Single Source Shortest Path

Transcription

1 Single Source Shortest Path A directed graph G = (V, E) and a pair of nodes s, d is given. The edges have a real-valued weight W i. This time we are looking for the weight and the shortest path from s to d. The weight of the shortest path from s to d is called δ(s, d). We still have the property that a shortest path between u and v contains shortest path between intermediate nodes. The shortest path between s and d might not be defined if the graph contains cycles with negative weight. A shortest path can not contain s positive weight cycle. Hence, path length is at most n 1. SSSP 1

2 Shortest Path Tree A shortest path tree G = (V, E ) is a directed subgraph of G (V V and E E), such that 1. V is the set of nodes reachable from s in G. 2. G forms a rooted tree with root s, and 3. For all v V, the unique simple path from s to v in G is a shortest path from s to v in G. Shortest paths and shortest path trees are not unique! SSSP 2

3 Relaxation For each vertex v V we maintain an attribute d[v], which is an upper bound on the weight of a shortest path from source s to v. The attribute is called shortest path estimate. In the beginning, d[v] = for v s and d[s] = 0. We also use variables π[v] storing the predecessor of v. In a relaxation step we may decrease the shortest path estimate and update the π values. SSSP 3

4 Relaxation (u, v, w) If d[v] > d[u] + w[u, v], then d[v] := d[u] + w(u, v) and π[v] = u The path that uses u as a last edge is shorter than the one we found previously. Bellman-Ford(G, w, s) Initialise-Single-Source(G, s) for i 1 to V [G] 1 do for each edge (u, v) E[G] do Relax(u, v, w) for each edge (u, v) E[G] do if d[v] > d[u] + w(u, v) then return False return True SSSP 4

5 Runtime and Correctness The runtime is easy. In every iteration of the for-loop we relax every edge. We need V 1 passes over all edges. This gives a runtime of O(V E). First we show the following: if there are no negative-weight cycles, the algorithm computes correct shortest path weights for all vertices reachable from s. But to do that we need a couple of other results,... Lemma 1 Let G = (V, E) be a weighted directed graph with source s and weight function w : E R, and assume that G contains no negative-weight cycle that is reachable from s. Then, after V 1 iterations of the first for-loop of the Bellman-Ford, we have d[v] = δ(s, v) for all vertices v that are reachable from s. SSSP 5

6 Upper bound property Lemma 2 Let G = (V, E) be a weighted, directed graph with weight function w : E R, and let s V be a source vertex. Assume G is initialised as we defined. Then d[v] δ(s, v) for all v V, and this invariant is maintained over any sequence of relaxation steps. Moreover, once d[v] achieves its lower bound δ(s, v), it never changes. SSSP 6

7 Proof of Lemma 2 We prove the invariant d[v] δ(s, v) by induction over the number of relaxation steps. After the in initialisation d[v] δ(s, v) is certainly true. For the inductive step, consider the relaxation of an edge (u, v). By the inductive hypothesis, d[x] δ(s, x) for all x V prior to the relaxation. If d[v] changes we have d[v] = d[u] + w(u, v) δ(s, u) + w(u, v) δ(s, v). To see that the value of d[v] never changes once d[v] = δ(s, v) have a look at the above equations again. We have just shown d[v] δ(s, v). It can not increase since relaxation steps do not increase values. SSSP 7

8 Lemma 3: Convergence property Let G = (V, E) be a weighted, directed graph with weight function w : E R, and let s V be a source vertex. Let s = u v be a shortest path in G for some vertices u, v V. Suppose that G is initialised as we defined. Then, a sequence of relaxation steps that includes the call RELAX(u, v, w) is executed on the edges of G. If 4d[u] = δ(s, u) at any time prior to the call, then d[v] = δ(s, v) at all times after the call. SSSP 8

9 Proof of Lemma 3 By the upper bound property: if d[u] = δ(s, u) at some point prior to relaxing edge (u, v), then this equality holds thereafter. After relaxing edge (u, v) we have we have d[v] d[u] + w(u, v) = δ(s, u) + w(u, v) = δ(s, v). The first inequality is due to the fact that, after relaxing edge (u, v), we have d[v] d[u] + w(u, v). The last equality is due to the fact that subpath of shortest path are also shortest paths. By the upper bound property, d[v] δ(s, v), from which we can conclude that d[v] = δ(s, v), and this equality is maintained thereafter. SSSP 9

10 Path-relaxation Property Lemma 4 Let G = (V, E) be a weighted, directed graph with weight function w : E R, and let s V be a source vertex. Consider any shortest path p = (v 1, v 2,..., v k ) from s = v 0 to v k. If G is initialised as we defined, and then a sequence of relaxation steps occurs that includes, in order, relaxations of the edges (v 0, v 1 ), (v 1, v 2 ),... (v k 1, v k ), then d[v k ] = δ(s, v k ). This holds no matter which other edge relaxations occur. SSSP 10

11 Proof of Lemma 4 We show by induction that after the ith edge of path p is relaxed, we have d[v i ] = δ(s, v i ). For the basis, i = 0 we have from the initialisation d[v 0 ] = d[s] = 0 = δ(s, s). By the upper bound property, the value of d[s] is never changes after initialisation. For the inductive step, we assume that d[v i 1 ] = δ(s, v i 1 ), and we examine the relation of edge (v i 1, v i ). By the convergence property, we have after the relation d[v i ] = δ(v, v i ). Again, this equality is maintained thereafter. SSSP 11

12 Proof of Lemma 1 Consider a vertex v that is reachable from s. Let p = (v 0, v 1,... v k ) (with v 0 = s and v k = v) be a shortest acyclic path from s to v. The path has at most V 1 edges, giving k V 1. Each of the iterations of the for loop relaxes all edges of E. Among the edges relaxed in the ith iteration we have the edge (v i 1, v i ). By the relaxation property we have d[v] = d[v k ] = δ(s, v k ) = δ(v, v) SSSP 12

13 Predecessor-subpath property Lemma 5 Let G = (V, E) be a weighted, directed graph with weight function w : E R, and let s V be a source vertex. Assume that G does not contain a negative cycle reachable from s. Then, if we execute relaxation steps resulting in d[v] = δ(s, v) for all v V. Then the predecessor subgraph G π is a shortest-path tree rooted at s. SSSP 13

14 Proof of Lemma 5 We have to show that all three shortest-paths tree properties hold. The vertices that are reachable from s are the ones with finite d[v]. A vertex v V {s} has s finite value d[v] if and only if π[v] nil. Thus, the vertices in V π are those reachable from s. Third property not here,... It remains to show that the unique path in G π are shortest paths. Let p = (s = v 0, v 1,... v k = v) be a path in G π. For i = 1, 2,... k we have both d[v i ] = δ(s, v i ) and d[v i ] d[v i 1 ] + w(v i 1, v i ). Hence, w(v i 1, v i ) δ(s, v i ) δ(s, v i 1 ). Summing the weights along p yields w(p) = k i=1 w(v i 1, v i ) k i=1 (δ(s, v i) δ(s, v i 1 ))) = δ(s, v k ) δ(s, v 0 ) = δ(s, v k ) SSSP 14

15 Thus, w(p) δ(s, v k ) and we can conclude w(p) = δ(s, v k ) since δ(s, v k ) is a lower bound. p is a shortest path. SSSP 15

16 Correctness of Bellman-Ford Lemma 6 Let G = (V, E) be a weighted, directed graph with weight function w : E R, and let s V be a source vertex. Consider any shortest path p = (v 1, v 2,..., v k ) from s = v 0 to v k. 1. If G contains no negative cycles that are reachable from s, then the algorithm returns true and we have d[v] = δ(s, v) for all vertices v V and the Bellman-Ford algorithm computes a shortest path tree rooted at s. 2. If G does contains a negative weight cycle reachable from s, then the algorithm returns false. SSSP 16

17 Proof of Lemma 6 Suppose the graph does not contain a negative cycle that is reachable for s. If a vertex v is reachable from s, then Lemma 1 shows that d[v] = δ(s, v). In v is not reachable from v, d[v] = δ(s, v) =. This follows from the upper-bound property since we always have = δ(u, v) d[v], Hence, we have d[v] =. The predecessor-subgraph property implies that G π is a shortest-path tree. It remains to show that the Bellman-Ford returns true. At termination we have for all edges (u, v) E d[v] = δ(s, v) δ(s, u) + w(u, v) = d[u] + w(u, v) SSSP 17

18 Proof of Lemma 6 cont Suppose the graph does contain a negative cycle c = (v 0, v 1,..., v k ) (v 0 = v k ) that is reachable for s. Assume for contradiction that Bellman-Ford returns true. Thus, d[v i ] d[v i 1 ] + w(v i 1, v i ) for 1 i k and k i=1 w(v i 1, v i ) < 0. k i=1 d[v i] k i=1 (d[v i 1] + w(v i 1, v i )) = k i=1 d[v i 1] + k i=1 w(v i 1, v i ). Each vertex appears exactly once in each of the above summations and k i=1 d[v i] = k i=1 d[v i 1]. This gives 0 < k i=1 w(v i 1, v i ) and a contradiction. SSSP 18

19 Dijkstra s Algorithm The algorithm solves the single-source shortest path problem for the case of non-negative edge weight. It maintains a set of vertices S whose final shortest path has already been determined. It repeatedly selects the vertex u V S with minimum shortest pats estimate adds it to S, and relaxes all edges leaving u. We will use a min-priority queue Q for the vertices, keyed by their d values. SSSP 19

20 Dijkstra s Algorithm Dijkstra(G, w, s) Initialise-Single-Source(G, s) S := and Q := V [G] while Q do u := Extract Min(Q) S := S {u} for each vertex v adjacent to u do Relax(u,v,w) SSSP 20

21 Idea of Dijkstra s Algorithm The algorithm maintains the invariant that Q = V S that holds at the start of each iteration of the while-loop. The invariant is certainly true at the beginning. Each time we go through the while loop, a vertex u is extracted from Q = V S and added to S. u has the smallest shortest path estimate d[u]. Then each outgoing edge (u, v) is relaxed. The shortest path estimate is updated if the shortest path to v can be improved by going through u. Because Dijkstra s algorithm always choses the lightest vertex to add to set S, it is a greedy strategy. Dijkstra calculates an optimal solution. SSSP 21

22 Idea of Dijkstra s Algorithm II The key of the correctness proof is to show that each time a vertex u is added to S we have d[u] = δ(s, u). The runtime is O((V + E) log V ). SSSP 22