2D Delaunay, Divide and Conquer - Rare bug - c++

I have been working on a two dimensional Divide and Conquer algorithm written in c++, a language I am rather new to. I used this page to base my algorithm on. The Y-axis is inverted in my example images.
I create a grid with a chosen width and height. The program then selects a random coordinate within each cell, these are the points for the Delaunay Triangulation.
The algorithm works as follows
Sort the coordinates in ascending order on the x-axis (similar x values; sort on y)
Divide into subgroups, until each group contains no more than 3 coordinates. Connect each coordinate with a line (I call them 'Edges'). This is in a binary tree structure. The Leaves contain a vector of these edges.
Recursively merge the leaves upwards
# Select the starting edge: select the coordinates, L and R, with the lowest y value from the left and right subgroup. If the edge L-R does not intersect with any edges of either group: select that edge. Else, select the next coordinate. (If the L-R edge intersects with an edge from the left subgroup select the next left coordinate. Ditto with the right group)
Based on the starting edge ('base' in my code) select the next candidate from both groups, left_C and right_C (the following is the left subgroup selection process, the right subgroup is the same but only mirrored):
Select the neighboring coordinates: all coordinates that have a direct connection to the base edge
Sort the neighbors by increasing clockwise angles.
# The candidate selection process: Select first (i = 0) coordinate as the candidate C, the second (i + 1) as the next candidate nC.
Check if the angle between the L-R and C is bigger than 180° or pi. If so: no candidate is selected, else move on to the next condition.
& Check if, the nC is within the circumcircle defined by triangle L-R-C (distance between circumcircle center and nC < radius circumcircle). delete the edge defined by L-C. nC becomes the new candidate C, the candidate with the next highest angle becomes nC, repeat check.
return candidate at the position i
If neither left nor right groups return a candidate, then the merge is complete.
Else if only one group returns a candidate, obviously that is the to-be-used candidate
Else both subgroups return a candidate, then select the candidate where the circumcircle defined by it and the base edge does not contain the other.
A triangle is created using the selected candidate and the base edge: an edge is created between the candidate and base-coordinate from the other subgroup ( eg. candidate of the left-subgroup: R-left_C ). This becomes the base edge for the next iteration.
iterate until the merge is complete (neither subgroup offers a candidate)
And so the tree is merged recursively.
So TL;DR.
The issue that I am having is two steps of the D&C algorithm coming into conflict. the two are the selection of the starting base edge (see step #) and the selection of the candidate coordinates (see step # and &)
The bug (read: feature) occurs with this orientation of the subgroups:
Animated Gif
It skips one of the coordinates of the right group in the first merger. This is because the algorithm specifies that you take the lowest y values of both groups that do not intersect with any other edges as the base edge. Then all the right group candidates, which are attached to the base edge, are selected with their counter-clockwise angle, the left with the clockwise angles. This skips one coordinate. leading to the following result.
I have no idea how to solve this.
Here is my code:
triangulate():
Node* Branch::triangulate() {
if (l != nullptr && r == nullptr) {
return l;
}
else {
// recursive merge
Node* left = l->triangulate();
Node* right = r->triangulate();
vector<Edge> left_edges = left->getEdges();
vector<Edge> right_edges = right->getEdges();
Edge base = selectBase(left_edges, right_edges);
bool merge = true;
vector<Edge> mergedEdges;
vector<Edge> newEdges;
newEdges.push_back(base);
while (merge) {
Coordinate left_candidate = selectCandidateFromSubgroup(left_edges, base, true);
Coordinate right_candidate = selectCandidateFromSubgroup(right_edges, base, false);
Coordinate null(DBL_MIN, DBL_MIN);
if (left_candidate == null && right_candidate == null) {
merge = false;
}
else {
Edge new_base;
if (right_candidate == null) {
new_base = Edge(left_candidate, *base.p2());
}
else if (left_candidate == null) {
new_base = Edge(*base.p1(), right_candidate);
}
else {
double left_a, left_b, left_c,
base_a, base_b, base_c;
functionFromEdge(base, base_a, base_b, base_c);
Edge left_edge(left_candidate, *base.p2());
functionFromEdge(left_edge, left_a, left_b, left_c);
perpendicularBisector(base, base_a, base_b, base_c);
perpendicularBisector(left_edge, left_a, left_b, left_c);
Coordinate left_circumcircle_center = intersection(left_a, left_b, left_c,
base_a, base_b, base_c);
new_base = (Coordinate::distance(left_circumcircle_center, left_candidate)
< Coordinate::distance(left_circumcircle_center, right_candidate))
? left_edge : Edge(*base.p1(), right_candidate);
}
newEdges.push_back(new_base);
base = new_base;
}
}
mergedEdges.reserve(newEdges.size() + left_edges.size() + right_edges.size());
for (Edge e : left_edges) {
mergedEdges.push_back(e);
}
for (Edge e : right_edges) {
mergedEdges.push_back(e);
}
for (Edge e : newEdges) {
mergedEdges.push_back(e);
}
return new Leaf(mergedEdges,i);
}
}
selectBase():
vector<Coordinate> left_points = extractCoordinates(left);
vector<Coordinate> right_points = extractCoordinates(right);
Edge* out = nullptr;
bool found = false;
int i = 0;
int j = 0;
while (!found) {
Coordinate l = left_points.at(i), r = right_points.at(j);
Edge base(l, r);
out = &base;
for (Edge e : left) {
if (out!= nullptr && e.intersects(base)) {
//cout << e << " intersects with " << *out << " i++" << endl;
i++;
out = nullptr;
}
if (out != nullptr && left_points.size() ==2 && (*e.p1() == *base.p1() || *e.p2() == *base.p1())) {
Coordinate other = (*e.p1() == *base.p1()) ? *e.p2() : *e.p1();
if (Angle::calcClockAngle(l, r, other, true).rad() > M_PI) {
out = nullptr; i++;
}
}
}
for (Edge e : right) {
if (out != nullptr && e.intersects(base)) {
//cout << e << " intersects with " << *out << " j++" << endl;
j++;
out = nullptr;
}
if (out != nullptr && right_points.size() == 2 && (*e.p1() == *base.p2() || *e.p2() == *base.p2())) {
Coordinate other = (*e.p1() == *base.p2()) ? *e.p2() : *e.p1();
if (Angle::calcClockAngle(r, l, other, false).rad() > M_PI) {
out = nullptr; j++;
}
}
}
if (out != nullptr) found = true;
}
return Edge(left_points.at(i), right_points.at(j));
}
In the above function I tried to make a check that checks if there is a skipped point, by checking in subgroups of size 2 if there is an angle larger than pi.
It worked for smaller groups. But the issue still occurs.
selectCandidateFromSubgroup:
Coordinate selectCandidateFromSubgroup(vector<Edge>& es, Edge& base, bool left) {
// select neighboring edges from base
Coordinate* shared_point = (left)? base.p1() : base.p2();
Coordinate* other_point = (left)? base.p2() : base.p1();
vector<Coordinate> candidates;
for (Edge e : es) {
if (*shared_point == *e.p2() || *shared_point == *e.p1()) {
Coordinate c = (*shared_point == *e.p2()) ? *e.p1() : *e.p2();
candidates.push_back(c);
}
}
if (candidates.size() != 0) {
algorithm::quicksort(candidates, 0, candidates.size() -1,
[&shared_point,&other_point,&left](Coordinate a, Coordinate b) {
double angle = Angle::calcClockAngle(*shared_point, *other_point, a, left).getDegrees();
double angle2 = Angle::calcClockAngle(*shared_point, *other_point, b, left).getDegrees();
return (angle < angle2 || (angle == angle2 && a.y() < b.y()));
});
double base_a, base_b, base_c;
functionFromEdge(base, base_a, base_b, base_c);
perpendicularBisector(base, base_a, base_b, base_c);
Coordinate* candidate=nullptr;
Coordinate* next_candidate = nullptr;
int i = 0;
bool checking_candidates = true;
while (checking_candidates) {
candidate = &candidates.at(i);
double angle = Angle::calcClockAngle(*shared_point,
*other_point,
*candidate,
left).rad();
if (angle < M_PI) {
Edge candidate_edge;
int cei = 0;
int j = 0;
for (Edge e : es) {
if ((*e.p1() == *shared_point && *e.p2() == *candidate) || (*e.p2() == *shared_point && *e.p1() == *candidate)) {
candidate_edge = e;
cei = j;
break;
}
j++;
}
if (i + 1 == candidates.size()) return *candidate;
next_candidate= &candidates.at(i+1);
double candidate_edge_a, candidate_edge_b, candidate_edge_c;
functionFromEdge(candidate_edge, candidate_edge_a,
candidate_edge_b, candidate_edge_c);
perpendicularBisector(candidate_edge, candidate_edge_a,
candidate_edge_b, candidate_edge_c);
Coordinate circumcicle_center = intersection(base_a, base_b, base_c,
candidate_edge_a, candidate_edge_b, candidate_edge_c);
double radius = Coordinate::distance(circumcicle_center, *shared_point);
double distance_center_nCandidate = Coordinate::distance(circumcicle_center, *next_candidate);
if (radius < distance_center_nCandidate) {
return *candidate;
}
else {
i++;
es.erase(es.begin() + cei);
}
}
else {
return Coordinate(DBL_MIN, DBL_MIN);
}
}
}
return Coordinate(DBL_MIN, DBL_MIN);
}
extractCoordinates(vector) extracts and sorts all coordiates on the y axis ascendingly.
functionFromEdge(Edge,double a,double b,double c) takes an edge and makes a functional representation of the lines like 'ax+by=c', which is used by
perpendicularBisector(Edge, double a, double b, double c) converts the a, b, and c values to represent the function of the perpendicular bisector of the Edge so that the algorithm can calulate the
intersection(double a1, double b1, double c1, double a2, double b2, double c2) of two functions.
the calcClockAngle(Coordinate a, Coordinate b, Coordinate c, bool clockwise) calulates the angle in a bewteen the vectors |AB| and |AC|, depending on the clockwise boolean, it calculates the cloclwise or counter clockwise angle.
If I forgot something please let me know.
Sorry for the long question. but I am seriously at a loss here.
I get results like this:

Related

Water in a falling sand simulation

I am currently working on a very simple 'Falling Sand' simulation game in C++ and SDL2, and am having problems with getting water to flow in a more realistic manner. I basically have a grid of cells that I iterate through bottom-to-top, left-to-right and if I find a water cell, I just check below, down to left, down to the right, left then right for empty cells and it moves into the first one its finds (it makes a random choice if both diagonal cells or both horizontal cells are free). I then mark the cell it moved into as processed so that it is not checked again for the rest of that loop.
My problem is a sort of 'left-bias' in how the particles move; if I spawn a square of water cells above a barrier, they will basically all shift to left without moving once the particles begin to reach the barrier, while the cells on the right will run down in the proper way. So instead of forming a nice triangular shape flowing out evenly to both sides, the whole shape will just move to the left. This effect is reversed whenever I iterate left-to-right, so I know it's something to do with that but so far I've been stumped trying to fix it. I initially thought it was a problem with how I marked the cells as processed but I've found no obvious bugs with that system in many hours of testing. Has anyone faced any similar challeneges in developing a simulation like this, or knows something that I'm missing? Any help would be very much appreciated.
EDIT:
Ok so I've made a little progress, however I've ran into another bug that seems to be unrelated to iteration, since now I save a copy of the old cells and read from that to decide an update, then update the original cells and display that. This already made the sand work better, however water, which checks horizontally for free cells, now 'disappears' when it does move horizontally. I've been testing it all morning and have yet to find a solution, I thought it might've been someting to do with how I was copying the arrays over, but it seems to work as far as I can tell.
New snippets:
Simulation.cpp
void Simulation::update()
{
copyStates(m_cells, m_oldCells); // so now oldcells is the last new state
for(int y = m_height - 1; y>= 0; y--)
for(int x = 0; x < m_width; x++)
{
Cell* c = getOldCell(x, y); // check in the old state for possible updates
switch(c->m_type)
{
case EMPTY:
break;
case SAND:
if(c->m_visited == false) update_sand(x, y);
break;
case WATER:
if(c->m_visited == false) update_water(x, y);
break;
default:
break;
}
}
}
void Simulation::update_water(int x, int y)
{
bool down = (getOldCell(x, y+1)->m_type == EMPTY) && checkBounds(x, y+1) && !getOldCell(x, y+1)->m_visited;
bool d_left = (getOldCell(x-1, y+1)->m_type == EMPTY) && checkBounds(x-1, y+1) && !getOldCell(x-1, y+1)->m_visited;
bool d_right = (getOldCell(x+1, y+1)->m_type == EMPTY) && checkBounds(x+1, y+1) && !getOldCell(x+1, y+1)->m_visited ;
bool left = (getOldCell(x-1, y)->m_type == EMPTY) && checkBounds(x-1, y) && !getOldCell(x-1, y)->m_visited ;
bool right = (getOldCell(x+1, y)->m_type == EMPTY) && checkBounds(x+1, y) && !getOldCell(x+1, y)->m_visited ;
// choose random dir if both are possible
if(d_left && d_right)
{
int r = rand() % 2;
if(r) d_right = false;
else d_left = false;
}
if(left && right)
{
int r = rand() % 2;
if(r) right = false;
else left = false;
}
if(down)
{
getCell(x, y+1)->m_type = WATER; // we now update the new state
getOldCell(x, y+1)->m_visited = true; // mark as visited so it will not be checked again in update()
} else if(d_left)
{
getCell(x-1, y+1)->m_type = WATER;
getOldCell(x-1, y+1)->m_visited = true;
} else if(d_right)
{
getCell(x+1, y+1)->m_type = WATER;
getOldCell(x+1, y+1)->m_visited = true;
} else if(left)
{
getCell(x-1, y)->m_type = WATER;
getOldCell(x-1, y)->m_visited = true;
} else if(right)
{
getCell(x+1, y)->m_type = WATER;
getOldCell(x+1, y)->m_visited = true;
}
if(down || d_right || d_left || left || right) // the original cell is now empty; update the new state
{
getCell(x, y)->m_type = EMPTY;
}
}
void Simulation::copyStates(Cell* from, Cell* to)
{
for(int x = 0; x < m_width; x++)
for(int y = 0; y < m_height; y++)
{
to[x + y * m_width].m_type = from[x + y * m_width].m_type;
to[x + y * m_width].m_visited = from[x + y * m_width].m_visited;
}
}
Main.cpp
sim.update();
Uint32 c_sand = 0xedec9a00;
for(int y = 0; y < sim.m_height; y++)
for(int x = 0; x < sim.m_width; x++)
{
sim.getCell(x, y)->m_visited = false;
if(sim.getCell(x, y)->m_type == 0) screen.setPixel(x, y, 0);
if(sim.getCell(x, y)->m_type == 1) screen.setPixel(x, y, c_sand);
if(sim.getCell(x, y)->m_type == 2) screen.setPixel(x, y, 0x0000cc00);
}
screen.render();
I've attached a gif showing the problem, hopefully this might help make it a little clearer. You can see the sand being placed normally, then the water and the strange patterns it makes after being placed (notice how it moves off to the left when it's spawned, unlike the sand)
You also have to mark the destination postion as visited to stop multiple cells moving in to the same place.

Error reconstructing path in a grid using BFS

The problem I am facing is the following:
I have a function based on the BFS search algorithm that I use in a NxM grid, the mission of this function is to return the following Direction from a set of possible Directions = {Up, Down, Left , Right} (No diagonal moves!)to which a player has to move, so that in each "round / frame" where there is a type of item of a game (For example, in this specific case, a bazooka) is closer to the item.
To address the problem, I have created a Map class made of vector <vector <Cell> > where vector is from the standard library and Cell is what the grid is made of and has some consulting methods on what is in one of the NxM cells (if there is a building, an enemy, a Bazooka, etc.)
So, for implementing a solution for this, I made a struct TrackingBFS to reconstruct the path of the BFS search:
struct TrackingBFS {
pair <int,int> anterior;
bool visited;
};
And this is the BFS search implementation:
//Pre: origen is a valid position on the grid where the player is
//Post:Returns a pair of bool and a direction to the closest bazooka. If we have access to a bazooka, then we will return a pair (true,Dir) where Dir is a direction to take to be closer to the bazooka else a pair (false, Dir) where dir is just the same direction as origen.
pair <bool,Dir> direction_to_closest_gun (const Pos& origen) {
//R = board_rows() C = board_cols()
//m = mapa
//sr,sc = origin positions
int sr = origen.i;
int sc = origen.j;
//Variables para mantener tracking de los pasos cogidos
queue <int> rq; //Queue of x-row coordinates
queue <int> cq; //Queue of y-row coordinates
int move_count = 0; //Number of steps
int nodes_left_in_layer = 1; //How many nodes we need to de-queue before doing a step
int nodes_in_next_layer = 0; //How many nodes we add to the expansio of the BFS so we can upgrade nodes_left_in_layer in the next iteration
bool arma_mejor_encontrada = false;
//Visited is a MxN matrix of TrackingBFS that for every i,j stores the following information:
//If we visited the cell at position visited [i][j], then the TrackingBFS struct will be visited = true and last_node = (k,l) where k and l are the components of the LAST cell on the grid we visited in the BFS search.
//Else if we didn't visited the cell at [i][j], the TrackingBFS struct will be visited = true and last_node (-1,-1).
TrackingBFS aux;
aux.last_node = make_pair(-1,-1);
aux.visited = false;
//We create a Matrix of TrackingBFS named visited of NxM filled with the unvisited cells
vector < vector <TrackingBFS> > visited (board_rows(), vector <TrackingBFS> (board_cols(),aux));
//--------------------------------SOLVE---------------------------------
rq.push(sr);
cq.push(sc);
visited[sr][sc].visited = true;
visited[sr][sc].last_node = make_pair(sr,sc);
int xfound;
int yfound;
while (!rq.empty()) {
int r = rq.front();
int c = cq.front();
rq.pop();
cq.pop();
if (mapa[r][c].weapon == Bazooka) {
arma_mejor_encontrada = true;
xfound = r;
yfound = c;
break;
}
//Explore neighbours
Pos p (r,c);
for (Dir d : dirs) {
Pos searching = p + d;
int rr = searching.i;
int cc = searching.j;
//If the position we are searching is out of range or it's been visited before or there is a obstacle then continue
if (!pos_ok(searching) or visited[rr][cc].visited or mapa[rr][cc].type == Building or mapa[rr][cc].resistance != -1 or mapa[rr][cc].id != -1) {
//NADA
}
//Else we add the visited node to the queue, and fill the information on visited[rr][cc] with the node we are coming and mark it as visited
else {
rq.push(rr);
cq.push(cc);
visited[rr][cc].visited = true;
visited[rr][cc].last_node = make_pair (r,c);
nodes_in_next_layer++;
}
}
nodes_left_in_layer--;
if (nodes_left_in_layer == 0) {
nodes_left_in_layer = nodes_in_next_layer;
nodes_in_next_layer = 0;
move_count++;
}
}
//If we found the Bazooka
if (arma_mejor_encontrada) {
//Return the pair (true,next direction of the player at position (sr,sc) to approach the bazooka)
return make_pair(arma_mejor_encontrada, reconstruir_camino(visited,xfound,yfound,sr,sc));
}
else {
//If there is no bazooka we return (false,Up (this second component is meaningless))
return make_pair(arma_mejor_encontrada, Up);
}
}
The reconstruir_camino (recosntruct_path in english) implementation:
//This function is made to reconstruct the path from where we found de bazooka (x,y) to where our player is (ox,oy), whe are only interested in the next position of because this is run each frame, so we are constantly refreshing the state of the map.
Dir reconstruir_camino(const vector < vector <TrackingBFS> >& visited, const int& x, const int& y, const int& ox, const int& oy) {
//In v we save the pair of coordinates of our path, this was done only for debuging and in debug_matriz_tracking(visited) is also only for debuging
vector <pair<int,int>> path;
debug_matriz_tracking(visited);
//
int i = visited[x][y].last_node.first;
int j = visited[x][y].last_node.second;
while (not (i == ox and j == oy)) { //While the next node is not iqual as the original node we started de search (The one where our player is)
path.push_back(make_pair(i,j)); //Just for debuging
i = visited[i][j].last_node.first;
j = visited[i][j].last_node.second;
}
//So now in path we have the start and end nodes of every edge on the path
int siguiente_x = path.back().first;
int siguiente_y = path.back().second;
debug_camino(path);
return direccion_contiguos(ox,oy,siguiente_x,siguiente_y);
}
And direccion_contiguos (contiguous direction / relative direction in english) implementation:
//Returns the RELATIVE direction to go if we want to go from (ox, oy) to (dx, dy) being these two contiguous positions, that is, (dx, dy) is in Up, Down, Left or Right with respect to (ox, oy) IT WORKS FINE, NOTHING WRONG WITH THIS DON'T TOUCH
Dir direccion_contiguos (const int& ox, const int& oy, const int& dx, const int& dy) {
Pos o (ox,oy);
Pos d (dx,dy);
if (o + Up == d) {
return Up;
}
else if (o + Down == d){
return Down;
}
else if (o + Left == d) {
return Left;
}
else if (o + Right == d) {
return Right;
}
return Down;
}
So now in visited, we have the information to reconstruct the path, in fact I debuged it (it's kinda messy i know, sorry), in a visual way so this is what I got for a Player in origen = (7,10) and bazooka at position = (4,11):
[Imgur link of the Visual representation of the Matrix for reconstructing the path from origin to bazooka][1]
To read this image, at the top and left there are the coordinates of every cell of the visited matrix, the ones with green font, are the ones that have been visited, and they store THE NEXT cell/vertex of the path, and the ones with (-1,-1) are the ones that have not been visited by the BFS algorithm and thus they don't have any previous node and are in white.
So, NICE! It seems to work, at least the visited matrix.
My problem is when I debug the vector of edges/directions of the graph/grid, this is what I used in the example of the image:
void debug_camino(const vector <pair<int,int>>& v) {
cerr << "----------------------CAMINO-------------------- DEBUG_CAMINO" << endl;
for (long unsigned int i = 0; i < v.size(); ++i) {
cerr << "(" << v[i].first << "," << v[i].second << "),";
}
cerr << endl;
}
And when I executed the program, this is the path that I got with debug_camino():
If you see the image attached you can see that that's almost the path but not quite yet.
(4,12),(4,13),(4,14),(3,15),(3,16),(4,16),(5,16),(6,16),(7,15),(7,14),(7,13),(7,12),(7,11)
These ones bolded are not real (even valid because they are diagonal moves) reconstructions of the path and I don't really know WHY this is happening, but it's provoking my player to not following right path, and I want to fix the error and I'm kinda desperate because I don't really know where the error is and I've been trying for days :( ! I hope somebody can help me with this. Thanks for reading all this and sorry if the code is in some parts in Spanish or if it's not all that readable.
[1]: https://i.stack.imgur.com/vZ2Go.png
Okay, I actually managed to fix this error, i was overwriting the i variable so that was causing the error.
//This function is made to reconstruct the path from where we found de bazooka (x,y) to where our player is (ox,oy), whe are only interested in the next position of because this is run each frame, so we are constantly refreshing the state of the map.
Dir reconstruir_camino(const vector < vector <TrackingBFS> >& visited, const int& x, const int& y, const int& ox, const int& oy) {
//In v we save the pair of coordinates of our path, this was done only for debuging and in debug_matriz_tracking(visited) is also only for debuging
vector <pair<int,int>> path;
debug_matriz_tracking(visited);
//
int i = visited[x][y].last_node.first;
int j = visited[x][y].last_node.second;
while (not (i == ox and j == oy)) { //While the next node is not iqual as the original node we started de search (The one where our player is)
path.push_back(make_pair(i,j)); //Just for debuging
**i** = visited[i][j].last_node.first;
j = visited[**i**][j].last_node.second;
}
//So now in path we have the start and end nodes of every edge on the path
int siguiente_x = path.back().first;
int siguiente_y = path.back().second;
debug_camino(path);
return direccion_contiguos(ox,oy,siguiente_x,siguiente_y);
}
It's already fixed

How to get the left chain of points of a polygon?

I am trying to get the left polygonal chain given a set of consecutive points. (NOTE: edges are non-intersecting.)
Image 1. Sample polygon and its bound.
What I did was:
Get the minY, maxY and minX. (Bound.)
Find the point that contains minY (or maxY) then save it as the first point.
Save any points until point with minY or maxY is found while checking for point with minX.
If the same Y is found first, save it as the new first point and repeat from #3.
If other Y is found first and the saved points has minX, this is the chain. Otherwise, save as the new first point and repeat from #3.
Image 2. The left chain of points.
But using this steps might give wrong result for some polygon, like this:
Since one point is (minX, maxY), either of the side will be returned.
EDIT:
With the idea of the left-bottom- and left-top-most points, here is the current code that I am using:
Get the min (left-bottom-most) and max (left-top-most) point.
std::vector<Coord> ret;
size_t i = 0;
Coord minCoord = poly[i];
Coord maxCoord = poly[i];
size_t minIdx = -1;
size_t maxIdx = -1;
size_t cnt = poly.size();
i++;
for (; i < cnt; i++)
{
Coord c = poly[i];
if (c.y < minCoord.y) // new bottom
{
minCoord = c;
minIdx = i;
}
else if (c.y == minCoord.y) // same bottom
{
if (c.x < minCoord.x) // left most
{
minCoord = c;
minIdx = i;
}
}
if (c.y > maxCoord.y) // new top
{
maxCoord = c;
maxIdx = i;
}
else if (c.y == maxCoord.y) // same top
{
if (c.x < maxCoord.x) // left most
{
maxCoord = c;
maxIdx = i;
}
}
}
Get the points connected to the max point.
i = maxIdx;
Coord mid = poly[i];
Coord ray1 = poly[(i + cnt - 1) % cnt];
Coord ray2 = poly[(i + 1) % cnt];
Get which has smallest angle. This will be the path we will follow.
double rad1 = Pts2Rad(mid, ray1);
double rad2 = Pts2Rad(mid, ray2);
int step = 1;
if (rad1 < rad2)
step = cnt - 1;
Save the points.
while (i != minIdx)
{
ret.push_back(poly[i]);
i = (i + step) % cnt;
}
ret.push_back(poly[minIdx]);
To be specific, I am assuming that no vertex is duplicated and define the "left chain" as the sequence of vertices from the original polygon loop that goes from the leftmost vertex in the top side of the bounding box, to the leftmost vertex in the bottom side of the bounding box. [In case the top and bottom sides coincide, these two vertices also coincide; I leave it to you what to return in this case.]
To obtain these, you can scan all vertices and keep the left-topmost so far and left-bottommost so far. Then compare to the next vertex. If above the left-topmost, becomes the new lef-topmost. If at the same level and to the left, becomes the new left-topmost. Similarly for the left-bottommost.

Multi-dimensional bounding box collision detection

There are answers for 2D/3D bounding box collision detection, however my problem is how to develop the situation to multi-dimension (4D or more) ?
Here is the code on 3D case.
template<typename T, typename _Prd>
bool BoundingBox3<T,_Prd>::Collision( const BoundingBox3<T,_Prd>& cube ) const
{
Point3<T,_Prd> min_1 = center - Point3<T,_Prd>(length,depth,height)/2.0;
Point3<T,_Prd> max_1 = center + Point3<T,_Prd>(length,depth,height)/2.0;
Point3<T,_Prd> min_2 = cube.Center() - Point3<T,_Prd>(cube.Length(),cube.Depth(),cube.Height())/2.0;
Point3<T,_Prd> max_2 = cube.Center() + Point3<T,_Prd>(cube.Length(),cube.Depth(),cube.Height())/2.0;
if(Volume()<cube.Volume())
{
Vector3D::Swap(min_1,min_2);
Vector3D::Swap(max_1,max_2);
}
if(min_1[0]<=min_2[0] && min_1[1]<=min_2[1] && min_1[2]<=min_2[2]
&& max_1[0]>=min_2[0] && max_1[1]>=min_2[1] && max_1[2]>=min_2[2])
return true;
if(min_1[0]<=max_2[0] && min_1[1]<=max_2[1] && min_1[2]<=max_2[2]
&& max_1[0]>=max_2[0] && max_1[1]>=max_2[1] && max_1[2]>=max_2[2])
return true;
return false;
};
One box collides with another if there is an intersecting range in all dimensions.
The intersection of range [a,b] with range [c,d] is [max(a,c),min(b,d)].
In the case of the ranges not intersecting, the result will be an invalid range, with the start of the range greater than the end of the range.
So the collision can be done like this:
bool collides(const Box &box1,const Box &box2)
{
assert(box1.numberOfDimensions()==box2.numberOfDimensions());
for (size_t i=0; i!=box1.numberOfDimensions(); ++i) {
float a = max(box1.minForDimension(i),box2.minForDimension(i));
float b = min(box1.maxForDimension(i),box2.maxForDimension(i));
if (a>b) return false;
}
return true;
}

Algorithm for edge intersection?

Given Polygon P which I have its verticies in order. and I have a rectangle R with 4 verticies how could I do this:
If any edge of P (line between adjacent vertexes) intersects an edge of R, then return TRUE, otherwise return FALSE.
Thanks
* *
* *
What you want is a quick way to determine if a line-segment intersects an axis-aligned rectangle. Then just check each line segment in the edge list against the rectangle. You can do the following:
1) Project the line onto the X-axis, resulting in an interval Lx.
2) Project the rectangle onto the X-axis, resulting in an interval Rx.
3) If Lx and Rx do not intersect, the line and rectangle do not intersect.
[Repeat for the Y-axis]:
4) Project the line onto the Y-axis, resulting in an interval Ly.
5) Project the rectangle onto the Y-axis, resulting in an interval Ry.
6) If Ly and Ry do not intersect, the line and rectangle do not intersect.
7) ...
8) They intersect.
Note if we reach step 7, the shapes cannot be separated by an axis-aligned line. The thing to determine now is if the line is fully outside the rectangle. We can determine this by checking that all the corner points on the rectangle are on the same side of the line. If they are, the line and rectangle are not intersecting.
The idea behind 1-3 and 4-6 comes from the separating axis theorem; if we cannot find a separating axis, they must be intersecting. All these cases must be tested before we can conclude they are intersecting.
Here's the matching code:
#include <iostream>
#include <utility>
#include <vector>
typedef double number; // number type
struct point
{
number x;
number y;
};
point make_point(number pX, number pY)
{
point r = {pX, pY};
return r;
}
typedef std::pair<number, number> interval; // start, end
typedef std::pair<point, point> segment; // start, end
typedef std::pair<point, point> rectangle; // top-left, bottom-right
namespace classification
{
enum type
{
positive = 1,
same = 0,
negative = -1
};
}
classification::type classify_point(const point& pPoint,
const segment& pSegment)
{
// implicit line equation
number x = (pSegment.first.y - pSegment.second.y) * pPoint.x +
(pSegment.second.x - pSegment.first.x) * pPoint.y +
(pSegment.first.x * pSegment.second.y -
pSegment.second.x * pSegment.first.y);
// careful with floating point types, should use approximation
if (x == 0)
{
return classification::same;
}
else
{
return (x > 0) ? classification::positive :classification::negative;
}
}
bool number_interval(number pX, const interval& pInterval)
{
if (pInterval.first < pInterval.second)
{
return pX > pInterval.first && pX < pInterval.second;
}
else
{
return pX > pInterval.second && pX < pInterval.first;
}
}
bool inteveral_interval(const interval& pFirst, const interval& pSecond)
{
return number_interval(pFirst.first, pSecond) ||
number_interval(pFirst.second, pSecond) ||
number_interval(pSecond.first, pFirst) ||
number_interval(pSecond.second, pFirst);
}
bool segment_rectangle(const segment& pSegment, const rectangle& pRectangle)
{
// project onto x (discard y values)
interval segmentX =
std::make_pair(pSegment.first.x, pSegment.second.x);
interval rectangleX =
std::make_pair(pRectangle.first.x, pRectangle.second.x);
if (!inteveral_interval(segmentX, rectangleX))
return false;
// project onto y (discard x values)
interval segmentY =
std::make_pair(pSegment.first.y, pSegment.second.y);
interval rectangleY =
std::make_pair(pRectangle.first.y, pRectangle.second.y);
if (!inteveral_interval(segmentY, rectangleY))
return false;
// test rectangle location
point p0 = make_point(pRectangle.first.x, pRectangle.first.y);
point p1 = make_point(pRectangle.second.x, pRectangle.first.y);
point p2 = make_point(pRectangle.second.x, pRectangle.second.y);
point p3 = make_point(pRectangle.first.x, pRectangle.second.y);
classification::type c0 = classify_point(p0, pSegment);
classification::type c1 = classify_point(p1, pSegment);
classification::type c2 = classify_point(p2, pSegment);
classification::type c3 = classify_point(p3, pSegment);
// test they all classify the same
return !((c0 == c1) && (c1 == c2) && (c2 == c3));
}
int main(void)
{
rectangle r = std::make_pair(make_point(1, 1), make_point(5, 5));
segment s0 = std::make_pair(make_point(0, 3), make_point(2, -3));
segment s1 = std::make_pair(make_point(0, 0), make_point(3, 0));
segment s2 = std::make_pair(make_point(3, 0), make_point(3, 6));
segment s3 = std::make_pair(make_point(2, 3), make_point(9, 8));
std::cout << std::boolalpha;
std::cout << segment_rectangle(s0, r) << std::endl;
std::cout << segment_rectangle(s1, r) << std::endl;
std::cout << segment_rectangle(s2, r) << std::endl;
std::cout << segment_rectangle(s3, r) << std::endl;
}
Hope that makes sense.
I think your problem is equivalent to convex polygon intersection, in which case this might help. See also: How do I determine if two convex polygons intersect?
Untested, obviously, but in rough pseudocode:
// test two points against an edge
function intersects ( side, lower, upper, pt1Perp, pt1Par, pt2Perp, pt2Par )
{
if ( ( pt1Perp < side and pt2Perp > side ) or ( pt1Perp > side and pt2Perp < side )
{
intersection = (side - pt1Perp) * (pt2Par - pt1Par) / (pt2Perp - pt1Perp);
return (intersection >= lower and intersection <= higher);
}
else
{
return false;
}
}
// left, right, bottom, top are the bounds of R
for pt1, pt2 adjacent in P // don't forget to do last,first
{
if ( intersects ( left, bottom, top, pt1.x, pt1.y, pt2.x, pt2.y )
or intersects ( right, bottom, top, pt1.x, pt1.y, pt2.x, pt2.y )
or intersects ( top, left, right, pt1.y, pt1.x, pt2.y, pt2.x )
or intersects ( bottom, left, right, pt1.y, pt1.x, pt2.y, pt2.x ) )
{
return true;
}
}
Basically, if two adjacent P vertices are on opposite sides of one of the R's edges, check whether the intersection point falls in range.
Just FYI, geometrictools is a great resource for such things (especially the Math section)