CS 361 Data Structures & Algs Lecture 13. Prof. Tom Hayes University of New Mexico

Transcription

1 CS 361 Data Structures & Algs Lecture 13 Prof. Tom Hayes University of New Mexico

2 Last Time Spanning Trees BFS DFS Testing Bipartiteness 2

3 Today Inside BFS Running Time Implementation Degrees & Degree sums Properties of BFS and DFS trees Identifying trees in the wild 3

4 Breadth-First Search BFS finds the vertices in the connected component of the start vertex s one level at a time. Level Li = {vertices whose distance to s equals i} distance(v,w) = number of edges in the shortest path from v to w. Question: how to implement? 4

5 Plan of Attack We will first talk about a general search framework (GSF), broad enough to include BFS and some other search algorithms. Then focus on BFS in more detail. 5

6 GSF (includes BFS) Each node has 3 different states : U, A, I Initially: everything marked U (unfound). Change s to A (active). for (v : A) for (w : Adj[v]) if (w is U) change w to A change v to I (inactive) 6

7 Analysis of GSF Each node has 3 different states : U, A, I Initially: everything marked U (unfound). Change s to A (active). for (v : A) for (w : Adj[v]) How many times? if (w is U) change w to A change v to I (inactive) 7

8 Analysis of GSF Each node has 3 different states : U, A, I Initially: everything marked U (unfound). Change s to A (active). for (v : A) for (w : Adj[v]) How many times? if (w is U) change w to A 2 (#edges) change v to I (inactive) 8

9 Analysis of GSF Lemma: Each node is initially U, then A for some time, then I, and never goes back to an earlier label. Proof. By inspection, there is no line of code which could turn a node marked A or I into U, or I into A. Corollary: Loop variable v never repeats an earlier value. 9

10 Degree of a node Let v be a vertex in a graph G=(V,E). The degree of v equals the number of neighbors of v, that is, the length of the adjacency list of v. Answer: What is the sum, v V degree(v) 10

11 Degree of a node Let v be a vertex in a graph G=(V,E). The degree of v equals the number of neighbors of v, that is, the length of the adjacency list of v. Theorem: v V degree(v) = 2 E 11

12 Analysis of GSF Each node has 3 different states : U, A, I Initially: everything marked U (unfound). Change s to A (active). 1 (#nodes) for (v : A) for (w : Adj[v]) if (w is U) change w to A change v to I (inactive) 2 (#edges) (#nodes) Total: O(#nodes + #edges) 12

13 Linear Time Algorithms We say an algorithm runs in linear time if the running time is O(input size). For a graph in adjacency list format, the input size is Θ(#nodes + #edges). So, for algorithms where the input is a graph, linear time means O( V + E ). In particular, BFS is a linear-time algorithm. 13

14 From GSF to BFS BFS wants to explore level by level. GSF iterates through active set A in any order. BFS wants to explore from all nodes in level i before any in level (i+1). Insight: BFS also discovers all nodes in level i before any in level (i+1). Solution: Store A in a queue (FIFO)! In Java: AbstractQueue. 14

15 GSF (includes BFS) Each node has 3 different states : U, A, I Initially: everything marked U (unfound). Change s to A (active). for (v : A) for (w : Adj[v]) if (w is U) change w to A change v to I (inactive) 15

16 BFS Initially: A = empty queue. Initially: U[each vertex] = true A.add(s) (enqueue s) while (A nonempty) v = A.remove() for (w : Adj[v]) if (U[w]) {U[w]=false, A.add(w)} 16

17 Notes Originally, state U,A,I, was a single attribute of each node. Helped prevent conflicts (a state marked both U and A). When A became a Queue, we made U into a boolean array, then dropped I entirely. Care needed to avoid conflicts. Didn t have to keep track of current Level explicitly, since Queue keeps the nodes in the right order. Suppose we want to. How can we do it? 17

18 Edges in BFS Theorem: Each time BFS looks at an edge, it is either: (a) joining a found node (level Li) to an unfound node (level Li+1). Becomes an edge in the tree, or (b) joining a found node (level Li) to an already found node (level Li or Li+1). Why only these? 18

19 Rephrased: Theorem: If T is a BFS tree for a graph G, then every edge in G either: (a) is in T, and joins adjacent levels, or (b) is not in T, and joins nodes in the same or adjacent levels. Level: equal distance from the root. 19

20 Is this a BFS tree? 20

21 Is this a BFS tree? Where is the root? 21

22 Is this a BFS tree? Yes, and root must be here! 22

23 Is this a BFS tree? 2 23

24 Is this a BFS tree? 2 Not even a tree! (cycle) 24

25 Is this a BFS tree? 3 25

26 Is this a BFS tree? 3 Where is the root? 26

27 Is this a BFS tree? 3 root can only be here or here (level 1 must include all neighbors of root) 27

28 Is this a BFS tree? 3 can root be here? 28

29 Is this a BFS tree? 3 can root be here? No! Blue node would be in level 2 of tree. 29

30 Is this a BFS tree? 3 can root be here? 30

31 Is this a BFS tree? 3 can root be here? No! Blue node would be in level 2 of tree. 31

32 Is this a BFS tree? 3 Not a BFS tree! 32

33 Is this a BFS tree? 4 33

34 Is this a BFS tree? 4 Where is the root? 34

35 Is this a BFS tree? 4 Check: Yes, with 3 possible roots. 35

36 Depth-First Search DFS: explores fully from each vertex, before backing up to try another one. DFS(s): Mark s as found. For each unfound neighbor v of s: add edge (v,s) to T. DFS(v) 36

37 DFS Trees DFS: explores fully from each vertex, before backing up to try another one. Theorem: Each time DFS looks at an edge, it either: (a) is added to the tree, or (b) is a back-edge to an ancestor of the current node. (see (3.7) on page 85) Lemma: Active nodes are always a path from the root. 37

38 DFS example 38

39 DFS example 39

40 DFS example 40

41 DFS example 41

42 DFS example 42

43 DFS example 43

44 DFS example 44

45 DFS example 45

46 DFS example 46

47 DFS example 47

48 DFS example 48

49 DFS example 49

50 DFS example 50

51 DFS example 51

52 Is this a DFS tree? 52

53 Is this a DFS tree? Where is the root? 53

54 Is this a DFS tree? Non-edges must join a node to its ancestor. 54

55 Is this a DFS tree? Yes, a DFS tree. Root must be here 55