I am aware of the fact that I have to apply Dijkstra's algorithm to get an answer.The entire algorithm is explained in depth in one of the answers .
However why do we need to apply Dijkstra's algorithm to this problem.According to my knowledge Dijkstra will find the shortest distance path.
But the problem setter has clearly asked for minimum cost path.Considering this should'nt we apply Prim's algorithm to the question and find the MST for the entire chess board.
Here is the link to the problem.
Dijkstra's algorithm is indeed for finding the shortest distance path. However, note that "distance" does not have to mean distance measured in the normal way (i.e. with a ruler).
In fact Dijkstra's algorithm will also work for finding the shortest cost path in any network (providing that all the costs are greater than or equal to zero). All you need to do is to define the distance between any two nodes to be equal to the cost of the corresponding edge.
So, in this problem, when they search for the shortest path, they are defining distance in terms of the cost function defined in the problem.
Related
Can we use Dijkstra's algorithm to find cycles???
Negative cycles
Positive cycles
If we can what, are the changes we have to do?
1) Dijkstra's doesn't work on graphs with negative edges because you can (possibly) find a minimum distance of negative infinity.
2) Um, you normally run it on graphs with cycles (otherwise, you might as well be traversing a tree), so it can handle them just fine.
If your real question is just about finding cycles, look at Finding all cycles in graph
No We cant use Dijkstra algorithm if negative cycles exist as the algorithm works on the shortest path and for such graphs it is undefined.Once you get to a negative cycle, you can bring the cost of your "shortest path" as low as you wish by following the negative cycle multiple times.
This type of restriction is applicable to all sort of algorithms to find shortest path in a graph and this is the same reason that prohibits all negative edges in Dijkstra.
You may modify Dijkstra to find the cycles in the graph but i do not think it is the best practice.
Rather you can possibility Use:
Tarjan's strongly connected components algorithm(Time complexity -O(|E| + |V|))
or Kosaraju's algorithm (uses DFS and is a linear time alogoritham)
or you may follow this link for better idea:
https://en.wikipedia.org/wiki/Strongly_connected_component
Hope I answered your question.
I know Bellman Ford algorithm works well with negative weighted Graph, But I've developed a code of Dijkstra Algorithm that works very well. But it fails when I insert negative weighted edges. Any Solution?
I think we can't do that, as Dijkstra algorithm will not find a final way to reach at destination vertex because it may get stuck in loops, it is not made for negative weighted graphs, you should go for bellman ford algorithm
Actually, it is possible in a special case. Dijkstra algorithm will fail, if there is an edge of a negative length. However, if all edges are of negative length, then you may inverse the lengths of all edges and use the algorithm to find the longest path between two vertices of the graph (the path will represent the shortest path in the original graph).
But in a general case, as you have stated, it is not possible to use Dijkstra. If there is no cycle of negative length, than you shall use Bellman-Ford algorithm. If you cannot guarantee that there no cycle of negative length, than the problem is NP-complete and there is no known polynomial algorithm.
As you stated, Bellman Ford is the algorithm of choice for finding the shortest path in a graph with negative weights. The issue with using Dijkstra's Algorithm in this scenario is that Dijkstra's assumes that all possible subpaths from s to t in a graph must be smaller than the goal subpath, which is not necessarily true when negative edge weights are added.
If all edge weights are positive, this guarantees that adding more edges to a path makes it longer. Knowing this, Dijkstra's algorithm will discard any paths that are longer than the shortest one it has found to some vertex since there is no chance of the long path becoming shorter than the short path.
However, this assumption is not true if there are negative edge weights since we could have an extremely long path P that we discarded, but somewhere down the road, P can pass through a very negative edge and become shorter than the shortest path you currently have. Therefore, we can't guarantee that a path we've found is the shortest at any stage in Dijkstra's algorithm, which the algorithm relies on.
I am trying to do c++ program.I am trying to do problem in which i have numbers of points. Now i need to find the path that goes through all the points. This is not actually TSP because as per my knowledge in TSP it is possible to travel from all points to every other points. But in my case the path network between the points is fixed and i just need to find the suitable path that goes through all the points provided that all points may not have connection to every other point..so what algorithm am i supposed to follow.
It seems you are looking for a way to traverse a graph? If so have you tried Breadth first search http://en.wikipedia.org/wiki/Breadth-first_search or Depth first search http://en.wikipedia.org/wiki/Depth-first_search to traverse your graph.
You want to find a Hamiltonian path for a graph.
In the mathematical field of graph theory, a Hamiltonian path (or
traceable path) is a path in an undirected graph that visits each
vertex exactly once. A Hamiltonian cycle (or Hamiltonian circuit) is a
Hamiltonian path that is a cycle. Determining whether such paths and
cycles exist in graphs is the Hamiltonian path problem, which is
NP-complete.
Some techniques that exist :
There are n! different sequences of vertices that might be Hamiltonian
paths in a given n-vertex graph (and are, in a complete graph), so a
brute force search algorithm that tests all possible sequences would
be very slow. There are several faster approaches. A search procedure
by Frank Rubin divides the edges of the graph into three classes:
those that must be in the path, those that cannot be in the path, and
undecided. As the search proceeds, a set of decision rules classifies
the undecided edges, and determines whether to halt or continue the
search. The algorithm divides the graph into components that can be
solved separately. Also, a dynamic programming algorithm of Bellman,
Held, and Karp can be used to solve the problem in time O(n2 2n). In
this method, one determines, for each set S of vertices and each
vertex v in S, whether there is a path that covers exactly the
vertices in S and ends at v. For each choice of S and v, a path exists
for (S,v) if and only if v has a neighbor w such that a path exists
for (S − v,w), which can be looked up from already-computed
information in the dynamic program.
Andreas Björklund provided an alternative approach using the
inclusion–exclusion principle to reduce the problem of counting the
number of Hamiltonian cycles to a simpler counting problem, of
counting cycle covers, which can be solved by computing certain matrix
determinants. Using this method, he showed how to solve the
Hamiltonian cycle problem in arbitrary n-vertex graphs by a Monte
Carlo algorithm in time O(1.657n); for bipartite graphs this algorithm
can be further improved to time O(1.414n).
For graphs of maximum degree three, a careful backtracking search can
find a Hamiltonian cycle (if one exists) in time O(1.251n).
My graph contains no such edges which connect a vertex to itself. There is only one edge between two vertices. From Wikipedia i got to know about some algorithm which are used for calculating shortest path based on the given conditions. One of the most famous algorithm is Dijkstra's algorithm, which finds a shortest paths from source vertex to all other vertices in the graph.
But by using Dijkstra's algorithm, i am unnecessary exploring all the vertices, however my goal is just to find shortest path from single source to single destination. Which strategy should i use here? So that i need not to explore all other vertices.
One of my approach is to use bidirectional bfs. By bidirectional bfs i mean to apply two bfs one from source node, another one from destination node. As soon as for the first time when i find any same child in both tree,i can stop both bfs .Now the path from source to that child union path from child to destination would be my shortest path from source to destination.
But out of all the approaches described by Wikipedia and bidirectional bfs, which one suits best for my graph?
It depends if there's any apparent heuristic function that you could use or if you don't have any further usable information about your graph.
Your options are:
BFS: in general case or if you don't care about computation time that much.
Dijkstra (Dijkstra "is" BFS with priority queue): if your edges have weights/prices (non negative) and you care about CPU time.
A* (A* "is" Dijkstra with heuristic function - e.g. distance on a city map): if you want it to be really fast in average case, but you have to provide good heuristic function.
For some graph problems it may be possible to use Dynamic programming or other algorithmic tools. It depends on a situation. Further information can be found in tutorials from http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=alg_index ...
From Introduction to Algorithms (CLRS) second edition, page 581 :
Find a shortest path from u to v for a given vertices u and v. If we solve the single-source problem with source vertex u, we solve this problem also. Moreover, no algorithms for this problem are known that run asymptotically faster than the best single-source algorithms in the worst case.
So, stick to Dijkstra's algorithm :)
The Wikipedia article spells out the answer for you:
If we are only interested in a shortest path between vertices source and target, we can terminate the search at line 13 if u = target.
You could use Dijkstra's algorithm and optimize it in the way that you stop exploring paths that are already longer than the shortest you found so far. Because those are not going to be shorter (provided that you don't have negative weighs on your edges).
How is Dijkstra algorithm better tham A* algorithm for finding shortest path?
It is not better at finding the shortest path. As long as you have an admissible heuristic for A* it will find the shorted path quicker than Dijkstra's would.
And as Mehrad pointed out in the comments, A* degrades into Dijktras if you give it a heuristic function that returns 0.
The wikipedia article for A* has a ton of good information on all of this.
If you want to find the shortest paths from a given source node to all nodes in a graph Dijkstra is better than A* since Dijkstra covers the whole graph anyway if you don't stop at a specific target. In contrast to simple Dijkstra A* is a goal-oriented algorithms and therefore needs to know both source and target node. If you wanted to cover the whole graph with N nodes with an A* algorithm you basically have to run the algorithm N times from the source node.
Dijkstra might also be better for finding the shortest paths from a source node to many targets of the graph. Depending on the location of the targets (especially distance from the source), number of targets M, size of the graph, quality of the A* heuristic there will be some break-even point where running one Dijkstra is better or less performant than running M times the A* algorithm.
Edit: Inspired by Matthew's correct critical comment I rephrase a bit and add remarks:
Dijkstra is not better in finding the shortest paths to all nodes than A*. Correct would be: It's not worse than A*.
To find the shortest paths to all nodes with A* one would set the function which estimates the costs to the target node to zero (since there is no target node). Setting the function which estimates the costs to the target node to zero makes A* identical with Dijkstra; Dijkstra is a special case of A*. (It's questionable if one should call the algorithm still "A*" (and not simply "Dijkstra") if the function is set to zero, because having a non-zero function is the core of A*.)
When we try to find the shortest path to all nodes, we could also say: Dijkstra is sufficient. The refinement of A* is not necessary and doesn't help us here.
The last remark is also true for searching in a graph, for instance: Find the node marked with a specific flag which is closest to a given source node. Since you don't know the target in advance it's not possible to define a function which estimates the costs to the searched node, thus A* (in the narrower sense of a non-zero function) is not applicable.