Pixel by pixel Bézier Curve - algorithm

The quadratic/cubic bézier curve code I find via google mostly works by subdividing the line into a series of points and connects them with straight lines. The rasterization happens in the line algorithm, not in the bézier one. Algorithms like Bresenham's work pixel-by-pixel to rasterize a line, and can be optimized (see Po-Han Lin's solution).
What is a quadratic bézier curve algorithm that works pixel-by-pixel like line algorithms instead of by plotting a series of points?

A variation of Bresenham's Algorithm works with quadratic functions like circles, ellipses, and parabolas, so it should work with quadratic Bezier curves too.
I was going to attempt an implementation, but then I found one on the web: http://members.chello.at/~easyfilter/bresenham.html.
If you want more detail or additional examples, the page mentioned above has a link to a 100 page PDF elaborating on the method: http://members.chello.at/~easyfilter/Bresenham.pdf.
Here's the code from Alois Zingl's site for plotting any quadratic Bezier curve. The first routine subdivides the curve at horizontal and vertical gradient changes:
void plotQuadBezier(int x0, int y0, int x1, int y1, int x2, int y2)
{ /* plot any quadratic Bezier curve */
int x = x0-x1, y = y0-y1;
double t = x0-2*x1+x2, r;
if ((long)x*(x2-x1) > 0) { /* horizontal cut at P4? */
if ((long)y*(y2-y1) > 0) /* vertical cut at P6 too? */
if (fabs((y0-2*y1+y2)/t*x) > abs(y)) { /* which first? */
x0 = x2; x2 = x+x1; y0 = y2; y2 = y+y1; /* swap points */
} /* now horizontal cut at P4 comes first */
t = (x0-x1)/t;
r = (1-t)*((1-t)*y0+2.0*t*y1)+t*t*y2; /* By(t=P4) */
t = (x0*x2-x1*x1)*t/(x0-x1); /* gradient dP4/dx=0 */
x = floor(t+0.5); y = floor(r+0.5);
r = (y1-y0)*(t-x0)/(x1-x0)+y0; /* intersect P3 | P0 P1 */
plotQuadBezierSeg(x0,y0, x,floor(r+0.5), x,y);
r = (y1-y2)*(t-x2)/(x1-x2)+y2; /* intersect P4 | P1 P2 */
x0 = x1 = x; y0 = y; y1 = floor(r+0.5); /* P0 = P4, P1 = P8 */
}
if ((long)(y0-y1)*(y2-y1) > 0) { /* vertical cut at P6? */
t = y0-2*y1+y2; t = (y0-y1)/t;
r = (1-t)*((1-t)*x0+2.0*t*x1)+t*t*x2; /* Bx(t=P6) */
t = (y0*y2-y1*y1)*t/(y0-y1); /* gradient dP6/dy=0 */
x = floor(r+0.5); y = floor(t+0.5);
r = (x1-x0)*(t-y0)/(y1-y0)+x0; /* intersect P6 | P0 P1 */
plotQuadBezierSeg(x0,y0, floor(r+0.5),y, x,y);
r = (x1-x2)*(t-y2)/(y1-y2)+x2; /* intersect P7 | P1 P2 */
x0 = x; x1 = floor(r+0.5); y0 = y1 = y; /* P0 = P6, P1 = P7 */
}
plotQuadBezierSeg(x0,y0, x1,y1, x2,y2); /* remaining part */
}
The second routine actually plots a Bezier curve segment (one without gradient changes):
void plotQuadBezierSeg(int x0, int y0, int x1, int y1, int x2, int y2)
{ /* plot a limited quadratic Bezier segment */
int sx = x2-x1, sy = y2-y1;
long xx = x0-x1, yy = y0-y1, xy; /* relative values for checks */
double dx, dy, err, cur = xx*sy-yy*sx; /* curvature */
assert(xx*sx <= 0 && yy*sy <= 0); /* sign of gradient must not change */
if (sx*(long)sx+sy*(long)sy > xx*xx+yy*yy) { /* begin with longer part */
x2 = x0; x0 = sx+x1; y2 = y0; y0 = sy+y1; cur = -cur; /* swap P0 P2 */
}
if (cur != 0) { /* no straight line */
xx += sx; xx *= sx = x0 < x2 ? 1 : -1; /* x step direction */
yy += sy; yy *= sy = y0 < y2 ? 1 : -1; /* y step direction */
xy = 2*xx*yy; xx *= xx; yy *= yy; /* differences 2nd degree */
if (cur*sx*sy < 0) { /* negated curvature? */
xx = -xx; yy = -yy; xy = -xy; cur = -cur;
}
dx = 4.0*sy*cur*(x1-x0)+xx-xy; /* differences 1st degree */
dy = 4.0*sx*cur*(y0-y1)+yy-xy;
xx += xx; yy += yy; err = dx+dy+xy; /* error 1st step */
do {
setPixel(x0,y0); /* plot curve */
if (x0 == x2 && y0 == y2) return; /* last pixel -> curve finished */
y1 = 2*err < dx; /* save value for test of y step */
if (2*err > dy) { x0 += sx; dx -= xy; err += dy += yy; } /* x step */
if ( y1 ) { y0 += sy; dy -= xy; err += dx += xx; } /* y step */
} while (dy < 0 && dx > 0); /* gradient negates -> algorithm fails */
}
plotLine(x0,y0, x2,y2); /* plot remaining part to end */
}
Code for antialiasing is also available on the site.
The corresponding functions from Zingl's site for cubic Bezier curves are
void plotCubicBezier(int x0, int y0, int x1, int y1,
int x2, int y2, int x3, int y3)
{ /* plot any cubic Bezier curve */
int n = 0, i = 0;
long xc = x0+x1-x2-x3, xa = xc-4*(x1-x2);
long xb = x0-x1-x2+x3, xd = xb+4*(x1+x2);
long yc = y0+y1-y2-y3, ya = yc-4*(y1-y2);
long yb = y0-y1-y2+y3, yd = yb+4*(y1+y2);
float fx0 = x0, fx1, fx2, fx3, fy0 = y0, fy1, fy2, fy3;
double t1 = xb*xb-xa*xc, t2, t[5];
/* sub-divide curve at gradient sign changes */
if (xa == 0) { /* horizontal */
if (abs(xc) < 2*abs(xb)) t[n++] = xc/(2.0*xb); /* one change */
} else if (t1 > 0.0) { /* two changes */
t2 = sqrt(t1);
t1 = (xb-t2)/xa; if (fabs(t1) < 1.0) t[n++] = t1;
t1 = (xb+t2)/xa; if (fabs(t1) < 1.0) t[n++] = t1;
}
t1 = yb*yb-ya*yc;
if (ya == 0) { /* vertical */
if (abs(yc) < 2*abs(yb)) t[n++] = yc/(2.0*yb); /* one change */
} else if (t1 > 0.0) { /* two changes */
t2 = sqrt(t1);
t1 = (yb-t2)/ya; if (fabs(t1) < 1.0) t[n++] = t1;
t1 = (yb+t2)/ya; if (fabs(t1) < 1.0) t[n++] = t1;
}
for (i = 1; i < n; i++) /* bubble sort of 4 points */
if ((t1 = t[i-1]) > t[i]) { t[i-1] = t[i]; t[i] = t1; i = 0; }
t1 = -1.0; t[n] = 1.0; /* begin / end point */
for (i = 0; i <= n; i++) { /* plot each segment separately */
t2 = t[i]; /* sub-divide at t[i-1], t[i] */
fx1 = (t1*(t1*xb-2*xc)-t2*(t1*(t1*xa-2*xb)+xc)+xd)/8-fx0;
fy1 = (t1*(t1*yb-2*yc)-t2*(t1*(t1*ya-2*yb)+yc)+yd)/8-fy0;
fx2 = (t2*(t2*xb-2*xc)-t1*(t2*(t2*xa-2*xb)+xc)+xd)/8-fx0;
fy2 = (t2*(t2*yb-2*yc)-t1*(t2*(t2*ya-2*yb)+yc)+yd)/8-fy0;
fx0 -= fx3 = (t2*(t2*(3*xb-t2*xa)-3*xc)+xd)/8;
fy0 -= fy3 = (t2*(t2*(3*yb-t2*ya)-3*yc)+yd)/8;
x3 = floor(fx3+0.5); y3 = floor(fy3+0.5); /* scale bounds to int */
if (fx0 != 0.0) { fx1 *= fx0 = (x0-x3)/fx0; fx2 *= fx0; }
if (fy0 != 0.0) { fy1 *= fy0 = (y0-y3)/fy0; fy2 *= fy0; }
if (x0 != x3 || y0 != y3) /* segment t1 - t2 */
plotCubicBezierSeg(x0,y0, x0+fx1,y0+fy1, x0+fx2,y0+fy2, x3,y3);
x0 = x3; y0 = y3; fx0 = fx3; fy0 = fy3; t1 = t2;
}
}
and
void plotCubicBezierSeg(int x0, int y0, float x1, float y1,
float x2, float y2, int x3, int y3)
{ /* plot limited cubic Bezier segment */
int f, fx, fy, leg = 1;
int sx = x0 < x3 ? 1 : -1, sy = y0 < y3 ? 1 : -1; /* step direction */
float xc = -fabs(x0+x1-x2-x3), xa = xc-4*sx*(x1-x2), xb = sx*(x0-x1-x2+x3);
float yc = -fabs(y0+y1-y2-y3), ya = yc-4*sy*(y1-y2), yb = sy*(y0-y1-y2+y3);
double ab, ac, bc, cb, xx, xy, yy, dx, dy, ex, *pxy, EP = 0.01;
/* check for curve restrains */
/* slope P0-P1 == P2-P3 and (P0-P3 == P1-P2 or no slope change) */
assert((x1-x0)*(x2-x3) < EP && ((x3-x0)*(x1-x2) < EP || xb*xb < xa*xc+EP));
assert((y1-y0)*(y2-y3) < EP && ((y3-y0)*(y1-y2) < EP || yb*yb < ya*yc+EP));
if (xa == 0 && ya == 0) { /* quadratic Bezier */
sx = floor((3*x1-x0+1)/2); sy = floor((3*y1-y0+1)/2); /* new midpoint */
return plotQuadBezierSeg(x0,y0, sx,sy, x3,y3);
}
x1 = (x1-x0)*(x1-x0)+(y1-y0)*(y1-y0)+1; /* line lengths */
x2 = (x2-x3)*(x2-x3)+(y2-y3)*(y2-y3)+1;
do { /* loop over both ends */
ab = xa*yb-xb*ya; ac = xa*yc-xc*ya; bc = xb*yc-xc*yb;
ex = ab*(ab+ac-3*bc)+ac*ac; /* P0 part of self-intersection loop? */
f = ex > 0 ? 1 : sqrt(1+1024/x1); /* calculate resolution */
ab *= f; ac *= f; bc *= f; ex *= f*f; /* increase resolution */
xy = 9*(ab+ac+bc)/8; cb = 8*(xa-ya);/* init differences of 1st degree */
dx = 27*(8*ab*(yb*yb-ya*yc)+ex*(ya+2*yb+yc))/64-ya*ya*(xy-ya);
dy = 27*(8*ab*(xb*xb-xa*xc)-ex*(xa+2*xb+xc))/64-xa*xa*(xy+xa);
/* init differences of 2nd degree */
xx = 3*(3*ab*(3*yb*yb-ya*ya-2*ya*yc)-ya*(3*ac*(ya+yb)+ya*cb))/4;
yy = 3*(3*ab*(3*xb*xb-xa*xa-2*xa*xc)-xa*(3*ac*(xa+xb)+xa*cb))/4;
xy = xa*ya*(6*ab+6*ac-3*bc+cb); ac = ya*ya; cb = xa*xa;
xy = 3*(xy+9*f*(cb*yb*yc-xb*xc*ac)-18*xb*yb*ab)/8;
if (ex < 0) { /* negate values if inside self-intersection loop */
dx = -dx; dy = -dy; xx = -xx; yy = -yy; xy = -xy; ac = -ac; cb = -cb;
} /* init differences of 3rd degree */
ab = 6*ya*ac; ac = -6*xa*ac; bc = 6*ya*cb; cb = -6*xa*cb;
dx += xy; ex = dx+dy; dy += xy; /* error of 1st step */
for (pxy = &xy, fx = fy = f; x0 != x3 && y0 != y3; ) {
setPixel(x0,y0); /* plot curve */
do { /* move sub-steps of one pixel */
if (dx > *pxy || dy < *pxy) goto exit; /* confusing values */
y1 = 2*ex-dy; /* save value for test of y step */
if (2*ex >= dx) { /* x sub-step */
fx--; ex += dx += xx; dy += xy += ac; yy += bc; xx += ab;
}
if (y1 <= 0) { /* y sub-step */
fy--; ex += dy += yy; dx += xy += bc; xx += ac; yy += cb;
}
} while (fx > 0 && fy > 0); /* pixel complete? */
if (2*fx <= f) { x0 += sx; fx += f; } /* x step */
if (2*fy <= f) { y0 += sy; fy += f; } /* y step */
if (pxy == &xy && dx < 0 && dy > 0) pxy = &EP;/* pixel ahead valid */
}
exit: xx = x0; x0 = x3; x3 = xx; sx = -sx; xb = -xb; /* swap legs */
yy = y0; y0 = y3; y3 = yy; sy = -sy; yb = -yb; x1 = x2;
} while (leg--); /* try other end */
plotLine(x0,y0, x3,y3); /* remaining part in case of cusp or crunode */
}
As Mike 'Pomax' Kamermans has noted, the solution for cubic Bezier curves on the site is not complete; in particular, there are issues with antialiasing cubic Bezier curves, and the discussion of rational cubic Bezier curves is incomplete.

You can use De Casteljau's algorithm to subdivide a curve into enough pieces that each subsection is a pixel.
This is the equation for finding the [x,y] point on a Quadratic Curve at interval T:
// Given 3 control points defining the Quadratic curve
// and given T which is an interval between 0.00 and 1.00 along the curve.
// Note:
// At the curve's starting control point T==0.00.
// At the curve's ending control point T==1.00.
var x = Math.pow(1-T,2)*startPt.x + 2 * (1-T) * T * controlPt.x + Math.pow(T,2) * endPt.x;
var y = Math.pow(1-T,2)*startPt.y + 2 * (1-T) * T * controlPt.y + Math.pow(T,2) * endPt.y;
To make practical use of this equation, you can input about 1000 T values between 0.00 and 1.00. This results in a set of 1000 points guaranteed to be along the Quadratic Curve.
Calculating 1000 points along the curve is probably over-sampling (some calculated points will be at the same pixel coordinate) so you will want to de-duplicate the 1000 points until the set represents unique pixel coordinates along the curve.
There is a similar equation for Cubic Bezier curves.
Here's example code that plots a Quadratic Curve as a set of calculated pixels:
var canvas=document.getElementById("canvas");
var ctx=canvas.getContext("2d");
var points=[];
var lastX=-100;
var lastY=-100;
var startPt={x:50,y:200};
var controlPt={x:150,y:25};
var endPt={x:250,y:100};
for(var t=0;t<1000;t++){
var xyAtT=getQuadraticBezierXYatT(startPt,controlPt,endPt,t/1000);
var x=parseInt(xyAtT.x);
var y=parseInt(xyAtT.y);
if(!(x==lastX && y==lastY)){
points.push(xyAtT);
lastX=x;
lastY=y;
}
}
$('#curve').text('Quadratic Curve made up of '+points.length+' individual points');
ctx.fillStyle='red';
for(var i=0;i<points.length;i++){
var x=points[i].x;
var y=points[i].y;
ctx.fillRect(x,y,1,1);
}
function getQuadraticBezierXYatT(startPt,controlPt,endPt,T) {
var x = Math.pow(1-T,2) * startPt.x + 2 * (1-T) * T * controlPt.x + Math.pow(T,2) * endPt.x;
var y = Math.pow(1-T,2) * startPt.y + 2 * (1-T) * T * controlPt.y + Math.pow(T,2) * endPt.y;
return( {x:x,y:y} );
}
body{ background-color: ivory; }
#canvas{border:1px solid red; margin:0 auto; }
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<h4 id='curve'>Q</h4>
<canvas id="canvas" width=350 height=300></canvas>

The thing to realise here is that "line segments", when created small enough, are equivalent to pixels. Bezier curves are not linearly traversible curves, so we can't easily "skip ahead to the next pixel" in a single step, like we can for lines or circular arcs.
You could, of course, take the tangent at any point for a t you already have, and then guess which next value t' will lie a pixel further. However, what typically happens is that you guess, and guess wrong because the curve does not behave linearly, then you check to see how "off" your guess was, correct your guess, and then check again. Repeat until you've converged on the next pixel: this is far, far slower than just flattening the curve to a high number of line segments instead, which is a fast operation.
If you pick the number of segments such that they're appropriate to the curve's length, given the display it's rendered to, no one will be able to tell you flattened the curve.
There are ways to reparameterize Bezier curves, but they're expensive, and different canonical curves require different reparameterization, so that's really not faster either. What tends to be the most useful for discrete displays is to build a LUT (lookup table) for your curve, with a length that works for the size the curve is on the display, and then using that LUT as your base data for drawing, intersection detection, etc. etc.

First of all, I'd like to say that the fastest and the most reliable way to render bezier curves is to approximate them by polyline via adaptive subdivision, then render the polyline. Approach by #markE with drawing many points sampled on the curve is rather fast, but it can skip pixels. Here I describe another approach, which is closest to line rasterization (though it is slow and hard to implement robustly).
I'll treat usually curve parameter as time. Here is the pseudocode:
Put your cursor at the first control point, find the surrounding pixel.
For each side of the pixel (four total), check when your bezier curves intersects its line by solving quadratic equations.
Among all the calculated side intersection times, choose the one which will happen strictly in future, but as early as possible.
Move to neighboring pixel depending on which side was best.
Set current time to time of that best side intersection.
Repeat from step 2.
This algorithm works until time parameter exceeds one. Also note that it has severe issues with curves exactly touching a side of a pixel. I suppose it is solvable with a special check.
Here is the main code:
double WhenEquals(double p0, double p1, double p2, double val, double minp) {
//p0 * (1-t)^2 + p1 * 2t(1 - t) + p2 * t^2 = val
double qa = p0 + p2 - 2 * p1;
double qb = p1 - p0;
double qc = p0 - val;
assert(fabs(qa) > EPS); //singular case must be handled separately
double qd = qb * qb - qa * qc;
if (qd < -EPS)
return INF;
qd = sqrt(max(qd, 0.0));
double t1 = (-qb - qd) / qa;
double t2 = (-qb + qd) / qa;
if (t2 < t1) swap(t1, t2);
if (t1 > minp + EPS)
return t1;
else if (t2 > minp + EPS)
return t2;
return INF;
}
void DrawCurve(const Bezier &curve) {
int cell[2];
for (int c = 0; c < 2; c++)
cell[c] = int(floor(curve.pts[0].a[c]));
DrawPixel(cell[0], cell[1]);
double param = 0.0;
while (1) {
int bc = -1, bs = -1;
double bestTime = 1.0;
for (int c = 0; c < 2; c++)
for (int s = 0; s < 2; s++) {
double crit = WhenEquals(
curve.pts[0].a[c],
curve.pts[1].a[c],
curve.pts[2].a[c],
cell[c] + s, param
);
if (crit < bestTime) {
bestTime = crit;
bc = c, bs = s;
}
}
if (bc < 0)
break;
param = bestTime;
cell[bc] += (2*bs - 1);
DrawPixel(cell[0], cell[1]);
}
}
Full code is available here.
It uses loadbmp.h, here it is.

Related

Grid - Obstacle / How to Calculate "Field of view"

My english is not my native language, I have no idea how title it, how to explain it clearly and not sure if it's the right term. I've tried to search on google first but for the reason quoted above, I couldn't find anything related.
Could you guys first please check the imgur album : https://imgur.com/a/4mMuCil
So...
Black square is an "obstacle"
Red Square is a "player"
Grey square are "area where player is not able to see"
Depending on the distance of the player from the obstacle, the player can see more or less "things"
Is there a general formula to determine the area that he can see or can't see ?
Or I have to write a unique formula depending on the player position relative to the obstacle
I'm sorry If what I wrote doesn't make sense
Thanks for your help
EDIT :
point player(5, 0);
point obstacle(4, 2);
.....o...
.........
....#....
.........
.....#...
.....##..
.....###.
.....####
......###
This needs a little work. I'll say the steps to make it work:
(1) Let's define the player position p and the obstacle position o.
We know we want to paint a "triangle" after the obstacle.
(2) Let's define the angle according to proximity.
The nearer, the bigger the angle is, so I set the angle to 45 if the distance is 1 and it decreases as the player gets further from the obstacle. The angle is 5 + (max(0, 50 - distance * 10)). You can tune this angle.
(3) Let's build a big triangle. The first vertex is the obstacle. Then, throw a big line from the player through the obstacle. Rotate this line around the obstacle half angle clockwise (to get the second vertex) and half angle anti-clockwise (to get the third vertex), as shown in the image:
(4) Lastly, iterate to the matrix and for each position, ask if that coordinates are inside the triangle.
#include <bits/stdc++.h>
using namespace std;
struct point{
float x, y;
point(){}
point(float x, float y){this->x = x; this->y = y;}
//point(int x, int y){this->x = x; this->y = y;}
};
point rotate(point pivot, float angle, point p, bool clockwise){
float s = sin(angle);
float c = cos(angle);
p.x -= pivot.x;
p.y -= pivot.y;
if(clockwise){
return point(p.x * c + p.y * s + pivot.x, -p.x * s + p.y * c + pivot.y);
}
else{
return point(p.x * c - p.y * s + pivot.x, p.x * s + p.y * c + pivot.y);
}
}
float triangleArea(point p1, point p2, point p3) { //find area of triangle formed by p1, p2 and p3
return abs((p1.x*(p2.y-p3.y) + p2.x*(p3.y-p1.y)+ p3.x*(p1.y-p2.y))/2.0);
}
bool inside(point p1, point p2, point p3, point p) { //check whether p is inside or outside
float area = triangleArea (p1, p2, p3); //area of triangle ABC
float area1 = triangleArea (p, p2, p3); //area of PBC
float area2 = triangleArea (p1, p, p3); //area of APC
float area3 = triangleArea (p1, p2, p); //area of ABP
return abs(area - area1 + area2 + area3) < 1; //when three triangles are forming the whole triangle
}
char m[9][9];
point player(4, 0);
point obstacle(4, 2);
float angle(){
float dist = sqrt(pow(player.x - obstacle.x, 2) + pow(player.y - obstacle.y, 2));
cout<<"dist: "<<(5.0 + max(0.0, 50.0 - 10.0 * dist))<<endl;
return (5.0 + max(0.0, 50.0 - 10.0 * dist)) * 0.0174533;
}
void print(){
for(int i = 0; i < 9; i++){
for(int j = 0; j < 9; j++){
cout<<m[i][j];
}
cout<<endl;
}
}
int main(){
for(int i = 0; i < 9; i++){
for(int j = 0; j < 9; j++){
m[i][j] = '.';
}
}
m[(int)player.y][(int)player.x] = 'o';
m[(int)obstacle.y][(int)obstacle.x] = '#';
float rad = angle();
point end(20.0 * (obstacle.x - player.x) + obstacle.x, 20.0 * (player.y - obstacle.y) + obstacle.y);
point p2 = rotate(obstacle, rad / 2.0, end, true);
point p3 = rotate(obstacle, rad / 2.0, end, false);
for(int i = 0; i < 9; i++){
for(int j = 0; j < 9; j++){
if(i == (int) player.y && j == (int) player.x) continue;
if(i == (int) obstacle.y && j == (int) obstacle.x) continue;
if(inside(obstacle, p2, p3, point(j, i))) m[i][j] = '#';
}
}
print();
return 0;
}
OUTPUT:
....o....
.........
....#....
....#....
....#....
....#....
...###...
...###...
...###...
Suppose the player is at (0,0), and the obstacle is at (j,k), with j>0 and k>=0.
Then a square at (x, y) will be visible if either (2j-1)y >= (2k+1)x or (2k-1)x >= (2j+1)y.
Applying this rule to the other three quadrants is straightforward.

Drawing ellipse with Bresenham's algorithm

Hello,
I'm trying to draw an ellipse, which is parallel to the orthogonal system, using Bresenham's algorithm. I want to draw the top-left (W,SW,S) quarter of the ellipse, and then deduce others.
To do this, i'm using an incremental algorithm with the second-order logic. I did it from another algorithm that draw the top-right quarter first, but what i'm doing isn't working.
The problem appears when the 2nd region is drawing, and I don't know where it comes from.
You can see what I have (black), and what I expect (green):
(center of the ellipse (xc, yc) and the upper right button (x2,y2) that is ~(xc+30,yc+20) in this example)
(a is abs(x2-xc), and b is abs(y2-yc))
The first parameter is the middle of the ellipse (xc, yc), the second is the upper right point established the x and y radius. You can see the ellipse goes too far (2 points on the left and on the right). You can see an other example
(center of the ellipse (xc, yc) and the upper right button (x2,y2) that is ~(xc+15,yc+18) in this example)
The algorithm is deduced from the incremental algorithm with the second-order logic.
Here is my code, (a is abs(x2-xc), and b is abs(y2-yc))
ellipse(int a, int b, int xc, int yc) {
int a2 = a*a, b2 = b*b;
int x = 0, y = b; //Starting point
int incSW = b2*2 + a2*2;
int deltaW = b2*(-2*x + 3); //deduced from incremental algorithm with the second-order logic
int deltaS = a2*(-2*y + 3);
int deltaSW = deltaW + deltaS;
int d1 = b2 - a2*b + a2/4; //dp starting value in the first region
int d2 = b2*(x - 0.5)*(x - 0.5) + a2*(y - 1)*(y - 1) - a2*b2; //dp starting value in the second region
//First region
while(a2*(y-0.5) >= b2*(-x-1)) {
DrawPixel(g,-x+xc, -y+yc); // 1st case
DrawPixel(g,-x+xc, y+yc); // 2nd case
DrawPixel(g,x+xc, y+yc); // 3rd case
DrawPixel(g,x+xc, -y+yc); // 4th case
if(d1>0) {
d1+=deltaSW;
deltaW+=b2*2;
deltaSW+=incSW;
y--;
}
else {
d1+=deltaW;
deltaW+=2*b2;
deltaSW+=2*b2;
}
x--;
}
deltaSW = b2*(2 - 2*x) + a2*(-2*y + 3);
//Second region
while(y>=0) {
DrawPixel(g,-x+xc, -y+yc); // 1st case
DrawPixel(g,-x+xc, y+yc); // 2nd case
DrawPixel(g,x+xc, y+yc); // 3rd case
DrawPixel(g,x+xc, -y+yc); // 4th case
if(d2>0) {
d2+=deltaS;
deltaS+=a2*2;
deltaSW+=a2*2;
}
else {
d2+=deltaSW;
deltaSW+=incSW;
deltaS+=a2*2;
x--;
}
y--;
}
}
I hope you can help me, thanks.
Using the error term e = a x^2 + b y^2 - r^2, it's pretty easy to show that a step from (x,y) to (x,y+1) changes the error by 2by + b, a step to (x+1,y+1) by 2ax + a + 2by + b, and a step to (x+1,y) by 2ax + a.
Starting from a point (-x0, 0), choose the least absolute error step from these three. The first two cases are the norm for the "first region" as you call it.
The first time a step right, (x,y) to (x+1,y), produces least error, you know you're in the second region. At this point the first case is no longer needed. The quarter ellipse can be finished using only the second two cases.
Note this check avoids the floating point operations you've used. The whole point of Bresenham-ish algorithms is to avoid floating point.
The last bit to notice is that you don't want to compute 2ax or 2by each iteration. The multiplications can be avoided by maintaining variables, say dx=2ax and dy=2by, and updating them. A step from x to x+1 increments dx by 2a, a constant. Similarly a step from y to y+1 increments dy by 2b.
Putting all this together, you get the (rough) code below.
Note that you can check the incremental error computation by verifying it against the original error term. If (x0,0) is the initial point, then you know x0^2 = r^2. So the actual error in every iteration is a * x^2 + b * y^2 - x0^2. This ought to equal e in the code below, and it does.
import static java.lang.Math.abs;
import java.util.Arrays;
import java.util.function.BiConsumer;
public class EllipseTracer {
static char [] [] raster = new char[51][101];
static void trace(int x, int y, int a, int b, BiConsumer<Integer, Integer> emitter) {
emitter.accept(x, y);
int e = 0;
int dx = 2 * a * x;
int dy = 2 * b * y;
// First region: stepping north and northeast.
while (x < 0) {
int dxa = dx + a;
int dyb = dy + b;
int eUp = e + dyb;
int eRt = e + dxa;
int eDg = e + dxa + dyb;
if (abs(eUp) < abs(eDg)) {
emitter.accept(x, ++y);
e = eUp;
dy += 2 * b;
} else {
if (abs(eRt) < abs(eDg)) {
// Step east is least error. Found second region.
emitter.accept(++x, y);
e = eRt;
dx += 2 * a;
break;
}
emitter.accept(++x, ++y);
e = eDg;
dy += 2 * b;
dx += 2 * a;
}
}
// Second region: step northeast and east.
while (x < 0) {
int dxa = dx + a;
int dyb = dy + b;
int eRt = e + dxa;
int eDg = e + dxa + dyb;
if (abs(eRt) < abs(eDg)) {
emitter.accept(++x, y);
e = eRt;
dx += 2 * a;
} else {
emitter.accept(++x, ++y);
e = eDg;
dy += 2 * b;
dx += 2 * a;
}
}
}
static void emit(int x, int y) {
raster[y][x + 100] = '*';
}
public static void main(String [] args) {
for (int i = 0; i < raster.length; ++i) {
Arrays.fill(raster[i], ' ');
}
trace(-100, 0, 1, 4, EllipseTracer::emit);
for (int i = 0; i < raster.length; ++i) {
System.out.println(raster[i]);
}
}
}
You can add more tricks to avoid the absolute values, but I'll let you look for those.

Obtaining orientation map of fingerprint image using OpenCV

I'm trying to implement the method of improving fingerprint images by Anil Jain. As a starter, I encountered some difficulties while extracting the orientation image, and am strictly following those steps described in Section 2.4 of that paper.
So, this is the input image:
And this is after normalization using exactly the same method as in that paper:
I'm expecting to see something like this (an example from the internet):
However, this is what I got for displaying obtained orientation matrix:
Obviously this is wrong, and it also gives non-zero values for those zero points in the original input image.
This is the code I wrote:
cv::Mat orientation(cv::Mat inputImage)
{
cv::Mat orientationMat = cv::Mat::zeros(inputImage.size(), CV_8UC1);
// compute gradients at each pixel
cv::Mat grad_x, grad_y;
cv::Sobel(inputImage, grad_x, CV_16SC1, 1, 0, 3, 1, 0, cv::BORDER_DEFAULT);
cv::Sobel(inputImage, grad_y, CV_16SC1, 0, 1, 3, 1, 0, cv::BORDER_DEFAULT);
cv::Mat Vx, Vy, theta, lowPassX, lowPassY;
cv::Mat lowPassX2, lowPassY2;
Vx = cv::Mat::zeros(inputImage.size(), inputImage.type());
Vx.copyTo(Vy);
Vx.copyTo(theta);
Vx.copyTo(lowPassX);
Vx.copyTo(lowPassY);
Vx.copyTo(lowPassX2);
Vx.copyTo(lowPassY2);
// estimate the local orientation of each block
int blockSize = 16;
for(int i = blockSize/2; i < inputImage.rows - blockSize/2; i+=blockSize)
{
for(int j = blockSize / 2; j < inputImage.cols - blockSize/2; j+= blockSize)
{
float sum1 = 0.0;
float sum2 = 0.0;
for ( int u = i - blockSize/2; u < i + blockSize/2; u++)
{
for( int v = j - blockSize/2; v < j+blockSize/2; v++)
{
sum1 += grad_x.at<float>(u,v) * grad_y.at<float>(u,v);
sum2 += (grad_x.at<float>(u,v)*grad_x.at<float>(u,v)) * (grad_y.at<float>(u,v)*grad_y.at<float>(u,v));
}
}
Vx.at<float>(i,j) = sum1;
Vy.at<float>(i,j) = sum2;
double calc = 0.0;
if(sum1 != 0 && sum2 != 0)
{
calc = 0.5 * atan(Vy.at<float>(i,j) / Vx.at<float>(i,j));
}
theta.at<float>(i,j) = calc;
// Perform low-pass filtering
float angle = 2 * calc;
lowPassX.at<float>(i,j) = cos(angle * pi / 180);
lowPassY.at<float>(i,j) = sin(angle * pi / 180);
float sum3 = 0.0;
float sum4 = 0.0;
for(int u = -lowPassSize / 2; u < lowPassSize / 2; u++)
{
for(int v = -lowPassSize / 2; v < lowPassSize / 2; v++)
{
sum3 += inputImage.at<float>(u,v) * lowPassX.at<float>(i - u*lowPassSize, j - v * lowPassSize);
sum4 += inputImage.at<float>(u, v) * lowPassY.at<float>(i - u*lowPassSize, j - v * lowPassSize);
}
}
lowPassX2.at<float>(i,j) = sum3;
lowPassY2.at<float>(i,j) = sum4;
float calc2 = 0.0;
if(sum3 != 0 && sum4 != 0)
{
calc2 = 0.5 * atan(lowPassY2.at<float>(i, j) / lowPassX2.at<float>(i, j)) * 180 / pi;
}
orientationMat.at<float>(i,j) = calc2;
}
}
return orientationMat;
}
I've already searched a lot on the web, but almost all of them are in Matlab. And there exist very few ones using OpenCV, but they didn't help me either. I sincerely hope someone could go through my code and point out any error to help. Thank you in advance.
Update
Here are the steps that I followed according to the paper:
Obtain normalized image G.
Divide G into blocks of size wxw (16x16).
Compute the x and y gradients at each pixel (i,j).
Estimate the local orientation of each block centered at pixel (i,j) using equations:
Perform low-pass filtering to remove noise. For that, convert the orientation image into a continuous vector field defined as:
where W is a two-dimensional low-pass filter, and w(phi) x w(phi) is its size, which equals to 5.
Finally, compute the local ridge orientation at (i,j) using:
Update2
This is the output of orientationMat after changing the mat type to CV_16SC1 in Sobel operation as Micka suggested:
Maybe it's too late for me to answer, but anyway somebody could read this later and solve the same problem.
I've been working for a while in the same algorithm, same method you posted... But there's some writting errors when the papper was redacted (I guess). After fighting a lot with the equations I found this errors by looking other similar works.
Here is what worked for me...
Vy(i, j) = 2*dx(u,v)*dy(u,v)
Vx(i,j) = dx(u,v)^2 - dy(u,v)^2
O(i,j) = 0.5*arctan(Vy(i,j)/Vx(i,j)
(Excuse me I wasn't able to post images, so I wrote the modified ecuations. Remeber "u" and "v" are positions of the summation across the BlockSize by BlockSize window)
The first thing and most important (obviously) are the equations, I saw that in different works this expressions were really different and in every one they talked about the same algorithm of Hong et al.
The Key is finding the Least Mean Square (First 3 equations) of the gradients (Vx and Vy), I provided the corrected formulas above for this ation. Then you can compute angle theta for the non overlapping window (16x16 size recommended in the papper), after that the algorithm says you must calculate the magnitud of the doubled angle in "x" and "y" directions (Phi_x and Phi_y).
Phi_x(i,j) = V(i,j) * cos(2*O(i,j))
Phi_y(i,j) = V(i,j) * sin(2*O(i,j))
Magnitud is just:
V = sqrt(Vx(i,j)^2 + Vy(i,j)^2)
Note that in the related work doesn't mention that you have to use the gradient magnitud, but it make sense (for me) in doing it. After all this corrections you can apply the low pass filter to Phi_x and Phi_y, I used a simple Mask of size 5x5 to average this magnitudes (something like medianblur() of opencv).
Last thing is to calculate new angle, that is the average of the 25ith neighbors in the O(i,j) image, for this you just have to:
O'(i,j) = 0.5*arctan(Phi_y/Phi_x)
We're just there... All this just for calculating the angle of the NORMAL VECTOR TO THE RIDGES DIRECTIONS (O'(i,j)) in the BlockSize by BlockSize non overlapping window, what does it mean? it means that the angle we just calculated is perpendicular to the ridges, in simple words we just calculated the angle of the riges plus 90 degrees... To get the angle we need, we just have to substract to the obtained angle 90°.
To draw the lines we need to have an initial point (X0, Y0) and a final point(X1, Y1). For that imagine a circle centered on (X0, Y0) with a radious of "r":
x0 = i + blocksize/2
y0 = j + blocksize/2
r = blocksize/2
Note we add i and j to the first coordinates becouse the window is moving and we are gonna draw the line starting from the center of the non overlaping window, so we can't use just the center of the non overlaping window.
Then to calculate the end coordinates to draw a line we can just have to use a right triangle so...
X1 = r*cos(O'(i,j)-90°)+X0
Y1 = r*sin(O'(i,j)-90°)+Y0
X2 = X0-r*cos(O'(i,j)-90°)
Y2 = Y0-r*cos(O'(i,j)-90°)
Then just use opencv line function, where initial Point is (X0,Y0) and final Point is (X1, Y1). Additional to it, I drawed the windows of 16x16 and computed the oposite points of X1 and Y1 (X2 and Y2) to draw a line of the entire window.
Hope this help somebody.
My results...
Main function:
Mat mat = imread("nwmPa.png",0);
mat.convertTo(mat, CV_32F, 1.0/255, 0);
Normalize(mat);
int blockSize = 6;
int height = mat.rows;
int width = mat.cols;
Mat orientationMap;
orientation(mat, orientationMap, blockSize);
Normalize:
void Normalize(Mat & image)
{
Scalar mean, dev;
meanStdDev(image, mean, dev);
double M = mean.val[0];
double D = dev.val[0];
for(int i(0) ; i<image.rows ; i++)
{
for(int j(0) ; j<image.cols ; j++)
{
if(image.at<float>(i,j) > M)
image.at<float>(i,j) = 100.0/255 + sqrt( 100.0/255*pow(image.at<float>(i,j)-M,2)/D );
else
image.at<float>(i,j) = 100.0/255 - sqrt( 100.0/255*pow(image.at<float>(i,j)-M,2)/D );
}
}
}
Orientation map:
void orientation(const Mat &inputImage, Mat &orientationMap, int blockSize)
{
Mat fprintWithDirectionsSmoo = inputImage.clone();
Mat tmp(inputImage.size(), inputImage.type());
Mat coherence(inputImage.size(), inputImage.type());
orientationMap = tmp.clone();
//Gradiants x and y
Mat grad_x, grad_y;
// Sobel(inputImage, grad_x, CV_32F, 1, 0, 3, 1, 0, BORDER_DEFAULT);
// Sobel(inputImage, grad_y, CV_32F, 0, 1, 3, 1, 0, BORDER_DEFAULT);
Scharr(inputImage, grad_x, CV_32F, 1, 0, 1, 0);
Scharr(inputImage, grad_y, CV_32F, 0, 1, 1, 0);
//Vector vield
Mat Fx(inputImage.size(), inputImage.type()),
Fy(inputImage.size(), inputImage.type()),
Fx_gauss,
Fy_gauss;
Mat smoothed(inputImage.size(), inputImage.type());
// Local orientation for each block
int width = inputImage.cols;
int height = inputImage.rows;
int blockH;
int blockW;
//select block
for(int i = 0; i < height; i+=blockSize)
{
for(int j = 0; j < width; j+=blockSize)
{
float Gsx = 0.0;
float Gsy = 0.0;
float Gxx = 0.0;
float Gyy = 0.0;
//for check bounds of img
blockH = ((height-i)<blockSize)?(height-i):blockSize;
blockW = ((width-j)<blockSize)?(width-j):blockSize;
//average at block WхW
for ( int u = i ; u < i + blockH; u++)
{
for( int v = j ; v < j + blockW ; v++)
{
Gsx += (grad_x.at<float>(u,v)*grad_x.at<float>(u,v)) - (grad_y.at<float>(u,v)*grad_y.at<float>(u,v));
Gsy += 2*grad_x.at<float>(u,v) * grad_y.at<float>(u,v);
Gxx += grad_x.at<float>(u,v)*grad_x.at<float>(u,v);
Gyy += grad_y.at<float>(u,v)*grad_y.at<float>(u,v);
}
}
float coh = sqrt(pow(Gsx,2) + pow(Gsy,2)) / (Gxx + Gyy);
//smoothed
float fi = 0.5*fastAtan2(Gsy, Gsx)*CV_PI/180;
Fx.at<float>(i,j) = cos(2*fi);
Fy.at<float>(i,j) = sin(2*fi);
//fill blocks
for ( int u = i ; u < i + blockH; u++)
{
for( int v = j ; v < j + blockW ; v++)
{
orientationMap.at<float>(u,v) = fi;
Fx.at<float>(u,v) = Fx.at<float>(i,j);
Fy.at<float>(u,v) = Fy.at<float>(i,j);
coherence.at<float>(u,v) = (coh<0.85)?1:0;
}
}
}
} ///for
GaussConvolveWithStep(Fx, Fx_gauss, 5, blockSize);
GaussConvolveWithStep(Fy, Fy_gauss, 5, blockSize);
for(int m = 0; m < height; m++)
{
for(int n = 0; n < width; n++)
{
smoothed.at<float>(m,n) = 0.5*fastAtan2(Fy_gauss.at<float>(m,n), Fx_gauss.at<float>(m,n))*CV_PI/180;
if((m%blockSize)==0 && (n%blockSize)==0){
int x = n;
int y = m;
int ln = sqrt(2*pow(blockSize,2))/2;
float dx = ln*cos( smoothed.at<float>(m,n) - CV_PI/2);
float dy = ln*sin( smoothed.at<float>(m,n) - CV_PI/2);
arrowedLine(fprintWithDirectionsSmoo, Point(x, y+blockH), Point(x + dx, y + blockW + dy), Scalar::all(255), 1, CV_AA, 0, 0.06*blockSize);
// qDebug () << Fx_gauss.at<float>(m,n) << Fy_gauss.at<float>(m,n) << smoothed.at<float>(m,n);
// imshow("Orientation", fprintWithDirectionsSmoo);
// waitKey(0);
}
}
}///for2
normalize(orientationMap, orientationMap,0,1,NORM_MINMAX);
imshow("Orientation field", orientationMap);
orientationMap = smoothed.clone();
normalize(smoothed, smoothed, 0, 1, NORM_MINMAX);
imshow("Smoothed orientation field", smoothed);
imshow("Coherence", coherence);
imshow("Orientation", fprintWithDirectionsSmoo);
}
seems nothing forgot )
I have read your code thoroughly and found that you have made a mistake while calculating sum3 and sum4:
sum3 += inputImage.at<float>(u,v) * lowPassX.at<float>(i - u*lowPassSize, j - v * lowPassSize);
sum4 += inputImage.at<float>(u, v) * lowPassY.at<float>(i - u*lowPassSize, j - v * lowPassSize);
instead of inputImage you should use a low pass filter.

convert bezier curve to polygonal chain?

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!!

An algorithm to find bounding box of closed bezier curves?

I'm looking for an algorithm to find bounding box (max/min points) of a closed quadratic bezier curve in Cartesian axis:
input: C (a closed bezier curve)
output: A B C D points
Image http://www.imagechicken.com/uploads/1270586513022388700.jpg
Note: above image shows a smooth curve. it could be not smooth. (have corners)
Ivan Kuckir's DeCasteljau is a brute force, but works in many cases. The problem with it is the count of iterations. The actual shape and the distance between coordinates affect to the precision of the result. And to find a precise enough answer, you have to iterate tens of times, may be more. And it may fail if there are sharp turns in curve.
Better solution is to find first derivative roots, as is described on the excellent site http://processingjs.nihongoresources.com/bezierinfo/. Please read the section Finding the extremities of the curves.
The link above has the algorithm for both quadratic and cubic curves.
The asker of question is interested in quadratic curves, so the rest of this answer may be irrelevant, because I provide codes for calculating extremities of Cubic curves.
Below are three Javascript codes of which the first (CODE 1) is the one I suggest to use.
** CODE 1 **
After testing processingjs and Raphael's solutions I find they had some restrictions and/or bugs. Then more search and found Bonsai and it's bounding box function, which is based on NISHIO Hirokazu's Python script. Both have a downside where double equality is tested using ==. When I changed these to numerically robust comparisons, then script succeeds 100% right in all cases. I tested the script with thousands of random paths and also with all collinear cases and all succeeded:
Various cubic curves
Random cubic curves
Collinear cubic curves
The code is as follows. Usually left, right, top and bottom values are the all needed, but in some cases it's fine to know the coordinates of local extreme points and corresponding t values. So I added there two variables: tvalues and points. Remove code regarding them and you have fast and stable bounding box calculation function.
// Source: http://blog.hackers-cafe.net/2009/06/how-to-calculate-bezier-curves-bounding.html
// Original version: NISHIO Hirokazu
// Modifications: Timo
var pow = Math.pow,
sqrt = Math.sqrt,
min = Math.min,
max = Math.max;
abs = Math.abs;
function getBoundsOfCurve(x0, y0, x1, y1, x2, y2, x3, y3)
{
var tvalues = new Array();
var bounds = [new Array(), new Array()];
var points = new Array();
var a, b, c, t, t1, t2, b2ac, sqrtb2ac;
for (var i = 0; i < 2; ++i)
{
if (i == 0)
{
b = 6 * x0 - 12 * x1 + 6 * x2;
a = -3 * x0 + 9 * x1 - 9 * x2 + 3 * x3;
c = 3 * x1 - 3 * x0;
}
else
{
b = 6 * y0 - 12 * y1 + 6 * y2;
a = -3 * y0 + 9 * y1 - 9 * y2 + 3 * y3;
c = 3 * y1 - 3 * y0;
}
if (abs(a) < 1e-12) // Numerical robustness
{
if (abs(b) < 1e-12) // Numerical robustness
{
continue;
}
t = -c / b;
if (0 < t && t < 1)
{
tvalues.push(t);
}
continue;
}
b2ac = b * b - 4 * c * a;
sqrtb2ac = sqrt(b2ac);
if (b2ac < 0)
{
continue;
}
t1 = (-b + sqrtb2ac) / (2 * a);
if (0 < t1 && t1 < 1)
{
tvalues.push(t1);
}
t2 = (-b - sqrtb2ac) / (2 * a);
if (0 < t2 && t2 < 1)
{
tvalues.push(t2);
}
}
var x, y, j = tvalues.length,
jlen = j,
mt;
while (j--)
{
t = tvalues[j];
mt = 1 - t;
x = (mt * mt * mt * x0) + (3 * mt * mt * t * x1) + (3 * mt * t * t * x2) + (t * t * t * x3);
bounds[0][j] = x;
y = (mt * mt * mt * y0) + (3 * mt * mt * t * y1) + (3 * mt * t * t * y2) + (t * t * t * y3);
bounds[1][j] = y;
points[j] = {
X: x,
Y: y
};
}
tvalues[jlen] = 0;
tvalues[jlen + 1] = 1;
points[jlen] = {
X: x0,
Y: y0
};
points[jlen + 1] = {
X: x3,
Y: y3
};
bounds[0][jlen] = x0;
bounds[1][jlen] = y0;
bounds[0][jlen + 1] = x3;
bounds[1][jlen + 1] = y3;
tvalues.length = bounds[0].length = bounds[1].length = points.length = jlen + 2;
return {
left: min.apply(null, bounds[0]),
top: min.apply(null, bounds[1]),
right: max.apply(null, bounds[0]),
bottom: max.apply(null, bounds[1]),
points: points, // local extremes
tvalues: tvalues // t values of local extremes
};
};
// Usage:
var bounds = getBoundsOfCurve(532,333,117,305,28,93,265,42);
console.log(JSON.stringify(bounds));
// Prints: {"left":135.77684049079755,"top":42,"right":532,"bottom":333,"points":[{"X":135.77684049079755,"Y":144.86387466397255},{"X":532,"Y":333},{"X":265,"Y":42}],"tvalues":[0.6365030674846626,0,1]}
CODE 2 (which fails in collinear cases):
I translated the code from http://processingjs.nihongoresources.com/bezierinfo/sketchsource.php?sketch=tightBoundsCubicBezier to Javascript. The code works fine in normal cases, but not in collinear cases where all points lie on the same line.
For reference, here is the Javascript code.
function computeCubicBaseValue(a,b,c,d,t) {
var mt = 1-t;
return mt*mt*mt*a + 3*mt*mt*t*b + 3*mt*t*t*c + t*t*t*d;
}
function computeCubicFirstDerivativeRoots(a,b,c,d) {
var ret = [-1,-1];
var tl = -a+2*b-c;
var tr = -Math.sqrt(-a*(c-d) + b*b - b*(c+d) +c*c);
var dn = -a+3*b-3*c+d;
if(dn!=0) { ret[0] = (tl+tr)/dn; ret[1] = (tl-tr)/dn; }
return ret;
}
function computeCubicBoundingBox(xa,ya,xb,yb,xc,yc,xd,yd)
{
// find the zero point for x and y in the derivatives
var minx = 9999;
var maxx = -9999;
if(xa<minx) { minx=xa; }
if(xa>maxx) { maxx=xa; }
if(xd<minx) { minx=xd; }
if(xd>maxx) { maxx=xd; }
var ts = computeCubicFirstDerivativeRoots(xa, xb, xc, xd);
for(var i=0; i<ts.length;i++) {
var t = ts[i];
if(t>=0 && t<=1) {
var x = computeCubicBaseValue(t, xa, xb, xc, xd);
var y = computeCubicBaseValue(t, ya, yb, yc, yd);
if(x<minx) { minx=x; }
if(x>maxx) { maxx=x; }}}
var miny = 9999;
var maxy = -9999;
if(ya<miny) { miny=ya; }
if(ya>maxy) { maxy=ya; }
if(yd<miny) { miny=yd; }
if(yd>maxy) { maxy=yd; }
ts = computeCubicFirstDerivativeRoots(ya, yb, yc, yd);
for(i=0; i<ts.length;i++) {
var t = ts[i];
if(t>=0 && t<=1) {
var x = computeCubicBaseValue(t, xa, xb, xc, xd);
var y = computeCubicBaseValue(t, ya, yb, yc, yd);
if(y<miny) { miny=y; }
if(y>maxy) { maxy=y; }}}
// bounding box corner coordinates
var bbox = [minx,miny, maxx,miny, maxx,maxy, minx,maxy ];
return bbox;
}
CODE 3 (works in most cases):
To handle also collinear cases, I found Raphael's solution, which is based on the same first derivative method as the CODE 2. I added also a return value dots, which has the extrema points, because always it's not enough to know bounding boxes min and max coordinates, but we want to know the exact extrema coordinates.
EDIT: found another bug. Fails eg. in 532,333,117,305,28,93,265,42 and also many other cases.
The code is here:
Array.max = function( array ){
return Math.max.apply( Math, array );
};
Array.min = function( array ){
return Math.min.apply( Math, array );
};
var findDotAtSegment = function (p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y, t) {
var t1 = 1 - t;
return {
x: t1*t1*t1*p1x + t1*t1*3*t*c1x + t1*3*t*t * c2x + t*t*t * p2x,
y: t1*t1*t1*p1y + t1*t1*3*t*c1y + t1*3*t*t * c2y + t*t*t * p2y
};
};
var cubicBBox = function (p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y) {
var a = (c2x - 2 * c1x + p1x) - (p2x - 2 * c2x + c1x),
b = 2 * (c1x - p1x) - 2 * (c2x - c1x),
c = p1x - c1x,
t1 = (-b + Math.sqrt(b * b - 4 * a * c)) / 2 / a,
t2 = (-b - Math.sqrt(b * b - 4 * a * c)) / 2 / a,
y = [p1y, p2y],
x = [p1x, p2x],
dot, dots=[];
Math.abs(t1) > "1e12" && (t1 = 0.5);
Math.abs(t2) > "1e12" && (t2 = 0.5);
if (t1 >= 0 && t1 <= 1) {
dot = findDotAtSegment(p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y, t1);
x.push(dot.x);
y.push(dot.y);
dots.push({X:dot.x, Y:dot.y});
}
if (t2 >= 0 && t2 <= 1) {
dot = findDotAtSegment(p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y, t2);
x.push(dot.x);
y.push(dot.y);
dots.push({X:dot.x, Y:dot.y});
}
a = (c2y - 2 * c1y + p1y) - (p2y - 2 * c2y + c1y);
b = 2 * (c1y - p1y) - 2 * (c2y - c1y);
c = p1y - c1y;
t1 = (-b + Math.sqrt(b * b - 4 * a * c)) / 2 / a;
t2 = (-b - Math.sqrt(b * b - 4 * a * c)) / 2 / a;
Math.abs(t1) > "1e12" && (t1 = 0.5);
Math.abs(t2) > "1e12" && (t2 = 0.5);
if (t1 >= 0 && t1 <= 1) {
dot = findDotAtSegment(p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y, t1);
x.push(dot.x);
y.push(dot.y);
dots.push({X:dot.x, Y:dot.y});
}
if (t2 >= 0 && t2 <= 1) {
dot = findDotAtSegment(p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y, t2);
x.push(dot.x);
y.push(dot.y);
dots.push({X:dot.x, Y:dot.y});
}
// remove duplicate dots
var dots2 = [];
var l = dots.length;
for(var i=0; i<l; i++) {
for(var j=i+1; j<l; j++) {
if (dots[i].X === dots[j].X && dots[i].Y === dots[j].Y)
j = ++i;
}
dots2.push({X: dots[i].X, Y: dots[i].Y});
}
return {
min: {x: Array.min(x), y: Array.min(y)},
max: {x: Array.max(x), y: Array.max(y)},
dots: dots2 // these are the extrema points
};
};
Well, I would say you start by adding all endpoints to your bounding box. Then, you go through all the bezier elements. I assume the formula in question is this one:
From this, extract two formulas for X and Y, respectively. Test both for extrema by taking the derivative (zero crossings). Then add the corresponding points to your bounding box as well.
Use De Casteljau algorithm to approximate the curve of higher orders. Here is how it works for cubic curve
http://jsfiddle.net/4VCVX/25/
function getCurveBounds(ax, ay, bx, by, cx, cy, dx, dy)
{
var px, py, qx, qy, rx, ry, sx, sy, tx, ty,
tobx, toby, tocx, tocy, todx, tody, toqx, toqy,
torx, tory, totx, toty;
var x, y, minx, miny, maxx, maxy;
minx = miny = Number.POSITIVE_INFINITY;
maxx = maxy = Number.NEGATIVE_INFINITY;
tobx = bx - ax; toby = by - ay; // directions
tocx = cx - bx; tocy = cy - by;
todx = dx - cx; tody = dy - cy;
var step = 1/40; // precision
for(var d=0; d<1.001; d+=step)
{
px = ax +d*tobx; py = ay +d*toby;
qx = bx +d*tocx; qy = by +d*tocy;
rx = cx +d*todx; ry = cy +d*tody;
toqx = qx - px; toqy = qy - py;
torx = rx - qx; tory = ry - qy;
sx = px +d*toqx; sy = py +d*toqy;
tx = qx +d*torx; ty = qy +d*tory;
totx = tx - sx; toty = ty - sy;
x = sx + d*totx; y = sy + d*toty;
minx = Math.min(minx, x); miny = Math.min(miny, y);
maxx = Math.max(maxx, x); maxy = Math.max(maxy, y);
}
return {x:minx, y:miny, width:maxx-minx, height:maxy-miny};
}
I believe that the control points of a Bezier curve form a convex hull that encloses the curve. If you just want a axis-aligned bounding box, I think you need to find the min and max of each (x, y) for each control point of all the segments.
I suppose that might not be a tight box. That is, the box might be slightly larger than it needs to be, but it's simple and fast to compute. I guess it depends on your requirements.
I think the accepted answer is fine, but just wanted to offer a little more explanation for anyone else trying to do this.
Consider a quadratic Bezier with starting point p1, ending point p2 and "control point" pc. This curve has three parametric equations:
pa(t) = p1 + t(pc-p1)
pb(t) = pc + t(p2-pc)
p(t) = pa(t) + t*(pb(t) - pa(t))
In all cases, t runs from 0 to 1, inclusive.
The first two are linear, defining line segments from p1 to pc and from pc to p2, respectively. The third is quadratic once you substitute in the expressions for pa(t) and pb(t); this is the one that actually defines points on the curve.
Actually, each of these equations is a pair of equations, one for the horizontal dimension, and one for the vertical. The nice thing about parametric curves is that the x and y can be handled independently of one another. The equations are exactly the same, just substitute x or y for p in the above equations.
The important point is that the line segment defined in equation 3, that runs from pa(t) to pb(t) for a specific value of t is tangent to the curve at the corresponding point p(t). To find the local extrema of the curve, you need to find the parameter value where the tangent is flat (i.e., a critical point). For the vertical dimension, you want to find the value of t such that ya(t) = yb(t), which gives the tangent a slope of 0. For the horizontal dimension, find t such that xa(t) = xb(t), which gives the tangent an infinite slope (i.e., a vertical line). In each case, you can just plug the value of t back into equation 1 (or 2, or even 3) to get the location of that extrema.
In other words, to find the vertical extrema of the curve, take just the y-component of equations 1 and 2, set them equal to each other and solve for t; plug this back into the y-component of equation 1, to get the y-value of that extrema. To get the complete y-range of the curve, find the minimum of this extreme y value and the y-components of the two end points, and likewise find the maximum of all three. Repeat for x to get the horizontal limits.
Remember that t only runs in [0, 1], so if you get a value outside of this range, it means there is no local extrema on the curve (at least not between your two endpoints). This includes the case where you end up dividing by zero when solving for t, which you will probably need to check for before you do it.
The same idea can be applied to higher-order Beziers, there are just more equations of higher degree, which also means there are potentially more local extrema per curve. For instance, on a cubic Bezier (two control points), solving for t to find the local extrema is a quadratic equation, so you could get 0, 1, or 2 values (remember to check for 0-denominators, and for negative square-roots, both of which indicate that there are no local extrema for that dimension). To find the range, you just need to find the min/max of all the local extrema, and the two end points.
I answered this question in Calculating the bounding box of cubic bezier curve
this article explain the details and also has a live html5 demo:
Calculating / Computing the Bounding Box of Cubic Bezier
I found a javascript in Snap.svg to calculate that: here
see the bezierBBox and curveDim functions.
I rewrite a javascript function.
//(x0,y0) is start point; (x1,y1),(x2,y2) is control points; (x3,y3) is end point.
function bezierMinMax(x0, y0, x1, y1, x2, y2, x3, y3) {
var tvalues = [], xvalues = [], yvalues = [],
a, b, c, t, t1, t2, b2ac, sqrtb2ac;
for (var i = 0; i < 2; ++i) {
if (i == 0) {
b = 6 * x0 - 12 * x1 + 6 * x2;
a = -3 * x0 + 9 * x1 - 9 * x2 + 3 * x3;
c = 3 * x1 - 3 * x0;
} else {
b = 6 * y0 - 12 * y1 + 6 * y2;
a = -3 * y0 + 9 * y1 - 9 * y2 + 3 * y3;
c = 3 * y1 - 3 * y0;
}
if (Math.abs(a) < 1e-12) {
if (Math.abs(b) < 1e-12) {
continue;
}
t = -c / b;
if (0 < t && t < 1) {
tvalues.push(t);
}
continue;
}
b2ac = b * b - 4 * c * a;
if (b2ac < 0) {
continue;
}
sqrtb2ac = Math.sqrt(b2ac);
t1 = (-b + sqrtb2ac) / (2 * a);
if (0 < t1 && t1 < 1) {
tvalues.push(t1);
}
t2 = (-b - sqrtb2ac) / (2 * a);
if (0 < t2 && t2 < 1) {
tvalues.push(t2);
}
}
var j = tvalues.length, mt;
while (j--) {
t = tvalues[j];
mt = 1 - t;
xvalues[j] = (mt * mt * mt * x0) + (3 * mt * mt * t * x1) + (3 * mt * t * t * x2) + (t * t * t * x3);
yvalues[j] = (mt * mt * mt * y0) + (3 * mt * mt * t * y1) + (3 * mt * t * t * y2) + (t * t * t * y3);
}
xvalues.push(x0,x3);
yvalues.push(y0,y3);
return {
min: {x: Math.min.apply(0, xvalues), y: Math.min.apply(0, yvalues)},
max: {x: Math.max.apply(0, xvalues), y: Math.max.apply(0, yvalues)}
};
}
Timo-s first variant adapted to Objective-C
CGPoint CubicBezierPointAt(CGPoint p1, CGPoint p2, CGPoint p3, CGPoint p4, CGFloat t) {
CGFloat x = CubicBezier(p1.x, p2.x, p3.x, p4.x, t);
CGFloat y = CubicBezier(p1.y, p2.y, p3.y, p4.y, t);
return CGPointMake(x, y);
}
// array containing TopLeft and BottomRight points for curve`s enclosing bounds
NSArray* CubicBezierExtremums(CGPoint p1, CGPoint p2, CGPoint p3, CGPoint p4) {
CGFloat a, b, c, t, t1, t2, b2ac, sqrtb2ac;
NSMutableArray *tValues = [NSMutableArray new];
for (int i = 0; i < 2; i++) {
if (i == 0) {
a = 3 * (-p1.x + 3 * p2.x - 3 * p3.x + p4.x);
b = 6 * (p1.x - 2 * p2.x + p3.x);
c = 3 * (p2.x - p1.x);
}
else {
a = 3 * (-p1.y + 3 * p2.y - 3 * p3.y + p4.y);
b = 6 * (p1.y - 2 * p2.y + p3.y);
c = 3 * (p2.y - p1.y);
}
if(ABS(a) < CGFLOAT_MIN) {// Numerical robustness
if (ABS(b) < CGFLOAT_MIN) {// Numerical robustness
continue;
}
t = -c / b;
if (t > 0 && t < 1) {
[tValues addObject:[NSNumber numberWithDouble:t]];
}
continue;
}
b2ac = pow(b, 2) - 4 * c * a;
if (b2ac < 0) {
continue;
}
sqrtb2ac = sqrt(b2ac);
t1 = (-b + sqrtb2ac) / (2 * a);
if (t1 > 0.0 && t1 < 1.0) {
[tValues addObject:[NSNumber numberWithDouble:t1]];
}
t2 = (-b - sqrtb2ac) / (2 * a);
if (t2 > 0.0 && t2 < 1.0) {
[tValues addObject:[NSNumber numberWithDouble:t2]];
}
}
int j = (int)tValues.count;
CGFloat x = 0;
CGFloat y = 0;
NSMutableArray *xValues = [NSMutableArray new];
NSMutableArray *yValues = [NSMutableArray new];
while (j--) {
t = [[tValues objectAtIndex:j] doubleValue];
x = CubicBezier(p1.x, p2.x, p3.x, p4.x, t);
y = CubicBezier(p1.y, p2.y, p3.y, p4.y, t);
[xValues addObject:[NSNumber numberWithDouble:x]];
[yValues addObject:[NSNumber numberWithDouble:y]];
}
[xValues addObject:[NSNumber numberWithDouble:p1.x]];
[xValues addObject:[NSNumber numberWithDouble:p4.x]];
[yValues addObject:[NSNumber numberWithDouble:p1.y]];
[yValues addObject:[NSNumber numberWithDouble:p4.y]];
//find minX, minY, maxX, maxY
CGFloat minX = [[xValues valueForKeyPath:#"#min.self"] doubleValue];
CGFloat minY = [[yValues valueForKeyPath:#"#min.self"] doubleValue];
CGFloat maxX = [[xValues valueForKeyPath:#"#max.self"] doubleValue];
CGFloat maxY = [[yValues valueForKeyPath:#"#max.self"] doubleValue];
CGPoint origin = CGPointMake(minX, minY);
CGPoint bottomRight = CGPointMake(maxX, maxY);
NSArray *toReturn = [NSArray arrayWithObjects:
[NSValue valueWithCGPoint:origin],
[NSValue valueWithCGPoint:bottomRight],
nil];
return toReturn;
}
Timo's CODE 2 answer has a small bug: the t parameter in computeCubicBaseValue function should be last. Nevertheless good job, works like a charm ;)
Solution in C# :
double computeCubicBaseValue(double a, double b, double c, double d, double t)
{
var mt = 1 - t;
return mt * mt * mt * a + 3 * mt * mt * t * b + 3 * mt * t * t * c + t * t * t * d;
}
double[] computeCubicFirstDerivativeRoots(double a, double b, double c, double d)
{
var ret = new double[2] { -1, -1 };
var tl = -a + 2 * b - c;
var tr = -Math.Sqrt(-a * (c - d) + b * b - b * (c + d) + c * c);
var dn = -a + 3 * b - 3 * c + d;
if (dn != 0) { ret[0] = (tl + tr) / dn; ret[1] = (tl - tr) / dn; }
return ret;
}
public double[] ComputeCubicBoundingBox(Point start, Point firstControl, Point secondControl, Point end)
{
double xa, ya, xb, yb, xc, yc, xd, yd;
xa = start.X;
ya = start.Y;
xb = firstControl.X;
yb = firstControl.Y;
xc = secondControl.X;
yc = secondControl.Y;
xd = end.X;
yd = end.Y;
// find the zero point for x and y in the derivatives
double minx = Double.MaxValue;
double maxx = Double.MinValue;
if (xa < minx) { minx = xa; }
if (xa > maxx) { maxx = xa; }
if (xd < minx) { minx = xd; }
if (xd > maxx) { maxx = xd; }
var ts = computeCubicFirstDerivativeRoots(xa, xb, xc, xd);
for (var i = 0; i < ts.Length; i++)
{
var t = ts[i];
if (t >= 0 && t <= 1)
{
var x = computeCubicBaseValue(xa, xb, xc, xd,t);
var y = computeCubicBaseValue(ya, yb, yc, yd,t);
if (x < minx) { minx = x; }
if (x > maxx) { maxx = x; }
}
}
double miny = Double.MaxValue;
double maxy = Double.MinValue;
if (ya < miny) { miny = ya; }
if (ya > maxy) { maxy = ya; }
if (yd < miny) { miny = yd; }
if (yd > maxy) { maxy = yd; }
ts = computeCubicFirstDerivativeRoots(ya, yb, yc, yd);
for (var i = 0; i < ts.Length; i++)
{
var t = ts[i];
if (t >= 0 && t <= 1)
{
var x = computeCubicBaseValue(xa, xb, xc, xd,t);
var y = computeCubicBaseValue(ya, yb, yc, yd,t);
if (y < miny) { miny = y; }
if (y > maxy) { maxy = y; }
}
}
// bounding box corner coordinates
var bbox = new double[] { minx, miny, maxx, maxy};
return bbox;
}

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