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Diagnosability of a multiprocessor system G is one important measure of the reliability of interconnection networks. In 2016, Zhang
<i>
et al</i>.
proposed the
<i>
g</i>-
extra diagnosability of
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G</i>
, which restrains that every component of
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G – S</i>
has at least
(g + 1)
vertices. The
locally twisted cube
<i>
LTQ<sub>n</sub></i>
is applie
d widely. In this paper, we show that
<i>
LTQ<sub>n</sub></i>
is tightly
(4n – 9)
super 3-extra connected for
n
≥
6
and the 3-extra diagnosability of
<i>
LTQ<sub>n</sub></i>
under the PMC model and MM^{*} model is
4n - 6
for
n
≥
5
and
n
≥
7
, respectively.

At present, semiconductor technology has been widely applied in various fields of large-scale computer systems. But processors or communication links failures of a multiprocessor system give our live a lot of troubles. How to find out the faulty processors accurately and timely becomes the primary problem when the system is in operation. The diagnosis of the system is the process of identifying the faulty processors from the fault-free ones.

There are two well-known diagnosis models, one is the PMC diagnosis model, introduced by Preparata et al. [^{*} model, in which each node must test another two neighbors.

In 1996, the g-extra connectivity κ ˜ ( g ) ( G ) of an interconnection network G was introduced by Fàbrega and Fiol [

In 2012, Peng et al. [^{*} model. In 2016, Zhang et al. [^{*} model. Ren et al. [^{*} model. In 2017, Wang and Yang [^{*} model.

In this paper, we show that L T Q n is tightly ( 4 n − 9 ) super 3-extra con- nected for n ≥ 6 and the 3-extra diagnosability of L T Q n under the PMC model and MM^{*} model is 4 n − 6 for n ≥ 5 and n ≥ 7 , respectively.

A multiprocessor system is modeled as an undirected simple graph G = ( V , E ) , whose vertices (nodes) represent processors and edges (links) represent com- munication links. Suppose that V ′ is a nonempty vertex subset of V. The in- duced subgraph by V ′ in G, denoted by G [ V ′ ] , is a graph, whose vertex set is V ′ and whose edge set consists of all the edges of G with both endpoints in V ′ . The degree d G ( v ) of a vertex v in G is the number of edges incident with v. We denote by δ ( G ) the minimum degree of vertices of G. For any vertex v, we define the neighborhood N G ( v ) of v in G to be the set of vertices adjacent to v. u is called a neighbor vertex or a neighbor of v for u ∈ N G ( v ) . Let S ⊆ V ( G ) . We denote by N G ( S ) the set ∪ v ∈ S N G ( v ) \ S . For neighborhoods and degrees, we will usually omit the subscript for the graph when no confusion arises. A graph G is said to be k-regular if d G ( v ) = k for any vertex v ∈ V . A bipartite graph is one whose each edge has one end in subsets of vertex X and one end in subsets of vertex Y; such a partition ( X , Y ) is called a bipartition of the graph. A complete bipartite graph is a simple bipartite graph with bipartition ( X , Y ) in which each vertex of X is joined to each vertex of Y; if | X | = m and | Y | = n , such a graph is denoted by K m , n . The connectivity κ ( G ) of a connected graph G is the minimum number of vertices whose removal results in a disconnected graph or only one vertex left. Let F 1 and F 2 be two distinct subsets of V, and let the symmetric difference F 1 Δ F 2 = ( F 1 \ F 2 ) ∪ ( F 2 \ F 1 ) . For graph-theoretical terminology and notation not defined here we follow [

Let G = ( V , E ) be a connected graph. A faulty set F ⊆ V is called a g- good-neighbor faulty set if | N ( v ) ∩ ( V \ F ) | ≥ g for every vertex v in V \ 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 κ ( g ) ( G ) . A faulty set F ⊆ V is called a g-extra faulty set if every component of G − F has at least ( g + 1 ) vertices. A g-extra cut of G is a g-extra faulty set F such that G − F is dis- connected. The minimum cardinality of g-extra cuts is said to be the g-extra connectivity of G, denoted by κ ˜ ( g ) ( G ) .

Proposition 1. ( [

Proposition 2. ( [

Definition 3. ( [

For the PMC model and MM^{*} model, we follow [

Similar to the PMC model, we can define a subset of vertices F ⊆ V ( G ) is consistent with a given syndrome σ * and two distinct sets F 1 and F 2 in V ( G ) are indistinguishable (resp. distinguishable) under the MM^{*} model.

In a system G = ( V , E ) , a faulty set F ⊆ V is called a g-extra faulty set if every component of G − F has more than g nodes. G is g-extra t-diagnosable if and only if for each pair of distinct faulty g-extra vertex subsets F 1 , F 2 ⊆ V ( G ) such that | F i | ≤ t , F 1 and F 2 are distinguishable. The g-extra diagnosability of G, denoted by t ˜ g ( G ) , is the maximum value of t such that G is g-extra t- diagnosable.

Proposition 4. [

For an integer n ≥ 1 , a binary string of length n is denoted by u 1 u 2 ⋯ u n , where u i ∈ { 0,1 } for any integer i ∈ { 1,2, ⋯ , n } . The n-dimensional locally twisted cube, denoted by L T Q n , is an n-regular graph of 2 n vertices and n 2 n − 1 edges, which can be recursively defined as follows [

Definition 5. ( [

1) L T Q 2 is a graph consisting of four nodes labeled with 00, 01, 10 and 11, respectively, connected by four edges {00, 01}, {01, 11}, {11, 10} and {10, 00}.

2) For n ≥ 3 , L T Q n is built from two disjoint copies of L T Q n − 1 according to the following steps. Let 0 L T Q n − 1 denote the graph obtained from one copy of L T Q n − 1 by prefixing the label of each node with 0. Let 1 L T Q n − 1 denote the graph obtained from the other copy of L T Q n − 1 by prefixing the label of each node with 1. Connect each node 0 u 2 u 3 ⋯ u n of 0 L T Q n − 1 to the node 1 ( u 2 + u n ) u 3 ⋯ u n of 1 L T Q n − 1 with an edge, where “+” represents the modulo 2 addition.

The edges whose end vertices in different i L T Q n − 1 s are called to be cross- edges. Figures 1-3 show four examples of locally twisted cubes. The locally twisted cube can also be equivalently defined in the following non-recursive fashion.

Definition 6. ( [

1) u i = v i ¯ and u i + 1 = ( v i + 1 + v n ) ( m o d 2 ) for some 1 ≤ i ≤ n − 2 , n ≥ 3 and u j = v j for all the remaining bits;

2) u i = v i ¯ for i ∈ { n − 1, n } , n ≥ 2 and u j = v j for all the remaining bits.

Proposition 7. ( [

Lemma 1. ( [

Lemma 2. ( [

Lemma 3. ( [

Lemma 4. ( [

Lemma 5. Let L T Q n be the locally twisted cube. If P = u v w x is a 3-path in L T Q n and u x ∉ E ( L T Q n ) for n ≥ 3 , | N ( V ( P ) ) | ≥ 4 n − 9 .

Proof. We decompose L T Q n into 0 L T Q n − 1 and 1 L T Q n − 1 . Then 0 L T Q n − 1 and 1 L T Q n − 1 are isomorphic to L T Q n − 1 . Without loss of generality, we have the following cases.

Case 1. u , x ∈ V ( 0 L T Q n − 1 ) and v , w ∈ V ( 1 L T Q n − 1 ) .

Since u ∈ V ( 0 L T Q n − 1 ) , v ∈ V ( 1 L T Q n − 1 ) and u , v are adjacent, by Propo- sition 7, u , v have no the common neighbor vertex. Similarly, x , w have no the common neighbor vertex and v , w have no the common neighbor vertex. Since u ∈ V ( 0 L T Q n − 1 ) , w ∈ V ( 1 L T Q n − 1 ) , u , w are not adjacent, v is a com- mon neighbor vertex of u , w , x ∈ V ( 0 L T Q n − 1 ) and x is a neighbor vertex of w, by Lemma 3, | ( N ( u ) ∩ N ( w ) ) \ { v } | = 0 . Similarly, | ( N ( x ) ∩ N ( v ) ) \ { w } | = 0 . Since u and x are not adjacent, by proposition 7, | N ( u ) ∩ N ( x ) | ≤ 2 . Therefore, | N ( V ( P ) ) | ≥ 2 ( n − 1 ) + 2 ( n − 2 ) − 2 = 4 n − 8 .

Case 2. u ∈ V ( 0 L T Q n − 1 ) and v , w , x ∈ V ( 1 L T Q n − 1 ) .

Since u , v are adjacent, by Proposition 7, | N ( u ) ∩ N ( v ) | = 0 . Similarly, | N ( v ) ∩ N ( w ) | = 0 , | N ( x ) ∩ N ( w ) | = 0 . And since u ∈ V ( 0 L T Q n − 1 ) , w ∈ V ( 1 L T Q n − 1 ) , u , w are not adjacent and v is the common neighbor vertex of u and w, by Lemma 3, | ( N ( u ) ∩ N ( w ) ) \ { v } | ≤ 1 . Since u , x are not adjacent, u ∈ V ( 0 L T Q n − 1 ) , x ∈ V ( 1 L T Q n − 1 ) , by Lemma 3, | N ( u ) ∩ N ( x ) | ≤ 1 . Since w is the common neighbor vertex of v and x and v , x are not adjacent, by pro- position 7, | ( N ( v ) ∩ N ( x ) ) \ { w } | ≤ 1 . Therefore, | N ( P ) | ≥ 2 ( n − 1 ) + 2 ( n − 2 ) − 3 = 4 n − 9 .

Case 3. u , v ∈ V ( 0 L T Q n − 1 ) and w , x ∈ V ( 1 L T Q n − 1 ) .

Since u , v are adjacent, by Proposition 7, | N ( u ) ∩ N ( v ) | = 0 . Similarly, | N ( u ) ∩ N ( w ) | = 0 , | N ( w ) ∩ N ( x ) | = 0 . Since u ∈ V ( 0 L T Q n − 1 ) , x ∈ V ( 1 L T Q n − 1 ) and u, x are not adjacent, by proposition 7, | N ( u ) ∩ N ( x ) | ≤ 2 . If | ( N ( u ) ∩ N ( w ) ) \ { v } | = 1 , then, by Lemma 3, | N ( u ) ∩ N ( x ) | ≤ 1 . If | ( N ( u ) ∩ N ( w ) ) \ { v } | = 0 , then, by Lemma 3, | N ( u ) ∩ N ( x ) | ≤ 2 . Therefore, | N ( V ( P ) ) | ≥ 2 ( n − 1 ) + 2 ( n − 2 ) − 2 = 4 n − 8 .

Case 4. u , v , w , x ∈ V ( 1 L T Q n − 1 ) .

This case is clear.

In conclusion, | N ( V ( P ) ) | ≥ 4 n − 9 .

Lemma 6. Let L T Q n be the locally twisted cube. If L T Q n [ { u , v , w , x } ] is isomorphic to K 1,3 for n ≥ 3 and d ( u ) = 3 , then | N ( V ( L T Q n [ { u , v , w , x } ] ) ) | ≥ 4 n − 9 .

Proof. Since d ( u ) = 3 and L T Q n [ { u , v , w , x } ] is isomorphic to K 1,3 , we have d ( v ) = 1 , d ( w ) = 1 and d ( x ) = 1 . Since v , w are not adjacent and u is a common neighbor vertex of v, w, by Proposition 7, | ( N ( v ) ∩ N ( w ) ) \ { u } | ≤ 1 . Similarly, | ( N ( v ) ∩ N ( x ) ) \ { u } | ≤ 1 , | ( N ( w ) ∩ N ( x ) ) \ { u } | ≤ 1 . Therefore, | N ( V ( L T Q n [ { u , v , w , x } ] ) ) | ≥ 3 ( n − 1 ) + ( n − 3 ) − 3 = 4 n − 9 .

If L T Q n [ { u , v , w , x } ] is a 4-cycle, then | N ( V ( L T Q n [ { u , v , w , x } ] ) ) | = 4 n − 8 . Combining this with Lemmas 5 and 6, we have the following corollary.

Corollary 1. Let L T Q n be the locally twisted cube and let H be a connected subgraph of L T Q n . If | V ( H ) | ≥ 4 , then | N ( V ( H ) ) | ≥ 4 n − 9 .

Lemma 7. Let A = { 0 ⋯ 0001,0 ⋯ 0111,0 ⋯ 0101,0 ⋯ 0100 } and let L T Q n be the locally twisted cube with n ≥ 4 . If F 1 = N L T Q n ( A ) , F 2 = F 1 ∪ A , where n ≥ 4 , then | F 1 | = 4 n − 9 , | F 2 | = 4 n − 5 , F 1 is a 3-extra cut of L T Q n , L T Q n − F 1 has two components L T Q n − F 2 and L T Q n [ A ] , | V ( L T Q n − F 2 ) | ≥ 4 , and | A | ≥ 4 .

Proof. According to the definition, L T Q n [ A ] is a 3-path and | A | = 4 . By Lemma 5, | F 1 | ≥ 4 n − 9 . From

To prove L T Q n − F 2 has two components and | V ( L T Q n − F 2 ) | ≥ 4 , we have the following discussion.

Claim 1. L T Q n − F 2 is connected for n ≥ 4 .

The proof is by induction on n. For n = 4 , A = { 0001 , 0111 , 0101 , 0100 } , F 1 = { 0000 , 0011 , 0110 , 1001 , 1011 , 1101 , 1100 } . It is easy to see that L T Q 4 − F 2 is connected (See

F 2 1 = { 11001 , 11110 , 11111 , 10100 } (See

By Claim 1, L T Q n − F 1 has two components L T Q n − F 2 and L T Q n [ A ] for n ≥ 4 . Then | V ( L T Q n − F 2 ) | = 2 n − ( 4 n − 5 ) ≥ 4 for n ≥ 4 . And since | A | = 4 , F 1 is a 3-extra cut of L T Q n .

Lemma 8. ( [

1) L T Q n − F has three components, two of which are isolated vertices;

2) L T Q n − F has two components, one of which is an isolated vertex;

3) L T Q n − F has two components, one of which is a K 2 ;

4) L T Q n − F is connected.

Theorem 8. ( [

Lemma 9. Let L T Q n be the locally twisted cube. If | F | = 10 for n = 5 , then L T Q 5 − F satisfies one of the following conditions:

1) L T Q 5 − F has four components, three of which are isolated vertices;

2) L T Q 5 − F has three components, one of which is isolated vertices and one of which is a K 2 ;

3) L T Q 5 − F has three components, two of which are isolated vertices;

4) L T Q 5 − F has two components, one of which is a path of length two;

5) L T Q 5 − F has two components, one of which is an isolated vertex;

6) L T Q 5 − F has two components, one of which is a K 2 ;

7) L T Q 5 − F is connected.

Proof. We decompose L T Q 5 into 0 L T Q 4 and 1 L T Q 4 . Then 0 L T Q 4 and 1 L T Q 4 are isomorphic to L T Q 4 . Suppose that F i = F ∩ V ( i L T Q 4 ) , i ∈ { 0,1 } . Without loss of generality, let | F 0 | ≥ | F 1 | . And since | F | = 10 , 5 ≤ | F 0 | ≤ 10 , 0 ≤ | F 1 | ≤ 5 . Let C i be the maximum component of i L T Q 4 − F i , i ∈ { 0,1 } . We consider the following cases.

Case 1. | F 0 | = 5 .

Since | F 0 | = 5 and | F | = 10 , | F 1 | = 10 − 5 = 5 . By Lemmas 1 and 2, both 0 L T Q 4 − F 0 and 1 L T Q 4 − F 1 are connected or has two components, one of which is an isolated vertex. Since 2 5 − 1 − 6 − 2 ≥ 1 , by Lemma 3, L T Q n [ V ( C 0 ) ∪ V ( C 1 ) ] is connected. Thus, L T Q 5 − F satisfies one of con- ditions:

1) L T Q 5 − F has three components, two of which are isolated vertices;

2) L T Q 5 − F has two components, one of which is an isolated vertex;

3) L T Q 5 − F has two components, one of which is a K 2 ;

4) L T Q 5 − F is connected.

Case 2. | F 0 | = 6 .

Since | F 0 | = 6 and | F | = 10 , | F 1 | = 10 − 6 = 4 . By Lemmas 1 and 2, 1 L T Q 4 − F 1 is connected or has two components, one of which is an isolated vertex. Since | F 0 | = 6 , by Lemma 8, 0 L T Q 4 − F 0 satisfies one of the following conditions:

1) 0 L T Q 4 − F 0 has three components, two of which are isolated vertices;

2) 0 L T Q 4 − F 0 has two components, one of which is an isolated vertex;

3) 0 L T Q 4 − F 0 has two components, one of which is a K 2 ;

4) 0 L T Q 4 − F 0 is connected.

Then L T Q 5 − F satisfies one of the conditions (1)-(7).

Case 3. | F 0 | ≥ 7 .

Since | F 0 | ≥ 7 and | F | = 10 , | F 1 | ≤ 10 − 7 = 3 . By Lemma 1, 1 L T Q 4 − F 1 is connected.

Suppose that 0 L T Q 4 − F 0 is connected. Since 2 5 − 1 − 10 ≥ 1 , by Lemma 3, L T Q n − F is connected.

Suppose that 0 L T Q 4 − F 0 is not connected. Let the components in 0 L T Q 4 − F 0 be G 1 , G 2 , ⋯ , G k for k ≥ 2 and | V ( G 1 ) | ≤ | V ( G 2 ) | ≤ ⋯ ≤ | V ( G k ) | . If | V ( G r ) | ≥ 4 ( 1 ≤ r ≤ k − 1 ) , by Lemma 3, | N ( V ( G r ) ) ∩ V ( 1 L T Q 4 ) | ≥ 4 . Combining this with | F 1 | ≤ 3 , we have that L T Q 5 [ V ( G r ) ∪ V ( 1 L T Q 4 − F 1 ) ] is connected. Therefore, G r is not a component of L T Q 5 − F for | V ( G r ) | ≥ 4 . Therefore, L T Q 5 − F is connected. The following we discuss G r is a com- ponent of L T Q 5 − F with | V ( G r ) | ≤ 3 ( 1 ≤ r ≤ k − 1 ) .

If k = 5 , by Lemma 3, | N ( V ( G 1 ) ) ∪ N ( V ( G 2 ) ) ∪ ⋯ ∪ N ( V ( G k − 1 ) ) ∩ V ( 1 L T Q 4 ) | ≥ 4 . Combining this with | F 1 | ≤ 3 , there is one G r ( 1 ≤ r ≤ k − 1 ) such that L T Q 5 [ V ( G r ) ∪ V ( 1 L T Q 4 − F 1 ) ] is connected. Thus, k ≤ 4 . Since | F 1 | = 3 , k ≤ 4 , and | V ( G r ) | ≤ 3 ( 1 ≤ r ≤ k − 1 ) , L T Q 5 − F satisfies one of the conditions (1)-(7).

Lemma 10. Let L T Q n be the locally twisted cube. If 3 n − 5 ≤ | F | ≤ 4 n − 10 for n ≥ 5 , then L T Q n − F satisfies one of the following conditions:

1) L T Q n − F has four components, three of which are isolated vertices;

2) L T Q n − F has three components, one of which is isolated vertices and one of which is a K 2 ;

3) L T Q n − F has three components, two of which are isolated vertices;

4) L T Q n − F has two components, one of which is a path of length two;

5) L T Q n − F has two components, one of which is an isolated vertex;

6) L T Q n − F has two components, one of which is a K 2 ;

7) L T Q n − F is connected.

Proof. By Lemma 9, the result holds for n = 5 . We proceed by induction on n. Assume n ≥ 6 and the result holds for L T Q n − 1 , i.e., if 3 n − 5 ≤ | F | ≤ 4 ( n − 1 ) − 10 = 4 n − 14 , then L T Q n − 1 − F satisfies one of the con- ditions (1)-(7) in Lemma 10. The following we prove L T Q n − F satisfies one of the conditions (1)-(7).

We decompose L T Q n into 0 L T Q n − 1 and 1 L T Q n − 1 . Then 0 L T Q n − 1 and 1 L T Q n − 1 are isomorphic to L T Q n − 1 . Suppose that F i = F ∩ V ( i L T Q n − 1 ) , i ∈ { 0,1 } . Without loss of generality, let | F 0 | ≥ | F 1 | . And since

3 n − 5 ≤ | F | ≤ 4 n − 10 , n ≤ ⌈ 3 n − 5 2 ⌉ ≤ | F 0 | ≤ 4 n − 10 , 0 ≤ | F 1 | ≤ ⌊ 4 n − 10 2 ⌋ ≤ 2 n − 5 .

Let C i be the maximum component of i L T Q n − 1 − F i , i ∈ { 0,1 } . We consider the following cases.

Case 1. n ≤ | F 0 | ≤ 3 ( n − 1 ) − 6 = 3 n − 9 .

Since | F 0 | ≥ | F 1 | and | F | ≤ 4 n − 10 ,

( 4 n − 10 ) − ( 3 n − 9 ) = n − 1 ≤ | F 1 | ≤ ⌊ 4 n − 10 2 ⌋ = 2 n − 5 . By Lemmas 1 and 2,

1 L T Q n − 1 − F 1 is connected or has two components, one of which is an isolated vertex. Since n ≤ | F 0 | ≤ 3 ( n − 1 ) − 6 = 3 n − 9 , by lemma 8, 0 L T Q n − 1 − F 0 satisfies one of the following conditions: 1) 0 L T Q n − 1 − F 0 has three components, two of which are isolated vertices; 2) 0 L T Q n − 1 − F 0 has two components, one of which is an isolated vertex; 3) 0 L T Q n − 1 − F 0 has two components, one of which is a K 2 ; 4) 0 L T Q n − 1 − F 0 is connected. Since 2 n − 1 − ( 4 n − 10 ) − 3 ≥ 1 , by Lemma 3, L T Q n [ V ( C 0 ) ∪ V ( C 1 ) ] is connected. Thus, L T Q n − F satisfies one of con- ditions (1)-(7) in Lemma 10.

Case 2. 3 n − 8 ≤ | F 0 | ≤ 4 n − 14 .

Since | F 0 | ≥ | F 1 | and | F | ≤ 4 n − 10 , | F 1 | ≤ ( 4 n − 10 ) − ( 3 n − 8 ) = n − 2 . By Le- mma 1, 1 L T Q n − 1 − F 1 is connected. Since 3 n − 8 ≤ | F 0 | ≤ 4 n − 14 , according to inductive hypothesis, 0 L T Q n − 1 − F 0 satisfies one of the following conditions:

1) 0 L T Q n − 1 − F 0 has four components, three of which are isolated vertices;

2) 0 L T Q n − 1 − F 0 has three components, one of which is isolated vertices and one of which is a K 2 ;

3) 0 L T Q n − 1 − F 0 has three components, two of which are isolated vertices;

4) 0 L T Q n − 1 − F 0 has two components, one of which is a path of length two;

5) 0 L T Q n − 1 − F 0 has two components, one of which is an isolated vertex;

6) 0 L T Q n − 1 − F 0 has two components, one of which is a K 2 ;

7) 0 L T Q n − 1 − F 0 is connected.

Thus, L T Q n − F satisfies one of the conditions (1)-(7) in Lemma 10.

Case 3. 4 n − 13 ≤ | F 0 | ≤ 4 n − 10 .

Since 4 n − 13 ≤ | F 0 | ≤ 4 n − 10 and | F | ≤ 4 n − 10 , | F 1 | ≤ ( 4 n − 10 ) − ( 4 n − 13 ) = 3 . By Lemma 1, 1 L T Q n − 1 − F 1 is connected.

Suppose that 0 L T Q n − 1 − F 0 is connected. Since 2 n − 1 − ( 4 n − 10 ) ≥ 1 , by Le- mma 3, L T Q n − F is connected.

Suppose that 0 L T Q n − 1 − F 0 is not connected. Let the components in

0 L T Q n − 1 − F 0 be G 1 , G 2 , ⋯ , G k for k ≥ 2 and | V ( G 1 ) | ≤ | V ( G 2 ) | ≤ ⋯ ≤ | V ( G k ) | . If | V ( G r ) | ≥ 4 ( 1 ≤ r ≤ k − 1 ) , by Lemma 3, | N ( V ( G r ) ) ∩ V ( 1 L T Q n − 1 ) | ≥ 4 . Combining this with | F 1 | ≤ ( 4 n − 10 ) − ( 4 n − 13 ) = 3 , we have that L T Q n [ V ( G r ) ∪ V ( 1 L T Q n − 1 − F 1 ) ] is connected. Therefore, G r is not a com- ponent of L T Q n − F for | V ( G r ) | ≥ 4 . Therefore, L T Q n − F is connected. The following we discuss G r is a component of L T Q n − F with | V ( G r ) | ≤ 3 ( 1 ≤ r ≤ k − 1 ) .

If k = 5 , by Lemma 3, | N ( V ( G 1 ) ) ∪ N ( V ( G 2 ) ) ∪ ⋯ ∪ N ( V ( G k − 1 ) ) ∩ V ( 1 L T Q n − 1 ) | ≥ 4 . Combining this with | F 1 | ≤ 3 , there is one G r ( 1 ≤ r ≤ k − 1 ) such that L T Q n [ V ( G r ) ∪ V ( 1 L T Q n − 1 − F 1 ) ] is connected. Thus, k ≤ 4 . Since | F 1 | ≤ 3 , | V ( G r ) | ≤ 3 ( 1 ≤ r ≤ k − 1 ) and k ≤ 4 , L T Q n − F satisfies one of the conditions (1)-(7).

A connected graph G is super g-extra connected if every minimum g-extra cut F of G isolates one connected subgraph of order g + 1 . In addition, if G − F has two components, one of which is the connected subgraph of order g + 1 , then G is tightly | F | super g-extra connected.

Theorem 9. Let L T Q n be the locally twisted cube for n ≥ 6 . Then L T Q n is tightly ( 4 n − 9 ) super 3-extra connected.

Proof. By Theorem 8, we know for any minimum 3-extra cut F ⊂ V ( L T Q n ) , | F | = 4 n − 9 . We decompose L T Q n into 0 L T Q n − 1 and 1 L T Q n − 1 . Then 0 L T Q n − 1 and 1 L T Q n − 1 are isomorphic to L T Q n − 1 . Suppose that

F i = F ∩ V ( i L T Q n − 1 ) , i ∈ { 0,1 } . Without loss of generality, let | F 0 | ≥ | F 1 | . And

since | F | = 4 n − 9 , 2 n − 4 ≤ ⌈ 4 n − 9 2 ⌉ ≤ | F 0 | ≤ 4 n − 9 , 0 ≤ | F 1 | ≤ ⌊ 4 n − 9 2 ⌋ ≤ 2 n − 5 .

Let C i be the maximum component of i L T Q n − 1 − F i , i ∈ { 0,1 } . We consider the following cases.

Case 1. 2 n − 4 ≤ | F 0 | ≤ 3 ( n − 1 ) − 6 = 3 n − 9 .

Since | F 0 | ≥ | F 1 | and | F | = 4 n − 9 , | F 1 | ≤ 2 n − 5 holds.

By Lemmas 1 and 2, 1 L T Q n − 1 − F 1 is connected or has two components, one of which is an isolated vertex. Since 2 n − 4 ≤ | F 0 | ≤ 3 ( n − 1 ) − 6 = 3 n − 9 , by lemma 8, 0 L T Q n − 1 − F 0 satisfies one of the following conditions: 1) 0 L T Q n − 1 − F 0 has three components, two of which are isolated vertices; 2) 0 L T Q n − 1 − F 0 has two components, one of which is an isolated vertex; 3) 0 L T Q n − 1 − F 0 has two com- ponents, one of which is a K 2 ; 4) 0 L T Q n − 1 − F 0 is connected. Since 2 n − 1 − ( 4 n − 9 ) − 3 ≥ 1 , by Lemma 3, L T Q n [ V ( C 0 ) ∪ V ( C 1 ) ] is connected. Then L T Q n − F satisfies one of the following conditions:

1) L T Q n − F has four components, three of which are isolated vertices;

2) L T Q n − F has three components, one of which is isolated vertices and one of which is a K 2 ;

3) L T Q n − F has three components, two of which are isolated vertices;

4) L T Q n − F has two components, one of which is a path of length two;

5) L T Q n − F has two components, one of which is an isolated vertex;

6) L T Q n − F has two components, one of which is a K 2 ;

7) L T Q n − F is connected.

Thus, in this case, F is not a minimum 3-extra cut of L T Q n , a contradiction.

Case 2. | F 0 | = 3 n − 8 .

Since | F 0 | = 3 n − 8 and | F | = 4 n − 9 , we have | F 1 | = ( 4 n − 9 ) − ( 3 n − 8 ) = n − 1 . By Lemmas 1 and 2, 1 L T Q n − 1 − F 1 is connected or has two components, one of which is an isolated vertex. Since | F 0 | = 3 n − 8 , by Lemma 10, 0 L T Q n − 1 − F 0 satisfies one of the following conditions:

1) 0 L T Q n − 1 − F 0 has four components, three of which are isolated vertices;

2) 0 L T Q n − 1 − F 0 has three components, one of which is isolated vertices and the other of which is a K 2 ;

3) 0 L T Q n − 1 − F 0 has three components, two of which are isolated vertices;

4) 0 L T Q n − 1 − F 0 has two components, one of which is a path of length two;

5) 0 L T Q n − 1 − F 0 has two components, one of which is an isolated vertex;

6) 0 L T Q n − 1 − F 0 has two components, one of which is a K 2 ;

7) 0 L T Q n − 1 − F 0 is connected.

If 0 L T Q n − 1 − F 0 satisfies the condition (4), i.e., 0 L T Q n − 1 − F 0 has two com- ponents, one of which is a path of length two, denoted by P = u v w , 1 L T Q n − 1 − F 1 has two components, one of which is an isolated vertex x, and | N ( x ) ∩ V ( P ) | = 1 , ( N ( V ( P ) ) ∩ V ( 1 L T Q n − 1 ) ) \ { x } ⊆ F 1 , then, by Lemma 3, L T Q n − F has one component which is a 3-path or a K 1,3 . Since 2 n − 1 − ( 4 n − 9 ) − 3 ≥ 1 for n ≥ 6 , L T Q n [ C 0 ∪ C 1 ] is connected. Thus, L T Q n − F exactly has two components. Then the other component C satisfies | C | = 2 n − ( 4 n − 9 ) − 4 > 4 for n ≥ 6 . Otherwise, F is not a minimum 3-extra cut of L T Q n , a contradiction.

Case 3. 3 n − 7 ≤ | F 0 | ≤ 4 n − 14 .

Since | F 0 | ≥ | F 1 | and | F | ≤ 4 n − 9 , | F 1 | ≤ ( 4 n − 9 ) − ( 3 n − 7 ) = n − 2 . By Le- mma 1, 1 L T Q n − 1 − F 1 is connected. Since 3 n − 7 ≤ | F 0 | ≤ 4 n − 14 , by Lemma 10, 0 L T Q n − 1 − F 0 satisfies one of the following conditions:

1) 0 L T Q n − 1 − F 0 has four components, three of which are isolated vertices;

2) 0 L T Q n − 1 − F 0 has three components, one of which is isolated vertices and the other of which is a K 2 ;

3) 0 L T Q n − 1 − F 0 has three components, two of which are isolated vertices;

4) 0 L T Q n − 1 − F 0 has two components, one of which is a path of length two;

5) 0 L T Q n − 1 − F 0 has two components, one of which is an isolated vertex;

6) 0 L T Q n − 1 − F 0 has two components, one of which is a K 2 ;

7) 0 L T Q n − 1 − F 0 is connected.

Thus, L T Q n − F satisfies one of the following conditions:

1) L T Q n − F has four components, three of which are isolated vertices;

2) L T Q n − F has three components, one of which is isolated vertices and one of which is a K 2 ;

3) L T Q n − F has three components, two of which are isolated vertices;

4) L T Q n − F has two components, one of which is a path of length two;

5) L T Q n − F has two components, one of which is an isolated vertex;

6) L T Q n − F has two components, one of which is a K 2 ;

7) L T Q n − F is connected.

In this case, F is not a minimum 3-extra cut of L T Q n , a contradiction.

Case 4. | F 0 | = 4 n − 13 .

Since | F 0 | = 4 n − 13 and | F | = 4 n − 9 for n ≥ 6 , | F 1 | = ( 4 n − 9 ) − ( 4 n − 13 ) = 4 . By Lemma 1, 1 L T Q n − 1 − F 1 is connected.

If there exists a 3-path P in 0 L T Q n − 1 − F 0 , then N ( V ( P ) ) ∩ V ( 0 L T Q n − 1 ) ⊆ F 0 . By Corollary 1, | N ( V ( P ) ) | ≥ 4 n − 13 = | F 0 | in 0 L T Q n − 1 − F 0 . Therefore, N ( V ( P ) ) = F 0 in 0 L T Q n − 1 − F 0 . Note that 2 n − 1 − ( 4 n − 9 ) − 4 ≥ 1 for n ≥ 6 , by Lemma 3, then L T Q n [ V ( C 0 ) ∪ V ( C 1 ) ] is connected. Then L T Q n − F just has two components, one of which is a 3-path.

If there exists a component K 1,3 in 0 L T Q n − 1 − F 0 , then N 0 L T Q n − 1 ( V ( K 1,3 ) ) ⊆ F 0 . By Corollary 1, | N ( V ( K 1 , 3 ) ) | ≥ 4 n − 13 = | F 0 | in 0 L T Q n − 1 − F 0 . Therefore, N ( V ( K 1 , 3 ) ) = F 0 in 0 L T Q n − 1 − F 0 . Note that 2 n − 1 − ( 4 n − 9 ) − 4 ≥ 1 for n ≥ 6 , by Lemma 3, L T Q n − F just has two com- ponents, one of which is a K 1,3 .

If there exists a 4-cycle C in 0 L T Q n − 1 − F 0 , then N 0 L T Q n − 1 ( C ) ∩ V ( 0 L T Q n − 1 ) ⊆ F 0 . By Proposition 7, | N 0 L T Q n − 1 ( V ( C ) ) | ≥ 4 ( n − 1 − 2 ) = 4 n − 12 > 4 n − 13 = | F 0 | , a contradiction to | F 0 | = 4 n − 13 . Therefore, 0 L T Q n − 1 − F 0 has not a 4-cycle.

Case 5. 4 n − 12 ≤ | F 0 | ≤ 4 n − 9 .

Since 4 n − 12 ≤ | F 0 | ≤ 4 n − 9 and | F | ≤ 4 n − 9 , | F 1 | ≤ ( 4 n − 9 ) − ( 4 n − 12 ) = 3 . By Lemma 1, 1 L T Q n − 1 − F 1 is connected.

Suppose that 0 L T Q n − 1 − F 0 is connected. Since 2 n − 1 − ( 4 n − 9 ) ≥ 1 , by Le- mma 3, L T Q n − F is connected, a contradiction.

Suppose that 0 L T Q n − 1 − F 0 is not connected. Let the components in 0 L T Q n − 1 − F 0 be G 1 , G 2 , ⋯ , G k for k ≥ 2 and | V ( G 1 ) | ≤ | V ( G 2 ) | ≤ ⋯ ≤ | V ( G k ) | . If | V ( G r ) | ≥ 4 ( 1 ≤ r ≤ k − 1 ) , by Lemma 3, | N ( V ( G r ) ) ∩ V ( 1 L T Q n − 1 ) | ≥ 4 . If k ≥ 5 , by Lemma 3, | N ( V ( G 1 ) ) ∪ N ( V ( G 2 ) ) ∪ … ∪ N ( V ( G k − 1 ) ) ∩ V ( 1 L T Q n − 1 ) | ≥ 4 . Combining this with | F 1 | ≤ ( 4 n − 9 ) − ( 4 n − 12 ) = 3 , we have that L T Q n − F satisfies one of the following conditions:

1) L T Q n − F has four components, three of which are isolated vertices;

2) L T Q n − F has three components, one of which is isolated vertices and one of which is a K 2 ;

3) L T Q n − F has three components, two of which are isolated vertices;

4) L T Q n − F has two components, one of which is a path of length two;

5) L T Q n − F has two components, one of which is an isolated vertex;

6) L T Q n − F has two components, one of which is a K 2 ;

7) L T Q n − F is connected.

In this case, F is not a minimum 3-extra cut of L T Q n , a contradiction.

In this section, we shall show the 3-extra diagnosability of locally twisted cubes under the PMC model.

Theorem 10. ( [

Lemma 11. Let n ≥ 4 . Then the 3-extra diagnosability of the locally twisted cube L T Q n under the PMC model is less than or equal to 4 n − 6 , i.e., t ˜ 3 ( L T Q n ) ≤ 4 n − 6 .

Proof. Let A be defined in Lemma 7, and let F 1 = N L T Q n ( A ) , F 2 = A ∪ N L T Q n ( A ) . By Lemma 7, | F 1 | = 4 n − 9 , | F 2 | = | A | + | F 1 | = 4 n − 5 , | V ( L T Q n [ A ] ) | ≥ 4 and | V ( L T Q n − F 2 ) | ≥ 4 , F 1 is a 3-extra cut of L T Q n . Therefore, F 1 and F 2 are 3-extra faulty sets of L T Q n with | F 1 | = 4 n − 9 and | F 2 | = 4 n − 5 . Since A = F 1 Δ F 2 and N L T Q n ( A ) = F 1 ⊂ F 2 , there is no edge of L T Q n between V ( L T Q n ) \ ( F 1 ∪ F 2 ) and F 1 Δ F 2 . By Theorem 10, we can deduce that L T Q n is not 3-extra ( 4 n − 5 ) -diagnosable under PMC model. Hence, by the definition of 3-extra diagnosability, we conclude that the 3-extra diagnosability of L T Q n is less than 4 n − 5 , i.e., t ˜ 3 ( L T Q n ) ≤ 4 n − 6 .

Lemma 12. Let n ≥ 5 . Then the 3-extra diagnosability of the locally twisted cube L T Q n under the PMC model is more than or equal to 4 n − 6 , i.e., t ˜ 3 ( L T Q n ) ≥ 4 n − 6 .

Proof. By the definition of 3-extra diagnosability, it is sufficient to show that L T Q n is 3-extra ( 4 n − 6 ) -diagnosable. By Theorem 10, to prove L T Q n is 3- extra ( 4 n − 6 ) -diagnosable, it is equivalent to prove that there is an edge u v ∈ E ( L T Q n ) with u ∈ V ( L T Q n ) \ ( F 1 ∪ F 2 ) and v ∈ F 1 Δ F 2 for each distinct pair of 3-extra faulty subsets F 1 and F 2 of V ( L T Q n ) with | F 1 | ≤ 4 n − 6 and | F 2 | ≤ 4 n − 6 .

Suppose, by way of contradiction, that there are two distinct 3-extra faulty subsets F 1 and F 2 of L T Q n with | F 1 | ≤ 4 n − 6 and | F 2 | ≤ 4 n − 6 , but the vertex set pair ( F 1 , F 2 ) is not satisfied with the condition in Theorem 10, i.e., there are no edges between V ( L T Q n ) \ ( F 1 ∪ F 2 ) and F 1 Δ F 2 . Without loss of generality, assume that F 2 \ F 1 ≠ ∅ .

Assume V ( L T Q n ) = F 1 ∪ F 2 . Since n ≥ 5 , we have that 2 n = | V ( L T Q n ) | = | F 1 ∪ F 2 | = | F 1 | + | F 2 | − | F 1 ∩ F 2 | ≤ | F 1 | + | F 2 | ≤ ( 4 n − 6 ) + ( 4 n − 6 ) = 8 n − 12 , a contra- diction. Therefore, V ( L T Q n ) ≠ F 1 ∪ F 2 .

The following we discuss the case when F 2 \ F 1 ≠ ∅ and V ( L T Q n ) ≠ F 1 ∪ F 2 .

Since there are no edges between V ( L T Q n ) \ ( F 1 ∪ F 2 ) and F 1 Δ F 2 , and F 1 is a 3-extra faulty set, L T Q n − F 1 has two parts L T Q n − F 1 − F 2 and L T Q n [ F 2 \ F 1 ] . Thus, every component G i of L T Q n − F 1 − F 2 satisfies | V ( G i ) | ≥ 4 and every component C i of L T Q n [ F 2 \ F 1 ] satisfies | V ( C i ) | ≥ 4 . Similarly, every component C ′ i of L T Q n [ F 1 \ F 2 ] satisfies | V ( C ′ i ) | ≥ 4 when F 1 \ F 2 ≠ ∅ . Therefore, F 1 ∩ F 2 is also a 3-extra faulty set. Since there are no edges between V ( L T Q n − F 1 − F 2 ) and F 1 Δ F 2 , F 1 ∩ F 2 is also a 3-extra cut. When F 1 \ F 2 = ∅ , F 1 ∩ F 2 = F 1 is also a 3-extra faulty set. Since there are no edges between V ( L T Q n − F 1 − F 2 ) and F 1 Δ F 2 , F 1 ∩ F 2 is a 3-extra cut. By Theorem 8, | F 1 ∩ F 2 | ≥ 4 n − 9 . Therefore, | F 2 | = | F 2 \ F 1 | + | F 1 ∩ F 2 | ≥ 4 + 4 n − 9 = 4 n − 5 , which contradicts with that | F 2 | ≤ 4 n − 6 . So L T Q n is 3-extra ( 4 n − 6 ) -diagnosable. By the definition of t ˜ 3 ( L T Q n ) , t ˜ 3 ( L T Q n ) ≥ 4 n − 6 . The proof is complete.

Combining Lemmas 11 and 12, we have the following theorem.

Theorem 11. Let n ≥ 5 . Then the 3-extra diagnosability of the locally twisted cubes L T Q n under the PMC model is 4 n − 6 .

Before discussing the 3-extra diagnosability of the locally twisted cube L T Q n under the MM^{*} model, we first give an existing result.

Theorem 12 ( [^{*} model if and only if for each distinct pair of g-extra faulty sub- sets F 1 and F 2 of V with | F 1 | ≤ t and | F 2 | ≤ t satisfies one of the following conditions.

1) There are two vertices u , w ∈ V \ ( F 1 ∪ F 2 ) and there is a vertex v ∈ F 1 Δ F 2 such that u w ∈ E and v w ∈ E .

2) There are two vertices u , v ∈ F 1 \ F 2 and there is a vertex w ∈ V \ ( F 1 ∪ F 2 ) such that u w ∈ E and v w ∈ E .

3) There are two vertices u , v ∈ F 2 \ F 1 and there is a vertex w ∈ V \ ( F 1 ∪ F 2 ) such that u w ∈ E and v w ∈ E .

Lemma 13. Let n ≥ 4 . Then the 3-extra diagnosability of the locally twisted cube L T Q n under the MM^{*} model is less than or equal to 4 n − 6 , i.e., t ˜ 3 ( L T Q n ) ≤ 4 n − 6 .

Proof. Let A be defined in Lemma 7, and let F 1 = N L T Q n ( A ) , F 2 = A ∪ N L T Q n ( A ) . By Lemma 7, | F 1 | = 4 n − 9 , | F 2 | = | A | + | F 1 | = 4 n − 5 , | V ( L T Q n [ A ] ) | ≥ 4 and | V ( L T Q n − F 2 ) | ≥ 4 , F 1 is a 3-extra cut of L T Q n . Therefore, F 1 and F 2 are 3-extra faulty sets of L T Q n with | F 1 | = 4 n − 9 and | F 2 | = 4 n − 5 . Since A = F 1 Δ F 2 and N L T Q n ( A ) = F 1 ⊂ F 2 , there is no edge of L T Q n between V ( L T Q n ) \ ( F 1 ∪ F 2 ) and F 1 Δ F 2 . By Theorem 12, we can deduce that L T Q n is not 3-extra ( 4 n − 5 ) -diagnosable under MM^{*} model. Hence, by the definition of 3-extra diagnosability, we conclude that the 3-extra diagnosability of L T Q n is less than 4 n − 5 , i.e., t ˜ 3 ( L T Q n ) ≤ 4 n − 6 .

A component of a graph G is odd or even according as it has an odd or even number of vertices. We denote by o ( G ) the number of odd components of G.

Lemma 14. ( [

Lemma 15. Let n ≥ 7 . Then the 3-extra diagnosability of the locally twisted cube L T Q n under the MM^{*} model is more than or equal to 4 n − 6 , i.e., t ˜ 3 ( L T Q n ) ≥ 4 n − 6 .

Proof. By the definition of the 3-extra diagnosability, it is sufficient to show that L T Q n is 3-extra ( 4 n − 6 ) -diagnosable.

By Theorem 12, suppose, by way of contradiction, that there are two distinct 3-extra faulty subsets F 1 and F 2 of L T Q n with | F 1 | ≤ 4 n − 6 and | F 2 | ≤ 4 n − 6 , but the vertex set pair ( F 1 , F 2 ) is not satisfied with any one condition in Theorem 12. Without loss of generality, assume that F 2 \ F 1 ≠ ∅ . Similarly to the discussion on V ( L T Q n ) = F 1 ∪ F 2 in Lemma 12, we can deduce V ( L T Q n ) ≠ F 1 ∪ F 2 . Therefore, we have the following discussion for V ( L T Q n ) ≠ F 1 ∪ F 2 .

Claim 1. L T Q n − F 1 − F 2 has no isolated vertex.

Suppose, by way of contradiction, that L T Q n − F 1 − F 2 has at least one isolated vertex w. Since F 1 is a 3-extra faulty set, there is at least one vertex u ∈ F 2 \ F 1 such that u are adjacent to w. Since the vertex set pair ( F 1 , F 2 ) is not satisfied with any one condition in Theorem 12, by the condition (3) of Theorem 12, there is at most one vertex u ∈ F 2 \ F 1 such that u is adjacent to w. Therefore, there is just a vertex u is adjacent to w.

Case 1. F 1 \ F 2 = ∅ .

If F 1 \ F 2 = ∅ , then F 1 ⊆ F 2 . Since F 2 is a 3-extra faulty set, every com- ponent G i of L T Q n − F 1 − F 2 has | V ( G i ) | ≥ 4 . Thus, L T Q n − F 1 − F 2 has no isolated vertex.

Case 2. F 1 \ F 2 ≠ ∅ .

Similarly, since F 1 \ F 2 ≠ ∅ , by the condition (2) of Theorem 12 and the hypothesis, we can deduce that there is just a vertex v ∈ F 1 \ F 2 such that v is adjacent to w.

Let W ⊆ V ( L T Q n ) \ ( F 1 ∪ F 2 ) be the set of isolated vertices in L T Q n [ V ( L T Q n ) \ ( F 1 ∪ F 2 ) ] , and H be the induced subgraph by the vertex set V ( L T Q n ) \ ( F 1 ∪ F 2 ∪ W ) . Then for any w ∈ W , there are ( n − 2 ) neighbors in F 1 ∩ F 2 . By Lemmas 14 and 3, | W | ≤ o ( L T Q n − ( F 1 ∪ F 2 ) ) ≤ | F 1 ∪ F 2 | = | F 1 | + | F 2 | − | F 1 ∩ F 2 | ≤ ( 4 n − 6 ) + ( 4 n − 6 ) − ( n − 2 ) = 7 n − 10 . Assume V ( H ) = ∅ . Then 2 n = | V ( L T Q n ) | = | F 1 ∪ F 2 | + | W | = | F 1 | + | F 2 | − | F 1 ∩ F 2 | ≤ ( 4 n − 6 ) + ( 4 n − 6 ) − ( n − 2 ) + ( 7 n − 10 ) = 14 n − 20 , a contradiction to that n ≥ 7 . So V ( H ) ≠ ∅ .

The following we discuss the case when F 1 \ F 2 ≠ ∅ , F 2 \ F 1 ≠ ∅ and V ( H ) ≠ ∅ .

Since the vertex set pair ( F 1 , F 2 ) is not satisfied with the condition (1) of Theorem 12, and there are not isolated vertices in H , we induce that there is no edge between V ( H ) and F 1 Δ F 2 . Note that F 2 \ F 1 ≠ ∅ . If F 1 ∩ F 2 = ∅ , then this is a contradiction to that L T Q n is connected. Therefore, F 1 ∩ F 2 ≠ ∅ . Thus, F 1 ∩ F 2 is a vertex cut of L T Q n . Since F 1 is a 3-extra faulty set of L T Q n , we have that every component H i of H has | V ( H i ) | ≥ 4 and every component C i of L T Q n [ W ∪ ( F 2 \ F 1 ) ] has | V ( C i ) | ≥ 4 . Since F 2 also is a 3-extra faulty set of L T Q n , we have that every component C ′ i of L T Q n [ W ∪ ( F 1 \ F 2 ) ] has | V ( C ′ i ) | ≥ 4 . Note that L T Q n − ( F 1 ∩ F 2 ) has two parts: H and L T Q n [ W ∪ ( F 1 Δ F 2 ) ] . Let b i ∈ V ( L T Q n [ W ∪ ( F 1 Δ F 2 ) ] ) . If b i ∈ W , then b i has two neighbors u ∈ V ( C i ) and v ∈ V ( C ′ i ) . Then b i ∈ V ( C i ∪ C ′ i ) and | V ( C i ∪ C ′ i ) | ≥ 4 . Thus, F 1 ∩ F 2 is a 3-extra cut of L T Q n . By Theorem 8, | F 1 ∩ F 2 | ≥ 4 n − 9 . Since | V ( C i ) | ≥ 4 , | F 2 \ F 1 | ≥ 3 . Since | F 1 ∩ F 2 | = | F 2 | − | F 2 \ F 1 | ≤ ( 4 n − 6 ) − 3 = 4 n − 9 , we have | F 1 ∩ F 2 | = 4 n − 9 . Then | F 2 \ F 1 | = 3 and | F 2 | = 4 n − 6 . Similarly, | F 1 \ F 2 | = 3 , | F 1 | = 4 n − 6 . By Lemma 9, the locally twisted cube L T Q n is tightly ( 4 n − 9 ) super 3-extra connected, i.e., L T Q n − ( F 1 ∩ F 2 ) has two components, one of which is a subgraph of or- der 4. Noted that | W | ≤ 7 n − 10 . 2 n = | V ( L T Q n ) | = | F 1 \ F 2 | + | F 2 \ F 1 | + | F 1 ∩ F 2 | + | V ( H ) | + | W | ≤ 3 + 3 + ( 4 n − 9 ) + 4 + ( 7 n − 10 ) = 11 n − 9 , a contradiction to n ≥ 7 . Therefore, L T Q n − F 1 − F 2 has no isolated vertex when F 1 \ F 2 ≠ ∅ , F 2 \ F 1 ≠ ∅ and V ( H ) ≠ ∅ . The proof of Claim 1 is complete.

Let u ∈ V ( L T Q n ) \ ( F 1 ∪ F 2 ) . By Claim 1, u has at least one neighbor vertex in L T Q n − F 1 − F 2 . Since the vertex set pair ( F 1 , F 2 ) is not satisfied with any one condition in Theorem 12, by the condition (1) of Theorem 12, for any pair of adjacent vertices u , w ∈ V ( L T Q n ) \ ( F 1 ∪ F 2 ) , there is no vertex v ∈ F 1 Δ F 2 such that u w ∈ E ( L T Q n ) and u v ∈ E ( L T Q n ) . It follows that u has no neighbor vertex in F 1 Δ F 2 . By the arbitrariness of u, there is no edge between V ( L T Q n ) \ ( F 1 ∪ F 2 ) and F 1 Δ F 2 . Since F 2 \ F 1 ≠ ∅ and F 1 is a 3-extra faulty set, | F 2 \ F 1 | ≥ 4 and | V ( L T Q n − F 2 − F 1 ) | ≥ 4 . Since F 1 also is 3-extra faulty sets, | F 1 \ F 2 | ≤ 4 and | V ( L T Q n − F 1 − F 2 ) | ≥ 4 . Then F 1 ∩ F 2 is a 3- extra cut of L T Q n . By Theorem 8, we have | F 1 ∩ F 2 | ≥ 4 n − 9 . Therefore, | F 2 | = | F 2 \ F 1 | + | F 1 ∩ F 2 | ≥ 4 + ( 4 n − 9 ) = 4 n − 5 , which contradicts | F 2 | ≤ 4 n − 6 . Therefore, L T Q n is 3-extra ( 4 n − 6 ) -diagnosable and t ˜ 3 ( L T Q n ) ≥ 4 n − 6 . The proof is complete.

Combining Lemmas 13 and 15, we have the following theorem.

Theorem 13. Let n ≥ 7 . Then the 3-extra diagnosability of the locally twisted cube L T Q n under the MM^{*} model is 4 n − 6 .

This work is supported by the National Science Foundation of China (61370001).

Wang, M., Ren, Y.X., Lin, Y.Q. and Wang, S.Y. (2017) The Tightly Super 3-Extra Connectivity and Diag- nosability of Locally Twisted Cubes. American Journal of Computational Mathematics, 7, 127-144. https://doi.org/10.4236/ajcm.2017.72011