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문제출처
광주과학기술원 GIST 최정옥 교수님의 “그래프 이론” 수업, 2024년 가을학기 중 Homework 5
Solution
(c) Construct a simple graph with radius $r$ and diameter $d$
Construction:
Consider the cycle $C_{2r}$ defined on $V = {0, 1, 2, \dots, 2r-1}$.
- In this case, we have:
$$
\text{rad}(C_{2r}) = r.
$$ - When $d = r$, the graph satisfies:
$$
\text{diam}(C_{2r}) = \text{rad}(C_{2r}) = r.
$$
- In this case, we have:
Now assume $2r \geq d > r$.
- Define a path $P_{d-r}$ starting from a vertex $r \in V$, with a length of $d-r$, consisting of vertices $r - v_1 - v_2 - \dots - v_{d-r}$, where $v_i \notin V$.
- Define $G = C_{2r} \cup P_{d-r}$.
Verification:
Starting from vertex $0$, a path of length $d$ can be created to vertex $v_{d-r}$. Therefore:
$$
\text{diam}(G) = d.
$$The shortest path from any vertex to $r$ must include $r$. Thus:
$$
\epsilon_G(v) = \epsilon_G(r) = r.
$$Therefore:
$$
\text{rad}(G) = r.
$$
Thus, $G$ is a simple graph with radius $r$ and diameter $d$ satisfying $r \leq d \leq 2r$.
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