Why doesn t Dijkstra s algorithm work for negative weight edges

Question

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

User · Answer

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.

User · Answer

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 -- gt 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- gt B- gt 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

User · Answer

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   lt   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

User · Answer

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

User · Answer

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

User · Answer

Dijkstra s Algorithm assumes that all edges are positive weighted and this assumption helps the algorithm run faster   O V 2    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 where A is the source vertex   Also  Dijkstra s Algorithm may sometimes give correct solution even if there are negative edges  Following is an example of such a case   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

User · Answer

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

User · Answer

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 lt 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  lt 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

User · Answer

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

[algorithm] Why doesn't Dijkstra's algorithm work for negative weight edges?

Examples related to algorithm

Examples related to shortest-path

Examples related to dijkstra