If $G$ is not a complete graph, by Lemma 2.8, $d(G^{c})=d(G)$. Meanwhile, by Lemma 2.7, $m_{G}(-1)=m_{G^{c}}(-1)+n(G)-n(G^{c})$. So, $m_{G}(-1)=n(G)-d(G)-1$ if and only if $m_{G^{c}}(-1)+(n(G)-n(G^{c})=n(G)-d(G)-1=n(G)-d(G^{c})-1)$. Then we have $m_{G}(-1)=n(G)-d(G)-1$ if and only if $m_{G^{c}}(-1)=n(G^{c})-d(G^{c})-1$. ∎

If -1 is not an eigenvalue of $P_{d+1}$, then by the interlacing theorem, we have $rank(H+I)=rank(P_{d+1})$. If -1 is an eigenvalue of $P_{d+1}$, then $rank(H+I)=rank(P_{d+1}+I)$ or $rank(P_{d+1}+I)+1$. This completes the proof. ∎

Since -1 is not an eigenvalue of $P$, we can deduce from Lemma 2.1 that $m_{G}(-1)\leq n-d-1$, with equality holding if and only if $X=V(G-P)$ forms a star set. From Lemma 2.4, we know that $P$ is a dominating set for $G$, which implies $N_{G}(P)=V(G-P)$. Let $N_{G}(P)=\cup_{i\geq 1}N_{i}$ and $N_{i}=\{x:|N_{P}(x)|=i\}$.

Claim 1: $N_{G}(P)=N_{1}\cup N_{2}$

If $i>3$ and $x\in N_{i}$, where $x$ is adjacent to $v_{j_{1}}$, $v_{j_{2}}$, $v_{j_{3}}$, $v_{j_{4}}$ with $j_{1}<j_{2}<j_{3}<j_{4}$, then the path $v_{1}v_{2}\ldots v_{j_{1}}xv_{j_{4}}\ldots v_{d+1}$ forms a path shorter than $P$, which contradicts the fact that $P$ is a diameter path. If $i=3$ and $x\in N_{3}$, with $x$ being adjacent to $v_{j_{1}}$, $v_{j_{2}}$, $v_{j_{3}}$, it follows that $j_{3}=j_{1}+2=j_{2}+1$. If this condition is not satisfied, then $v_{1}v_{2}\ldots v_{j_{1}}xv_{j_{3}}\ldots v_{d+1}$ would again represent a path shorter than $P$, leading to a contradiction. When $j_{3}=j_{1}+2=j_{2}+1$, we find that $N_{P}(x)=N_{P}[v_{j_{2}}]$. According to Lemma 2.5, this implies $N_{G}[x]=N_{G}[v_{j_{2}}]$, which contradicts the assumption that $G=G^{c}$. Therefore, we conclude that $N_{G}(P)=N_{1}\cup N_{2}$.

Claim 2: If $x,y\in V(G-P)$, then $N_{P}(x)\neq N_{P}(y)$.

If $x\sim y$, then $N_{P}[x]=N_{P}[y]$. According to Lemmas 2.6 and 2.11, it follows that $N_{G}[x]=N_{G}[y]$, which contradicts the condition $G=G^{c}$. If $x\nsim y$, consider $P+x+y$. By Lemma 2.13, we have $Fe_{x}=Fe_{y}$, which contradicts the fact that $Fe_{x}$ and $Fe_{y}$ are linearly independent. Therefore, we conclude that $N_{P}(x)\neq N_{P}(y)$.

Since $m_{P}(-1)=0$, we can infer from Lemma 2.2 that $d\equiv 0$ or $2\ (\text{mod}\ 3)$ . Therefore, we will discuss two cases in the following analysis.

Case 1: $d\equiv 2\ (\text{mod}\ 3)$

Claim 1.1: $N_{1}=\emptyset$.

First, by the interlacing theorem, we know that $m_{P+u}(-1)=1$.

If $N_{1}\neq\emptyset$, let $x\in N_{1}$ such that $x\sim v_{m}$. If $m\equiv 0$ or $1\mod 3$, then by Lemma 2.10, we have $m_{P+x}(-1)=0$, leading to a contradiction. If $m\equiv 2\mod 3$, it follows that $m_{P+x}(-1)=m_{K_{1,3}}=0$, which also leads to a contradiction. Therefore, we conclude that $N_{1}=\emptyset$.

Claim 1.2: If $u\in N_{2}$ and $u$ is adjacent to $v_{i}$ and $v_{j}$ with $i<j$, then it follows that $j=i+1$.

If $j-i>2$, then the path $v_{1}v_{2}\ldots v_{i}uv_{j}\ldots v_{d+1}$ is shorter than $P$, leading to a contradiction.

When $j=i+2$ and $i\equiv 0\ (\text{mod}\ 3)$, then $m_{P+u}(-1)=0$, resulting in another contradiction.

When $j=i+2$ and $i\equiv 1\ (\text{mod}\ 3)$, then $m_{P+u}(-1)=m_{H_{4}}(-1)=0$, resulting in yet another contradiction.

The case for $j=i+2$ and $i\equiv 2\ (\text{mod}\ 3)$ can be similarly analyzed, leading to a contradiction. In summary, we conclude that $j=i+1$.

Claim 1.3: If $u\in N_{2}$ and $u$ is adjacent to $v_{m}$ and $v_{m+1}$, then $m\equiv 0$ or $1\ (\text{mod}\ 3)$.

When $m\equiv 0\ (\text{mod}\ 3)$, then by using Lemma 2.10, we obtain $m_{P+u}(-1)=0$, which leads to a contradiction.

When $m\equiv 1\ (\text{mod}\ 3)$, then $m_{P+u}(-1)=1$, satisfying the condition.

When $m\equiv 2\ (\text{mod}\ 3)$, then $m_{P+u}(-1)=1$, satisfying the condition. Thus, the proof is established.

Claim 1.4: If $u,v\in N_{2}$ with $u\sim v_{m},v_{m+1}$ and $v\sim v_{h},v_{h+1}$, we may assume $m<h$. Then, $u\not\sim v$ and $P\leq G\leq G_{1}$.

First, let us assume that $u\sim v$. When $m\equiv 1\ (\text{mod}\ 3)$, since $P$ is a diameter path, we only need to consider the case where $h=m+1$. In this case, $m_{P+u+v}(-1)=1$, which leads to a contradiction. Similarly, when $m\equiv 2\ (\text{mod}\ 3)$, we only need to consider the case where $h=m+2$. In this situation, $m_{P+u+v}(-1)=m_{H_{5}}(-1)=0$, which also results in a contradiction. Thus, it follows that $u\not\sim v$.

If $m\equiv 1\ (\text{mod}\ 3)$ and $h\equiv 2\ (\text{mod}\ 3)$, then $m_{P+u+v}(-1)=2$, which satisfies the conditions. If $m\equiv 2\ (\text{mod}\ 3)$ and $h\equiv 1\ (\text{mod}\ 3)$, then $m_{P+u+v}(-1)=m_{H_{2}}(-1)=0$, leading to a contradiction. If both $m\equiv 1\ (\text{mod}\ 3)$ and $h\equiv 1\ (\text{mod}\ 3)$ (or if $m\equiv 2\ (\text{mod}\ 3)$ and $h\equiv 2\ (\text{mod}\ 3))$, we can apply Lemmas 2.7 and 2.10, which yield $m_{P+u+v}(-1)=2$, satisfying the conditions. Therefore, based on the above discussion, $G_{1}$ is a maximal graph that satisfies the given conditions. Furthermore, by reapplying Lemmas 2.7 and 2.10, we obtain $m_{G_{1}}(-1)=t=n-d-1$, which concludes the proof.

We will now begin to consider the case where $d\equiv 0\ (\text{mod}\ 3)$, while omitting certain proofs that are similar to those in Case 1.

Caes2: $d\equiv 0\ (\text{mod}\ 3)$

Claim 2.1: If $x\in N_{1}$ and $x\sin v_{m}$, then $m\equiv 1\ (\text{mod}\ 3)$.

If $m\equiv 0$ or $2$ $\ (\text{mod}\ 3)$, then $m_{P+u}(-1)=0$, leading to a contradiction.

If $m\equiv 1\ (\text{mod}\ 3)$, then $m_{P+u}(-1)=1$, satisfying the condition.

Claim 2.2: If $x,y\in N_{1}$, then $x\sim y$. Therefore, $|N_{1}|\leq 2$.

Let $x\sim v_{m}$ and $y\sim v_{h}$. According to Claim 2.1, we know that $m=3t+1$ and $h=3l+1$. If $x\not\sim y$, then $m_{P+x+y}(-1)=m_{P_{3}}(-1)=0$, which leads to a contradiction. If $x\sim y$, we may assume $m<h$, thus $h=m+3$. Otherwise, the sequence $v_{1}v_{2}\ldots v_{m}xyv_{h}\ldots v_{d+1}$ would constitute a path shorter than $P$. In this case, we have $m_{P+x+y}(-1)=m_{C_{6}}(-1)=2$, which satisfies the conditions.

Claim 2.3: If $u\in N_{2}$ and $u$ is adjacent to $v_{i}$ and $v_{j}$ with $i<j$, then it follows that $j=i+1$.

Claim 2.4: If $u\in N_{2}$ and $u$ is adjacent to $v_{m}$ and $v_{m+1}$, $m\equiv 0$ or $1\ (\text{mod}\ 3)$.

When $m\equiv 0\ (\text{mod}\ 3)$, then by using Lemma 2.10, we obtain $m_{P+u}(-1)=m_{P_{2}}(-1)=1$, satisfying the condition.

When $m\equiv 1\ (\text{mod}\ 3)$, then $m_{P+u}(-1)=m_{P_{2}}(-1)=1$, satisfying the condition.

When $m\equiv 2\ (\text{mod}\ 3)$, then $m_{P+u}(-1)=m_{H_{1}}=0$, leading to a contradiction;

Claim 2.5: Let $u,v\in N_{2}$ such that $u\sim v_{3a},v_{3a+1}$, and $v\sim v_{3b+1},v_{3b+2}$. If $b\geq a$, then it follows that $u\sim v$ and $b=a$. Conversely, if $b<a$, then $u\nsim v$.

If $b\geq a$ and $u\not\sim v$, then according to Lemma 2.10, we have $m_{P+u+v}(-1)=0<2$, which leads to a contradiction. Furthermore, since $P$ is a diameter path, we can conclude that $a=b$.

If $b<a$ and $u\sim v$, considering that $P$ is a diameter path, we have $b=a-1$. Thus, $m_{P+u+v}(-1)=m_{H_{6}}(-1)=0<2$, which leads to a contradiction. Therefore, it follows that $u\nsim v$.

Claim 2.6: If $x\in N_{1}$, $u\in N_{2}$, $x\sim v_{i}$, $u\sim v_{j},v_{j+1}$, and $u\not\sim x$, then if $j>i$, it follows that $j\equiv 0\ (\text{mod}\ 3)$. Conversely, by symmetry, if $j<i$, then $j\equiv 1\ (\text{mod}\ 3)$.

From Claim 2.4, we know that $j\equiv 0$ or $1\ (\text{mod}\ 3)$. If $j>i$ and $j\equiv 1\ (\text{mod}\ 3)$, then $m_{P+x+u}(-1)=m_{P_{3}}(-1)=0$, which leads to a contradiction. On the other hand, if $j>i$ and $j\equiv 0\ (\text{mod}\ 3)$, then $m_{P+x+u}(-1)=m_{C_{3}}(-1)=2$, satisfying the conditions. By symmetry, the proof is completed.

Claim 2.7: Let $x\in N_{1}$ and $u\in N_{2}$, with $x\sim v_{i}$, $u\sim v_{j}$, $v_{j+1}$. If $x\sim u$, then it follows that $j=i$ or $j=i-1$.

By symmetry, it is sufficient to prove that when $j\geq i$, then $j=i$. If $j\equiv 1\ (\text{mod}\ 3)$, then $i=j$; otherwise, $v_{1}v_{2}\ldots v_{i}xuv_{j+1}\ldots v_{d+1}$ would form a path shorter than $P$ ; when $i=j$, we have $m_{P+u+v}(-1)=m_{C_{3}}(-1)=2$, which satisfies the conditions. If $j\equiv 0\ (\text{mod}\ 3)$, then $j=i+2$; otherwise, a path shorter than $P$ would emerge. When $j=i+2$, we find $m_{P+u+v}(-1)=m_{H_{3}}(-1)=0$, leading to a contradiction. Therefore, the conclusion is established.

Claim 2.8: Let $N_{1}=\{x,y\}$. If there are two vertices $u,v\in N_{2}$ that are adjacent to vertices in $N_{1}$, then $P+u+x+y\cong H_{7}$.

According to claims 2.6 and 2.7, when there are two vertices $u,v\in N_{2}$ that are adjacent to vertices in $N_{1}$, it follows that $P+u+v+x+y\cong H_{7}$. Based on Lemmas 2.7 and 2.10, we have $m_{H_{7}}(-1)=2+m_{C_{6}}(-1)=4$, which satisfies the conditions and thereby concludes the proof.

Claim 2.9: When $d\equiv 0\ (\text{mod}\ 3)$, $P\leq G\leq G_{2}$ or $G_{3}$.

Based on the above discussion, it can be concluded that $G_{2}$ or $G_{3}$ are maximal graphs that satisfy the given conditions. Furthermore, according to Lemmas 2.7 and 2.10, we obtain $m_{G_{2}}(-1)=m_{G_{3}}(-1)=i+k+4=n-d-1$, which satisfies the conditions. The proof is thereby established by the alternating theorem. ∎

Initially, since $m_{P}(-1)=1$, we conclude from Lemma 2.2 that $d=3k+1$. Furthermore, there cannot exist a point $z$ that is at a distance of 3 from $P$. To illustrate this, suppose ( z ) is a point at a distance of 3 from $P$, $y$ is a vertex at a distance of 2 from $P$, and $x$ is adjacent to $P$. According to Lemma 2.10, we then have $m_{P+x+y+z}(-1)=m_{P}(-1)=1<2$, which leads to a contradiction.

Claim 1: If $x\in N_{1}$ and $x\sim v_{i}$, then $m_{P+x}(-1)=1$ and $i\equiv 0\mod{3}$.

According to Lemma 2.11, we have $m_{P+x}(-1)\leq 1+\dim(V(G,-1)\cap Z(v_{1})$. Since $\dim(V(G,-1)\cap Z(v_{1}))=0$, it follows that $m_{P+x}(-1)\leq 1$. If $m_{P+u}(-1)=0$, then it would imply that $m_{G}(-1)\leq n-d-2$, which leads to a contradiction. Therefore, we conclude that $m_{P+u}(-1)=1$. This also indicates that $V(G-P-x)$ is a star set.

If $i\equiv 0\ (\text{mod}\ 3)$, then by utilizing Lemma 2.10, we obtain $m_{P+x}(-1)=m_{P_{2}}(-1)=1$, which satisfies the condition. If $m\equiv 1$ or $2\ (\text{mod}\ 3)$, then $m_{P+u}(-1)=0$, leading to a contradiction.

Claim 2: If $x,y\in N_{1}$ and $x\sim v_{3a}$, $y\sim v_{3b}$, then $a\neq b$ and $u\sim v$. Hence, it follows that $|N_{1}|\leq 2$.

First, we demonstrate that $a\neq b$. If $a=b$ and $x\sim y$, then $N_{P}(x)=N_{P}(y)$. By Lemma 2.6, this implies $N_{G}[x]=N_{G}[y]$, which leads to a contradiction. Alternatively, if $a=b$ and $x\not\sim y$, then $m_{P+x+y}(-1)=1<2$, resulting in another contradiction.

If $x\not\sim y$, then by Claim 1 and Lemma 2.10, we have $m_{P+x+y}=m_{P_{2}}(-1)=1<2$, which leads to a contradiction. Hence, it follows that $x\sim y$. Furthermore, considering that $P$ is a diameter path, we conclude that $|N_{1}|\leq 2$.

Claim 3: Let $u\in N_{2}$ and $u\sim v_{m},v_{m+1}$. Then $m\equiv 1\ (\text{mod}\ 3)$ and $m_{P+u}(-1)=2$.

First, we know that $1\leq m_{P+u}(-1)\leq 2$. If $m\equiv 0\ (\text{mod}\ 3)$ or $m\equiv 2\ (\text{mod}\ 3)$ , then $m_{P+u}(-1)=0$, which leads to a contradiction. If $m\equiv 1\ (\text{mod}\ 3)$, then $m_{P+u}(-1)=m_{C_{3}}(-1)=2$, thus proving the claim. Similar to Claim 2, for $u,v\in N_{2}$, we have $N_{P}(u)\neq N_{P}(v)$.

Claim 4: Let $x\in N_{1}$ and $u,v\in N_{2}$. Then $x\not\sim N_{2}$ and $u\not\sim v$.

Assume $u\sim v$. According to Claim 3, we may assume $u\sim v_{3a+1}$, $v_{3a+2}$, and $v\sim v_{3b+1}$, $v_{3b+2}$. Considering that $P$ is a diameter path, we have $|b-a|=1$. Thus, $m_{P+u+v}(-1)=m_{H_{8}}(-1)=1<2$, which leads to a contradiction. If $x\sim N_{2}$, we may assume $x\sim v_{3c}$ and $u$. Given that $P$ is a diameter path, we have $|b-c|=1$. Therefore, $m_{P+x+u}(-1)=1<2$, which results in another contradiction.

Next, we begin to consider the vertices that are at a distance of 2 from $P$.

Claim 5: If $d(z,P)=2$, then $z\sim N_{1}$ and $z\not\sim N_{2}$.

If $x\in N_{1}$ and $z\sim x$, then $m_{P+x+z}(-1)=2$, satisfying the condition. On the other hand, if $u\in N_{2}$ and $z\sim u$, then $m_{P+u+z}(-1)=1<2$, which leads to a contradiction.

Claim 6: Let $N^{2}=\{z\mid d(z,P)=2\}$. Then it follows that $|N^{2}|\leq 1$.

If $z,w\in N$, by Claim 5, we know there exist $x,y\in N_{1}$ such that $x\sim z$ and $y\sim w$. If $x=y$ and $z\sim w$, let $H=P+x$; then $N_{H}(z)=N_{H}(w)$. By Lemma 2.5, we have $N_{G}[z]=N_{G}[w]$, which leads to a contradiction. If $x=y$ and $z\not\sim w$, then $m_{P+x+z+w}(-1)=1<3$, which is also a contradiction. Thus, we conclude that $x\neq y$. At this point, based on Claim 1 and Claim 2, we can assume without loss of generality that $x\sim v_{3a}$ and $y\sim v_{3a+3}$. If $z\not\sim y$, then $m_{P+x+y+z}(-1)=1<2$, which is again a contradiction. If $z\sim y$, then $m_{P{x+y+z}}(-1)=1+m_{C_{3}}(-1)=3$, satisfying the condition, which implies $z\sim x,y$. Similarly, we can conclude that $w\sim x,y$. If $z\sim w$, let $H=P+x+y)$; then $N_{H}(z)=N_{H}(w)$. By Lemma 2.6, we know $N_{G}[z]=N_{G}[w]$, leading to a contradiction. If $z\not\sim w$, then $m_{P+x+y+z+w}(-1)=2<4$, which is also a contradiction. In summary, we have $|N^{2}|\leq 1$. If $z\in N^{2}$ and $x,y\in N_{1}$, then it follows that $z\sim x,y$.