CS781 Lecture 2 January 13, Graph Traversals, Search, and Ordering

Transcription

1 CS781 Lecture 2 January 13, 2010 Graph Traversals, Search, and Ordering

2 Review of Lecture 1 Notions of Algorithm Scalability Worst-Case and Average-Case Analysis Asymptotic Growth Rates: Big-Oh Prototypical Example Algorithms: linear, nlogn, quadrayic, cubic, O(n^k) Graph Representations Sample Graph Applications Today: Graph Traversals visit all nodes in specified order 2

3 Graph Traversal via Breadth First Search BFS intuition. Explore outward from s in all possible directions, adding nodes one "level" at a time. s L 1 L 2 L n-1 BFS algorithm. L 0 = { s }. L 1 = all neighbors of L 0. 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. L i+1 = all nodes that do not belong to an earlier level, and that have an edge to a node in L i. Theorem. The ith level 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 level. 3

4 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. If y is first visited via x in BFS then (x,y) belongs to T. L 0 L 1 L 2 L 3 4

5 Breadth First Search Listing of level graphs: [{1: set([2, 3])}, {2: set([4, 5]), 3: set([8, 5, 7])}, {8: set([]), 4: set([]), 5: set([6]), 7: set([])}, {6: set([])}] L 0 L 1 L 2 L 3 5

6 def BreadthFirstLevels(G,root): visited = set() currentlevel = [root] while currentlevel!= []: for v in currentlevel: visited.add(v) nextlevel = set() layergraph = dict([(v,set()) for v in currentlevel]) for v in currentlevel: for w in G[v]: if w not in visited: layergraph[v].add(w); nextlevel.add(w) yield layergraph currentlevel = nextlevel Breadth First Search: Implementation The code generates a sequence of layer graphs, which is useful for processing graphs one level at a time. Note the use of the yield statement in python. G={ 1:[2,3], 2:[1,3,4,5], 3:[1,2,5,7,8], 4:[2,5], 5:[2,3,4,6],6:[5],7:[3],8:[3]} print [levelgraph for levelgraph in BreadthFirstLevels(G, 1)] 6

7 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. Easy to prove O(m n) running time: at most n level lists and n-1 layer graphs each node occurs on at most one level list; so while loop runs n times Each layer contains at most m edges, and we spend O(1) processing each edge 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 Σ u V deg(u) = 2m 7

8 Connected Component Connected component. Find all nodes reachable from s. Connected component containing node1 = { 1, 2, 3, 4, 5, 6, 7, 8 }. 8

9 Application of BFS: Flood Fill Flood fill. Given a selected pixel in an image, change color of entire blob of like-colored pixels to different color. Node: pixel. Edge: two neighboring pixels of same color. Blob: connected component of like-colored pixels. recolor lime green blob to blue 9

10 Connected Components Connected component. Find all nodes reachable from s. s R u v it's safe to add v Theorem. Upon termination, R is the connected component containing s. Two efficient O(m + n) algorithms: BFS = explore in order of distance from s DFS = explore in a different way, moving faster away from s towards the boundary of connected component. 10

11 Depth First Search (Recursive Algorithm) def dfs(g,parent,visited): visited.add(parent) for child in G[parent]: if child not in visited: print parent, child, 'forward' dfs(g,child,visited) visited = set() dfs(g,1,visited) G={ 1:[2,3], 2:[1,3,4,5], 3:[1,2,5,7,8], 4: [2,5], 5:[2,3,4,6], 6:[5], 7:[3,8], 8:[3,7]} >>> 1 2 forward 2 3 forward 3 5 forward 5 4 forward 5 6 forward 3 7 forward 7 8 forward >>> 11

12 Depth First Search (Iterative Implementation) def dfsearch(g,root): visited = set() visited.add(root) stack = [(root,iter(g[root]))] while stack: parent,children = stack[-1] try: child = children.next() if child in visited: yield parent,child,'nontree' else: yield parent,child,'forward' visited.add(child) stack.append((child,iter(g[child]))) except StopIteration: stack.pop() if stack: yield parent, stack[-1][0],'reverse' >>>for move in dfsearch(g,1): print move (1, 2, 'forward') (2, 1, 'nontree') (2, 3, 'forward') (3, 1, 'nontree') (3, 2, 'nontree') (3, 5, 'forward') (5, 2, 'nontree') (5, 3, 'nontree') (5, 4, 'forward') (4, 2, 'nontree') (4, 5, 'nontree') (4, 5, 'reverse') (5, 6, 'forward') (6, 5, 'nontree') (6, 5, 'reverse') (5, 3, 'reverse') (3, 7, 'forward') (7, 3, 'nontree') (7, 8, 'forward') (8, 3, 'nontree') (8, 7, 'nontree') (8, 7, 'reverse') (7, 3, 'reverse') (3, 8, 'nontree') (3, 2, 'reverse') (2, 4, 'nontree') (2, 5, 'nontree') (2, 1, 'reverse') (1, 3, 'nontree') 12

13 Testing Bipartiteness

14 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. Structure of each Layer in BFS Matching buyers and sellers Assigning clients and servers Scheduling machines and jobs a bipartite graph 14

15 Testing Bipartiteness Testing bipartiteness. Given a graph G, is it bipartite? Many graph problems become: easier if the underlying graph is bipartite (matching) tractable if the underlying graph is bipartite (independent set) Before attempting to design an algorithm, we need to understand structure of bipartite graphs. v 2 v 3 v 2 v 4 v 1 v 6 v 5 v 4 v 5 v 3 v 6 v 7 v 1 v 7 a bipartite graph G another drawing of G 15

16 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) 16

17 Bipartite Graphs Lemma. Let G be a connected graph, and let L 0,, L k be the levels produced by BFS starting at node s. Exactly one of the following holds. (i) No edge of G joins two nodes of the same levels, 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). L 1 L 2 L 3 L 1 L 2 L 3 Case (i) Case (ii) 17

18 Bipartite Graphs Lemma. Let G be a connected graph, and let L 0,, L k be the levels 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) Suppose no edge joins two nodes in the same level. This implies all edges join nodes on adjacent levels. Bipartition: red = nodes on odd levels, blue = nodes on even levels. L 1 L 2 L 3 Case (i) 18

19 Bipartite Graphs Lemma. Let G be a connected graph, and let L0,, Lk be the levels 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) Suppose (x, y) is an edge with x, y both in same level Lj. Let z = lca(x, y) = lowest common ancestor. Let Li be level containing z. Consider cycle that takes edge from x to y, then path from y to z, then path from z to x. 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 19

20 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) 20

21 Connectivity in Directed Graphs

22 Directed Graphs Directed graph. G = (V, E) Edge (u, v) goes from node u to node v. Ex. Web graph - hyperlink points from one web page to another. Directedness of graph is crucial. Web search engines exploit hyperlink structure to rank web pages by importance. Some college football rankings use the link structure induced by the win-lose outcomes of games. 22

23 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. Problem: one page may have many different names. 23

24 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 u v 24

25 Strong Connectivity: Algorithm Theorem. Can determine if G is strongly connected in O(m + n) time. Pf. Pick any node s. Run BFS from s in G. Run BFS from s in G rev reverse orientation of every edge in G. Return true iff all nodes reached in both BFS executions. Correctness follows immediately from previous lemma. strongly connected not strongly connected 25

26 DAGs and Topological Ordering

27 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. v 2 v 3 v 6 v 5 v 4 v 1 v 2 v 3 v 4 v 5 v 6 v 7 v 7 v 1 a DAG a topological ordering 27

28 Precedence Constraints Precedence constraints. Edge (v i, v j ) means task v i must occur before v j. Applications. Course prerequisite graph: course v i must be taken before v j. 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. v 1 v 2 v 3 v 4 v 5 v 6 v 7 a topological ordering 28

29 Directed Acyclic Graphs Lemma. If G has a topological order, then G is a DAG. Pf. (by contradiction) 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. Let v i be the lowest-indexed node in C, and let v j be the node just before v i in C - thus (v j, v i ) is an edge. By our choice of i, we have i < j. 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. the directed cycle C v 1 v i v j v n the supposed topological order: v 1,, v n 29

30 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? Lemma. If G is a DAG, then G has a source node with no incoming edges. Also, G has a sink node with no outgoing edges. 30

31 Directed Acyclic Graphs Lemma. If G is a DAG, then G has a source node. Pf. (by contradiction) Suppose that G is a DAG and every node has at least one incoming edge. Let's see what happens. 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. Then, since u has at least one incoming edge (x, u), we can walk backward to x. Repeat until we visit a node, say w, twice. Let C denote the sequence of nodes encountered between successive visits to w. C is a directed cycle. w x u v 31

32 Directed Acyclic Graphs Lemma. If G is a DAG, then G has a topological ordering. Pf. (by induction on n) Base case: true if n = 1. Given DAG on n > 1 nodes, find a node v with no incoming edges. G - { v } is a DAG, since deleting v cannot create cycles. By inductive hypothesis, G - { v } has a topological ordering. Place v first in topological ordering; then append nodes of G - { v } in topological order. This is valid since v has no incoming edges. DAG v 32

33 Topological Sorting Algorithm: Running Time Theorem. Algorithm finds a topological order in O(m + n). Pf. Maintain the following information: count[w] = remaining number of incoming edges S = set of remaining nodes with no incoming edges Initialize count and S: O(m + n) via single scan through graph. Update operation: remove any v from S and mark it as next in order. Decrement count[w] for all edges from v to w, and add w to S if count[w] hits 0 this is O(1) per edge 33

34 Topological Ordering Algorithm: Example v 2 v 3 v 6 v 5 v 4 v 7 v 1 Topological order: 34

35 Topological Ordering Algorithm: Example v 2 v 3 v 6 v 5 v 4 v 7 Topological order: v 1 35

36 Topological Ordering Algorithm: Example v 3 v 6 v 5 v 4 v 7 Topological order: v 1, v 2 36

37 Topological Ordering Algorithm: Example v 6 v 5 v 4 v 7 Topological order: v 1, v 2, v 3 37

38 Topological Ordering Algorithm: Example v 6 v 5 v 7 Topological order: v 1, v 2, v 3, v 4 38

39 Topological Ordering Algorithm: Example v 6 v 7 Topological order: v 1, v 2, v 3, v 4, v 5 39

40 Topological Ordering Algorithm: Example v 7 Topological order: v 1, v 2, v 3, v 4, v 5, v 6 40

41 Topological Ordering Algorithm: Example v 2 v 3 v 6 v 5 v 4 v 1 v 2 v 3 v 4 v 5 v 6 v 7 v 7 v 1 Topological order: v 1, v 2, v 3, v 4, v 5, v 6, v 7. 41

42 def topological_sort(graph): incount = { } for node in graph: incount[node] = 0 for node in graph: for successor in graph[node]: incount[successor] += 1 ready = [ node for node in graph if incount[node] == 0 ] result = [ ] while ready: node = ready.pop(-1) result.append(node) for successor in graph[node]: incount[successor] -= 1 if incount[successor] == 0: ready.append(successor) return result G= { 1:[4,5,7], 2:[3,5,6], 3: [4,5], 4: [5], 5:[6,7], 6:[7], 7:[]} print topological_sort(g) >>> [2, 3, 1, 4, 5, 6, 7] 42

43 Next Time Greedy Method Paradigm Minimum Spanning Tree Alogrithms and Analysis 43