Graps 2
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The document discusses breadth-first search (BFS) algorithms for graphs. It explains that BFS runs in O(n^2) time on adjacency matrices due to checking all elements in each row, but in O(n+m) time on adjacency lists where m is the number of edges. It then describes how to modify BFS to record the shortest path between nodes using a predecessor array. The rest of the document provides examples of how BFS works and applications such as finding connected components and directed acyclic graphs.
![Shortest Path RecordingBFS we saw only tells us whether a path exists from source s, to other vertices v.It doesn’t tell us the path!We need to modify the algorithm to record the path.How can we do that?Note: we do not know which vertices lie on this path until we reach v!Efficient solution:Use an additional array pred[0..n-1]Pred[w] = v means that vertex w was visited from v](https://image.slidesharecdn.com/graps2-110225222926-phpapp01/85/Graps-2-4-320.jpg)
![BFS + Path Finding initialize all pred[v] to -1Record where you came from](https://image.slidesharecdn.com/graps2-110225222926-phpapp01/85/Graps-2-5-320.jpg)
![OutdegreeAll of the arcs going “out” from vSimple to computeScan through list Adj[v] and count the arcsWhat is the total outdegree? (m=#edges)](https://image.slidesharecdn.com/graps2-110225222926-phpapp01/85/Graps-2-37-320.jpg)
![IndegreeAll of the arcs coming “in” to vNot as simple to compute as outdegreeFirst, initialize indegree[v]=0 for each vertex vScan through adj[v] list for each vFor each vertex w seen, indegree[w]++;Running time: O(n+m)What is the total indegree?](https://image.slidesharecdn.com/graps2-110225222926-phpapp01/85/Graps-2-38-320.jpg)
Graps 2
- 2. Graph
- 3. Time Complexity of BFS(Using adjacency matrix)Assume adjacency listn = number of vertices m = number of edgesO(n2)Finding the adjacent vertices of v requires checking all elements in the row. This takes linear time O(n).Summing over all the n iterations, the total running time is O(n2).So, with adjacency matrix, BFS is O(n2) independent of number of edges m. With adjacent lists, BFS is O(n+m); if m=O(n2) like a dense graph, O(n+m)=O(n2).
- 4. Shortest Path RecordingBFS we saw only tells us whether a path exists from source s, to other vertices v.It doesn’t tell us the path!We need to modify the algorithm to record the path.How can we do that?Note: we do not know which vertices lie on this path until we reach v!Efficient solution:Use an additional array pred[0..n-1]Pred[w] = v means that vertex w was visited from v
- 5. BFS + Path Finding initialize all pred[v] to -1Record where you came from
- 6. 0829173645ExampleAdjacency ListVisited Table (T/F)sourcePredInitialize visitedtable (all False)Initialize Pred to -1{ }Q = Initialize Q to be empty
- 7. 0829173645ExampleAdjacency ListVisited Table (T/F)sourcePredFlag that 2 has been visited.{ 2 }Q = Place source 2 on the queue.
- 8. 0829173645ExampleAdjacency ListVisited Table (T/F)NeighborssourcePredMark neighborsas visited.Record in Predthat we came from 2.{2} -> { 8, 1, 4 }Q = Dequeue 2. Place all unvisited neighbors of 2 on the queue
- 9. 0829173645ExampleAdjacency ListVisited Table (T/F)sourceNeighborsPredMark new visitedNeighbors.Record in Predthat we came from 8.{ 8, 1, 4 } -> { 1, 4, 0, 9 } Q = Dequeue 8. -- Place all unvisited neighbors of 8 on the queue. -- Notice that 2 is not placed on the queue again, it has been visited!
- 10. 0829173645ExampleAdjacency ListVisited Table (T/F)NeighborssourcePredMark new visitedNeighbors.Record in Predthat we came from 1.{ 1, 4, 0, 9 } -> { 4, 0, 9, 3, 7 } Q = Dequeue 1. -- Place all unvisited neighbors of 1 on the queue. -- Only nodes 3 and 7 haven’t been visited yet.
- 11. 0829173645ExampleAdjacency ListVisited Table (T/F)NeighborssourcePred{ 4, 0, 9, 3, 7 } -> { 0, 9, 3, 7 } Q = Dequeue 4. -- 4 has no unvisited neighbors!
- 12. 0829173645ExampleAdjacency ListVisited Table (T/F)NeighborssourcePred{ 0, 9, 3, 7 } -> { 9, 3, 7 } Q = Dequeue 0. -- 0 has no unvisited neighbors!
- 13. 0829173645ExampleAdjacency ListVisited Table (T/F)sourceNeighborsPred{ 9, 3, 7 } -> { 3, 7 } Q = Dequeue 9. -- 9 has no unvisited neighbors!
- 14. 0829173645ExampleAdjacency ListVisited Table (T/F)NeighborssourcePredMark new visitedVertex 5.Record in Predthat we came from 3.{ 3, 7 } -> { 7, 5 } Q = Dequeue 3. -- place neighbor 5 on the queue.
- 15. 0829173645ExampleAdjacency ListVisited Table (T/F)sourceNeighborsPredMark new visitedVertex 6.Record in Predthat we came from 7.{ 7, 5 } -> { 5, 6 } Q = Dequeue 7. -- place neighbor 6 on the queue.
- 16. 0829173645ExampleAdjacency ListVisited Table (T/F)sourceNeighborsPred{ 5, 6} -> { 6 } Q = Dequeue 5. -- no unvisited neighbors of 5.
- 17. 0829173645ExampleAdjacency ListVisited Table (T/F)sourceNeighborsPred{ 6 } -> { } Q = Dequeue 6. -- no unvisited neighbors of 6.
- 18. 0829173645ExampleAdjacency ListVisited Table (T/F)sourcePredPred now can be traced backwardto report the path!{ } STOP!!! Q is empty!!!Q =
- 19. Path reportingnodesvisited fromTry some examples, report path from s to v:Path(0) ->Path(6) ->Path(1) ->The path returned is the shortest from s to v (minimum number of edges).
- 20. BFS treeThe paths found by BFS is often drawn as a rooted tree (called BFS tree), with the starting vertex as the root of the tree.BFS tree for vertex s=2.Question: What would a “level” order traversal tell you?
- 21. How do we record the shortest distances?d(v) = ;d(s) = 0;d(w)=d(v)+1;
- 22. Application of BFS One application concerns how to find connected components in a graph If a graph has more than one connected components, BFS builds a BFS-forest (not just BFS-tree)!Each tree in the forest is a connected component.
- 24. Application 1: ConnectivityG =PQNLROMsDECAHow do we tell if two vertices are connected?FBKGA connected to F?A connected to L?H
- 25. ConnectivityA graph isconnectedif and only if there exists a path between every pair of distinct vertices.A graph is connected if and only if there exists a simple path between every pair of distinct vertices (since every non-simple path contains a cycle, which can be bypassed)How to check for connectivity?Run BFS or DFS (using an arbitrary vertex as the source)If all vertices have been visited, the graph is connected.Running time? O(n + m)
- 27. Subgraphs
- 28. Connected ComponentsFormally stated:A connected component is a maximal connected subgraph of a graph.The set of connected components is unique for a given graph.
- 29. Finding Connected ComponentsFor each vertexIf not visitedCall DFSThis will findall verticesconnectedto “v” => oneconnected componentBasic DFS algorithm.
- 30. Time ComplexityRunning time for each i connected componentQuestion:Can two connected components have the same edge?Can two connected components have the same vertex?So:
- 31. TreesTree arises in many computer science applicationsA graph G is a tree if and only if it is connected and acyclic (Acyclic means it does not contain any simple cycles)The following statements are equivalentG is a treeG is acyclic and has exactly n-1 edgesG is connected and has exactly n-1 edges
- 32. Tree as a (directed) GraphIs it a graph?Does it contain cycles? (in other words, is it acyclic)How many vertices?How many edges?15618833016
- 33. Directed GraphA graph is directed if direction is assigned to each edge. We call the directed edges arcs.An edge is denoted as an ordered pair (u, v) Recall: for an undirected graphAn edge is denoted {u,v}, which actually corresponds to two arcs (u,v) and (v,u)
- 34. RepresentationsThe adjacency matrix and adjacency list can be used
- 35. Directed Acyclic GraphA directed path is a sequence of vertices (v0, v1, . . . , vk)Such that (vi, vi+1) is an arcA directed cycleis a directed path such that the first and last vertices are the same.A directed graph is acyclic if it does not contain any directed cycles
- 36. Indegree and Outdegree Since the edges are directedWe can’t simply talk about Deg(v)Instead, we need to consider the arcs coming “in” and going “out”Thus, we define termsIndegree(v)Outdegree(v)
- 37. OutdegreeAll of the arcs going “out” from vSimple to computeScan through list Adj[v] and count the arcsWhat is the total outdegree? (m=#edges)
- 38. IndegreeAll of the arcs coming “in” to vNot as simple to compute as outdegreeFirst, initialize indegree[v]=0 for each vertex vScan through adj[v] list for each vFor each vertex w seen, indegree[w]++;Running time: O(n+m)What is the total indegree?
- 39. Indegree + OutdegreeEach arc (u,v) contributes count 1 to the outdegree of u and count 1 to the indegree of v.
- 40. Example3680729154Indeg(2)?Indeg(8)?Outdeg(0)?Num of Edges?Total OutDeg?Total Indeg?
- 41. Directed Graphs UsageDirected graphs are often used to represent order-dependent tasksThat is we cannot start a task before another task finishesWe can model this task dependent constraint using arcsAn arc (i,j) means task j cannot start until task i is finishedClearly, for the system not to hang, the graph must be acyclic. jiTask j cannot start until task i is finished
- 42. University ExampleCS departments course structure Any directed cycles?104180171151221342252211251271M132M111201231272361381303343341327334336362332How many indeg(171)?How many outdeg(171)?