Applied Mathematical Sciences, Vol. 9, 2015, no. 146, 7247-7254 HIKARI Ltd, www.m-hikari.com http://dx.doi.or/10.12988/ams.2015.510679 The -Extra Conditional Dianosability of Folded Hypercubes Weipin Han Collee of Mathematics and Information Science Henan Normal University, China Shiyin Wan Collee of Mathematics and Information Science Henan Normal University, China Copyriht c 2015 Weipin Han and Shiyin Wan. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the oriinal work is properly cited. Abstract Dianosability plays a crucial role in measurin the reliability and fault tolerance of interconnection networks. The -extra conditional dianosability of multiprocessor systems is a new dianosability, which is more accurate than the classical dianosability. In this paper, we show that the -extra conditional dianosability t (F Q n ) of the folded hypercube F Q n is ( + 1)n ( + 1 under the PMC model and the MM model in some cases, which is several times larer than the classical dianosability of folded hypercubes. Mathematics Subject Classification: 05C90 Keywords: interconnection networks, folded hypercubes, dianosability 1 Introduction As an underlyin topoloy of a multiprocessor system, an interconnection network is usually modeled by a connected raph G, whose vertices represent processors and edes represent communication links. The process of identifyin faulty processors is called the dianosis of the system. Dianosability is
7248 Weipin Han and Shiyin Wan defined as the maximum number of faulty processors which the system can uarantee to identify, which is an important parameter to measure the reliability and fault-tolerance of a multiprocessor system. The hypercube structure [8] is a well-known interconnection network model for multiprocessor systems. As a variant of the hypercube, the folded hypercube [3] can be constructed from an n-dimensional hypercube by addin 2 n 1 edes, called complementary edes. The n-dimensional folded hypercube F Q n has recently received considerable attention [3], [4], [10], [12], [13]. The PMC model [7] and the MM model [6] are two widely adopted as the fault dianosis model. In 2015, Zhan et al. [11] defined -extra conditional dianosability under assumption that every component of G F has at least + 1 vertices, where F is a faulty subset of G. Moreover, they also obtained -extra conditional dianosability results for hypercubes under the PMC model and the MM model. In this paper, we show that the -extra conditional dianosability t (F Q n ) of the folded hypercube F Q n is ( + 1)n ( + 1 under the PMC model and the MM model in some cases, which is several times larer than the classical dianosability of folded hypercubes. For raph-theoretical terminoloy and notation not defined here we follow [1]. Definition 1.1 ([11]) A system G is -extra conditionally t-dianosable if and only if for each pair of distinct faulty -extra vertex subsets F 1, F 2 V (G) such that F 1 t, F 2 t, F 1 and F 2 are distinuishable. The -extra conditional dianosability of G, denoted as t (G), is the maximum value of t such that G is -extra conditionally t-dianosable. 2 Results Lemma 2.1 ([10]) Let A be a subraph of F Q n with V (A) = +1 n 1 for n 4 and 0 n 4. Then N F Qn (V (A)) ( + 1)(n + 1) 2 (. Lemma 2.2 ([10]) Let F Q n = D 0 D1, where n 8 and 0 n 4, and let A D 1 and A = K 1,. Then the followin two statements hold: (i) N F Qn (V (A)) = (+1)(n+1) 2 (, (ii) F Qn C F Qn (V (A)) is a connected subraph of F Q n with at least + 1 vertices. Lemma 2.3 ([10]) κ (F Q n ) = ( + 1)(n + 1) 2 (, for n 8 and 0 n 4. Lemma 2.4 ([12]) Any two vertices in V (F Q n ) exactly have two common neihbors for n 4 if they have. Lemma 2.5 ([2]) For a system G and any two distinct subsets F 1 and F 2 of V, F 1 and F 2 are distinuishable under the PMC model if and only if there exist two vertices u V \(F 1 F 2 ) and v F 1 F 2 such that (u, v) E(G).
The -extra conditional dianosability of folded hypercubes 7249 Lemma 2.6 ([9]) For a system G and any two distinct subsets F 1 and F 2 of V, F 1 and F 2 are distinuishable under the MM model if and only if at least one of the followin conditions is satisfied: (i). There are three vertices u, w V (G) \ (F 1 F 2 ) and v F 1 F 2 such that uw E(G), vw E(G). (ii). There are three vertices u, v F 1 \ F 2 and w V (G) \ (F 1 F 2 ) such that uw E(G), vw E(G). (iii). There are three vertices u, v F 2 \ F 1 and w V (G) \ (F 1 F 2 ) such that uw E(G), vw E(G). Lemma 2.7 Assume that n 8 and 0 n 4. Then t (F Q n ) ( + 1)n ( + 1 under the PMC model and the MM model. Proof. Suppose that A is a connected subraph of F Q n, A D 1 and A = K 1,. Let F 1 = N F Qn (V (A)) and F 2 = C F Qn (V (A)). Then, by Lemma 2.2, F 1 = (+1)(n+1) 2 ( ) 2, F2 = (+1)n ( +2, and F Qn C F Qn (V (A)) is a connected subraph of F Q n with at least + 1 vertices. Thus, F 1 ( + 1)n ( ) 2 + 2, F2 ( + 1)n ( + 2, and both F1 and F 2 are -extra vertex subsets. Since there is no ede between F 1 F 2 and V (F Q n )\(F 1 F 2 ), by Lemmas 2.5 and 2.6, F 1 and F 2 are indistinuishable under the PMC model and the MM model. By Definition 1.1, t (F Q n ) ( + 1)n ( + 1 under the PMC model and the MM model. This completes the proof. Lemma 2.8 Assume that n 8 and 0 n 4. Then t (F Q n ) ( + 1)n ( + 1 under the PMC model. Proof. By contradiction. Suppose that t (F Q n ) ( + 1)n (. Let F1 and F 2 be two distinct -extra faulty vertex subsets of F Q n, such that F 1 ( +1)n ( ) 2 +1, F1 ( +1)n ( +1, and F1 and F 2 are indistinuishable under the PMC model. Let Φ n () = ( + 1)n ( + 1, where n 8 and 0 n 4. Then Φ n () is strictly monotonic increasin for 0 n 4. Therefore, V (F Q n ) F 1 F 2 = V (F Q n ) F 1 F 2 + F 1 F 2 2 n 2[(+ 1)n ( ) 2 +1] 2 n 2[(n 4+1)n ( ) n 4 2 +1] = 2 n (n 2 +3n 18). Whereby n 8, then V (F Q n ) F 1 F 2 > 0. Therefore, V (F Q n )\(F 1 F 2 ). Since F 1 and F 2 are indistinuishable under the PMC model, and by Lemma 2.5, there is no ede between F 1 F 2 and V (F Q n )\(F 1 F 2 ). So all neihbors of vertices in F 1 F 2 and V (F Q n )\(F 1 F 2 ) are located in F 1 F 2. Thus, the neihbors of all these vertices in the components of F Q n [F 1 F 2 ] and F Q n [V (F Q n )\(F 1 F 2 )] are located in F 1 F 2. By the definitions of F 1 and F 2, we conclude that every component of F Q n F 1 and F Q n F 2 has at least + 1 vertices. If F 1 F 2 is deleted, then F Q n F 1 F 2 will be disconnected and all the components of F Q n F 1 F 2 have at least + 1 vertices. Thus, F 1 F 2 is a -extra vertex cut of F Q n. By Lemma 2.3, we conclude that
7250 Weipin Han and Shiyin Wan F 1 F 2 ( + 1)(n + 1) 2 (. Notin that F1 F 2, without loss of enerality, we assume that F 2 \F 1. Observin that F 1 is a -extra vertex cut, thus every component of F Q n F 1 has at least +1 vertices. Since F 2 \F 1 is not connected with V (F Q n )\(F 1 F 2 ), every component of F Q n [F 2 \F 1 ] has at least + 1 vertices, i.e., F 2 \F 1 + 1. Hence F 2 = F 2 \F 1 + F 1 F 2 + 1 + ( + 1)(n + 1) 2 ( ( = ( + 1)n + 2, which contradicts to F 2 ( + 1)n ( + 1. This completes the proof. Lemma 2.9 Let n 32, 3 n and F be a -extra faulty vertex subset 4 of F Q n. If F Q n F has at most n vertices of deree 1, then t 2 (F Q n ) ( + 1)n ( + 1 under the MM model. Proof. By contradiction. Suppose that the -extra conditional dianosability of F Q n t (F Q n ) ( + 1)n ( under the MM model. Assume that there exist two distinct -extra faulty vertex subsets F 1 and F 2, such that F 1 ( + 1)n ( ) 2 + 1, F2 ( + 1)n ( + 1, every component of F Qn F 1 and F Q n F 2 has at most n vertices of deree 1, and F 2 1 and F 2 are indistinuishable under the MM model. Whereby n 32, then n < n 4. Since Φ 4 n() = (+1)n ( +1 is strictly monotonic increasin for 0 n 4, V (F Qn ) F 1 F 2 = V (F Q n ) F 1 F 2 + F 1 F 2 2 n 2[( + 1)n ( + 1] > 2 n 2[(n 4 + 1)n ( ) n 4 2 + 1] > 0. So V (F Qn )\(F 1 F 2 ). Claim 1. There is no isolated vertex in F Q n (F 1 F 2 ). Assume that S is a set of all the isolated vertices in F Q n (F 1 F 2 ). For any vertex w S, N F Qn (w) F 1 F 2. We prove that F 1 \F 2 and F 2 \F 1. If F 2 \F 1 =, then N F Qn (w) F 1. Then w is an isolated vertex in F Q n F 1 and so = 0, a contradiction. Similarly, F 1 \F 2. Assume that there exist two distinct vertices u 1, v 1 F 2 \F 1 such that u 1 w, v 1 w E(F Q n ), by Lemma 2.6 (iii), F 1 and F 2 are distinuishable, a contradiction. If there exists no vertex u 1 F 2 \F 1 such that u 1 w E(F Q n ), then N F Qn (w) F 1 and so w is an isolated vertex of F Q n F 1. Thus = 0, a contradiction. Therefore, N F Qn (w) (F 2 \F 1 ) = 1. Similarly, N F Qn (w) (F 1 \F 2 ) = 1. Since F Q n is (n + 1) reular, N F Qn (w) (F 1 F 2 ) = N F Qn (w) N F Qn (w) (F 1 \F 2 ) N F Qn (w) (F 2 \F 1 ) = n + 1 1 1 = n 1. Thus F 1 F 2. If F Q n F 1 has at most n vertices of deree 1, then S n. Let F 2 2 1\F 2 = p, F 2 \F 1 = q and V = (F 1 F 2 ) S. Then V = p + q + S and p, q 1. We prove that N F Qn (V ) F 1 F 2. If F 1 F 2 S = V (F Q n ), then N F Qn (V ) F 1 F 2. If F 1 F 2 S V (F Q n ), then let R = V (F Q n )\(F 1 F 2 S), then R. Observin that all the isolated vertices of F Q n F 1 F 2 are included in S, so there is no isolated vertex in F Q n [R]. We claim that there is no ede between F 1 F 2 and R. Otherwise, notin that F Q n [R] has no isolated vertex, thus there exists an ede uv in F Q n [R] such that u(v) is adjacent to some vertex in F 1 F 2. By Lemma 2.6 (i), F 1 and F 2 are distinuishable, a contradiction. Since S is a set of all the isolated vertices in F Q n F 1 F 2, we conclude that
The -extra conditional dianosability of folded hypercubes 7251 there is no ede between S and R. Furthermore, there is no ede between V and R. Therefore, N F Qn (V ) F 1 F 2 Case 1. S > 2. Subcase 1.1. Either p + 1 or q + 1. Without loss of enerality, suppose p + 1. For an arbitrary vertex w S, N F Qn (w) (F 1 F 2 ) = N F Qn (w) N F Qn (w) (F 1 \F 2 ) N F Qn (w) (F 2 \F 1 ) = (n+1) 1 1 = n 1. By Lemma 2.4, F 1 F 2 (n 1) S 2 ( ) S 2. Let S = t, then h n ( S ) = h n (t) = (n 1)t 2 ( t = t 2 + nt, and h n ( S ) is strictly monotonic increasin for 2+1 S n. Thus, F 2 1 F 2 h n (2+1). Since F 2 = F 2 \F 1 + F 1 F 2 = q + F 1 F 2 and F 2 ( + 1)n ( + 1, F 1 F 2 (+1)n ( ( +1 q (+1)n ( +1 (+1) = (+1)n. Therefore, h n (2 + 1) F 1 F 2 ( + 1)n (. If n 32 and 3 n 2n 8, then n. Thus, h 4 7 4 n(2 + 1) [( + 1)n ( ] = (n 1)(2 + 1) 2 ( ) ( 2+1 2 [( + 1)n ] = (2n 7 7) 1 3 1 1 > 0, 2 2 a contradiction. Subcase 1.2. p and q. Since V = (F 1 F 2 ) S = F 1 \F 2 + F 2 \F 1 + S = p + q + S and S n, V 1 = p + q + S 1 + + n (n 1 = 2 +. Whereby 2 2 2 3 n (n, i.e., 4 n 0, then (2 + ) (n 1) = 1 (4 n) 0. 4 2 2 So V 1 2 + (n n 1. Since p 1, q 1 and S > 2, 2 V 1 = p+q+ S 1 > p+q+2 1 2+1 > 2. Then 2 < V 1 n 1. Let f n () = (+1)(n+1) 2 (, then fn () is strictly monotonic increasin when 0 n 1. Since N F Qn (V ) F 1 F 2, by Lemma 2.1, F 1 F 2 N F Qn (V ) ( V 1+1)(n+1) 2( V 1) ( ) V 1 2 > (2+1)(n+1) 4 ( 2 ) 2. Since n 32 and 3 n < (2n 3), F 4 3 1 F 2 [( + 1)n ( + 1] > [(2 + 1)(n + 1) 4 ( ) ( 2 2 ] [( + 1)n ) 2 + 1] = (2n 3 3) > 0. Therefore, 2 F 2 = F 2 \F 1 + F 1 F 2 = q+ F 1 F 2 > 1+(+1)n ( ( +1 > (+1)n +1, a contradiction. Case 2. S 2. If n 32, then 3 n < n 4. Let I 4 n() = ( + 1)n + ( + 1, then I n () is strictly monotonic increasin for 3 < n 4. Therefore, V (F Q n ) F 1 F 2 S 2 n 2[(+1)n ( +1] 2 > 2 n 2[(n 4+1)n+ n 4 ( ) n 4 2 +1] = 2 n (n 2 +5n 26) > 0. So R = V (F Q n )\(F 1 F 2 S). Since all the isolated vertices are in S, there is no isolated vertex in F Q n [R]. Then there exists one ede uv E(F Q n [R]). If u(v) is adjacent to some vertex in F 1 F 2, then by Lemma 2.6 (i), F 1 and F 2 are distinuishable, a contradiction. If there is a vertex u(v) R which is adjacent to some vertex w in S, and let u 0 N F1 F 2 (w). By Lemma 2.6 (i), F 1 and F 2 are distinuishable, a contradiction. Thus there exists no ede between V and R. Moreover, there exists no ede between (F 1 \F 2 ) S and R, and there exists no ede between (F 2 \F 1 ) S and R. Therefore, N F Qn (R) F 1 F 2. Since every component
7252 Weipin Han and Shiyin Wan of F Q n F 1 and F Q n F 2 has at least + 1 vertices, we conclude that every component of F Q n [R], F Q n [(F 1 \F 2 ) S] and F Q n [(F 2 \F 1 ) S] has at least + 1 vertices. We claim that every component of F Q n [V ] has at least + 1 vertices. If every component of F Q n [F 2 \F 1 ] has at least +1 vertices, then the claim holds because every component of F Q n [(F 1 \F 2 ) S] has at least + 1 vertices. Suppose that there is a component C 0 of F Q n [F 2 \F 1 ] with less than + 1 vertices. If there exists no ede between V (C 0 ) and (F 1 \F 2 ) S, then C 0 is also a component of F Q n [(F 2 \F 1 ) S], a contradiction. Thus there exists at least one ede between V (C 0 ) and (F 1 \F 2 ) S. Without loss of enerality, we assume that uv is an ede between V (C 0 ) and (F 1 \F 2 ) S, such that u V (C 0 ) and v V (C 1 ), where C 1 is a component of F Q n [(F 1 \F 2 ) S]. Since every component of F Q n [(F 1 \F 2 ) S] has at least + 1 vertices, there are at least + 1 vertices in C 1. Assume that C 2 is an arbitrary component of F Q n [V ], such that F Q n [V (C 0 ) V (C 1 )] is a connected subraph of C 2, then there are at least + 1 vertices in C 2. Therefore, every component of F Q n [V ] has at least + 1 vertices. Observin that every component of F Q n [V ] and F Q n [R] has at least + 1 vertices, and there is no ede between V and R, thus F 1 F 2 is a -extra vertex cut of F Q n. By Lemma 2.3, we have F 1 F 2 ( + 1)(n + 1) 2 (. Assume q = F 2 \F 1 + 1, then F 2 = F 2 \F 1 + F 1 F 2 = q + F 1 F 2 + 1 + ( + 1)(n + 1) 2 ( ( = ( + 1)n ( + 2 > ( + 1)n + 1, a contradiction. So q. Similarly, p. Since there exists no ede between (F 2 \F 1 ) S and R, and every component of F Q n F 1 has at least +1 vertices, every component of F Q n [(F 2 \F 1 ) S] has at least +1 vertices. Thus, +1 (F 2 \F 1 ) S = F 2 \F 1 + S = q + S. Similarly, +1 p+ S. Let r = min{p, q}, then + 1 min{p + S, q + S } = r + S. So r S 1. Since N F Qn (V ) F 1 F 2 and V = p + q + S, by Lemma 2.1, F 1 F 2 N F Qn (V ) (p+q+ S 1+1)(n+1) 2(p+q+ S 1) ( ) p+q+ S 1 2. Notin that n and S 2, thus + r = r + r + ( r) p + q + r 4 p+q+ S 1 2+ S 1 4 1 n 1. Since f n () = (+1)(n+1) 2 ( ) 2 is strictly monotonic increasin for 0 n 1, F 1 F 2 (p + q + S 1 + 1)(n+1) 2(p+q + S 1) ( ) ( p+q+ S 1 2 ( +r +1)(n+1) 2( +r) +r ) 2. Since F 2 (+1)n ( ) 2 +1, 0 F2 [(+1)n ( +1] = F1 F 2 + F 2 \F 1 [(+1)n ( ) ( 2 +1] (+r+1)(n+1) 2(+r) +r ) ( 2 +r (+1)n+ ) 2 1 = nr 1 2 r2 r + 1r = (n+ 1)r [(1+r)+ 1 2 2 2 r2 ]. So (1+r)+ 1 2 r2 (n+ 1)r, 2 i.e., 2 + 2r + r 2 2nr + r. Whereby 3 n, then n 4. So 4 2 + 2r + r 2 2 4r + r, which implies that r 2 + 2 6r + r. Notin that p, q and r = min{p, q}, so r. Since r + 2 r 2 + 2 6r + r, 2 5r+r = (5+1)r, 2 r 5+1 = 5+ 1. Whereby r 1, then 2 r 2 < 5+ 1, a contradiction. Therefore S =. This completes the proof of Claim 1. By Claim 1, there is no isolated vertex in F Q n [V (F Q n )\(F 1 F 2 )]. That is to say, for any vertex u 1 V (F Q n )\(F 1 F 2 ), there exists one vertex
The -extra conditional dianosability of folded hypercubes 7253 v 1 V (F Q n )\(F 1 F 2 ) which is adjacent to u 1. Assume that there exists one vertex w F 1 F 2 that is adjacent to u 1. By Lemma 2.6 (i), F 1 and F 2 are distinuishable, a contradiction. So there exists no ede between F 1 F 2 and V (F Q n )\(F 1 F 2 ). Since both F 1 and F 2 are -extra vertex subsets, every component of F Q n [F 1 \F 2 ], F Q n [F 2 \F 1 ] and F Q n [V (F Q n )\(F 1 F 2 )] has at least + 1 vertices. Hence, all the neihbors of the vertices in F 1 F 2 and V (F Q n )\(F 1 F 2 ) are located in F 1 F 2. If we delete F 1 F 2, then F Q n F 1 F 2 will be disconnected and all components of F Q n F 1 F 2 have at least + 1 vertices. Therefore, F 1 F 2 is a -extra vertex cut of F Q n. By Lemma 2.3, F 1 F 2 ( + 1)(n + 1) 2 (. Since every component of F Q n [F 2 \F 1 ] has at least + 1 vertices, F 2 \F 1 + 1. Hence, F 2 = F 2 \F 1 + F 1 F 2 +1+( +1)(n+1) 2 ( = ( +1)n ( +2, a contradiction. The proof is completed. By Lemmas 2.7, 2.8, 2.9, we have the -extra conditional dianosability of F Q n as follows: Theorem 2.10 Assume that n 8 and 0 n 4. Then t (F Q n ) = ( + 1)n ( + 1 under the PMC model. Theorem 2.11 Assume that n 32 and 3 n. If F Q 4 n F has at most n vertices of deree 1, where F is a -extra vertex subset of F Q 2 n, then t (F Q n ) = ( + 1)n ( + 1 under the MM model. 3 Conclusion In this study, we show that the -extra conditional dianosability t (F Q n ) of the folded hypercube F Q n is ( + 1)n ( + 1 under the PMC model and the MM model in some cases. ( + 1)n ( + 1 is several times larer than the conditional dianosability of the folded hypercube iven by Zhu et al. [13] under the PMC model and the conditional dianosability of the folded hypercube iven by Hsieh et al. [4] under the MM model. References [1] J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, The Macmillan Press Ltd, New York, 1976. [2] A.T. Dahbura, G.M. Masson, An O(n 2.5 ) fault identification alorithm for dianosable systems, IEEE Transactions on Computers, C-33 (1984), 486-492. http://dx.doi.or/10.1109/tc.1984.1676472
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