CSCI 2670
Fall 2005
HW 1
August 30, 2005
0.3
a) no
b) yes
c) {x,y,z}
d) {x,y}
e) {(x,x),(x,y),(y,x),(y,y),(z,x),(z,y)}
f) {
∅
,{x},{y},{x,y}}
0.7 a) {(x,y) | x,y are stings of 0’s and 1’s the same number of 0’s OR 1’s}
b) {(x,y) | x ≥ y}
c)
0.12 Proof by contradiction.
Let G be a graph with n≥ 2 nodes and assume all nodes of G have different degrees.
Since G contains n nodes, the degree of each node must be some value in the set
D={1,2,3,…,n-1}. Note that the number of values in D is n. Therefore, for each value
d
∈
D, there is exactly one node v in G such that the degree of v is d. In particular, there is
some node with degree equal to n-1. This node must have an edge to every other node in
G. But this means that there can be no node with degree equal to 0. This is a contradiction
since 0
∈
D. Therefore, our assumption that all nodes have different degrees must be
incorrect. I.e., there are at least two nodes with the same degree.