPointF c(num[0] + offsetX, num[1] + offsetY);
QPointF e(num[2] + offsetX, num[3] + offsetY);
num += 4;
count -= 4;
path.quadTo(c, e);
ctrlPt = c;
x = e.x();
y = e.y();
break;
}
case 'Q': {
if (count < 4) {
num += count;
count = 0;
break;
}
QPointF c(num[0], num[1]);
QPointF e(num[2], num[3]);
num += 4;
count -= 4;
path.quadTo(c, e);
ctrlPt = c;
x = e.x();
y = e.y();
break;
}
case 't': {
if (count < 2) {
num += count;
count = 0;
break;
}
QPointF e(num[0] + offsetX, num[1] + offsetY);
num += 2;
count -= 2;
QPointF c;
if (lastMode == 'q' || lastMode == 'Q' ||
lastMode == 't' || lastMode == 'T')
c = QPointF(2*x-ctrlPt.x(), 2*y-ctrlPt.y());
else
c = QPointF(x, y);
path.quadTo(c, e);
ctrlPt = c;
x = e.x();
y = e.y();
break;
}
case 'T': {
if (count < 2) {
num += count;
count = 0;
break;
}
QPointF e(num[0], num[1]);
num += 2;
count -= 2;
QPointF c;
if (lastMode == 'q' || lastMode == 'Q' ||
lastMode == 't' || lastMode == 'T')
c = QPointF(2*x-ctrlPt.x(), 2*y-ctrlPt.y());
else
c = QPointF(x, y);
path.quadTo(c, e);
ctrlPt = c;
x = e.x();
y = e.y();
break;
}
case 'a': {
if (count < 7) {
num += count;
count = 0;
break;
}
qreal rx = (*num++);
qreal ry = (*num++);
qreal xAxisRotation = (*num++);
qreal largeArcFlag = (*num++);
qreal sweepFlag = (*num++);
qreal ex = (*num++) + offsetX;
qreal ey = (*num++) + offsetY;
count -= 7;
qreal curx = x;
qreal cury = y;
pathArc(path, rx, ry, xAxisRotation, int(largeArcFlag),
int(sweepFlag), ex, ey, curx, cury);
x = ex;
y = ey;
}
break;
case 'A': {
if (count < 7) {
num += count;
count = 0;
break;
}
qreal rx = (*num++);
qreal ry = (*num++);
qreal xAxisRotation = (*num++);
qreal largeArcFlag = (*num++);
qreal sweepFlag = (*num++);
qreal ex = (*num++);
qreal ey = (*num++);
count -= 7;
qreal curx = x;
qreal cury = y;
pathArc(path, rx, ry, xAxisRotation, int(largeArcFlag),
int(sweepFlag), ex, ey, curx, cury);
x = ex;
y = ey;
}
break;
default:
return false;
}
lastMode = pathElem.toLatin1();
}
}
return true;
}
One question, i doesn't find Q_PI constant in the standard qt headers and i replace it with M_PI hope is OK!!