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MCA 520: Graph Theory Instructor Neelima Gupta

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1 MCA 520: Graph Theory Instructor Neelima Gupta

2 Table of Contents Directed Graphs

3 Simple Digraph May contain one loop at each vertex. Distance: we say that a vertex y is at a distance d from a vertex x, if d is the length of a shortest path from x to y.

4 Motivation Finite State Machine – For Example: a bulb controlled by two switches often called “three-way switch”. Prediction Graphs Functional Graphs

5 Underlying Graph Weekly and Strongly Connected Components.

6 Subgraph, Isomorphism, decomposition and Union are same as that in undirected graphs Incidence and Adjacency Matrices

7 Kernel

8 Theorem: Richardson Every graph having no odd cycle contains a kernel. Proof: First prove for strongly connected di- graphs D. Let y be any vertex in D. Let S be the set of all vertices at even distance from y. – Show that there is no edge between the vertices in S…left as exercise. – Let x be in V \ S. Then x is at odd distance say d_1 from y. Since D is strongly connected there is a path from x to y. Then the length of this path must be odd for else we have an odd walk and hence an odd cycle. Let x’ be the first vertex after x on this path. Then x’ must be at even distance from y for if there exists a (shorter) path of odd length from y to x’, then again we get an odd cycle…y to x’ to y..since x to y was odd, x’ to y is even and thus y to x’ to y is odd.

9 An example to show that this is not true if there is an odd cycle.

10 Proof continued by induction Let n(D) =1, single vertex graph, this itself is a kernel. n(D) > 1, IH: true for a graph with v2 -> v3 -> v4. Then D has 4 SCC, each consisting of a singleton and {v4} has no out going edge. Since D’ is SCC, it has a kernel say S’. Remove S’ and its predecessors from the graph. Let D” be the remaining graph. By IH, D” has a kernel say S”. Claim: S’ union S” is a kernel for D

11 Vertex Degrees In-deg, Out-deg. In-neighborhood/predecessor set Out-neighborhood/successor set Degree-Sum Result: Sum of indegrees = Sum of Outdegrees = number of edges

12 Degree Sequence Is a list of degree pairs (d + (v i ), d - (v i )). Proposition: A list of pairs of non-negative integers is realizable as degree pairs of a digraph iff Σ i d + (v i ) = Σ i d - (v i ) This is true when multiple edges are allowed Proof: let m = sum. Consider m points and label d + (v i ) points with label I and d - (v i ) points with –j. For each point receiving label i and –j put an edge from I to j. Note that the resulting di-graph may contain loops and muti-edges….so not a simple di-graph

13 Characterization of DS for simple graphs (recall loops are allowed) Split of a digraph: A constructive algorithm similar to the one for undirected graphs can be given to test whether a list of pairs of non-negative integers is realizable as degree pairs of a digraph or not. The algorithm uses split of a digraph……assignment.

14 Eulerian Di-graphs Definitions of Trail, Walk, Circuit/Cycle remain the same with directions on them. A di-graph is Eulerian if it has an Eulerian Cycle. Lemma: If G is a di-graph with d + (G) > 1 (or d - (G) > 1 ), then it contains a cycle. A digraph G is Eulerian iff d + (v) = d - (v) for all v and the underlying graph has at most one non-trivial component.

15 Orientations and Tournaments 2 n^2 simple di-graphs with n vertices – n 2 ordered pairs including loops, each is present or absent…..(0,1) choices Orientation: for each edge in a simple undirected graph G (no loops), choose the direction ….three choices (0,+1,-1)…so …3 nC2 Tournament: Orientation of a complete simple simple graph…no loops, two choices (+1, -1) …so …2 nC2

16 Representing n-teams league match by an orientation of K n Let (u,v) be an edge if u wins. Score(u) = number of matches u wins = outdeg(u). Thus out-degree sequence is also called score sequence in a tournament. Indegree can be determined from outdegree (how?)

17 How to define the winner? Suppose we define winner to be the team that has not lost to any other team…..there may be no winner, we may have a cycle. Def2: that has won against maximum number of teams.

18 King of a tournament Claim: A team(/vertex) with maximum score (/out- degree) beats every other team either directly or by a path of length 2. Definition: In a di-graph, a king is a vertex from which every vertex is reachable by a path of length at most 2. Theorem: Every tournament has a king. Proof: here we’ll essentially prove our claim. Note:There may be several vertices with maximum out-degree i.e. maximum score and there may still be no clear winner. Worst is that all the vertices may have same out- degree. The only thing we can claim is that there is at least one winner.

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