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Floyd-Warshall Algorithm PROF. NAVRATI SAXENA ASSISTANT PROFESSOR, DEPT. OF COMPUTER SCIENCE SAN JOSE STATE UNIVERSITY, CA.

Floyd-Warshall Algorithm A weighted, directed graph is a collection vertices connected by weighted edges (where the weight is some real number). ◦ One of the most common examples of a graph in the real world is a road map.  Each location is a vertex and each road connecting locations is an edge.  We can think of the distance traveled on a road from one location to another as the weight of that edge. SFO San Jose LA SFO 0 1.7 3.5 San Jose 1.5 0 ∞ LA 4 2.5 0 SFO San Jose LA 3.5 4 1.5 1.7 2.5

Storing a Weighted, Directed Graph  Adjacency Matrix: ◦ Let D be an edge-weighted graph in adjacency-matrix form • D( i,j ) is the weight of edge (i, j), or  if there is no such edge. • Update matrix D , with the shortest path through immediate vertices. 0 1 2 3 0 0 6 5 ∞ 1 ∞ 0 4 3 2 ∞ ∞ 0 2 3 ∞ ∞ ∞ 0 2 1 3 0 6 5 4 3 2 D =

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Floyd-Warshall Algorithm  Given a weighted graph, we want to know the shortest path from one vertex in the graph to another. ◦ The Floyd-Warshall algorithm determines the shortest path between all pairs of vertices in a graph. ◦ What is the difference between Floyd-Warshall and Dijkstra’s?

Floyd-Warshall vs. Dijkstra  Dijkstra’s runs in  ( V 2 ), where V: number of vertices ◦ We can just call Dijkstra |V| times , passing a different source vertex each time. ◦  ( V  V 2 ) =  ( V 3 ) ◦ (runtime of Floyd-Warshall Algorithm is the same)  BUT , Dijkstra’s doesn’t work with negative -weight edges .  Floyd Warshall can work with negative edges.

Floyd Warshall Algorithm: Basics  Working Principle of Floyd Warshall ◦ Vertices in a graph: Numbered from 1 … n . ◦ Consider the subset {1,2,…, k} of these n vertices. ◦ Imagine finding the shortest path from vertex i to vertex j using vertices in the set {1,2,…,k} only. ◦ There are two situations: 1) k is an intermediate vertex on the shortest path . 2) k is not an intermediate vertex on the shortest path . j k i

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Floyd Warshall Algorithm – Example Consider Vertex 3: Nothing changes. Consider Vertex 2: D(1,3) = D(1,2) + D(2,3) Consider Vertex 1: D(3,2) = D(3,1) + D(1,2) Original weights.

Floyd Warshall Algorithm: Pseudocode  Looking at this example, come up with the following: ◦ Let D store the matrix with the initial graph edge information initially, and update D with the calculated shortest paths. For k=1 to n { For i=1 to n { For j=1 to n D[i,j] = min(D[i,j],D[i,k]+D[k,j]) } } ◦ The final D matrix will store all the shortest paths.

Floyd Warshall Algorithm  Example…

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Floyd Warshall – Path Reconstruction (1/3)  The path matrix will store the last vertex visited on the path from i to j. ◦ So path[i][j] = k means that in the shortest path from vertex i to vertex j , the LAST vertex on that path before reaching j is k .  Based on this definition, initialize the path matrix as follows: ◦ path[i][j] = i if (i!=j and there exists an edge from i to j ) ◦ = NIL ; otherwise

Floyd Warshall – Path Reconstruction (2/3)  Before running Floyd’s, initialize distance matrix D and path matrix P to indicate use of no immediate vertices. ◦ (Thus, you are only allowed to traverse direct paths between vertices.)  At each step find out whether or not using vertex k will improve an estimate of the distances between vertex i, j.  If it does improve the estimate what you need to record?

Floyd Warshall – Path Reconstruction (3/3)  If it does improve the estimate here’s what you need to record: 1) Record the new shortest path weight between i and j  No need to change our path and we do not update the path matrix 2) Record that the shortest path between i and j goes through k  Store the last vertex from the shortest path from vertex k to vertex j. This will NOT necessarily be k, but rather, it will be path[k][j]. This gives us the following update to our algorithm: if (D[i][k]+D[k][j] < D[i][j]) { // Update is necessary to use k as intermediate vertex D[i][j] = D[i][k]+D[k][j]; path[i][j] = path[k][j]; }

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Path Reconstruction  Example…

Path Reconstruction (1/2)  Now, the once this path matrix is computed, we have all the information necessary to reconstruct the path. ◦ Consider the following path matrix (indexed from 1 to 5 instead of 0 to 4):  How to Reconstruct the path from vertex1 to vertex 2? NIL 3 4 5 1 4 NIL 4 2 1 4 3 NIL 2 1 4 3 4 NIL 1 4 3 4 5 NIL

Path Reconstruction (2/2) • Now, look at path[1][3], this stores a 4. Thus, we find the last vertex visited on the path from 1 to 3 is 4. So, our path now: 1…4 ->3->2 . • So, we must now look at path[1][4]. This stores a 5, thus, we know our path is 1…5 ->4->3->2 . • When we finally look at path[1][5], we find 1, which means our path is: 1->5->4->3->2 . NIL 3 4 5 1 4 NIL 4 2 1 4 3 NIL 2 1 4 3 4 NIL 1 4 3 4 5 NIL • First look at path[1][2] = 3. • This signifies that on the path from 1 to 2, 3 is the last vertex visited before 2. • Thus, the path is now, 1…3 ->2. 1 2 3

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