Why doesn't the A* algorithm devide the edge's cost by the amount of distance one gets closer to the end node? As far as I know A* adds the distance to the end node to the weight of an edge to sort the possible nodes. But wouldn't it be more efficient to devide an edges cost by the amount of distance one gets closer to the end node?
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Suppose we have a set of N points on the cartesian plane (x_i and y_i). Suppose we connect those points with lines.
Is there any way like using a graph and something like a shortest path algorithm or minimum spanning tree so that we can reach any point starting from any point but minimizing the total length of the lines??
I though that maybe I could set the cost of the edges with the distance of a graph and use a shortest path algorithm but I'm not sure if this is possible.
Any ideas ?
I'm not 100% sure what you want, so I go for two algorithms.
First: you just want a robust algorithm, use dijkstras algorithm. The only challange left is to define the edge cost. Which would be 1 for neighboring nodes, I assume.
Second: you want to use heuristics to estimate the next best node and optimize time consumption. Use A*, but you need to write a heuristic which under estimates the distance. You could use the euclidean distance to do so. The edge problematic stays the same.
"Given a connected undirected weighted graph, find the maximum weight of the edge in the path from s to t where the maximum weight of the edges in the path is the minimum."
This seems like a Floyd–Warshall algorithm problem. Is there an approach faster than O(V^3)?
I submit that a breadth first search (BFS) to determine whether s is connected to t by any path can run in O(V) time, since each vertex need only be visited once.
I propose therefore, that the solution to this problem is to sort the edges by weight, and then
while the BFS succeeds in finding a path from s to t
remove the highest weighted edge from the graph
The last edge to be removed before the BFS fails is the maximum weighted edge that you're looking for.
Total running time is O(E log E) to sort the edges by weight, plus O(VE) to run the BFS until removing an edge disconnects the graph.
My question is with regard to the increase/decrease of the diameter of a network. I'm thinking that as one adds more nodes to an existing network, the density should effectively increase and the probability of the edges created by the new nodes could result in higher degree of clustering. If this is the case, my assumption is that the diameter of the network should decrease as we add more nodes, owing to the probability that shorter geodesic paths can now exist and become the new diameter. Am I wrong with this logic? Or is there a better explanation or perhaps something I'm missing?
Work by Leskovec, Kleinberg, and Faloutsos has examined this question specifically [1,2]. They find:
"First, graphs densify over time, with the number of edges crowing super-linearly in the number of nodes. Second, the average distance between nodes often shrinks over time, in contrast to the conventional wisdom that such distance parameters should increase slowly as a function of the number of nodes."
Is it possible to modify A* to return the shortest path with the least number of turns?
One complication: Nodes can no longer be distinguished solely by their location, because their parent node is relevant in determining future turns, so they have to have a direction associated with them as well.
But the main problem I'm having, is how to work number of turns into the partial path cost (g). If I multiply g by the number of turns taken (t), weird things are happening like: A longer path with N turns near the end is favored over a shorter path with N turns near the beginning.
Another less optimal solution I'm considering is: After calculating the shortest path, I could run a second A* iteration (with a different path cost formula), this time bounded within the x/y range of the shortest path, and return the path with the least turns. Any other ideas?
The current "state" of the search is actually represented by two things: The node you're in, and the direction you're facing. What you want is to separate each of those states into different nodes.
So, for each node in the initial graph, split it into E separate nodes, where E is the number of incoming edges. Each of these new nodes represents the old node, but facing in different directions. The outgoing edges of these new nodes will all be the same as the old outgoing edges, but with a different weight. If the old weight was w, then...
If the edge doesn't represent a turn, make the new weight w as well
If the edge does represent a turn, make the new weight w + ε, where ε is some number significantly smaller than the smallest weight.
Then just do a normal A* search. Since none of the weights have decreased, your heuristic will still be admissible, so you can still use the same heuristic.
If you really want to minimize the number of turns (as opposed to finding a nice tradeoff between turns and path length), why not transform your problem space by adding an edge for every pair of nodes connected by an unobstructed straight line; these are the pairs you can travel between without a turn. There are O(n) such edges per node, so the new graph is O(n3) in terms of edges. That makes A* solutions as much as O(n3) in terms of time.
Manhattan distance might be a good heuristic for A*.
Is it possible to modify A* to return the shortest path with the least number of turns?
It is most likely not possible. The reason being that it is an example of the weight-constrained shortest path problem. It is therefore NP-Complete and cannot be solved efficiently.
You can find papers that discuss solving this problem e.g. http://web.stanford.edu/~shushman/math15_report.pdf
I just want to make sure this would work. Could you find the greatest path using Dijkstra's algorithm? Would you have to initialize the distance to something like -1 first and then change the relax subroutine to check if it's greater?
This is for a problem that will not have any negative weights.
This is actually the problem:
Suppose you are given a diagram of a telephone network, which is a graph
G whose vertices represent switches centers, and whose edges represent communication
lines between two centers. The edges are marked by their bandwidth of its lowest
bandwidth edge. Give an algorithm that, given a diagram and two switches centers a
and b, will output the maximum bandwidth of a path between a and b.
Would this work?
EDIT:
I did find this:
Hint: The basic subroutine will be very similar to the subroutine Relax in Dijkstra.
Assume that we have an edge (u, v). If min{d[u],w(u, v)} > d[v] then we should update
d[v] to min{d[u],w(u, v)} (because the path from a to u and then to v has bandwidth
min{d[u],w(u, v)}, which is more than the one we have currently).
Not exactly sure what that's suppose to mean though since all distance are infinity on initialization. So, i don't know how this would work. any clues?
I'm not sure Djikstra's is the way to go. Negative weights do bad, bad things to Djikstra's.
I'm thinking that you could sort by edge weight, and start removing the lowest weight edge (the worst bottleneck), and seeing if the graph is still connected (or at least your start and end points). The point at which the graph is broken is when you know you took out the bottleneck, and you can look at that edge's value to get the bandwidth. (If I'm not mistaken, each iteration takes O(E) time, and you will need O(E) iterations to find the bottleneck edge, so this is an O(E2) algorithm.
Edit: you have to realize that the greatest path isn't necessarily the highest bandwidth: you're looking to maximize the value of min({edges in path}), not sum({edges in path}).
You can solve this easily by modifying Dijkstra's to calculate maximum bandwidth to all other vertices.
You do not need to initialize the start vertex to -1.
Algorithm: Maximum Bandwidth(G,a)
Input: A simple undirected weighted graph G with non -ve edge weights, and a distinguished vertex a of G
Output: A label D[u], for each vertex u of G, such that D[u] is the maximum bandwidth available from a to u.
Initialize empty queue Q;
Start = a;
for each vertex u of G do,
D[u] = 0;
for all vertices z adjacent to Start do{ ---- 1
If D[Start] => D[z] && w(start, z) > D[z] {
Q.enqueue(z);
D[z] = min(D[start], D[z]);
}
}
If Q!=null {
Start = Q.dequeue;
Jump to 1
}
else
finish();
This may not be the most efficient way to calculate the bandwidth, but its what I could think of for now.
calculating flow may be more applicable, however flow allows for multiple paths to be used.
Just invert the edge weights. That is, if the edge weight is d, consider it instead as d^-1. Then do Dijkstra's as normal. Initialize all distances to infinity as normal.
You can use Dijkstra's algorithm to find a single longest path but since you only have two switch centers I don't see why you need to visit each node as in Dijkstra's. There is most likes a more optimal way of going this, such as a branch and bound algorithm.
Can you adjust some of the logic in algorithm AllPairsShortestPaths(Floyd-Warshall)?
http://www.algorithmist.com/index.php/Floyd-Warshall%27s_Algorithm
Initialize unconnected edges to negative infinity and instead of taking the min of the distances take the max?