If we are given a graph, Now from source we are to calculate the shortest path. Now , If an edge has a negative weight , but there is edge to back-edge to get back to that edge while reaching the destination I mean if there is no cycle, then we don't have a negative cycle. But the here in Wikipedia the given algorithm which runs from source again thus it detects a negative edge weight but not a negative cycle. My Question is, How to determine a negative cycle?
A negative weight cycle is a cycle with weights that sum to a negative number. The Bellman-Ford algorithm propagates correct distance estimates to all nodes in a graph in V-1 steps, unless there is a negative weight cycle. If there is a negative weight cycle, you can go on relaxing its nodes indefinitely. Therefore, the ability to relax an edge after V-1 steps is a test for the presence of a negative weight cycle, as seen in the Wikipedia algorithm. So the Bellman-Ford algorithm tests for negative weight cycles.
You can refer this link for determining actual negative cycle.
https://cp-algorithms.com/graph/finding-negative-cycle-in-graph.html
In last iteration of the algo . We only get to know about the effected node .
affected because of negative cycle.
The actual negative cycle forms a infinite loop of parent[parent[parent].
Second loop of the code is like throwing a pinball from the top, after some time the ball revolves around a circular maze infinitely. We find that circular maze.
Related
I'm making an implementation of the Floyd-Warshall algorithm, and I have one question:
If I have a loop in my graph (I mean, the cost of going from A to A is 1), what should the algorithm output, 0 (because the cost of going from any node to the same node is 0), or 1 (because there is an edge from A to A of cost 1?
I don't include any code because that's simply that question.
In the Wikipedia article on the Floyd-Warhsall algorithm there is a paragraph discussing explicitly how to deal with cycles of negative length as follows.
A negative cycle is a cycle whose edges sum to a negative value. There is no shortest path between any pair of vertices i, j which form part of a negative cycle, because path-lengths from i to j can be arbitrarily small (negative). For numerically meaningful output, the Floyd–Warshall algorithm assumes that there are no negative cycles. Nevertheless, if there are negative cycles, the Floyd–Warshall algorithm can be used to detect them.
Including the details, finally the inner workings of the algorithm can be utilized as follows.
Hence, to detect negative cycles using the Floyd–Warshall algorithm, one can inspect the diagonal of the path matrix, and the presence of a negative number indicates that the graph contains at least one negative cycle.
I study on Greedy Algorithm. summarize some important aspects about Dijkstra's Algorithms, that will be TRUE. i suspect about (4) and (1), anyone could help me?
I) if all edges weight be negative, the Dijkstra's algorithm, works well.
II) if in graph we have a negative cycle, Dijkstra's get into a infinite loop and never end.
III) if a graph has a one edge with negative weight, but hasn't a negative cycle, the algorithm doesn't works well.
IV) if graph hasn't a negative cycle, the algorithms work well.
Dijkstra's algorithm only works on graphs with non-negative edges. This is because it assumes that the first time a node is popped off the queue we have found the shortest path to that node and this is not necessarily true as soon as you have even one negative weight.
Therefore I is false, II is false (because the negative cycle may not necessarily be reachable), III is true, IV is false (it may still have a negative edge even without a negative cycle).
If I remember it correctly Dijkstra's algorithm (at least the classic version of it) has a built in assumption that all weights in a graph are non-negative. So we have no guarantee that it will work correctly if we have negative edges in our graph.
I) In the first case we will most likely get a longest possible chain of edges with highest absolute values of weights as the shortest path, technically this will be the correct shortest path. (I assume the graph has no negative cycles)
II) False as negative cycle may be unreachable (as Peter de Rivaz correctly said)
III-IV) I assume that the 3rd paragraph is about graph having one edge with negative weight, but hasn't a negative cycle. In this case there is a possibility of algorithm stucking in a loop, because it can find a cycle with a negative-weight edge that will seem to an algorithm as a good path prolongation beacuse at some point in algorithm we decide on the next step on the weight of one edge. The negative one will always be the first choice no doubt, so even with a non-negative cycle we getting a possibility of infinite looping.
I have a directed graph that has all non-negative edges except the edge(s) that leave the source (S). There are no edges from any other vertices to the source. To find the shortest distance from source (S) to a vertex (T) in the graph, can I use Dijkstra's shortest path algorithm even though the edges leaving the source is negative?
Assuming only source-adjecent edges can have negative weights and there is no path back to the source from any of the source-adjecent nodes (as mentioned in the comment), you can just add a constant C onto all edges leaving the source to make them all non-negative. Then subtract C from the final result.
On a more general note, Dijkstra can be used to solve shortest-path in any graph with negative edge weights (but no negative cycles) after applying Johnson's reweighting algorithm (which is essentially Bellman-Ford, but needs to be performed only once).
Yes, you can use Dijkstra on that type of directed graph.
If you use already finished alghoritm for Dijsktra and it cannot use negative values, it can be good practise to find the lowest negative edge and add that number to all starting edges, therefore there is no-negative number at all. You substract that number after finishing.
If you code it yourself (which is acutally pretty easy and I recommend it to you), you almost does not change anything, just start with lowest value (as usual for Dijkstra) and allow it, that lowest value can be negative. It will work in your case.
The reason you generally can't use Dijkstra's algorithm for (directed) graphs with negative links is that Dijkstra's algorithm is greedy. It assumes that once you pick a vertex with minimum distance, there is no way it can later be reached by a smaller paths.
In your particular graph, after the very first step, you traverse all possible negative edges and Dijkstra's assumption actually holds from now on. Regardless of the fact that those vertices directly connected to start now have negative values, once you identify which has the minimum distance, it can never be reached again with a smaller distance (since all edges you would traverse from this point on would have a positive distance).
If you think about the conditions that dijkstra's algorithm puts upon the edges for the algorithm to work it is only that they are never decreasing after initialisation.
Thus, it actually doesn't matter if the first step is negative as from those several points onwards the function is constantly increasing and thus the correct output will be found (provided there is no way to get back to the start square.).
I learned that the Bellman-Ford Algorithm has a running time of O(|E|*|V|), in which the E is the number of edges and V the number of vertices. Assume the graph does not have any negative weighted cycles.
My first question is that how do we prove that within (|V|-1) iterations (every iteration checks every edge in E), it updates the shortest path to every possible node, given a particular start node? Is it possible that we have iterated (|V|-1) times but still not ending up with shortest paths to every node?
Assume the correctness of the algorithm, can we actually do better than that? It occurs to me that not all edges are negatively weighted in a particular graph. The Bellman-Ford Algorithm seems expensive, as every iteration it goes through every edges.
The longest possible path from the source to any vertice would involve at most all the other vertices in the graph. In other words - you won't have a path that goes through the same vertice more than once, since that would necessarily increase the weights (this is true only thanks to the fact there are no negative cycles).
On each iteration you would update the shortest path weight on the next vertice in this path, until after |V|-1 iterations your updates would have to reach the end of that path. After that there won't be any vertices with non-tight values, since your update has covered all shortest paths up to that length.
This complexity is tight (at least for BF), think of a long line of connected vertices. Pick the leftmost as the source - your updating process would have to work its way from there to the other side once vertice at a time. Now you might argue that you don't have to check each edge that way, so let's throw in a few random edges with a very large weight (N > |V|*max-weight) - they can't help you, but your algorithm can't know that for sure, so if has to go through the process of updating the vertices with these weights (they're still better than the initial infinity).
Given a weighted graph (directed or undirected) I need to find the cycle of the graph with the maximum weight.
The weight of a cycle being the sum of the weight of the edges of the graph.
It can be any cycle, not just base cycle for which we can
find all base cycle (see Algorithms to Identify All the Cycle Bases in a UnDirected Graph )
compute the weight of each base cycle and find the maximum
I could try to enumerate all cycles of the graph and then compute the maximum but the total number of cycles can be really big (if the graph is complete then any sequence of vertices where the first and last one are identical is a cycle).
Do you have any idea to find that maximum weight cycle without enumerating all cycles ?
If you need hypothesis on the graph (positives weights for example) please indicates them.
This is NP-Hard.
Hamiltonian Cycle problem can be reduced to this.
Given a graph for which we need to check if there exists a Hamiltonian Cycle or not, assign weight 1 to each edge.
Now run your algorithm to get the maximum weight cycle. If the weight is < n, then the original graph has no Hamiltonian cycle, otherwise it does.
If you can find the minimum weighted path in your specific case, just reverse the signs of all the weights and apply your algorithm. Of course you are making some unstated assumptions because Moron's argument is correct (no pun intended). The assumptions you are making could be positive weights or no negative weight cycles. I think you should make an effort to state them instead of letting people search in the infinite space of possible assumptions. As to hardness results, this is also hard to approximate in a number of way, check out this paper. The same paper contains several positive results for important types of graphs, but it's concerned with longest unweighted paths so my guess is that most algorithms in the paper won't directly help in your case. If you search for "Heavy cycles" you will find a number of interesting papers, but they are more mathematical in character. If your weights are small integers (up to a polynomial in the size of the graph), you can try and replace every edge with an unweighted path to reduce your problem to the unweighted case. I hope this helps to some degree, but you might have an open research problem on your hands.