# Shortest Paths Definitions Single Source Algorithms –Bellman Ford –DAG shortest path algorithm –Dijkstra All Pairs Algorithms –Using Single Source Algorithms.

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Shortest Paths Definitions Single Source Algorithms –Bellman Ford –DAG shortest path algorithm –Dijkstra All Pairs Algorithms –Using Single Source Algorithms –Matrix multiplication –Floyd-Warshall Both of above use adjacency matrix representation and dynamic programming –Johnson’s algorithm Uses adjacency list representation

Single Source Definition Input –Weighted, connected directed graph G=(V,E) Weight (length) function w on each edge e in E –Source node s in V Task –Compute a shortest path from s to all nodes in V

All Pairs Definition Input –Weighted, connected directed graph G=(V,E) Weight (length) function w on each edge e in E We will typically assume w is represented as a matrix Task –Compute a shortest path from all nodes in V to all nodes in V

Comments If edges are not weighted, then BFS works for single source problem Optimal substructure –A shortest path between s and t contains other shortest paths within it No known algorithm is better at finding a shortest path from s to a specific destination node t in G than finding the shortest path from s to all nodes in V

Negative weight edges Negative weight edges can be allowed as long as there are no negative weight cycles If there are negative weight cycles, then there cannot be a shortest path from s to any node t (why?) If we disallow negative weight cycles, then there always is a shortest path that contains no cycles

Relaxation technique For each vertex v, we maintain an upper bound d[v] on the length of shortest path from s to v d[v] initialized to infinity Relaxing an edge (u,v) –Can we shorten the path to v by going through u? –If d[v] > d[u] + w(u,v), d[v] = d[u] + w(u,v) –This can be done in O(1) time

Bellman-Ford Algorithm Bellman-Ford (G, w, s) –Initialize-Single- Source(G,s) –for (i=1 to V-1) for each edge (u,v) in E –relax(u,v); –for each edge (u,v) in E if d[v] > d[u] + w(u,v) –return NEGATIVE WEIGHT CYCLE 3 5 7 -5 3 2 1 s 3 5 -5 3 2 1 s G1G1 G2G2

Running Time for (i=1 to V-1) –for each edge (u,v) in E relax(u,v); The above takes (V-1)O(E) = O(VE) time for each edge (u,v) in E –if d[v] > d[u] + w(u,v) return NEGATIVE WEIGHT CYCLE The above takes O(E) time

Proof of Correctness Theorem: If there is a shortest path from s to any node v, then d[v] will have this weight at end of Bellman-Ford algorithm Theorem restated: –Define links[v] to be the minimum number of edges on a shortest path from s to v –After i iterations of the Bellman-Ford for loop, nodes v with links[v] ≤ i will have their d[v] value set correctly Prove this by induction on links[v] Base case: links[v] = 0 which means v is s –d[s] is initialized to 0 which is correct unless there is a negative weight cycle from s to s in which case there is no shortest path from s to s. Induction hypothesis: After k iterations for k ≥ 0, d[v] must be correct if links[v] ≤ k Inductive step: Show after k+1 iterations, d[v] must be correct if links[v] ≤ k+1 –If links[v] ≤ k, then by IH, d[v] is correctly set after kth iteration –If links[v] = k+1, let p = (e 1, e 2, …, e k+1 ) = (v 0, v 1, v 2, …, v k+1 ) be a shortest path from s to v s = v 0, v = v k+1, e i = (v i-1, v i ) –In order for links[v] = k+1, then links[v k ] = k. –By the inductive hypothesis, d[v k ] will be correctly set after the kth iteration. –During the k+1st iteration, we relax all edges including edge e k+1 = (v k,v) –Thus, at end of the k+1 st iteration, d[v] will be correct

Negative weight cycle for each edge (u,v) in E –if d[v] > d[u] + w(u,v) return NEGATIVE WEIGHT CYCLE If no neg weight cycle, d[v] ≤ d[u] + w(u,v) for all (u,v) If there is a negative weight cycle C, for some edge (u,v) on C, it must be the case that d[v] > d[u] + w(u,v). –Suppose this is not true for some neg. weight cycle C –sum these (d[v] ≤ d[u] + w(u,v)) all the way around C –We end up with Σ v in C d[v] ≤ (Σ u in C d[u]) + weight(C) This is impossible unless weight(C) = 0 But weight(C) is negative, so this cannot happen –Thus for some (u,v) on C, d[v] > d[u] + w(u,v)

DAG shortest path algorithm DAG-SP (G, w, s) –Initialize-Single- Source(G,s) –Topologically sort vertices in G –for each vertex u, taken in sorted order for each edge (u,v) in E –relax(u,v); 3 5 -4 3 8 1 s 5 4

Running time Improvement O(V+E) for the topological sorting We only do 1 relaxation for each edge: O(E) time –for each vertex u, taken in sorted order for each edge (u,v) in E –relax(u,v); Overall running time: O(V+E)

Proof of Correctness If there is a shortest path from s to any node v, then d[v] will have this weight at end Let p = (e 1, e 2, …, e k ) = (v 1, v 2, v 3, …, v k+1 ) be a shortest path from s to v –s = v 1, v = v k+1, e i = (v i, v i+1 ) Since we sort edges in topological order, we will process node v i (and edge e i ) before processing later edges in the path.

Dijkstra’s Algorithm Dijkstra (G, w, s) /* Assumption: all edge weights non-negative */ –Initialize-Single- Source(G,s) Completed = {}; ToBeCompleted = V; –While ToBeCompleted is not empty u =EXTRACT- MIN(ToBeCompleted); Completed += {u}; for each edge (u,v) relax(u,v); 15 5 7 4 3 2 1 s

Running Time Analysis While ToBeCompleted is not empty –u =EXTRACT-MIN(ToBeCompleted); –Completed += {u}; –for each edge (u,v) relax(u,v); Each edge relaxed at most once: O(E) –Need to decrease-key potentially once per edge Need to extract-min once per node –The node’s d[v] is then complete

Running Time Analysis cont’d Priority Queue operations –O(E) decrease key operations –O(V) extract-min operations Three implementations of priority queues –Array: O(V 2 ) time decrease-key is O(1) and extract-min is O(V) –Binary heap: O(E log V) time assuming E ≥ V decrease-key and extract-min are O(log V) –Fibonacci heap: O(V log V + E) time decrease-key is O(1) amortized and extract-min is O(log V)

Compare Dijkstra’s algorithm to Prim’s algorithm for MST Dijsktra –Priority Queue operations O(E) decrease key operations O(V) extract-min operations Prim –Priority Queue operations O(E) decrease-key operations O(V) extract-min operations Is this a coincidence or is there something more here?

Proof of Correctness Assume that Dijkstra’s algorithm fails to compute length of all shortest paths from s Let v be the first node whose shortest path length is computed incorrectly Let S be the set of nodes whose shortest paths were computed correctly by Dijkstra prior to adding v to the processed set of nodes. Dijkstra’s algorithm has used the shortest path from s to v using only nodes in S when it added v to S. The shortest path to v must include one node not in S Let u be the first such node. suv Only nodes in S

Proof of Correctness The length of the shortest path to u must be at least that of the length of the path computed to v. –Why? The length of the path from u to v must be < 0. –Why? No path can have negative length since all edge weights are non-negative, and thus we have a contradiction. suv Only nodes in S

Computing paths (not just distance) Maintain for each node v a predecessor node p(v) p(v) is initialized to be null Whenever an edge (u,v) is relaxed such that d(v) improves, then p(v) can be set to be u Paths can be generated from this data structure

All pairs algorithms using single source algorithms Call a single source algorithm from each vertex s in V O(V X) where X is the running time of the given algorithm –Dijkstra linear array: O(V 3 ) –Dijkstra binary heap: O(VE log V) –Dijkstra Fibonacci heap: O(V 2 log V + VE) –Bellman-Ford: O(V 2 E) (negative weight edges)

Two adjacency matrix based algorithms Matrix-multiplication based algorithm –Let L m (i,j) denote the length of the shortest path from node i to node j using at most m edges What is our desired result in terms of L m (i,j)? What is a recurrence relation for L m (i,j)? Floyd-Warshall algorithm –Let L k (i,j) denote the length of the shortest path from node i to node j using only nodes within {1, …, k} as internal nodes. What is our desired result in terms of L k (i,j)? What is a recurrence relation for L k (i,j)?

Conceptual pictures ij Shortest path using at most 2m edges Shortest path using at most m edges k Try all possible nodes k k ij Shortest path using nodes 1 through k Shortest path using nodes 1 through k-1 OR

Running Times Matrix-multiplication based algorithm –O(V 3 log V) log V executions of “matrix-matrix” multiplication –Not quite matrix-matrix multiplication but same running time Floyd-Warshall algorithm –O(V 3 ) V iterations of an O(V 2 ) update loop The constant is very small, so this is a “fast” O(V 3 )

Johnson’s Algorithm Key ideas –Reweight edge weights to eliminate negative weight edges AND preserve shortest paths –Use Bellman-Ford and Dijkstra’s algorithms as subroutines –Running time: O(V 2 log V + VE) Better than earlier algorithms for sparse graphs

Reweighting Original edge weight is w(u,v) New edge weight: –w’(u,v) = w(u,v) + h(u) – h(v) –h(v) is a function mapping vertices to real numbers Key observation: –Let p be any path from node u to node v –w’(p) = w(p) + h(u) – h(v)

Computing vertex weights h(v) Create a new graph G’ = (V’, E’) by –adding a new vertex s to V –adding edges (s,v) for all v in V with w(s,v) = 0 Set h(v) to be the length of the shortest path from this new node s to node v –This is well-defined if G’ does not contain negative weight cycles –Note that h(v) ≤ h(u) + w(u,v) for all (u,v) in E’ –Thus, w’(u,v) = w(u,v) + h(u) – h(v) ≥ 0 3 4 7 -5 3 -7 1

Algorithm implementation Run Bellman-Ford on G’ from new node s –If no negative weight cycle, then use h(v) values from Bellman-Ford –Now compute w’(u,v) for each edge (u,v) in E Now run Dijkstra’s algorithm using w’ –Use each node as source node –Modify d[u,v] at end by adding h(v) and subtracting h(u) to get true path weight Running time: –O(VE) [from one run of Bellman-Ford] + –O(V 2 log V + VE) [from V runs of Dijkstra]

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