1. Introduction
An interconnection network is typically modeled as an undirected graph
, where the vertex and edge correspond to the processor and the link between two distinct processors, respectively. In the network, the failure of processors or links is inevitable. How to find the fault processors or links becomes the crucial problem to maintain the stability of such system. On 9 February 2001, the submarine cable between China and the United States was broken by the anchor hook of a fishing boat, which made a large number of domain names inaccessible. On 27 May 2015, an optical fiber was dug in Hangzhou, resulting in a large area of Alipay paralysis. In order to enable the system to operate normally when local faults occur, the concept of fault tolerance is proposed. Fault tolerance is the ability of the network to tolerate and recover from simultaneous failures of components and connections under normal operating conditions. Friedrich et al. [
1] studied a strong network and fault tolerance setting. Fault diagnosis is the process of identifying faulty processors in a system. Many diagnostic models have been studied, such as PMC, BGM and MM (see references [
2,
3,
4,
5,
6]). Among these models, the PMC model and the MM model are well known. In 1967, Preparata et al. [
6] first introduced the PMC model. In this model, any two adjacent processors can be tested each other. The MM model, where the diagnosis is performed by sending the same task from one processor to its two adjacent processors and comparing the feedback results, was introduced by Maeng and Malek [
4]. Sengupta and Dahbura [
7] introduced a modification of the MM model, called the MM* model, in which each processor must test any pair of its neighboring processors. In recent years, some new fault-tolerant measures have been introduced. Zhang et al. [
8], in 2016, proposed the
g-extra diagnosability of a system, which requires that every component of
has more than
g processors. Moreover, they demonstrated the
g-extra diagnosability of hypercubes under the PMC model and MM* model. There are many research works about the
g-extra diagnosability (see [
8,
9,
10,
11]).
Let
. If
is disconnected or contains just one vertex, we call
F a vertex cut. The connectivity of a graph
G is defined as
and
F is a vertex cut}. The definition of connectivity implicitly assumes that the likelihood of any subset of the interconnection network experiencing simultaneous failures is the same. However, when a processor error occurs, not all processors adjacent to the processor will simultaneously lose their running ability in practical applications. In order to overcome this deficiency, Harary [
12] introduced the concept of conditional connectivity by adding some restrictions on connected components. Additionally, the concept of
g-extra connectivity was proposed by Fàbrega and Fiol [
13], which restrains that each remaining component has at least
vertices after removing the faulty vertex set. For recent results on the
g-extra connectivity of the interconnection network, see [
14,
15,
16,
17,
18].
As a generalization of an
n-dimensional bubble-sort graph
, the
-bubble-sort graph, namely
, was proposed by Shavash [
19]. Considering the above ideas, we mainly study the
g-extra connectivity of
-bubble-sort network
that is
for
and
. Afterwards, we show that
g-extra diagnosability of
under the PMC model
and
and the MM* model
and
is
, respectively.
3. The -Extra Connectivity of -Bubble-Sort Network
In this section, we attempt to determine the size of g-extra connectivity of an -bubble-sort network .
Lemma 3. for and .
Proof. We need to find a subset
such that
. Obviously,
X is located in the same subgraph. Without a loss of generality, let
. Then, we get
Note that
is disconnected and
is a vertex-cut. Via Proposition 1 (4), every vertex in
has exactly one outside neighbor. Since
, we can find that the external neighborhoods of induced subgraph by
X are within the same subgraph, denoted by
. Via Proposition 1 (3),
,
is connected for any
. Let the rest subgraphs in
, excluding subgraphs
and
, be
; then,
and
are connected. Via Proposition 1 (1), for any
,
, there is an edge connecting
and
. Thus,
is connected. For any vertex
,
x has an outgoing neighbor, say
, and
. Thus,
x is connected to
. By the arbitrariness of
x,
is connected. Therefore,
has two components, namely
X and
. Hence,
is a
g-extra cut of
. So,
. □
Lemma 4. For . Let F be the set of vertices of . If , then is either connected or has two components, one of which is a singleton.
Proof. This lemma is proved via induction on k. Let us consider the base case for removing vertices from .
is connected for all i because is -connected. Since only vertices are removed and s has n choices, there exists s such that . Thus, for any . We can see that at most, vertices are deleted in . There are edges between and , and since , one of these edges is retained in . By the arbitrariness of i, there is an edge between and for all , so is connected.
Since and , one faulty vertex is outside of , at most. Hence, has a component Y that includes every for . According to Proposition 1 (4), each vertex in has an outside neighbor, so it also belongs to Y unless that outside neighbor is in . There is at most one vertex in , so all but one of the vertices of belong to Y. Thus, the base case has been proved.
Now suppose that for all pairs (), and for , the statement holds, and consider removing up to vertices from , where . We will consider two cases.
Since is -connected, is connected for all i. For any , there are independent edges between and , but only a total of vertices are removed, leaving one edge between and ; thus, is connected.
Assume there exists
such that
. Then,
is a contradiction. Then, we get
for any
, which implies that
is connected. All the
,
, belong to a large component, say
Y, in
due to the same argument as Case 1. If
, according to the inductive hypothesis,
is either connected, which is similar to Case 1, or there are two components, one of which is a singleton. Since, at most,
vertices were deleted from
, and only one vertex remained in the small component, there is at least
vertices in the large component; in other words, at least
independent edges are present. Since
, the large component of
is part of
Y, and, at most, one vertex does not belong to
Y. If
or
, due to the fact that only one or zero vertex outside of
can be removed, the proof is similar to Case 2 of the base case. Thus, the proof is complete. □
Lemma 5 ([
23])
. Let . If F is a set of vertices of with , then is connected or has a large component and small components with, at most, two vertices in total. Lemma 6. For any subset with where . Then, is connected or has a large component and small components having, at most, two vertices in total.
Proof. When , we can obtain a bubble-sort graph . Lemma 5 holds due to the removal of up to vertices from . Now let us consider . We proceed via induction on k. Firstly, we consider the base case of deleting n vertices from . Let us consider the following two cases:
In this case, all of the are connected because is -connected. For all i, if , then all the are part of one large component. Otherwise, since , there exists some s such that . Then, we obtain for any . There are edges between and by Proposition 1 (1), so an edge remains between and . Hence, is connected.
Since and , two faulty vertices, at most, are outside of . Hence, has a component Y that includes all except possibly the components of . According to Proposition 1 (4), each vertex in has an external neighbor. Conversely, assuming that at least three vertices in do not belong to Y, then each of these outer neighbors must be in F. This means that , which is a contradiction. Thus, two vertices, at most, do not belong to Y, and the base case is proved.
Now suppose that for all pairs (), and for , the statement holds, and consider removing up to vertices from , where . We will consider three cases.
Thus, every is connected. There are independent edges between and . Since implies that , we can obtain . Then, for any pair , there is an edge between and . Hence, is connected.
It means that for , is connected and every belongs to a large component Y in . If , then ; so, at most, two vertices are not in Y regarding Case 2 of the base case. If , then we can apply the inductive hypothesis to get that is connected or has a big component and small components with at most two vertices in total. If it is connected, we follow the method of Case 1; then, is connected. Otherwise, we find that the large component will be part of a Y. Since the number of edges outside , . Since , is greater than the number of vertices removed plus the number of vertices in small components, which is at most . Therefore, at most, two vertices do not belong to Y, and the proof is complete.
Assume there exist three integers
such that
. Then
which is a contradiction. Therefore, there are exactly two positive integers
, such that
and
are connected for
. All of the
belong to a large component
Y in
for
. If
or
, then
At most,
vertices are removed, which is a contradiction. Thus,
. Hence, according to Lemma 4 for
and
, each has at most one vertex that is not connected to
Y. According to the same argument as in Case 2, the large components of both
and
are part of
Y; so, at most, two vertices do not belong to
Y, and our statement is proven. □
Lemma 7. Let and . If with , then is connected or contains a large component and small components with g vertices at most.
Proof. First, we solve the base case of removing up to vertices. We have two cases:
Every is connected in this case. There are edges between and for . Thus, if for all , then there will be an edge between and . If for some i, then we cannot get for all because . Therefore, there exists such that . For every , we have . If for all , then there exists s such that ; otherwise, is a contradiction. We consider this and then for all . In summary, regardless of the type, we can get for all ; thus, there is an edge between and for every , which shows that is connected.
Since and for some i, then we will get . Thus, all the belong to a large component of , say Y. Since each vertex in has an outside neighbor and at most g vertices are removed outside , at most g vertices do not belong to Y. This case is proven.
Now we suppose . We use induction on g. When and , then is connected by Proposition 1 (3). When , the cases are true via Lemmas 4 and 6, respectively. Suppose that the statement holds for all for some . Let us prove the case of via induction on k. The base case is completed above, so suppose that the declaration is true for all for some and consider that when reached, vertices are removed from . There are two cases:
In this case, . Thus, there is a large component Y in containing all the for . Since every vertex in has an outside neighbor and , at most vertices do not belong to Y, as desired.
We will construct the function .
As
, we can get inequality
Thus, at least one of the
n subgraphs has a number of faulty vertices less than or equal to
. Moreover, there exists some
s such that
. Since
is (
n − 2)-connected, we can find that
is connected. We can still use the inductive hypothesis on
and
because
and
; thus, for any
,
consists of a large component and small components containing at most
vertices in total. Hence, we require
However,
, so
. Despite this,
We use two inequalities, and . Therefore, this equality is true if and only if these two inequalities are true. However, the first inequality takes an equal sign, which implies that , and then the second takes an equal sign, which means that . Then, going back to the first inequality, we can get , which is contradictory to the hypothesis that . So, this inequality strictly holds. By the arbitrariness of i, there is an edge between the large component of and for all . Then, all these large components belong to a large component of .
Next, we prove that the small components contain at most vertices in total. Suppose, on the contrary, that at least vertices are in the small components. Let us construct another function . Assume is disconnected for any i and let be the total number of vertices in the small components of . At most, vertices are disconnected in any , so . Since and , then at least two of these are positive. So, we let () be at least 1. According to the induction hypothesis, we can get that if , then for any integer , there are at most vertices in the small components of . Thus, for all integers , we have . Otherwise, ; thus, at most, vertices are in the small components of , which contradicts to the fact the small components have a total of vertices. Since , we can choose for , such that and for all i. This means that for , we can get .
For
, the sum of the
does not exceed the total number of removed vertices, so
Since
, this implies
. Then,
Equivalently,
Using
and
, we can find that
, where
. Since
and
, we get
, which is a contradiction. Therefore, there are at most
vertices in the small components. The proof of this lemma is complete. □
According to Lemma 7, we can easily obtain the following Lemma.
Lemma 8. for and .
Theorem 1. The g-extra connectivity of -bubble-sort network is for and .