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 . □