Math236 Discrete Maths with Applications P. Ittmann UKZN, Pietermaritzburg Semester 1, 2012 Ittmann (UKZN PMB) Math236 2012 1 / 19
Degree Sequences Let G be a graph with vertex set V (G) = {v 1, v 2, v 3,..., v n } The degree sequence of G is the sequence of numbers deg v 1, deg v 2, deg v 3,..., deg v n Ittmann (UKZN PMB) Math236 2012 2 / 19
Degree Sequences (cont.) Suppose we wish to nd the degree sequence of the following graph There are two vertices of degree 1, two of degree 3 and one of degree 2 Hence the degree sequence is 3, 3, 2, 1, 1 Note we usually write degree sequences in non-increasing order Ittmann (UKZN PMB) Math236 2012 3 / 19
Degree Sequences (cont.) Can 3, 3, 3, 3, 3, 3, 3 be the degree sequence of some graph G? No it cannot Since the graph would have 7 vertices with odd degree and deg v 2 m v V Ittmann (UKZN PMB) Math236 2012 4 / 19
Degree Sequences (cont.) Denition A sequence d 1, d 2,..., d n of non-negative integer is graphical if it is a degree sequence of some graph Ittmann (UKZN PMB) Math236 2012 5 / 19
Degree Sequences (cont.) Consider the following sequences 3, 3, 3, 3, 1 2, 2, 2, 0 4, 4, 4 5, 5, 4, 2, 0 Are these graphical? We now introduce a powerful tool to determine whether a particular sequence is graphical due to Havel and Hakimi Ittmann (UKZN PMB) Math236 2012 6 / 19
Havel-Hakimi Theorem Let D be the sequence d 1, d 2,..., d n with d 1 d 2 d n and n 2 Let D be the sequence obtained from D by 1 Discarding d 1, and, 2 Subtracting 1 from each of the next d 1 entries of D That is, D is the sequence d 2 1, d 3 1,..., d d1 +1 1, d d1 +2,..., d n Then, D is graphical if and only if D is graphical We give the proof later Ittmann (UKZN PMB) Math236 2012 7 / 19
Consider the sequence 5, 3, 3, 3, 2, 2, 1, 1 Is it graphical? Applying the Havel-Hakimi theorem 5, 3, 3, 3, 2, 2, 1, 1 is graphical i 2, 2, 2, 1, 1, 1, 1 is graphical i 1, 1, 1, 1, 1, 1 is graphical This last sequence is graphical! Hence, 5, 3, 3, 3, 2, 2, 1, 1 is graphical Ittmann (UKZN PMB) Math236 2012 8 / 19
Is the sequence 6, 5, 5, 5, 4, 4, 2, 1 graphical? Applying the Havel-Hakimi theorem 6, 5, 5, 5, 4, 4, 2, 1 is graphical i 4, 4, 4, 3, 3, 1, 1 is graphical i 3, 3, 2, 2, 1, 1 is graphical i 2, 1, 1, 1, 1 is graphical i 0, 0, 1, 1 is graphical This last sequence is graphical! Hence, 6, 5, 5, 5, 4, 4, 2, 1 is graphical Ittmann (UKZN PMB) Math236 2012 9 / 19
Is the the sequence 8, 7, 6, 6, 5, 3, 2, 2, 2, 1 graphical? Applying the Havel-Hakimi theorem 8, 7, 6, 6, 5, 3, 2, 2, 2, 1 is graphical i 6, 5, 5, 4, 2, 1, 1, 1, 1 is graphical i 4, 4, 3, 1, 0, 0, 1, 1 is graphical i 4, 4, 3, 1, 1, 1, 0, 0 is graphical (re-arranging) i 3, 2, 0, 0, 1, 0, 0 is graphical i 3, 2, 1, 0, 0, 0, 0 is graphical (re-arranging) i 1, 0, 1, 0, 0, 0 is graphical Ittmann (UKZN PMB) Math236 2012 10 / 19
This last sequence is clearly not graphical Hence, 8, 7, 6, 6, 5, 3, 2, 2, 2, 1 is not graphical Ittmann (UKZN PMB) Math236 2012 11 / 19
Given a sequence D, it is not necessarily easy to obtain a graph G with degree sequence D We can use the Havel-Hakimi theorem in reverse Suppose D is the sequence formed by H-H and we know a graph G with degree sequence D We can generate G by adding a vertex to G and adding d 1 edges as necessary Ittmann (UKZN PMB) Math236 2012 12 / 19
Let D be the sequence 4, 4, 4, 3, 3, 2 We prove that D is graphical and then draw a graph G with degree sequence D Ittmann (UKZN PMB) Math236 2012 13 / 19
Applying the Havel-Hakimi theorem D : 4, 4, 4, 3, 3, 2 is graphical i D 1 : 3, 3, 2, 2, 2 is graphical i D 2 : 2, 1, 1, 2 is graphical i D 2 : 2, 2, 1, 1 is graphical i D 3 : 1, 0, 1 is graphical i D 3 : 1, 1, 0 is graphical i D 4 : 0, 0 is graphical Ittmann (UKZN PMB) Math236 2012 14 / 19
Clearly 0, 0 is graphical and hence 4, 4, 4, 3, 3, 2 is graphical We now work in reverse to obtain a graph G with degree sequence D D 3 D 2 D 1 D Thus, we have found a graph with degree sequence 4, 4, 4, 3, 3, 2 Ittmann (UKZN PMB) Math236 2012 15 / 19
We now prove the Havel-Hakimi theorem Proof. Suppose that the sequence D is graphical Let G 1 be a graph of order n 1 with degree sequence D Then the vertices of G 1 can be labelled as v 2, v 3,..., v n in such a way that deg G (v i ) = d i 1 if 2 i d 1 + 1 and d(v i ) = d i if d 1 + 2 i n We can now construct a new graph G from G 1 by adding a new vertex v 1 and then joining v 1 with an edge to each of v 2, v 3,..., v d1 +1 Ittmann (UKZN PMB) Math236 2012 16 / 19
Proof. The degree of v 1 is d 1 The degree of the other vertices are the remaining values of D Thus, we have constructed a graph with degree sequence D, and so D is graphical We show next that if D is graphical, then D is graphical Assume that D is a graphical sequence Therefore, there are one or more graphs of order n with degree sequence D Ittmann (UKZN PMB) Math236 2012 17 / 19
Proof. Among all such graphs, let G be one such that V (G) = {v 1, v 2,..., v n }, where deg G (v i ) = d i for 1 i n, and the sum of the degrees of the vertices adjacent with v 1 is maximum We show that in G, the vertex v 1 must be adjacent with vertices having degrees d 2, d 3,..., d d1 +1 Suppose this is not the case There must exist two vertices v j and v k with d j > d k such that v 1 is adjacent to v k, but not to v j Since the degree of v j exceeds that of v k, there must be some vertex v l such that v l is adjacent to v j, but not to v k Ittmann (UKZN PMB) Math236 2012 18 / 19
Proof. Removing the edges v 1 v k and v j v l and adding the edges v 1 v j and v k v l produces a new graph H that also has degree sequence D However, in H the sum of the degrees of the vertices adjacent with v 1 is larger than that in G, contradicting our choice of G This contradiction argument veries that our initial assumption that v 1 is not adjacent with vertices having degrees d 2, d 3,..., d d1 +1) was false Thus, as claimed, v 1 must be adjacent with vertices having degrees d 2, d 3,..., d d1 +1 Hence, the graph obtained from G by removing v 1, together with all the edges incident with v 1, produces a graph with degree sequence D, so D is graphical Ittmann (UKZN PMB) Math236 2012 19 / 19