I'm trying to build a MKPolygon using the outer boundary of a set of coordinates.
From what I can tell, there is no delivered functionality to achieve this in Xcode (the MKPolygon methods would use all points to build the polygon, including interior points).
After some research I've found that a convex-hull solves this problem.
After looking into various algorithms, the one I can best wrap my head around to implement is QuickHull.
This takes the outer lat coords and draws a line between the two. From there, you split your points based on that line into two subsets and process distance between the outer lats to start building triangles and eliminating points within until you are left with the outer boundary.
I can find the outer points just by looking at min/max lat and can draw a line between the two (MKPolyline) - but how would I determine whether a point falls on one side or the other of this MKPolyline?
A follow up question is whether there is a hit test to determine whether points fall within an MKPolygon.
Thanks!
I ended up using a variation of the gift wrap algorithm. Certainly not a trivial task.
Having trouble with formatting of the full code so I'll have to just put my steps (probably better because I have some clean up to do!)
I started with an array of MKPointAnnotations
1) I got the lowest point that is furthest left. To do this, I looped through all of the points and compared lat/lng to get lowest point. This point will definitely be in the convex hull, so add it to a NSMutableArray that will store our convex hull points (cvp)
2) Get all points to the left of the lowest point and loop through them, calculating the angle of the cvp to the remaining points on the left. Whichever has the greatest angle, will be the point you need to add to the array.
atan(cos(lat1)sin(lat2)-sin(lat1)*cos(lat2)*cos(lon2-lon1), sin(lon2-lon1)*cos(lat2))
For each point found, create a triangle (by using lat from new point and long from previous point) and create a polygon. I used this code to do a hit test on my polygon:
BOOL mapCoordinateIsInPolygon = CGPathContainsPoint(polygonView.path, NULL, polygonViewPoint, NO);
If anything was found in the hit test, remove it from the comparison array (all those on the left of the original array minus the hull points)
Once you have at least 3 points in your cvp array, build another polygon with all of the cvp's in the array and remove anything within using the hit test.
3) Once you've worked through all of the left points, create a new comparison array of the remaining points that haven't been eliminated or added to the hull
4) Use the same calculations and polygon tests to remove points and add the cvp's found
At the end, you're left with a list of points in that make up your convex hull.
Related
I'm trying to code a trivial pursuit game. I want to give an id for every space of the board in order to use them for the movements. I need to know for every space which is next to which/match with each other.
But because of the geometry of the board(extern circle + radii), I didn't find the right logic behind this.
I am thinking of an ID based on 7 numbers (for the 6 radii + the circle). For example :
//this is not my code, i'm just trying to show example of IDs
center = [0][0][0][0][0][0][-2]
one on the "2nd radius" = [0][3][0][0][0][0][-2]
one the circle and the "3rd radius" = [0][0][6][0][0][0][22]
one on the circle = [0][0][0][0][0][0][21]
I have no idea if it's gonna work or if it's optimal, i will try and see.
If some of you have any better idea for name the ID, i would be happy to listen to them.
Here is an image of the board.
enter image description here
Thank you for helping!
OK, seeems you are inventing some coordinate system for this wheel for easy addressing and easy transtions between cells. System with many indices looks too complex.
Perhaps two-index scheme would be appropriate. Resembles polar coordinates:
The first index 0..6 as distance from the center.
The second one 1..42 - angular position.
So center cell is A[0][0] (the second index is not defined, we can choose any)
It's neighbors are A[1][1], A[1][8], A[1][15] ..A[1][36] (marked with 1 at your picture)
Similar for the next cells at the rays A[2][1], A[2][8], A[2][15] ..A[2][36] and so on
Wheel cells are A[6][1], A[6][2]..A[6][42]
Now neighbor cells have coordinates where one index differs by 1 (except for central cell, extra case)
Is this scheme suitable?
I am trying to model the interior of an epithelial space and am stuck on movement around the interior edges of a cylindrical space. Basically, I'm trying to implement StickyBorders and keep agents on those borders in a cylindrical space that I am creating.
Is there a way to use cylindrical coordinates in Repast Simphony? I found this example (https://www.researchgate.net/publication/259695792_An_Agent-Based_Model_of_Vascular_Disease_Remodeling_in_Pulmonary_Arterial_Hypertension) where they seem to have done something similar, but the paper doesn't explain methods in much depth, and I don't believe this is an example in the repast simphony models.
Currently, I have a class of epithelial cells that are set up to form a cylinder and other agents start just inside that cylinder. To move, they are choosing their most desired spot (similar to the Zombie code) then pointing to a new location in the direction of that desired location within one grid square of that original location. They check that new point before moving to it and make sure that there are at least two other epithelial cells in the immediate moore neighborhood, to ensure they stay against the wall.
GridPoint intendedpt = new GridPoint((int)Math.rint(alongX),(int)Math.rint(alongY),(int)Math.rint(alongZ));
GridCellNgh<EpithelialCell> nearEpithelium = new GridCellNgh<EpithelialCell>(mac_grid, intendedpt, EpithelialCell.class, 1,1,1);
List<GridCell<EpithelialCell>> EpiCells = nearEpithelium.getNeighborhood(false);
int nearbyEpiCellsCount=0;
for (GridCell<EpithelialCell> cell: EpiCells) {
nearbyEpiCellsCount++;
}
if (nearbyEpiCellsCount<2) {
System.out.println(this + " leaving epithelial wall /r");
RunEnvironment.getInstance().pauseRun();
//TODO: where to go if false
}
I am wondering if there is a way to either set the boundaries of the space to be a cylinder or to check which side of the agent is against the wall and restrict its movement in that direction.
The sticky border code (StickyBorders.java) essentially just checks if the point that the agent moves to is beyond any of the space's dimensions, and if so the point is clamped to that dimension. So, for example, if the space is 3x4 and an agent's movement would take it to 4,2, then that point becomes 3,2 and the agent is placed there. Can you do something like that in this case? If not, can you edit your question to explain why not and maybe that will help us understand better.
The approach we took in that model was to use a 3D grid space with custom borders and query methods. The space itself was still Cartesian - we just visualized it as a cylinder using custom display code. Using the Cartesian grid was an reasonable approximation for this application since the cell dimensions were significantly smaller that the vessel radius, so curvature effects were neglected. The boundary conditions on the vessel space were wrap around in the angular dimension, so that cells could move continuously around the circumference of the vessel, and the axial boundary conditions were also wrapped, as we assumed a long enough vessel length that this would be reasonable. The wall thickness dimension had hard boundaries at the basement membrane (y=0) and at the fluid interface (y=wall thickness).
Depending on which type of space you are using, you will need to implement a PointTranslator or GridPointTranslator that performs the border functions. If you want specific examples of the code I suggest you reach out to the author's directly.
I'm working in 2D.
My user can draw some lines & line-segments, which I store in a custom object class, but basically in startX-Y & endX-Y.
Now I want to find, actually only the polygon where the ball is inside, but after reading and researching about some algoirthms etc. I think I have to find all polygons and serach after that the right one.
Is there anyone with some example code, c#, java objective-c !?
I tried several times with some pseudo-code explanations, I don't get it.
Overview
There are a number of different interesting questions at play here:
1. Given you have a set of lines on the screen, and the user places their finger from an arbitrary point and drags it to an arbitrary end point, we need to form a new line which does not cross over lines and where the start and end point actually lie on existing lines or on borders.
2. The next question is how do we maintain an active set of "relevant" line segments which form the convex hull which the ball resides in.
3. Finally given we have an active set of line segments how do we find the area of this shape.
Now It seems you've already answered part 1. So we focus on parts 2 and 3.
For part 2, we will also be interested in asking:
4. If we have a convex hull and a point, how do we determine if that point is in the hull.
We refer to here for the solution to 4 Find if a point is inside a convex hull for a set of points without computing the hull itself
The full implementation can be done efficiently if you are careful with the data structures you are using. This is left as a simple exercise. Let me know if I have done something incorrect here or you do not understand something. I look forward to playing your game when its ready!
Solution to Part 3
In fact 3 is easy from 2, since if we know the active set of line segments containing the ball (this is a list of pairs of tuples ((x_1start,y_1start), (x_1end, y1_end))), there are two ways to compute the area. The first is to do a straightforward algorithm to compute the area of the convex hull formed by all start and end points in this list of tuples. Look up more sophisticated algorithms for area, but if you cannot find one, note that the hull with n sides has n-2 non-overlapping triangles, and the area of triangles is easy to compute (http://www.mathopenref.com/polygontriangles.html). Note:
area = abs((x1-x2)*(y1-y3)-(y1-y2)*(x1-x3)) for triangle (x1,y1)(x2,y2)(x3,y3)
Now if you sort all (x,y) points by their angle about the positive x axis, then we simply fix the first point, and consecutively walk through the remaining pairs and find the areas of those triangles. Left as an easy exercise why this sorting step is required (hint: draw a picture).
Solution to Part 2
Now the tough part is 2. Given that we have an active set of line segments enclosing the ball, how do we add a new line, and then adjust the size and number of line segments inside our active set. Note that at the beginning of the game there are precisely 4 lines in the active set which are the borders of the screen (this will be our base case if you like induction).
Now suppose we have an arbitrary active set containing the ball, and the user adds a new line, we assume it hits exactly two existing line segments and does not cross any line segments (by part 1 algorithm). Now we need to modify the active set. By algorithm 1, we know which line segments are hit by this point. So the only way that the active set can change is if both line segments hit by this point are in the active set (draw a picture to see this fact).
Now assume that the new line segment hits two lines inside the active set (this means it essentially splits the active set into two convex hulls). If we maintain our active set in a rotated order (that is to say we ensure that it is always sorted by angle about the positive x axis), we simply need to add our new points in a way that maintains this sorted ordering. So for instance suppose our points of the active set (collapsing line segments to single lines) are:
[(x1, y1), (x2, y2), (x3, y3), ..., (xn, yn), (x1, y1)]
And we want to add the new line segment ((x', y'), (x'', y'')), and our new set looks like:
[(x1, y1), (x2, y2), (x', y'), (x3, y3), ..., (xn, yn), (x'', y''), (x1, y1)]
Then there are now two convex hulls that are formed:
1. [(x1, y1), (x2, y2), (x', y'), (x'', y''), (x1, y1)]
2. [(x', y'), (x3, y3), ..., (xn, yn), (x'', y'')]
So the only remaining question is which is our new active set. Simply use algorithm 4.
Data Structures for Part 2
First we need the classes line segment and point (im avoiding implementing things like equals, hashcode for simplicity):
class Point {
public int x;
public int y;
}
class LineSegment {
public Point lineStart;
public Point lineEnd;
}
Now we can represent our active set datastructure:
class ActiveSet {
public List<Line> thesortedlist;
}
Now we first initialize our active set with four lines and denote the center of the screen as (0,0):
LineSegment TopBorder = new LineSegment(new Point(10, 10), new Point(-10, 10));
LineSegment LftBorder = new LineSegment(new Point(-10, 10), new Point(-10, -10));
LineSegment BtmBorder = new LineSegment(new Point(-10, -10), new Point(10, -10));
LineSegment RightBorder = new LineSegment(new Point(10, -10), new Point(10, 10));
ActiveSet activeset = new ActiveSet();
activeSet.theActiveLines.add(TopBorder);
activeSet.theActiveLines.add(LftBorder);
activeSet.theActiveLines.add(BtmBorder);
activeSet.theActiveLines.add(RightBorder);
Now say the user adds a line from point (0, 10) to (10, 0), so this is a diagonal (call it newSegment) and the new active set will look like:
------
- -
- -
- -
- -
- -
- B -
- -
-----------
Note the cut in the upper right corner, and B denotes the ball. Now by algorithm 1 we know that lines TopBorder and RightBorder are hit. Now both of these are inside the activeset (we can test membership faster by using a hashmap). Now we need to form two activesets as follows:
ActiveSet activesetLeft = new ActiveSet();
ActiveSet activesetRight = new ActiveSet();
Now in order to build these sets proceed as follows:
List<LineSegment> leftsegments = activeset.GetSegmentsBetween(TopBorder,
RightBorder);
List<RightSegment> rightsegments = activeset.GetSegmentsBetween(RightBorder,
TopBorder);
This function GetSegmentsBetween(LineSegment firstline, LineSegment secondline) should just locate firstline in the list and then return all elements until it finds secondline (this may need to do two passes through the list). It should not include these firstline or secondline in its return value. Now suppose we have activesetLeft and activesetright, we need to do the following:
activesetLeft.append(new Line(secondLine.lineStart, newSegment.lineStart));
activesetLeft.append(newSegment);
activesetLeft.append(new Line(newSegment.lineEnd, firstLine.lineEnd));
activesetRight.append(new Line(firstLine.lineStart, newSegment.lineEnd));
activesetRight.append(new Line(newSegment.lineEnd, newSegment.lineStart));
activesetRight.append(new Line(newSegment.lineStart, secondLine.lineEnd));
It is really hard to understand in code, but the order of everything above is important, sicne we want to maintain sorted going in counterclockwise order.
It is left as an exercise how you can speed this up (in fact you dont need to build two active sets, just first figure out if the ball is above or below the new segment and build the left or right activeset accordingly).
I have several boxes (x,y,width,height) randomly scattered around, and some of them need to be linked from point (x1,y1) in box1 to point (x2,y2) in box2 by drawing a line. I am trying to figure a way to make such line avoid passing through any other boxes (other than box1 and box2) by drawing several straight interconnected lines to go around any box in the way (if it is not possible to go with one straight line). The problem is that I don't know an algorithm for such thing (let alone having a technical/common name for it). Would appreciate any help in the form of algorithm or expressed ideas.
Thanks
Assuming that the lines can't be diagonal, here's one simple way. It's based on BFS and will also find the shortest line connecting the points:
Just create a graph, containing one vertex for each point (x, y) and for each point the edges:
((x,y),(x+1,y)) ((x,y),(x-1,y)) ((x,y),(x,y+1)) ((x,y),(x,y-1))
But each of this edges must be present only if it doesn't overlap a box.
Now just do a plain BFS from point (x1,y1) to (x2,y2)
It's really easy to obtain also diagonal lines the same way but you will need 8 edges for each vertex, that are, in addition to the previouses 4:
((x,y),(x-1,y+1)) ((x,y),(x-1,y-1)) ((x,y),(x+1,y-1)) ((x,y),(x+1,y+1))
Still, each edge must be present only if it doesn't overlap a box.
EDIT
If you can't consider space divided into a grid, here's another possibility, it won't give you the very shortest path, though.
Create a graph, in which each box is a vertex and has an edge to any other box that can be reached without the line to overlap a third box. Now find the shortet path using dijkstra between box1 and box2 containing the two points.
Now consider each box to have a small countour that doesn't overlap any other box. This way you can link the entering and the exiting point of each box in the path found through dijistra, passing through the countour.
Put all (x,y) coords of the corners of the boxes in a set V
Add the start- and end coordinates to V.
Create a set of edges E connecting each corner that does not cross any box-side (except for the diagonals in the boxes).
How to check if a line crosses a box side can be done with this algorithm
Now use a path-finding algorithm of your choice, to find a path in the graph (V, E).
If you need a simple algorithm that finds the shortest path, just go with a BFS.
(This will produce a path that goes along the sides of some boxes. If this is undesirable, you could in step 1 put the points at some distance delta from the actual corners.)
If the edges may not be diagonal:
Create a large grid of lines that goes between the boxes.
Throw away the grid-edges that cross a box-side.
Find a path in the grid using a path-finding algorithm of your choice.
I have a projectile that I would like to pass through specific coordinates at the apex of its path. I have been using a superb equation that giogadi outlined here, by plugging in the velocity values it produces into chipmunk's cpBodyApplyImpulse function.
The equation has one drawback that I haven't been able to figure out. It only works when the coordinates that I want to hit have a y value higher than the cannon (where my projectile starts). This means that I can't shoot at a downward angle.
Can anybody help me find a suitable equation that works no matter where the target is in relation to the cannon?
As pointed out above, there isn't any way to make the apex be lower than the height of the cannon (without making gravity work backwards). However, it is possible to make the projectile pass through a point below the cannon; the equations are all here. The equation you need to solve is:
angle = arctan((v^2 [+-]sqrt(v^4 - g*(x^2+2*y*v^2)))/g*x)
where you choose a velocity and plug in the x and y positions of the target - assuming the cannon is at (0,0). The [+-] thing means that you can choose either root. If the argument to the square root function is negative (an imaginary root) you need a larger velocity. So, if you are "in range" you have two possible angles for any particular velocity (other than in the maximum range 45 degree case where the two roots should give the same answer).
I suspect one trajectory will tend to 'look' much more sensible than the other, but that's something to play around with once you have something working. You may want to stick with the apex grazing code for the cases where the target is above the cannon.