cooldude832 Posted August 16, 2007 Share Posted August 16, 2007 You would have to develop some sort of alogrthim that would say from any point (x,y) you can get to any other point (x1,y1) in a straight line path. If that path leads to a path that is pointless it will not work. Like I said a quality structure for each point would need to be developed to rate a point. Quote Link to comment Share on other sites More sharing options...
btherl Posted August 17, 2007 Share Posted August 17, 2007 That sounds like Dijkstra's algorithm to me. Or perhaps A* (Dijkstra's algorithm augmented with heuristic evaluations for each point). Quote Link to comment Share on other sites More sharing options...
cooldude832 Posted August 17, 2007 Share Posted August 17, 2007 yeah,except its a lot more linear because that method invovles an infinite number of points not finite (we have only counting numbers as possible solutions) Quote Link to comment Share on other sites More sharing options...
btherl Posted August 17, 2007 Share Posted August 17, 2007 An infinite number of points? How many computers do you know that can handle an infinite number of points? I'm sure that those algorithms are intended for finite graphs only. Quote Link to comment Share on other sites More sharing options...
cooldude832 Posted August 17, 2007 Share Posted August 17, 2007 that is the differential aspect to it. Differentials and calculus have stayed away from consumer level computers because of the large amount of processing needed, however I have seen programs handle the inifinte idea (using differentials) such as maple mathematics lab. The distance as I said is an integration of the sum of each weighted value (in this case an area of dx/dy or dy/dx depending on our summing mechanics) however since the weight of each node is equal (a 1x1 square of area) we can solve using a finite number of points. Quote Link to comment Share on other sites More sharing options...
btherl Posted August 17, 2007 Share Posted August 17, 2007 What exactly are you integrating here? We have a set V of vertexes, E of edges (equal weighted). We can find sums of paths, sure. And we can think about directions, since we're on a grid. Where does integration come into that? The direct distance from one point to another is always sqrt(abs(x1-x2)^2 + abs(y1-y2)^2), the high school algorithm. The actual distance is what we are calculating. No shortcuts there. Quote Link to comment Share on other sites More sharing options...
cooldude832 Posted August 17, 2007 Share Posted August 17, 2007 Yes we are calculating distances and the distance formula is a linearization of a basic rule of integration. Distance is the definition of the anti-derrivative of dv/dt (in a motion sense). I think we are getting off hte php point of this though. Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.