distance

I'm working on a 2D game that has a huge amount of dynamic entities. For fun's sake, let's call them soldiers, and let's say there are 50000 of them (which I just randomly thought up, it might be much more or much less :)). All these soldiers are moving every frame according to rules - think boids / flocking / steering behaviour. For each soldier, to update it's movement I need the X soldiers that are closest to the one I'm processing. What would be the best spatial hierarchy to store them to facilitate calculations like this without too much overhead ? (All entities are updated/moved every

A basic/common class in R is called "dist" , and is a relatively efficient representation of a symmetric distance matrix. Unlike a "matrix" object, however, there does not seem to be support for manipulating an "dist" instance by index pairs using the "[" operator. For example, the following code returns nothing, NULL , or an error: # First, create an example dist object from a matrix mat1 <- matrix(1:100, 10, 10) rownames(mat1) <- 1:10 colnames(mat1) <- 1:10 dist1 <- as.dist(mat1) # Now try to access index features, or index values names(dist1) rownames(dist1) row.names(dist1) colnames(dist1)

What would be the best approach to compare two hexadecimal file signatures against each other for similarities. More specifically, what I would like to do is to take the hexadecimal representation of an .exe file and compare it against a series of virus signature. For this approach I plan to break the file (exe) hex representation into individual groups of N chars (ie. 10 hex chars) and do the same with the virus signature. I am aiming to perform some sort of heuristics and therefore statistically check whether this exe file has X% of similarity against the known virus signature. The simplest

I have a collection of stores with a geospacial index on the location propery. What I am trying to do is given the user's latitude, latitude and a search radius (mi), I want to return the list of stores that are within those parameters. I saw the following example on the MongoDB documentation ( http://www.mongodb.org/display/DOCS/Geospatial+Indexing ), but it looks like the distance is in radians. center = [50, 50] radius = 10 db.places.find({"loc" : {"$within" : {"$center" : [center, radius]}}}) So, what is the formula to convert miles to radians? Solution The awesome people at mongodb-user

In PHP, I have the following code for calculating the distance between two locations: <?php function distance($lat1, $long1, $lat2, $long2) { // DEGREE TO RADIAN $latitude1 = $lat1/180*pi(); $longitude1 = $long1/180*pi(); $latitude2 = $lat2/180*pi(); $longitude2 = $long2/180*pi(); // FORMULA: e = ARCCOS ( SIN(Latitude1) * SIN(Latitude2) + COS(Latitude1) * COS(Latitude2) * COS(Longitude2-Longitude1) ) * EARTH_RADIUS $distance = acos(sin($latitude1)*sin($latitude2)+cos($latitude1)*cos($latitude2)*cos($longitude2-$longitude1))*6371; return $distance; } echo distance(9.9921962, 53.5534074, 9

I have a distance as a float and I'm looking for a way to format it nicely for human readers. Ideally, I'd like it to change from m to km as it gets bigger, and to round the number nicely. Converting to miles would be a bonus. I'm sure many people have had a need for one of these and I'm hoping that there's some code floating around somewhere. Here's how I'd like the formats: 0-100m: 47m (as a whole number) 100-1000m: 325m or 320m (round to the nearest 5 or 10 meters) 1000-10000m: 1.2km (round to nearest with one decimal place) 10000m +: 21km If there's no code available, how can I write my

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! I doubt there is better solution than calculating