Develop Reference

How can I draw the points of a route equidistant of one another?

I am writing a program that outputs the shortest route between two points on a map. The problem is that if the route is too long and it has many points that define its path it slows the program a lot and I am looking for a way to draw just some points instead of all the points in the array.
My approach goes as follows: the map has a zoom, each time the zoom changes check which points overlaps with the others. All the points that doesn't overlap go into the routeToDraw list and then it is drawn.
To check if the points overlap or not I have the following function:
//route is a list of latitude and longitude points
LinkedList<Point.Double> route = MapPanel.this.getGlassPane().getRoute();
LinkedList<Point.Double> routeToDraw = new LinkedList<Point.Double>();
int ovalSize = 8;
boolean compareMorePoints;
for(int i = 0; i < route.size(); i++) {
Point p1 = getScreenCoordinates(route.get(i).x, route.get(i).y);
compareMorePoints = true;
int j = i + 1;
while (j < route.size() && compareMorePoints == true) {
Point p2 = getScreenCoordinates(route.get(j).x, route.get(j).y);
if (Math.sqrt(Math.pow(p1.x - p2.x, 2) + Math.pow(p1.y - p2.y, 2)) > ovalSize ) {
routeToDraw.add(route.get(i));
compareMorePoints = false;
}
j++;
}
}
MapPanel.this.getGlassPane().setRouteToDraw(routeToDraw);
The problem is that this function is quite expensive and although it does reduce the amount of points to draw and I seem to obtain some speed after calculating routeToDraw I don't think it is worth the wait each time I zoom in or out.
The ideal solution would be something like Google Maps' does when routing, drawing a series of equidistant points that modify each time you zoom in or out and look quite nice.

Two suggestions...
(old trick)... Don't do unnecessary math inside of a loop. You can and should eliminate the sqrt function, which is an "expensive" math operation when doing distances. Just compare to the square of ovalSize. It is mathematically equivalent.
Is your list sorted in any way? If there were a convenient point in your program to sort your list (or a copy of it) before displaying, then you could very quickly:
Lop off the first and last part that is outside your zoom window in one of the coordinates (say X, if you sorted by X) by doing a binary search for the window boundary
Tighten up your loop to only look at neighbors within a window of concern, and do a sliding window instead of all-compared-to-all.

Issues with Raytracing triangles (orientation and coloring)

EDIT: I found out that all the pixels were upside down because of the difference between screen and world coordinates, so that is no longer a problem.
EDIT: After following a suggestion from #TheVee (using absolute values), my image got much better, but I'm still seeing issues with color.
I having a little trouble with ray-tracing triangles. This is a follow-up to my previous question about the same topic. The answers to that question made me realize that I needed to take a different approach. The new approach I took worked much better, but I'm seeing a couple of issues with my raytracer now:
There is one triangle that never renders in color (it is always black, even though it's color is supposed to be yellow).
Here is what I am expecting to see:
But here is what I am actually seeing:
Addressing debugging the first problem, even if I remove all other objects (including the blue triangle), the yellow triangle is always rendered black, so I don't believe that it is an issues with my shadow rays that I am sending out. I suspect that it has to do with the angle that the triangle/plane is at relative to the camera.
Here is my process for ray-tracing triangles which is based off of the process in this website.
Determine if the ray intersects the plane.
If it does, determine if the ray intersects inside of the triangle (using parametric coordinates).
Here is the code for determining if the ray hits the plane:
private Vector getPlaneIntersectionVector(Ray ray)
{
double epsilon = 0.00000001;
Vector w0 = ray.getOrigin().subtract(getB());
double numerator = -(getPlaneNormal().dotProduct(w0));
double denominator = getPlaneNormal().dotProduct(ray.getDirection());
//ray is parallel to triangle plane
if (Math.abs(denominator) < epsilon)
{
//ray lies in triangle plane
if (numerator == 0)
{
return null;
}
//ray is disjoint from plane
else
{
return null;
}
}
double intersectionDistance = numerator / denominator;
//intersectionDistance < 0 means the "intersection" is behind the ray (pointing away from plane), so not a real intersection
return (intersectionDistance >= 0) ? ray.getLocationWithMagnitude(intersectionDistance) : null;
}
And once I have determined that the ray intersects the plane, here is the code to determine if the ray is inside the triangle:
private boolean isIntersectionVectorInsideTriangle(Vector planeIntersectionVector)
{
//Get edges of triangle
Vector u = getU();
Vector v = getV();
//Pre-compute unique five dot-products
double uu = u.dotProduct(u);
double uv = u.dotProduct(v);
double vv = v.dotProduct(v);
Vector w = planeIntersectionVector.subtract(getB());
double wu = w.dotProduct(u);
double wv = w.dotProduct(v);
double denominator = (uv * uv) - (uu * vv);
//get and test parametric coordinates
double s = ((uv * wv) - (vv * wu)) / denominator;
if (s < 0 || s > 1)
{
return false;
}
double t = ((uv * wu) - (uu * wv)) / denominator;
if (t < 0 || (s + t) > 1)
{
return false;
}
return true;
}
Is think that I am having some issue with my coloring. I think that it has to do with the normals of the various triangles. Here is the equation I am considering when I am building my lighting model for spheres and triangles:
Now, here is the code that does this:
public Color calculateIlluminationModel(Vector normal, boolean isInShadow, Scene scene, Ray ray, Vector intersectionPoint)
{
//c = cr * ca + cr * cl * max(0, n \dot l)) + cl * cp * max(0, e \dot r)^p
Vector lightSourceColor = getColorVector(scene.getLightColor()); //cl
Vector diffuseReflectanceColor = getColorVector(getMaterialColor()); //cr
Vector ambientColor = getColorVector(scene.getAmbientLightColor()); //ca
Vector specularHighlightColor = getColorVector(getSpecularHighlight()); //cp
Vector directionToLight = scene.getDirectionToLight().normalize(); //l
double angleBetweenLightAndNormal = directionToLight.dotProduct(normal);
Vector reflectionVector = normal.multiply(2).multiply(angleBetweenLightAndNormal).subtract(directionToLight).normalize(); //r
double visibilityTerm = isInShadow ? 0 : 1;
Vector ambientTerm = diffuseReflectanceColor.multiply(ambientColor);
double lambertianComponent = Math.max(0, angleBetweenLightAndNormal);
Vector diffuseTerm = diffuseReflectanceColor.multiply(lightSourceColor).multiply(lambertianComponent).multiply(visibilityTerm);
double angleBetweenEyeAndReflection = scene.getLookFrom().dotProduct(reflectionVector);
angleBetweenEyeAndReflection = Math.max(0, angleBetweenEyeAndReflection);
double phongComponent = Math.pow(angleBetweenEyeAndReflection, getPhongConstant());
Vector phongTerm = lightSourceColor.multiply(specularHighlightColor).multiply(phongComponent).multiply(visibilityTerm);
return getVectorColor(ambientTerm.add(diffuseTerm).add(phongTerm));
}
I am seeing that the dot product between the normal and the light source is -1 for the yellow triangle, and about -.707 for the blue triangle, so I'm not sure if the normal being the wrong way is the problem. Regardless, when I added made sure the angle between the light and the normal was positive (Math.abs(directionToLight.dotProduct(normal));), it caused the opposite problem:
I suspect that it will be a small typo/bug, but I need another pair of eyes to spot what I couldn't.
Note: My triangles have vertices(a,b,c), and the edges (u,v) are computed using a-b and c-b respectively (also, those are used for calculating the plane/triangle normal). A Vector is made up of an (x,y,z) point, and a Ray is made up of a origin Vector and a normalized direction Vector.
Here is how I am calculating normals for all triangles:
private Vector getPlaneNormal()
{
Vector v1 = getU();
Vector v2 = getV();
return v1.crossProduct(v2).normalize();
}
Please let me know if I left out anything that you think is important for solving these issues.
EDIT: After help from #TheVee, this is what I have at then end:
There are still problems with z-buffering, And with phong highlights with the triangles, but the problem I was trying to solve here was fixed.

It is an usual problem in ray tracing of scenes including planar objects that we hit them from a wrong side. The formulas containing the dot product are presented with an inherent assumption that light is incident at the object from a direction to which the outer-facing normal is pointing. This can be true only for half the possible orientations of your triangle and you've been in bad luck to orient it with its normal facing away from the light.
Technically speaking, in a physical world your triangle would not have zero volume. It's composed of some layer of material which is just thin. On either side it has a proper normal that points outside. Assigning a single normal is a simplification that's fair to take because the two only differ in sign.
However, if we made a simplification we need to account for it. Having what technically is an inwards facing normal in our formulas gives negative dot products, which case they are not made for. It's like light was coming from the inside of the object or that it hit a surface could not possibly be in its way. That's why they give an erroneous result. The negative value will subtract light from other sources, and depending on the magnitude and implementation may result in darkening, full black, or numerical underflow.
But because we know the correct normal is either what we're using or its negative, we can simply fix the cases at once by taking a preventive absolute value where a positive dot product is implicitly assumed (in your code, that's angleBetweenLightAndNormal). Some libraries like OpenGL do that for you, and on top use the additional information (the sign) to choose between two different materials (front and back) you may provide if desired. Alternatively, they can be set to not draw the back faces for solid object at all because they will be overdrawn by front faces in solid objects anyway (known as face culling), saving about half of the numerical work.

Searching a grid of points, visting each point once only

I have a grid of (d) dimensions, all dimensions are partitioned using delta = 0.25
An example of this grid is this figure (d here is 2, and each dimension is normalized 0-1):
Each intersection represents 1 point, for example, the point in the middle is represented as:
double[] A={0.5, 0.5};
My question is:
I want to search this grid, starting from an input point A and its neighbors. Then continue doing that. With one condition: Each point is visited only once.
To clarify more, consider this example:
The starting point is:
double[] A={0.5, 0.5};
So, A is checked first, then its neighbours are generated and inserted into a Queue (the queue is ordered based on a function f).
Here point A is the dark circle . Its neighbors are the green circles:
{0.25, 0.25}
{0.25, 0.5}
{0.25, 0.75}
..
..
..
{0.75, 0.75}
Now, the algorithm loops until the Queue becomes empty.
In the loop, the top point is removed (pop()), then checked, then its neighbors are added to the Queue.
For example, in the first loop, the blue circle happened to be the top point in the Queue.
It is removed from the Queue, then checked, then its neighbors (red circles) are generated and added to the Queue.
The problem here is, my code that generates the neighbors points, does not know if a previous point is visited before or not.
and I cannot keep a list of previously visited points, and check it every time a new point is generated ( with high dimensions and high resolution e.g., d=8 and delta= 0.03125, it will take for ever!)
This is the search algorithm:
public void search(int d, double delta, double[] inpoint)
{
Queue queue = new Queue();
queue.push(inpoint);
while( !queue.isEmpty() )
{
// remove top point from Queue
double[] a = queue.pop();
// Check point a and do some calculations
// ;;;;;;;;;;;;;;;;
// Generate neighbours of current point:
ArrayList<double[]> neighbors = new ArrayList<double[]>();
nextMoves(a, new double[d], delta, d, neighbors);
// For each point in neighbors, push to Queue:
for( int i=0 ; i < neighbors.size(), i++ )
queue.push(neighbors.get(i));
}
}
And this is the algorithm for generating the neighbors, it is a recursive algorithm.
public static void nextMoves(double[] inatt, double[] outatt, double delta, int d, ArrayList<double[]> neighbors)
{
if( d == inatt.length )
{
if( !outOfBound(outatt,d) )
{
moves.add(outatt);
}
}
else
{
// first case: add delta:
outatt[d] = inatt[d]+delta;
nextMoves( inatt, outatt, delta, d+1, moves);
// second case: subtract delta:
outatt[d] = inatt[d]-delta;
nextMoves( inatt, outatt, delta, d+1, moves);
// third case: no change:
outatt[d] = inatt[d];
nextMoves( inatt, outatt, delta, d+1, moves);
}
}
As I mentioned before, keeping a list of previously visited points is not a possible solution.
If I do this, then the list becomes very large when I have high dimensionality and high resolution.
And this list will have to be searched linearly each time a point is created!
Perhaps I should organize this list in a spatial index? I don't know...
I would appreciate your input.

There are a couple of potential shortcuts you could consider:
You could remove points from your 'previously visited' list once all surrounding points have been added. The reasoning is simple: they can only be added from surrounding points. This means only the surface of the visited volume needs to be in the list. At higher dimensions this would be a substantial saving.
More complicated would be to store the shape of the volume rather than the points. This would mean recording each inflection point on the surface of the volume and testing if each new point is on that surface.
The first is simpler but not as efficient. I'd suggest starting with that and seeing if it's enough.

You dont have to store all visited points. Going around and collect first all neighbours and build a hashkey out of this points to store this in a hashmap. Then validate all this points with your code and collect the next circle of neighbours. In the worst case you started in the middle. Then the calculationtime of the hashkeys are high at the end of the algortihm. To solve this problem you can build cluster and search within the cluster first.
You can start with clustering too. Build up a cluster layout and start to search within this cluster with the above circle method.

This sounds similar to a pathfinding problem where you might also parse a grid and don't unnecessarily want to revisit nodes.
If it works for you, set a boolean on each visited node, so if the algorithm happens upon it again, it knows it doesn't need to check it.
Look at Djikstras Algorithm for inspiration.

Combining 1x1 tiles with walls in grid into larger rectangular polygons [duplicate]

I have a set of rectangles and I would like to "reduce" the set so I have the fewest number of rectangles to describe the same area as the original set. If possible, I would like it to also be fast, but I am more concerned with getting the number of rectangles as low as possible. I have an approach now which works most of the time.
Currently, I start at the top-left most rectangle and see if I can expand it out right and down while keeping it a rectangle. I do that until it can't expand anymore, remove and split all intersecting rectangles, and add the expanded rectangle back in the list. Then I start the process again with the next top-left most rectangle, and so on. But in some cases, it doesn't work. For example:
With this set of three rectangles, the correct solution would end up with two rectangles, like this:
However, in this case, my algorithm starts by processing the blue rectangle. This expand downwards and splits the yellow rectangle (correctly). But then when the remainder of the yellow rectangle is processed, instead of expanding downwards, it expands right first and takes back the portion that was previously split off. Then the last rectangle is processed and it can't expand right or down, so the original set of rectangles is left. I could tweak the algorithm to expand down first and then right. That would fix this case, but it would cause the same problem in a similar scenario that was flipped.
Edit: Just to clarify, the original set of rectangles do not overlap and do not have to be connected. And if a subset of rectangles are connected, the polygon which completely covers them can have holes in it.

Despite the title to your question, I think you’re actually looking for the minimum dissection into rectangles of a rectilinear polygon. (Jason’s links are about minimum covers by rectangles, which is quite a different problem.)
David Eppstein discusses this problem in section 3 of his 2010 survey article Graph-Theoretic Solutions to Computational Geometry Problems, and he gives a nice summary in this answer on mathoverflow.net:
The idea is to find the maximum number of disjoint axis-parallel diagonals that have two concave vertices as endpoints, split along those, and then form one more split for each remaining concave vertex. To find the maximum number of disjoint axis-parallel diagonals, form the intersection graph of the diagonals; this graph is bipartite so its maximum independent set can be found in polynomial time by graph matching techniques.
Here’s my gloss on this admirably terse description, using figure 2 from Eppstein’s article. Suppose we have a rectilinear polygon, possibly with holes.
When the polygon is dissected into rectangles, each of the concave vertices must be met by at least one edge of the dissection. So we get the minimum dissection if as many of these edges as possible do double-duty, that is, they connect two of the concave vertices.
So let’s draw the axis-parallel diagonals between two concave vertices that are contained entirely within the polygon. (‘Axis-parallel’ means ‘horizontal or vertical’ here, and a diagonal of a polygon is a line connecting two non-adjacent vertices.) We want to use as many of these lines as possible in the dissection as long as they don’t intersect.
(If there are no axis-parallel diagonals, the dissection is trivial—just make a cut from each concave vertex. Or if there are no intersections between the axis-parallel diagonals then we use them all, plus a cut from each remaining concave vertex. Otherwise, read on.)
The intersection graph of a set of line segments has a node for every line segment, and an edge joins two nodes if the lines cross. Here’s the intersection graph for the axis-parallel diagonals:
It’s bipartite with the vertical diagonals in one part, and the horizontal diagonals in the other part. Now, we want to pick as many of the diagonals as possible as long as they don’t intersect. This corresponds to finding the maximum independent set in the intersection graph.
Finding the maximum independent set in a general graph is an NP-hard problem, but in the special case of a bipartite graph, König’s theorem shows that it’s equivalent to the problem of finding a maximum matching, which can be solved in polynomial time, for example by the Hopcroft–Karp algorithm. A given graph can have several maximum matchings, but any of them will do, as they all have the same size. In the example, all the maximum matchings have three pairs of vertices, for example {(2, 4), (6, 3), (7, 8)}:
(Other maximum matchings in this graph include {(1, 3), (2, 5), (7, 8)}; {(2, 4), (3, 6), (5, 7)}; and {(1, 3), (2, 4), (7, 8)}.)
To get from a maximum matching to the corresponding minimum vertex cover, apply the proof of König’s theorem. In the matching shown above, the left set is L = {1, 2, 6, 7}, the right set is R = {3, 4, 5, 8}, and the set of unmatched vertices in L is U = {1}. There is only one alternating path starting in U, namely 1–3–6, so the set of vertices in alternating paths is Z = {1, 3, 6} and the minimum vertex cover is thus K = (L \ Z) ∪ (R ∩ Z) = {2, 3, 7}, shown in red below, with the maximum independent set in green:
Translating this back into the dissection problem, this means that we can use five axis-parallel diagonals in the dissection:
Finally, make a cut from each remaining concave vertex to complete the dissection:

Today I found O(N^5) solution for this problem, and I will share it here.
For the first step, you need to find a way to get the sum of any rectangle in a matrix, with complexity O(1). It's pretty easy to do.
Now for the second step, you need to know dynamic programming. The idea is to store a rectangle and break it into smaller pieces. If the rectangle is empty, you can return 0. And if it's filled, return 1.
There are N^4 states to store the rectangle, plus the O(N) complexity for each state... So you will get an O(N^5) algorithm.
Here's my code. I think it will help.
The input is simple. N, M (size of matrix)
After that, the following N lines will have 1s and 0s.
Example:
4 9
010000010
111010111
101111101
000101000
#include <bits/stdc++.h>
#define MAX 51
int tab[MAX][MAX];
int N,M;
int sumed[MAX][MAX];
int t(int x,int y) {
if(x<0||y<0)return 0;
return sumed[x][y];
}
int subrec(int x1,int y1,int x2,int y2) {
return t(x2,y2)-t(x2,y1-1)-t(x1-1,y2)+t(x1-1,y1-1);
}
int resp[MAX][MAX][MAX][MAX];
bool exist[MAX][MAX][MAX][MAX];
int dp(int x1,int y1,int x2,int y2) {
if(exist[x1][y1][x2][y2])return resp[x1][y1][x2][y2];
exist[x1][y1][x2][y2]=true;
int soma = subrec(x1,y1,x2,y2);
int area = (x2-x1+1)*(y2-y1+1);
if(soma==area){return resp[x1][y1][x2][y2]=1;}
if(!soma) {return 0;}
int best = 1000000;
for(int i = x1;i!=x2;++i) {
best = std::min(best,dp(x1,y1,i,y2)+dp(i+1,y1,x2,y2));
}
for(int i = y1;i!=y2;++i) {
best = std::min(best,dp(x1,y1,x2,i)+dp(x1,i+1,x2,y2));
}
return resp[x1][y1][x2][y2]=best;
}
void backtracking(int x1,int y1,int x2,int y2) {
int soma = subrec(x1,y1,x2,y2);
int area = (x2-x1+1)*(y2-y1+1);
if(soma==area){std::cout<<x1+1<<" "<<y1+1<<" "<<x2+1<<" "<<y2+1<<"\n";return;}
if(!soma) {return;}
int best = 1000000;
int obj = resp[x1][y1][x2][y2];
for(int i = x1;i!=x2;++i) {
int ans = dp(x1,y1,i,y2)+dp(i+1,y1,x2,y2);
if(ans==obj){
backtracking(x1,y1,i,y2);
backtracking(i+1,y1,x2,y2);
return;
}
}
for(int i = y1;i!=y2;++i) {
int ans = dp(x1,y1,x2,i)+dp(x1,i+1,x2,y2);
if(ans==obj){
backtracking(x1,y1,x2,i);
backtracking(x1,i+1,x2,y2);
return;
}
}
}
int main()
{
std::cin >> N >> M;
for(int i = 0; i != N;++i) {
std::string s;
std::cin >> s;
for(int j = 0; j != M;++j) {
if(s[j]=='1')tab[i][j]++;
}
}
for(int i = 0; i != N;++i) {
int val = 0;
for(int j = 0; j != M;++j) {
val += tab[i][j];
sumed[i][j]=val;
if(i)sumed[i][j]+=sumed[i-1][j];
}
}
std::cout << dp(0,0,N-1,M-1) << std::endl;
backtracking(0,0,N-1,M-1);
}

How to apply vector2s for direction/velocity of object

I have a line segment that represents a direction and magnitude (length), when i draw the segment it works as it should. the value getAimArmsRotation is being pulled from another class that contains a touchpad value.
if (player.getFacingRight()) {
lOriginX = (player.getPosition().x + Player.SIZEw/2);
lOriginY = (player.getPosition().y + Player.SIZEh/1.5f);
//lEndX = lOriginX + (float)Math.cos((player.getAimArmsRotation())/57) * 15f;
//lEndY = lOriginY + (float)Math.sin((player.getAimArmsRotation())/57) * 15f;
laserO = new Vector2(lOriginX, lOriginY);
laserE = new Vector2(lEndX, lEndY);
However if I use the Vectors or floats from this calculation and apply them to a model's velocity vector it does not move the model along the line segment as I would think it should.
EDIT: Sorry meant to attach this picture when I created the question. Fig 1 is how my line segment looks, when I set the velocity values that make up the line segment to my object it moves in the direction that fig 2 shows.
getAimArmsRotation() is just a method that sets a sprite's rotation with a value from the touchpad in another class. I don't think that the values should matter since these floats are what i've used in order to give the line segment it's length and direction, I would think that giving an object a velocity of the x and y floats would give it the same direction as the line?

Thanks for the DV, jerks.
I wasn't taking into account the origin position of the object when trying to send it along the desired path. I was only using the LineEnd values, I needed to give the object it's origin point to correctly calculate the trajectory or path.
for (GameObject go: gObjects) {
if (go.getType() == PROJECTILE_ID) {
go.getVelocity().x = player.getLineEndX() - player.getLineOrgX();
go.getVelocity().y = player.getLineEndY() - player.getLineOrgY();
System.out.println(go.getVelocity().x);
System.out.println(go.getVelocity().y);
}
}