I wrote a program which implements Bellman-Ford, and identifies when negative weight cycles are present in a graph. However what I'm actually interested in, is given some starting vertex and a graph, which path do I actually trace to get to the original vertex having traveled a negative amount.

So to be clear say I have a graph with vertexes, a, b, c, and d and there is a negative cycle between a, b, and d, then when I check for negative weight cycles

// Step 1: initialize graph
   for each vertex v in vertices:
       if v is source then distance[v] := 0
       else distance[v] := infinity
       predecessor[v] := null

   // Step 2: relax edges repeatedly
   for i from 1 to size(vertices)-1:
       for each edge (u, v) with weight w in edges:
           if distance[u] + w < distance[v]:
               distance[v] := distance[u] + w
               predecessor[v] := u

   // Step 3: check for negative-weight cycles
   for each edge (u, v) with weight w in edges:
       if distance[u] + w < distance[v]:
           "Graph contains a negative-weight cycle"

Instead of it just telling me that a negative cycle is there, I would like it to tell me, go from a -> b -> d -> a. After the relaxing step what do I have to change in my check for negative weight cycles to get it to output this information?

  • Here is the best information I've been able to find, but I'm still having trouble making sense of it.

  • Also this which suggests that I need to run breadth first search on the predecessor array to find the information, but I'm not exactly sure where to start (what do I queue first?)

  • Here is a stack overflow question which shows how to find one of the nodes in the path.

  • $\begingroup$ Just to clarify, is $a\xrightarrow{1} b \xrightarrow{-2} c \xrightarrow{-1} b \xrightarrow{1} a$ a (negative) cycle? And are you looking for the most efficient algorithm or just a working algorithm (but with a not so good complexity) is enough? $\endgroup$
    – wece
    May 21, 2013 at 14:03
  • $\begingroup$ I would prefer the most efficient, but it dons't have to be. At the same if it's like $O(n^2)$ time on top of bellman ford then I dont want that either. I know that the predisessor array has the information, so really I'm asking how do I extract it. And more like a->b = 1, b->d = -3, d->a = 1, but really just any negative weight cycle $\endgroup$
    – Loourr
    May 21, 2013 at 14:33
  • $\begingroup$ We already have a question on this topic: Getting negative cycle using Bellman Ford. Does that thread answer your question? If not, please edit your question to state what you still need answered. $\endgroup$ May 22, 2013 at 22:33

1 Answer 1


Per Kleinberg–Tardos, you want to run Bellman–Ford for n iterations and find a cycle in the predecessor array.

To find a cycle in the predecessor array, start by coloring every node white. For each node u in an arbitrary order, set v := u, and, while v is white and has a predecessor, recolor v gray and set v := predecessor[v]. Upon exiting the loop, if v is gray, we found a cycle; loop through again to read it off. Otherwise, none of the gray nodes are involved in a cycle; loop through again to recolor them black.

  • $\begingroup$ Some points of confusion. by "every node u" you mean, for every node u in predecessor? What do you mean by "while v is white and has a predecessor"? and what does it mean to set v := predecessor[v]? Thanks for the answering - @David Eisenstat $\endgroup$
    – Loourr
    May 21, 2013 at 23:02
  • $\begingroup$ := is the assignment operator. The while condition is meaningful because v changes. I meant every node u, though it would suffice to consider only the ones with non-null predecessors. $\endgroup$ May 21, 2013 at 23:09
  • $\begingroup$ I was not questioning what := means but rather what predecessor[v] is, and why its garnered to have a place in the predecessor array? $\endgroup$
    – Loourr
    May 21, 2013 at 23:33
  • $\begingroup$ @Loourr predecessor[v] is the tail of the arc most recently relaxed among those having head v, just like in your pseudocode for B–F. $\endgroup$ May 21, 2013 at 23:35
  • $\begingroup$ So in the step for i from 1 to size(vertices)-1: do I want to be doing for i from 1 to size(vertices)? and I'm still not sure what you mean by v "has a predecessor". $\endgroup$
    – Loourr
    May 22, 2013 at 3:02

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