The Maximum Dimensional Fault-Free Subcube Allocatable in Faulty Hypercube

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1 The Maximum Dimensional FaultFree Subcube Allocatable in Faulty Hypercube Hiroshi Masuyama Information and Knowledge Engineering, Tottori 0 Japan Toshihiko Sasama Graduate School, Tottori 0 Japan Hiroyuki Hashimoto Graduate School Tottori 0 Japan Abstrust The maximum dimensional subcube located in faulty hypercubes is studied in this paper. Most parallel algorithms can be formulated with the dimension n of the hypercube being a parameter of the algorithm. The reconfiguration problem in a hypercube reduces to finding the maximum dimensional faultfree subcube in the hypercube, that is, helps in achieving graceful and the most effective degradation of the system. Thai paper presents the maximum, number of faults on an n (n 2)subcube. key words: Hypercube, Fault Tolerance, Allocation, The Maximum Dimemional FaultFree Subcube 1 Introduction Different but related approaches to hypercube faulttolerance have been reported. Chang and Bhuyan [l] present,ed a modified subcube partitioning strategy to construct faultfree subcubes in a hypercube with faults. The technique selects a regular (n 1)cube and then the faulty node in the selected (n 1)cube is replaced with its corresponding nonfaulty node in the other (n 1)cube. They showed that [n/21 faults ca.n be tolerated in the worst case by the modified subcube partitioning technique while maintaining a faultfree (n 1)cube. Yang and Raghavendra [2] presented a distributed scheme for the reconfiguration of an embedded binary tree into a hypercube with faults. In this scheme, the faulty node in the selected binary tree is replaced with a nonfaulty node which is out of the tree and at Hamming Distance 1. Horng and Kleinrock [3] proposed a set of techniques to restore the regularity of a Boolean ncube network in the presence of node failures. The shortest path between any node pair in the modified faultfree network holds the shortest property also in the nchbe. Bruck et al. [] studied a technique to obtain tolerant partitions by which partitioned m cube S is guaranteed such that S conta,ins a connected component of 2m1 + 1 or more nonfaulty nodes. Most parallel algorithms can be formulated with the dimension n of the hypercube being a parameter of the algorithm []. Since a subcube is a subset of a hypercube which preserves the properties of hypercube, the reconfiguration problem in a faulty hypercube re duces to find the maximum dimensional faultfree subcube. B.Becker and H.U.Simon [] presented simple procedures to find a maximal dimension of a faultfree subcube. F.Ozgiiner and C.Aykanat [B] also presented a procedure to find the maximum dimensional faultfree subcube where the principle of inclusionexclusion is used. This procedure can always give the maximum dimension d, the number of faultfree dsubcubes, and the complete set of faultfree dsubcubes. Another algebraic technique, called ATARIC 11, was reported. ATARIC uses two algebraic operators # and $ to identify the maximum dimensional faultfree subcube. These presented procedures are all based on the given addresses of faulty nodes. In this paper, we will try to formulate the maximum dimension of a faultfree subcube located in a faulty hypercube when the number of faults is given. It will be assumed throughout this paper that all faults are static and are known. Both nodes and edges may be faulty. However, we will only consider node faults, as an edge fault can be tolerated by assuming that one of the node s incidence upon it is faulty. It will be assumed that faulty nodes can neither perform computation nor communication. 2 Any Dimensional Subcube Allocation The ncube network consists of 2 nodes, each of which are numbered from 0 to 2 1 by nbit binary numbers an1a,2... a0, called the address of the node. Any two nodes on an ncube are adjacent iff their addresses differ by exactly one bit. A subcube of dimension m, denoted by an msubcube, is addressed by a string of n symbols drawn from the set {0,1, *}, where * is a don t care symbol. Coordinate values 0 and 1 can be referred to as the fixed or bounded coordinates and * as a free coordinate. An msubcube on an ncube has (n m) bounded coordinates and m free coordinates. For example, in a cube, nodes 0000,0001,0010 and 0011 form a 2cube addressed by 0 0 ~. [Property 11: The number of faults on an n(> 3) (n 2)subcube is not over n. 0110/9 $ IEEE 220

2 Proof: We first assume any index represented in the binary system with n bits. Let f' be a set of indices which includes different kinds of ordered pairs (l,l),(l,o),(o,l) and (0,O) for every bit pair i and j (i # j ) in n (2 3) bits. Then, F is given in the following equation: F= {(anlan2an3... azalao), (Un]Qngang *..z2z]eo), (enlan2an3 ''.izzlco),... (cnlcnzcn3 '.. Uzalzr3), (izn1zn2izn3...e2zl'lao)} j means that there exists no faultfree (n2)subcube when F is assumed to be a set of faulty nodes on n cube. The existence of proves Property 1. Q.E.D. The hypercube we consider is assumed to be n 2 3, which is satisfied with the property mentioned in the above proof. Next, we will try to extend an faultfree (n 2) subcube allocation up to an arbitrary msubcuhe allocation. [Property 21: The number of faults on an n( 2 3) (n 3)subcube is not over (2n 1). Proof: In this proof, in order to evaluate faultfree (n 3)subcubes, the condition of faultfree n 2) subcube to exist on ncube is utilized. We wil r divide an ncube into two n 1)subcubes and discuss the minimum number o I faults such that there exists no faultfree (n 3)subcube in an (n 1)cube. Namely, we will show that 2n faults on an ncube are sufficient to exist exactly one fault in every (n 3)subcube. In Property 1, we have introduced a set F of faults such that there exists no faultfree (n 2)subcube on an ncube. We will apply n 1 + n in F to obtain a set Fnl of faults such that there exists no faultfree (n S)subcube in an (n 1)cube. Then, Fnl is given in the following equation: pn1 = { (an2an3 ' '. a]uo), (anzan3 ' ' ' also), (zn2an3 ' ' alzo),... (znzzn3.' ' alzf)), (zn2zn3 ' ' 'zlao)} We obtain ljnl 1 = n. Thus, if each (n 1)subcube includes at least n faults, then there not always exist one faultfree (n 3)subcube on ncube. In other words, 2n is the number of faults such that there exists at least one fault in every (n 3)subcube on an n cube. According to Property 1, we will discuss the lnumber of faultfree disjoint (n3)subcubes on an ncube when the number of faults is 2n or more. Assume that there are 2n faults on an ncube. On the condition that the number of faultfree (n 3)subcubes is zero on the ncube, we can consider only one case such that one (n 1)subcube includes n faults and the other includes n faults. That is, by applying Property 1 to an (n 1)cube, we obtain that there not always exists one faultfree (n 3)subcube in the both (n 1) cubes which has at least n faults in each. Therefore, Property 2 is proven. Q. E.D. [Property 31: The number of faults on an n dimensional hypercube on which there always exists at least one faultfree (n i 2)subcube is not over P(n i + 1) 1. Proof: We will prove that 2'(n i+ 1) is the number of faults such that there not always exists faultfree (n i 2)subcube on an ncube. By using Properties 1 and 2, we will obtain the number of faults such that all disjoint (n i 2)subcubes in each one of 2i disjoint (n i)subcubes are able to be not faultfree. The case of i = 0: An (n i = n)subcube is an ncube. From Property 1, we obtain that (n + 1) is the number of faults such that all (n i 2 = n 2) subcubes are able to be not faultfree on the ncube. Then, the address set of faulty nodes is F. The case of i = 1: Iin the proof of Property 2, we obtain that n is the inumber of faults such that all disjoint (n i 2 = 2 3)subcubes are able to be not faultfree in each (n i = n 1)subcube. Then, the address set of faulty nodes is obtained in the following discussion. The addresses of any 2l n 1)subcubes are obtained by rewriting address I UnlU,2...aj+1ajaj1...alal)) to (**...*1*...**) and (**...*O*...**), where * means don't care. Let n + n 1 be in F and alpply F to (n 1) bits with *. We obtain an address set of faulty nodes such that all disjoint (n 3)subcubes are able to be not faultfree in an (n 1)subcube (* *... * 1 *... * *). We can perform the same discussion in the case of another (n 1)subcube (* *... * 0 *... * *). The total number of faults in these two (n 1)subcubes is clearly twice n. The case of i 2 2: Let's evaluate the addresses of faulty nodes such that all (disjoint (n i 2)subcutbes are able to be not faultfree in any (n i)subcube. In the case of i = 1, we could induce the address set of faulty nodes from F by taking n 1 + n. In the case where all disjoint (n i 2)subcubes are able to be not faultfree in any (n 1')cube which is set i bits to fixed values, the address set of faulty nodes is induced by taking n i t n. Since the number of faulty nodes is (n i + 1) in each (n i)subcube, then we obtain 2"n i + 1). Q.E.D. Let us remember that, the minimum number of faults on an n(> 3)cube on which there always exists no faultfree (n m)subcube is the minimum number E of indices such that every E mbit binary numbers composed of m bits in each of E indices can cover all 2 binary numbers. 3 (n 2)Dimensional Subcube Allocat ion In this section, we will focus on faultfree (n 2) subcubes and try to bring the number 2*(n i+ 1) 1 presented in Property 3 close to the optimum value. Let be the minimum set of indices which includes different kinds of ordered pairs (1, l), (1,0),(0, l), and (0,O) for every bit pair i and j (i # j)in n(> 3) bits. Let Fk be F when n = IC. = {(a2alao),(a2~1~1), (i2alib), ( ~ ~ i a o ) } Since IF1 2, then F3 = F when n = 3. We can cal 221

3 culate F when n = in an exhaustive method. Now, we estimate the time complexity to calculate F with = 1 in this method. The calculation procedure is as follows : Select 1 indices which make a new combination in a set of 2" indices, and judge whether there exist different kinds of ordered pairs or not for every bit pair in the 1 indices. Therefore, the upper bound of the time complexity can be given as follows : M {(2"/2)2"/(//2)'((2" 1)/2pnl)}. 21' {(n 1)n/2} Then, the order of the upper bound of the time complexity is O(2'". 11'. n2). The time complexity to calculate F becomes tremendous as n becomes very large, but F can be calculated within the realistic range, that is, n 1. The calculated values of F when n 1 are shown in Table 1. When n = 1, the fact F # is calculated. Next, we will formalize an approximate value F( 2 F) of F. Let Fk and Fk be F and F when n = k, respectively. Fk can be calculated by using FkI( > k'). Since IF/ = is known in an exhaustive method, we get F = F = F. J.Bruck, et al. [] said in Corollary that F =. For n 2, we will calculate Fk without Fk. For the expression~l simplicity, we express one index by one row, and Fk with IFkI = 1 by a matrix with 1 rows and k columns. Fk can be expressed by using two 's, three f i t 's, and six &j's as shown in Fig.1 (a), (b), and (c), respectively. The point to express F k shown in Fig.1 (a), (b, and (c) is as follows : Since there always exist all kin d s of ordered pairs for every bit pair in one &I and there always exist only 2 ordered pairs (0,O) and (1,l) for every bit pair whose 2 bits are in different Fk's each other, we need the minimum number of indices to make 2 ordered pairs ( 1,O) and (0,l) for every bit pair whose 2 bits are in different Fk each other. These minimum numbers of required indices are 2, 3, and when the numbers of basic PkT's are 2, 3, and, respectively, as shown in Fig.l(a), (b), and (cb When mk' # k, in order to set n =, eliminate any its from every index with mk' bits. The important point is to select the combination lead to IFk I = minimum. The expressions of Fk shown in Fig.1 (a), (b), and (c) will be referred to as types 2&, 3fi), and fi1, respectively. Let us calculate F. =. By using type 2F3, we get F with six bit indices. <.lfgi.,fs with six bit indices can be obhined by eliminating any one bit from F. Let_ us next calculate Fg. By using type 3F3, we get Fg with seven 9bit indices. IF91 < 1Fgl. F with indices and F with indices can be eliminating any 2 bits and 1 bit, respectively, from every index in Fg. _Though F can be also obtained by using type 2F, b0t.h types 3F3 and 2F require indices. Any F k can obtained by extending this method. On condition that the number of required indices must be as _small aspossible, though single type 2F3,3F3,F3,2F, or 3F is adequate to express any F, for n 1, a hierarchical design by using several different types can be also introduced. From the above discussion, IFnI can be formalized. In the case where type mf3(m = 2,3,) is used, since 2"'.3a2."3(> n) = 3.2"'.3uz1."3(cu2 2 1), then types 2F3,3F3, and F3 must be used al,(a2 l), and 0 3 times, respectively. This fact leads to IFnI = + 2al+ 3(az 1) On the other hand, in the case where type mf is used, since 2a1.3aa."3(> n) =. 2a12 ' + 3"2 a3(012 a), then types 2F,3, and F must be used (a1 2),02, and 03, respectively. This fact leads to IFn I = + 2(al 2) The final formula of IFn I is defined as follows: For a1, az, and a3 given from 3 ~ < 2a1. 3a2. a3, lfnl = min( + 2a1 + 3a2 + a3 (for a2 2 I), +2((~1'2)+3a2+a3 (for a1 2 2)) I obtained from the above equation is presented IFn in Table 1. From the above discussion, we can present the following theorem without the proof. [Theorem 11: The number of faults on an n(2 3) (n 2)subcube is not over IFn/. We will present the other theorem. This theorem depends on the following easy lemmas. [Lemma 11: For all n 2, given at most 1 faulty Table.1 n I The number of required indices. () () () IF" I I employed type 2B 2B 3B or 2B 3% or 2Fi 3% g or 3B % or 3F 3 or 3i$ B F3 % B B 222

4 node in an (n 1)subcube, there exist two disjoint faultfree (n 2) and (n 3)subcubes which can take any dimensions. Proof: Obvious. pemma 21: For all n 2, given at most LIFrF;II.I/2J fau ty nodes in an (n 1)subcube, there exist two disjoint faultfree (n 3)subcubes. Proof: Obvious. The above two lemmas lead to that, given at most 1 and LIPn 1/21 faulty nodes in an (n 1) and the other n 1)subcubes respectively, there exist two (n 2)subcubes. [Theorem 21: For all n 2, given any set of F consisting of lfnl [lfnl/2] or fewer faulty nodes on an ncube, there exist at least two faultfree disjoint (n 2)subcube. Proof: lfnl is the minimum number of faulls on an ncube on which there always exists at least, one fault on every disjoint (n 2)subcube. Given IFn/ faulty nodes in an ncube, the distributions of the faulty nodes are classified into three types : Types 1, 2, and 3 are that the ncube has zero, one, and two or more faultfree disjoint (n 2)subcubes respectively. Since the last type satisfies this theorem, let us prove this theorem in the case of Types 1 and 2. Type 1 : On condition that an ncube has no fault free (n 2)subcube, the ncube has two (n 1) subcubes on which faults must be well bal anced in the faulty nodes distribution, t,hat is, one (n 1)subcube is occupied by [li; n1/21 faults and the other by llfn 1/21 faults. Therefore, if LIPn 1/21 faults are removed, then the distributions of the remaining (lfnl LIFnl/2J) faulty nodes can be classified into the following two cases : Every two disjoint (n 1) subcubes have one and L1pn1/2 faulty nodes respectively, and every two disjoint (!I 1)subcubes have zero and [IFn1/21 faulty nodes respectively. In the former case, this theorem can be proven by Lemmas 1 and 2. In the latter case, this theorem can be proven by the result that one (n 1)subcube becomes faultfree. Type 2 : On condition that an ncube ha3 only one faultfree (n 2)subcube, the ncube has two (n 1)subcubes on which IFn[ faults must be idso well balanced in the faulty nodes distribution. Therefore, if LlFn1/21 faults are removed, there always errist two disjoint (n 1)subcubes which have at most one and at most klfnl/21 faults respectively. In this case, the theorem can be also proven by Lemmas P and 2. Q.E.D. Fig.2 shows a cube which possesses two disjoint faultfree 3subcubes. Next, from a different angle, we will formalize a different approximate value F (1 p) of P. Let F be also a set of indices which include different kinds of ordered pairs (1,1>, (l,o), (0, l), and (0,O) of 2 different bits i and j for every pair i and j in n(> 3) bits, as shown in Fig.3 (a) and (b) where F is shown by a matrix in the same way as F. In this figure, F has Q indices whose bit len th is alcj where 1 j [(a 1)/21 1 in fa), [(a 1)/2] + 1 j cy 1 in (b). A set of indices in the square solid line in Fig.3 (a) always includes three kinds of ordered pairs (1, Ci), (0, l), and (0,O) for every bit pair, but not always includes 1,1>. Therefore, a special index ( ) is prepare for the lacked ordered pair (1,l). On the other hand, a set of indices in the square solid line in Fig.S(b) always includes three kinds of ordered pairs (1,0),(0, l), and (1,l) for every bit pair, brit not always includes (0,O). A special index ( ) is prepared for the lacked ordered pair (0,O). Notice that F in the case j = 1 conforms to F, and that there exists a bit complement relation between two F s in Fig.3 (a) and (b) The optimum F 223

5 is F defined by the minimum a. We will present the following property for F shown in Fig.3 (a). [Property 1: The optimum F is given by a on condition that (o1)c(r(a1)/211) is minimum but equal or greater than n. Proof: The optimum F is given by the minimum a which satisfies the condition alcj 2 n. The minimum a which makes alc, maximum is given on condition that l(a 1)/2 jl = minimum. Q.E.D. For n 1, IF I = lfnl. Therefore, F is the minimum set of indices which includes different kinds of ordered pair (1, l), (1, 0), (0, l), and (0,O) for every bit pair in n 1. Conclusion In this paper, the maximum dimension of faultfree subcube located in a faulty hypercube when the number of faults was given was formulated. This paper is leaving the formalization of the maximum number of faults located on an ncube on which there always exists at least one faultfree m(< n 2) subcube. The time complexity to calculate the maximum number in an exhaustive method is tremendous as n becomes large, then the formalization of the maximum number is significant. References [l] Y. Chang and L.N. Bhuyan, Fault Tolerant Subcube Allocation in Hypercubes, Proc. Int I Conf. on Parallel Processing, pp.i , [2] P.J. Yang and C.S. Raghavendra, Reconfiguration of Binary Trees in Faulty Hypercubes, Proc. th Int Parallel Processing Symp., pp.010, April [3] M. Horng and L. Kleinrock, FaultTolerant Routing with Regularity Restoration in Boolean ncube Interconnection Networks, Proc. IEEE Symp. on Parallel and Distributed Processing, pp., Dec [] J. Bruck et al., Tolerating Faults in Hypercubes Using Subcube Partitioning, IEEE Tkans. Comput., Vo1.1, No., pp.990, May [] B.Becker and H.U.Simon, How Robust is the n Cube?, Proc. 2th IEEE Symposium on Foundations of Computer Science, pp , Oct. 19. [] F.0zgiiner and C.Aykanat, A Reconfiguration Algorithm for Fault Tolerance in a Hypercube Multiprocessor, Information Processing Letters, Vo1.29, pp.22, Nov. 19. [] S.Rai and J.L.Trahan, ATARIC : An Algebraic Technique to Analyze Reconfiguration for Fault Tolerance in a Hypercube, Proc. 3rd IEEE Symposium on Parallel and Distributed Processing, pp., Dec [] J.Brruck et al., Embedding CubeConnected Cycles Graphs into Faulty Hypercubes, IEEE Trans. Comput., vo1.3, No.10, pp , Oct a 1 lj (a) R column vector with weight j a c ~ J column vectors with weight j aicj A (ha1) /2k 1 s j sa2)... a1 (b) a column vector with weight j I c J column vectors with weight j Fig.2 A cube which possesses two disjoint Fig.3 A method to make F. faultfree 3subcubes. 22