






Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
Various topics in graph theory as presented in lecture no. 20. Topics include the adjacency polynomial of a graph, orientations of graphs, weight of sequences, and list coloring. The document also discusses the combinatorial nullstellen szatz and its implications for graph coloring.
Typology: Slides
1 / 12
This page cannot be seen from the preview
Don't miss anything!
Let G be a graph with vertex set V = {v 1 , v 2 ,... , vn}. Set X = (x 1 ,... , xn). The adjacency polynomial of G is the multivariate polynomial
A(G , X ) = Πi<j {(xi − xj ) : vi vj ∈ E }
Let D be an orientation of G. Then σ(D) = Π{σ(a) : a ∈ A(D)} where σ(a) = +1 if a = (vi , vj ) with i < j and σ(a) = − 1 if a = (vi , vj ) with i > j.
Setting xd^ = Πni=1xi di
A(G , X ) = Σd w (d)xd
Let f be a nonzero polynomial over a field F in the variables X = (x 1 , x 2 ,... , xn), of degree di in xi , for 1 ≤ i ≤ n. Let Li be a set of di + 1 elements of F , 1 ≤ i ≤ n. Then there exists t ∈ L 1 ×... × Ln such that f (t) 6 = 0.
Suppose G has an orientation D such that its outdegree sequence is d, then:
1 If D′^ is an orientation of G with outdegree sequence d then σ(D′) = σ(D) if and only if |A(D) − A(D′)| is even. 2 If D has no directed odd cycles, then all orientations of G with outdegree sequence d have the same sign.
Let G be a graph and let D be an orientation of G without directed odd cycles. Then G is (d + 1)-list colorable, where d is the outdegree sequence of G.
For any loopless graph G , there exists a polynomial P(G , x) such that P(G , k) = C (G , k) for all non-negative integers k. (Here C (G , k) is the number of distinct proper k-colorings of a graph G ) Moreover, if G is simple, and e is any edge of G , then P(G , x) satisfies the recursion formula:
P(G , x) = P(G − e, x) − P(G /e, x)
The polynomial P(G , x) is of degree n, with integer coefficients which alternate in sign, leading term xn^ and constant term 0.