I've got a triangular mesh class which contains a list of nodes (2d in my case but that shouldn't matter) and a list of faces. Each face is a triangle and it only contains the indices into the node array. The mesh comes out of a Delaunay algorithm so it's very clean.
For every node in the mesh I need to find which nodes are connected to it with a single edge. What would be a fast way to construct and search this topology database?
Much obliged,
David Rutten
There are two somewhat-standard data structs that facilitate mesh topology-queries. One is Winged Edges (commonly referred to also as half-edge), and the other is Directed Edges. Google around and you'd get kajillions of details, and various-level intros into each one.
Don't know enough about your scenario to recommend one of them. E.g., directed edges is storage-optimized, and best suited for very large meshes. Winged edges is considered a 'classic', and is a good starting point for more advanced flavours.
Actually if you're certain that's the only query you'd need, then both are an overkill and you'd do just fine with a single hash. If, however, you find yourself in need of efficient answers to queries like -
Which faces use this vertex?
Which edges use this vertex?
Which faces border this edge?
Which edges border this face?
Which faces are adjacent to this
face?
You should consider diving into one of them.
I think I've stared myself blind on HashTables, Dictionaries and Sorted Lists... The following is probably the easiest and fastest:
Public Sub SolveConnectivity(ByVal nodes As Node2List, ByVal faces As List(Of Face))
m_map = New List(Of List(Of Int32))(nodes.Count)
'Create blank lists
For i As Int32 = 0 To nodes.Count - 1
m_map.Add(New List(Of Int32)(6))
Next
'Populate connectivity diagram
For i As Int32 = 0 To faces.Count - 1
Dim F As Face = faces(i)
m_map(F.A).Add(F.B)
m_map(F.A).Add(F.C)
m_map(F.B).Add(F.A)
m_map(F.B).Add(F.C)
m_map(F.C).Add(F.A)
m_map(F.C).Add(F.B)
Next
End Sub
Related
So basically im doing this for my minecraft spigot plugin (java). I know there are already some land claim plugins but i would like to make my own.
For this claim plugin i'd like to know how to get if a point (minecraft block) is inside a region (rectangle). i know how to check if a point is inside a rectangle, the main problem is how to check as quickly as possible when there are like lets say 10.000 rectangles.
What would be the most efficient way to check 10.000 or even 100.000 without having to manually loop through all of them and check every single rectangle?
Is there a way to add a logical test when the rectangles get generated in a way that checks if they hold that point? In that case you could set a boolean to true if they contain that point when generated, and then when checking for that minecraft block the region (rectangle) replies with true or false.
This way you run the loops or checks when generating the rectangles, but when running the game the replies should happen very fast, just check if true or false for bool ContainsPoint.
If your rectangles are uniformly placed neighbors of each other in a big rectangle, then finding which rectangle contains point is easy:
width = (maxX-minX)/num_rectangles_x;
height = same but for y
idx = floor( (x - minX)/width );
idy = floor( (y - minY)/height );
id_square = idx + idy*num_rectangles_x;
If your rectangles are randomly placed, then you should use a spatial acceleration structure like octree. Then check if point is in root, then check if point is in one of its nodes, repeat until you find a leaf that includes the point. 10000 tests per 10milliseconds should be reachable on cpu. 1 million tests per 10ms should be ok for a gpu. But you may need to implement a sparse version of the octree and a space filling curve order for leaf nodes to have better caching, to reach those performance levels.
I had the idea of creating a fantasy city, and to avoid having the same house over and over, but not have to manually create hundreds of houses I was thinking on creating collections like "windows", "doors", "roofs", etc, and then create objects with vertex's assigned to specific groups with the same names ("windows" vertex groups, "doors" vertex groups, etc), and then have blender pick for each instance of a house a random window for each of the vertex in the group, same for doors, roofs, etc.
Is there a way of doing this? (couldn't find anything online), or do I need to create a custom addon? If so, any good reference or starting point where something close to this is done?
I've thought of particle systems, or child objects, but couldn't find a way to attach to the vertex a random part of the collection. Also thought of booleans, but it doesn't have an option to attach to specific vertex, nor to use collections. So I'm out of ideas of how to approach this.
What I have in mind:
Create basic shape, and assign vertex to the "windows" vertex group:
https://i.imgur.com/DAkgDR3.png
And then have random objects within the "Windows" collection attached to those vertex, as either a particle or modifier:
https://i.imgur.com/rl5BDQL.png
Thanks for any help :)
Ok, I've found a way of doing this.
I'm using 3 particle systems (doors, roofs and windows), each using vertex as emitters, and using vector groups to define where to display one of each the different options.
To avoid the particle emitter to put more than one object per vertex, I created a small script that counts the number of vertex of each vertex group and updates each of the particle system Emission number accordingly.
import bpy
o = bpy.data.objects["buildings"]
groups = ["windows", "doors", "roofs"]
for group in groups:
vid = o.vertex_groups.find(group)
vectors = [ v for v in o.data.vertices if vid in [ vg.group for vg in v.groups ] ]
bpy.data.particles[group].count = len(vectors)
I've used someone's code from stack overflow for counting the number of vectors in a vector group, but can't find again the link to the specific question, so if you see your code here, please do comment and I'll update my answer with the proper credit.
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 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.