Talk:Floyd–Warshall algorithm

 Question: Is the main function correct?

Where it appears:

   next[i][j] ← next[i][k]

Should it be?:

   next[i][j] ← next[k][j]

Clearly, it seems to be a problem for many people. In fact, as in the web reference that goes with the pseudocode, the modification of the array next is path[i][j] := path[k][j]. I did the modification yesterday without looking the talk page (my mistake) thinking it was a minor error, but my modification was reverted by @MfR: with this explanation :

Pseudocode in this page computes the second node of the path from i to j, not the penultimate (as in reference).

which I don't really understand... In Introduction to Algorithms, Cormen et al., MIT Press, Third Edition at page 697, again, we see [k][j]... Raphaelbwiki (talk) 13:49, 4 June 2019 (UTC)

Can confirm that this doesn't work if implemented as written in the article right now. I am busy doing something and won't be fixing the article; whoever's reverting it when other people fix it needs to stop. — Preceding unsigned comment added by 174.34.23.247 (talk) 23:11, 29 November 2019 (UTC)

Is there a bug in the main loop?

Should the main loop for updating the 'next' matrix instead of looking like:

   next[i][j] ← next[i][k]

be

   next[i][j] ← next[i][k]
   next[j][i] ← next[j][k]

? I have been random fuzz testing the algorithm in Python and occasionally get a wrong path which appears fixed with the additional line. — Preceding unsigned comment added by Shane.magrath (talk • contribs) 08:03, 17 April 2018 (UTC)

Litmus Test

My correction of the Floyd Allgorthm are correct. The entire artcle needs a rewrite to be both understandalb eand correct. As it remains, it is confusing my students and I spend a lot of time correcting their errors gleened from this article. This is a simple to undertand allgorithm when explained clearly. Instead we don't only have mathmatical snobbery, but truly inacuate information. I am SICK of wikipeadia inability to just tell wrong from write. Mathmatical truth is not a matter of a VOTE. When k = 1, it is not equal to zero. Fixing this page requires a complete rewrite because the graph and the agorthms don't match and the explaination doesn't inform the reader of facts. Not having fixed these problems in the article has been called not passing a litmus test. The only litmus test here is if the article is informatative and educational. It is NOT. — Preceding unsigned comment added by 96.57.23.82 (talk) 02:17, 17 May 2015 (UTC)

It was stated when correcting something that was obviously incorrect in the path allgorithm, one should consider understanding the poorly written article as a "litmus test."

There is no litmus test. The article should be written clearly. It should be understandable and correct. It is not secret code for a select few. If this is not understood, David Esptein, then it is time for you to take some time off of wikipedia editing and do something else, like create the secret society of the Knights templer, or whatever.

Do not reverse the correction without a clear explanation of the ERROR that fixed the original version. If you feel passionate enough to insult people, then feel passionate enough to fix the error and to make the correction in the algorithm clear. Stop vandalizing the page — Preceding unsigned comment added by 96.57.23.82 (talk • contribs)

Please see WP:COMPETENCE and please stop editing this article until you actually understand the algorithm. To be specific, the kth stage of the algorithm finds paths whose interior vertices form a subset of the set of numbers from 1 to k. The first stage finds paths whose interior vertices belong to the singleton set {1}, the second stage finds paths whose interior vertices belong to the set {1,2}, the third stage finds paths whose interior vertices belong to the set {1,2,3}, etc. For some reason this IP editor insists on replacing the set {1,2,3} by the number 3 in this description. It is an incorrect change, the article is correct as-is, and my repeated attempts at explanation (in the edit summaries) have fallen on deaf ears. —David Eppstein (talk) 22:58, 16 May 2015 (UTC)

Your not listening. The problem isn't that the set of vertices is wrong, the problem is that the material is NOT consistant and therefore confuses the STUDENT. The Graph has to match the Allgorithm and mathc the equation. You have to SAY what you mean. In this case it does not SAY what it means, nor does it say what you said.

It is just a pile of nonsense. — Preceding unsigned comment added by 96.57.23.82 (talk) 06:07, 17 May 2015 (UTC)

I actually understand the article very well and there is no excuse for your attitude or for your confusing readers. I was teaching this algorithm probably when you were in High School. If you can't write it clearly, then take time off and stop vandalizing the article. — Preceding unsigned comment added by 96.57.23.82 (talk) 23:04, 16 May 2015 (UTC)

Actually, while we are the topic of High School, there is nothing inherently difficult about this topic. This graph vertex algorithm can be understand by any high school student, and most junior high school students. There problem with this article is that it is written sloppily and like garbage. It is confusing, full of unnecessarily jargon, and an impedance to actually understanding the topic. The entire thing needs a rewrite.

Dear non-account user, a request: before going back and making the same edit again, could we instead try to find common understanding? In particular, what do you think the symbols "{1,2,3}" represent? --JBL (talk) 01:21, 17 May 2015 (UTC)

there is no need to "know" what it means because it says what it is, aand what it says is WRONG. In fact, the __entire__ article is wrong.

1 let dist be a |V| × |V| array of minimum distances initialized to ∞ (infinity) 2 for each vertex v 3 dist[v][v] ← 0 4 for each edge (u,v) 5 dist[u][v] ← w(u,v) // the weight of the edge (u,v) 6 for k from 1 to |V| 7 for i from 1 to |V| 8 for j from 1 to |V| 9 if dist[i][j] > dist[i][k] + dist[k][j] 10 dist[i][j] ← dist[i][k] + dist[k][j] 11 end if

Example

The algorithm above is executed on the graph on the left below:

Floyd-Warshall example.svg

Note - you have an EXMAMPLE pf K=0 K NEVER equals zero ... K starts at 1.

Prior to the first iteration of the outer loop, labeled k=0 above,

Yeah - that si WRONG, k is 1.

the only known paths correspond to the single edges in the graph. At k=1, paths that go through the vertex 1 are found: in particular, the path 2→1→3 is found, replacing the path 2→3 which has fewer edges but is longer. At k=2, paths going through the vertices {1,2} are found. The red and blue boxes show how the path 4→2→1→3 is assembled from the two known paths 4→2 and 2→1→3 encountered in previous iterations, with 2 in the intersection. The path 4→2→3 is not considered, because 2→1→3 is the shortest path encountered so far from 2 to 3. At k=3, paths going through the vertices {1,2,3} are found.

WRONG, the paths do NOT go through that set.

Finally, at k=4, all shortest paths are found.

So the whole thing needs a rewrite. Try starting with Cormen.... — Preceding unsigned comment added by 96.57.23.82 (talk) 01:30, 17 May 2015 (UTC)

Your post is not really comprehensible -- possibly, if you want your comments to be taken seriously, you should take some time and write a few clear sentences laying out exactly what you think the incorrect statement is and why you believe it to be incorrect. As it is I am not able to understand either of these things from your comments. (Also, it really might help if you would answer the question that I asked.) --JBL (talk) 01:37, 17 May 2015 (UTC)

Since it quotes the article, it is not surprising that it is hard to understand. As it is, it is clear. The algorithm in the code doesn't match diagram, and the diagram doesn't match the text. This whole article needs a rewrite. Perhaps the esteemed Professor of Univ Irvine can donate his class notes. — Preceding unsigned comment added by 96.57.23.82 (talk) 01:41, 17 May 2015 (UTC)

Follow Cormen if you need to The structure of a shortest path In the Floyd-Warshall algorithm, we use a different characterization of the structure of a shortest path than we used in the matrix-multiplication-based all-pairs algorithms. The algorithm considers the "intermediate" vertices of a shortest path, where an intermediate vertex of a simple path p = 〈 v , v , . . . , v 〉 is any vertex of p other than 1 2 lv or v , that is, any vertex in the set {v ,v , . . . , v }. 1 l 2 3 l-1The Floyd-Warshall algorithm is based on the following observation. Let the vertices of G be V = {1, 2, . . . , n}, and consider a subset {1, 2, . . . , k} of vertices for some k. For any pair of vertices i, j ∈ V, consider all paths from i to j whose intermediate vertices are all drawn from {1, 2, . . . , k}, and let p be a minimum-weight path from among them. (Path p is simple, since we assume that G contains no negative-weight cycles.) The Floyd- Warshall algorithm exploits a relationship between path p and shortest paths from i to j with all intermediate vertices in the set {1, 2, . . . , k - 1}. The relationship depends on whether or not k is an intermediate vertex of path p. Figure 26.3 Path p is a shortest path from vertex i to vertex j, and k is the highest-numbered intermediate vertex of p. Path p1 , the portion of path p from vertex i to vertex k, has all intermediate vertices in the set {1, 2, . . . , k – 1}. The same holds for path p2 from vertex k to vertex j. • If k is not an intermediate vertex of path p, then all intermediate vertices of path p are in the set {1, 2, . . . , k – 1}. Thus, a shortest path from vertex i to vertex j with all intermediate vertices in the set {1, 2, . . . , k – 1} is also a shortest path from i to j with all intermediate vertices in the set {1, 2, . . . , k}. • If k is an intermediate vertex of path p, then we break p down into

as shown in Figure 26.3

. By Lemma 25.1, p1 is a shortest path from i to k with all intermediate vertices in the set {1, 2, . . . , k}. In fact, vertex k is not an intermediate vertex of path p1 , and so p1 is a shortest path from i to k with all intermediate vertices in the set {1, 2, . . . , k - 1}. Similarly, p2 is a shortest path from vertex k to vertex j with all intermediate vertices in the set {1, 2, . . . , k - 1}.

A recursive solution to the all-pairs shortest-paths problem Based on the above observations, we define a different recursive formulation of shortest-path estimates than we did in Section 26.1. Let

be the weight of a shortest path from vertex i to vertex j with all intermediate vertices in

the set {1, 2, . . . , k}. When k = 0, a path from vertex i to vertex j with no intermediate vertex numbered higher than 0 has no intermediate vertices at all. It thus has at most one edge, and hence

. A recursive definition

is given by (26.5) The matrix

gives the final answer—

are in the set {1, 2, . . . , n}. for all i, j ∈ Vbecause all intermediate vertices Computing the shortest-path weights bottom up Based on recurrence (26.5), the following bottom-up procedure can be used to compute the values

in order of

increasing values of k. Its input is an n × n matrix W defined as in equation (26.1). The procedure returns the matrix D(n) of shortest-path weights. Figure 26.4 shows a directed graph and the matrices D (k) computed by the Floyd-Warshall algorithm. The running time of the Floyd-Warshall algorithm is determined by the triply nested for loops of lines 3-6. Each execution of line 6 takes O(1) time. The algorithm thus runs in time Θ(n3). As in the final algorithm in Section 26.1 , the code is tight, with no elaborate data structures, and so the constant hidden in the Θ -notation is small. Thus, the Floyd-Warshall algorithm is quite practical for even moderate-sized input graphs. Constructing a shortest path There are a variety of different methods for constructing shortest paths in the Floyd-Warshall algorithm. One way is to compute the matrix D of shortest-path weights and then construct the predecessor matrix Π from the D matrix. This method can be implemented to run in O(n3 ) time (Exercise 26.1-5). Given the predecessor matrix Π, the PRINT- ALL- PAIRS- SHORTEST- PATH procedure can be used to print the vertices on a given shortest path. We can compute the predecessor matrix Π "on-line" just as the Floyd-Warshall algorithm computes the matrices D (k) . Specifically, we compute a sequence of matrices Π(0) , Π(1) , . . . , Π(n), where Π = Π(n) and

is

defined to be the predecessor of vertex j on a shortest path from vertex i with all intermediate vertices in the set {1, 2, . . . , k}. We can give a recursive formulation of vertices at all. Thus, . When k = 0, a shortest path from i to j has no intermediate Figure 26.4 The sequence of matrices D (k) and Π(k) computed by the Floyd-Warshall algorithm for the graph in Figure 26.1 . — Preceding unsigned comment added by 96.57.23.82 (talk) 02:07, 17 May 2015 (UTC)

I am new to wikipedia authoring, but whether or not the article is clear the pseudo-code is incorrect. On line 9 there is a > symbol where there should be a < symbol. Being new I thought I would talk about it on this page before diving in and correcting it Philsrice (talk) 08:28, 14 August 2015 (UTC)

Pseudocode contains end-ifs but no end-fors

The pseudocode contains end-ifs but no end-fors:

I think it makes sense to have it be consistent: either no end-fors and no end-ifs or every for-loop terminated with an end-for and every if-statement terminated with end-if

1 let dist be a |V| × |V| array of minimum distances initialized to ∞ (infinity)
2 for each edge (u,v)
3    dist[u][v] ← w(u,v)  // the weight of the edge (u,v)
4 for each vertex v
5    dist[v][v] ← 0
6 for k from 1 to |V|
7    for i from 1 to |V|
8       for j from 1 to |V|
9          if dist[i][j] > dist[i][k] + dist[k][j] 
10             dist[i][j] ← dist[i][k] + dist[k][j]
11         end if  — Preceding unsigned comment added by 2601:282:8001:2E4A:845:D12F:594F:7A77 (talk) 22:43, 4 October 2018 (UTC) 

Could Use an Example Graph Not Containing a Negative Cycle

The Floyd-Warshall algorithm only finds shortest paths if there are no negative cycles. It is illustrative to see what happens if there is a negative cycle, so the existing example provided in the article is useful. However, it would also be nice to have a graph not containing any negative-cycles.

Below are some notes on a step-through of the algorithm for the given example:

upper-bound on cost from node 2 to node 1 is +4.
upper-bound on cost from node 1 to node 3 is -2.
cost on the path from 2 to 3 (going through 1 has) new upper bound of -2.
old upper-bound on cost from node 2 to 3 was +3

Next path: (4 --> 2 --> 3)
OBSERVE cost(4, 2) <= -1
OBSERVE cost(2, 3) <= -2
UPDATE cost(4, 3) <= -3
old cost(4, 3) was +inf

Next path: (1 --> 3 --> 4)
OBSERVE cost(1, 3) <= -2
OBSERVE cost(3, 4) <= +2
UPDATE cost(1, 4) <= 0

Next path: (2 --> 3 --> 4)
OBSERVE cost(2, 3) <= -2
OBSERVE cost(3, 4) <= +2
UPDATE cost(2, 4) <= 0

Next path: (4 --> 3 --> 4)
OBSERVE cost(4, 3) <= -3
OBSERVE cost(3, 4) <= +2
UPDATE cost(4, 4) <= -1

Negative cycle found. can leave 4 and return to 4 with net negative cost