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MCA 520: Graph Theory Instructor Neelima Gupta

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Table of Contents Directed Graphs

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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.

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

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Underlying Graph Weekly and Strongly Connected Components.

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

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Kernel

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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.

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

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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

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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

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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

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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.

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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.

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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

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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?)

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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.

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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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