Graph Theory problem about Regular Graphs. [1] Problem: Let $G$ be a graph of order $n$ in which every vertex has degree equal to $d.$

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*How large must $d$ be in order to guarantee that $G$ is connected?

*How large must $d$ be in order to guarantee that $G$ is $2-$ connected?

My Attempt:
i. By the Handshaking lemma $$\frac{nd}{2}=e\geq n-1.\implies d\geq \frac{2(n-1)}{n}.$$ However the answer is $$\lceil\frac{n-1}{2}\rceil.$$ I am not sure as to what reasoning was employed to come up with this solution.
ii. A graph is said to be $2$-connected if for every vertex $v\in V(G)$ the graph $G-v$ is connected. For this part, I am not even sure as to how should proceed.
Any hints/suggestions will be helpful.
 A: Your answer by the handshake lemma implies that $d$ needs to be at least $\frac{2(n-1)}{n}$ (which means at least $2$) for the graph to be connected: otherwise you won't have the necessary $n-1$ edges.
But even if $d$ is that large, it's still possible to get a disconnected graph. For example, when $n=10$, you can go up to $d=4$ and still be disconnected: 

You can generalize this construction:


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*When $n$ is even, take two complete graphs on $\frac n2$ vertices; this is regular with degree $\frac n2 -1$.

*When $n$ is odd, take a complete graph on $\frac{n-1}{2}$ vertices, and a complete-graph-without-a-perfect-matching on $\frac{n+1}{2}$ vertices; this is regular with degree $\frac{n-1}{2}-1$.


The content of part 1 of the problem is that you can't do better than these two examples: as soon as you're regular with degree $\frac{n-1}{2}$ (when $n$ is odd) or $\frac n2$ (when $n$ is even), you're connected.
We can prove this by contradiction. Suppose the graph is not connected; then it has $2$ components (at least) with no edges between them. One of them has size at most $\frac n2$, and the degrees in that component can be at most $\frac{n}{2}-1 < \frac{n-1}{2}$. 

Part 2 of the problem can be solved similarly. Suppose that $G$ is $d$-regular and connected but not $2$-connected. Then there is a vertex $v$ we can remove to get two components. 
The smallest component has size $k$, $k \le \lfloor \frac{n-1}{2}\rfloor$. In that component of $G-v$, all degrees are at most $k-1$, so in $G$ the degrees are at most $k$. In fact, even in $G$, one of those degrees must be $k-1$: the vertex $v$ would have too high degree if it were adjacent to every vertex of the component of size $k$, and a vertex of every other component. So if $G$ is not $2$-connected, we have $d \le \lfloor \frac{n-1}{2}\rfloor - 1$. In other words, if $d \ge \lfloor \frac{n-1}{2} \rfloor$ and connected, then $G$ is $2$-connected.
