Discrete Math Notes:
Chapter 5: Functions and Relations
5.10 Graph powers and the transitive closure
power of G, the graph G* is defined to be the directed graph whose edge set is E* and is called the kth
transitive closure of R, relation R
transitive closure of G, If G is a directed graph, then G*
Induction, starts by showing that a theorem is true for k = 1
Overview:
Theorem 5.10.1: The Graph Power Theorem, states in assuming G is a directed graph. Let u and v be any two vertices in G. There is an edge from u to v in Gk if and only if there is a walk of length k from u to v in G.
Example of Graph powers and Walks
G=Gl
G3=GoG2 walks of length 3 = wak (a,* a, d edge a,d) in G3 =wak (b *,d,a edge(b,a in G3 wak (c,*,a,d) edge (c,d) in G3 = wak (d,*, d, a edge (d,a in G3
walks of length 2
walk (a *, a) edge (a,a) in G2 walk(b,*,d) edge(b,d) in G2 walk (c,*, a) edge (c,a) in G2 walk {d,*,d) edge (d,d) in G2
+ edge (a,c)
+ edge (d,a)
+ edge (a,d)
+ edge (d,a)
Captions ^ 1. The walk a, d, a in G implies that (aa) is an edge in G2 2. The walk (b, a, d) in G implies that (b,d) is an edge in G2 3. The walk (c, b, a) in G implies that (c,a) is an edge in G2 4. (c, d, a) is a walk in G, but (c,a) is already present in G2 5. The walk (d, a, d) in G implies that (d,d) is an edge in G2 6. G3 represents walks of length 3 in G and is obtained by composing G with G2 7. The walk (a,*,a) in G (represented by edge (a,a) in G2) and the edge (a,d) in G imply an edge a,d) in G3 8. The walk (b,*,d) in G (represented by edge (b, d) in G2) and the edge (d, a) in G imply an edge (b, a) in G3 9. The walk (c,*,a) in G (represented by edge (c, a) in G2) and the edge (a, d) in G imply an edge (c, d) in G3 10. The walk (d,*,d) in G (represented by edge (d, d) in G2) and the edge (d, a) in G imply an edge (d, a) in G3
Proof 5.10.1: Proof of the Graph Power Theorem.
Theorem: Let G be a directed graph. Let u and v be any two vertices in G. There is an edge from u to v in Gk if and only if there is a walk of length k from u to v in G Proof. By induction on k Base case: k = 1. G1 = G, by definition. Moreover an edge (u, v) is a walk (u, v) of length 1. Therefore, there is an edge (u, v) in G1 if and onl