Graph Degree Edges 6Ad0E8

1. **State the problem:** We have an undirected graph with vertices $a, b, c, d, e, f$ and edges as described. We need to find the degree of each vertex, the total number of edges, and prove the Handshaking Lemma for this graph. 2. **Recall the Handshaking Lemma:** The Handshaking Lemma states that the sum of the degrees of all vertices in an undirected graph equals twice the number of edges: $$\sum_{v \in V} \deg(v) = 2|E|$$ 3. **Find the degree of each vertex:** - Vertex $a$ is connected to $b$ by two parallel edges and to $f$ by one edge. - Each edge counts once for degree. - So, $\deg(a) = 2 + 1 = 3$ - Vertex $b$ is connected to $a$ by two edges, to $c$ by one edge, and to $f$ by one edge. - So, $\deg(b) = 2 + 1 + 1 = 4$ - Vertex $c$ is connected to $b$ by one edge and to $e$ by one edge. - So, $\deg(c) = 1 + 1 = 2$ - Vertex $d$ is isolated, so no edges. - So, $\deg(d) = 0$ - Vertex $e$ is connected to $c$ by one edge. - So, $\deg(e) = 1$ - Vertex $f$ is connected to $a$ by one edge, to $b$ by one edge, and has one loop. - A loop contributes $2$ to the degree. - So, $\deg(f) = 1 + 1 + 2 = 4$ 4. **Calculate the sum of degrees:** $$\sum \deg = 3 + 4 + 2 + 0 + 1 + 4 = 14$$ 5. **Find the number of edges:** Count edges: - Between $a$ and $b$: 2 edges - Between $b$ and $c$: 1 edge - Between $a$ and $f$: 1 edge - Between $b$ and $f$: 1 edge - Between $c$ and $e$: 1 edge - Loop at $f$: 1 edge - $d$ isolated: 0 edges Total edges: $$|E| = 2 + 1 + 1 + 1 + 1 + 1 = 7$$ 6. **Verify the Handshaking Lemma:** $$\sum \deg = 14 = 2 \times 7 = 2|E|$$ This confirms the Handshaking Lemma is true for this graph. **Final answers:** - Degrees: $\deg(a)=3$, $\deg(b)=4$, $\deg(c)=2$, $\deg(d)=0$, $\deg(e)=1$, $\deg(f)=4$ - Number of edges: $7$ - Handshaking Lemma verified: $\sum \deg = 2|E|$