Shortest Path Problems Algorithms Shortest Path Problems

G = (V, E) weighted directed graph w: ER weight function Weight of a path p = <v0, v1,. . ., vn> Shortest path weight from u to v Shortest path from u to v: Any path from u to v with w(p) = (u,v) [v] predecessor of v on a path

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Variants Single-source shortest paths: find shortest paths from source vertex to every other vertex Single-destination shortest paths: find shortest paths to a destination from every vertex Single-pair shortest-path find shortest path from u to v All pairs shortest paths

Lemma 25.1 Subpaths of shortest paths are shortest paths. Given G=(G,E) w: E  R Let p = p = <v1, v2,. . ., vk> be a shortest path from v1 to vk For any i,j such that 1 i j k, let pij be a subpath from vi to vj.. Then pij is a shortest path from vi to vj.

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Corollary 25.2 Let G = (V,E) w: E  R Suppose shortest path p from a source s to vertex v can be decomposed into p’ s u v for vertex u and path p’. Then weight of the shortest path from s to v is (s,v) = (s,u) + w(u,v)

Lemma 25.3 Let G = (V,E) w: E  R Source vertex s For all edges (u,v)E (s,v) (s,u) + w(u,v)

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Relaxation Shortest path estimate d[v] is an attribute of each vertex which is an upper bound on the weight of the shortest path from s to v Relaxation is the process of incrementally reducing d[v] until it is an exact weight of the shortest path from s to v

INITIALIZE-SINGLE-SOURCE(G, s) 1. for each vertex v V(G) 2. do d[v]  3. [v]  nil 4. d[s]  0

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Relaxing an Edge (u,v) Question: Can we improve the shortest path to v found so far by going through u? If yes, update d[v] and [v]

RELAX(u,v,w) 1. if d[v] > d[u] + w(u,v) 2. then d[v]  d[u] + w(u,v) 3. [v]  u

EXAMPLE 1 s  s  Relax       v u v u

EXAMPLE 2 s  s  Relax       v u v u

Dijkstra’s Algorithm Problem: Solve the single source shortest-path problem on a weighted, directed graph G(V,E) for the cases in which edge weights are non-negative

Dijkstra’s Algorithm Approach maintain a set S of vertices whose final shortest path weights from the source s have been determined. repeat select vertex from V-S with the minimum shortest path estimate insert u in S relax all edges leaving u

DIJKSTRA(G,w,s) 1. INITIALIZE-SINGLE-SOURCE(G,s) 2. S  3. Q  V[G] 4. while Q  5. do u  EXTRACT-MIN(Q) 6. S  S {u} 7. for each vertex v Adj[u] 8. do RELAX(u,v,w)

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Analysis of Dijkstra’s Algorithm Suppose priority Q is: an ordered (by d) linked list Building the Q O(V lg V) Each EXTRACT-MIN O(V) This is done V times O(V2) Each edge is relaxed one time O(E) Total time O(V2 + E) = O(V2)

Analysis of Dijkstra’s Algorithm Suppose priority Q is: a binary heap BUILD-HEAP O(V) Each EXTRACT-MIN O(lg V) This is done V times O(V lg V) Each edge is relaxation O(lg V) Each edge relaxed one time O(E lg V) Total time O(V lg V + E lg V))

Properties of Relaxation Lemma 25.4 G=(V,E) w: E  R (u,v)  E After relaxing edge (u,v) by executing RELAX(u,v,w) we have d[v]  d[u] + w(u,v)

Lemma 25.5 Given: G=(V,E) w: E  R source s  V Graph initialized by INITIALIZE-SINGLE-SOURCE(G,s) then d[v]  (s,v) for all v V and this invariant is maintained over all relaxation steps Once d[v] achieves a lower bound (s,v), it never changes

Corollary 25.6 Given: G=(V,E) w: E  R source s  V No path connects s to given v then after initialization d[v]  (s,v) and this inequality is maintained over all relaxation steps.

Lemma 25.7 Given: G=(V,E) w: E  R source s  V Let s - - u v be the shortest path in G for all vertices u and v. Suppose G initialized by INITIALIZE-SINGLE-SOURCE is followed by a sequence of relaxations including RELAX(u,v,w) Then d[u] = (s,u) prior to call implies that d[u] = (s,u) after the call

Bottom Line Therefore, relaxation causes the shortest path estimates to descend monotonically toward the actual shortest-path weights.

Shortest-Paths Tree of G(V,E) The shortest-paths tree at S of G(V,E) is a directed subgraph G’-(V’,E’), where V’ V, E’E, such that V’ is the set of vertices reachable from S in G G’ forms a rooted tree with root s, and for all v V’, the unique simple path from s to v in G’ is a shortest path from s to v in G

Goal We want to show that successive relaxations will yield a shortest-path tree

Lemma 25.8 Given: G=(V,E) w: E  R source s  V Assume that G contains no negative-weight cycles reachable from s. Then after the graph is initialized with INITIALIZE-SINGLE-SOURCE • the predecessor subgraph G forms a rooted tree with root s, and • any sequence of relaxation steps on edges in G maintains this property as an invariant.

Algorithms Bellman-Ford Algorithm Directed-Acyclic Graphs All Pairs-Shortest Path Algorithm

Why does Dijkstra’s greedy algorithm work? Because we know that when we add a node u to the set S, the value d is the length of the shortest path from s to u. But, this only works if the edges of the graph are nonnegative.

a 10 7 b c -4 Would Dikjstra’s Algorithm work with this graph?

a 6 7 b c -14 What is the length of the shortest path from a to c in this graph?

Bellman-Ford Algorithm The Bellman-Ford algorithm can be used to solve the general single source shortest path problem Negative weights are allowed The algorithm detects negative cycles and returns false if the graph contains one.

BELLMAN-FORD(G,w,s) 1 INITIALIZE-SINGLE-SOURCE(G,s) 2 for i  1 to |V[G]| -1 3 do for each edge (u,v)  E[G] 4 do RELAX(u,v,w) 5 for each edge (u,v)  E[G] 6 do if d[v] > d[u] + w(u,v) 7 then return false 8 return true

When there are no cycles of negative length, what is maximum number of edges in a shortest path when the graph has |V[G]| vertices and |E[G]| edges?

Dynamic Programming Formulation The following recurrence shows how the Bellman Ford algorithm computes the d values for paths of length k.

a 10 Processing order 1 2 3 (a,c) (b,a) (c,b) (c,d) (d,b) 7 b c -4 -5 9 d

Complexity of the Bellman Ford Algorithm Time complexity: Performance can be improved by adding a test to the loop to see if any d values were updated on the previous iteration, or maintain a queue of vertices whose d value changed on the previous iteration-only process these on the next iteration

Single-source shortest paths in directed acyclic graphs Topological sorting is the key to efficient algorithms for many DAG applications. A topological sort of a DAG is a linear ordering of all of its vertices such that if G contains an edge (u,v), then u appears before v in the ordering.

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DAG-SHORTEST-PATHS(G,w,s) 1 topologically sort the vertices of G 2 INITIALIZE-SINGLE-SOURCE(G,s) 3 for each vertex u taken in topological order 4 do for each vertex v  Adj[u] 5 do RELAX(u,v,w)

Topological Sorting TOPOLOGICAL-SORT(G) 1 call DFS(G) to compute finishing times f[v] for each vertex v 2 as each vertex is finished, insert it onto the front of a linked list 3 return the linked list of vertices

Depth-first search Goal: search all edges in the graph one time Strategy: Search deeper in the graph whenever possible Edges are explored out of the most recently discovered vertex v that still has unexplored edges leaving it. Backtrack when a dead end is encountered

Predecessor subgraph The predecessor subgraph of a depth first search forms a depth-first forest composed of depth-first trees The edges in E are called tree edges

Vertex coloring scheme All vertices are initially white A vertex is colored gray when it is discovered A vertex is colored black when it is finished (all vertices adjacent to the vertex have been examined completely)

Time Stamps Each vertex v has two time-stamps For every vertex u d[v] records when v is first discovered (and grayed) f[v] records when the search finishes examining its adjacency list (and is blackened) For every vertex u d[u] < f[u]

Color and Time Stamp Summary Vertex u is white before d[u] gray between d[u] and f[u] black after f[u] Time is a global variable in the pseudocode

DFS(G) 1 for each vertex u  V[G] 2 do color[u] white 3 [u] nil 4 time 0 5 do for each vertex u  Adj[u] 6 do if color[u] = white 7 then DFS-VISIT(u)

DFS-VISIT(u) 1 color[u] gray 2 d[u] time time time +1 3 for each vertex v  Adj[u] 4 do if color[v] = white 5 then [v] u 6 DFS-VISIT(v) 7 color[u] black 8 f[u] time time time +1

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b 2/ e 3/ 4/ 1/ c a g f d

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b 2/7 e 3/6 8/9 4/5 1/ c a g 10/ f d

b 2/7 e 3/6 8/9 4/5 1/ c a g 10/ 11/ f d

b 2/7 e 3/6 8/9 4/5 1/ c a g 11/12 10/ f d

b 2/7 e 3/6 8/9 4/5 1/ c a g 11/12 10/13 f d

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Running time of DFS lines 1-3 of DFS lines 5-7 of DFS lines 2-6 of DFS-VISIT

Running Time of Topological Sort DFS Insertion in linked list

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Running Time for DAG-SHORTEST-PATHS Topological sort Initialize-single source 3-5 each edge examined one time