I'm trying to create a smooth "wave" when the mouse moves over isometric logo shape.
I've created in in processing now I'm trying to recreate it in THREE.js
The shape acts strangely - the shape doesn't look as smooth when elevated compared to the processing sketch. If you look at the edges you can see segments that are not supposed to be there. I'm not sure what causes this.
Basically the shape is created through a loops that goes over 2 arrays:
for (var i = 0; i < xpos0.length; i++) {
shape.lineTo(xpos0[i], ypos0[i]);
}
Then it animates through another for loop that checks the distance between verteces[i].x and mouse location intersection with the ground
for (let p = 0; p < mesh.geometry.vertices.length; p=p+1) {
let delta = Math.abs(mesh.geometry.vertices[p].x - intersects[0].point.x);
mesh.geometry.vertices[p].z = bump2(-2, 2000, -1, 2, delta);
}
z value is calculated through this function:
function bump2(a,b,c,d,xval) {
xval = parseFloat(xval);
// console.log(typeof xval);
return Math.exp(a / (-xval * xval / b + c) + d) * -1;
}
https://codepen.io/NotYetDesignLab/pen/JjYaqRJ
How it looks on THREE.JS
notice how some segments appear "broken", like it's made of stiff parts instead of the many points that make up the segment in the array and give the illusion of "paper".
THIS IS HOW IT'S SUPPOSED TO LOOK: (Processing/java)
This has been done using Processing. Notice how the elevation of the edges is smooth and not broken.
I have the following problem as shown in the figure. I have point cloud and a mesh generated by a tetrahedral algorithm. How would I carve the mesh using the that algorithm ? Are landmarks are the point cloud ?
Pseudo code of the algorithm:
for every 3D feature point
convert it 2D projected coordinates
for every 2D feature point
cast a ray toward the polygons of the mesh
get intersection point
if zintersection < z of 3D feature point
for ( every triangle vertices )
cull that triangle.
Here is a follow up implementation of the algorithm mentioned by the Guru Spektre :)
Update code for the algorithm:
int i;
for (i = 0; i < out.numberofpoints; i++)
{
Ogre::Vector3 ray_pos = pos; // camera position);
Ogre::Vector3 ray_dir = (Ogre::Vector3 (out.pointlist[(i*3)], out.pointlist[(3*i)+1], out.pointlist[(3*i)+2]) - pos).normalisedCopy(); // vertex - camea pos ;
Ogre::Ray ray;
ray.setOrigin(Ogre::Vector3( ray_pos.x, ray_pos.y, ray_pos.z));
ray.setDirection(Ogre::Vector3(ray_dir.x, ray_dir.y, ray_dir.z));
Ogre::Vector3 result;
unsigned int u1;
unsigned int u2;
unsigned int u3;
bool rayCastResult = RaycastFromPoint(ray.getOrigin(), ray.getDirection(), result, u1, u2, u3);
if ( rayCastResult )
{
Ogre::Vector3 targetVertex(out.pointlist[(i*3)], out.pointlist[(3*i)+1], out.pointlist[(3*i)+2]);
float distanceTargetFocus = targetVertex.squaredDistance(pos);
float distanceIntersectionFocus = result.squaredDistance(pos);
if(abs(distanceTargetFocus) >= abs(distanceIntersectionFocus))
{
if ( u1 != -1 && u2 != -1 && u3 != -1)
{
std::cout << "Remove index "<< "u1 ==> " <<u1 << "u2 ==>"<<u2<<"u3 ==> "<<u3<< std::endl;
updatedIndices.erase(updatedIndices.begin()+ u1);
updatedIndices.erase(updatedIndices.begin()+ u2);
updatedIndices.erase(updatedIndices.begin()+ u3);
}
}
}
}
if ( updatedIndices.size() <= out.numberoftrifaces)
{
std::cout << "current face list===> "<< out.numberoftrifaces << std::endl;
std::cout << "deleted face list===> "<< updatedIndices.size() << std::endl;
manual->begin("Pointcloud", Ogre::RenderOperation::OT_TRIANGLE_LIST);
for (int n = 0; n < out.numberofpoints; n++)
{
Ogre::Vector3 vertexTransformed = Ogre::Vector3( out.pointlist[3*n+0], out.pointlist[3*n+1], out.pointlist[3*n+2]) - mReferencePoint;
vertexTransformed *=1000.0 ;
vertexTransformed = mDeltaYaw * vertexTransformed;
manual->position(vertexTransformed);
}
for (int n = 0 ; n < updatedIndices.size(); n++)
{
int n0 = updatedIndices[n+0];
int n1 = updatedIndices[n+1];
int n2 = updatedIndices[n+2];
if ( n0 < 0 || n1 <0 || n2 <0 )
{
std::cout<<"negative indices"<<std::endl;
break;
}
manual->triangle(n0, n1, n2);
}
manual->end();
Follow up with the algorithm:
I have now two versions one is the triangulated one and the other is the carved version.
It's not not a surface mesh.
Here are the two files
http://www.mediafire.com/file/cczw49ja257mnzr/ahmed_non_triangulated.obj
http://www.mediafire.com/file/cczw49ja257mnzr/ahmed_triangulated.obj
I see it like this:
So you got image from camera with known matrix and FOV and focal length.
From that you know where exactly the focal point is and where the image is proected onto the camera chip (Z_near plane). So any vertex, its corresponding pixel and focal point lies on the same line.
So for each view cas ray from focal point to each visible vertex of the pointcloud. and test if any face of the mesh hits before hitting face containing target vertex. If yes remove it as it would block the visibility.
Landmark in this context is just feature point corresponding to vertex from pointcloud. It can be anything detectable (change of intensity, color, pattern whatever) usually SIFT/SURF is used for this. You should have them located already as that is the input for pointcloud generation. If not you can peek pixel corresponding to each vertex and test for background color.
Not sure how you want to do this without the input images. For that you need to decide which vertex is visible from which side/view. May be it is doable form nearby vertexes somehow (like using vertex density points or corespondence to planar face...) or the algo is changed somehow for finding unused vertexes inside mesh.
To cast a ray do this:
ray_pos=tm_eye*vec4(imgx/aspect,imgy,0.0,1.0);
ray_dir=ray_pos-tm_eye*vec4(0.0,0.0,-focal_length,1.0);
where tm_eye is camera direct transform matrix, imgx,imgy is the 2D pixel position in image normalized to <-1,+1> where (0,0) is the middle of image. The focal_length determines the FOV of camera and aspect ratio is ratio of image resolution image_ys/image_xs
Ray triangle intersection equation can be found here:
Reflection and refraction impossible without recursive ray tracing?
If I extract it:
vec3 v0,v1,v2; // input triangle vertexes
vec3 e1,e2,n,p,q,r;
float t,u,v,det,idet;
//compute ray triangle intersection
e1=v1-v0;
e2=v2-v0;
// Calculate planes normal vector
p=cross(ray[i0].dir,e2);
det=dot(e1,p);
// Ray is parallel to plane
if (abs(det)<1e-8) no intersection;
idet=1.0/det;
r=ray[i0].pos-v0;
u=dot(r,p)*idet;
if ((u<0.0)||(u>1.0)) no intersection;
q=cross(r,e1);
v=dot(ray[i0].dir,q)*idet;
if ((v<0.0)||(u+v>1.0)) no intersection;
t=dot(e2,q)*idet;
if ((t>_zero)&&((t<=tt)) // tt is distance to target vertex
{
// intersection
}
Follow ups:
To move between normalized image (imgx,imgy) and raw image (rawx,rawy) coordinates for image of size (imgxs,imgys) where (0,0) is top left corner and (imgxs-1,imgys-1) is bottom right corner you need:
imgx = (2.0*rawx / (imgxs-1)) - 1.0
imgy = 1.0 - (2.0*rawy / (imgys-1))
rawx = (imgx + 1.0)*(imgxs-1)/2.0
rawy = (1.0 - imgy)*(imgys-1)/2.0
[progress update 1]
I finally got to the point I can compile sample test input data for this to get even started (as you are unable to share valid data at all):
I created small app with hard-coded table mesh (gray) and pointcloud (aqua) and simple camera control. Where I can save any number of views (screenshot + camera direct matrix). When loaded back it aligns with the mesh itself (yellow ray goes through aqua dot in image and goes through the table mesh too). The blue lines are casted from camera focal point to its corners. This will emulate the input you got. The second part of the app will use only these images and matrices with the point cloud (no mesh surface anymore) tetragonize it (already finished) now just cast ray through each landmark in each view (aqua dot) and remove all tetragonals before target vertex in pointcloud is hit (this stuff is not even started yet may be in weekend)... And lastly store only surface triangles (easy just use all triangles which are used just once also already finished except the save part but to write wavefront obj from it is easy ...).
[Progress update 2]
I added landmark detection and matching with the point cloud
as you can see only valid rays are cast (those that are visible on image) so some points on point cloud does not cast rays (singular aqua dots)). So now just the ray/triangle intersection and tetrahedron removal from list is what is missing...
I am trying to write a C source code visualization program which I expect to draw out function hierarchy using spheres inside spheres.
For a simple example, in code like this:
#include "stdio.h"
int pow(int base, int power) {
while(--power) {
base*=base;
}
return base;
}
int checkOdd(int i) {
if (i%2==0) return 0;
else return 1;
}
int checkPrime(int i) {
int j = i;
if (!checkOdd(i)) {
return 0;
}
for(j=1; j<i/2; j++) {
if (i%j==0) {
return 0;
}
}
return 1;
}
int main() {
int a = 3;
int b = 2;
int res = pow(a,b);
int bl = checkPrime(res);
printf("%d",res);
return 1;
}
The largest sphere which is the main function has two functions inside it, pow and checkPrime, which are two spheres inside main's sphere. Checkprime function's sphere has checkOdd's sphere in it.
I would like it so that the children spheres are somewhat clumped together in one side of the larger sphere, because I would like some space left out for putting in other things. However, the spheres must not touch each other, and the radius of the spheres are pre-determined and can not change to accomodate drawing accurately. I need an algorithm which determines the perfect center coordinates which would allow the smaller spheres not to touch each other while still being collected to one side of the larger sphere.
I have a 3D vector which shows the direction at which the smaller spheres must be concentrated in, looking from the center of the larger sphere. In the case of the picture, my 3D vector is facing bottom-down in a 2D-sense looking at the screen.
Currently my algorithm produces this by dispersing the sphere's center around a circle projected onto the larger sphere's surface, but it fails as the spheres overlap. Can anyone enlighten me with a way to disperse smaller spheres to one direction within the larger sphere?
I am using OpenGL and I prepared a function which can draw those transparent mesh spheres by feeding center coordinate and radius. I have knowledge of parent sphere's center and radius in designing this algorithm.
I have n circles that must be perfectly surrounding an ellipse as shown in the picture here :
In this picture I need to find out the position of each circle around the ellipse, and also be able to calculate the ellipse that will fit perfectly inside those surrounding circles.
The information i know is the radius of each circles (all same), and the number of circles.
Hopefully this time the post is clear.
Thanks for your help.
Please let me know if you need more explanation.
OK as i understand you know common radius of circles R0 and their number N and want to know inside ellipse parameters and positions of everything.
If we convert ellipse to circle then we get this:
const int N=12; // number of satelite circles
const double R=10.0; // radius of satelite circles
struct _circle { double x,y,r; } circle[N]; // satelite circles
int i;
double x,y,r,l,a,da;
x=0.0; // start pos of first satelite circle
y=0.0;
r=R;
l=r+r; // distance ang angle between satelite circle centers
a=0.0*deg;
da=divide(360.0*deg,N);
for (i=0;i<N;i++)
{
circle[i].x=x; x+=l*cos(a);
circle[i].y=y; y+=l*sin(a);
circle[i].r=r; a+=da;
}
// inside circle params
_circle c;
r=divide(0.5*l,sin(0.5*da))-R;
c.x=circle[i].x;
c.y=circle[i].y+R+r;
c.r=r;
[Edit 1]
For ellipse its a whole new challenge (took me two hours to find all quirks out)
const int N=20; // number of satelite circles
const double R=10.0; // satelite circles radius
const double E= 0.7; // ellipse distortion ry=rx*E
struct _circle { double x,y,r; _circle() { x=0; y=0; r=0.0; } } circle[N];
struct _ellipse { double x,y,rx,ry; _ellipse() { x=0; y=0; rx=0.0; ry=0.0; } } ellipse;
int i,j,k;
double l,a,da,m,dm,x,y,q,r0;
l=double(N)*R; // circle cener lines polygon length
ellipse.x =0.0; // set ellipse parameters
ellipse.y =0.0;
r0=divide(l,M_PI*sqrt(0.5*(1.0+(E*E))))-R;// aprox radius to match ellipse length for start
l=R+R; l*=l;
m=1.0; dm=1.0; x=0.0;
for (k=0;k<5;k++) // aproximate ellipse size to the right size
{
dm=fabs(0.1*dm); // each k-iteration layer is 10x times more accurate
if (x>l) dm=-dm;
for (;;)
{
ellipse.rx=r0 *m;
ellipse.ry=r0*E*m;
for (a=0.0,i=0;i<N;i++) // set circle parameters
{
q=(2.0*a)-atanxy(cos(a),sin(a)*E);
circle[i].x=ellipse.x+(ellipse.rx*cos(a))+(R*cos(q));
circle[i].y=ellipse.y+(ellipse.ry*sin(a))+(R*sin(q));
circle[i].r=R;
da=divide(360*deg,N); a+=da;
for (j=0;j<5;j++) // aproximate next position to match 2R distance from current position
{
da=fabs(0.1*da); // each j-iteration layer is 10x times more accurate
q=(2.0*a)-atanxy(cos(a),sin(a)*E);
x=ellipse.x+(ellipse.rx*cos(a))+(R*cos(q))-circle[i].x; x*=x;
y=ellipse.y+(ellipse.ry*sin(a))+(R*sin(q))-circle[i].y; y*=y; x+=y;
if (x>l) for (;;) // if too far dec angle
{
a-=da;
q=(2.0*a)-atanxy(cos(a),sin(a)*E);
x=ellipse.x+(ellipse.rx*cos(a))+(R*cos(q))-circle[i].x; x*=x;
y=ellipse.y+(ellipse.ry*sin(a))+(R*sin(q))-circle[i].y; y*=y; x+=y;
if (x<=l) break;
}
else if (x<l) for (;;) // if too short inc angle
{
a+=da;
q=(2.0*a)-atanxy(cos(a),sin(a)*E);
x=ellipse.x+(ellipse.rx*cos(a))+(R*cos(q))-circle[i].x; x*=x;
y=ellipse.y+(ellipse.ry*sin(a))+(R*sin(q))-circle[i].y; y*=y; x+=y;
if (x>=l) break;
}
else break;
}
}
// check if last circle is joined as it should be
x=circle[N-1].x-circle[0].x; x*=x;
y=circle[N-1].y-circle[0].y; y*=y; x+=y;
if (dm>0.0) { if (x>=l) break; }
else { if (x<=l) break; }
m+=dm;
}
}
Well I know its a little messy code so here is some info:
first it try to set as close ellipse rx,ry axises as possible
ellipse length should be about N*R*2 which is polygon length of lines between circle centers
try to compose circles so they are touching each other and the ellipse
I use iteration of ellipse angle for that. Problem is that circles do not touch the ellipse in their position angle thats why there is q variable ... to compensate around ellipse normal. Look for yellowish-golden lines in image
after placing circles check if the last one is touching the first
if not interpolate the size of ellipse actually it scales the rx,ry by m variable up or down
you can adjust accuracy
by change of the j,k fors and/or change of dm,da scaling factors
input parameter E should be at least 0.5 and max 1.0
if not then there is high probability of misplacing circles because on very eccentric ellipses is not possible to fit circles (if N is too low). Ideal setting is 0.7<=E<=1.0 closser to 1 the safer the algorithm is
atanxy(dx,dy) is the same as `atan(dy/dx)
but it handles all 4 quadrants like atan2(dy,dx) by sign analysis of dx,dy
Hope it helps
I have a 2D image randomly and sparsely scattered with pixels.
given a point on the image, I need to find the distance to the closest pixel that is not in the background color (black).
What is the fastest way to do this?
The only method I could come up with is building a kd-tree for the pixels. but I would really want to avoid such expensive preprocessing. also, it seems that a kd-tree gives me more than I need. I only need the distance to something and I don't care about what this something is.
Personally, I'd ignore MusiGenesis' suggestion of a lookup table.
Calculating the distance between pixels is not expensive, particularly as for this initial test you don't need the actual distance so there's no need to take the square root. You can work with distance^2, i.e:
r^2 = dx^2 + dy^2
Also, if you're going outwards one pixel at a time remember that:
(n + 1)^2 = n^2 + 2n + 1
or if nx is the current value and ox is the previous value:
nx^2 = ox^2 + 2ox + 1
= ox^2 + 2(nx - 1) + 1
= ox^2 + 2nx - 1
=> nx^2 += 2nx - 1
It's easy to see how this works:
1^2 = 0 + 2*1 - 1 = 1
2^2 = 1 + 2*2 - 1 = 4
3^2 = 4 + 2*3 - 1 = 9
4^2 = 9 + 2*4 - 1 = 16
5^2 = 16 + 2*5 - 1 = 25
etc...
So, in each iteration you therefore need only retain some intermediate variables thus:
int dx2 = 0, dy2, r2;
for (dx = 1; dx < w; ++dx) { // ignoring bounds checks
dx2 += (dx << 1) - 1;
dy2 = 0;
for (dy = 1; dy < h; ++dy) {
dy2 += (dy << 1) - 1;
r2 = dx2 + dy2;
// do tests here
}
}
Tada! r^2 calculation with only bit shifts, adds and subtracts :)
Of course, on any decent modern CPU calculating r^2 = dx*dx + dy*dy might be just as fast as this...
As Pyro says, search the perimeter of a square that you keep moving out one pixel at a time from your original point (i.e. increasing the width and height by two pixels at a time). When you hit a non-black pixel, you calculate the distance (this is your first expensive calculation) and then continue searching outwards until the width of your box is twice the distance to the first found point (any points beyond this cannot possibly be closer than your original found pixel). Save any non-black points you find during this part, and then calculate each of their distances to see if any of them are closer than your original point.
In an ideal find, you only have to make one expensive distance calculation.
Update: Because you're calculating pixel-to-pixel distances here (instead of arbitrary precision floating point locations), you can speed up this algorithm substantially by using a pre-calculated lookup table (just a height-by-width array) to give you distance as a function of x and y. A 100x100 array costs you essentially 40K of memory and covers a 200x200 square around the original point, and spares you the cost of doing an expensive distance calculation (whether Pythagorean or matrix algebra) for every colored pixel you find. This array could even be pre-calculated and embedded in your app as a resource, to spare you the initial calculation time (this is probably serious overkill).
Update 2: Also, there are ways to optimize searching the square perimeter. Your search should start at the four points that intersect the axes and move one pixel at a time towards the corners (you have 8 moving search points, which could easily make this more trouble than it's worth, depending on your application's requirements). As soon as you locate a colored pixel, there is no need to continue towards the corners, as the remaining points are all further from the origin.
After the first found pixel, you can further restrict the additional search area required to the minimum by using the lookup table to ensure that each searched point is closer than the found point (again starting at the axes, and stopping when the distance limit is reached). This second optimization would probably be much too expensive to employ if you had to calculate each distance on the fly.
If the nearest pixel is within the 200x200 box (or whatever size works for your data), you will only search within a circle bounded by the pixel, doing only lookups and <>comparisons.
You didn't specify how you want to measure distance. I'll assume L1 (rectilinear) because it's easier; possibly these ideas could be modified for L2 (Euclidean).
If you're only doing this for relatively few pixels, then just search outward from the source pixel in a spiral until you hit a nonblack one.
If you're doing this for many/all of them, how about this: Build a 2-D array the size of the image, where each cell stores the distance to the nearest nonblack pixel (and if necessary, the coordinates of that pixel). Do four line sweeps: left to right, right to left, bottom to top, and top to bottom. Consider the left to right sweep; as you sweep, keep a 1-D column containing the last nonblack pixel seen in each row, and mark each cell in the 2-D array with the distance to and/or coordinates of that pixel. O(n^2).
Alternatively, a k-d tree is overkill; you could use a quadtree. Only a little more difficult to code than my line sweep, a little more memory (but less than twice as much), and possibly faster.
Search "Nearest neighbor search", first two links in Google should help you.
If you are only doing this for 1 pixel per image, I think your best bet is just a linear search, 1 pixel width box at time outwards. You can't take the first point you find, if your search box is square. You have to be careful
Yes, the Nearest neighbor search is good, but does not guarantee you'll find the 'nearest'. Moving one pixel out each time will produce a square search - the diagonals will be farther away than the horizontal / vertical. If this is important, you'll want to verify - continue expanding until the absolute horizontal has a distance greater than the 'found' pixel, and then calculate distances on all non-black pixels that were located.
Ok, this sounds interesting.
I made a c++ version of a soulution, I don't know if this helps you. I think it works fast enough as it's almost instant on a 800*600 matrix. If you have any questions just ask.
Sorry for any mistakes I've made, it's a 10min code...
This is a iterative version (I was planing on making a recursive one too, but I've changed my mind).
The algorithm could be improved by not adding any point to the points array that is to a larger distance from the starting point then the min_dist, but this involves calculating for each pixel (despite it's color) the distance from the starting point.
Hope that helps
//(c++ version)
#include<iostream>
#include<cmath>
#include<ctime>
using namespace std;
//ITERATIVE VERSION
//picture witdh&height
#define width 800
#define height 600
//indexex
int i,j;
//initial point coordinates
int x,y;
//variables to work with the array
int p,u;
//minimum dist
double min_dist=2000000000;
//array for memorising the points added
struct point{
int x;
int y;
} points[width*height];
double dist;
bool viz[width][height];
// direction vectors, used for adding adjacent points in the "points" array.
int dx[8]={1,1,0,-1,-1,-1,0,1};
int dy[8]={0,1,1,1,0,-1,-1,-1};
int k,nX,nY;
//we will generate an image with white&black pixels (0&1)
bool image[width-1][height-1];
int main(){
srand(time(0));
//generate the random pic
for(i=1;i<=width-1;i++)
for(j=1;j<=height-1;j++)
if(rand()%10001<=9999) //9999/10000 chances of generating a black pixel
image[i][j]=0;
else image[i][j]=1;
//random coordinates for starting x&y
x=rand()%width;
y=rand()%height;
p=1;u=1;
points[1].x=x;
points[1].y=y;
while(p<=u){
for(k=0;k<=7;k++){
nX=points[p].x+dx[k];
nY=points[p].y+dy[k];
//nX&nY are the coordinates for the next point
//if we haven't added the point yet
//also check if the point is valid
if(nX>0&&nY>0&&nX<width&&nY<height)
if(viz[nX][nY] == 0 ){
//mark it as added
viz[nX][nY]=1;
//add it in the array
u++;
points[u].x=nX;
points[u].y=nY;
//if it's not black
if(image[nX][nY]!=0){
//calculate the distance
dist=(x-nX)*(x-nX) + (y-nY)*(y-nY);
dist=sqrt(dist);
//if the dist is shorter than the minimum, we save it
if(dist<min_dist)
min_dist=dist;
//you could save the coordinates of the point that has
//the minimum distance too, like sX=nX;, sY=nY;
}
}
}
p++;
}
cout<<"Minimum dist:"<<min_dist<<"\n";
return 0;
}
I'm sure this could be done better but here's some code that searches the perimeter of a square around the centre pixel, examining the centre first and moving toward the corners. If a pixel isn't found the perimeter (radius) is expanded until either the radius limit is reached or a pixel is found. The first implementation was a loop doing a simple spiral around the centre point but as noted that doesn't find the absolute closest pixel. SomeBigObjCStruct's creation inside the loop was very slow - removing it from the loop made it good enough and the spiral approach is what got used. But here's this implementation anyway - beware, little to no testing done.
It is all done with integer addition and subtraction.
- (SomeBigObjCStruct *)nearestWalkablePoint:(SomeBigObjCStruct)point {
typedef struct _testPoint { // using the IYMapPoint object here is very slow
int x;
int y;
} testPoint;
// see if the point supplied is walkable
testPoint centre;
centre.x = point.x;
centre.y = point.y;
NSMutableData *map = [self getWalkingMapDataForLevelId:point.levelId];
// check point for walkable (case radius = 0)
if(testThePoint(centre.x, centre.y, map) != 0) // bullseye
return point;
// radius is the distance from the location of point. A square is checked on each iteration, radius units from point.
// The point with y=0 or x=0 distance is checked first, i.e. the centre of the side of the square. A cursor variable
// is used to move along the side of the square looking for a walkable point. This proceeds until a walkable point
// is found or the side is exhausted. Sides are checked until radius is exhausted at which point the search fails.
int radius = 1;
BOOL leftWithinMap = YES, rightWithinMap = YES, upWithinMap = YES, downWithinMap = YES;
testPoint leftCentre, upCentre, rightCentre, downCentre;
testPoint leftUp, leftDown, rightUp, rightDown;
testPoint upLeft, upRight, downLeft, downRight;
leftCentre = rightCentre = upCentre = downCentre = centre;
int foundX = -1;
int foundY = -1;
while(radius < 1000) {
// radius increases. move centres outward
if(leftWithinMap == YES) {
leftCentre.x -= 1; // move left
if(leftCentre.x < 0) {
leftWithinMap = NO;
}
}
if(rightWithinMap == YES) {
rightCentre.x += 1; // move right
if(!(rightCentre.x < kIYMapWidth)) {
rightWithinMap = NO;
}
}
if(upWithinMap == YES) {
upCentre.y -= 1; // move up
if(upCentre.y < 0) {
upWithinMap = NO;
}
}
if(downWithinMap == YES) {
downCentre.y += 1; // move down
if(!(downCentre.y < kIYMapHeight)) {
downWithinMap = NO;
}
}
// set up cursor values for checking along the sides of the square
leftUp = leftDown = leftCentre;
leftUp.y -= 1;
leftDown.y += 1;
rightUp = rightDown = rightCentre;
rightUp.y -= 1;
rightDown.y += 1;
upRight = upLeft = upCentre;
upRight.x += 1;
upLeft.x -= 1;
downRight = downLeft = downCentre;
downRight.x += 1;
downLeft.x -= 1;
// check centres
if(testThePoint(leftCentre.x, leftCentre.y, map) != 0) {
foundX = leftCentre.x;
foundY = leftCentre.y;
break;
}
if(testThePoint(rightCentre.x, rightCentre.y, map) != 0) {
foundX = rightCentre.x;
foundY = rightCentre.y;
break;
}
if(testThePoint(upCentre.x, upCentre.y, map) != 0) {
foundX = upCentre.x;
foundY = upCentre.y;
break;
}
if(testThePoint(downCentre.x, downCentre.y, map) != 0) {
foundX = downCentre.x;
foundY = downCentre.y;
break;
}
int i;
for(i = 0; i < radius; i++) {
if(leftWithinMap == YES) {
// LEFT Side - stop short of top/bottom rows because up/down horizontal cursors check that line
// if cursor position is within map
if(i < radius - 1) {
if(leftUp.y > 0) {
// check it
if(testThePoint(leftUp.x, leftUp.y, map) != 0) {
foundX = leftUp.x;
foundY = leftUp.y;
break;
}
leftUp.y -= 1; // moving up
}
if(leftDown.y < kIYMapHeight) {
// check it
if(testThePoint(leftDown.x, leftDown.y, map) != 0) {
foundX = leftDown.x;
foundY = leftDown.y;
break;
}
leftDown.y += 1; // moving down
}
}
}
if(rightWithinMap == YES) {
// RIGHT Side
if(i < radius - 1) {
if(rightUp.y > 0) {
if(testThePoint(rightUp.x, rightUp.y, map) != 0) {
foundX = rightUp.x;
foundY = rightUp.y;
break;
}
rightUp.y -= 1; // moving up
}
if(rightDown.y < kIYMapHeight) {
if(testThePoint(rightDown.x, rightDown.y, map) != 0) {
foundX = rightDown.x;
foundY = rightDown.y;
break;
}
rightDown.y += 1; // moving down
}
}
}
if(upWithinMap == YES) {
// UP Side
if(upRight.x < kIYMapWidth) {
if(testThePoint(upRight.x, upRight.y, map) != 0) {
foundX = upRight.x;
foundY = upRight.y;
break;
}
upRight.x += 1; // moving right
}
if(upLeft.x > 0) {
if(testThePoint(upLeft.x, upLeft.y, map) != 0) {
foundX = upLeft.x;
foundY = upLeft.y;
break;
}
upLeft.y -= 1; // moving left
}
}
if(downWithinMap == YES) {
// DOWN Side
if(downRight.x < kIYMapWidth) {
if(testThePoint(downRight.x, downRight.y, map) != 0) {
foundX = downRight.x;
foundY = downRight.y;
break;
}
downRight.x += 1; // moving right
}
if(downLeft.x > 0) {
if(testThePoint(upLeft.x, upLeft.y, map) != 0) {
foundX = downLeft.x;
foundY = downLeft.y;
break;
}
downLeft.y -= 1; // moving left
}
}
}
if(foundX != -1 && foundY != -1) {
break;
}
radius++;
}
// build the return object
if(foundX != -1 && foundY != -1) {
SomeBigObjCStruct *foundPoint = [SomeBigObjCStruct mapPointWithX:foundX Y:foundY levelId:point.levelId];
foundPoint.z = [self zWithLevelId:point.levelId];
return foundPoint;
}
return nil;
}
You can combine many ways to speed it up.
A way to accelerate the pixel lookup is to use what I call a spatial lookup map. It is basically a downsampled map (say of 8x8 pixels, but its a tradeoff) of the pixels in that block. Values can be "no pixels set" "partial pixels set" "all pixels set". This way one read can tell if a block/cell is either full, partially full or empty.
scanning a box/rectangle around the center may not be ideal because there are many pixels/cells which are far far away. I use a circle drawing algorithm (Bresenham) to reduce the overhead.
reading the raw pixel values can happen in horizontal batches, for example a byte (for a cell size of 8x8 or multiples of it), dword or long. This should give you a serious speedup again.
you can also use multiple levels of "spatial lookup maps", its again a tradeoff.
For the distance calculatation the mentioned lookup table can be used, but its a (cache)bandwidth vs calculation speed tradeoff (I dunno how it performs on a GPU for example).
Another approach I have investigated and likely will stick to: Utilizing the Bresenham circle algorithm.
It is surprisingly fast as it saves you any sort of distance comparisons!
You effectively just draw bigger and bigger circles around your target point so that when the first time you encounter a non-black pixel you automatically know it is the closest, saving any further checks.
What I have not verified yet is whether the bresenham circle will catch every single pixel but that wasn't a concern for my case as my pixels will occur in blobs of some sort.
I would do a simple lookup table - for every pixel, precalculate distance to the closest non-black pixel and store the value in the same offset as the corresponding pixel. Of course, this way you will need more memory.