CS 361 – Chapter 13

1. CS 361 – Chapter 13 Graph Review • Purpose • Representation • Traversal • Comparison with tree

2. Graph • Definition, assumptions • Why would we use a graph data structure? • To model something that “looks like” a graph • Relationships between objects we can quantify/measure • Representation • Adjacency list • Adjacency matrix (implement as array or ArrayList?) • What about weighted or directed graph?

3. Possible design • Graph Attributes • # vertices, # edges • List of vertices • Adjacency matrix • Graph Operations • Create empty graph • Add vertex; add edge • getVertex(key) • edgeExists(v1, v2) • degree(v) • Neighbor iterator(v) • BFS & DFS iterators (v) • Is connected, etc. • Design of Vertex depends on graph’s application… • Vertex attributes • Name • Marked • Level • Others depending on app. • Vertex operations • Mark / unmark / is marked • Get and set level • equals

4. Example • In this graph, how would we determine: • Degree and neighbors • BFS • DFS • Is the graph connected? • Is the graph bipartite? • Is there a cycle? • What does it mean if matrix is symmetric?

5. DFS applications • If we can perform a DFS on a graph, doing so also gives us further information : • “discovery edges” form a spanning tree • Can tell if graph is connected • Can tell if graph is cyclic • Can find a path between two vertices • BFS also gives you this information. BFS is good at finding a shortest path.

6. Directed graphs • What does it mean to be cyclic; connected? • DFS Traversal • Transitive closure • Provides quick way to determine if vertex is reachable • Topological ordering • Relevant for Directed Acyclic Graphs • Find a way to appropriately number the vertices

7. Properties • In a directed graph, every edge is a 1-way street. • Adjacency matrix • Not necessarily symmetric • Each “1” counts as a single directed edge • Slightly different way to determine total number of edges compared to undirected graph • To convert undirected to directed, need to double each edge • Raises the bar on what counts as “cyclic” • must have directed cycle • Smallest cycle is round trip to adjacent vertex. • “Strongly” connected: analogous to undirected version: For all u, v: each is reachable from the other.

8. DFS • Analogous to undirected case. We only travel in the indicated direction. • 4 kinds of edges when we’re done: • Discovery = in the DFS spanning tree • Back = to ancestor in the DFS tree • Forward = to descendant in the DFS tree • Cross = to neither ancestor nor descendant • Can we create a directed graph having all 4 types of edges? • As before, DFS can help answer connectivity questions.

9. Transitive closure • We’d like to make shortcuts: Add more edges, so that if v was reachable from u, now we can get there immediately. • Creating this 2nd graph can speed up future queries about reachability. • Floyd-Warshall algorithm (p. 377) • Number the vertices 1 to n (can be arbitrary). • We are going to modify the graph n times, once per vertex. • For each vertex vk, we look for all edges of the form (vi, vk) and (vk, vj). If we find a match, create edge (vi, vj). • What is the run-time complexity?

10. Algorithm FloydWarshall(G): Number vertices v1, v2, ..., vn. G[0] = G for k = 1 to n: G[k] = G[k-1] for i = 1 to n except k: for j = 1 to n except i,k: if  (vi,vk) and (vk,vj) add (vi,vj) to G[k] if not  already // G[n] is the transitive closure return G[n]

11. DAG • Directed Acyclic Graph • Not quite a tree • We can have (undirected) cycle, but not a directed one. • Useful for identifying pre-requisite actions in a series of tasks. • Another application: function call graph in a computer program (as long as there’s no recursion) • Topological ordering • Number vertices, telling us the order we may correctly visit the vertices. • For every edge (vi, vj), i < j. • There could be more than 1 possible ordering. • Useful fact:  topological ordering  the directed graph is acyclic.

12. Proof  (by contradiction) • Suppose  topological ordering, and  cycle. • The cycle includes edges (vk, __) (__, __) (__, __) … (__, vk) and we’d conclude that k < k. Contradiction  no cycle.  (algorithm) • Suppose acyclic. Here’s how to order the vertices.  • Let v1 be a vertex with in-degree 0. There must be such a vertex. • Remove v1 and its out edges. Resulting graph still acyclic. • Can then find v2 with in-degree 0. Remove, continue…

13. Biconnected components • Problem: given a graph, are there potential bottlenecks? Are there places where there is only 1 path versus 2+ paths between arbitrary vertices? • Essential edge (or separation edge) = an edge whose removal disconnects the graph. • Essential vertex (or separation vertex). • Biconnected means a graph has 2 disjoint paths between any two vertices. • Disjoint paths cannot share any edges or vertices except endpoints. One path can act as detour of the other. • Biconnected component = a subgraph that is either: • Just an essential edge with its endpoints (degenerate case), or • Biconnected in a maximal sense: adding any more vertices or edges renders it no longer biconnected. See figure on p. 386

14. Look at edges • We want to find the biconnected components in a graph. • The key is to look at the edges. • Define a link relation on edges: 2 edges are “linked” if • They are the same edge, or • Some simple cycle* contains both edges. * Simple cycle: the only vertices that repeat are the endpoints. • This link relation is an equivalence relation. • Equivalence classes correspond to biconnected components. • Justification is Lemma 13.23: • Being biconnected  any 2 vertices lie along a simple cycle. • (Being biconnected  no essential vertices or edges.)

15. Approach • Theoretically, to find the biconnected components (i.e. “link components”), we can find the equivalence classes of the edges. See which edges belong to the same cycle. (p. 388) • How to do it: • Find the DFS of the original graph G. • Create a 2nd graph “B” based on G. Edges of G become the vertices of B! Edges of B? For each back edge e in G’s DFS spanning tree, find the discovery edges fi in same simple cycle as e. Create an edge in B (e, fi) for each fi. • Because the simple cycles in G probably don’t overlap, we’d expect B to be disconnected. Return the connected components of B. • We can save time by noting that we don’t need all of G’s edges.

16. Algorithm Find biconnected/link components of G: Find G’s DFS traversal. Take note of discovery and back edges, and the order of visiting vertices. Create 2nd graph F. Discovery edges in G become vertices of F. Label all vertices in F as “unlinked”. /* Note: # discovery edges = n-1. Names of edges should be based on the order of direction as we created discovery/back edge. For example, “JF” would be a back edge. For book’s example, A-M visited in that order, so every back edge’s name is in reverse alpha order. */

17. Algorithm (2) for each vertex v in G: for each back edge e = (u,v): Add e as new vertex in F. while u ≠ v, we march up: Find vertex f  F matching discovery edge: u’s parent, u) in DFS tree Add edge (e,f) to F, where e = current back edge and f is discovery edge just found. if f unlinked, mark f as “linked” u = parent(u) else break // avoid repetition

18. Example • Book’s example G has vertices A-M. Corresponding graph F has 12 vertices: GH, HI, DE, FG, HJ, CD, CF, BC, CM, CK, KL, AB. • Begin nested loop! • Vertex A has back edges KA and MA to it. Look at KA first. • In DFS tree we see A  B  C  K. So, v=A and u=K. • Add KA to graph F. • u=K and v=A not equal. Discovery edge is f = CK. Draw edge in F from KA to CK. Mark CK linked. • u=C and v=A not equal. Discovery edge is f = BC. Draw edge in F from KA to BC. Mark BC linked. • u=B and v=A not equal. Discovery edge is f = AB. Draw edge in F from KA to AB. Mark AB linked. • u=A and v=A equal. Exit loop. • At this point, we look at other back edge MA. • … Final answer (p. 391) shows 5 components.