Sometimes, many links in a multiple CPU system may be missing. A missing edge implies that a link between 2 CPUs that was faulty. The existence of missing edges in a system may affect the diagnosability of the whole system, and degrees and the local diagnosability of some nodes. Especially, in a regular graph, nodes that are adjacent to missing edges have lower degrees than others. Hence, those nodes may not keep the strong local diagnosability feature, and the graph may not keep the strong local diagnosability feature again. Those new degrees can be used to determine whether the incomplete graph keeps the strong local diagnosability feature or not. Next, we demonstrate that an n-dimensional wheel network keeps the strong local diagnosability feature even with up to missing edges.
Proof. By Proposition 4, is node transitive. Therefore, can be chosen as the root of an extended star , i.e., , where is the identity element of the symmetric group . can be partitioned into n disjoint subgraphs , where each node has a fixed integer i in the last position for . Clearly, is isomorphic to , where is the bubble-sort star graph with the dimension .
Let for , and , then . For convenience, let , and denote , and three outside neighbors of . To prove this lemma, we need to discuss whether is an empty set or not. When , the first step is to discover an extended star in ; the second step is to find a 3-path in , a 3-path in and a 3-path in ; the third step is to connect , and to x. Then, an extended star is found in at x, (with written simply as B). For the case of , we obtain an extended star to satisfy the lemma by removing any 4-path that contains any link of and starts at x from B.
It should be noted that removing a 4-path P from a graph G means removing all nodes and links of the 4-path except y from a graph G in this paper, where y is the only common node of P and G.
Firstly, we give two claims as follows.
Claim 1. If , then there exists an extended star in at for . In the extended star , there exist at least and at most 4-paths (resp. 3-paths) with just common node .
Proof. Notice that each is isomorphic to and , by Lemma 9, then there exists an extended star in at for and . Therefore, , and we can find at least and at most 4-paths (resp. 3-paths) with just common node in the combining the definition of the extended star. In particular, in the extended star , there exist just 4-paths (resp. 3-paths) with just common node if and only if each of edges in is incident with . □
Claim 2. If , then there exists at least a 3-path in starting at for .
Proof. By Propositions 1–3, we have that . Since each is isomorphic to , we have that . If , then is connected. By Proposition 6 and for . So a 3-path starting at can be found in for . □
Then we can find the extended star by discussing and as follows.
Case 1. .
By Claim 1, there exists an extended star in at x.
Case 1.1. .
Here, for . By Claim 2, there exists a 3-path (resp. , ) in (resp. , ). We connect , and to x. Combining them with A, an extended star is found in at x, (with written simply as B). If , then B satisfies the lemma. If , then we find an extended star to satisfy the lemma by removing any 4-path that contains any edge of and starts at x from B.
Case 1.2. .
If each of , and is less than , then we can complete the proof as Case 1.1. Clearly, at most one of , and is equal to , without loss of generality, we consider . Then . We choose a 4-path . Clearly, is in . By Claim 1 and , there exist 3-paths that do not contain a missing edge in (resp. ). Notice that , and for , so there exists a 3-path starting at v in such that and have no common node. We can choose a 3-path starting at w in . Notice that any two of and have no common node, and each of and does not contain a missing edge. We connect , and to x. Combining them with A, we can get an extended star in .
Case 1.3. .
If each of , and is less than , then we can complete the proof as Case 1.1. Clearly, at most one of , and is equal to or . Without loss of generality, let , or . If , then the proof for can be completed as Case 1.2. If , then .
Case 1.3.1. .
Without loss of generality, let . Here, . We choose a 4-path . Clearly, is in . Combining Claim 1, there exist at least 3-paths in (resp. ). Notice that , and for , so there exists a 3-path starting at w in such that and have no common node. Choose a 3-path starting at v in . Notice that any two of and have no common node, and each of and does not contain a missing edge. We connect , and to x. Combining them with A, we can get an extended star in .
Case 1.3.2. .
Notice that . If , then we choose a 4-path . Clearly, is in . By Claim 1 and , there exist at least 3-paths in (resp. ). Notice that , and for , so there exists a 3-path starting at w in such that and have no common node. Choose a 3-path starting at v in . Notice that any two of and have no common node, and each of and does not contain a missing edge. We connect , and to x. Combining them with A, an extended star is found in at x, (with written simply as B), and B satisfies the lemma. If , then we find an extended star to satisfy the lemma by removing any 4-path that contains any link of and starts at x from B.
Case 2. .
Here, . By Claim 1, we can choose a 3-path (resp. , ) from at least 3-paths that do not contain a missing edge in (resp. , ). Let f be an arbitrary link of , and let . Then , there exists an extended star in by Claim 1. Let .
Case 2.1. .
Here, is one of extended star at x of . Notice that , so we can connect , and to x. Combining them with , an extended star can be obtained in .
Case 2.2. .
Let be a 4-path containing f and starting at x in . A graph is obtained by removing from , denoted by A.
Case 2.2.1. f is incident with x.
Notice that , so we connect , and to x. Combining them with A, an extended star can be obtained in .
Case 2.2.2. f is not incident with x.
Let be a 3-path containing f and starting at a, and then a is adjacent to x. Next, we consider for . Without loss of generality, let . Notice that , we can choose a 2-path that does not contain a missing edge in , where . By Claim 1 and , we can find 3-paths in , where . Note that, and for , so we can find a 3-path in that does not contain a missing edge, and and have no common node. At the same time, we can find a 3-path (resp. ) in (resp. ), and connect to a to obtain a 3-path . Notice that each of , , and do not contain a missing edge, and any two of them have no common node. We connect , , and to x. Combining them with A, an extended star can be got in . The case of for can be proved similarly.
Case 3. .
Notice that . By Claim 1, we can choose a 3-path (resp. , ) from at least 3-paths that do not contain a missing edge in (resp. , ). Let f be an arbitrary edge of , and let . Then . Combining Claim 1, there exists an extended star in . Let .
Case 3.1. .
Here, is one of extended star at x in . We connect , and to x. Combining them with , an extended star is found in at x, (with written simply as B). Notice that . If , then B satisfies the lemma. If , then we find an extended star to satisfy the lemma by removing any path that contains any edge of and starts at x from B.
Case 3.2. .
Let be a 4-path containing f and starting at x in . A graph is obtained by removing from , denoted by A.
Case 3.2.1. f is incident with x.
Connect , and to x. Combining them with A, an extended star is found in at x, (with written simply as B). Notice that . If , then B satisfies the lemma. If , then we find an extended star to satisfy the lemma by removing any path that contains any edge of and starts at x from B.
Case 3.2.2. f is not incident with x.
Let be a 3-path starting at a, and let it contain f, then a is adjacent to x.
Case 3.2.2.1. .
Here, . Without loss of generality, let . Now we consider for . Without loss of generality, let . We can choose a 2-path in , where . Notice that , so does not contain a missing edge. By Claim 1 and for , we can find 3-paths that do not contain a missing edge in , and any two of these 3-paths have no common node except u (resp. v, w). Notice that , and for , so we can find a 3-path that does not contain a missing edge in , and and have no common node. We can find a 3-path (resp. ) that does not contain a missing edge in (resp. ) as well. We can connect to a to obtain . Notice that each of , , and does not contain a missing edge, and any two of them have no common node. We connect , , , to x. Combining them with A, an extended star can be got in . The case of for can be proved similarly.
Case 3.2.2.2. .
Here, . Now we consider for . Without loss of generality, let . We can find in and in , where and . Since , we can choose a 2-path from and that does not contain a missing edge. Notice that and are in different ’s and , and we can connect to a to obtain a 3-path that does not contain a missing edge. By Claim 1 and for , we can find 3-paths that do not contain a missing edge in , and any two of these 3-paths have no common node except u (resp. v, w). Notice that , and for . If , then we can find a 3-path that does not contain a missing edge in , and and have no common node, and we can find a 3-path (resp. ) that does not contain a missing edge in (resp. ) as well. If , then we can find a 3-path that does not contain a missing edge in , and and have no common node, and we can find a 3-path (resp. ) that does not contain a missing edge in (resp. ) as well. We connect , , , to x. Notice that each of , , and does not contain a missing edge, and any two of them have no common node. Combining them with A, an extended star is found in at x, (with written simply as B). Notice that . If , then B satisfies the lemma. If , then we find an extended star to satisfy the lemma by removing any path that contains any link of and starts at x from B. The case of for can be proved similarly.
Case 4. .
Here, . By Claim 1, we can choose a 3-path (resp. , ) from at least 3-paths that do not contain a missing edge in (resp. , ). Let f be an arbitrary link of , and let . Then , so there exists an extended star in by Claim 1. Let .
Case 4.1. .
Here, is one of extended star at x in . We connect , and to x. Combining them with , an extended star is found in at x, (with written simply as B). Notice that . If , then B satisfies the lemma. If , then we find an extended star to satisfy the lemma by removing any path that contains any link of and starts at x from B.
Case 4.2. .
Let be a 4-path containing f and starting at x in . A graph is obtained by removing from , denoted by A.
Case 4.2.1. f is incident with x.
Connect , and to x. Combining them with A, Then, an extended star is found in at x, (with written simply as B). Notice that . If , then B satisfies the lemma. If , then we find an extended star to satisfy the lemma by removing any path that contains any link of and starts at x from B.
Case 4.2.2. f is not incident with x.
Let be a 3-path starting at a and containing f, then a is adjacent to x.
Case 4.2.2.1. .
Here, .
Case 4.2.2.1.1. for some .
Without loss of generality, let . We can complete the proof as Case 3.2.2.1.
Case 4.2.2.1.2. and for some , .
Without loss of generality, let and . Now we consider for . Without loss of generality, let . We can choose a 2-path in , where . Notice that , so does not contain a missing edge. By Claim 1 and for , we can find 3-paths that do not contain a missing edge in , and any two of these 3-paths have no common node except u (resp. v, w). Notice that , and for , so we can find a 3-path that does not contain a missing edge in , and and have no common node. We can find a 3-path (resp. ) that does not contain a missing edge in (resp. ). Connect to a to obtain a 3-path . Notice that each of , , and does not contain a missing edge, and any two of them have no common node. We connect , , , to x. Combining them with A, we can get the extended star in . The case of for can be proved similarly.
Case 4.2.2.2. .
Here, . Without loss of generality, let . Next, we consider for . Without loss of generality, let . We can find in and in , where and . Since , and do not contain a missing edge. We can choose a 2-path from and . Since and are in different ’s and , connect to a to obtain a 3-path that does not contain a missing edge. By Claim 1 and for , we can find 3-paths that do not contain a missing edge in , and any two of these 3-paths have no common node except u (resp. v, w). Notice that , and for . If , then we can find a 3-path that does not contain a missing edge in , and and have no common node, and we can find a 3-path (resp. ) that does not contain a missing edge in (resp. ) as well. If , then we can find a 3-path that does not contain a missing edge in , and and have no common node, and we can find a 3-path (resp. ) that does not contain a missing edge in (resp. ) as well. We connect , , , to x. Notice that each of , , and does not contain a missing edge, and any two of them have no common node. Combining them with A, an extended star is found in at x, (with written simply as B). Notice that . If , then B satisfies the lemma. If , then we find an extended star to satisfy the lemma by removing any path that contains any edge of and starts at x from B. The case of for can be proved similarly.
Case 4.2.2.3. .
Here, . Now we consider for . Without loss of generality, let . We can find in , in and in , where , and . Since , then , and do not contain a missing edge. Notice that , and are in different ’s and , then we can choose a 2-path from , and , and attach the 2-path to a to obtain a 3-path that does not contain a missing edge. By Claim 1 and for , we can find 3-paths that do not contain a missing edge in , and any two of these 3-paths have no common node except u (resp. v, w). Notice that , and for . If , then we can find a 3-path that does not contain a missing edge in such that and have no common node, and we can find a 3-path (resp. ) that does not contain a missing edge in (resp. ) as well. If , then we can find a 3-path in that does not contain a missing edge, and and have no common node, and we can find a 3-path (resp. ) that does not contain a missing edge in (resp. ) as well. If , then we can find a 3-path in that does not contain a missing edge, and and have no common node, and we can find a 3-path (resp. ) that does not contain a missing edge in (resp. ). Notice that each of , , and does not contain a missing edge, and any two of them have no common node. We connect , , , to x. Combining them with A, an extended star is found in at x, (with written simply as B). Notice that . If , then B satisfies the lemma. If , then we find an extended star to satisfy the lemma by removing any path that contains any link of and starts at x from B. The case of for can be proved similarly.
Case 5. .
Here, . By Claim 1, we can choose a 3-path (resp. , ) from at least 3-paths that do not contain a missing edge in (resp. , ). Let be any two elements of and . Then , so there exists an extended star in by Claim 1. Let .
Case 5.1. Neither f nor belongs to .
Here, is one of extended star at x in . Notice that , so we connect , and to x. Combining them with , we can get an extended star in .
Case 5.2. contains f or .
Without loss of generality, we assume that contains only f. Let be a 4-path containing f and starting at x in . A graph is obtained by removing from , denoted by A. Next we discuss whether f is incident with x or not. If f is incident with x, then it can be proved as Case 2.2.1. If f is not incident with x, then it can be proved as Case 2.2.2.
Case 5.3. contains f and .
Case 5.3.1. f and are both incident with x.
Let (resp. ) be a 4-path containing f (resp. ) and starting at x in . A graph is obtained by removing and from , denoted by A. Notice that , so we connect , and to x. Combining them with A, we can get an extended star in .
Case 5.3.2. Just one of f and is incident with x.
Without loss of generality, assumes that only is incident with x. Let (resp. ) be a 4-path containing f (resp. ) and starting at x in . A graph is obtained by removing and from , denoted by A. Let be a 3-path containing f and starting at a, and then a is adjacent to x. The next proof can be completed as Case 2.2.2.
Case 5.3.3. Neither f nor is incident with x.
Next we discuss whether f and belong to the same path or not.
Case 5.3.3.1. f and belong to the same path in .
Then we can complete the proof as Case 2.2.2.
Case 5.3.3.2. f and belong to different paths in .
Let (resp. ) be a 3-path starting at a (resp. b), and it contains f (resp. ). Then a and b are both incident with x. Notice that . It is easy to find a 3-path (resp. ) that does not contain a missing edge in (resp. ), and (resp. ) and (resp. ) have no common vertices. Connecting (resp. ) to a (resp. b), we can obtain a 3-path (resp. ). Notice that , so we connect , , , and to x. Combining them with A, we can get an extended star in .
Case 6. .
Here, . By Claim 1, we can choose a 3-path (resp. , ) from at least 3-paths that do not contain a missing edge in (resp. , ). Let be any two elements of and . Then , there exists an extended star in by Claim 1. Let .
Case 6.1. Neither f nor belongs to .
Here, is one of extended star at x in . Connect , and to x. Combining them with , an extended star is found in at x, (with written simply as B). Notice that . If , then B satisfies the lemma. If , then we find an extended star to satisfy the lemma by removing any path that contains any link of and starts at x from B.
Case 6.2. contains f or .
Without loss of generality, we suppose that contains only f. Let be a 4-path containing f and starting at x in . A graph is obtained by removing from , denoted by A. Next we consider whether f is incident with x or not. If f is incident with x, then the next proof can be completed as Case 3.2.1. If f is not incident with x, then the next proof can be completed as Case 3.2.2.
Case 6.3. contains f and .
Case 6.3.1. f and are both incident with x.
Let (resp. ) be a 4-path containing f (resp. ) and starting at x in . A graph is obtained by removing and from , denoted by A. Connect , and to x. Combining them with A, an extended star at x is found in , (with written simply as B). Notice that . If , then B satisfies the lemma. If , then we find an extended star to satisfy the lemma by removing any path that contains any link of and starts at x from B.
Case 6.3.2. Just one of f and is incident with x.
Without loss of generality, assume that only is incident with x. Let (resp. ) be a 4-path containing f (resp. ) and starting at x in . A graph is obtained by removing and from , denoted by A. Let be a 3-path containing f and starting at a, and then a is adjacent to x. We can complete the proof as in Case 3.2.2.
Case 6.3.3. Neither f nor is incident with x.
Then we consider whether f and belong to the same path or not.
Case 6.3.3.1. f and belong to the same path in .
Then we can complete the proof as in Case 3.2.2.
Case 6.3.3.2. f and belong to different paths in .
Let (resp. ) be a 3-path starting at a (resp. b), and it contains f (resp. ). Then a and b are both incident with x. Notice that , then . Note that , . There is a 2-path (resp. ) that do not contain a missing edge in , or or , and the edge (resp. ), where (resp. ) is an outside neighbor of a (resp. b). Connecting (resp. ) to a (resp. b), we can obtain a 3-path (resp. ). Connect , , , and to x. Combining them with A, we can get an extended star in .
Case 7. .
Here, . By Claim 1, we can choose a 3-path (resp. , ) from at least 3-paths that do not contain a missing edge in (resp. , ). Let be any three elements of and . Then , by Claim 1, there exists an extended star in . Let .
Case 7.1. None of f, and belongs to .
We can complete the proof as in Case 5.1.
Case 7.2. contains just one of f, and .
We can complete the proof as in Case 5.2.
Case 7.3. contains just two of f, and .
We can complete the proof as in Case 5.3.
Case 7.4. contains f, and .
Case 7.4.1. f, and are all incident with x.
Let (resp. , ) be a 4-path containing f (resp. , ) and starting at x in . A graph is obtained by removing , and from , denoted by A. Notice that , so we connect , and to x. Combining them with A, we can get an extended star in .
Case 7.4.2. Just two of f, and are incident with x.
Without loss of generality, assumes that and are incident with x. Let (resp. , ) be a 4-path containing f (resp. , ) and starting at x in . A graph is obtained by removing , and from , denoted by A. Let be a 3-path containing f and starting at a, and then a is adjacent to x. The next proof can be completed as in Case 5.3.2.
Case 7.4.3. Just one of f, and is incident with x.
Without loss of generality, assumes that only is incident with x. Let (resp. , ) be a 4-path containing f (resp. , ) and starting at x in . A graph is obtained by removing , and from , denoted by A. Let be a 3-path containing f and starting at a, and then a is adjacent to x. The next proof can be completed as in Case 5.3.3.
Case 7.4.4. None of f, and is incident with x.
Then we consider whether f, and belong to the same path or not.
Case 7.4.4.1. f, and belong to the same path in .
Then we can complete the proof as in Case 2.2.2.
Case 7.4.4.2. f, and belong to different paths in .
Case 7.4.4.2.1. Just two of f, and belong to a path in .
Without loss of generality, assumes is a 3-path containing f and and starting at a, is a 3-path containing and starting at b. Then a and b are both incident with x. Then we can complete the proof as in Case 5.3.3.
Case 7.4.4.2.2. Each of f, and belong to a path in separately. Let (resp. , ) be a 3-path starting at a (resp. b, c), and it contains f (resp. , ). Then a, b and c are all incident with x. Notice that . It is easy to find a 3-path (resp. , ) that does contain a missing edge in (resp. , ), and (resp. , ) and (resp. , ) have no common vertices. Connecting (resp. , ) to a (resp. b, c), we can obtain a 3-path (resp. , ). Notice that , so we connect , , , , and to x. Combining them with , we can get an extended star in . □