Notes slides from before lecture. CSE 21, Winter 2017, Section A00. Lecture 9 Notes. Class URL:

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1 Notes slides from before lecture CSE 21, Winter 2017, Section A00 Lecture 9 Notes Class URL:

2 Notes slides from before lecture Notes February 8 (1) HW4 is due on FRIDAY February 10 11:59pm HW5 should be posted by tomorrow (due on Thursday, February 16, 11:59pm) HW3 grades are published (or, will be published) as of today 72-hour regrade window If you used unauthorized resources in doing HW3, please instructors ASAP. Other cases will be referred to colleges. Discussion worksheets: partial credit is pro-rated from 7/10 = 5% credit MT1 should be back to you this week (target = Wednesday night) Today: Day of posted slides Any logistic, other issues?

3 Connected Components Disconnected graphs can be broken up into pieces where each is connected. Each (maximal) connected piece of the graph is a connected component. maximal means that if you make it any larger, it will lose the given (in this case, connected ) property.

4 Eulerian Tours and Fleury s Algorithm CSE21 Winter 2017, Day 12 (B00), Day 8 (A00) February 8,

5 Vocabulary Path (or walk): describes a route from one vertex to another (v 1, e 1, v 2, e 2,, v k ) Length of path: number of edges Simple path: path that doesn t repeat vertices Circuit (or cycle or closed walk): path that starts and ends at the same vertex, with length greater than zero Loop (or self-loop): an edge from a vertex to itself

6 Consider only undirected graphs. Finding Eulerian tours Path that includes each edge once 1 st goal: Determine whether a given undirected graph G has an Eulerian tour. 2 nd goal: Actually find an Eulerian tour in an undirected graph G, when possible.

7 Connected Components Disconnected graphs can be broken up into pieces where each is connected. Each connected piece of the graph is a connected component. Does this graph have an Eulerian tour?

8 Finding Eulerian tours Let G = (V,E) be an - undirected - connected graph with n vertices. 1 st goal: Determine whether G has an Eulerian tour. 2 nd goal: If yes, find the tour itself.

9 Finding Eulerian tours Observation: If v is an intermediate* vertex on a path p, then p must enter v the same number of times it leaves v. * not the start vertex, not the end vertex. v

10 Finding Eulerian tours Observation: If v is an intermediate* vertex on a path p, then p must enter v the same number of times it leaves v. An Eulerian tour contains all edges incident on v. So, half of the edges are used to enter v, half to leave. * not the start vertex, not the end vertex. v

11 Recall: Degree The degree of a vertex in an undirected graph is the total number of edges incident with it, except that a loop contributes twice. Rosen p. 652

12 Finding Eulerian tours Observation: If v is an intermediate* vertex on a path p, then p must enter v the same number of times it leaves v. An Eulerian tour contains all edges incident on v. So, half of the edges are used to enter v, half to leave. Degree of v must be even! v * not the start vertex, not the end vertex.

13 (Summary of) Observation: Finding Eulerian tours In an Eulerian tour, any intermediate vertex has even degree. In a circuit, all vertices are intermediate so all have even degree. In a tour starting and ending at different vertices (not a circuit), the starting and ending vertices will have odd degree; all others will have even degree.

14 [Side note: remember contrapositive?] p q p q q p T T T T T F F F T T F T F F T T

15 Finding Eulerian tours Theorem: If G has an Eulerian tour, G has at most two odd-degree vertices. Which of the following statements is equivalent to the theorem (using the facts we know so far about graphs)? A. If the number of odd-degree vertices in G is anything other than 0 or 2, then G has no Eulerian tour. B. If G has three or more odd-degree vertices, then G does not have an Eulerian tour. C. If G has an Eulerian tour, then G has either all vertices with even degree or G has exactly two vertices with odd degree. D. All of the above. E. None of the above.

16 Finding Eulerian tours Theorem: If G has an Eulerian tour, G has at most two odd-degree vertices. Which of the following statements is the converse to the theorem? A. If G does not have an Eulerian tour, then G does not have at most two odd-degree vertices. B. If G has at most two odd-degree vertices, then G has an Eulerian tour. C. If G does not have at most two odd-degree vertices, then G does not have an Eulerian tour. D. More than one of the above. E. None of the above.

17 Finding Eulerian tours Theorem: If G has an Eulerian tour, G has at most two odd-degree vertices. Question: Is the converse also true? I.e., If G has at most two odd-degree vertices, then must G have an Eulerian tour?

18 Finding Eulerian tours Theorem: If G has an Eulerian tour, G has at most two odd-degree vertices. Question: is the converse also true? i.e If G has at most two odd degree vertices, then must G have an Eulerian tour? Answer: give algorithm to build the Eulerian tour! We'll develop some more graph theory notions along the way.

19 Eulerian tour? Finding Eulerian tours

20 Finding Eulerian tours Eulerian tour? Start at 4. Where should we go next? A. 2 or 3. B. 2 or 5. C. 3 or 5. D. 2 or 3 or 5.

21 Bridges A bridge is an edge, which, if removed, would cause G to be disconnected. A D B C E Which of the edges in this graph are bridges? A. D, E B. C, D C. D only D. None of the above.

22 Bridges A bridge is an edge, which, if removed, would cause G to be disconnected. Connection with Eulerian tours: In an Eulerian tour, we have to visit every edge on one side of the bridge before we cross it (because there's no coming back). Do you see divide & conquer in here?

23 Eulerian Tours: HOW (Fleury's Algorithm) 1. Check that G has at most 2 odd-degree vertices. 2. Start at vertex v, an odd-degree vertex if possible. 3. While there are still edges in G, 4. If there is more than one edge incident on v 5. Cross any edge incident on v that is not a bridge 6. Else, cross the only edge available from v. 7. Delete the edge just crossed from G, update v.

24 Eulerian Tours: HOW (Fleury's Algorithm) 1. Check that G has at most 2 odd-degree vertices. 2. Start at vertex v, an odd-degree vertex if possible. 3. While there are still edges in G, 4. If there is more than one edge incident on v 5. Cross any edge incident on v that is not a bridge 6. Else, cross the only edge available from v. 7. Delete the edge just crossed from G, update v. What Eulerian tour do you get when following Fleury's algorithm, starting at vertex 4?

25 Eulerian Tours: WHY (Fleury's Algorithm) 1. Check that G has at most 2 odd-degree vertices. 2. Start at vertex v, an odd-degree vertex if possible. 3. While there are still edges in G, 4. If there is more than one edge incident with v 5. Cross any edge incident with v that is not a bridge 6. Else, cross the only edge available from v. 7. Delete the edge just crossed from G, update v. Will there always be such an edge? Will go through each edge at most once, so if the while loop iterates E times, get an Eulerian tour.

26 Eulerian Tours: WHY (Fleury's Algorithm) 1. Check that G has at most 2 odd-degree vertices. 2. Start at vertex v, an odd-degree vertex if possible. 3. While there are still edges in G, 4. If there is more than one edge incident with v 5. Cross any edge incident with v that is not a bridge 6. Else, cross the only edge available from v. 7. Delete the edge just crossed from G, update v. Will there always be such an edge? Need to show (loop invariant): Connected graph with no more than one bridge from v. If we enter the while loop, then after the t th iteration of the while loop, the (remaining) graph is still connected, AND there is at most one other odd-degree vertex in the (remaining) graph other than v, AND across every bridge from v there is an odd-degree vertex (and hence there is at most one bridge incident with v).

27 Fleury s algorithm: invariants 1. Check that G has at most 2 odd-degree vertices. 2. Start at vertex v, an odd-degree vertex if possible. 3. While there are still edges in G, 4. If there is more than one edge incident with v 5. Cross any edge incident with v that is not a bridge 6. Else, cross the only edge available from v. 7. Delete the edge just crossed from G, update v. Will there always be such an edge? (1) If v has odd degree then there is exactly one other vertex w of odd degree. If v has even degree, all other vertices have even degree. (2) The (remaining) graph G stays connected (deleting edges as we use them).

28 Fleury s algorithm: invariants (1) If v has odd degree then there is exactly one other vertex w of odd degree. If v has even degree, all other vertices have even degree. Proof by induction B.C.: Since G has at most 2 odd-degree vertices and we pick one as v, there must be at most one other. I.H.: Suppose that after k iterations, the current vertex has odd degree and one other vertex has odd degree, or else all vertices have even degree

29 Fleury s algorithm: invariants (1) If v has odd degree then there is exactly one other vertex w of odd degree. If v has even degree, all other vertices have even degree. I.H.: Suppose that after k iterations, the current vertex has odd degree and one other vertex has odd degree, or else all vertices have even degree

30 Fleury s algorithm: invariants (2) The (remaining) graph G stays connected (deleting edges as we use them). Proof by induction: B.C. G starts as being connected. I.H. After k iterations, G is still connected. I.S.. We remove the current vertex v and all its incident edges. What can happen to the graph? Removing the current vertex v and its incident edges, the graph could split into connected components C 1, C 2,, C m

31 Fleury s algorithm: invariants (2) The (remaining) graph G stays connected (deleting edges as we use them). We remove the current vertex v and all its incident edges. What can happen to the graph?

32 Fleury s algorithm: invariants (2) The (remaining) graph G stays connected (deleting edges as we use them). We remove the current vertex v and all its incident edges. What can happen to the graph? Put another way: Let s see what the structure of G can possibly be. Three cases There is a single component. I.e., m = 1 and there is only one edge from v. G stays connected since we delete v after the one edge is removed.

33 Fleury s algorithm: invariants (2) The (remaining) graph G stays connected (deleting edges as we use them). We remove the current vertex v and all its incident edges. What can happen to the graph? Put another way: Let s see what the structure of G can possibly be. Three cases There is a component C i (shown here: C 1 ) with at least 2 edges from v. Neither is a bridge, so Fleury s algorithm can take either edge and the graph will stay connected.

34 Fleury s algorithm: invariants (2) The (remaining) graph G stays connected (deleting edges as we use them). Put another way: Let s see what the structure of G can possibly be. Three cases We remove the current vertex v and all its incident edges. What can happen to the graph? First, note that aside from Cases I, II and III, there are no other cases. Next, we re going to show that Case III cannot happen. If two connected components each have a bridge from v, only one of them contains w (= the other odd-degree vertex). Let s look at the subgraph of G that includes v and a different component (shown in the blue outline). In this subgraph, the degree of v is 1 (purple edge). But the handshake lemma then implies that there must be another odd-degree vertex w in this other component, which is impossible. (Since this component didn t contain w, all of its vertices must have even degree.)

35 Eulerian Tours WHY Fleury's Algorithm 1. Check that G has at most 2 odd-degree vertices. 2. Start at vertex v, an odd-degree vertex if possible. 3. While there are still edges in G, 4. If there is more than one edge incident with v 5. Cross any edge incident with v that is not a bridge 6. Else, cross the only edge available from v. 7. Delete the edge just crossed from G, update v. Will there always be such an edge? Why is more than one bridge from v bad? v

36 Seven Bridges of Konigsberg Is there a path where each edge occurs exactly once? Eulerian tour No! Rosen p. 693

37 Seven Bridges of Konigsberg redux Which of these puzzles can you draw without lifting your pencil off the paper? A. No B. No C. No D. Yes

38 Eulerian Tours: recap Consider only undirected graphs. 1 st goal: Determine whether a given undirected graph G has an Eulerian tour. G has an Eulerian tour if and only if G has at most 2 odd-degree vertices. 2 nd goal: Actually find an Eulerian tour in an undirected graph G, when possible. Fleury's Algorithm: don't burn your bridges.

39 Eulerian Tours: recap Number of odd-degree vertices Is such a graph possible? ,5,7,... 4,6,8,... yes no yes no yes Is an Eulerian tour possible? yes, a circuit yes, not a circuit no

40 DAGs and Graph Search CSE21 Winter 2017, Day 13 (B00), Day 9 (A00) February 10,

41 Today s plan Definition of a DAG. Ordering algorithm on a DAG. Graph search and reachability. In the textbook: Sections 10.4 and 10.5

42 (some) Prerequisites for (some) CSE classes year plan: Take classes in some order. If course A is prerequisite for course B, must take course A before we take course B

43 (some) Prerequisites for (some) CSE classes Which of the following orderings are ok? A. 30, 145, 151, 100. B. 110, 105, 21, 101. C. 21, 105, 130. D. More than one of the above. E. None of the above

44 (some) Prerequisites for (some) CSE classes What if we want to include all vertices (i.e. courses)? Is this possible for any graph?

45 (some) Prerequisites for (some) CSE classes What if we want to include all vertices (i.e. courses)? Is this possible for any graph? Classify graphs for which it is possible. 2. For those, find a good ( legal ) ordering

46 (some) Prerequisites for (some) CSE classes What if we want to include all vertices (i.e. courses)? Is this possible for any graph? Classify graphs for which it is possible. 2. For those, find a good ( legal ) ordering

47 Barriers to ordering Which of the following graphs have good orderings? A. C. B. D. E. None of the above.

48 Barriers to ordering A can't be first (because B is before it). B can't be first (because C is before it). C can't be first (because D is before it). D can't be first (because A is before it). A B D C Cycle Whenever there is a cycle, can't find a good ordering.

49 Directed Acyclic Graphs Directed graphs with no cycles are called directed acyclic graphs (DAGs).

50 Directed Acyclic Graphs Directed graphs with no cycles are called directed acyclic graphs (DAGs). ABGCEHDFI A topological ordering of a graph is an (ordered) list of all its vertices such that, for each (directed) edge (v,w) in the graph, v comes before w in the list.

51 Topological ordering ABGCEHDFI Two algorithmic questions: 1. Given an (ordered) list of all vertices in the graph, is it a topological ordering? 2. Given a graph, produce a topological ordering (if one exists).

52 Topological ordering ABGCEHDFI 1. Given an (ordered) list of all vertices in the graph, is it a topological ordering? How would you do this?

53 Topological ordering A, B, G, C, E, H, D, F, I At what vertex should we start? 2. Given a graph, produce a topological ordering (if one exists). A. Any vertex is okay. B. We must start at A. C. Choose any vertex with at least one outgoing edge. D. Choose any vertex with no incoming edges. E. None of the above.

54 Sources of a DAG In a DAG, vertices with no incoming edges are called sources. Which of these vertices are sources? A. Only A and G. B. Only A. C. Only I. D. Only I and F. E. None of the above.

55 Sources of a DAG Lemma 1: Every DAG has a (at least one) source How would you prove this?

56 Sources of a DAG Lemma 1: Every DAG has a (at least one) source How would you prove this? Not a source look at incoming edges

57 Sources of a DAG Lemma 1: Every DAG has a (at least one) source How would you prove this? Not a source look at incoming edges

58 Sources of a DAG Lemma 1: Every DAG has a (at least one) source How would you prove this? Found a source!

59 Sources of a DAG Lemma 1: Every DAG has a (at least one) source Let G be a DAG. We want to show that G has a source vertex. In a proof by contradiction (aka indirect proof), what should we assume? A. G has a source vertex. B. All the vertices in G are sources. C. No vertex in G is a source. D. G has at least one source vertex and at least one vertex that's not a source. E. None of the above.

60 Sources of a DAG Lemma 1: Every DAG has a (at least one) source Proof of Lemma 1: Let G be a DAG with n >1 vertices. We want to show that G has a source vertex. Assume towards a contradiction that no vertex in G is a source. Let v 0 be a vertex in G. Since v 0 is not a source (by assumption), it has an incoming edge. Let v 1 be a vertex in G such that (v 1, v 0 ) is an edge in G. Since v 1 is also not a source, let v 2 be a vertex in G such that (v 2, v 1 ) is an edge in G. Keep going to find v 0, v 1, v 2,, v n vertices. There must be a repeated vertex in this list (Pigeonhole Principle). Contradiction (contradicts the given fact that G is acyclic)!!!

61 Sources of a DAG Notation: G-v is the graph that results when we remove v and all of its outgoing edges from G. Lemma 2: If v is a source vertex of G, then G is a DAG if and only if G-v is a DAG. Look at the proof (following slides) we ll pick up with this on Monday!