2 1
1----------2---------4
| | |
|3 |3 |1
| 6 | |
3---------5 ---------
Ok, so this is the graph. My source node is 1 and destination node is 5
My question is.
Are both the algorithms going to give the same output or not?
That is, will both return 1->2->4->5? (Except that negative weights are not allowed in dijkstra's)
Thanks in advance for help.
Bellman-Ford algorithm is a single-source shortest path algorithm, which allows for negative edge weight and can detect negative cycles in a graph.
Dijkstra algorithm is also another single-source shortest path algorithm. However, the weight of all the edges must be non-negative.
For your case, as far as the total cost is concerned, there will be no difference, since the edges in the graph have non-negative weight. However, Dijkstra's algorithm is usually used, since the typical implementation with binary heap has Theta((|E|+|V|)log|V|) time complexity, while Bellman-Ford algorithm has O(|V||E|) complexity.
If there are more than 1 path that has minimum cost, the actual path returned is implementation dependent (even for the same algorithm).
The vertexes in Dijkstra’s algorithm contain the whole information of a network. There is no such thing that every vertex only cares about itself and its neighbors. On the other hand, Bellman-Ford algorithm’s nodescontain only the information that are related to. This information allows that node just to know about which neighbor nodes can it connect and the node that the relation come from, mutually. Dijkstra’s algorithm is faster than Bellman-Ford’s algorithm however the second algorithm can be more useful to solve some problems, such as negative weights of paths.
Djikstra algorithm is a greedy technique where as for implementing bellmanford algorithm we require dynamic approach.
In djikstra algo, we do relaxation of every node/vertices in a loop where as in bellmanford algo we perform relaxation only|v-1| times .
The Dijkstra algo is suitable for the graph when it's edge-weight are positive, and fails in the case of negative-edged graph where as the bellmanford algo have advantage over djikstra that it can implement even when the edge's weight are assign negatively.
the bellmanford can find whether the graph solution exists or not (ie given directed graph has negative weight cycle or not) where as the djikstra algo fails to do the same.
the time complexity of djikstra algo is O(v^2) when implemented by linear array,O(e.log v)when implemented with binay heap or Fibonacci heap where as bellmanford algo has O(|v| |e|) time complexity .
Related
Let be G a given undirected simple graph, with edge-weights w. Does there exist an algorithm with time complexity O((n+m)log^*(n+m)) that counts the number of node pairs (u,v) for which there exists a path from u to v with total weight under some given constant W? Looking for either an algorithm or a proof that no such algorithm exists.
I’ve tried union find + DFS yet it doesn’t seem that will only use up to n+m calls to Find/Union... I’ve also tried dis-proving the existence on an algorithm by solving APSP with time complexity lower than the lower bound, yet to no avail.
Successfully shown no such algorithm exists:
Assume by contradiction that such an algorithm exists.
Let G be a given undirected un-weighted graph, we’ll compute the diameter of the graph as follows:
Assign weight 1 to every edge on the graph. Thus the weighted diameter is equal to un-weighted diameter.
We find that the diameter of the graph is bounded by m
Perform a binary search using the algorithm to find the diameter of the graph (check for each value whether or not the count returns zero).
Overall, we find the diameter of G in O((n+m)log(n)log*(n+m)). The current best algorithms for finding a graph diameter are O(min(nm, ~n^2.4)). Thus, at the very least, if this algorithm succeeded then the time complexity of computing a diameter would decrease dramatically. Not entirely a proof, but good enough for the purpose I needed it for.
I'm trying to solve the standard bipartization problem, i.e., find a subset of the edges such that the output graph is a bipartite graph.
My additional constraints are:
The number of vertices on each side must be equal.
Each vertex has exactly 1 edge.
In fact, it would suffice to know whether such a subset exists at all - I don't really need the construction itself.
Optimally, the algorithm should be fast as I need to run it for O(400) nodes repeatedly.
If each vertex is to be incident on exactly one edge, it seems what you want is a matching. If so, Edmonds's blossom algorithm will do the job. I haven't used an implementation of the algorithm to recommend. You might check out http://www.algorithmic-solutions.com/leda/ledak/index.htm
I was wondering what's the difference between uniform-cost search and Dijkstra's algorithm. They seem to be the same algorithm.
Dijkstra's algorithm, which is perhaps better-known, can be regarded
as a variant of uniform-cost search, where there is no goal state and
processing continues until all nodes have been removed from the
priority queue, i.e. until shortest paths to all nodes (not just a
goal node) have been determined
http://en.wikipedia.org/wiki/Uniform-cost_search#Relationship_to_other_algorithms
Dijkstra's algorithm searches for shortest paths from root to every other node in a graph, whereas uniform-cost searches for shortest paths in terms of cost to a goal node.
Also, uniform cost has less space requirements, whereas the priority queue is filled "lazily" opposed to Dijkstra's, which adds all nodes to the queue on start with an infinite cost.
Compilation of other answers by NotAUser, dreaMone and Bruno Calza
Dijkstra's Algorithm finds the shortest path from the root node to every other node. uniform cost searches for shortest paths in terms of cost from the root node to a goal node. Uniform Cost Search is Dijkstra's Algorithm which is focused on finding a single shortest path to a single finishing point rather than the shortest path to every point.
UCS does this by stopping as soon as the finishing point is found. For Dijkstra, there is no goal state and processing continues until all nodes have been removed from the priority queue, i.e. until shortest paths to all nodes (not just a goal node) have been determined.
UCS has fewer space requirements, where the priority queue is filled gradually as opposed to Dijkstra's, which adds all nodes to the queue on start with an infinite cost.
As a result of the above points, Dijkstra is more time consuming than UCS
UCS is usually formulated on trees while Dijkstra is used on general graphs
Djikstra is only applicable in explicit graphs where the entire graph is given as input. UCS starts with the source vertex and gradually traverses the necessary parts of the graph. Therefore, it is applicable for both explicit graphs and implicit graphs (where states/nodes are generated).
There's a paper that talk about the similarities and differences about both.
http://www.aaai.org/ocs/index.php/SOCS/SOCS11/paper/view/4017/4357
The main difference is that Dijkstra's algorithm is defined when numbers of vertices is finite. It says to put all the vertices in a queue. But we can not put all the vertices in a queue when numbers of vertices tend to infinite. Uniform Cost Search is defined in a situation like this, where numbers of vertices are unknown.
I have a directed acyclic weighted graph which I want to traverse.
The constraints for a valid solution route are:
The sum of the weights of all edges traversed in the route must be the highest possible in the graph, taking in mind the second constraint.
Exactly N vertices must have been visited in the chosen route (including the start and end vertex).
Typically the graph will have a high amount of vertices and edges, so trying all possibilities is not an option, and requires quite an efficient algorithm.
Looking for some pointers or a suitable algorithm for this problem. I know the first condition is easily fulfilled using Dijkstra's algorithm, but I am not sure how to incorporate the second condition, or even where to begin to look.
Please let me know if any additional information is needed.
I'm not sure if you are interested in any path of length N in the graph or just path between two specific vertices; I suspect the latter, but you did not mention that constraint in your question.
If the former, the solution should be a trivial Dijkstra-like algorithm where you sort all edges by their potential path value that starts at the edge weight and gets adjusted by already built adjecent paths. In each iteration, take the node with the best potential path value and add it to an adjecent path. Stop when you get a path of length N (or longer that you cut off at the sides). There are some other technical details esp. wrt. creating long paths, but I won't go into details as I suspect this is not what you are interested in. :-)
If you have fixed source and sink, I think there is no deep magic involved - just run a basic Dijkstra where a path will be associated with each vertex added to the queue, but do not insert vertices with path length >= N into the queue and do not insert sink into the queue unless its path length is N.
I am looking for a Dijkstra's algorithm implementation, that also takes into consideration the number of nodes traversed.
What I mean is, a typical Dijkstra's algorithm, takes into consideration the weight of the edges connecting the nodes, while calculating the shortest route from node A to node B. I want to insert another parameter into this. I want the algorithm to give some weightage to the number of nodes traversed, as well.
So that the shortest route computed from A to B, under certain values, may not necessarily be the Shortest Route, but the route with the least number of nodes traversed.
Any thoughts on this?
Cheers,
RD
Edit :
My apologies. I should have explained better. So, lets say, the shortest route from
(A, B) is A -> C -> D -> E -> F -> B covering a total of 10 units
But I want the algorithm to come up with the route A -> M -> N -> B covering a total of 12 units.
So, what I want, is to be able to give some weightage to the number of nodes as well, not just the distance of the connected nodes.
Let me demonstrate that adding a constant value to all edges can change which route is "shortest" (least total weight of edges).
Here's the original graph (a triangle):
A-------B
\ 5 /
2 \ / 2
\ /
C
Shortest path from A to B is via C. Now add constant 2 to all edges. The shortest path becomes instead the single step from A directly to B (owing to "penalty" we've introduced for using additional edges).
Note that the number of edges used is (excluding the node you start from) the same as the number of nodes visited.
The way you can do that is adapt the weights of the edges to always be 1, so that you traverse 5 nodes, and you've gone a distance of "5". The algorithm would be the same at that point, optimizing for number of nodes traversed rather than distance traveled.
If you want some sort of hybrid, you need to determine how much importance to give to traversing a node and the distance. The weight used in calculations should look something like:
weight = node_importance * 1 + (1 - node_importance) * distance
Where node_importance would be a percentage which gauges how much distance is a factor and how much minimum node traversal is important. Though you may have to normalize the distances to be an average of 1.
I'm going to go out on a limb here, but have you tried the A* algorithm? I may have understood your question wrong, but it seems like A* would be a good starting point for what you want to do.
Check out: http://en.wikipedia.org/wiki/A*_search_algorithm
Some pseudo code there to help you out too :)
If i understood the question correctly then its best analogy would be that used to find the best network path.
In network communication a path may not only be selected because it is shortest but has many hop counts(node), thus may lead to distortion, interference and noise due to node connection.
So the best path calculation contains the minimizing the function of variables as in your case Distance and Hop Count(nodes).
You have to derive a functional equation that could relate the distance and node counts with quality.
so something as suppose
1 hop count change = 5 unit distance (which means the impact is same for 5unit distace or 1 node change)
so to minimize the loss you can use it in the linear equation.
minimize(distance + hopcount);
where hopcount can be expressed as distance.