1. 1 Graphs - Definitions - Breadth First Search - Depth First Search - Connectivity - Topological Ordering

2. 3.1 Basic Definitions and Applications

3. 3 Undirected Graphs Undirected graph. G = (V, E) n V = nodes. n E = edges between pairs of nodes. n Captures pairwise relationship between objects. n Graph size parameters: n = |V|, m = |E|. V = { 1, 2, 3, 4, 5, 6, 7, 8 } E = { 1-2, 1-3, 2-3, 2-4, 2-5, 3-5, 3-7, 3-8, 4-5, 5-6 } n = 8 m = 11

4. 4 Some Graph Applications transportation Graph street intersections NodesEdges highways communicationcomputersfiber optic cables World Wide Webweb pageshyperlinks socialpeoplerelationships food webspeciespredator-prey software systemsfunctionsfunction calls schedulingtasksprecedence constraints circuitsgateswires

5. 5 World Wide Web Web graph. n Node: web page. n Edge: hyperlink from one page to another. cnn.com cnnsi.com novell.comnetscape.com timewarner.com hbo.com sorpranos.com

6. 6 Ecological Food Web Food web graph. n Node = species. n Edge = from prey to predator. Reference: http://www.twingroves.district96.k12.il.us/Wetlands/Salamander/SalGraphics/salfoodweb.giff

7. 7 Graph Representation: Adjacency Matrix Adjacency matrix. n-by-n matrix with A uv = 1 if (u, v) is an edge. n Two representations of each edge. n Space proportional to n 2. n Checking if (u, v) is an edge takes  (1) time. n Identifying all edges takes  (n 2 ) time. 1 2 3 4 5 6 7 8 1 0 1 1 0 0 0 0 0 2 1 0 1 1 1 0 0 0 3 1 1 0 0 1 0 1 1 4 0 1 0 1 1 0 0 0 5 0 1 1 1 0 1 0 0 6 0 0 0 0 1 0 0 0 7 0 0 1 0 0 0 0 1 8 0 0 1 0 0 0 1 0

8. 8 Graph Representation: Adjacency List Adjacency list. Node indexed array of lists. n Two representations of each edge. n Space proportional to m + n. n Checking if (u, v) is an edge takes O(deg(u)) time. n Identifying all edges takes  (m + n) time. 1 23 2 3 4 25 5 6 7 38 8 1345 125 8 7 2346 5 degree = number of neighbors of u 37

9. 9 Paths and Connectivity Def. A path in an undirected graph G = (V, E) is a sequence P of nodes v 1, v 2, …, v k-1, v k with the property that each consecutive pair v i, v i+1 is joined by an edge in E. Def. A path is simple if all nodes are distinct. Def. An undirected graph is connected if for every pair of nodes u and v, there is a path between u and v.

10. 10 Cycles Def. A cycle is a path v 1, v 2, …, v k-1, v k in which v 1 = v k, k > 2, and the first k-1 nodes are all distinct. cycle C = 1-2-4-5-3-1

11. 11 Trees Def. An undirected graph is a tree if it is connected and does not contain a cycle. Theorem. Let G be an undirected graph on n nodes. Any two of the following statements imply the third. n G is connected. n G does not contain a cycle. n G has n-1 edges.

12. 12 Rooted Trees Rooted tree. Given a tree T, choose a root node r and orient each edge away from r. Importance. Models hierarchical structure. a tree the same tree, rooted at 1 v parent of v child of v root r

13. 13 Phylogeny Trees Phylogeny trees. Describe evolutionary history of species.

14. 14 GUI Containment Hierarchy Reference: http://java.sun.com/docs/books/tutorial/uiswing/overview/anatomy.html GUI containment hierarchy. Describe organization of GUI widgets.

15. 3.2 Graph Traversal

16. 16 Connectivity s-t connectivity problem. Given two node s and t, is there a path between s and t? s-t shortest path problem. Given two node s and t, what is the length of the shortest path between s and t? Applications. n Friendster. n Maze traversal. n Kevin Bacon number. n Fewest number of hops in a communication network.

17. 17 Breadth First Search BFS intuition. Explore outward from s in all possible directions, adding nodes one "layer" at a time. BFS algorithm. n L 0 = { s }. n L 1 = all neighbors of L 0. n L 2 = all nodes that do not belong to L 0 or L 1, and that have an edge to a node in L 1. n L i+1 = all nodes that do not belong to an earlier layer, and that have an edge to a node in L i. Theorem. For each i, L i consists of all nodes at distance exactly i from s. There is a path from s to t iff t appears in some layer. s L1L1 L2L2 L n-1

18. 18 Breadth First Search Algorithm

19. 19 Breadth-First Search - Idea Given a graph G = (V, E), start at the source vertex “s” and discover which vertices are reachable from s n At any time there is a “frontier” of vertices that have been discovered, but not yet processed Next pick the nodes in the frontier in sequence and discover their neighbors, forming a new “frontier” n Breadth-first search is so named because it visits vertices across the entire breadth of this frontier before moving on s s s 1 1 1 2 2 2 2 2 2

20. 20 BFS - Continued Represent the final result as follows: For each vertex v  V, we will store d[v] which is the distance (length of shortest path) from s to v – Distance between a vertex “v” and “s” is defined to be the minimum number of edges on a path from “s” to “v” – Note that d[s] = 0 We will also store a predecessor (or parent) pointer pred[v] or  v], which indicates the first vertex along the shortest path if we walk from v backwards to s – We will let prev[s] or  s] = 0 – Notice that these predecessor pointers are sufficient to reconstruct the shortest path to any vertex

21. 21 BFS – Implementation Initially all vertices (except the source) is colored white, meaning they have not been discovered just yet When a vertex is first discovered, it is colored gray (and is part of the frontier) When a gray vertex is processed, it becomes black s s s 1 1 1 2 2 2 2 2 2

22. 22 BFS - Implementation The search makes use of a FIFO queue, Q We also maintain arrays n color[u], which holds the color of vertex u – either white, gray, black n pred[u], which points to the predecessor of u – The vertex that discovered u n d[u], the distance from s to u s s s 1 1 1 2 2 2 2 2 2

23. 23 BFS –Another way of Implemenattion BFS(G, s){ for each u in V- {s} { // Initialization color[u] = white; d[u] = INFINITY; pred[u] = NULL; } //end-for color[s] = GRAY; // initialize source s d[s] = 0; pred[s] = NULL; Q = {s}; // Put s in the queue while (Q is nonempty){ u = Dequeue(Q); // u is the next vertex to visit for each v in Adj[u] { if (color[v] == white){ // if neighbor v undiscovered color[v] = gray; // … mark is discovered d[v] = d[u] + 1; // … set its distance pred[v] = u; // … set its predecessor Enqueue(v); //… put it in the queue } //end-if } //end-for color[u] = black; // we are done with u } //end-while } //end-BFS

24. 24 BFS - Analysis Let n = |V| and e = |E| Observe that the initial portion requires  (n) The real meat is through the traversal loop Since we never visit a vertex twice, the number of times we go through the loop os at most n (exactly n assuming each vertex is reachable from the source) The number of iterations through the inner loop is proportional to deg(u) Summing up over all vertices we have T(n) = n + Sum_{v  V} deg[v] = n + 2e =  (n+e)

25. 25 Breadth First Search Property. Let T be a BFS tree of G = (V, E), and let (x, y) be an edge of G. Then the level of x and y differ by at most 1. L0L0 L1L1 L2L2 L3L3

26. 26 Breadth First Search: Analysis Theorem. The above implementation of BFS runs in O(m + n) time if the graph is given by its adjacency representation. Pf. n Easy to prove O(n 2 ) running time: – at most n lists L[i] – each node occurs on at most one list; for loop runs  n times – when we consider node u, there are  n incident edges (u, v), and we spend O(1) processing each edge n Actually runs in O(m + n) time: – when we consider node u, there are deg(u) incident edges (u, v) – total time processing edges is  u  V deg(u) = 2m ▪ each edge (u, v) is counted exactly twice in sum: once in deg(u) and once in deg(v)

27. 27

28. 28 Depth First Search : The Concept Consider searching the way out from a maze. n As you enter a location of the maze, paint some graffiti on the wall to remind yourself that you were there n Successively travel from room to room as long as you come to a place you have not already been n When you return to the same room, try a different door (assuming it goes somewhere you have not been before) n When all doors have been tried in a room, backtrack n Same idea in trying to find a door out of a puzzle

29. 29 DFS Algorithm Recursive Version 1

30. 30

31. 31 DFS Stack Version 2

32. 32 DFS – Version 3 (CLR Book) Assume you are given a digraph G = (V, E) n The same algorithm works for undirected graphs but the resulting structure imposed on the graph is different We use 4 auxiliary arrays n color[u] – White – undiscovered – Gray – discovered but not yet processed – Black – finished processing n pred[u], which points to the predecessor of u – The vertex that discovered u n 2 timestamps: Purpose will be explained later – d[u]: Time at which the vertex was discovered  Not to be confused with distance of u in BFS! – f[u]: Time at which the processing of the vertex was finished

33. 33 DFS – Version 3 DFS(G, s){ for each u in V { // Initialization color[u] = white; pred[u] = NULL; } //end-for time = 0; for each u in V if (color[u] == white) // Found an undiscovered vertex DFSVisit(u); // Start a new search there } // end-DFS DFSVisit(u){ // Start a new search at u color[u] = gray; // Mark u visited d[u] = ++time; for each v in Adj[u] { if (color[v] == white){ // if neighbor v undiscovered pred[v] = u; // … set its predecessor DFSVisit(v); // …visit v } //end-if } //end-for color[u] = black; // we are done with u f[u] = ++time; } //end-while } //end-DFSVisit

34. 34 DFS - Example d e f a b c g 1/10 a 2/5 6/9 12/13 11/14 3/4 7/8 b c f g C F B d e C C DFS imposes a tree structure (actually a collection of trees or a forest) on the structure of the graph This is just the recursion tree, where the edge (u, v) arises when processing vertex “u” we call DFSVisit(v) for some neighbor v

35. 35 Connected Component Connected component. Find all nodes reachable from s. Connected component containing node 1 = { 1, 2, 3, 4, 5, 6, 7, 8 }.

36. 36 Flood Fill Flood fill. Given lime green pixel in an image, change color of entire blob of neighboring lime pixels to blue. n Node: pixel. n Edge: two neighboring lime pixels. n Blob: connected component of lime pixels. recolor lime green blob to blue

37. 37 Flood Fill Flood fill. Given lime green pixel in an image, change color of entire blob of neighboring lime pixels to blue. n Node: pixel. n Edge: two neighboring lime pixels. n Blob: connected component of lime pixels. recolor lime green blob to blue

38. 38 Connected Component Connected component. Find all nodes reachable from s. Theorem. Upon termination, R is the connected component containing s. n BFS = explore in order of distance from s. n DFS = explore in a different way. s uv R it's safe to add v

39. 3.4 Testing Bipartiteness

40. 40 Bipartite Graphs Def. An undirected graph G = (V, E) is bipartite if the nodes can be colored red or blue such that every edge has one red and one blue end. Applications. n Stable marriage: men = red, women = blue. n Scheduling: machines = red, jobs = blue. a bipartite graph

41. 41 Testing Bipartiteness Testing bipartiteness. Given a graph G, is it bipartite? n Many graph problems become: – easier if the underlying graph is bipartite (matching) – tractable if the underlying graph is bipartite (independent set) n Before attempting to design an algorithm, we need to understand structure of bipartite graphs. v1v1 v2v2 v3v3 v6v6 v5v5 v4v4 v7v7 v2v2 v4v4 v5v5 v7v7 v1v1 v3v3 v6v6 a bipartite graph Ganother drawing of G

42. 42 An Obstruction to Bipartiteness Lemma. If a graph G is bipartite, it cannot contain an odd length cycle. Pf. Not possible to 2-color the odd cycle, let alone G. bipartite (2-colorable) not bipartite (not 2-colorable)

43. 43 Bipartite Graphs Lemma. Let G be a connected graph, and let L 0, …, L k be the layers produced by BFS starting at node s. Exactly one of the following holds. (i) No edge of G joins two nodes of the same layer, and G is bipartite. (ii) An edge of G joins two nodes of the same layer, and G contains an odd-length cycle (and hence is not bipartite). Case (i) L1L1 L2L2 L3L3 Case (ii) L1L1 L2L2 L3L3

44. 44 Bipartite Graphs Lemma. Let G be a connected graph, and let L 0, …, L k be the layers produced by BFS starting at node s. Exactly one of the following holds. (i) No edge of G joins two nodes of the same layer, and G is bipartite. (ii) An edge of G joins two nodes of the same layer, and G contains an odd-length cycle (and hence is not bipartite). Pf. (i) n Suppose no edge joins two nodes in the same layer. n By previous lemma, this implies all edges join nodes on same level. n Bipartition: red = nodes on odd levels, blue = nodes on even levels. Case (i) L1L1 L2L2 L3L3

45. 45 Bipartite Graphs Lemma. Let G be a connected graph, and let L 0, …, L k be the layers produced by BFS starting at node s. Exactly one of the following holds. (i) No edge of G joins two nodes of the same layer, and G is bipartite. (ii) An edge of G joins two nodes of the same layer, and G contains an odd-length cycle (and hence is not bipartite). Pf. (ii) n Suppose (x, y) is an edge with x, y in same level L j. n Let z = lca(x, y) = lowest common ancestor. n Let L i be level containing z. n Consider cycle that takes edge from x to y, then path from y to z, then path from z to x. n Its length is 1 + (j-i) + (j-i), which is odd. ▪ z = lca(x, y) (x, y)path from y to z path from z to x

46. 46 Obstruction to Bipartiteness Corollary. A graph G is bipartite iff it contain no odd length cycle. 5-cycle C bipartite (2-colorable) not bipartite (not 2-colorable)

47. 3.5 Connectivity in Directed Graphs

48. 48 Directed Graphs Directed graph. G = (V, E) n Edge (u, v) goes from node u to node v. Ex. Web graph - hyperlink points from one web page to another. n Directedness of graph is crucial. n Modern web search engines exploit hyperlink structure to rank web pages by importance.

49. 49 Graph Search Directed reachability. Given a node s, find all nodes reachable from s. Directed s-t shortest path problem. Given two node s and t, what is the length of the shortest path between s and t? Graph search. BFS extends naturally to directed graphs. Web crawler. Start from web page s. Find all web pages linked from s, either directly or indirectly.

50. 50 Strong Connectivity Def. Node u and v are mutually reachable if there is a path from u to v and also a path from v to u. Def. A graph is strongly connected if every pair of nodes is mutually reachable. Lemma. Let s be any node. G is strongly connected iff every node is reachable from s, and s is reachable from every node. Pf.  Follows from definition. Pf.  Path from u to v: concatenate u-s path with s-v path. Path from v to u: concatenate v-s path with s-u path. ▪ s v u ok if paths overlap

51. 51 Strong Connectivity: Algorithm Theorem. Can determine if G is strongly connected in O(m + n) time. Pf. n Pick any node s. n Run BFS from s in G. n Run BFS from s in G rev. n Return true iff all nodes reached in both BFS executions. n Correctness follows immediately from previous lemma. ▪ reverse orientation of every edge in G strongly connected not strongly connected

52. 3.6 DAGs and Topological Ordering

53. 53 Directed Acyclic Graphs Def. An DAG is a directed graph that contains no directed cycles. Ex. Precedence constraints: edge (v i, v j ) means v i must precede v j. Def. A topological order of a directed graph G = (V, E) is an ordering of its nodes as v 1, v 2, …, v n so that for every edge (v i, v j ) we have i < j. a DAG a topological ordering v2v2 v3v3 v6v6 v5v5 v4v4 v7v7 v1v1 v1v1 v2v2 v3v3 v4v4 v5v5 v6v6 v7v7

54. 54 Precedence Constraints Precedence constraints. Edge (v i, v j ) means task v i must occur before v j. Applications. n Course prerequisite graph: course v i must be taken before v j. n Compilation: module v i must be compiled before v j. Pipeline of computing jobs: output of job v i needed to determine input of job v j.

55. 55 Directed Acyclic Graphs Lemma. If G has a topological order, then G is a DAG. Pf. (by contradiction) n Suppose that G has a topological order v 1, …, v n and that G also has a directed cycle C. Let's see what happens. n Let v i be the lowest-indexed node in C, and let v j be the node just before v i ; thus (v j, v i ) is an edge. n By our choice of i, we have i < j. n On the other hand, since (v j, v i ) is an edge and v 1, …, v n is a topological order, we must have j < i, a contradiction. ▪ v1v1 vivi vjvj vnvn the supposed topological order: v 1, …, v n the directed cycle C

56. 56 Directed Acyclic Graphs Lemma. If G has a topological order, then G is a DAG. Q. Does every DAG have a topological ordering? Q. If so, how do we compute one?

57. 57 Directed Acyclic Graphs Lemma. If G is a DAG, then G has a node with no incoming edges. Pf. (by contradiction) n Suppose that G is a DAG and every node has at least one incoming edge. Let's see what happens. n Pick any node v, and begin following edges backward from v. Since v has at least one incoming edge (u, v) we can walk backward to u. n Then, since u has at least one incoming edge (x, u), we can walk backward to x. n Repeat until we visit a node, say w, twice. n Let C denote the sequence of nodes encountered between successive visits to w. C is a cycle. ▪ wxuv

58. 58 Directed Acyclic Graphs Lemma. If G is a DAG, then G has a topological ordering. Pf. (by induction on n) n Base case: true if n = 1. n Given DAG on n > 1 nodes, find a node v with no incoming edges. n G - { v } is a DAG, since deleting v cannot create cycles. n By inductive hypothesis, G - { v } has a topological ordering. n Place v first in topological ordering; then append nodes of G - { v } n in topological order. This is valid since v has no incoming edges. ▪ DAG v

59. 59 Topological Sorting Algorithm: Running Time Theorem. Algorithm finds a topological order in O(m + n) time. Pf. n Maintain the following information: – count[w] = remaining number of incoming edges – S = set of remaining nodes with no incoming edges n Initialization: O(m + n) via single scan through graph. n Update: to delete v – remove v from S – decrement count[w] for all edges from v to w, and add w to S if c count[w] hits 0 – this is O(1) per edge ▪

60. 60 Topological Ordering Example

61. 61 v1v1 Topological Ordering Algorithm: Example Topological order: v2v2 v3v3 v6v6 v5v5 v4v4 v7v7 v1v1

62. 62 v2v2 Topological Ordering Algorithm: Example Topological order: v 1 v2v2 v3v3 v6v6 v5v5 v4v4 v7v7

63. 63 v3v3 Topological Ordering Algorithm: Example Topological order: v 1, v 2 v3v3 v6v6 v5v5 v4v4 v7v7

64. 64 v4v4 Topological Ordering Algorithm: Example Topological order: v 1, v 2, v 3 v6v6 v5v5 v4v4 v7v7

65. 65 v5v5 Topological Ordering Algorithm: Example Topological order: v 1, v 2, v 3, v 4 v6v6 v5v5 v7v7

66. 66 v6v6 Topological Ordering Algorithm: Example Topological order: v 1, v 2, v 3, v 4, v 5 v6v6 v7v7

67. 67 v7v7 Topological Ordering Algorithm: Example Topological order: v 1, v 2, v 3, v 4, v 5, v 6 v7v7

68. 68 Topological Ordering Algorithm: Example Topological order: v 1, v 2, v 3, v 4, v 5, v 6, v 7. v2v2 v3v3 v6v6 v5v5 v4v4 v7v7 v1v1 v1v1 v2v2 v3v3 v4v4 v5v5 v6v6 v7v7