Polar Coordinate Map Generation - algorithm

I am currently work some sort of map generation algorithm for my game. I have a basic understanding on what I want it to do and how it would generate the map.
I want to use the Polar Coordinate system. I want a circular graph so that each player would spawn on the edge of the circle, evenly spread out.
The algorithm should generate "cities" spread out from across the circle (but only inside the circle). Each city should be connected some form of way.
The size of the circle should depends on the number of players.
Everything should be random, meaning if I run
GenerateMap()
two times, it should not give the same results.
Here is a picture showing what I want: img
The red arrows are pointing to the cities and the lines are the connections between the cities.
How would I go about creating an algorithm based on the above?
Update: Sorry the link was broken. Fixed it.

I see the cities like this:
compute sizes and constants from N
as your cities should have constant average density then the radius can be computed from it directly. as it scales linearly with average or min city distance.
loop N (cities) times
generate random (x,y) with uniform distribution
throw away iterations where (x,y) is outside circle
throw away iterations where (x,y) is too near to already generated city
The paths are similar just generate all possible paths (non random) and throw away:
paths much longer then average or min distance between cities (connects jutst neighbors)
paths that intersect already generated path
In C++ code it could look like this:
//---------------------------------------------------------------------------
// some globals first
const int _max=128; // just max limit for cities and paths
const int R0=10; // city radius
const int RR=R0*R0*9; // min distance^2 between cities
int N=20; // number of cities
int R1=100; // map radius
struct _city { int x,y; }; // all the data you need for city
_city city[_max]; // list of cities
struct _path { int i0,i1; };// index of cities to join with path
_path path[_max]; // list of paths
int M=0; // number of paths in the list
//---------------------------------------------------------------------------
bool LinesIntersect(float X1,float Y1,float X2,float Y2,float X3,float Y3,float X4,float Y4)
{
if (fabs(X2-X3)+fabs(Y2-Y3)<1e-3) return false;
if (fabs(X1-X4)+fabs(Y1-Y4)<1e-3) return false;
float dx1,dy1,dx2,dy2;
dx1 = X2 - X1;
dy1 = Y2 - Y1;
dx2 = X4 - X3;
dy2 = Y4 - Y3;
float s,t,ax,ay,b;
ax=X1-X3;
ay=Y1-Y3;
b=(-(dx2*dy1)+(dx1*dy2)); if (fabs(b)>1e-3) b=1.0/b; else b=0.0;
s = (-(dy1*ax)+(dx1*ay))*b;
t = ( (dx2*ay)-(dy2*ax))*b;
if ((s>=0)&&(s<=1)&&(t>=0)&&(t<=1)) return true;
return false; // No collision
}
//---------------------------------------------------------------------------
// here generate n cities into circle at x0,y0
// compute R1,N from R0,RR,n
void genere(int x0,int y0,int n)
{
_city c;
_path p,*q;
int i,j,cnt,x,y,rr;
Randomize(); // init pseudo random generator
// [generate cities]
R1=(sqrt(RR*n)*8)/10;
rr=R1-R0; rr*=rr;
for (cnt=50*n,i=0;i<n;) // loop until all cities are generated
{
// watchdog
cnt--; if (cnt<=0) break;
// pseudo random position
c.x=Random(R1+R1)-R1; // <-r,+r>
c.y=Random(R1+R1)-R1; // <-r,+r>
// ignore cities outside R1 radius
if ((c.x*c.x)+(c.y*c.y)>rr) continue;
c.x+=x0; // position to center
c.y+=y0;
// ignore city if closer to any other then RR
for (j=0;j<i;j++)
{
x=c.x-city[j].x;
y=c.y-city[j].y;
if ((x*x)+(y*y)<=RR) { j=-1; break; }
}
if (j<0) continue;
// add new city to list
city[i]=c; i++;
}
N=i; // just in case watch dog kiks in
// [generate paths]
for (M=0,p.i0=0;p.i0<N;p.i0++)
for (p.i1=p.i0+1;p.i1<N;p.i1++)
{
// ignore too long path
x=city[p.i1].x-city[p.i0].x;
y=city[p.i1].y-city[p.i0].y;
if ((x*x)+(y*y)>5*RR) continue; // this constant determine the path density per city
// ignore intersecting path
for (q=path,i=0;i<M;i++,q++)
if ((q->i0!=p.i0)&&(q->i0!=p.i1)&&(q->i1!=p.i0)&&(q->i1!=p.i1))
if (LinesIntersect(
city[p.i0].x,city[p.i0].y,city[p.i1].x,city[p.i1].y,
city[q->i0].x,city[q->i0].y,city[q->i1].x,city[q->i1].y
)) { i=-1; break; }
if (i<0) continue;
// add path to list
if (M>=_max) break;
path[M]=p; M++;
}
}
//---------------------------------------------------------------------------
Here overview of generated layout:
And Growth of N:
The blue circles are the cities, the gray area is the target circle and Lines are the paths. The cnt is just watch dog to avoid infinite loop if constants are wrong. Set the _max value properly so it is high enough for your N or use dynamic allocation instead. There is much more paths than cities so they could have separate _max value to preserve memory (was too lazy to add it).
You can use the RandSeed to have procedural generated maps ...
You can rescale output to better match circle layout after the generation simply by finding bounding box and rescale to radius R1.
Some constants are obtained empirically so play with them to achieve best output for your purpose.

Related

How to Compute OBB of Multiple Curves?

Given a number of curves,include line segments and circular arcs, how to compute the total OBB of all curves?
It seems that the union of each OBB of the individual curves does not right, it's not the minimal coverage.
Check this picture, how to compute the red box?
you should also add the input in vector form so we can test on your data ... I would approach like this:
find center of axis aligned bbox O(n)
compute max distance in each angle O(n)
just create table for enough m angles (like 5 deg step so m = 360/5) where for each angle section you remember max distant point distance only.
compute max perpendicular distance for each rotation O(m^2)
so for each angle section compute value that is:
value[actual_section] = max(distance[i]*cos(section_angle[i]-section_angle[actual_section]))
where i covers +/- 90 deg around actual section angle so now you got max perpendicular distances for each angle...
pick best solution O(m)
so look all rotations from 0 to 90 degrees and remember the one that has minimal OBB area. Just to be sure the OBB is aligned to section angle and size of axises is the value of that angle and all the 90 deg increments... around center
This will not result in optimal solution but very close to it. To improve precision you can use more angle sections or even recursively search around found solution with smaller and smaller angle step (no need to compute the other angle areas after first run.
[Edit1]
I tried to code this in C++ as proof of concept and use your image (handled as set of points) as input so here the result so you got something to compare to (for debugging purposes)
gray are detected points from your image, green rectangle is axis aligned BBox the red rectangle is found OBBox. The aqua points are found max distance per angle interval and green dots are max perpendicular distance for +/-90deg neighbor angle intervals. I used 400 angles and as you can see the result is pretty close ... 360/400 deg accuracy so this approach works well ...
Here C++ source:
//---------------------------------------------------------------------------
struct _pnt2D
{
double x,y;
// inline
_pnt2D() {}
_pnt2D(_pnt2D& a) { *this=a; }
~_pnt2D() {}
_pnt2D* operator = (const _pnt2D *a) { *this=*a; return this; }
//_pnt2D* operator = (const _pnt2D &a) { ...copy... return this; }
};
struct _ang
{
double ang; // center angle of section
double dis; // max distance of ang section
double pdis; // max perpendicular distance of +/-90deg section
// inline
_ang() {}
_ang(_ang& a) { *this=a; }
~_ang() {}
_ang* operator = (const _ang *a) { *this=*a; return this; }
//_ang* operator = (const _ang &a) { ...copy... return this; }
};
const int angs=400; // must be divisible by 4
const int angs4=angs>>2;
const double dang=2.0*M_PI/double(angs);
const double dang2=0.5*dang;
_ang ang[angs];
List<_pnt2D> pnt;
_pnt2D bbox[2],obb[4],center;
//---------------------------------------------------------------------------
void compute_OBB()
{
_pnt2D ppp[4];
int i,j; double a,b,dx,dy;
_ang *aa,*bb;
_pnt2D p,*pp; DWORD *q;
// convert bmp -> pnt[]
pnt.num=0;
Graphics::TBitmap *bmp=new Graphics::TBitmap;
bmp->LoadFromFile("in.bmp");
bmp->HandleType=bmDIB;
bmp->PixelFormat=pf32bit;
for (p.y=0;p.y<bmp->Height;p.y++)
for (q=(DWORD*)bmp->ScanLine[int(p.y)],p.x=0;p.x<bmp->Width;p.x++)
if ((q[int(p.x)]&255)<20)
pnt.add(p);
delete bmp;
// axis aligned bbox
bbox[0]=pnt[0];
bbox[1]=pnt[0];
for (pp=pnt.dat,i=0;i<pnt.num;i++,pp++)
{
if (bbox[0].x>pp->x) bbox[0].x=pp->x;
if (bbox[0].y>pp->y) bbox[0].y=pp->y;
if (bbox[1].x<pp->x) bbox[1].x=pp->x;
if (bbox[1].y<pp->y) bbox[1].y=pp->y;
}
center.x=(bbox[0].x+bbox[1].x)*0.5;
center.y=(bbox[0].y+bbox[1].y)*0.5;
// ang[] table init
for (aa=ang,a=0.0,i=0;i<angs;i++,aa++,a+=dang)
{
aa->ang=a;
aa-> dis=0.0;
aa->pdis=0.0;
}
// ang[].dis
for (pp=pnt.dat,i=0;i<pnt.num;i++,pp++)
{
dx=pp->x-center.x;
dy=pp->y-center.y;
a=atan2(dy,dx);
j=floor((a/dang)+0.5); if (j<0) j+=angs; j%=angs;
a=(dx*dx)+(dy*dy);
if (ang[j].dis<a) ang[j].dis=a;
}
for (aa=ang,i=0;i<angs;i++,aa++) aa->dis=sqrt(aa->dis);
// ang[].adis
for (aa=ang,i=0;i<angs;i++,aa++)
for (bb=ang,j=0;j<angs;j++,bb++)
{
a=fabs(aa->ang-bb->ang);
if (a>M_PI) a=(2.0*M_PI)-a;
if (a<=0.5*M_PI)
{
a=bb->dis*cos(a);
if (aa->pdis<a) aa->pdis=a;
}
}
// find best oriented bbox (the best angle is ang[j].ang)
for (b=0,j=0,i=0;i<angs;i++)
{
dx =ang[i].pdis; i+=angs4; i%=angs;
dy =ang[i].pdis; i+=angs4; i%=angs;
dx+=ang[i].pdis; i+=angs4; i%=angs;
dy+=ang[i].pdis; i+=angs4; i%=angs;
a=dx*dy; if ((b>a)||(i==0)) { b=a; j=i; }
}
// compute endpoints for OBB
i=j;
ppp[0].x=ang[i].pdis*cos(ang[i].ang);
ppp[0].y=ang[i].pdis*sin(ang[i].ang); i+=angs4; i%=angs;
ppp[1].x=ang[i].pdis*cos(ang[i].ang);
ppp[1].y=ang[i].pdis*sin(ang[i].ang); i+=angs4; i%=angs;
ppp[2].x=ang[i].pdis*cos(ang[i].ang);
ppp[2].y=ang[i].pdis*sin(ang[i].ang); i+=angs4; i%=angs;
ppp[3].x=ang[i].pdis*cos(ang[i].ang);
ppp[3].y=ang[i].pdis*sin(ang[i].ang); i+=angs4; i%=angs;
obb[0].x=center.x+ppp[0].x+ppp[3].x;
obb[0].y=center.y+ppp[0].y+ppp[3].y;
obb[1].x=center.x+ppp[1].x+ppp[0].x;
obb[1].y=center.y+ppp[1].y+ppp[0].y;
obb[2].x=center.x+ppp[2].x+ppp[1].x;
obb[2].y=center.y+ppp[2].y+ppp[1].y;
obb[3].x=center.x+ppp[3].x+ppp[2].x;
obb[3].y=center.y+ppp[3].y+ppp[2].y;
}
//---------------------------------------------------------------------------
I used mine dynamic list template so:
List<double> xxx; is the same as double xxx[];
xxx.add(5); adds 5 to end of the list
xxx[7] access array element (safe)
xxx.dat[7] access array element (unsafe but fast direct access)
xxx.num is the actual used size of the array
xxx.reset() clears the array and set xxx.num=0
xxx.allocate(100) preallocate space for 100 items
You can ignore the // convert bmp -> pnt[] VCL part as you got your data already.
I recommend to also take a look at my:
3D OBB approximation

Infinite Blue Noise

I am looking for an algorithm which produces a point placement result similar to blue noise.
However, it needs to work for an infinite plane. Where a bounding box is given, and it returns the locations of all points which fall inside. Any help would be appreciated. I've done a lot of research, and found nothing that suits my needs.
Finally I've managed to get results.
One way of generating point distributions with blue noise properties
is by means of a Poisson-Disk Distribution
Following the algorithm from the paper Fast Poisson disk sampling in
arbitrary dimensions, Robert Bridson I've got:
The steps mentioned in the paper are:
Step 0. Initialize an n-dimensional background grid for storing
samples and accelerating spatial searches. We pick the cell size to be
bounded by r/sqrt(n), so that each grid cell will contain at most one
sample, and thus the grid can be implemented as a simple n-dimensional
array of integers: the default −1 indicates no sample, a non-negative
integer gives the index of the sample located in a cell
Step 1. Select the initial sample, x0, randomly chosen uniformly from
the domain. Insert it into the background grid, and initialize the
“active list” (an array of sample indices) with this index (zero).
Step 2. While the active list is not empty, choose a random index from
it (say i). Generate up to k points chosen uniformly from the
spherical annulus between radius r and 2r around xi. For each point in
turn, check if it is within distance r of existing samples (using the
background grid to only test nearby samples). If a point is adequately
far from existing samples, emit it as the next sample and add it to
the active list. If after k attempts no such point is found, instead
remove i from the active list.
Note that for simplicity I've skipped step 0. Despite that the run-time is still reasonable. It's < .5s. Implementing this step would most definitely increase the performance.
Here's a sample code in Processing. It's a language built on top of Java so the syntax is very similar. Translating it for your purposes shouldn't be hard.
import java.util.List;
import java.util.Collections;
List<PVector> poisson_disk_sampling(int k, int r, int size)
{
List<PVector> samples = new ArrayList<PVector>();
List<PVector> active_list = new ArrayList<PVector>();
active_list.add(new PVector(random(size), random(size)));
int len;
while ((len = active_list.size()) > 0) {
// picks random index uniformly at random from the active list
int index = int(random(len));
Collections.swap(active_list, len-1, index);
PVector sample = active_list.get(len-1);
boolean found = false;
for (int i = 0; i < k; ++i) {
// generates a point uniformly at random in the sample's
// disk situated at a distance from r to 2*r
float angle = 2*PI*random(1);
float radius = random(r) + r;
PVector dv = new PVector(radius*cos(angle), radius*sin(angle));
PVector new_sample = dv.add(sample);
boolean ok = true;
for (int j = 0; j < samples.size(); ++j) {
if (dist(new_sample.x, new_sample.y,
samples.get(j).x, samples.get(j).y) <= r)
{
ok = false;
break;
}
}
if (ok) {
if (0 <= new_sample.x && new_sample.x < size &&
0 <= new_sample.y && new_sample.y < size)
{
samples.add(new_sample);
active_list.add(new_sample);
len++;
found = true;
}
}
}
if (!found)
active_list.remove(active_list.size()-1);
}
return samples;
}
List<PVector> samples;
void setup() {
int SIZE = 500;
size(500, 500);
background(255);
strokeWeight(4);
noLoop();
samples = poisson_disk_sampling(30, 10, SIZE);
}
void draw() {
for (PVector sample : samples)
point(sample.x, sample.y);
}
However, it needs to work for an infinite plane.
You control the size of the box with the parameter size. r controls the relative distance between the points. k controls how many new sample should you try before rejecting the current. The paper suggests k=30.

Algorithm for shape calculation (Ellipse)

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

How to smooth the blocks of a 3D voxel world?

In my (Minecraft-like) 3D voxel world, I want to smooth the shapes for more natural visuals. Let's look at this example in 2D first.
Left is how the world looks without any smoothing. The terrain data is binary and each voxel is rendered as a unit size cube.
In the center you can see a naive circular smoothing. It only takes the four directly adjacent blocks into account. It is still not very natural looking. Moreover, I'd like to have flat 45-degree slopes emerge.
On the right you can see a smoothing algorithm I came up with. It takes the eight direct and diagonal neighbors into account in order to come up with the shape of a block. I have the C++ code online. Here is the code that comes up with the control points that the bezier curve is drawn along.
#include <iostream>
using namespace std;
using namespace glm;
list<list<dvec2>> Points::find(ivec2 block)
{
// Control points
list<list<ivec2>> lines;
list<ivec2> *line = nullptr;
// Fetch blocks, neighbours start top left and count
// around the center block clock wise
int center = m_blocks->get(block);
int neighs[8];
for (int i = 0; i < 8; i++) {
auto coord = blockFromIndex(i);
neighs[i] = m_blocks->get(block + coord);
}
// Iterate over neighbour blocks
for (int i = 0; i < 8; i++) {
int current = neighs[i];
int next = neighs[(i + 1) % 8];
bool is_side = (((i + 1) % 2) == 0);
bool is_corner = (((i + 1) % 2) == 1);
if (line) {
// Border between air and ground needs a line
if (current != center) {
// Sides are cool, but corners get skipped when they don't
// stop a line
if (is_side || next == center)
line->push_back(blockFromIndex(i));
} else if (center || is_side || next == center) {
// Stop line since we found an end of the border. Always
// stop for ground blocks here, since they connect over
// corners so there must be open docking sites
line = nullptr;
}
} else {
// Start a new line for the border between air and ground that
// just appeared. However, corners get skipped if they don't
// end a line.
if (current != center) {
lines.emplace_back();
line = &lines.back();
line->push_back(blockFromIndex(i));
}
}
}
// Merge last line with first if touching. Only close around a differing corner for air
// blocks.
if (neighs[7] != center && (neighs[0] != center || (!center && neighs[1] != center))) {
// Skip first corner if enclosed
if (neighs[0] != center && neighs[1] != center)
lines.front().pop_front();
if (lines.size() == 1) {
// Close circle
auto first_point = lines.front().front();
lines.front().push_back(first_point);
} else {
// Insert last line into first one
lines.front().insert(lines.front().begin(), line->begin(), line->end());
lines.pop_back();
}
}
// Discard lines with too few points
auto i = lines.begin();
while (i != lines.end()) {
if (i->size() < 2)
lines.erase(i++);
else
++i;
}
// Convert to concrete points for output
list<list<dvec2>> points;
for (auto &line : lines) {
points.emplace_back();
for (auto &neighbour : line)
points.back().push_back(pointTowards(neighbour));
}
return points;
}
glm::ivec2 Points::blockFromIndex(int i)
{
// Returns first positive representant, we need this so that the
// conditions below "wrap around"
auto modulo = [](int i, int n) { return (i % n + n) % n; };
ivec2 block(0, 0);
// For two indices, zero is right so skip
if (modulo(i - 1, 4))
// The others are either 1 or -1
block.x = modulo(i - 1, 8) / 4 ? -1 : 1;
// Other axis is same sequence but shifted
if (modulo(i - 3, 4))
block.y = modulo(i - 3, 8) / 4 ? -1 : 1;
return block;
}
dvec2 Points::pointTowards(ivec2 neighbour)
{
dvec2 point;
point.x = static_cast<double>(neighbour.x);
point.y = static_cast<double>(neighbour.y);
// Convert from neighbour space into
// drawing space of the block
point *= 0.5;
point += dvec2(.5);
return point;
}
However, this is still in 2D. How to translate this algorithm into three dimensions?
You should probably have a look at the marching cubes algorithm and work from there. You can easily control the smoothness of the resulting blob:
Imagine that each voxel defines a field, with a high density at it's center, slowly fading to nothing as you move away from the center. For example, you could use a function that is 1 inside a voxel and goes to 0 two voxels away. No matter what exact function you choose, make sure that it's only non-zero inside a limited (preferrably small) area.
For each point, sum the densities of all fields.
Use the marching cubes algorithm on the sum of those fields
Use a high resolution mesh for the algorithm
In order to change the look/smoothness you change the density function and the threshold of the marching cubes algorithm. A possible extension to marching cubes to create smoother meshes is the following idea: Imagine that you encounter two points on an edge of a cube, where one point lies inside your volume (above a threshold) and the other outside (under the threshold). In this case many marching cubes algorithms place the boundary exactly at the middle of the edge. One can calculate the exact boundary point - this gets rid of aliasing.
Also I would recommend that you run a mesh simplification algorithm after that. Using marching cubes results in meshes with many unnecessary triangles.
As an alternative to my answer above: You could also use NURBS or any algorithm for subdivision surfaces. Especially the subdivision surfaces algorithms are spezialized to smooth meshes. Depending on the algorithm and it's configuration you will get smoother versions of your original mesh with
the same volume
the same surface
the same silhouette
and so on.
Use 3D implementations for Biezer curves known as Biezer surfaces or use the B-Spline Surface algorithms explained:
here
or
here

Finding closest non-black pixel in an image fast

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.

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