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I have an algorithmic problem where there's a straightforward solution, but it seems wasteful. I'm wondering if there's a more efficient way to do the same thing.
Here's the problem:
Input: A large graph G with non-negative edge weights (interpreted as lengths), a list of vertices v, and a list of distances d the same length as v.
Output: The subgraph S of G consisting of all of the vertices that are at a distance of at most d[i] from v[i] for some i.
The obvious solution is to use Dijkstra's algorithm starting from each v[i], modified so that it bails out after hitting a distance of d[i], and then taking the union of the subgraphs that each search traverses. However, in my use case it's frequently going to be the case that the search trees from the v[i]s overlap substantially. That means the Dijkstra approach will wastefully traverse the vertices in the overlap multiple times before I take the union.
In the case that there is only one vertex in v, the Dijkstra approach runs in O(|S|log|S|), taking |S| to be the number of vertices (my graph is sparse, so I ignore the edges term). Is it possible to achieve the same asymptotic run time when v has more than one vertex?
My first idea was to combine the searches out of each v[i] into the same priority queue, but the "bail out" condition mentioned above complicates this approach. Sometimes a vertex will be reached in a shorter distance from one v[i], but you would still want to search through it from another v[j] if the second vertex has a larger d[j] allotted to it.
Thanks!
You can solve this with the complexity of a single Dijkstra run.
Let D be the maximum of the distances in d.
Define a new start vertex, and give it edges to each of the vertices in v.
The length of the edge between start and v[i] should be set to D-d[i].
Then in this new graph, S is given by all vertices within a length D of the start vertex, so apply Dijkstra to the start vertex.
Consider a directed graph with n nodes and m edges. Each edge is weighted. There is a start node s and an end node e. We want to find the path from s to e that has the maximum number of nodes such that:
the total distance is less than some constant d
starting from s, each node in the path is closer than the previous one to the node e. (as in, when you traverse the path you are getting closer to your destination e. in terms of the edge weight of the remaining path.)
We can assume there are no cycles in the graph. There are no negative weights. Does an efficient algorithm already exist for this problem? Is there a name for this problem?
Whatever you end up doing, do a BFS/DFS starting from s first to see if e can even be reached; this only takes you O(n+m) so it won't add to the complexity of the problem (since you need to look at all vertices and edges anyway). Also, delete all edges with weight 0 before you do anything else since those never fulfill your second criterion.
EDIT: I figured out an algorithm; it's polynomial, depending on the size of your graphs it may still not be sufficiently efficient though. See the edit further down.
Now for some complexity. The first thing to think about here is an upper bound on how many paths we can actually have, so depending on the choice of d and the weights of the edges, we also have an upper bound on the complexity of any potential algorithm.
How many edges can there be in a DAG? The answer is n(n-1)/2, which is a tight bound: take n vertices, order them from 1 to n; for two vertices i and j, add an edge i->j to the graph iff i<j. This sums to a total of n(n-1)/2, since this way, for every pair of vertices, there is exactly one directed edge between them, meaning we have as many edges in the graph as we would have in a complete undirected graph over n vertices.
How many paths can there be from one vertex to another in the graph described above? The answer is 2n-2. Proof by induction:
Take the graph over 2 vertices as described above; there is 1 = 20 = 22-2 path from vertex 1 to vertex 2: (1->2).
Induction step: assuming there are 2n-2 paths from the vertex with number 1 of an n vertex graph as described above to the vertex with number n, increment the number of each vertex and add a new vertex 1 along with the required n edges. It has its own edge to the vertex now labeled n+1. Additionally, it has 2i-2 paths to that vertex for every i in [2;n] (it has all the paths the other vertices have to the vertex n+1 collectively, each "prefixed" with the edge 1->i). This gives us 1 + Σnk=2 (2k-2) = 1 + Σn-2k=0 (2k-2) = 1 + (2n-1 - 1) = 2n-1 = 2(n+1)-2.
So we see that there are DAGs that have 2n-2 distinct paths between some pairs of their vertices; this is a bit of a bleak outlook, since depending on weights and your choice of d, you may have to consider them all. This in itself doesn't mean we can't choose some form of optimum (which is what you're looking for) efficiently though.
EDIT: Ok so here goes what I would do:
Delete all edges with weight 0 (and smaller, but you ruled that out), since they can never fulfill your second criterion.
Do a topological sort of the graph; in the following, let's only consider the part of the topological sorting of the graph from s to e, let's call that the integer interval [s;e]. Delete everything from the graph that isn't strictly in that interval, meaning all vertices outside of it along with the incident edges. During the topSort, you'll also be able to see whether there is a
path from s to e, so you'll know whether there are any paths s-...->e. Complexity of this part is O(n+m).
Now the actual algorithm:
traverse the vertices of [s;e] in the order imposed by the topological
sorting
for every vertex v, store a two-dimensional array of information; let's call it
prev[][] since it's gonna store information about the predecessors
of a node on the paths leading towards it
in prev[i][j], store how long the total path of length (counted in
vertices) i is as a sum of the edge weights, if j is the predecessor of the
current vertex on that path. For example, pres+1[1][s] would have
the weight of the edge s->s+1 in it, while all other entries in pres+1
would be 0/undefined.
when calculating the array for a new vertex v, all we have to do is check
its incoming edges and iterate over the arrays for the start vertices of those
edges. For example, let's say vertex v has an incoming edge from vertex w,
having weight c. Consider what the entry prev[i][w] should be.
We have an edge w->v, so we need to set prev[i][w] in v to
min(prew[i-1][k] for all k, but ignore entries with 0) + c (notice the subscript of the array!); we effectively take the cost of a
path of length i - 1 that leads to w, and add the cost of the edge w->v.
Why the minimum? The vertex w can have many predecessors for paths of length
i - 1; however, we want to stay below a cost limit, which greedy minimization
at each vertex will do for us. We will need to do this for all i in [1;s-v].
While calculating the array for a vertex, do not set entries that would give you
a path with cost above d; since all edges have positive weights, we can only get
more costly paths with each edge, so just ignore those.
Once you reached e and finished calculating pree, you're done with this
part of the algorithm.
Iterate over pree, starting with pree[e-s]; since we have no cycles, all
paths are simple paths and therefore the longest path from s to e can have e-s edges. Find the largest
i such that pree[i] has a non-zero (meaning it is defined) entry; if non exists, there is no path fitting your criteria. You can reconstruct
any existing path using the arrays of the other vertices.
Now that gives you a space complexity of O(n^3) and a time complexity of O(n²m) - the arrays have O(n²) entries, we have to iterate over O(m) arrays, one array for each edge - but I think it's very obvious where the wasteful use of data structures here can be optimized using hashing structures and other things than arrays. Or you could just use a one-dimensional array and only store the current minimum instead of recomputing it every time (you'll have to encapsulate the sum of edge weights of the path together with the predecessor vertex though since you need to know the predecessor to reconstruct the path), which would change the size of the arrays from n² to n since you now only need one entry per number-of-nodes-on-path-to-vertex, bringing down the space complexity of the algorithm to O(n²) and the time complexity to O(nm). You can also try and do some form of topological sort that gets rid of the vertices from which you can't reach e, because those can be safely ignored as well.
Can somebody tell me why Dijkstra's algorithm for single source shortest path assumes that the edges must be non-negative.
I am talking about only edges not the negative weight cycles.
Recall that in Dijkstra's algorithm, once a vertex is marked as "closed" (and out of the open set) - the algorithm found the shortest path to it, and will never have to develop this node again - it assumes the path developed to this path is the shortest.
But with negative weights - it might not be true. For example:
A
/ \
/ \
/ \
5 2
/ \
B--(-10)-->C
V={A,B,C} ; E = {(A,C,2), (A,B,5), (B,C,-10)}
Dijkstra from A will first develop C, and will later fail to find A->B->C
EDIT a bit deeper explanation:
Note that this is important, because in each relaxation step, the algorithm assumes the "cost" to the "closed" nodes is indeed minimal, and thus the node that will next be selected is also minimal.
The idea of it is: If we have a vertex in open such that its cost is minimal - by adding any positive number to any vertex - the minimality will never change.
Without the constraint on positive numbers - the above assumption is not true.
Since we do "know" each vertex which was "closed" is minimal - we can safely do the relaxation step - without "looking back". If we do need to "look back" - Bellman-Ford offers a recursive-like (DP) solution of doing so.
Consider the graph shown below with the source as Vertex A. First try running Dijkstra’s algorithm yourself on it.
When I refer to Dijkstra’s algorithm in my explanation I will be talking about the Dijkstra's Algorithm as implemented below,
So starting out the values (the distance from the source to the vertex) initially assigned to each vertex are,
We first extract the vertex in Q = [A,B,C] which has smallest value, i.e. A, after which Q = [B, C]. Note A has a directed edge to B and C, also both of them are in Q, therefore we update both of those values,
Now we extract C as (2<5), now Q = [B]. Note that C is connected to nothing, so line16 loop doesn't run.
Finally we extract B, after which . Note B has a directed edge to C but C isn't present in Q therefore we again don't enter the for loop in line16,
So we end up with the distances as
Note how this is wrong as the shortest distance from A to C is 5 + -10 = -5, when you go .
So for this graph Dijkstra's Algorithm wrongly computes the distance from A to C.
This happens because Dijkstra's Algorithm does not try to find a shorter path to vertices which are already extracted from Q.
What the line16 loop is doing is taking the vertex u and saying "hey looks like we can go to v from source via u, is that (alt or alternative) distance any better than the current dist[v] we got? If so lets update dist[v]"
Note that in line16 they check all neighbors v (i.e. a directed edge exists from u to v), of u which are still in Q. In line14 they remove visited notes from Q. So if x is a visited neighbour of u, the path is not even considered as a possible shorter way from source to v.
In our example above, C was a visited neighbour of B, thus the path was not considered, leaving the current shortest path unchanged.
This is actually useful if the edge weights are all positive numbers, because then we wouldn't waste our time considering paths that can't be shorter.
So I say that when running this algorithm if x is extracted from Q before y, then its not possible to find a path - which is shorter. Let me explain this with an example,
As y has just been extracted and x had been extracted before itself, then dist[y] > dist[x] because otherwise y would have been extracted before x. (line 13 min distance first)
And as we already assumed that the edge weights are positive, i.e. length(x,y)>0. So the alternative distance (alt) via y is always sure to be greater, i.e. dist[y] + length(x,y)> dist[x]. So the value of dist[x] would not have been updated even if y was considered as a path to x, thus we conclude that it makes sense to only consider neighbors of y which are still in Q (note comment in line16)
But this thing hinges on our assumption of positive edge length, if length(u,v)<0 then depending on how negative that edge is we might replace the dist[x] after the comparison in line18.
So any dist[x] calculation we make will be incorrect if x is removed before all vertices v - such that x is a neighbour of v with negative edge connecting them - is removed.
Because each of those v vertices is the second last vertex on a potential "better" path from source to x, which is discarded by Dijkstra’s algorithm.
So in the example I gave above, the mistake was because C was removed before B was removed. While that C was a neighbour of B with a negative edge!
Just to clarify, B and C are A's neighbours. B has a single neighbour C and C has no neighbours. length(a,b) is the edge length between the vertices a and b.
Dijkstra's algorithm assumes paths can only become 'heavier', so that if you have a path from A to B with a weight of 3, and a path from A to C with a weight of 3, there's no way you can add an edge and get from A to B through C with a weight of less than 3.
This assumption makes the algorithm faster than algorithms that have to take negative weights into account.
Correctness of Dijkstra's algorithm:
We have 2 sets of vertices at any step of the algorithm. Set A consists of the vertices to which we have computed the shortest paths. Set B consists of the remaining vertices.
Inductive Hypothesis: At each step we will assume that all previous iterations are correct.
Inductive Step: When we add a vertex V to the set A and set the distance to be dist[V], we must prove that this distance is optimal. If this is not optimal then there must be some other path to the vertex V that is of shorter length.
Suppose this some other path goes through some vertex X.
Now, since dist[V] <= dist[X] , therefore any other path to V will be atleast dist[V] length, unless the graph has negative edge lengths.
Thus for dijkstra's algorithm to work, the edge weights must be non negative.
Dijkstra's Algorithm assumes that all edges are positive weighted and this assumption helps the algorithm run faster ( O(E*log(V) ) than others which take into account the possibility of negative edges (e.g bellman ford's algorithm with complexity of O(V^3)).
This algorithm wont give the correct result in the following case (with a -ve edge) where A is the source vertex:
Here, the shortest distance to vertex D from source A should have been 6. But according to Dijkstra's method the shortest distance will be 7 which is incorrect.
Also, Dijkstra's Algorithm may sometimes give correct solution even if there are negative edges. Following is an example of such a case:
However, It will never detect a negative cycle and always produce a result which will always be incorrect if a negative weight cycle is reachable from the source, as in such a case there exists no shortest path in the graph from the source vertex.
Try Dijkstra's algorithm on the following graph, assuming A is the source node and D is the destination, to see what is happening:
Note that you have to follow strictly the algorithm definition and you should not follow your intuition (which tells you the upper path is shorter).
The main insight here is that the algorithm only looks at all directly connected edges and it takes the smallest of these edge. The algorithm does not look ahead. You can modify this behavior , but then it is not the Dijkstra algorithm anymore.
You can use dijkstra's algorithm with negative edges not including negative cycle, but you must allow a vertex can be visited multiple times and that version will lose it's fast time complexity.
In that case practically I've seen it's better to use SPFA algorithm which have normal queue and can handle negative edges.
Recall that in Dijkstra's algorithm, once a vertex is marked as "closed" (and out of the open set) -it assumes that any node originating from it will lead to greater distance so, the algorithm found the shortest path to it, and will never have to develop this node again, but this doesn't hold true in case of negative weights.
The other answers so far demonstrate pretty well why Dijkstra's algorithm cannot handle negative weights on paths.
But the question itself is maybe based on a wrong understanding of the weight of paths. If negative weights on paths would be allowed in pathfinding algorithms in general, then you would get permanent loops that would not stop.
Consider this:
A <- 5 -> B <- (-1) -> C <- 5 -> D
What is the optimal path between A and D?
Any pathfinding algorithm would have to continuously loop between B and C because doing so would reduce the weight of the total path. So allowing negative weights for a connection would render any pathfindig algorithm moot, maybe except if you limit each connection to be used only once.
So, to explain this in more detail, consider the following paths and weights:
Path | Total weight
ABCD | 9
ABCBCD | 7
ABCBCBCD | 5
ABCBCBCBCD | 3
ABCBCBCBCBCD | 1
ABCBCBCBCBCBCD | -1
...
So, what's the perfect path? Any time the algorithm adds a BC step, it reduces the total weight by 2.
So the optimal path is A (BC) D with the BC part being looped forever.
Since Dijkstra's goal is to find the optimal path (not just any path), it, by definition, cannot work with negative weights, since it cannot find the optimal path.
Dijkstra will actually not loop, since it keeps a list of nodes that it has visited. But it will not find a perfect path, but instead just any path.
Adding few points to the explanation, on top of the previous answers, for the following simple example,
Dijktra's algorithm being greedy, it first finds the minimum distance vertex C from the source vertex A greedily and assigns the distance d[C] (from vertex A) to the weight of the edge AC.
The underlying assumption is that since C was picked first, there is no other vertex V in the graph s.t. w(AV) < w(AC), otherwise V would have been picked instead of C, by the algorithm.
Since by above logic, w(AC) <= w(AV), for all vertex V different from the vertices A and C. Now, clearly any other path P that starts from A and ends in C, going through V , i.e., the path P = A -> V -> ... -> C, will be longer in length (>= 2) and total cost of the path P will be sum of the edges on it, i.e., cost(P) >= w(AV) >= w(AC), assuming all edges on P have non-negative weights, so that
C can be safely removed from the queue Q, since d[C] can never get smaller / relaxed further under this assumption.
Obviously, the above assumption does not hold when some.edge on P is negative, in a which case d[C] may decrease further, but the algorithm can't take care of this scenario, since by that time it has removed C from the queue Q.
In Unweighted graph
Dijkstra can even work without set or priority queue, even if you just use STACK the algorithm will work but with Stack its time of execution will increase
Dijkstra don't repeat a node once its processed becoz it always tooks the minimum route , which means if you come to that node via any other path it will certainly have greater distance
For ex -
(0)
/
6 5
/
(2) (1)
\ /
4 7
\ /
(9)
here once you get to node 1 via 0 (as its minimum out of 5 and 6)so now there is no way you can get a minimum value for reaching 1
because all other path will add value to 5 and not decrease it
more over with Negative weights it will fall into infinite loop
In Unweighted graph
Dijkstra Algo will fall into loop if it has negative weight
In Directed graph
Dijkstra Algo will give RIGHT ANSWER except in case of Negative Cycle
Who says Dijkstra never visit a node more than once are 500% wrong
also who says Dijkstra can't work with negative weight are wrong
I am trying to understand why Dijkstra's algorithm will not work with negative weights. Reading an example on Shortest Paths, I am trying to figure out the following scenario:
2
A-------B
\ /
3 \ / -2
\ /
C
From the website:
Assuming the edges are all directed from left to right, If we start
with A, Dijkstra's algorithm will choose the edge (A,x) minimizing
d(A,A)+length(edge), namely (A,B). It then sets d(A,B)=2 and chooses
another edge (y,C) minimizing d(A,y)+d(y,C); the only choice is (A,C)
and it sets d(A,C)=3. But it never finds the shortest path from A to
B, via C, with total length 1.
I can not understand why using the following implementation of Dijkstra, d[B] will not be updated to 1 (When the algorithm reaches vertex C, it will run a relax on B, see that the d[B] equals to 2, and therefore update its value to 1).
Dijkstra(G, w, s) {
Initialize-Single-Source(G, s)
S ← Ø
Q ← V[G]//priority queue by d[v]
while Q ≠ Ø do
u ← Extract-Min(Q)
S ← S U {u}
for each vertex v in Adj[u] do
Relax(u, v)
}
Initialize-Single-Source(G, s) {
for each vertex v V(G)
d[v] ← ∞
π[v] ← NIL
d[s] ← 0
}
Relax(u, v) {
//update only if we found a strictly shortest path
if d[v] > d[u] + w(u,v)
d[v] ← d[u] + w(u,v)
π[v] ← u
Update(Q, v)
}
Thanks,
Meir
The algorithm you have suggested will indeed find the shortest path in this graph, but not all graphs in general. For example, consider this graph:
Let's trace through the execution of your algorithm.
First, you set d(A) to 0 and the other distances to ∞.
You then expand out node A, setting d(B) to 1, d(C) to 0, and d(D) to 99.
Next, you expand out C, with no net changes.
You then expand out B, which has no effect.
Finally, you expand D, which changes d(B) to -201.
Notice that at the end of this, though, that d(C) is still 0, even though the shortest path to C has length -200. This means that your algorithm doesn't compute the correct distances to all the nodes. Moreover, even if you were to store back pointers saying how to get from each node to the start node A, you'd end taking the wrong path back from C to A.
The reason for this is that Dijkstra's algorithm (and your algorithm) are greedy algorithms that assume that once they've computed the distance to some node, the distance found must be the optimal distance. In other words, the algorithm doesn't allow itself to take the distance of a node it has expanded and change what that distance is. In the case of negative edges, your algorithm, and Dijkstra's algorithm, can be "surprised" by seeing a negative-cost edge that would indeed decrease the cost of the best path from the starting node to some other node.
Note, that Dijkstra works even for negative weights, if the Graph has no negative cycles, i.e. cycles whose summed up weight is less than zero.
Of course one might ask, why in the example made by templatetypedef Dijkstra fails even though there are no negative cycles, infact not even cycles. That is because he is using another stop criterion, that holds the algorithm as soon as the target node is reached (or all nodes have been settled once, he did not specify that exactly). In a graph without negative weights this works fine.
If one is using the alternative stop criterion, which stops the algorithm when the priority-queue (heap) runs empty (this stop criterion was also used in the question), then dijkstra will find the correct distance even for graphs with negative weights but without negative cycles.
However, in this case, the asymptotic time bound of dijkstra for graphs without negative cycles is lost. This is because a previously settled node can be reinserted into the heap when a better distance is found due to negative weights. This property is called label correcting.
TL;DR: The answer depends on your implementation. For the pseudo code you posted, it works with negative weights.
Variants of Dijkstra's Algorithm
The key is there are 3 kinds of implementation of Dijkstra's algorithm, but all the answers under this question ignore the differences among these variants.
Using a nested for-loop to relax vertices. This is the easiest way to implement Dijkstra's algorithm. The time complexity is O(V^2).
Priority-queue/heap based implementation + NO re-entrance allowed, where re-entrance means a relaxed vertex can be pushed into the priority-queue again to be relaxed again later.
Priority-queue/heap based implementation + re-entrance allowed.
Version 1 & 2 will fail on graphs with negative weights (if you get the correct answer in such cases, it is just a coincidence), but version 3 still works.
The pseudo code posted under the original problem is the version 3 above, so it works with negative weights.
Here is a good reference from Algorithm (4th edition), which says (and contains the java implementation of version 2 & 3 I mentioned above):
Q. Does Dijkstra's algorithm work with negative weights?
A. Yes and no. There are two shortest paths algorithms known as Dijkstra's algorithm, depending on whether a vertex can be enqueued on the priority queue more than once. When the weights are nonnegative, the two versions coincide (as no vertex will be enqueued more than once). The version implemented in DijkstraSP.java (which allows a vertex to be enqueued more than once) is correct in the presence of negative edge weights (but no negative cycles) but its running time is exponential in the worst case. (We note that DijkstraSP.java throws an exception if the edge-weighted digraph has an edge with a negative weight, so that a programmer is not surprised by this exponential behavior.) If we modify DijkstraSP.java so that a vertex cannot be enqueued more than once (e.g., using a marked[] array to mark those vertices that have been relaxed), then the algorithm is guaranteed to run in E log V time but it may yield incorrect results when there are edges with negative weights.
For more implementation details and the connection of version 3 with Bellman-Ford algorithm, please see this answer from zhihu. It is also my answer (but in Chinese). Currently I don't have time to translate it into English. I really appreciate it if someone could do this and edit this answer on stackoverflow.
you did not use S anywhere in your algorithm (besides modifying it). the idea of dijkstra is once a vertex is on S, it will not be modified ever again. in this case, once B is inside S, you will not reach it again via C.
this fact ensures the complexity of O(E+VlogV) [otherwise, you will repeat edges more then once, and vertices more then once]
in other words, the algorithm you posted, might not be in O(E+VlogV), as promised by dijkstra's algorithm.
Since Dijkstra is a Greedy approach, once a vertice is marked as visited for this loop, it would never be reevaluated again even if there's another path with less cost to reach it later on. And such issue could only happen when negative edges exist in the graph.
A greedy algorithm, as the name suggests, always makes the choice that seems to be the best at that moment. Assume that you have an objective function that needs to be optimized (either maximized or minimized) at a given point. A Greedy algorithm makes greedy choices at each step to ensure that the objective function is optimized. The Greedy algorithm has only one shot to compute the optimal solution so that it never goes back and reverses the decision.
Consider what happens if you go back and forth between B and C...voila
(relevant only if the graph is not directed)
Edited:
I believe the problem has to do with the fact that the path with AC* can only be better than AB with the existence of negative weight edges, so it doesn't matter where you go after AC, with the assumption of non-negative weight edges it is impossible to find a path better than AB once you chose to reach B after going AC.
"2) Can we use Dijksra’s algorithm for shortest paths for graphs with negative weights – one idea can be, calculate the minimum weight value, add a positive value (equal to absolute value of minimum weight value) to all weights and run the Dijksra’s algorithm for the modified graph. Will this algorithm work?"
This absolutely doesn't work unless all shortest paths have same length. For example given a shortest path of length two edges, and after adding absolute value to each edge, then the total path cost is increased by 2 * |max negative weight|. On the other hand another path of length three edges, so the path cost is increased by 3 * |max negative weight|. Hence, all distinct paths are increased by different amounts.
You can use dijkstra's algorithm with negative edges not including negative cycle, but you must allow a vertex can be visited multiple times and that version will lose it's fast time complexity.
In that case practically I've seen it's better to use SPFA algorithm which have normal queue and can handle negative edges.
I will be just combining all of the comments to give a better understanding of this problem.
There can be two ways of using Dijkstra's algorithms :
Marking the nodes that have already found the minimum distance from the source (faster algorithm since we won't be revisiting nodes whose shortest path have been found already)
Not marking the nodes that have already found the minimum distance from the source (a bit slower than the above)
Now the question arises, what if we don't mark the nodes so that we can find shortest path including those containing negative weights ?
The answer is simple. Consider a case when you only have negative weights in the graph:
)
Now, if you start from the node 0 (Source), you will have steps as (here I'm not marking the nodes):
0->0 as 0, 0->1 as inf , 0->2 as inf in the beginning
0->1 as -1
0->2 as -5
0->0 as -8 (since we are not relaxing nodes)
0->1 as -9 .. and so on
This loop will go on forever, therefore Dijkstra's algorithm fails to find the minimum distance in case of negative weights (considering all the cases).
That's why Bellman Ford Algo is used to find the shortest path in case of negative weights, as it will stop the loop in case of negative cycle.
What if the only negative edge costs are coming from the initial node? Will the algorithm still work?
I feel like yes, because I can't think of a counter-example, but I'm having trouble proving it. Is there a counter-example?
Negative edges are a problem for Dijkstra's because there's no guarantee that the edge you pick produces the shortest path if there is an edge you can pick later that is largely negatively weighted. But if the only negative edges are coming out of the initial node, I don't see the problem.
I'm not looking for an algorithm. I'm looking for some insight into the Dijkstra's.
I'm talking about a directed graph, if that makes a difference.
The trouble with having a negative-cost edge is that you can go back and forth along it as many times as you like.
If you impose a rule that an edge may not be used more than once, you still have a problem. Dijkstra's algorithm involves marking a node as "visited", when it's distance from the initial node is considered know once and for all. This happens before all of the edges have been examined; the shortest path from the initial node to node X has been found, all other paths from the initial node are already longer than that, nothing that is discovered later can make those paths shorter. But if there are negative-cost edges somewhere, then a later discovery can make a path shorter, so it may be that a shorter path exists which Dijkstra will not discover.
If only the edges that connect to the initial node may have negative costs, then you still have problem, because the shortest path might involve revisiting the initial node to take advantage of the negative costs, something Dijkstra cannot do.
If you impose another rule that a node may not be visited more than once, then Dijkstra's algorithm works. Notice that in Dijkstra's algorithm, the initial node is given an initial distance of zero. If you give it some other initial distance, the algorithm will still find the shortest path-- but all of the distances will be off by that same amount. (If you want the real distance at the end, you must subtract the value you put in.)
So take your graph, call it A, find the smallest cost of any edge connected to the initial node, call it k which will be negative in this case).
Make a new graph B which you get by subtracting k from the cost of each edge connected to the initial node. Note that all of these costs are now non-negative. So Dijkstra works on B. Also note that the shortest path in B is also the shortest path in A.
Assign the initial node of B the distance k, then run Dijkstra (this will give the same path as running with an initial distance of zero). Compare this to running Dijkstra naively on A: once you leave the initial node everything's the same in the two graphs. The distances are the same, the decisions are the same, the two will produce the same path. And in the case of A the distace will be correct, since it started at zero.
Counter-example:
Graph G = (V, E), with vertices V = {A, B}, edges E = {(A, B), (B, A)} and weight function w(A, B) = -2, w(B, A) = +1.
There's a negative weight cycle, hence minimum distances are undefined (even using A as initial node).
Dijkstra's algorithm doesn't produce correct answer for graph with negative edge weights (even if graph doesn't have any negative weight cycle). For e.g. it computes incorrect shortest path value between (A, C) for following graph with source vertex A,
A -> B : 6
A -> C : 5
B -> D : 2
B -> E : 1
D -> E : -5
E -> C : -2