What is the quickest way to find the shortest cart

问题:

I have 1 red polygon say and 50 randomly placed blue polygons - they are situated in geographical 2D space. What is the quickest/speediest algorithim to find the the shortest distance between a red polygon and its nearest blue polygon?

Bear in mind that it is not a simple case of taking the points that make up the vertices of the polygon as values to test for distance as they may not necessarily be the closest points.

So in the end - the answer should give back the closest blue polygon to the singular red one.

This is harder than it sounds!

回答1:

I doubt there is better solution than calculating the distance between the red one and every blue one and sorting these by length.

Regarding sorting, usually QuickSort is hard to beat in performance (an optimized one, that cuts off recursion if size goes below 7 items and switches to something like InsertionSort, maybe ShellSort).

Thus I guess the question is how to quickly calculate the distance between two polygons, after all you need to make this computation 50 times.

The following approach will work for 3D as well, but is probably not the fastest one:

Minimum Polygon Distance in 2D Space

The question is, are you willing to trade accuracy for speed? E.g. you can pack all polygons into bounding boxes, where the sides of the boxes are parallel to the coordinate system axes. 3D games use this approach pretty often. Therefor you need to find the maximum and minimum values for every coordinate (x, y, z) to construct the virtual bounding box. Calculating the distances of these bounding boxes is then a pretty trivial task.

Here's an example image of more advanced bounding boxes, that are not parallel to the coordinate system axes:

Oriented Bounding Boxes - OBB

However, this makes the distance calculation less trivial. It is used for collision detection, as you don't need to know the distance for that, you only need to know if one edge of one bounding box lies within another bounding box.

The following image shows an axes aligned bounding box:

Axes Aligned Bounding Box - AABB

OOBs are more accurate, AABBs are faster. Maybe you'd like to read this article:

Advanced Collision Detection Techniques

This is always assuming, that you are willing to trade precision for speed. If precision is more important than speed, you may need a more advanced technique.

回答2:

You might be able to reduce the problem, and then do an intensive search on a small set.

Process each polygon first by finding:

• Center of polygon
• Maximum radius of polygon (i.e., point on edge/surface/vertex of the polygon furthest from the defined center)

Now you can collect, say, the 5-10 closest polygons to the red one (find the distance center to center, subtract the radius, sort the list and take the top 5) and then do a much more exhaustive routine.

回答3:

For polygon shapes with a reasonable number of boundary points such as in a GIS or games application it might be quicker easier to do a series of tests.

For each vertex in the red polygon compute the distance to each vertex in the blue polygons and find the closest (hint, compare distance^2 so you don't need the sqrt() ) Find the closest, then check the vertex on each side of the found red and blue vertex to decide which line segments are closest and then find the closest approach between two line segments.

See http://local.wasp.uwa.edu.au/~pbourke/geometry/lineline3d/ (it's easy to simply for the 2d case)

回答4:

This screening technique is intended to reduce the number of distance computations you need to perform in the average case, without compromising the accuracy of the result. It works on convex and concave polygons.

Find the the minimum distance between each pair of vertexes such that one is a red vertex and one is a blue. Call it r. The distance between the polygons is at most r. Construct a new region from the red polygon where each line segment is moved outward by r and is joined to its neighbors by an arc of radius r is centered at the vertex. Find the distance from each vertex inside this region to every line segment of the opposite color that intersects this region.

Of course you could add an approximate method such as bounding boxes to quickly determine which of the blue polygons can't possibly intersect with the red region.

回答5:

Maybe the Frechet Distance is what your looking for?

Computing the Fréchet distance between two polygonal curves
Computing the Fréchet Distance Between Simple Polygons

回答6:

I know you said "the shortest distance" but you really meant the optimal solution or a "good/very good" solution is fine for your problem?

Because if you need to find the optimal solution, you have to calculate the distance between all of your source and destination poligon bounds (not only vertexes). If you are in 3D space then each bound is a plane. That can be a big problem (O(n^2)) depending on how many vertexes you have.

So if you have vertex count that makes that squares to a scarry number AND a "good/very good" solution is fine for you, go for a heuristic solution or approximation.

回答7:

You might want to look at Voronoi Culling. Paper and video here:

http://www.cs.unc.edu/~geom/DVD/

回答8:

I would start by bounding all the polygons by a bounding circle and then finding an upper bound of the minimal distance. Then i would simply check the edges of all blue polygons whose lower bound of distance is lower than the upper bound of minimal distance against all the edges of the red polygon.

upper bound of min distance = min {distance(red's center, current blue's center) + current blue's radius}

for every blue polygon where distance(red's center, current blue's center) - current blue's radius < upper bound of min distance
    check distance of edges and vertices

But it all depends on your data. If the blue polygons are relatively small compared to the distances between them and the red polygon, then this approach should work nicely, but if they are very close, you won't save anything (many of them will be close enough). And another thing -- If these polygons don't have many vertices (like if most of them were triangles), then it might be almost as fast to just check each red edge against each blue edge.

hope it helps

回答9:

As others have mentioned using bounding areas (boxes, circles) may allow you to discard some polygon-polygon interactions. There are several strategies for this, e.g.

1. Pick any blue polygon and find the distance from the red one. Now pick any other polygon. If the minimum distance between the bounding areas is greater than the already found distance you can ignore this polygon. Continue for all polygons.
2. Find the minimum distance/centroid distance between the red polygon and all the blue polygons. Sort the distances and consider the smallest distance first. Calculate the actual minimum distance and continue through the sorted list until the maximum distance between the polygons is greater than the minimum distance found so far.

Your choice of circles/axially aligned boxes, or oriented boxes can have a great affect on performance of the algorithm, dependent on the actual layout of the input polygons.

For the actual minimum distance calculation you could use Yang et al's 'A new fast algorithm for computing the distance between two disjoint convex polygons based on Voronoi diagram' which is O(log n + log m).

回答10:

Gotta run off to a funeral in a sec, but if you break your polygons down into convex subpolies, there are some optimizations you can do. You can do a binary searches on each poly to find the closest vertex, and then I believe the closest point should either be that vertex, or an adjacent edge. This means you should be able to do it in log(log m * n) where m is the average number of vertices on a poly, and n is the number of polies. This is kind of hastey, so it could be wrong. Will give more details later if wanted.

回答11:

You could start by comparing the distance between the bounding boxes. Testing the distance between rectangles is easier than testing the distance between polygons, and you can immediately eliminate any polygons that are more than nearest_rect + its_diagonal away (possibly you can refine that even more). Then, you can test the remaining polygons to find the closest polygon.

There are algorithms for finding polygon proximity - I'm sure Wikipedia has a good review of them. If I recall correctly, those that only allow convex polygons are substantially faster.

回答12:

I believe what you are looking for is the A* algorithm, its used in pathfinding.

回答13:

The naive approach is to find the distance between the red and 50 blue objects -- so you're looking at 50 3d Pythagorean calculations + sorting to find the answer. That would only really work for finding the distance between center points though.

If you want arbitrary polygons, maybe your best best is a raytracing solution that emits rays from the surface of the red polygon with respect to the normal, and reports when another polygon is hit.

A hybrid might work -- we could find the distance from the center points, assuming we had some notion of the relative size of the blue polygons, we could cull the result set to the closest among those, then use raytracing to narrow down the truly closest polygon(s).