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Nattee Niparnan

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Distance of nodes The distance between two nodes is the length of the shortest path between them S A 1 S C 1 S B 2

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DFS and Length DFS finds all nodes reachable from the starting node But it might not be “visited” according to the distance

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Ball and Strings We can compute distance by proceeding from “layer” to “layer”

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Shortest Path Problem (Undi, Unit) Input: A graph, undirected A starting node S Output: A label on every node, giving the distance from S to that node

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Breadth-First-Search Visit node according to its layer procedure bfs(G; s) //Input: Graph G = (V,E), directed or undirected; vertex s V //Output: visit[u] is set to true for all nodes u reachable from v for all u V : visit[u] = false visit[s] = true Queue Q = [s] (queue containing just s) while Q is not empty: u = eject(Q) previsit(u) visit[u] = true; for all edges (u,v) E: if visit[v] = false: visit[v] = true; inject(Q,v); postvisit(u)

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Distance using BFS procedure shortest_bfs(G, s) //Input: Graph G = (V,E), directed or undirected; vertex s V //Output: For all vertices u reachable from s, dist(u) is set to the distance from s to u. for all u V : dist[u] = -1 dist[s] = 0 Queue Q = [s] (queue containing just s) while Q is not empty: u = eject(Q); for all edges (u,v) E: if dist[v] = -1: inject(Q,v); dist[v] = dist[u] + 1; Use dist as visit

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DFS by Stack procedure dfs(G, s) //Input: Graph G = (V,E), directed or undirected; vertex s V //Output: visit[u] is set to true for all nodes u reachable from v for all u V : visit[u] = false visit[s] = true Stack S = [s] (queue containing just s) while S is not empty: u = pop(S) previsit(u) for all edges (u,v) E: if visit [v] = false: push(S,v) visit[v] = true; postvisit(u)

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DFS vs BFS DFS goes depth first Trying to go further if possible Backtrack only when no other possible way to go Using Stack BFS goes breadth first Trying to visit node by the distance from the starting node Using Queue

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Dijkstra’s Algorithm Graph with Length

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Edge with Length Length function l(a,b) = distance from a to b

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Finding Shortest Path BFS can give us the shortest path Just convert the length edge into unit edge However, this is very slow Imagine a case when the length is 1,000,000

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Alarm Clock Analogy No need to walk to every node Since it won’t change anything We skip to the “actual” node Set up the clock at alarm at the target node

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Alarm Clock Algorithm Set an alarm clock for node s at time 0. Repeat until there are no more alarms: Say the next alarm goes off at time T, for node u. Then: The distance from s to u is T. For each neighbor v of u in G: If there is no alarm yet for v, set one for time T + l(u, v). If v's alarm is set for later than T + l(u, v), then reset it to this earlier time.

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Dijkstra’s Algo from BFS procedure dijkstra(G, l, s) //Input: Graph G = (V;E), directed or undirected; vertex s V; positive edge lengths l // Output: For all vertices u reachable from s, dist[u] is set to the distance from s to u. for all u V : dist[u] = + prev(u) = nil dist[s] = 0 H = makequeue(V) (using dist-values as keys) while H is not empty: u = deletemin(H) for all edges (u; v) E: if dist[v] > dist[u] + l(u, v): dist[v] = dist[u] + l(u, v) prev[v] = u decreasekey(H, v)

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Another Implementation of Dijkstra’s Growing from Known Region of shortest path Given a graph and a starting node s What if we know a shortest path from s to some subset S’ V? Divide and Conquer Approach?

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Dijktra’s Algo #2 procedure dijkstra(G, l, s) //Input: Graph G = (V;E), directed or undirected; vertex s V; positive edge lengths l // Output: For all vertices u reachable from s, dist[u] is set to the distance from s to u. for all u V : dist[u] = + prev(u) = nil dist[s] = 0 R = {} // (the “known region”) while R ≠ V : Pick the node v R with smallest dist[] Add v to R for all edges (v,z) E: if dist[z] > dist[v] + l(v,z): dist[z] = dist[v] + l(v,z)

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Analysis There are |V| ExtractMin Need to check all edges At most |E|, if we use adjacency list Maybe |V 2 |, if we use adjacency matrix Value of dist[] might be changed Depends on underlying data structure

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Choice of DS Using simple array Each ExtractMin uses O(V) Each change of dist[] uses O(1) Result = O(V 2 + E) = O(V 2 ) Using binary heap Each ExtractMin uses O(lg V) Each change of dist[] uses O(lg V) Result = O( (V + E) lg V) Can be O (V 2 lg V) Might be V 2 Good when the graph is sparse

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Fibonacci Heap Using simple array Each ExtractMin uses O( lg V) (amortized) Each change of dist[] uses O(1) (amortized) Result = O(V lg V + E)

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Graph with Negative Edge Disjktra’s works because a shortest path to v must pass throught a node closer than v Shortest path to A pass through B which is… in BFS sense… is further than A

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Negative Cycle A graph with a negative cycle has no shortest path The shortest.. makes no sense.. Hence, negative edge must be a directed

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Key Idea in Shortest Path Update the distance if dist[z] > dist[v] + l(v,z): dist[z] = dist[v] + l(v,z) This is safe to perform now, a shortest path must has at most |V| - 1 edges

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Bellman-Ford Algorithm procedure BellmanFord(G, l, s) //Input: Graph G = (V;E), directed; vertex s V; edge lengths l (may be negative), no negative cycle // Output: For all vertices u reachable from s, dist[u] is set to the distance from s to u. for all u V : dist[u] = + prev(u) = nil dist[s] = 0 repeat |V| - 1 times: for all edges (a,b) E: if dist[b] > dist[a] + l(a,b): dist[b] = dist[a] + l(a,b)

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Shortest Path in DAG Path in DAG appears in linearized order procedure dag-shortest-path(G, l, s) //Input: DAG G = (V;E), vertex s V; edge lengths l (may be negative) // Output: For all vertices u reachable from s, dist[u] is set to the distance from s to u. for all u V : dist[u] = + prev(u) = nil dist[s] = 0 Linearize G For each u V, in linearized order: for all edges (u,v) E: if dist[v] > dist[u] + l(u,v): dist[v] = dist[u] + l(y,v)

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