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.
Related
I was testing the two algorithms and Bellman-Ford performed better on sparse graphs and looking at the big-O analysis of both, O(VE) for Bellman-Ford and O(E + V lg V) for Dijkstra's. I believe this is correct. I did some researching that said
Dijkstra's is always faster and Bellman-Ford is only used when negative weight cycles are present.
Is that really the case?
TRUE.
Wikipedia: However, Dijkstra's algorithm greedily selects the minimum-weight node
that has not yet been processed, and performs this relaxation process
on all of its outgoing edges; in contrast, the Bellman–Ford algorithm
simply relaxes all the edges, and does this |V | − 1 times, where |V |
is the number of vertices in the graph. In each of these repetitions,
the number of vertices with correctly calculated distances grows, from
which it follows that eventually all vertices will have their correct
distances. This method allows the Bellman–Ford algorithm to be applied
to a wider class of inputs than Dijkstra.
Bellman-Ford performs the check on all the vertices, Dijkstra only on the one with the best distance calculated so far. Again already noted, this improves the complexity of the Dijkstra approach, however it requires to compare all the vertices to find out the minimum distance value. Being this not necessary in the Bellman-Ford, it is easier to implement in a distributed environment. That's why it is used in Distance Vector routing protocols (e.g., RIP and IGRP), where mostly local information is used. To use Dijkstra in routing protocols, instead, it is necessary first to distribute the entire topology, and this is what happens in Link State protocols, such as OSPF and ISIS.
Asymptotically, for any graph where E ≥ V, the runtime of Dijkstra’s algorithm (O(E + V log V)) is smaller than that of Bellman-Ford (O(EV)). This means that, at least in theory, Dijkstra’s algorithm will outperform Bellman-Ford for large graphs of any density (assuming there’s at least as many edges as nodes). That’s borne out in practice, with Dijkstra’s algorithm usually running much faster.
Bellman-Ford is used, as you’ve mentioned, in cases where there are negative edges, which Dijkstra can’t handle. But there are other cases where Bellman-Ford is useful, too. For example, in network routing, where each network router needs to find shortest paths and there isn’t a central computer coordinating everything, the network routers can run a distributed version of Bellman-Ford to find shortest paths between computers due to how the computation only requires local updates to node distances. Dijkstra’s algorithm doesn’t work in this case.
There are also ways to improve the performance of Bellman-Ford in practice for many types of graphs. The Shortest Paths Faster Algorithm (SPFA) is a relatively simple optimization of Bellman-Ford that, while still retaining Bellman-Ford’s worst case runtime, is empirically faster in practice. Dijkstra’s algorithm, IIRC, is usually still faster than SPFA, but this does close the gap in some circumstances.
As you’ve mentioned
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.
I know that Dijkstra's algorithm can be used only on positive lengths of edges, and Bellman-Ford can be used when the graph also has negative ones.
Suppose we have a graph with only positive edges, though. Will Bellman-Ford give the same results as Dijkstra?
Yes, it will give the same results. It will run slower, though, as it could also have been used for graphs with negative edges (subject to the absence of negative cycles). If you look at the proof of BF's correctness, there is no assumption there that some of the edges are negative.
I want to add something to Ami Tavory's answer. Bellman-ford's algorithm can be made a little bit faster if you can detect that on any pass, there is no node value update, then return from there. If there is no node update then it proves that every node traversal is complete.
After a lot of Googling, I've found that most sources say that the Dijkstra algorithm is "more efficient" than the Bellman-Ford algorithm. But under what circumstances is the Bellman-Ford algorithm better than the Dijkstra algorithm?
I know "better" is a broad statement, so specifically I mean in terms of speed and also space if that applies. Surely there is some situation in which the Bellman-Ford approach is better than the Dijkstra approach.
Bellman-Ford algorithm is a single-source shortest path algorithm, so when you have negative edge weight then it can detect negative cycles in a graph.
The only difference between the two is that Bellman-Ford is also capable of handling negative weights whereas Dijkstra Algorithm can only handle positives.
From wiki
However, Dijkstra's algorithm greedily selects the minimum-weight node
that has not yet been processed, and performs this relaxation process
on all of its outgoing edges; in contrast, the Bellman–Ford algorithm
simply relaxes all the edges, and does this |V | − 1 times, where |V |
is the number of vertices in the graph. In each of these repetitions,
the number of vertices with correctly calculated distances grows, from
which it follows that eventually all vertices will have their correct
distances. This method allows the Bellman–Ford algorithm to be applied
to a wider class of inputs than Dijkstra.
Dijkstra is however generally considered better in the absence of negative weight edges, as a typical binary heap priority queue implementation has O((|E|+|V|)log|V|) time complexity [A Fibonacci heap priority queue gives O(|V|log|V| + |E|)], while the Bellman-Ford algorithm has O(|V||E|) complexity
As already stated in the chosen answer, Bellman-Ford performs the check on all the vertices, Dijkstra only on the one with the best distance calculated so far. Again already noted, this improves the complexity of the Dijkstra approach, however it requires to compare all the vertices to find out the minimum distance value. Being this not necessary in the Bellman-Ford, it is easier to implement in a distributed environment. That's why it is used in Distance Vector routing protocols (e.g., RIP and IGRP), where mostly local information is used. To use Dijkstra in routing protocols, instead, it is necessary first to distribute the entire topology, and this is what happens in Link State protocols, such as OSPF and ISIS.
There are 4 major difference among them I know:-
1. bellman time complexity is O(VE) and Dijkstra Algo has O(ElogV)in case of maxheap is used.
Bellman does relaxation for n-1 times and Dijkstra Algo only 1 time.
Bellman can handle negative weights but Dijkstra Algo can't.
Bellman visit a vertex more then once but Dijkstra Algo only once.
The only difference is that Dijkstra's algorithm cannot handle negative edge weights which Bellman-ford handles.And bellman-ford also tells us whether the graph contains negative cycle.
If graph doesn't contain negative edges then Dijkstra's is always better.
An efficient alternative for Bellman-ford is Directed Acyclic Graph (DAG) which uses topological sorting.
http://www.geeksforgeeks.org/shortest-path-for-directed-acyclic-graphs/
Dijkstra Algo
Dijkstra algo is not capable to differentiate between Negative edge weight cycle is present in graph or not
1. Positive edge weight:- Dijkstra always PASS if all edge weight in a graph is positive
2. Negative edge wt. and No -ve edge wt. cycle:- Dijkstra always PASS even if we have some edges weight as Negative but NO cycle/loop in graph having negative edge weight.
[i.e No Negative edge weight cycle is present]
3. Negative edge wt. and -ve edge wt. cycle:- Dijkstra may PASS/FAIL even if we have some edges weight as negative along with cycle/loop in graph having negative edge weight.
In a normal introduction to algorithm class, you will learn that the only difference between Dijkstra and Bell-man Ford, is that the latter works for negative edges at the cost of more computation time. The discussion on time complexity is already give in the accepted answer.
However I want to emphasize and add a bit more to #Halberdier's answer that in a distributed system, Bellman-Ford is implemented EVEN WHEN ALL EDGES ARE Positive. This is because in a Bellman-Ford algorithm, the entity S does not need to know every weight of every edge in the graph to compute the shortest distance to T - it only needs to know the shortest distance for all neighbors of S to T, plus the weight of S to all its neighbors.
A typical application of such algorithm is in Computer Networking, where you need to find the shortest route between two routers. Dijkstra is implemented in a centralized manner called link state Routing, while
Bellman-Ford allows each router to update themselves asynchronously, called distance-vector routing.
I believe no one explains better than Jim Kurose, the author of <Computer Network, a top-down approach>. See his youtube videos below.
Link State routing:
https://www.youtube.com/watch?v=bdh2kfgxVuw&list=TLPQMTIwNjIwMjLtHllygYsxMg&index=3
Distance Vector routing:
https://www.youtube.com/watch?v=jJU2AVX6gpU&list=TLPQMTIwNjIwMjLtHllygYsxMg&index=4
Bellman Ford’s Algorithm
Dijkstra’s Algorithm
Bellman Ford’s Algorithm works when there is a negative weight edge, it also detects the negative weight cycle.
Dijkstra’s Algorithm may or may not work when there is a negative weight edge. But will definitely not work when there is a negative weight cycle.
The result contains the vertices which contain the information about the other vertices they are connected to.
The result contains the vertices containing whole information about the network, not only the vertices they are connected to.
It can easily be implemented in a distributed way.
It can not be implemented easily in a distributed way.
It is more time-consuming than Dijkstra’s algorithm. Its time complexity is O(VE).
It is less time-consuming. The time complexity is O(E logV).
Dynamic Programming approach is taken to implement the algorithm.
Greedy approach is taken to implement the algorithm.
Bellman Ford’s Algorithm has more overheads than Dijkstra’s Algorithm.
Dijkstra’s Algorithm has less overheads than Bellman Ford’s Algorithm.
Bellman Ford’s Algorithm has less scalability than Dijkstra’s Algorithm.
Dijkstra’s Algorithm has more scalability than Bellman Ford’s Algorithm.
Source 1
I do not agree completely, difference is in implementation and complexity, Dijsktra's algorithm is faster (O(n^2)) but difficult to implement, while Bellman Ford complexity is O(n^3) but is easier to implement.
Could you please help me find out the time complexity of the Fleury' algorithm (which is used to get the Eulerian circuit)?
Fleury's algorithm isn't really complete until you specify how the bridge edges are identified. Tarjan gave a linear-time algorithm for identifying all bridges (see http://en.wikipedia.org/wiki/Bridge_(graph_theory) ), so a naive implementation that reran Tarjan's algorithm after each deleted edge would be O(E^2). There are probably better ways to recompute the set of bridges, but there is also a better O(E) algorithm. (See http://www.algorithmist.com/index.php/Euler_tour#Fleury.27s_algorithm ; not my site :))
Here:
http://roticv.rantx.com/book/Eulerianpathandcircuit.pdf
you can read among other things, that it is O(E), linear edge count.
Fleury algorithm involves following steps :
Make sure the graph has either 0 or 2 odd vertices.
If there are 0 odd vertices, start anywhere. If there are 2 odd vertices, start at one of them.
Follow edges one at a time. If you have a choice between a bridge and a non-bridge, always choose the non-bridge.
Stop when you run out of edges.
If bridges are found out by Tarjan's algorithm and these bridges are stored in an adjacency matrix then we need not run tarjan's algorithm every time to check whether an edge is a bridge or not. We can check it in O(1) time for all other bridge queries. Thus Flury's algorithm time complexity can be reduced to O(V+E) {as this is a DFS} but this method needs O(V2) extra space to store bridges.