I have been trying to find a way to draw a curved line/cubic bezier line using a custom function. However, all the examples and such found on the internet, differ a little from each other and usually produce different results, why? . None of the ones i have tried produce the same result as windows api PolyBezier which is what i need.
This is my current code for drawing cubic bezier lines:
double Factorial(int number)
{
double factorial = 1;
if (number > 1)
{
for (int count = 1; count <= number; count++) factorial = factorial * count;
}
return factorial;
}
double choose(double a, double b)
{
return Factorial(a) / (Factorial(b) * Factorial(a - b));
}
VOID MyPolyBezier(HDC hdc, PPOINT Pts, int Total)
{
float x, y;
MoveToEx(hdc, Pts[0].x, Pts[0].y, 0);
Total -= 1;
//for (float t = 0; t <= 1; t += (1./128.))
for (float t = 0; t <= 1; t += 0.0078125)
{
x = 0;
y = 0;
for (int I = 0; I <= Total; I++)
{
x += Pts[I].x * choose(Total, I) * pow(1 - t, Total - I) * pow(t, I);
y += Pts[I].y * choose(Total, I) * pow(1 - t, Total - I) * pow(t, I);
}
LineTo(hdc, x, y);
}
}
And here is the code for testing it.
POINT TestPts[4];
BYTE TestType[4] = {PT_MOVETO, PT_BEZIERTO, PT_BEZIERTO, PT_BEZIERTO};
//set x, y points for the curved line.
TestPts[0].x = 50;
TestPts[0].y = 200;
TestPts[1].x = 100;
TestPts[1].y = 100;
TestPts[2].x = 150;
TestPts[2].y = 200;
TestPts[3].x = 200;
TestPts[3].y = 200;
//Draw using custom function.
MyPolyBezier(hdc, TestPts, 4);
//Move the curve down some.
TestPts[0].y += 10;
TestPts[1].y += 10;
TestPts[2].y += 10;
TestPts[3].y += 10;
//Draw using windows api.
//PolyDraw(hdc, TestPts, TestType, 4); //PolyDraw gives the same result as PolyBezier.
PolyBezier(hdc, TestPts, 4);
And an attached image of my bad results:
Note: the bottom bezier line is windows(PolyBezier) version.
Edit:
the final goal, Windows(On the left) VS custom funtion. Hopefully this helps in some way.
So a cubic bezier is a mathematical curve. The cubic bezier is a specific case of a more general curve.
The cubic bezier is defined by 4 control points -- a start and end point, and 2 control points. In general, a bezier has n control points in order.
The line is drawn as a time parameter t goes from 0 to 1.
To find out where a general bezier of degree n is at time t:
For each adjacent pair of control points in your bezier, find the weighted average of them, as controlled by t. So at + b(1-t) for control points a before b.
Use these n-1 points to form a degree n-1 bezier.
Solve the new bezier at time t.
when you hit a degree 1 bezier, stop. That is your point.
Try writing an algorithm based off the true definition of bezier, and see where it differs from the windows curve. This may ne less frustrating than taking some approximation and having two sets of errors to reconcile.
Related
imageData = new double*[imageHeight];
for(int i = 0; i < imageHeight; i++) {
imageData[i] = new double[imageWidth];
for(int j = 0; j < imageWidth; j++) {
// compute the distance and angle from the swirl center:
double pixelX = (double)i - swirlCenterX;
double pixelY = (double)j - swirlCenterY;
double pixelDistance = pow(pow(pixelX, 2) + pow(pixelY, 2), 0.5);
double pixelAngle = atan2(pixelX, pixelY);
// double swirlAmount = 1.0 - (pixelDistance/swirlRadius);
// if(swirlAmount > 0.0) {
// double twistAngle = swirlTwists * swirlAmount * PI * 2.0;
double twistAngle = swirlTwists * pixelDistance * PI * 2.0;
// adjust the pixel angle and compute the adjusted pixel co-ordinates:
pixelAngle += twistAngle;
pixelX = cos(pixelAngle) * pixelDistance;
pixelY = sin(pixelAngle) * pixelDistance;
// }
(this)->setPixel(i, j, tempMatrix[(int)(swirlCenterX + pixelX)][(int)(swirlCenterY + pixelY)]);
}
}
I am trying to implement a c++ function (code above) based on the following pseudo-code
which is supposed to create a swirl on an image, but I have some continuity problems on the borders.
The function I have for the moment is able to apply the swirl on a disk of a given size and to deform it almost as I whished but its influence doesn't decrease as we get close to the borders. I tried to multiply the angle of rotation by a 1 - (r/R) factor (with r the distance between the current pixel in the function and the center of the swirl, and R the radius of the swirl), but this doesn't give the effect I hoped for.
Moreover, I noticed that at some parts of the border, a thin white line appears (which means that the values of the pixels there is equal to 1) and I can't exactly explain why.
Maybe some of the problems I have are linked to the atan2 C++ standard function.
I have a 13 x 13 array of pixels, and I am using a function to draw a circle onto them. (The screen is 13 * 13, which may seem strange, but its an array of LED's so that explains it.)
unsigned char matrix[13][13];
const unsigned char ON = 0x01;
const unsigned char OFF = 0x00;
Here is the first implementation I thought up. (It's inefficient, which is a particular problem as this is an embedded systems project, 80 MHz processor.)
// Draw a circle
// mode is 'ON' or 'OFF'
inline void drawCircle(float rad, unsigned char mode)
{
for(int ix = 0; ix < 13; ++ ix)
{
for(int jx = 0; jx < 13; ++ jx)
{
float r; // Radial
float s; // Angular ("theta")
matrix_to_polar(ix, jx, &r, &s); // Converts polar coordinates
// specified by r and s, where
// s is the angle, to index coordinates
// specified by ix and jx.
// This function just converts to
// cartesian and then translates by 6.0.
if(r < rad)
{
matrix[ix][jx] = mode; // Turn pixel in matrix 'ON' or 'OFF'
}
}
}
}
I hope that's clear. It's pretty simple, but then I programmed it so I know how it's supposed to work. If you'd like more info / explanation then I can add some more code / comments.
It can be considered that drawing several circles, eg 4 to 6, is very slow... Hence I'm asking for advice on a more efficient algorithm to draw the circles.
EDIT: Managed to double the performance by making the following modification:
The function calling the drawing used to look like this:
for(;;)
{
clearAll(); // Clear matrix
for(int ix = 0; ix < 6; ++ ix)
{
rad[ix] += rad_incr_step;
drawRing(rad[ix], rad[ix] - rad_width);
}
if(rad[5] >= 7.0)
{
for(int ix = 0; ix < 6; ++ ix)
{
rad[ix] = rad_space_step * (float)(-ix);
}
}
writeAll(); // Write
}
I added the following check:
if(rad[ix] - rad_width < 7.0)
drawRing(rad[ix], rad[ix] - rad_width);
This increased the performance by a factor of about 2, but ideally I'd like to make the circle drawing more efficient to increase it further. This checks to see if the ring is completely outside of the screen.
EDIT 2: Similarly adding the reverse check increased performance further.
if(rad[ix] >= 0.0)
drawRing(rad[ix], rad[ix] - rad_width);
Performance is now pretty good, but again I have made no modifications to the actual drawing code of the circles and this is what I was intending to focus on with this question.
Edit 3: Matrix to polar:
inline void matrix_to_polar(int i, int j, float* r, float* s)
{
float x, y;
matrix_to_cartesian(i, j, &x, &y);
calcPolar(x, y, r, s);
}
inline void matrix_to_cartesian(int i, int j, float* x, float* y)
{
*x = getX(i);
*y = getY(j);
}
inline void calcPolar(float x, float y, float* r, float* s)
{
*r = sqrt(x * x + y * y);
*s = atan2(y, x);
}
inline float getX(int xc)
{
return (float(xc) - 6.0);
}
inline float getY(int yc)
{
return (float(yc) - 6.0);
}
In response to Clifford that's actually a lot of function calls if they are not inlined.
Edit 4: drawRing just draws 2 circles, firstly an outer circle with mode ON and then an inner circle with mode OFF. I am fairly confident that there is a more efficient method of drawing such a shape too, but that distracts from the question.
You're doing a lot of calculations that aren't really needed. For example, you're calculating the angle of the polar coordinates, but never use it. The square root can also easily be avoided by comparing the square of the values.
Without doing anything fancy, something like this should be a good start:
int intRad = (int)rad;
int intRadSqr = (int)(rad * rad);
for (int ix = 0; ix <= intRad; ++ix)
{
for (int jx = 0; jx <= intRad; ++jx)
{
if (ix * ix + jx * jx <= radSqr)
{
matrix[6 - ix][6 - jx] = mode;
matrix[6 - ix][6 + jx] = mode;
matrix[6 + ix][6 - jx] = mode;
matrix[6 + ix][6 + jx] = mode;
}
}
}
This does all the math in integer format, and takes advantage of the circle symmetry.
Variation of the above, based on feedback in the comments:
int intRad = (int)rad;
int intRadSqr = (int)(rad * rad);
for (int ix = 0; ix <= intRad; ++ix)
{
for (int jx = 0; ix * ix + jx * jx <= radSqr; ++jx)
{
matrix[6 - ix][6 - jx] = mode;
matrix[6 - ix][6 + jx] = mode;
matrix[6 + ix][6 - jx] = mode;
matrix[6 + ix][6 + jx] = mode;
}
}
Don't underestimate the cost of even basic arithmetic using floating point on a processor with no FPU. It seems unlikely that floating point is necessary, but the details of its use are hidden in your matrix_to_polar() implementation.
Your current implementation considers every pixel as a candidate - that is also unnecessary.
Using the equation y = cy ± √[rad2 - (x-cx)2] where cx, cy is the centre (7, 7 in this case), and a suitable integer square root implementation, the circle can be drawn thus:
void drawCircle( int rad, unsigned char mode )
{
int r2 = rad * rad ;
for( int x = 7 - rad; x <= 7 + rad; x++ )
{
int dx = x - 7 ;
int dy = isqrt( r2 - dx * dx ) ;
matrix[x][7 - dy] = mode ;
matrix[x][7 + dy] = mode ;
}
}
In my test I used the isqrt() below based on code from here, but given that the maximum r2 necessary is 169 (132, you could implement a 16 or even 8 bit optimised version if necessary. If your processor is 32 bit, this is probably fine.
uint32_t isqrt(uint32_t n)
{
uint32_t root = 0, bit, trial;
bit = (n >= 0x10000) ? 1<<30 : 1<<14;
do
{
trial = root+bit;
if (n >= trial)
{
n -= trial;
root = trial+bit;
}
root >>= 1;
bit >>= 2;
} while (bit);
return root;
}
All that said, on such a low resolution device, you will probably get better quality circles and faster performance by hand generating bitmap lookup tables for each radius required. If memory is an issue, then a single circle needs only 7 bytes to describe a 7 x 7 quadrant that you can reflect to all three quadrants, or for greater performance you could use 7 x 16 bit words to describe a semi-circle (since reversing bit order is more expensive than reversing array access - unless you are using an ARM Cortex-M with bit-banding). Using semi-circle look-ups, 13 circles would need 13 x 7 x 2 bytes (182 bytes), quadrant look-ups would be 7 x 8 x 13 (91 bytes) - you may find that is fewer bytes that the code space required to calculate the circles.
For a slow embedded device with only a 13x13 element display, you should really just make a look-up table. For example:
struct ComputedCircle
{
float rMax;
char col[13][2];
};
Where the draw routine uses rMax to determine which LUT element to use. For example, if you have 2 elements with one rMax = 1.4f, the other = 1.7f, then any radius between 1.4f and 1.7f will use that entry.
The column elements would specify zero, one, or two line segments per row, which can be encoded in the lower and upper 4 bits of each char. -1 can be used as a sentinel value for nothing-at-this-row. It is up to you how many look-up table entries to use, but with a 13x13 grid you should be able to encode every possible outcome of pixels with well under 100 entries, and a reasonable approximation using only 10 or so. You can also trade off compression for draw speed as well, e.g. putting the col[13][2] matrix in a flat list and encoding the number of rows defined.
I would accept MooseBoy's answer if only he explained the method he proposes better. Here's my take on the lookup table approach.
Solve it with a lookup table
The 13x13 display is quite small, and if you only need circles which are fully visible within this pixel count, you will get around with a quite small table. Even if you need larger circles, it should be still better than any algorithmic way if you need it to be fast (and have the ROM to store it).
How to do it
You basically need to define how each possible circle looks like on the 13x13 display. It is not sufficient to just produce snapshots for the 13x13 display, as it is likely you would like to plot the circles at arbitrary positions. My take for a table entry would look like this:
struct circle_entry_s{
unsigned int diameter;
unsigned int offset;
};
The entry would map a given diameter in pixels to offsets in a large byte table containing the shape of the circles. For example for diameter 9, the byte sequence would look like this:
0x1CU, 0x00U, /* 000111000 */
0x63U, 0x00U, /* 011000110 */
0x41U, 0x00U, /* 010000010 */
0x80U, 0x80U, /* 100000001 */
0x80U, 0x80U, /* 100000001 */
0x80U, 0x80U, /* 100000001 */
0x41U, 0x00U, /* 010000010 */
0x63U, 0x00U, /* 011000110 */
0x1CU, 0x00U, /* 000111000 */
The diameter specifies how many bytes of the table belong to the circle: one row of pixels are generated from (diameter + 7) >> 3 bytes, and the number of rows correspond to the diameter. The output code of these can be made quite fast, while the lookup table is sufficiently compact to get even larger than the 13x13 display circles defined in it if needed.
Note that defining circles this way for odd and even diameters may or may not appeal you when output by a centre location. The odd diameter circles will appear to have a centre in the "middle" of a pixel, while the even diameter circles will appear to have their centre on the "corner" of a pixel.
You may also find it nice later to refine the overall method so having multiple circles of different apparent sizes, but having the same pixel radius. Depends on what is your goal: if you want some kind of smooth animation, you may get there eventually.
Algorithmic solutions I think mostly will perform poorly here, since with this limited display surface really every pixel's state counts for the appearance.
I'm trying to implement a very simple 1-dimensional gradient descent algorithm. The code I have does not work at all. Basically depending on my alpha value, the end parameters will either be wildly huge (like ~70 digits), or basically zero (~ 0.000). I feel like a gradient descent should not be nearly this sensitive in alpha (I'm generating small data in [0.0,1.0], but I think the gradient itself should account for the scale of the data, no?).
Here's the code:
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <vector>
using namespace std;
double a, b;
double theta0 = 0.0, theta1 = 0.0;
double myrand() {
return double(rand()) / RAND_MAX;
}
double f(double x) {
double y = a * x + b;
y *= 0.1 * (myrand() - 0.5); // +/- 5% noise
return y;
}
double h(double x) {
return theta1 * x + theta0;
}
int main() {
srand(time(NULL));
a = myrand();
b = myrand();
printf("set parameters: a = %lf, b = %lf\n", a, b);
int N = 100;
vector<double> xs(N);
vector<double> ys(N);
for (int i = 0; i < N; ++i) {
xs[i] = myrand();
ys[i] = f(xs[i]);
}
double sensitivity = 0.008;
double d0, d1;
for (int n = 0; n < 100; ++n) {
d0 = d1 = 0.0;
for (int i = 0; i < N; ++i) {
d0 += h(xs[i]) - ys[i];
d1 += (h(xs[i]) - ys[i]) * xs[i];
}
theta0 -= sensitivity * d0;
theta1 -= sensitivity * d1;
printf("theta0: %lf, theta1: %lf\n", theta0, theta1);
}
return 0;
}
Changing the value of alpha can produce the algorithm to diverge, so that may be one of the causes of what is happening. You can check by computing the error in each iteration and see if is increasing or decreasing.
In adition, it is recommended to set randomly the values of theta at the beginning in stead of assigning them to zero.
Apart from that, you should divide by N when you update the value of theta as follows:
theta0 -= sensitivity * d0/N;
theta1 -= sensitivity * d1/N;
I had a quick look at your implementation and it looks fine to me.
The code I have does not work at all.
I wouldn't say that. It seems to behave correctly for small enough values of sensitivity, which is a value that you just have to "guess", and that is how the gradient descent is supposed to work.
I feel like a gradient descent should not be nearly this sensitive in alpha
If you struggle to visualize that, remember that you are using gradient descent to find the minimum of the cost function of linear regression, which is a quadratic function. If you plot the cost function you will see why the learning rate is so sensitive in these cases: intuitively, if the parabola is narrow, the algorithm will converge more quickly, which is good, but then the learning rate is more "sensitive" and the algorithm can easily diverge if you are not careful.
The problem to solve is finding the floating status of a floating body, given its weight and the center of gravity.
The function i use calculates the displaced volume and center of bouyance of the body given sinkage, heel and trim.
Where sinkage is a length unit and heel/trim is an angle limited to a value from -90 to 90.
The floating status is found when displaced volum is equal to weight and the center of gravity is in a vertical line with center of bouancy.
I have this implemeted as a non-linear Newton-Raphson root finding problem with 3 variables (sinkage, trim, heel) and 3 equations.
This method works, but needs good initial guesses. So I am hoping to find either a better approach for this, or a good method to find the initial values.
Below is the code for the newton and jacobian algorithm used for the Newton-Raphson iteration. The function volume takes the parameters sinkage, heel and trim. And returns volume, and the coordinates for center of bouyancy.
I also included the maxabs and GSolve2 algorithms, I belive these are taken from Numerical Recipies.
void jacobian(float x[], float weight, float vcg, float tcg, float lcg, float jac[][3], float f0[]) {
float h = 0.0001f;
float temp;
float j_volume, j_vcb, j_lcb, j_tcb;
float f1[3];
volume(x[0], x[1], x[2], j_volume, j_lcb, j_vcb, j_tcb);
f0[0] = j_volume-weight;
f0[1] = j_tcb-tcg;
f0[2] = j_lcb-lcg;
for (int i=0;i<3;i++) {
temp = x[i];
x[i] = temp + h;
volume(x[0], x[1], x[2], j_volume, j_lcb, j_vcb, j_tcb);
f1[0] = j_volume-weight;
f1[1] = j_tcb-tcg;
f1[2] = j_lcb-lcg;
x[i] = temp;
jac[0][i] = (f1[0]-f0[0])/h;
jac[1][i] = (f1[1]-f0[1])/h;
jac[2][i] = (f1[2]-f0[2])/h;
}
}
void newton(float weight, float vcg, float tcg, float lcg, float &sinkage, float &heel, float &trim) {
float x[3] = {10,1,1};
float accuracy = 0.000001f;
int ntryes = 30;
int i = 0;
float jac[3][3];
float max;
float f0[3];
float gauss_f0[3];
while (i < ntryes) {
jacobian(x, weight, vcg, tcg, lcg, jac, f0);
if (sqrt((f0[0]*f0[0]+f0[1]*f0[1]+f0[2]*f0[2])/2) < accuracy) {
break;
}
gauss_f0[0] = -f0[0];
gauss_f0[1] = -f0[1];
gauss_f0[2] = -f0[2];
GSolve2(jac, 3, gauss_f0);
x[0] = x[0]+gauss_f0[0];
x[1] = x[1]+gauss_f0[1];
x[2] = x[2]+gauss_f0[2];
// absmax(x) - Return absolute max value from an array
max = absmax(x);
if (max < 1) max = 1;
if (sqrt((gauss_f0[0]*gauss_f0[0]+gauss_f0[1]*gauss_f0[1]+gauss_f0[2]*gauss_f0[2])) < accuracy*max) {
x[0]=x2[0];
x[1]=x2[1];
x[2]=x2[2];
break;
}
i++;
}
sinkage = x[0];
heel = x[1];
trim = x[2];
}
int GSolve2(float a[][3],int n,float b[]) {
float x,sum,max,temp;
int i,j,k,p,m,pos;
int nn = n-1;
for (k=0;k<=n-1;k++)
{
/* pivot*/
max=fabs(a[k][k]);
pos=k;
for (p=k;p<n;p++){
if (max < fabs(a[p][k])){
max=fabs(a[p][k]);
pos=p;
}
}
if (ABS(a[k][pos]) < EPS) {
writeLog("Matrix is singular");
break;
}
if (pos != k) {
for(m=k;m<n;m++){
temp=a[pos][m];
a[pos][m]=a[k][m];
a[k][m]=temp;
}
}
/* convert to upper triangular form */
if ( fabs(a[k][k])>=1.e-6)
{
for (i=k+1;i<n;i++)
{
x = a[i][k]/a[k][k];
for (j=k+1;j<n;j++) a[i][j] = a[i][j] -a[k][j]*x;
b[i] = b[i] - b[k]*x;
}
}
else
{
writeLog("zero pivot found in line:%d",k);
return 0;
}
}
/* back substitution */
b[nn] = b[nn] / a[nn][nn];
for (i=n-2;i>=0;i--)
{
sum = b[i];
for (j=i+1;j<n;j++)
sum = sum - a[i][j]*b[j];
b[i] = sum/a[i][i];
}
return 0;
}
float absmax(float x[]) {
int i = 1;
int n = sizeof(x);
float max = x[0];
while (i < n) {
if (max < x[i]) {
max = x[i];
}
i++;
}
return max;
}
Have you considered some stochastic search methods to find the initial value and then fine-tuning with Newton Raphson? One possibility is evolutionary computation, you can use the Inspyred package. For a physical problem similar in many ways to the one you describe, look at this example: http://inspyred.github.com/tutorial.html#lunar-explorer
What about using a damped version of Newton's method? You could quite easily modify your implementation to make it. Think about Newton's method as finding a direction
d_k = f(x_k) / f'(x_k)
and updating the variable
x_k+1 = x_k - L_k d_k
In the usual Newton's method, L_k is always 1, but this might create overshoots or undershoots. So, let your method chose L_k. Suppose that your method usually overshoots. A possible strategy consists in taking the largest L_k in the set {1,1/2,1/4,1/8,... L_min} such that the condition
|f(x_k+1)| <= (1-L_k/2) |f(x_k)|
is satisfied (or L_min if none of the values satisfies this criteria).
With the same criteria, another possible strategy is to start with L_0=1 and if the criteria is not met, try with L_0/2 until it works (or until L_0 = L_min). Then for L_1, start with min(1, 2L_0) and do the same. Then start with L_2=min(1, 2L_1) and so on.
By the way: are you sure that your problem has a unique solution? I guess that the answer to this question depends on the shape of your object. If you have a rugby ball, there's one angle that you cannot fix. So if your shape is close to such an object, I would not be surprised that the problem is difficult to solve for that angle.
I am trying to implement the Winding Number Algorithm to test if a point is within another polygon. Although the results from my algorithm are wrong and not consistent. I have been working on this for ages now and it has become a bit of a pain!
I have basically converted pseudo code from notes and websites, such as, softsurfer.com
I successfully detect if my player and building object bounding boxes overlap. I return the result to a struct, (BoxResult) which lets me know if there has been a collision and returns the box which it has collided with (Below)
struct BoxResult{
bool collide;
Building returned;
};
void buildingCollision(){
int wn = 0; //winding number count
BoxResult detect = boxDetection(); //detect if any bounding boxes intersect
if(detect.collide){ //If a bounding box has collided, excute Winding Number Algorithm.
for(int i = 0; i < player.getXSize(); i++){
Point p;
p.x = player.getXi(i);
p.y = player.getYi(i);
wn = windingNum(detect.returned,p);
cout << wn << endl;
//Continue code to figure out rebound reaction
}
}
}
I then test for a collision between the building and the player (Below). I have tried 5 different attempts and hours of debugging to understand where the error is occuring, however I am implementing the most ineffienct method which just uses maths (Below).
int windingNum(Building & b, Point & p){
int result = 0; //Winding number is one, if point is in poly
float total; //Counts the total angle between different vertexs
double wn;
for(int i = 0; i <= b.getXSize()-1;i++){
float acs, nom, modPV, modPV1, denom, angle;
if(i == 3){
//Create the different points PVi . PVi+1
Point PV, PV1;
PV.x = (b.getXi(i) + wx) * p.x;
PV.y = (b.getYi(i) + wy) * p.y;
PV1.x = (b.getXi(0) + wx) * p.x;
PV1.y = (b.getYi(0) + wy) * p.y;
modPV = sqrt( (PV.x * PV.x) + (PV.y * PV.y)); //Get the modulus of PV
modPV1 = sqrt( (PV1.x * PV1.x) + (PV1.y * PV1.y)); //Get modulus of PV1
nom = (PV1.x * PV.x) + (PV1.y * PV.y); //Dot product of PV and PV1
denom = modPV * modPV1; //denomintor of winding number equation
angle = nom / denom;
acs = acos(angle) * 180/PI; //find the angle between the different points
total = total + acs; //add this angle, to the total angle count
}
if(i < 3){
//Create the different points PVi . PVi+1
Point PV, PV1;
PV.x = (b.getXi(i) + wx) * p.x;
PV.y = (b.getYi(i) + wy) * p.y;
PV1.x = (b.getXi(i+1) +wx) * p.x;
PV1.y = (b.getYi(i+1) +wy) * p.y;
modPV = sqrt((PV.x * PV.x) + (PV.y * PV.y)); //Get the modulus of PV
modPV1 = sqrt((PV1.x * PV1.x) + (PV1.y * PV1.y)); //Get modulus of PV1
nom = (PV1.x * PV.x) + (PV1.y * PV.y); //Dot product of PV and PV1
denom = modPV * modPV1; //denomintor of winding number equation
angle = nom / denom;
acs = acos(angle) * 180/PI; //find the angle between the different points
total = total + acs; //add this angle, to the total angle count
}
}
wn = total;
if(wn < 360){
result = 0;}
if(wn == 360){
result = 1;}
return result;
}
For reasons I do not understand acs = acos(angle) always returns 1.#IND000.
Btw so you know, I am just testing the algorithm against another square, hence the two if statements if i == 3 and if i < 3.
Also incase you need to know these, wy and wx are the world co-ordinates which are translated. Thus moving the player around the world e.g. to move the player forward everything is translated by a minus number for wy.
Further, a Building object would look something like the following struct below:
struct Building {
vector<float> x; //vector storing x co-ords
vector<float> y; //vector storing y co-ords
float ymax, ymin, xmax, xmin //values for bounding box
vector<int> polygons; //stores the number points per polygon (not relevant to the problem)
}
If anyone can help I would amazingly grateful! I just wish I could see where it is all going wrong! (Something I am sure all programmers have said in there time lol) Thanks for readings...
The two lines calculating the modulus of PV and PV1 are incorrect. They should be
modPV = sqrt(PV.x * PV.x + PV.y * PV.y );
modPV1 = sqrt(PV1.x * PV1.x + PV1.y * PV1.y);
Does that fix the problem?
I probably don't understand your problem/question, but there's a simple and robust point in polygon test available here: PNPOLY.
As regards your implementation of the crossing number algorithm the first obvious mistake is that you are not looping over all the sides. You are one short. You should loop up to i < n and then define i plus one as
int ip1 = ( i + 1 ) % n;
This applies to the code in your original question too of course to save you having to have two copies of the code.
The second one is that
rem = cn % 1;
has no effect. The code on softsurfer is fine i.e.
rem = (cn&1);
It is trying to detect if cn is odd or even by testing if the last bit is set. If you want to the same test using the modulo operator % then you should write it as
rem = cn % 2;
as that assigns the remainder on division by two of cn to rem.
I haven't looked beyond that to see if there are any more issues.
I have given up with the winding number code, it really has got me! If anyone does find the solution I would still be amazingly grateful. I am now trying with point in poly detection using the crossing number algorithm. I kept the pesudo code in the comments, again from softsurfer....
int cn_PnPoly( Point P, Building & b, int n )
{
int cn = 0; // the crossing number counter
int rem = 0;
vector<float>x;
vector<float>y;
x.swap(b.getX());
y.swap(b.getY());
//// loop through all edges of the polygon
//for (int i=0; i<n; i++) { // edge from V[i] to V[i+1]
// if (((V[i].y <= P.y) && (V[i+1].y > P.y)) // an upward crossing
// || ((V[i].y > P.y) && (V[i+1].y <= P.y))) { // a downward crossing
// // compute the actual edge-ray intersect x-coordinate
// float vt = (float)(P.y - V[i].y) / (V[i+1].y - V[i].y);
// if (P.x < V[i].x + vt * (V[i+1].x - V[i].x)) // P.x < intersect
// ++cn; // a valid crossing of y=P.y right of P.x
// }
//}
//return (cn&1); // 0 if even (out), and 1 if odd (in)
// loop through all edges of the polygon
for (int i=0; i<n-1; i++) { // edge from V[i] to V[i+1]
if (((y.at(i) <= P.y) && (y.at(i+1) > P.y)) // an upward crossing
|| ((y.at(i) > P.y) && (y.at(i+1) <= P.y))) { // a downward crossing
// compute the actual edge-ray intersect x-coordinate
float vt = (float)(P.y - y.at(i)) / (y.at(i+1) - y.at(i));
if (P.x < x.at(i) + vt * (x.at(i+1) - x.at(i))) // P.x < intersect
++cn; // a valid crossing of y=P.y right of P.x
}
}
rem = cn % 1;
return (rem); // 0 if even (out), and 1 if odd (in)
}
Again this always returns zero, I am unsure why!?! Have I converted the algorithm incorrectly? Does it matter which direction the points are tested (i.e. clockwise, anti-clockwise)?
I have tried implementing PNPOLY as audris suggests. However this gives some funny results.
Below is the orginal C code, then below that is my conversion of that for my app...
Original C code...
int pnpoly(int nvert, float *vertx, float *verty, float testx, float testy)
{
int i, j, c = 0;
for (i = 0, j = nvert-1; i < nvert; j = i++) {
if ( ((verty[i]>testy) != (verty[j]>testy)) &&
(testx < (vertx[j]-vertx[i]) * (testy-verty[i]) / (verty[j]-verty[i]) + vertx[i]) )
c = !c;
}
return c;
}
My code....
Where wx and wy are the global co-ordinates.
int pnpoly(int nvert, vector<float> vertx, vector<float> verty, float testx, float testy)
{
int i, j, c = 0;
for (i = 0, j = nvert-1; i < nvert; j = i++) {
if ( (( (verty.at(i)+wy) > testy) != ( (verty.at(j)+wy) >testy)) &&
(testx < ((vertx.at(j)+wx) - (vertx.at(i)+wx) ) * (testy- (verty.at(i)+wy) ) / ( (verty.at(j)+wy) - (verty.at(i)+wy)) + (vertx.at(i)+wx)) )
c++;
}
return c;
}
I am testing the player object, against a 2D square building. This also returns strange results, when I hit bottom line (xmin,ymin to xmax,ymin) it works fine. If I hit ethier of the sides (xmin,ymin to xmin,ymax or xmax,ymin to xmax,ymax) it returns 1 only if the player is so far in its past the orgin point. Also on side (xmin,ymin to xmin,ymax) where the player enters the bounding box the algorithm returns 2 despite to hitting the polygon. On the top side, (xmin,ymax to xmax,ymax) it returns 1 only if the player is totally in the polygon.
Also i pass two vectors x and y which are from the Building object, and the vector size as int nvert. Could any of this be to do with the heading of the player object? How is the accounted for within the algorithm?
Hi have done as Troubadour has suggested concerning the crossing number algorithm and made several changes, however the if statement never returns true for some reason. I post of the new code is below. Btw thanks again for everyones replies :-)
int cn_PnPoly( Point P, Building & b, int n )
{
int cn = 0; // the crossing number counter
int rem = 0;
vector<float>x;
vector<float>y;
x.swap(b.getX());
y.swap(b.getY());
//// loop through all edges of the polygon
//for (int i=0; i<n; i++) { // edge from V[i] to V[i+1]
// if (((V[i].y <= P.y) && (V[i+1].y > P.y)) // an upward crossing
// || ((V[i].y > P.y) && (V[i+1].y <= P.y))) { // a downward crossing
// // compute the actual edge-ray intersect x-coordinate
// float vt = (float)(P.y - V[i].y) / (V[i+1].y - V[i].y);
// if (P.x < V[i].x + vt * (V[i+1].x - V[i].x)) // P.x < intersect
// ++cn; // a valid crossing of y=P.y right of P.x
// }
//}
//return (cn&1); // 0 if even (out), and 1 if odd (in)
// loop through all edges of the polygon
for (int i=0; i<n; i++) { // edge from V[i] to V[i+1]
int ip1 = (i +1) %n;
if (((y.at(i) <= P.y) && (y.at(ip1) > P.y)) // an upward crossing
|| ((y.at(i) > P.y) && (y.at(ip1) <= P.y))) { // a downward crossing
// compute the actual edge-ray intersect x-coordinate
float vt = (float)(P.y - y.at(i)) / (y.at(ip1) - y.at(i));
if (P.x < x.at(i) + vt * (x.at(ip1) - x.at(i))) // P.x < intersect
++cn; // a valid crossing of y=P.y right of P.x
}
}
rem = (cn&1);
return (rem); // 0 if even (out), and 1 if odd (in)
}
Below I corrected the code, I forgot to add the world co-ords into account. Yet another silly silly error...
int cn_PnPoly( Point P, Building & b, int n )
{
int cn = 0; // the crossing number counter
int rem = 0;
vector<float>x;
vector<float>y;
x.swap(b.getX());
y.swap(b.getY());
// loop through all edges of the polygon
for (int i=0; i<n; i++) { // edge from V[i] to V[i+1]
int ip1 = (i +1) %n;
if ((( (y.at(i)+wy) <= P.y) && ( (y.at(ip1)+wy) > P.y)) // an upward crossing
|| (( (y.at(i)+wy) > P.y) && ( (y.at(ip1)+wy) <= P.y))) { // a downward crossing
// compute the actual edge-ray intersect x-coordinate
float vt = (float)(P.y - (y.at(i)+wy) ) / ( (y.at(ip1)+wy) - (y.at(i)+wy) );
if (P.x < (x.at(i)+wx) + vt * ( (x.at(ip1)+wx) - (x.at(i)+wx) )) // P.x < intersect
++cn; // a valid crossing of y=P.y right of P.x
}
}
rem = (cn&1);
return (rem); // 0 if even (out), and 1 if odd (in)
}
Although this works to detect when a point is in a polygon, it does not take into account the current heading of the player.
If this doesn't make sense, in the 2D game I move the world map around the player by translating all the polygons by the world co-ordinates. These are wx and wy in the game.
Also I rotate the player about a heading varriable.
These are figured out within the draw function, however the collision detection function does not take the heading into account. To do this do I symply multiply the x and y co-ord given by the Building object by the heading? Unfortunately I am not very good at geometry.