## 2. Preliminaries

Under the PMC model [

5,

23], to diagnose a system

$G=(V(G),E(G))$, two adjacent nodes in

G can perform tests on each other. For two adjacent nodes

u and

v in

$V(G)$, the test performed by

u on

v is represented by the ordered pair

$(u,v).$ The outcome of a test

$(u,v)$ is 1 (respectively, 0) if

u evaluate

v as faulty (respectively, fault-free). We assume that the test result is reliable (respectively, unreliable) if the node

u is fault-free (respectively, faulty). A test assignment

T for

G is a collection of tests for every adjacent pair of vertices. It can be modeled as a directed testing graph

$T=(V(G),L$), where

$(u,v)\in L$ implies that

u and

v are adjacent in

G. The collection of all test results for a test assignment

T is called a syndrome. Formally, a syndrome is a function

$\sigma :L\mapsto \{0,1\}$. The set of all faulty processors in

G is called a faulty set. This can be any subset of

$V(G)$. For a given syndrome

$\sigma $, a subset of vertices

$F\subseteq V(G)$ is said to be consistent with

$\sigma $ if syndrome

$\sigma $ can be produced from the situation that, for any

$(u,v)\in L$ such that

$u\in V\backslash F$,

$\sigma (u,v)=1$ if and only if

$v\in F$. This means that

F is a possible set of faulty processors. Since a test outcome produced by a faulty processor is unreliable, a given set

F of faulty vertices may produce a lot of different syndromes. On the other hand, different faulty sets may produce the same syndrome. Let

$\sigma (F)$ denote the set of all syndromes which

F is consistent with. Under the PMC model, two distinct sets

${F}_{1}$ and

${F}_{2}$ in

$V(G)$ are said to be indistinguishable if

$\sigma ({F}_{1})\cap \sigma ({F}_{2})\ne \mathsf{\varnothing}$; otherwise,

${F}_{1}$ and

${F}_{2}$ are said to be distinguishable. Besides, we say

$({F}_{1},{F}_{2})$ is an indistinguishable pair if

$\sigma ({F}_{1})\cap \sigma ({F}_{2})\ne \mathsf{\varnothing}$; else,

$({F}_{1},{F}_{2})$ is a distinguishable pair.

In the MM model, a processor sends the same task to a pair of distinct neighbors and then compares their responses to diagnose a system G. The comparison scheme of $G=(V(G),E(G))$ is modeled as a multigraph, denoted by $M=(V(G),L)$, where L is the labeled-edge set. A labeled edge ${(u,v)}_{w}\in L$ represents a comparison in which two vertices u and v are compared by a vertex w, which implies $uw,vw\in E(G)$. We usually assume that the testing result is reliable (respectively, unreliable) if the node u is fault-free (respectively, faulty). If $u,v\in F$ and $w\in V(G)\backslash F$, then ${(u,v)}_{w}\to 1$. If $u\in F$ and $v,w\in V(G)\backslash F$, then ${(u,v)}_{w}\to 1$. If $v\in F$ and $u,w\in V(G)\backslash F$, then ${(u,v)}_{w}\to 1$. If $u,v,w\in V(G)\backslash F$, then ${(u,v)}_{w}\to 0$. The collection of all comparison results in $M=(V(G),L)$ is called the syndrome of the diagnosis, denoted by $\sigma $. If the comparison ${(u,v)}_{w}$ disagrees, then $\sigma ({(u,v)}_{w})=1$. Otherwise, $\sigma ({(u,v)}_{w})=0$. Hence, a syndrome is a function from L to $\{0,1\}$. The MM* is a special case of the MM model and each node must test its any pair of adjacent nodes, i.e., if $uw,vw\in E(G)$, then ${(u,v)}_{w}\in L$. The set of all faulty processors in the system is called a faulty set. This can be any subset of $V(G)$. For a given syndrome $\sigma $, a faulty subset of vertices $F\subseteq V(G)$ is said to be consistent with $\sigma $ if syndrome $\sigma $ can be produced from the situation that, for any ${(u,v)}_{w}\in L$ such that $w\in V\backslash F$, $\sigma {(u,v)}_{w}=1$ if and only if $u,v\in F$ or $u\in F$ or $v\in F$ under the MM${}^{*}$ model. Let $\sigma (F)$ denote the set of all syndromes which F is consistent with. Let ${F}_{1}$ and ${F}_{2}$ be two distinct faulty sets in $V(G)$. If $\sigma ({F}_{1})\cap \sigma ({F}_{2})\ne \mathsf{\varnothing}$, we say $({F}_{1},{F}_{2})$ is an indistinguishable pair under the MM${}^{*}$ model; else, $({F}_{1},{F}_{2})$ is a distinguishable pair under the MM${}^{*}$ model.

**Definition** **1.** A system $G=(V,E)$ is g-good-neighbor t-diagnosable if ${F}_{1}$ and ${F}_{2}$ are distinguishable under the PMC (MM${}^{*}$) model for each distinct pair of g-good-neighbor faulty subsets ${F}_{1}$ and ${F}_{2}$ of V with $|{F}_{1}|\le t$ and $|{F}_{2}|\le t$. The g-good-neighbor diagnosability ${t}_{g}(G)$ of G is the maximum value of t such that G is g-good-neighbor t-diagnosable under the PMC (MM${}^{*}$) model.

A multiprocessor system and network is modeled as an undirected simple graph

$G=(V,E)$, whose vertices (nodes) represent processors and edges (links) represent communication links. Given a nonempty vertex subset

${V}^{\prime}$ of

V, the induced subgraph by

${V}^{\prime}$ in

G, denoted by

$G\left[{V}^{\prime}\right]$, is a graph, whose vertex set is

${V}^{\prime}$ and the edge set is the set of all the edges of

G with both endpoints in

${V}^{\prime}$. For any vertex

v, we define the neighborhood

${N}_{G}(v)$ of

v in

G to be the set of vertices adjacent to

v. For

$u\in {N}_{G}(v)$,

u is called a neighbor vertex or a neighbor of

v. We denote the numbers of vertices and edges in

G by

$|V(G)|$ and

$|E(G)|$. The degree

${d}_{G}(v)$ of a vertex

v is the number of neighbors of

v in

G. The minimum degree of a vertex in

G is denoted by

$\delta (G)$. Let

$S\subseteq V$. We use

${N}_{G}(S)$ to denote the set

${\cup}_{v\in S}{N}_{G}(v)\backslash S$. For neighborhoods and degrees, we usually omit the subscript for the graph when no confusion arises. A path in

G is a sequence of vertices such that from each of its vertices there is an edge to the next vertex in the sequence. The path with a length of

n is denoted by

n-path. The length of a shortest path between

x and

y is called the distance between

x and

y, denoted by

${d}_{G}(x,y)$. A complete graph

${K}_{n}$ is a graph in which any two vertices are adjacent on

n vertices. A graph

${G}_{1}$ is isomorphic to another graph

${G}_{2}$ (denoted by

${G}_{1}\cong {G}_{2}$) if and only if there exists a bijection

$\phi :V({G}_{1})\to V({G}_{2})$ such that for any two vertices

$u,v\in V({G}_{1})$,

$uv\in E({G}_{1})$ if and only if

$\phi (u)\phi (v)\in E({G}_{2})$. A graph

G is said to be

k-regular if for any vertex

v,

${d}_{G}(v)=k$. Let

G be connected. The connectivity

$\kappa (G)$ of

G is the minimum number of vertices whose removal results in a disconnected graph or only one vertex left when

G is complete. Let

${F}_{1}$ and

${F}_{2}$ be two distinct subsets of

V, and let the symmetric difference

${F}_{1}\u2206{F}_{2}=({F}_{1}\backslash {F}_{2})\cup ({F}_{2}\backslash {F}_{1})$. For graph-theoretical terminology and notation not defined here, we follow [

33].

Let $G=(V,E)$ be connected. A fault set $F\subseteq V$ is called a g-good-neighbor faulty set if $|N(v)\cap (V\backslash F)|\ge g$ for every vertex v in $V\backslash F$. A g-good-neighbor cut of G is a g-good-neighbor faulty set F such that $G-F$ is disconnected. The minimum cardinality of g-good-neighbor cuts is said to be the g-good-neighbor connectivity of G, denoted by ${\kappa}^{(g)}(G)$. A connected graph G is said to be g-good-neighbor connected if G has a g-good-neighbor cut.

For two positive integers n and k, let $\langle n\rangle $ denote the set $\{1,2,\dots ,n\}$ and $\langle k\rangle $ denote the set $\{1,2,\dots ,k\}$. Let ${P}_{n,k}$ be a set of arrangements of k elements in $\langle n\rangle $, that is, ${P}_{n,k}=\{{p}_{1}{p}_{2}\cdots {p}_{k}$: ${p}_{i}\in \langle n\rangle $ for $1\le i\le k$ and ${p}_{s}\ne {p}_{t}$ for $1\le s,t\le k$, $s\ne t\}$.

**Definition** **2.** Given two positive integers n and k with $n>k\ge 1$. The $(n,k)$-arrangement graph, denoted by ${A}_{n,k}$, has vertex set $V({A}_{n,k})=\{p:p={p}_{1}\cdots {p}_{k}\in {P}_{n,k}\}$, and edge set $E({A}_{n,k})=\{(p,q):p,q\in V({A}_{n,k})$ with ${p}_{i}\ne {q}_{i}$ for some $i\in \langle k\rangle $ and ${p}_{j}={q}_{j}$ for all $j\in \langle k\rangle \backslash \left\{i\right\}\}$.

From the definition, we know that the vertices of

${A}_{n,k}$ are the arrangements of

k elements in

$\langle n\rangle $, and the edges of

${A}_{n,k}$ connect arrangements which differ in exactly one of their

k positions.

${A}_{n,k}$ is a regular graph of degree

$k(n-k)$ with

$\frac{n!}{(n-k)!}$ vertices.

Figure 1 shows the arrangement graph

${A}_{4,2}$.

**Definition** **3** ([

26]).

A graph is vertex-transitive if and only if for any pair of its vertices u and v, there exists an automorphism of the graph that maps u to v. A graph is edge-transitive if and only if for any pair of its edges $(u,v)$ and $(x,y)$, there exists an automorphism of the graph that maps $(u,v)$ to $(x,y)$.**Lemma** **1** ([

26]).

${A}_{n,k}$ is vertex-transitive and edge-transitive.**Lemma** **2** ([

26]).

$\kappa ({A}_{n,k})=k(n-k)$ for $n>k\ge 1$.**Lemma** **3** ([

28]).

$n\ge 3$ and $n\ne 4$, $k\in [2,n)$, ${\kappa}^{(1)}({A}_{n,k})=(2k-1)(n-k)-1$ and ${\kappa}^{(1)}({A}_{4,2})={\kappa}^{(1)}({A}_{4,3})(={\kappa}^{(1)}({S}_{4}))=4$.**Lemma** **4** ([

28]).

$n\ge 3$ and $n\ne 4$, $2\le k<n$, ${\kappa}^{(1)}({A}_{n,k})=(2k-1)(n-k)-1$ and ${\kappa}^{(1)}({A}_{4,2})={\kappa}^{(1)}({A}_{4,3})(={\kappa}^{(1)}({S}_{4}))=4$.**Lemma** **5** ([

28]).

For $n\ge 8$, ${\kappa}^{(2)}({A}_{n,2})=4n-12$, and, for $k\in \{i:i=3,\dots ,n-5\}\cup \{n-2,n-1\}$, ${\kappa}^{(2)}({A}_{n,k})=(3k-2)(n-k)-2$.**Lemma** **6** ([

28]).

Let $n,k,g$ be positive integers such that $n\ge 4$, $2\le k\le n-2,g\ge 3$. Then, An edge cut of a graph G is a set of edges whose removal makes the remaining graph no longer connected. The edge connectivity $\lambda (G)$ of G is the minimum cardinality of an edge cut of G.

**Lemma** **7** ([

33]).

$\kappa (G)\le \lambda (G)\le \delta (G)$.According to Lemmas 2 and 7, we get the following corollary.

**Corollary** **1.** The edge connectivity $\lambda ({A}_{n,k})=k(n-k)$ for $n>k\ge 1$.

For $i\in \langle n\rangle $, $j\in \langle k\rangle $, let $V({A}_{n,k}^{j:i})$ be the set of all vertices in ${A}_{n,k}$ with the jth position being i, that is, $V({A}_{n,k}^{j:i})=\{p:p={p}_{1}\cdots {p}_{j}\cdots {p}_{k}\in {P}_{n,k}$ with ${p}_{j}=i\}$. It is easy to check that each ${A}_{n,k}^{j:i}$ is a subgraph of ${A}_{n,k}$, and we say that ${A}_{n,k}$ is decomposed into n subgraphs ${A}_{n,k}^{j:i}$$(1\le i\le n)$ according to the jth position. For simplicity, we shall take j as the last position k, and use ${A}_{n,k}^{i}$ to denote ${A}_{n,k}^{k:i}$. Then, $V({A}_{n,k}^{i})=\{p:p={p}_{1}\cdots {p}_{k-1}i$ with ${p}_{j}\in \langle n\rangle \backslash \left\{i\right\}$ and ${p}_{s}\ne {p}_{t}$ for $1\le s,t\le k-1\}$ for $1\le i\le n$. It is easy to see that ${A}_{n,1}$ is a complete graph ${K}_{n}$.

**Proposition** **1** ([

34]).

Let $n>k\ge 2$. For each $j\in \langle k\rangle $, ${A}_{n,k}^{j:i}$ is isomorphic to ${A}_{n-1,k-1}$ where $1\le i\le n$.For any vertex $u\in V({A}_{n,k}^{i})$$(1\le i\le n)$, in this paper, we say that $N(u)\cap V({A}_{n,k}^{i})$ is the set of inner neighbors of u, which is denoted by $IN(u)$ and $N(u)\cap (V({A}_{n,k})\backslash V({A}_{n,k}^{i}))$ is the set of outer neighbors of u, which is denoted by $ON(u)$.

**Proposition** **2** ([

31]).

Let $n>k\ge 2$, $i\in \langle n\rangle $. For any two vertices u, v in the subgraph ${A}_{n,k}^{i}$, $ON(u)\cap ON(v)=\mathsf{\varnothing}$ and $|ON(u)|=n-k$. Furthermore, the vertices of $ON(u)$ are distributed in $(n-k)$ distinct subgraphs.**Proposition** **3.** For any vertex $u\in V({A}_{n,k}^{i})$$(1\le i\le n)$, let $ON(u)$ be the set of outer neighbors of u. Then, ${A}_{n,k}[\left\{u\right\}\cup ON(u)]$ is isomorphic to the complete graph ${K}_{n-k+1}$.

**Proof.** By Lemma 1, ${A}_{n,k}$ is vertex-transitive. Without loss of generality, let $u=(n-k+1)(n-k+2)\cdots n\in V({A}_{n,k}^{n})$. By the definition of arrangement graphs, $ON(u)=\{{u}_{j}:{u}_{j}=(n-k+1)(n-k+2)\cdots (n-1)j,j\in \{1,2,\dots ,n-k\}\}$. Then, $|ON(u)|=n-k$. Note that u, ${u}_{1}$,…,${u}_{n-k-1}$ and ${u}_{n-k}$ are only different in last position. By the definition of arrangement graphs, any pair of vertices of u ${u}_{1}$,…,${u}_{n-k-1}$ and ${u}_{n-k}$ are adjacent. Thus, ${A}_{n,k}[\left\{u\right\}\cup ON(u)]$ is a complete graph. Note that $|\{u\}\cup ON(u)|=n-k+1$. Thus, ${A}_{n,k}[\left\{u\right\}\cup ON(u)]$ is isomorphic to ${K}_{n-k+1}$. □

**Definition** **4.** Let $\langle n\rangle =\{1,2,\cdots ,n\}$, and let ${S}_{n}$ be the symmetric group on $\langle n\rangle $ containing all permutations $p={p}_{1}{p}_{2}\cdots {p}_{n}$ of $\langle n\rangle $. The alternating group ${A}_{n}$ is the subgroup of ${S}_{n}$ containing all even permutations. It is well known that $\{(12i),(1i2),3\le i\le n\}$ is a generating set for ${A}_{n}$. The n-dimensional alternating group graph $A{G}_{n}$ is the graph with vertex set $V(A{G}_{n})$ = ${A}_{n}$ in which two vertices u, v are adjacent if and only if $u=v(12i)$ or $u=v(1i2)$, $3\le i\le n$.

**Definition** **5.** The n-dimensional star graph denoted by ${S}_{n}$. The vertex set of ${S}_{n}$ is $\{{u}_{1}{u}_{2}\cdots {u}_{n}:{u}_{1}{u}_{2}\cdots {u}_{n}$ is a permutation of $\langle n\rangle \}$. Vertex adjacency is defined as follows: ${u}_{1}{u}_{2}\cdots {u}_{n}$ is adjacent to ${u}_{i}{u}_{2}\cdots {u}_{i-1}{u}_{1}{u}_{i+1}\cdots {u}_{n}$ for all $2\le i\le n$.

**Lemma** **8** ([

29]).

(1). The arrangement graph ${A}_{n,n-2}$ is isomorphic to the n-dimensional alternating group graph $A{G}_{n}$. (2). The arrangement graph ${A}_{n,n-1}$ is isomorphic to the n-dimensional star graph ${S}_{n}$.**Lemma** **9** ([

31]).

For any two distinct vertices u and v in the arrangement graph ${A}_{n,k}$, we have the following results:- 1.
If $d(u,v)=1$, then $|N(u)\cap N(v)|=n-k-1$;

- 2.
If $d(u,v)=2$ and $n=k+1$, then $|N(u)\cap N(v)|=1$;

- 3.
If $d(u,v)=2$ and $n\ge k+2$, then $|N(u)\cap N(v)|\le 2$; and

- 4.
If $d(u,v)\ge 3$, then $|N(u)\cap N(v)|=0$.

## 3. The g-Good-Neighbor Diagnosability of Arrangement Graphs under the PMC Model

In this section, we show the

g-good-neighbor diagnosability of arrangement graphs under the PMC model (

Figure 2).

**Theorem** **1** ([

23]).

A system $G=(V,E)$ is g-good-neighbor t-diagnosable under the $PMC$ model if and only if there is an edge $uv\in E$ with $u\in V\backslash ({F}_{1}\cup {F}_{2})$ and $v\in {F}_{1}\u2206{F}_{2}$ for each distinct pair of g-good-neighbor faulty subsets ${F}_{1}$ and ${F}_{2}$ of V with $|{F}_{1}|\le t$ and $|{F}_{2}|\le t$.**Lemma** **10** ([

28]).

For $n\ge 3$ and $n\ne 4$, $2\le k<n$, ${\kappa}^{(1)}({A}_{n,k})=(2k-1)(n-k)-1$ and ${\kappa}^{(1)}({A}_{4,2})={\kappa}^{(1)}({A}_{4,3})(={\kappa}^{(1)}({S}_{4}))=4$.**Lemma** **11** ([

28]).

For $n\ge 8$, ${\kappa}^{(2)}({A}_{n,2})=4n-12$, and, for $k\in \{i:i=3,\dots ,n-5\}\cup \{n-2,n-1\}$, ${\kappa}^{(2)}({A}_{n,k})=(3k-2)(n-k)-2$.**Lemma** **12** ([

27]).

Let $n\ge 7$ and let T be a subset of the vertices of ${A}_{n,2}$ such that $|T|\le 4n-12$. Then, ${A}_{n,2}-T$ is either connected or has a large component and small components with at most two vertices or $|T|=4n-12$ and ${A}_{n,2}-T$ has a large component and a four-cycle.**Lemma** **13** ([

28]).

Let $n,k,g$ be positive integers such that $n\ge 4,2\le k\le n-2,3\le g<n-k$. Then, Let $\alpha \in P(n,k-1)$, $\alpha ={p}_{1}\dots {p}_{k-1}$ and ${V}_{\alpha}=\{{p}_{1}\dots {p}_{k-1}i:i\in \langle n\rangle \backslash \{{p}_{1},\dots ,{p}_{k-1}\}\}$. Let $u=\alpha i={p}_{1}\dots {p}_{k-1}i$ and $v=\alpha j={p}_{1}\dots {p}_{k-1}j,\phantom{\rule{4pt}{0ex}}i\ne j$ and neither i nor j occurs in $\alpha $. Clearly, $u,v\in V({A}_{n,k})$, and $(u,v)\in E({A}_{n,k})$. Since any symbol that does not occur in $\alpha $ can serve as the last symbol in a vertex in ${V}_{\alpha}$, $|{V}_{\alpha}|=n-(k-1)$. Thus, the graph ${K}_{n+k+1}^{\alpha}$ induced by ${V}_{\alpha}$ is a complete graph of order $n-k+1$. Let $g\in [0,n-k)$ and $X\subseteq V({K}_{n+k+1}^{\alpha})$ such that $|X|=g+1$. Notice that $g+1=|X|<|V({K}_{n+k+1}^{\alpha})|=n-k+1$. Then, ${A}_{n,k}\left[X\right]$ is a complete graph ${K}_{g+1}$.

**Lemma** **14.** Let $n,k,g$ be positive integers such that $n\ge 3$, $2\le k<n$, $0\le g<n-k$, and let ${A}_{n,k}$ be the arrangement graph. Let X be defined as above, and let ${F}_{1}={N}_{{A}_{n,k}}(X)$, ${F}_{2}=X\cup {N}_{{A}_{n,k}}(X)$. Then, $|{F}_{1}|=[(g+1)(k-1)+1](n-k)-g$, $|{F}_{2}|=[(g+1)(k-1)+1](n-k)+1$, $\delta ({A}_{n,k}\left[X\right])\ge g$ and $\delta ({A}_{n,k}-{F}_{1}-{F}_{2})\ge g$.

**Proof.** Let

X be defined as above. By the process of the proof of Lemma 13 in [

28],

$N(X)$ is a

g-good-neighbor cut of

${A}_{n,k}$ and

$|N(X)|=|{F}_{1}|=[(g+1)(k-1)+1](n-k)-g$. Since

$|X|=g+1$,

$|{F}_{2}|=[(g+1)(k-1)+1](n-k)+1$. □

**Lemma** **15.** Let $n\ge 3$, $2\le k<n$ and $0\le g<n-k$. Then, the g-good-neighbor diagnosability of the arrangement graph ${A}_{n,k}$ under the PMC model is less than or equal to $[(g+1)(k-1)+1](n-k)$, i.e., ${t}_{g}({A}_{n,k})\le [(g+1)(k-1)+1](n-k)$.

**Proof.** Let X be defined as above, and let ${F}_{1}={N}_{{A}_{n,k}}(X)$, ${F}_{2}=X\cup {N}_{{A}_{n,k}}(X)$. By Lemma 14, $|{F}_{1}|=[(g+1)(k-1)+1](n-k)-g$, $|{F}_{2}|=|X|+|{F}_{1}|=[(g+1)(k-1)+1](n-k)+1$, $\delta ({A}_{n,k}-{F}_{1})\ge g$ and $\delta ({A}_{n,k}-{F}_{2})\ge g$. Therefore, ${F}_{1}$ and ${F}_{2}$ are g-good-neighbor faulty sets of ${A}_{n,k}$ with $|{F}_{1}|=[(g+1)(k-1)+1](n-k)-g$ and $|{F}_{2}|=[(g+1)(k-1)+1](n-k)+1$.

We prove that ${A}_{n,k}$ is not g-good-neighbor $([(g+1)(k-1)+1](n-k)+1)$-diagnosable. Since $X={F}_{1}\u2206{F}_{2}$ and ${N}_{{A}_{n,k}}(X)={F}_{1}\subset {F}_{2}$, there is no edge of ${A}_{n,k}$ between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$. By Theorem 1, we can show that ${A}_{n,k}$ is not g-good-neighbor $([(g+1)(k-1)+1](n-k)+1)$-diagnosable under the PMC model. Hence, by the definition of the g-good-neighbor diagnosability, we show that the g-good-neighbor diagnosability of ${A}_{n,k}$ is less than $[(g+1)(k-1)+1](n-k)+1$, i.e., ${t}_{g}({A}_{n,k})\le [(g+1)(k-1)+1](n-k)$. □

**Lemma** **16.** Let $n,k,g$ be positive integers such that $n\ge 4$, $2\le k\le n-2,3\le g<n-k$. Then, the arrangement graph ${A}_{n,k}$ is g-good-neighbor $(((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+1)$-diagnosable under the PMC model.

**Proof.** By Theorem 1, to prove ${A}_{n,k}$ is g-good-neighbor $(((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+1)$-diagnosable, it is equivalent to prove that there is an edge $uv\in E({A}_{n,k})$ with $u\in V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and $v\in {F}_{1}\u2206{F}_{2}$ for each distinct pair of g-good-neighbor faulty subsets ${F}_{1}$ and ${F}_{2}$ of $V({A}_{n,k})$ with $|{F}_{1}|\le ((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+1$ and $|{F}_{2}|\le ((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+1$.

We prove this statement by contradiction. Suppose that there are two distinct g-good-neighbor faulty subsets ${F}_{1}$ and ${F}_{2}$ of ${A}_{n,k}$ with $|{F}_{1}|\le ((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+1$ and $|{F}_{2}|\le ((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+1$, but the vertex set pair $({F}_{1},{F}_{2})$ is not satisfied with the condition in Theorem 1, i.e., there are no edges between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$. Without loss of generality, suppose that ${F}_{2}\backslash {F}_{1}\ne \mathsf{\varnothing}$.

Case 1. $V({A}_{n,k})={F}_{1}\cup {F}_{2}$.

Note that $(g+1)(k-2)+2-\frac{{(g+1)}^{2}}{2}=-\frac{{g}^{2}}{2}+(k-3)g+k-\frac{1}{2}$. Since $k\in [2,n-2]$, $-\frac{{g}^{2}}{2}+(k-3)g+k-\frac{1}{2}\le -\frac{{g}^{2}}{2}+(n-5)g+n-2-\frac{1}{2}$. Let $y=-\frac{{g}^{2}}{2}+(n-5)g+n-2-\frac{1}{2}$. Then, ${y}_{max}=\frac{1}{2}{n}^{2}-4n+10$ for $g=n-5$ and $-\frac{{g}^{2}}{2}+(k-3)g+k-\frac{1}{2}\le \frac{1}{2}{n}^{2}-4n+10$.

Assume $V({A}_{n,k})={F}_{1}\cup {F}_{2}$. We have that $\frac{n!}{(n-k)!}=|V({A}_{n,k})|=|{F}_{1}\cup {F}_{2}|=|{F}_{1}|+|{F}_{2}|-|{F}_{1}\cap {F}_{2}|\le |{F}_{1}|+|{F}_{2}|\le 2(((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+1)\le 2(((g+1)(k-2)+2-\frac{{(g+1)}^{2}}{2})(n-k)+g+1)\le 2((\frac{1}{2}{n}^{2}-4n+10)(n-k)+g+1)=({n}^{2}-8n+20)(n-2)+2(n-2)+2={n}^{3}-10{n}^{2}+34n-34$. When $k=3$, $\frac{n!}{(n-k)!}={n}^{3}-3{n}^{2}+2n$. Note ${n}^{3}-3{n}^{2}+2n\le \frac{n!}{(n-k)!}$ for $k\ge 3$. Thus, ${n}^{3}-3{n}^{2}+2n\le {n}^{3}-10{n}^{2}+36n-40$. In fact, ${n}^{3}-3{n}^{2}+2n>{n}^{3}-10{n}^{2}+36n-40$ when $n\ge 4$. This is a contradiction. Therefore, $V({A}_{n,k})\ne {F}_{1}\cup {F}_{2}$.

Case 2. $V({A}_{n,k})\ne {F}_{1}\cup {F}_{2}$

According to the hypothesis, there are no edges between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$. Since ${F}_{1}$ is a g-good-neighbor faulty set and ${A}_{n,k}-{F}_{1}$ has two parts ${A}_{n,k}-{F}_{1}-{F}_{2}$ and ${A}_{n,k}[{F}_{2}\backslash {F}_{1}]$, we have that $\delta ({A}_{n,k}-{F}_{1}-{F}_{2})\ge g$ and $\delta ({A}_{n,k}[{F}_{2}\backslash {F}_{1}])\ge g$. Similarly, $\delta ({A}_{n,k}[{F}_{1}\backslash {F}_{2}])\ge g$ when ${F}_{1}\backslash {F}_{2}\ne \mathsf{\varnothing}$. Therefore, ${F}_{1}\cap {F}_{2}$ is also a g-good-neighbor faulty set. Since there are no edges between $V({A}_{n,k}-{F}_{1}-{F}_{2})$ and ${F}_{1}\u2206{F}_{2}$, ${F}_{1}\cap {F}_{2}$ is also a g-good-neighbor cut. When ${F}_{1}\backslash {F}_{2}=\mathsf{\varnothing}$, ${F}_{1}\cap {F}_{2}={F}_{1}$ is also a g-good-neighbor faulty set. Since there are no edges between $V({A}_{n,k}-{F}_{1}-{F}_{2})$ and ${F}_{1}\u2206{F}_{2}$, ${F}_{1}\cap {F}_{2}$ is a g-good-neighbor cut. By Lemma 13, $|{F}_{1}\cap {F}_{2}|\ge ((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+1$. Since $\delta ({A}_{n,k}[{F}_{2}\backslash {F}_{1}])\ge g$, $|{F}_{2}\backslash {F}_{1}|\ge g+1$. Therefore, $|{F}_{2}|=|{F}_{2}\backslash {F}_{1}|+|{F}_{1}\cap {F}_{2}|\ge g+1+((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+1=((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+2$, which contradicts with that $|{F}_{2}|\le ((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+1$. Thus, ${A}_{n,k}$ is g-good-neighbor $(((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+1)$-diagnosable. By the definition of ${t}_{g}({A}_{n,k})$, ${t}_{g}({A}_{n,k})\ge ((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+1$. The proof is complete. □

Combining Lemmas 15 and 16, we have the following theorem.

**Theorem** **2.** Let $n,k,g$ be positive integers such that $n\ge 4$, $3\le k\le n-2,3\le g<n-k$. Then, $((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+1\le {t}_{g}({A}_{n,k})\le [(g+1)(k-1)+1](n-k)$ under the PMC model.

**Theorem** **3.** Let ${A}_{n,k}$ be the arrangement graph with $n>k\ge 2$. Then, the diagnosability $t({A}_{n,k})=k(n-k)$ under the PMC model.

**Proof.** Let $u\in V({A}_{n,k})$. Then, $N(u)$ is a cut of ${A}_{n,k}$ and $|N(u)|=k(n-k)$. Let ${F}_{1}=N(u)$, ${F}_{2}=\left\{u\right\}\cup N(u)$. Then, $|{F}_{1}|=k(n-k)$, $|{F}_{2}|=|X|+|{F}_{1}|=k(n-k)+1$, $\delta ({A}_{n,k}-{F}_{1})\ge 0$ and $\delta ({A}_{n,k}-{F}_{2})\ge 0$. Therefore, ${F}_{1}$ and ${F}_{2}$ are 0-good-neighbor faulty sets of ${A}_{n,k}$ with $|{F}_{1}|=k(n-k)$ and $|{F}_{2}|=k(n-k)+1$. We will prove ${A}_{n,k}$ is not 0-good-neighbor $(k(n-k)+1)$-diagnosable. Since $\left\{u\right\}={F}_{1}\u2206{F}_{2}$ and $N(u)={F}_{1}\subset {F}_{2}$, there is no edge of ${A}_{n,k}$ between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$. By Theorem 1, we can show that ${A}_{n,k}$ is not 0-good-neighbor $(k(n-k)+1)$-diagnosable under the PMC model. Hence, by the definition of the 0-good-neighbor diagnosability, we conclude that the 0-good-neighbor diagnosability of ${A}_{n,k}$ is less than $k(n-k)+1$, i.e., ${t}_{0}({A}_{n,k})\le k(n-k)$.

By Theorem 1, to prove ${A}_{n,k}$ is 0-good-neighbor $k(n-k)$-diagnosable, it is equivalent to prove that there is an edge $uv\in E({A}_{n,k})$ with $u\in V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and $v\in {F}_{1}\u2206{F}_{2}$ for each distinct pair of 0-good-neighbor faulty subsets ${F}_{1}$ and ${F}_{2}$ of $V({A}_{n,k})$ with $|{F}_{1}|\le k(n-k)$ and $|{F}_{2}|\le k(n-k)$.

We prove this statement by contradiction. Suppose that there are two distinct 0-good-neighbor faulty subsets ${F}_{1}$ and ${F}_{2}$ of ${A}_{n,k}$ with $|{F}_{1}|\le k(n-k)$ and $|{F}_{2}|\le k(n-k)$, but the vertex set pair $({F}_{1},{F}_{2})$ is not satisfied with the condition in Theorem 1, i.e., there are no edges between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$. Without loss of generality, suppose that ${F}_{2}\backslash {F}_{1}\ne \mathsf{\varnothing}$.

Assume $V({A}_{n,k})={F}_{1}\cup {F}_{2}$. We have that $\frac{n!}{(n-k)!}=|V({A}_{n,k})|=|{F}_{1}\cup {F}_{2}|=|{F}_{1}|+|{F}_{2}|-|{F}_{1}\cap {F}_{2}|\le |{F}_{1}|+|{F}_{2}|\le 2k(n-k)$. When $k=2$, ${n}^{2}-n=\frac{n!}{(n-2)!}\le 4n-8$, a contradiction. Therefore, $V({A}_{n,2})\ne {F}_{1}\cup {F}_{2}$. When $k=3$, $\frac{n!}{(n-k)!}={n}^{3}-3{n}^{2}+2n$. Note ${n}^{3}-3{n}^{2}+2n\le \frac{n!}{(n-k)!}$ and $2k(n-k)\le 2{n}^{2}-8n+6$ for $k\ge 3$. Thus, ${n}^{3}-3{n}^{2}+2n\le 2{n}^{2}-8n+6$. In fact, ${n}^{3}-3{n}^{2}+2n>2{n}^{2}-8n+6$ when $n\ge 4$. This is a contradiction. Therefore, $V({A}_{n,k})\ne {F}_{1}\cup {F}_{2}$.

According to the hypothesis, there are no edges between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$. Since ${F}_{1}$ is a 0-good-neighbor faulty set and ${A}_{n,k}-{F}_{1}$ has two parts ${A}_{n,k}-{F}_{1}-{F}_{2}$ and ${A}_{n,k}[{F}_{2}\backslash {F}_{1}]$, we have that $\delta ({A}_{n,k}-{F}_{1}-{F}_{2})\ge 0$ and $\delta ({A}_{n,k}[{F}_{2}\backslash {F}_{1}])\ge 0$. Similarly, $\delta ({A}_{n,k}[{F}_{1}\backslash {F}_{2}])\ge 0$ when ${F}_{1}\backslash {F}_{2}\ne \mathsf{\varnothing}$. Therefore, ${F}_{1}\cap {F}_{2}$ is also a 0-good-neighbor faulty set. Since there are no edges between $V({A}_{n,k}-{F}_{1}-{F}_{2})$ and ${F}_{1}\u2206{F}_{2}$, ${F}_{1}\cap {F}_{2}$ is also a 0-good-neighbor cut. When ${F}_{1}\backslash {F}_{2}=\mathsf{\varnothing}$, ${F}_{1}\cap {F}_{2}={F}_{1}$ is also a 0-good-neighbor faulty set. Since there are no edges between $V({A}_{n,k}-{F}_{1}-{F}_{2})$ and ${F}_{1}\u2206{F}_{2}$, ${F}_{1}\cap {F}_{2}$ is a 0-good-neighbor cut. By Lemma 2, $|{F}_{1}\cap {F}_{2}|\ge k(n-k)$. Since $\delta ({A}_{n,k}[{F}_{2}\backslash {F}_{1}])\ge 0$, $|{F}_{2}\backslash {F}_{1}|\ge 1$. Therefore, $|{F}_{2}|=|{F}_{2}\backslash {F}_{1}|+|{F}_{1}\cap {F}_{2}|\ge 1+k(n-k)$, which contradicts with that $|{F}_{2}|\le k(n-k)$. Thus, ${A}_{n,k}$ is 0-good-neighbor $k(n-k)$-diagnosable. By the definition of ${t}_{0}({A}_{n,k})$, ${t}_{0}({A}_{n,k})\ge k(n-k)$. Therefore, ${t}_{0}(G)=t(G)=k(n-k)$. □

**Lemma** **17.** Let $n\ge 5$ and $2\le k<n$. Then, ${t}_{1}({A}_{n,k})\ge (2k-1)(n-k)$ under the PMC model.

**Proof.** By Theorem 1, to prove ${A}_{n,k}$ is 1-good-neighbor $(2k-1)(n-k)$-diagnosable, it is equivalent to prove that there is an edge $uv\in E({A}_{n,k})$ with $u\in V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and $v\in {F}_{1}\u2206{F}_{2}$ for each distinct pair of g-good-neighbor faulty subsets ${F}_{1}$ and ${F}_{2}$ of $V({A}_{n,k})$ with $|{F}_{1}|\le (2k-1)(n-k)$ and $|{F}_{2}|\le (2k-1)(n-k)$.

We prove this statement by contradiction. Suppose that there are two distinct 1-good-neighbor faulty subsets ${F}_{1}$ and ${F}_{2}$ of ${A}_{n,k}$ with $|{F}_{1}|\le (3k-2)(n-k)$ and $|{F}_{2}|\le (3k-2)(n-k)$, but the vertex set pair $({F}_{1},{F}_{2})$ is not satisfied with the condition in Theorem 1, i.e., there are no edges between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$. Without loss of generality, assume that ${F}_{2}\backslash {F}_{1}\ne \mathsf{\varnothing}$.

Assume $V({A}_{n,k})={F}_{1}\cup {F}_{2}$. We have that $\frac{n!}{(n-k)!}=|V({A}_{n,k})|=|{F}_{1}\cup {F}_{2}|=|{F}_{1}|+|{F}_{2}|-|{F}_{1}\cap {F}_{2}|\le |{F}_{1}|+|{F}_{2}|\le 2(2k-1)(n-k)\le 2(2n-3)(n-2)=4{n}^{2}-14n+12$. When $k=3$, $\frac{n!}{(n-k)!}={n}^{3}-3{n}^{2}+2n$. Note ${n}^{3}-3{n}^{2}+2n\le \frac{n!}{(n-k)!}$ for $k\ge 3$. Thus, ${n}^{3}-3{n}^{2}+2n\le 4{n}^{2}-14n+12$. In fact, ${n}^{3}-3{n}^{2}+2n>4{n}^{2}-14n+12$ when $n\ge 5$. This is a contradiction. When $k=2$, ${n}^{2}-n=\frac{n!}{(n-k)!}\le 2(2k-1)(n-k)=6(n-2)$. In fact, ${n}^{2}-n>6(n-2)$ when $n\ge 5$. This is a contradiction. Therefore, $V({A}_{n,k})\ne {F}_{1}\cup {F}_{2}$.

According to the hypothesis, there are no edges between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$. Since ${F}_{1}$ is a 1-good-neighbor faulty set and ${A}_{n,k}-{F}_{1}$ has two parts ${A}_{n,k}-{F}_{1}-{F}_{2}$ and ${A}_{n,k}[{F}_{2}\backslash {F}_{1}]$, we have that $\delta ({A}_{n,k}-{F}_{1}-{F}_{2})\ge 1$ and $\delta ({A}_{n,k}[{F}_{2}\backslash {F}_{1}])\ge 1$. Similarly, $\delta ({A}_{n,k}[{F}_{1}\backslash {F}_{2}])\ge 1$ when ${F}_{1}\backslash {F}_{2}\ne \mathsf{\varnothing}$. Therefore, ${F}_{1}\cap {F}_{2}$ is also a 1-good-neighbor faulty set. Since there are no edges between $V({A}_{n,k}-{F}_{1}-{F}_{2})$ and ${F}_{1}\u2206{F}_{2}$, ${F}_{1}\cap {F}_{2}$ is also a 1-good-neighbor cut. When ${F}_{1}\backslash {F}_{2}=\mathsf{\varnothing}$, ${F}_{1}\cap {F}_{2}={F}_{1}$ is also a 1-good-neighbor faulty set. Since there are no edges between $V({A}_{n,k}-{F}_{1}-{F}_{2})$ and ${F}_{1}\u2206{F}_{2}$, ${F}_{1}\cap {F}_{2}$ is a 1-good-neighbor cut. By Lemma 10, $|{F}_{1}\cap {F}_{2}|\ge (2k-1)(n-k)-1$. Since $\delta ({A}_{n,k}[{F}_{2}\backslash {F}_{1}])\ge 1$, $|{F}_{2}\backslash {F}_{1}|\ge 2$. Therefore, $|{F}_{2}|=|{F}_{2}\backslash {F}_{1}|+|{F}_{1}\cap {F}_{2}|\ge 2+(2k-1)(n-k)-1=(2k-1)(n-k)+1$, which contradicts with that $|{F}_{2}|\le (2k-1)(n-k)$. Thus, ${A}_{n,k}$ is 1-good-neighbor $(2k-1)(n-k)$-diagnosable. By the definition of ${t}_{1}({A}_{n,k})$, ${t}_{1}({A}_{n,k})\ge (2k-1)(n-k)$. □

Combining Lemmas 15 and 17, we have the following theorem.

**Theorem** **4.** Let $n\ge 5$ and $2\le k<n$. Then, ${t}_{1}({A}_{n,k})=(2k-1)(n-k)$ under the PMC model.

**Lemma** **18.** Let $n\ge 8$ and $k\in \{i:i=3,\dots ,n-5\}\cup \{n-2,n-1\}$. Then, ${t}_{2}({A}_{n,k})\ge (3k-2)(n-k)$ under the PMC model.

**Proof.** By Theorem 1, to prove ${A}_{n,k}$ is 2-good-neighbor $(3k-2)(n-k)$-diagnosable, it is equivalent to prove that there is an edge $uv\in E({A}_{n,k})$ with $u\in V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and $v\in {F}_{1}\u2206{F}_{2}$ for each distinct pair of g-good-neighbor faulty subsets ${F}_{1}$ and ${F}_{2}$ of $V({A}_{n,k})$ with $|{F}_{1}|\le (3k-2)(n-k)$ and $|{F}_{2}|\le (3k-2)(n-k)$.

We prove this statement by contradiction. Suppose that there are two distinct g-good-neighbor faulty subsets ${F}_{1}$ and ${F}_{2}$ of ${A}_{n,k}$ with $|{F}_{1}|\le (3k-2)(n-k)$ and $|{F}_{2}|\le (3k-2)(n-k)$, but the vertex set pair $({F}_{1},{F}_{2})$ is not satisfied with the condition in Theorem 1, i.e., there are no edges between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$. Without loss of generality, assume that ${F}_{2}\backslash {F}_{1}\ne \mathsf{\varnothing}$.

Assume $V({A}_{n,k})={F}_{1}\cup {F}_{2}$. We have that $\frac{n!}{(n-k)!}=|V({A}_{n,k})|=|{F}_{1}\cup {F}_{2}|=|{F}_{1}|+|{F}_{2}|-|{F}_{1}\cap {F}_{2}|\le |{F}_{1}|+|{F}_{2}|\le 2(3k-2)(n-k)\le 2(3n-5)(n-3)=6{n}^{2}-28n+30$. When $k=3$, $\frac{n!}{(n-k)!}={n}^{3}-3{n}^{2}+2n$. Note ${n}^{3}-3{n}^{2}+2n\le \frac{n!}{(n-k)!}$ for $k\ge 3$. Thus, ${n}^{3}-3{n}^{2}+2n\le 6{n}^{2}-28n+30$. In fact, ${n}^{3}-3{n}^{2}+2n>6{n}^{2}-28n+30$ when $n\ge 8$. This is a contradiction. Therefore, $V({A}_{n,k})\ne {F}_{1}\cup {F}_{2}$.

According to the hypothesis, there are no edges between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$. Since ${F}_{1}$ is a 2-good-neighbor faulty set and ${A}_{n,k}-{F}_{1}$ has two parts ${A}_{n,k}-{F}_{1}-{F}_{2}$ and ${A}_{n,k}[{F}_{2}\backslash {F}_{1}]$, we have that $\delta ({A}_{n,k}-{F}_{1}-{F}_{2})\ge 2$ and $\delta ({A}_{n,k}[{F}_{2}\backslash {F}_{1}])\ge 2$. Similarly, $\delta ({A}_{n,k}[{F}_{1}\backslash {F}_{2}])\ge 2$ when ${F}_{1}\backslash {F}_{2}\ne \mathsf{\varnothing}$. Therefore, ${F}_{1}\cap {F}_{2}$ is also a 2-good-neighbor faulty set. Since there are no edges between $V({A}_{n,k}-{F}_{1}-{F}_{2})$ and ${F}_{1}\u2206{F}_{2}$, ${F}_{1}\cap {F}_{2}$ is also a 2-good-neighbor cut. When ${F}_{1}\backslash {F}_{2}=\mathsf{\varnothing}$, ${F}_{1}\cap {F}_{2}={F}_{1}$ is also a 2-good-neighbor faulty set. Since there are no edges between $V({A}_{n,k}-{F}_{1}-{F}_{2})$ and ${F}_{1}\u2206{F}_{2}$, ${F}_{1}\cap {F}_{2}$ is a 2-good-neighbor cut. By Lemma 11, $|{F}_{1}\cap {F}_{2}|\ge (3k-2)(n-k)-2$. Since $\delta ({A}_{n,k}[{F}_{2}\backslash {F}_{1}])\ge 2$, $|{F}_{2}\backslash {F}_{1}|\ge 3$. Therefore, $|{F}_{2}|=|{F}_{2}\backslash {F}_{1}|+|{F}_{1}\cap {F}_{2}|\ge 3+(3k-2)(n-k)-2=(3k-2)(n-k)+1$, which contradicts with that $|{F}_{2}|\le (3k-2)(n-k)$. Thus, ${A}_{n,k}$ is 2-good-neighbor $(3k-2)(n-k)$-diagnosable. By the definition of ${t}_{2}({A}_{n,k})$, ${t}_{2}({A}_{n,k})\ge (3k-2)(n-k)$. □

Combining Lemmas 15 and 18, we have the following theorem.

**Theorem** **5.** Let $n\ge 8$ and $k\in \{i:i=3,\dots ,n-5\}\cup \{n-2,n-1\}$. Then, ${t}_{2}({A}_{n,k})=(3k-2)(n-k)$ under the PMC model.

For $n\ge 8$, ${A}_{n,2}$ is decomposed into n subgraphs ${A}_{n,2}^{1},\dots ,{A}_{n,2}^{n}$. By Proposition 1, ${A}_{n,2}^{i}$ is isomorphic to ${K}_{n-1}$ for $i=1,2,\dots ,n$. Let $a=(1,n),b=(2,n),c=(1,n-1),d=(2,n-1)$. Then, $a,b\in V({A}_{n,2}^{n})$, $ab\in E({A}_{n,2}^{n})$, $c,d\in V({A}_{n,2}^{n-1})$, $cd\in E({A}_{n,2}^{n-1})$, $ac\in E({A}_{n,2})$ and $bd\in E({A}_{n,2})$, and $abdca$ is a 4-cycle of ${A}_{n,2}$.

**Lemma** **19.** For $n\ge 8$ and ${A}_{n,2}$, let $X=\{a,b,c,d\}$ be defined as above, and let ${F}_{1}={N}_{{A}_{n,2}}(X)$, ${F}_{2}=X\cup {N}_{{A}_{n,2}}(X)$. Then, $|{F}_{1}|=4n-12$, $|{F}_{2}|=4n-8$, $\delta ({A}_{n,2}\left[X\right])=2$ and $\delta ({A}_{n,2}-{F}_{1}-{F}_{2})\ge 2$.

**Proof.** Note that $|N(X)|=4(n-3)=4n-12$. Then, $|{F}_{2}|=4n-8$. Since $abdca$ is a four-cycle of ${A}_{n,2}$, $\delta ({A}_{n,2}\left[X\right])=2$. Since $n\ge 8$, $\delta ({A}_{n,2}^{n}-\{a,b\})\ge 2$ and $\delta ({A}_{n,2}^{n-1}-\{c,d\})\ge 2$. Thus, $\delta ({A}_{n,2}-{F}_{1}-{F}_{2})\ge \phantom{\rule{3.33333pt}{0ex}}2$. □

**Lemma** **20.** For $n\ge 8$, ${t}_{2}({A}_{n,2})\le 4n-9$ under the PMC model.

**Proof.** Let X be defined in Lemma 19, and let ${F}_{1}={N}_{{A}_{n,2}}(X)$, ${F}_{2}=X\cup {N}_{{A}_{n,2}}(X)$. By Lemma 19, $|{F}_{1}|=4n-12$, $|{F}_{2}|=|X|+|{F}_{1}|=4n-8$, $\delta ({A}_{n,2}-{F}_{1})\ge 2$ and $\delta ({A}_{n,2}-{F}_{2})\ge 2$. Therefore, ${F}_{1}$ and ${F}_{2}$ are 2-good-neighbor faulty sets of ${A}_{n,2}$ with $|{F}_{1}|=4n-12$ and $|{F}_{2}|=4n-8$.

We will prove ${A}_{n,2}$ is not 2-good-neighbor $(4n-8)$-diagnosable. Since $X={F}_{1}\u2206{F}_{2}$ and ${N}_{{A}_{n,k}}(X)={F}_{1}\subset {F}_{2}$, there is no edge of ${A}_{n,2}$ between $V({A}_{n,2})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$. By Theorem 1, we can deduce that ${A}_{n,2}$ is not 2-good-neighbor $(4n-8)$-diagnosable under the PMC model. Hence, by the definition of the 2-good-neighbor diagnosability, we conclude that the 2-good-neighbor diagnosability of ${A}_{n,2}$ is less than $4n-8$, i.e., ${t}_{g}({A}_{n,2})\le 4n-9$. □

**Lemma** **21.** For $n\ge 8$, ${t}_{2}({A}_{n,2})\ge 4n-9$ under the PMC model.

**Proof.** By Theorem 1, to prove ${A}_{n,2}$ is 2-good-neighbor $(4n-9)$-diagnosable, it is equivalent to prove that there is an edge $uv\in E({A}_{n,2})$ with $u\in V({A}_{n,2})\backslash ({F}_{1}\cup {F}_{2})$ and $v\in {F}_{1}\u2206{F}_{2}$ for each distinct pair of 2-good-neighbor faulty subsets ${F}_{1}$ and ${F}_{2}$ of $V({A}_{n,2})$ with $|{F}_{1}|\le 4n-9$ and $|{F}_{2}|\le 4n-9$.

We prove this statement by contradiction. Suppose that there are two distinct 2-good-neighbor faulty subsets ${F}_{1}$ and ${F}_{2}$ of ${A}_{n,2}$ with $|{F}_{1}|\le 4n-9$ and $|{F}_{2}|\le 4n-9$, but the vertex set pair $({F}_{1},{F}_{2})$ is not satisfied with the condition in Theorem 1, i.e., there are no edges between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$. Without loss of generality, assume that ${F}_{2}\backslash {F}_{1}\ne \mathsf{\varnothing}$.

Assume $V({A}_{n,2})={F}_{1}\cup {F}_{2}$. We have that ${n}^{2}-n=\frac{n!}{(n-2)!}=|V({A}_{n,2})|=|{F}_{1}\cup {F}_{2}|=|{F}_{1}|+|{F}_{2}|-|{F}_{1}\cap {F}_{2}|\le |{F}_{1}|+|{F}_{2}|\le 2(4n-9)=8n-18$, a contradiction to $n\ge 8$. Therefore, $V({A}_{n,k})\ne {F}_{1}\cup {F}_{2}$.

According to the hypothesis, there are no edges between $V({A}_{n,2})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$. Since ${F}_{1}$ is a 2-good-neighbor faulty set and ${A}_{n,2}-{F}_{1}$ has two parts ${A}_{n,k}-{F}_{1}-{F}_{2}$ and ${A}_{n,2}[{F}_{2}\backslash {F}_{1}]$, we have that $\delta ({A}_{n,2}-{F}_{1}-{F}_{2})\ge 2$ and $\delta ({A}_{n,2}[{F}_{2}\backslash {F}_{1}])\ge 2$. Similarly, $\delta ({A}_{n,2}[{F}_{1}\backslash {F}_{2}])\ge 2$ when ${F}_{1}\backslash {F}_{2}\ne \mathsf{\varnothing}$. Therefore, ${F}_{1}\cap {F}_{2}$ is also a 2-good-neighbor faulty set. Since there are no edges between $V({A}_{n,2}-{F}_{1}-{F}_{2})$ and ${F}_{1}\u2206{F}_{2}$, ${F}_{1}\cap {F}_{2}$ is also a 2-good-neighbor cut. When ${F}_{1}\backslash {F}_{2}=\mathsf{\varnothing}$, ${F}_{1}\cap {F}_{2}={F}_{1}$ is also a 2-good-neighbor faulty set. Since there are no edges between $V({A}_{n,2}-{F}_{1}-{F}_{2})$ and ${F}_{1}\u2206{F}_{2}$, ${F}_{1}\cap {F}_{2}$ is a 2-good-neighbor cut. By Lemma 11, $|{F}_{1}\cap {F}_{2}|\ge 4n-12$. If $|{F}_{1}\cap {F}_{2}|=4n-12$, then, by Lemma 12, $|{F}_{2}\backslash {F}_{1}|=4$. If $|{F}_{1}\cap {F}_{2}|=4n-11$ or $4n-10$, then $|{F}_{2}\backslash {F}_{1}|\le 2$, a contradiction to that $\delta ({A}_{n,2}[{F}_{1}\backslash {F}_{2}])\ge 2$. Therefore, $|{F}_{2}|=|{F}_{2}\backslash {F}_{1}|+|{F}_{1}\cap {F}_{2}|\ge 4+(4n-12)=4n-8$, which contradicts with that $|{F}_{2}|\le 4n-9$. Thus, ${A}_{n,k}$ is 2-good-neighbor $(4n-9)$-diagnosable. By the definition of ${t}_{2}({A}_{n,k})$, ${t}_{2}({A}_{n,2})\ge 4n-9$. □

Combining Lemmas 20 and 21, we have the following theorem.

**Theorem** **6.** Let $n\ge 8$. Then, ${t}_{2}({A}_{n,2})=4n-9$ under the PMC model.

## 4. The g-Good-Neighbor Diagnosability of Arrangement Graphs under the MM* Model

Before discussing the

g-good-neighbor diagnosability of the arrangement graph

${A}_{n,k}$ under the MM

${}^{*}$ model (

Figure 3), we first give an existing result.

**Theorem** **7** ([

1,

23]).

A system $G=(V,E)$ is g-good-neighbor t-diagnosable under the $M{M}^{*}$ model if and only if for each distinct pair of g-good-neighbor faulty subsets ${F}_{1}$ and ${F}_{2}$ of V with $|{F}_{1}|\le t$ and $|{F}_{2}|\le t$ satisfies one of the following conditions. (1) There are two vertices $u,w\in V\backslash ({F}_{1}\cup {F}_{2})$ and there is a vertex $v\in {F}_{1}\u2206{F}_{2}$ such that $uw\in E$ and $vw\in E$. (2) There are two vertices $u,v\in {F}_{1}\backslash {F}_{2}$ and there is a vertex $w\in V\backslash ({F}_{1}\cup {F}_{2})$ such that $uw\in E$ and $vw\in E$. (3) There are two vertices $u,v\in {F}_{2}\backslash {F}_{1}$ and there is a vertex $w\in V\backslash ({F}_{1}\cup {F}_{2})$ such that $uw\in E$ and $vw\in E$.**Lemma** **22.** Let $n\ge 3$, $2\le k<n$ and $0\le g<n-k$. Then, the g-good-neighbor diagnosability of the arrangement graph ${A}_{n,k}$ under the MM* model is less than or equal to $[(g+1)(k-1)+1](n-k)$, i.e., ${t}_{g}({A}_{n,k})\le [(g+1)(k-1)+1](n-k)$.

**Proof.** Let X be defined in Lemma 15, and let ${F}_{1}={N}_{{A}_{n,k}}(X)$, ${F}_{2}=X\cup {N}_{{A}_{n,k}}(X)$. By Lemma 14, $|{F}_{1}|=[(g+1)(k-1)+1](n-k)-g$, $|{F}_{2}|=|X|+|{F}_{1}|=[(g+1)(k-1)+1](n-k)+1$, $\delta ({A}_{n,k}-{F}_{1})\ge g$ and $\delta ({A}_{n,k}-{F}_{2})\ge g$. Therefore, ${F}_{1}$ and ${F}_{2}$ are g-good-neighbor faulty sets of ${A}_{n,k}$ with $|{F}_{1}|=[(g+1)(k-1)+1](n-k)-g$ and $|{F}_{2}|=[(g+1)(k-1)+1](n-k)+1$.

We will prove that ${A}_{n,k}$ is not g-good-neighbor $([(g+1)(k-1)+1](n-k)+1)$-diagnosable. Since $X={F}_{1}\u2206{F}_{2}$ and ${N}_{{A}_{n,k}}(X)={F}_{1}\subset {F}_{2}$, there is no edge of ${A}_{n,k}$ between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$. By Theorem 7, we can show that ${A}_{n,k}$ is not g-good-neighbor $([(g+1)(k-1)+1](n-k)+1)$-diagnosable under the MM* model. Hence, by the definition of the g-good-neighbor diagnosability, we show that the g-good-neighbor diagnosability of ${A}_{n,k}$ is less than $[(g+1)(k-1)+1](n-k)+1$, i.e., ${t}_{g}({A}_{n,k})\le [(g+1)(k-1)+1](n-k)$. □

**Lemma** **23.** Let $n,k,g$ be positive integers such that $n\ge 4$, $3\le k\le n-2,3\le g<n-k$. Then, the arrangement graph ${A}_{n,k}$ is g-good-neighbor $(((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+1)$-diagnosable under the MM* model.

**Proof.** By the definition of the g-good-neighbor diagnosability, it is sufficient to show that ${A}_{n,k}$ is g-good-neighbor $(((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+1)$-diagnosable for $3\le g\le n-k-1$.

By Theorem 7, suppose, on the contrary, that there are two distinct g-good-neighbor faulty subsets ${F}_{1}$ and ${F}_{2}$ of ${A}_{n,k}$ with $|{F}_{1}|\le ((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+1$ and $|{F}_{2}|\le ((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+1$, but the vertex set pair $({F}_{1},{F}_{2})$ is not satisfied with any condition in Theorem 7. Without loss of generality, assume that ${F}_{2}\backslash {F}_{1}\ne \mathsf{\varnothing}.$ Similar to the discussion on $V({A}_{n,k})={F}_{1}\cup {F}_{2}$ in Lemma 16, we can show $V({A}_{n,k})\ne {F}_{1}\cup {F}_{2}$.

Claim 1.${A}_{n,k}-{F}_{1}-{F}_{2}$ has no isolated vertex.

Since ${F}_{1}$ is a g-good neighbor faulty set, for an arbitrary vertex $u\in V({A}_{n,k})\backslash {F}_{1}$, $|{N}_{{A}_{n,k}-{F}_{1}}(u)|\ge g$. Suppose, on the contrary, that ${A}_{n,k}-{F}_{1}-{F}_{2}$ has at least one isolated vertex x. Since ${F}_{1}$ is a g-good neighbor faulty set and $g\ge 3$, there are at least two vertices $u,v\in {F}_{2}\backslash {F}_{1}$ such that $u,v$ are adjacent to x. According to the hypothesis, the vertex set pair $({F}_{1},{F}_{2})$ is not satisfied with any condition in Theorem 7, by Condition (3) of Theorem 7, a contradiction. Therefore, there are at most one vertex $u\in {F}_{2}\backslash {F}_{1}$ such that u are adjacent to x. Thus, $|{N}_{{A}_{n,k}-{F}_{1}}(x)|=1$, a contradiction to that ${F}_{1}$ is a g-good neighbor faulty set, where $g\ge 3$. Thus, ${A}_{n,k}-{F}_{1}-{F}_{2}$ has no isolated vertex. The proof of Claim 1 is complete.

Let $u\in V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$. By Claim 1, $\delta ({A}_{n,k}-{F}_{1}-{F}_{2})\ge 1$. Since the vertex set pair $({F}_{1},{F}_{2})$ is not satisfied with any condition in Theorem 7, by the condition (1) of Theorem 7, for any pair of adjacent vertices $u,w\in V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$, there is no vertex $v\in {F}_{1}\u2206{F}_{2}$ such that $uw\in E({A}_{n,k})$ and $uv\in E({A}_{n,k})$. It follows that u has no neighbor in ${F}_{1}\u2206{F}_{2}$. Since u is taken arbitrarily, there is no edge between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$.

Since ${F}_{2}\backslash {F}_{1}\ne \mathsf{\varnothing}$ and ${F}_{1}$ is a g-good-neighbor faulty set, we have that ${\delta}_{{A}_{n,k}}([{F}_{2}\backslash {F}_{1}])\ge g$, $\delta ({A}_{n,k}-{F}_{2}-{F}_{1})\ge g$ and $|{F}_{2}\backslash {F}_{1}|\ge g+1$. Since both ${F}_{1}$ and ${F}_{2}$ are g-good-neighbor faulty sets, and there is no edge between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$, ${F}_{1}\cap {F}_{2}$ is a g-good-neighbor cut of ${A}_{n,k}$. By Lemma 13, we have $|{F}_{1}\cap {F}_{2}|\ge ((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+1$. Therefore, $|{F}_{2}|=|{F}_{2}\backslash {F}_{1}|+|{F}_{1}\cap {F}_{2}|\ge g+1+((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+1=((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+2$, which contradicts $|{F}_{2}|\le ((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+1$. Therefore, ${A}_{n,k}$ is g-good-neighbor $(((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+1)$-diagnosable and ${t}_{g}({A}_{n,k})\ge ((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+1$. The proof is complete. □

Combining Lemmas 22 and 23, we have the following theorem.

**Theorem** **8.** Let $n,k,g$ be positive integers such that $n\ge 4$, $3\le k\le n-2,3\le g<n-k$. Then, $((g+1)(k-2)+2-\lfloor \frac{{(g+1)}^{2}}{2}\rfloor )(n-k)+g+1\le {t}_{g}({A}_{n,k})\le [(g+1)(k-1)+1](n-k)$ under the MM* model.

**Theorem** **9** ([

34]).

Let ${A}_{n,k}$ be an n-dimensional arrangement graph and $3\le k<n$. Then, the diagnosability of ${A}_{n,k}$ is $k(n-k)$, i.e., $t({A}_{n,k})=k(n-k)$ under the MM* model.**Lemma** **24** ([

30]).

${A}_{n,k}$ is hamiltonian for $1\le k\le n-1$.A component of a graph G is odd according as it has an odd number of vertices. We denote by $o(G)$ the number of odd component of G.

**Theorem** **10** ([

33]).

A graph $G=(V,E)$ has a perfect matching if and only if $o(G-S)\le |S|$ for all $S\subseteq V$.**Lemma** **25.** Let $n\ge 8$ and $2\le k<n$. Then, ${t}_{1}({A}_{n,k})\ge (2k-1)(n-k)$ under the MM${}^{*}$ model.

**Proof.** By the definition of 1-good-neighbor diagnosability, it is sufficient to show that ${A}_{n,k}$ is 1-good-neighbor $(2k-1)(n-k)$-diagnosable.

By Theorem 7, suppose, on the contrary, that there are two distinct 1-good-neighbor faulty subsets ${F}_{1}$ and ${F}_{2}$ of ${A}_{n,k}$ with $|{F}_{1}|\le (2k-1)(n-k)$ and $|{F}_{2}|\le (2k-1)(n-k)$, but the vertex set pair $({F}_{1},{F}_{2})$ is not satisfied with any condition in Theorem 7. Without loss of generality, suppose that ${F}_{2}\backslash {F}_{1}\ne \mathsf{\varnothing}.$ Assume $V({A}_{n,k})={F}_{1}\cup {F}_{2}$. We have that $\frac{n!}{(n-k)!}=|V({A}_{n,k})|=|{F}_{1}\cup {F}_{2}|=|{F}_{1}|+|{F}_{2}|-|{F}_{1}\cap {F}_{2}|\le |{F}_{1}|+|{F}_{2}|\le 2(2k-1)(n-k)$. When $k=2$, ${n}^{2}-n=\frac{n!}{(n-2)!}=|V({A}_{n,2})|=|{F}_{1}\cup {F}_{2}|\le 6n-12$, a contradiction to $n\ge 5$. Therefore, $V({A}_{n,2})\ne {F}_{1}\cup {F}_{2}$. When $k=3$, $\frac{n!}{(n-k)!}={n}^{3}-3{n}^{2}+2n$. Note ${n}^{3}-3{n}^{2}+2n\le \frac{n!}{(n-k)!}$ for $k\ge 3$. Thus, $2(2k-1)(n-k)\le 2(2(n-1)-1)(n-3)\le 4{n}^{2}-18n+18$. In fact, ${n}^{3}-3{n}^{2}+2n>4{n}^{2}-18n+18$ when $n\ge 5$. This is a contradiction. Therefore, $V({A}_{n,k})\ne {F}_{1}\cup {F}_{2}$.

Claim 1.${A}_{n,k}-{F}_{1}-{F}_{2}$ has no isolated vertex.

Suppose, on the contrary, that ${A}_{n,k}-{F}_{1}-{F}_{2}$ has at least one isolated vertex w. Since ${F}_{1}$ is a 1-good-neighbor faulty set, there is a vertex $u\in {F}_{2}\backslash {F}_{1}$ such that u is adjacent to w. Since the vertex set pair $({F}_{1},{F}_{2})$ is not satisfied with any condition in Theorem 7, there is at most one vertex $u\in {F}_{2}\backslash {F}_{1}$ such that u is adjacent to w. Thus, there is just a vertex $u\in {F}_{2}\backslash {F}_{1}$ such that u is adjacent to w. Similarly, we can show that there is just a vertex $v\in {F}_{1}\backslash {F}_{2}$ such that v is adjacent to w when ${F}_{1}\backslash {F}_{2}\ne \mathsf{\varnothing}$. Suppose ${F}_{1}\backslash {F}_{2}=\mathsf{\varnothing}$. Then, ${F}_{1}\subseteq {F}_{2}$. Since ${F}_{2}$ is a 1-good neighbor faulty set, ${A}_{n,k}-{F}_{2}={A}_{n,k}-{F}_{1}-{F}_{2}$ has no isolated vertex. Therefore, ${F}_{1}\backslash {F}_{2}\ne \mathsf{\varnothing}$ as follows. Let $W\subseteq V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ be the set of isolated vertices in ${A}_{n,k}[V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})]$, and let H be the subgraph induced by the vertex set $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2}\cup W)$. Then, for any $w\in W$, there are $(k(n-k)-2)$ neighbors in ${F}_{1}\cap {F}_{2}$. Since $|V({A}_{n,k})|$ is even and Lemma 24, ${A}_{n,k}$ has a perfect matching. By Theorem 10, $|W|\le o(G-({F}_{1}\cup {F}_{2}))\le |{F}_{1}\cup {F}_{2}|\le |{F}_{1}|+|{F}_{2}|-|{F}_{1}\cap {F}_{2}|\le 2(2k-1)(n-k)-(k(n-k)-2)=(n-k)(3k-2)+2\le 3{n}^{2}-11n+12$. In particular, $|W|\le 4n-6$ when $k=2$. When $k=2$, ${n}^{2}-n=|V({A}_{n,2})|=|{F}_{1}\cup {F}_{2}|+|W|\le 2(4n-6)=8n-12$. This is a contradiction to $n\ge 8$. Thus, $V(H)\ne \mathsf{\varnothing}$. When $k=3$, $\frac{n!}{(n-k)!}={n}^{3}-3{n}^{2}+2n$. Note ${n}^{3}-3{n}^{2}+2n\le \frac{n!}{(n-k)!}$ for $k\ge 3$. Note that ${n}^{3}-3{n}^{2}+2n=|V({A}_{n,k})|=|{F}_{1}\cup {F}_{2}|+|W|\le 2(3{n}^{2}-11n+12)=6{n}^{2}-22n+24$. This is a contradiction to $n\ge 8$. Thus, $V(H)\ne \mathsf{\varnothing}$. Since the vertex set pair $({F}_{1},{F}_{2})$ is not satisfied with the condition (1) of Theorem 7, and any vertex of $V(H)$ is not isolated in H, we show that there is no edge between $V(H)$ and ${F}_{1}\u2206{F}_{2}$. Thus, ${F}_{1}\cap {F}_{2}$ is a vertex cut of ${A}_{n,k}$ and $\delta ({A}_{n,k}-({F}_{1}\cap {F}_{2}))\ge 1$, i.e., ${F}_{1}\cap {F}_{2}$ is a 1-good-neighbor cut of ${A}_{n,k}$. By Lemma 10, $|{F}_{1}\cap {F}_{2}|\ge (2k-1)(n-k)-1$. Because $|{F}_{1}|\le (2k-1)(n-k)$, $|{F}_{2}|\le (2k-1)(n-k)$, and neither ${F}_{1}\backslash {F}_{2}$ nor ${F}_{2}\backslash {F}_{1}$ is empty, we have $|{F}_{1}\backslash {F}_{2}|=|{F}_{2}\backslash {F}_{1}|=1$. Let ${F}_{1}\backslash {F}_{2}=\left\{{v}_{1}\right\}$ and ${F}_{2}\backslash {F}_{1}=\left\{{v}_{2}\right\}$. Then, for any vertex $w\in W$, w are adjacent to ${v}_{1}$ and ${v}_{2}$. Suppose that ${v}_{1}$ is adjacent to ${v}_{2}$. Then, ${v}_{1}{v}_{2}v{v}_{1}$ is a three-cycle and $|N(\{{v}_{1},{v}_{2,}v\})|=3[(k-1)(n-k)-1]+n-k+1>(2k-1)(n-k)-1\ge |{F}_{1}\cap {F}_{2}|$, a contradiction. Therefore, suppose that ${v}_{1}$ is not adjacent to ${v}_{2}$. According to Lemma 9, there are at most two common neighbors for any pair of vertices in ${A}_{n,k}$, it follows that there are at most three isolated vertices in ${A}_{n,k}-{F}_{1}-{F}_{2}$, i.e., $|W|\le 2$.

Suppose that there is exactly one isolated vertex v in ${A}_{n,k}-{F}_{1}-{F}_{2}$. Let ${v}_{1}$ and ${v}_{2}$ be adjacent to v. Then, ${N}_{{A}_{n,k}}(v)\backslash \{{v}_{1},{v}_{2}\}\subseteq {F}_{1}\cap {F}_{2}$ and $|{N}_{{A}_{n,k}}(v)\cap ({F}_{1}\cap {F}_{2})|=k(n-k)-2$. Note that $|{N}_{{A}_{n,k}}({v}_{1})\cap ({F}_{1}\cap {F}_{2})|=k(n-k)-1$ and $|{N}_{{A}_{n,k}}({v}_{2})\cap ({F}_{1}\cap {F}_{2})|=k(n-k)-1$. By Lemma 9, $|{F}_{1}\cap {F}_{2}|\ge k(n-k)-2+k(n-k)-1+k(n-k)-1-2(n-k-1)-2=(3k-2)(n-k)-4.$ It follows that $|{F}_{2}|=|{F}_{2}\backslash {F}_{1}|+|{F}_{1}\cap {F}_{2}|\ge 1+(3k-2)(n-k)-4=(3k-2)(n-k)-3>(2k-1)(n-k)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}(n\ge 8)$, which contradicts $|{F}_{2}|\le (2k-1)(n-k)$.

Suppose that there are exactly two isolated vertices

v and

w in

${A}_{n,k}-{F}_{1}-{F}_{2}$. Let

${v}_{1}$ and

${v}_{2}$ be adjacent to

v and

w, respectively. Since

${v}_{1},{v}_{2}\in {N}_{{A}_{n,k}}(\{v,w\})$, by Lemma 9,

$|{N}_{{A}_{n,k}}(\{v,w\})\cap ({F}_{1}\cap {F}_{2})|=2(k(n-k)-2)$. Note that

$|{F}_{1}\cap {F}_{2}|\le (2k-1)(n-k)-1$. If

$n>k+3$, then

$2(k(n-k)-2)>(2k-1)(n-k)-1$, a contradiction. Thus,

$n\le k+3$. Since

$n\ge k+1$,

$k+1\le n\le k+3$. If

$n=k+1$, then, by Lemma 9, a contradiction to

$|W|=2$. Suppose that

$n=k+2$. Then,

${A}_{n,k}={A}_{n,n-2}$. By the proof of Lemma 3.2 ([

18]),

${A}_{n,n-2}-{F}_{1}-{F}_{2}$ has no isolated vertex. Suppose that

$n=k+3$. Then,

$2(k(n-k)-2)=(2k-1)(n-k)-1=6n-22$. By Lemma 1, let

${v}_{1}=(1,2,\dots ,n-4,n-3)$.Without loss of generality, suppose

$v=(1,2,\dots ,n-4,n)$ and

$w=(n,2,\dots ,n-4,n-3)$. Then, the vertex

${v}^{\prime}=(1,n-1,\dots ,n-4,n-3)$ is not adjacent to

v and

w. Thus,

$|{F}_{1}\cap {F}_{2}|>(2k-1)(n-k)-1$, a contradiction. The proof of Claim 1 is complete.

Let $u\in V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$. By Claim 1, u has at least one neighbor in ${A}_{n,k}-{F}_{1}-{F}_{2}$. Since the vertex set pair $({F}_{1},{F}_{2})$ is not satisfied with any condition in Theorem 7, by the condition (1) of Theorem 7, for any pair of adjacent vertices $u,w\in V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$, there is no vertex $v\in {F}_{1}\u2206{F}_{2}$ such that $uw\in E({A}_{n,k})$ and $vw\in E({A}_{n,k})$. It follows that u has no neighbor in ${F}_{1}\u2206{F}_{2}$. Since u is taken arbitrarily, there is no edge between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$. Since ${F}_{2}\backslash {F}_{1}\ne \mathsf{\varnothing}$ and ${F}_{1}$ is a 1-good-neighbor faulty set, ${\delta}_{{A}_{n,k}}([{F}_{2}\backslash {F}_{1}])\ge 1$ and $|{F}_{2}\backslash {F}_{1}|\ge 2$. Since both ${F}_{1}$ and ${F}_{2}$ are 1-good-neighbor faulty sets, and there is no edge between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$, ${F}_{1}\cap {F}_{2}$ is a 1-good-neighbor cut of ${A}_{n,k}$. By Lemma 10, we have $|{F}_{1}\cap {F}_{2}|\ge (2k-1)(n-k)-1$. Therefore, $|{F}_{2}|=|{F}_{2}\backslash {F}_{1}|+|{F}_{1}\cap {F}_{2}|\ge 2+((2k-1)(n-k)-1)=(2k-1)(n-k)+1$, which contradicts $|{F}_{2}|\le (3k-2)(n-k)$. Therefore, ${A}_{n,k}$ is 1-good-neighbor $(3k-2)(n-k)$-diagnosable and ${t}_{1}({A}_{n,k})\ge (3k-2)(n-k)$. The proof is complete. □

Combining Lemmas 22 and 25, we have the following theorem.

**Theorem** **11.** Let $n\ge 8$. Then, ${t}_{1}({A}_{n,k})=(2k-1)(n-k)$ under the MM${}^{*}$ model.

**Lemma** **26.** Let $n\ge 8$ and $k\in \{i:i=3,\dots ,n-5\}\cup \{n-2,n-1\}$. Then, ${t}_{2}({A}_{n,k})\ge (3k-2)(n-k)$ under the MM${}^{*}$ model.

**Proof.** By the definition of the 2-good-neighbor diagnosability, it is sufficient to show that ${A}_{n,k}$ is g-good-neighbor $(3k-2)(n-k)$-diagnosable.

By Theorem 7, suppose, on the contrary, that there are two distinct g-good-neighbor faulty subsets ${F}_{1}$ and ${F}_{2}$ of ${A}_{n,k}$ with $|{F}_{1}|\le (3k-2)(n-k)$ and $|{F}_{2}|\le (3k-2)(n-k)$, but the vertex set pair $({F}_{1},{F}_{2})$ is not satisfied with any condition in Theorem 7. Without loss of generality, suppose that ${F}_{2}\backslash {F}_{1}\ne \mathsf{\varnothing}.$ Similar to the discussion on $V({A}_{n,k})={F}_{1}\cup {F}_{2}$ in Lemma 18, we have $V({A}_{n,k})\ne {F}_{1}\cup {F}_{2}$.

Claim 1.${A}_{n,k}-{F}_{1}-{F}_{2}$ has no isolated vertex.

Since ${F}_{1}$ is a 2-good neighbor faulty set, for an arbitrary vertex $u\in V({A}_{n,k})\backslash {F}_{1}$, $|{N}_{{A}_{n,k}-{F}_{1}}(u)|\ge 2$. Suppose, on the contrary, that ${A}_{n,k}-{F}_{1}-{F}_{2}$ has at least one isolated vertex x. Since ${F}_{1}$ is a 2-good neighbor faulty set, there are at least two vertices $u,v\in {F}_{2}\backslash {F}_{1}$ such that $u,v$ are adjacent to x. According to the hypothesis, the vertex set pair $({F}_{1},{F}_{2})$ is not satisfied with any condition in Theorem 7, by the condition (3) of Theorem 7, a contradiction. Therefore, there are at most one vertex $u\in {F}_{2}\backslash {F}_{1}$ such that u are adjacent to x. Thus, $|{N}_{{A}_{n,k}-{F}_{1}}(x)|=1$, a contradiction to that ${F}_{1}$ is a 2-good neighbor faulty set. Thus, ${A}_{n,k}-{F}_{1}-{F}_{2}$ has no isolated vertex. The proof of Claim 1 is complete.

Let $u\in V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$. By Claim 1, $\delta ({A}_{n,k}-{F}_{1}-{F}_{2})\ge 1$. Since the vertex set pair $({F}_{1},{F}_{2})$ is not satisfied with any condition in Theorem 7, by the condition (1) of Theorem 7, for any pair of adjacent vertices $u,w\in V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$, there is no vertex $v\in {F}_{1}\u2206{F}_{2}$ such that $uw\in E({A}_{n,k})$ and $uv\in E({A}_{n,k})$. It follows that u has no neighbor in ${F}_{1}\u2206{F}_{2}$. Since u is taken arbitrarily, there is no edge between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$.

Since ${F}_{2}\backslash {F}_{1}\ne \mathsf{\varnothing}$ and ${F}_{1}$ is a 2-good-neighbor faulty set, we have that ${\delta}_{{A}_{n,k}}([{F}_{2}\backslash {F}_{1}])\ge 2$, $\delta ({A}_{n,k}-{F}_{2}-{F}_{1})\ge 2$ and $|{F}_{2}\backslash {F}_{1}|\ge 2+1=3$. Since both ${F}_{1}$ and ${F}_{2}$ are 2-good-neighbor faulty sets, and there is no edge between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$, ${F}_{1}\cap {F}_{2}$ is a 2-good-neighbor cut of ${A}_{n,k}$. By Lemma 11, we have $|{F}_{1}\cap {F}_{2}|\ge (3k-2)(n-k)-2$. Therefore, $|{F}_{2}|=|{F}_{2}\backslash {F}_{1}|+|{F}_{1}\cap {F}_{2}|\ge 3+(3k-2)(n-k)-2=(3k-2)(n-k)+1$, which contradicts $|{F}_{2}|\le (3k-2)(n-k)$. Therefore, ${A}_{n,k}$ is 2-good-neighbor $(3k-2)(n-k)$-diagnosable and ${t}_{2}({A}_{n,k})\ge (3k-2)(n-k)$. The proof is complete. □

Combining Lemmas 22 and 26, we have the following theorem.

**Theorem** **12.** Let $n\ge 8$ and $k\in \{i:i=3,\dots ,n-5\}\cup \{n-2,n-1\}$. Then, ${t}_{2}({A}_{n,k})=(3k-2)(n-k)$ under the MM${}^{*}$ model.

**Lemma** **27.** For $n\ge 8$, ${t}_{2}({A}_{n,2})\le 4n-9$ under the MM${}^{*}$ model.

**Proof.** Let X be defined in Lemma 19, and let ${F}_{1}={N}_{{A}_{n,2}}(X)$, ${F}_{2}=X\cup {N}_{{A}_{n,2}}(X)$. By Lemma 19, $|{F}_{1}|=4n-12$, $|{F}_{2}|=|X|+|{F}_{1}|=4n-8$, $\delta ({A}_{n,2}-{F}_{1})\ge 2$ and $\delta ({A}_{n,2}-{F}_{2})\ge 2$. Therefore, ${F}_{1}$ and ${F}_{2}$ are 2-good-neighbor faulty sets of ${A}_{n,2}$ with $|{F}_{1}|=4n-12$ and $|{F}_{2}|=4n-8$.

We will prove ${A}_{n,2}$ is not 2-good-neighbor $(4n-8)$-diagnosable. Since $X={F}_{1}\u2206{F}_{2}$ and ${N}_{{A}_{n,k}}(X)={F}_{1}\subset {F}_{2}$, there is no edge of ${A}_{n,2}$ between $V({A}_{n,2})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$. By Theorem 7, we show that ${A}_{n,2}$ is not 2-good-neighbor $(4n-8)$-diagnosable under the MM${}^{*}$ model. Hence, by the definition of the 2-good-neighbor diagnosability, we show that the 2-good-neighbor diagnosability of ${A}_{n,2}$ is less than $4n-8$, i.e., ${t}_{g}({A}_{n,2})\le 4n-9$. □

**Lemma** **28.** For $n\ge 8$, ${t}_{2}({A}_{n,2})\ge 4n-9$ under the MM${}^{*}$ model.

**Proof.** By the definition of the 2-good-neighbor diagnosability, it is sufficient to show that ${A}_{n,k}$ is 2-good-neighbor $(4n-9)$-diagnosable.

By Theorem 7, suppose, on the contrary, that there are two distinct 2-good-neighbor faulty subsets ${F}_{1}$ and ${F}_{2}$ of ${A}_{n,k}$ with $|{F}_{1}|\le 4n-9$ and $|{F}_{2}|\le 4n-9$, but the vertex set pair $({F}_{1},{F}_{2})$ is not satisfied with any condition in Theorem 7. Without loss of generality, suppose that ${F}_{2}\backslash {F}_{1}\ne \mathsf{\varnothing}.$ Similar to the discussion on $V({A}_{n,k})={F}_{1}\cup {F}_{2}$ in Lemma 21, we have $V({A}_{n,k})\ne {F}_{1}\cup {F}_{2}$.

Claim 1.${A}_{n,k}-{F}_{1}-{F}_{2}$ has no isolated vertex.

Since ${F}_{1}$ is a 2-good neighbor faulty set, for an arbitrary vertex $u\in V({A}_{n,k})\backslash {F}_{1}$, $|{N}_{{A}_{n,k}-{F}_{1}}(u)|\ge 2$. Suppose, on the contrary, that ${A}_{n,k}-{F}_{1}-{F}_{2}$ has at least one isolated vertex x. Since ${F}_{1}$ is a 2-good neighbor faulty set, there are at least two vertices $u,v\in {F}_{2}\backslash {F}_{1}$ such that $u,v$ are adjacent to x. According to the hypothesis, the vertex set pair $({F}_{1},{F}_{2})$ is not satisfied with any condition in Theorem 7, by the condition (3) of Theorem 7, a contradiction. Therefore, there are at most one vertex $u\in {F}_{2}\backslash {F}_{1}$ such that u are adjacent to x. Thus, $|{N}_{{A}_{n,k}-{F}_{1}}(x)|=1$, a contradiction to that ${F}_{1}$ is a 2-good neighbor faulty set. Thus, ${A}_{n,k}-{F}_{1}-{F}_{2}$ has no isolated vertex. The proof of Claim 1 is complete.

Let $u\in V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$. By Claim 1, $\delta ({A}_{n,k}-{F}_{1}-{F}_{2})\ge 1$. Since the vertex set pair $({F}_{1},{F}_{2})$ is not satisfied with any condition in Theorem 7, by the condition (1) of Theorem 7, for any pair of adjacent vertices $u,w\in V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$, there is no vertex $v\in {F}_{1}\u2206{F}_{2}$ such that $uw\in E({A}_{n,k})$ and $uv\in E({A}_{n,k})$. It follows that u has no neighbor in ${F}_{1}\u2206{F}_{2}$. Since u is taken arbitrarily, there is no edge between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$.

Since ${F}_{2}\backslash {F}_{1}\ne \mathsf{\varnothing}$ and ${F}_{1}$ is a 2-good-neighbor faulty set, we have that ${\delta}_{{A}_{n,k}}([{F}_{2}\backslash {F}_{1}])\ge 2$, $\delta ({A}_{n,k}-{F}_{2}-{F}_{1})\ge 2$ and $|{F}_{2}\backslash {F}_{1}|\ge 2+1=3$. Since both ${F}_{1}$ and ${F}_{2}$ are 2-good-neighbor faulty sets, and there is no edge between $V({A}_{n,k})\backslash ({F}_{1}\cup {F}_{2})$ and ${F}_{1}\u2206{F}_{2}$, ${F}_{1}\cap {F}_{2}$ is a 2-good-neighbor cut of ${A}_{n,k}$. By Lemma 11, we have $|{F}_{1}\cap {F}_{2}|\ge 4n-12$. If $|{F}_{1}\cap {F}_{2}|=4n-12$, then, by Lemma 12, $|{F}_{2}\backslash {F}_{1}|=4$. If $|{F}_{1}\cap {F}_{2}|=4n-11$ or $4n-10$, then $|{F}_{2}\backslash {F}_{1}|\le 2$, a contradiction to that $\delta ({A}_{n,2}[{F}_{1}\backslash {F}_{2}])\ge 2$. Therefore, $|{F}_{2}|=|{F}_{2}\backslash {F}_{1}|+|{F}_{1}\cap {F}_{2}|\ge 4+(4n-12)=4n-8$, which contradicts with that $|{F}_{2}|\le 4n-9$. Therefore, ${A}_{n,k}$ is 2-good-neighbor $(4n-9)$-diagnosable and ${t}_{2}({A}_{n,k})\ge 4n-9$. The proof is complete. □

Combining Lemmas 27 and 28, we have the following theorem.

**Theorem** **13.** Let $n\ge 8$. Then, ${t}_{2}({A}_{n,2})=4n-9$ under the MM${}^{*}$ model.