CODEWITHSUNDEEP
iphoneobjective-cgps
edrai
edrai

Reputation: 111

Finding the closest point to a given point

I have searched all over for this, but I can't seem to find the best approach to this. I have about 22000 lat/lon points and I want to find the closest one's to the current location of the iPhone. I've seen people ask about Quad Trees, Dijkstra's Algorithm, and spatial databases. Which is the best for the iPhone? Spatial databases seem easiest, but I am not sure.

EDIT: there are actually over 20,000 points. You think iterating through all of them is the way to do it? But thanks for you input.

Thanks.

Upvotes: 11

Views: 12679

Answers (6)

Noomak
Noomak

Reputation: 371

you must consider that to use Dijkstra you must know your node position in the graph, that is instead the problem you're trying to solve; you're not in the graph, but you want to know the closest point to you

so simply, as already Chaos told you, you must calculate all distances beetween your position and all 20.000 points, then sort them

Upvotes: 0

Roger
Roger

Reputation:

I think this algorithm works:

Create an array sorted by latitude Create an array sorted by longitude

To find the closest, first find the closest by latitude by doing a binary search in the latitude array. Do the same for the longitude array. Now you have 2 points, one closest by latitude, the other closest by longitude. Compute the distances to each point via the pythagorean theorem. The closest point wins.

Roger

Upvotes: -3

Jane Sales
Jane Sales

Reputation: 13546

Why not tile the globe into regions? (Think hexes.) Then, either when you add points to your list, or in one big pre-processing loop, for each point, store the region it is.

Then, when searching for points near point A in hex X, you only need to check points in hex X and a maximum of 3 neighbouring hexes.

If this is still too many points to check, add subregions.

Upvotes: 1

Jon Watte
Jon Watte

Reputation:

If you need better than O(N), you can only get that if you first pay N lg N for building a spatial hash of some sort (a quadtree, octree, hash grid, or similar). Then each test will be approximately O(lg N), and can be much better typically by caching the last location you checked, if there's a lot of coherency (generally, there is).

I would probably build an octree in Euler (geocentric, XYZ) space, because that allows me to get "true" distance, not "warped" lat/lon distance. However, in practice, a quad tree in lat/lon space will probably work well enough. Once you have a hit, you hold on to that tree node (assuming the tree isn't re-built at runtime), and the next query starts walking from that tree node, and only needs to worry about nodes that may be closer if the previous point moved further away from the previous answer.

Upvotes: 5

Rog
Rog

Reputation: 17170

As you are on iPhone, you can use CoreLoaction to perform the geographic distance - using CLLocation's – getDistanceFrom:

I would be tempted to use a brute force linear search though all 2k points nad, if that isn't fast enough, switch to something like GeoHash to store meta data against your points for search.

Upvotes: 2

Alex Taylor
Alex Taylor

Reputation: 7168

Actually, it is best to use Haversine (great circle) calculation for Lat/Long points, otherwise increasingly large distances will be wrong, especially if you use simple trig like in Jherico's answer.

A quick search provides this javascript example:

var R = 6371; // km Radius of earth
var dLat = (lat2-lat1).toRad();
var dLon = (lon2-lon1).toRad(); 
var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
        Math.cos(lat1.toRad()) * Math.cos(lat2.toRad()) * 
        Math.sin(dLon/2) * Math.sin(dLon/2); 
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); 
var d = R * c;

In terms of the datastructure, Geohash is worth looking at.

Upvotes: 5

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