Homework 1 Solutions:

1.1.16. Both sequence can be represented by simple graphs, for example,

(1,1,2,2,4): a triangle, one of its vertices has two new adjacent edges;

(2,2,2,2): C_4

1.1.19. Impossible. Since degree sum 3 * 9 = 27 is odd, contradicting the degree theorem of Euler. In general, every r-regular graph with odd r must have even number of vertices.

1.2.2. Suppose two partition sets have sizes k and m-k. Then e(G)= k (m-k) is maximize when k – (m-k) = 0 or 1. Hence e(G) = the floor of m^2 /4 (check!).

This can be seen by differentiating k(m-k) resp to k or (better) say that when k – (m-k) is at least 2, we can move one vertex from the large set to the small

set and therefore enlarge its degree and consequently e(G).

1.2.4. K_n is bipartite iff n = 1 or 2, because K_p (p>= 3) contains K_3, an odd cycle.

C_n is bipartite iff n is even; P_n (n>0) is always bipartite.

1.4.12. d (x, y)=4.

1.4.23. Assume the vertices are v_1 = u, v_2,…, v_n = v. Since G is connected, there exists a (x,y)-path between any two vertices x, y.

We can form a (u,v)-walk by linking (v_i, v_{i+1}) paths for i=1,…,n-1. (Note, we can not guarantee a (u,v)-path covering all the vertices)

1.5.2. (skipped)

1.5.7. Girth(Q_5) = 4, because, e.g, (00000), (01000), (11000), (10000) is a C_4.

On the other hand, hypercubes (not only Q_5) are bipartite and thus have no C_3.

Peterson graph has girth 5, since it contains a C_5 and no C_3, or C_4 as I noted in class.

15. 22 Diameter D and radius R of Hypercube Q_n are both n. D=R because Q_n is symmetric, i.e., every vertex play the same role.

D= R = how many steps we can reach another vertex from (00…0). Clearly (11…1) is furthest from (00…0), with distance n.