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?
Related
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.
This is more of an open discussion topic than anything else. Currently I'm storing 50 Float32 values in my NSMutableArray *voltageArray before I refresh my CPTPlot *plot. Every time I obtain 50 values, I remove the previous 50 from the voltageArray and repeat the process....always displaying the 50 values in "real time" on my plot.
However, the data I'm receiving (which is voltage coming from a Cypress BLE module equipped with a pressure transducer) is so quick that any variation (0.4 V to 4.0 V; no pressure to lots of pressure) cannot be seen on my graph. It just shows up as a straight line, varying up and down without showing increased or decreased slopes.
To show overall change, I wanted to take those 50 values, store them in the first index of another NSMutableArray *stampArray and use the index of stampArray to display information. Meanwhile, the numberOfRecordsForPlot: method would look like this:
- (NSUInteger)numberOfRecordsForPlot:(CPTPlot *)plotnumberOfRecords {
return (DATA_PER_STAMP * _stampCount);
}
This would initially be 50, then after 50 pieces of data are captured from the BLE module, _stampCount would increase by one, and the number of records for plot would increase by 50 (till about 2500-10000 range, then I'd refresh the whole the thing and restart the process.)
Is this the right approach? How would I be able to make the first 50 points stay on the graph, while building the next 50, etc.? Imagine an y = x^2 graph, and what the graph looks like when applying integration (the whole breaking the area under the curve into rectangles).
Look at the "Real Time Plot" demo in the Plot Gallery example app included with Core Plot. It starts off with an empty plot, adding a new point each cycle until reaching the maximum number of points. After that, one old point is removed for each new one added so the total number stays constant. The demo uses a timer to pass random data to the plot, but your app can of course collect data from anywhere. Be sure to always interact with the graph from the main thread.
I doubt you'll be able to display 10,000 data points on one plot (does your display have enough pixels to resolve that many points?). If not, you'll get much better drawing performance if you filter and/or smooth the data to remove some of the points before sending them to the plot.
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.
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.