Hamiltonicity and Fault Tolerance

Transcription

1 Hamiltonicit and Fault Tolerance in the k-ar n-cube B Clifford R. Haithcock Portland State Universit Department of Mathematics and Statistics 006 In partial fulfillment of the requirements of the degree of Master s of Science in Mathematics Advisor: John S. Caughman, IV

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3 Contents 1 Introduction 5 Preliminaries 7.1 Basic Definitions The Graph Q k n Conventions / Notation The Theorem of Ashir and Stewart The Induction Step Argument Overview Case i: Min deg in each Q i Case i-a: No Q i has 4n 4 faults Case i-b: Some Q i contains 4n 4 faults Case ii: Some Q i has min deg Case ii-a: A particular configuration holds in Q Case ii-b: No such configuration eists in Q Case iii: Some Q i has min deg The Base Case Q k, k Argument Overview Case i: All 3 faults in dimension Case ii: Onl faults lie in dimension A Lemma on Q Case i: Both dimensions contain faults Case i-a: A -bridge has health dim-1 links Case i-b: No such -bridge eists Case ii: All faults lie in one dimension

4 4 CONTENTS 6 The Base Case Q Case i: Two distinct dimensions have 3-ccles Case ii: All 3-ccles in a single dimension Case ii-a: Dim-1 has more than five faults case ii-b: Dim-1 has no more than five faults 46 7 Conclusion 51 8 Acknowledgements 53 Bibliograph 55

5 Chapter 1 Introduction This paper is about Hamiltonian ccles in the k-ar n-cube. While it is known that the k-ar n-cube Q k n is Hamiltonian for (k, n) (, 1), we will consider the Hamiltonicit of Q k n in the presence of missing edges. An obvious necessar condition for such a subgraph of Q k n to be Hamiltonian is that ever verte have degree at least two. Assuming this minimal condition is met, it is natural to ask how man faults Q k n can sustain and still necessaril contain a Hamiltonian ccle. In their 00 paper [1], Ashir and Stewart establish a sharp bound on this number of faults. The prove the following theorem. Theorem 1 Suppose k 4 and n, or k = 3 and n 3. If Q k n has no more than 4n 5 faults and ever node in Q k n is incident to at least two health links, then Q k n contains a Hamiltonian ccle. In this paper, we will work through the details of Ashir and Stewart s result. The appl an inductive argument in which the induction step is broken into several cases. We epand upon their eplanation and provide a number of diagrams to illuminate the argument. The motivation for such problems comes from analing the fault tolerance of massivel parallel computers. A ke factor in the architecture of a parallel computer is the communication network. The natural model for this network is a graph with vertices representing the computational units and edges representing the communication links. Fundamental results force us to onl consider communication networks based on highl structured graphs that we can more full anale. Tpical networks include rings, trees, meshes, and hpercubes in addition to the k-ar n-cube. The development of parallel algorithms depends upon the underling communication network. Porting an eisting algorithm to a new architecture involves embedding the original communication network within the new architecture s network (the guest and host networks respectivel). In particular, the eistence of a Hamiltonian ccle demonstrates the possibilit of embedding a ring in Q k n 5

6 6 CHAPTER 1. INTRODUCTION

7 Chapter Preliminaries.1 Basic Definitions Graph A graph is an ordered pair of disjoint sets (V, E) such that E V (), which denotes the set of unordered pairs of V. We refer to the elements of V as vertices and the elements of E as edges. An edge {, } is said to join the vertices and. We use the notation V (G) and E(G) to refer to the sets of vertices and edges of G respectivel. We alternativel use nodes and links as snonms for vertices and edges respectivel. These terms come from the application of parallel computing. Deletion ( G e ) Given a graph G and an edge e E(G), we define G e to be the graph with verte set V (G) and edge set E(G) e). Graph Instersection ( G H ) Given two graphs G and H, we define the intersection G H = (V (G) V (H), E(G) E(H)). Subgraph Given graphs G and H, we sa H is a subgraph of G if V (H) V (G) and E(H) E(G). Path A path P is a graph of the form V (P ) = { 1,,, n } and E(P ) = {{ 1, }, {, 3 },, { n 1, n }}. We refer to the vertices 1 and n as the endpoints of P, and we make use of the notation P : 1 n to indicate that P is a path with endpoints 1 and n. 7

8 8 CHAPTER. PRELIMINARIES Concatenating Paths If we have two paths P : and R : where V (S) V (R) = {} then we define P R R P to be the concatenation of the two paths. That is, V (P R) = V (R P ) = V (P ) V (R) and E(P R) = E(P ) E(R). Thus we have R P :. Ccle A ccle C is a graph of the form V (C) = { 1, 1,, n }, n 3 and E(C) = {{ 1, }, {, 3 },, { n 1, n }, { n, 1 }}. n-ccle An n-ccle is a ccle with n vertices. Hamiltonian Path Given a graph G and a path P, we sa P is a Hamiltonian path in G if P is a subgraph of G and V (P ) = V (G). Hamiltonian Ccle Given a graph G and a ccle C, we sa C is a Hamiltonian ccle in G if C is a subgraph of G and V (C) = V (G). Incidence We sa two edges are incident if the share an endpoint. Furthermore, we sa an edge e and a verte are incident to one another if is an endpoint of e. It will be clear from contet which definition we are appling. Adjacenc Given a graph G and vertices and, we sa and are adjacent if {, } E(G). Degree Given a graph G and a verte, we sa the degree of, denoted d(), is the number of edges in E(G) with as an endpoint. Bridge (-bridge / 3-bridge ) A -bridge is a graph G of the form V (G) = { 0, 1,, 3 } and E(G) = {{ 0, 1 }, {, 3 }, { 0, }, { 1, 3 }}

9 .. THE GRAPH Q K N A 3-bridge is a graph G of the form V (G) = { 0, 1,, 3, 4, 5 } and E(G) = {{ 0, 1 }, {, 3 }, { 4, 5 }, { 0, }, {, 4 }, { 1, 3 }, { 3, 5 }}. Star A star is a graph G of the form V (G) = { 0, 1,, n } and E(G) = {{ 0, 1 }, { 0, },, { 0, n }}. We refer to the node 0 as the center of the star. 1 3 n The Graph Q k n Given positive integers k and n, we define the graph Q k n known as the k-ar n- cube. The vertices are n-tuples in which each coordinate is an integer between 0 and k 1. Two vertices are adjacent if the two n-tuples differ in eactl one coordinate and that difference equals 1 mod k. Figure.1 shows all of the vertices adjacent to (0, 1, ) in the graph Q 3 3. (0,1,0) (,1,) (0,0,) (0,1,) (1,1,) (0,,) (0,1,1) Figure.1: Vertices adjacent to (0, 1, ) in Q 3 3

10 10 CHAPTER. PRELIMINARIES Note that ever verte in Q k n has degree n. If, are adjacent, then the differ in eactl one coordinate i. We then sa the edge {, } lies in dimension i. We can partition Q k n along a dimension i to get k copies of Q k n 1 connected b dimension i links. The vertices of the j th cop are the n-tuples whose i th coordinate equals j..3 Conventions / Notation Faults in a graph The focus of this paper is on the eistence of Hamiltonian ccles within Q k n in the presence of fault links. Throughout this work, there are numerous graphs depicted visuall. An given edge ma be known to be fault, known to be not fault (i.e. health), or its state unknown. If an edge is known to be fault, we will draw that edge with dashed lines. We do not consider fault edges to be deleted from a graph; we merel note them as fault. We do not visuall distinguish edges of unknown state from health edges. When we refer to a graph H as health, we understand that all of the edges in H are health. Indeing Conventions Furthermore, we will partition the graph Q k n along some dimension. We will label the Q k n 1 subgraphs Q 1, Q,..., Q k. We then strictl use subscripts in accordance with these labels. Thus b the subscripts it is understood that the nodes 1,, and w 3 are in Q 1, Q, and Q 3 respectivel. We also understand the nodes 1 and are adjacent via an edge in dimension 1. Similarl, the nodes k and 1 are adjacent via an edge in dimension 1. Joining ccles We will make such frequent use of the following constructions that it is worth covering them as a concept. Our general goal is to establish the eistence of Hamiltonian ccles. We will accomplish this b successivel joining ccles connected via a -bridge to get an ever larger ccle as depicted in Figure.3. Joining a path and a ccle or joining two paths is a clear etension of the given eample..4 The Theorem of Ashir and Stewart The theorem of Ashir and Stewart is concerned with guaranteeing the eistence of Hamiltonian ccles in Q k n under conditions of faulted links. Specificall the give a sharp result on the number of faults Q k n can sustain and still maintain the necessit of a Hamiltonian ccle.

11 .4. THE THEOREM OF ASHIR AND STEWART 11 n k Figure.: Eamples of Q k n

12 1 CHAPTER. PRELIMINARIES Figure.3: Joining Ccles Theorem 1 For k 4 and n or k = 3 and n 3, if Q k n has no more than 4n 5 faults and ever node in Q k n is incident to at least two health links, then Q k n contains a Hamiltonian ccle. The hpothesis that ever node is incident to at least two health links is a minimal condition for a graph to contain a Hamiltonian ccle. The proof follows b induction on n. This leaves as base cases the graph Q 3 3 and the famil of graphs Q k for k 4 (see Figure.4). Chapter 3 proves the induction step. Chapter 4 covers the base case Q k for k 4, and chapter 6 addresses the base case Q 3 3. k n Q 3 Q 4 Q 5 Q 6 Q 3 3 Q 4 3 Q 5 3 Q 6 3 Q 3 4 Q 4 4 Q 5 4 Q 6 4 Q 3 5 Q 4 5 Q 5 5 Q 6 5 Q 3 6 Q 4 6 Q 5 6 Q 6 6 Figure.4: Inductive Structure

13 Chapter 3 The Induction Step 3.1 Argument Overview To access the induction hpothesis, we partition the graph Q k n+1 across a dimension to get k copies of Q k n. Assume we have been given Q k n+1 with k 4 and n or k = 3 and n 3 with 4(n + 1) 5 = 4n 1 faults such that ever node is incident to two health links. With 4(n+1) 5 faults, we cannot guarantee that one of the n+1 dimensions has four faults. However, with n, we have 4(n+1) 5 > (n+1). Thus, b the pigeon hole principle, some dimension must have at least three faults. Without loss of generalit (WLOG) we assume dimension 1 has at least three faults. We now partition Q k n+1 across dimension 1. This leaves us with k copies of Q k n which we label Q 1, Q, Q k. We now have the required graph to appl the induction hpothesis. With at least 3 faults committed to dimension 1, we have at most 4n 4 faults distributed among the Q i s. The induction argument follows three principal cases: case (i): Within each Q i ever node is incident to two health links in Q i, case (ii): Some node i is incident to onl one health link in Q i, case (iii): Some node i is incident to ero health links in Q i. Note that these cases are distinct. The onl overlap to consider is between cases (ii) and (iii). Suppose i and j are nodes in Q k n+1 such that i is incident to onl one health link in Q i and j is incident to no health links in Q j. Then i must be incident to n 1 faults in Q i and j must be incident to n faults in Q j. It ma be the case i = j, and one of the faults incident to i is also incident to j. Thus we have at least (n 1) + (n) 1 = 4n faults distributed among the Q i s. Since we can have at most 4n 4 faults distributed among the Q i s, cases (ii) and (iii) cannot occur simultaneousl. We further break cases (i) and (ii) into sub-cases. 13

14 14 CHAPTER 3. THE INDUCTION STEP case i: Within each Q i ever node is incident with two health links in Q i case i-a: No Q i contains all 4n 4 faults distributed among the Q i s, case i-b: Some Q i contains 4n 4 faults. case ii: Some node i is incident to onl one health link in Q i, case ii-a: There is an edge {w i, i } in Q i that is fault with the edge {w i, w i+1 } health. case ii-b: There is no such edge {w i, i }. case iii: Some node i is incident to ero health links in Q i, 3. Case i: Within each Q i ever node is incident to two health links in Q i We have the hpotheses of the main theorem that Q k n+1 has 4n 1 faults such that ever node is incident to at least two health links. We also have the assumption that dimension 1 contains at least three faults Case i-a: No Q i has 4n 4 faults The graph Q 1 contains a Hamiltonian ccle C 1 We can relabel the Q i s such that Q 1 contains at least as man faults as an other Q i. Thus each of Q,, Q k contains no more than n faults which is half the maimum number of possible faults distributed among the Q i s. B our assumptions, Q 1 contains at most 4n 5 faults with ever node in Q 1 incident to at least two health links in Q 1. Thus, we can appl our induction hpothesis to Q 1 to get a Hamiltonian ccle C 1. There eists a health 3-bridge between C 1 and Q We start b using a counting argument to establish a health 3-bridge between Q 1 and Q k or a health 3-bridge between Q 1 and Q (see Figure 3.1 ). We partition C 1 into groups of three vertices to form desired configurations that onl overlap in health links in C 1 (See Figure 3.). Recall Q i = Q k n. So, there are k n nodes in C 1 which gives us kn 3 candidate 3-bridges. However the faults ma be distributed among Q, Q k, or dimension 1; so, we have at most 4n 1 candidate 3-bridges eliminated due to containing a fault edge. The following edge case calculations demonstrate that for k 4 and n or k = 3 and n 3 we have kn 3 > 4n 1.

15 3.. CASE I: MIN DEG IN EACH Q I 15 Q k Q 1 Q 1 Q k 1 1 k 1 or 1 k 1 1 Figure 3.1: Putative health 3-bridge Q Q Q k 1 Figure 3.: Partitioning C 1 k = 4, n = : k n 4 = = 5 = 10, 3 3 4n 1 = 4 1 = 7, k = 3, n = 3 : k n 3 3 = = 9 = 18, 3 3 4n 1 = = 11. So, there remains at least one desired health 3-bridge. B relabeling the Q i s if necessar, assume we have the right hand image in Figure 3.1. There is a ccle C in Q that contains the established 3-bridge

16 16 CHAPTER 3. THE INDUCTION STEP In order to connect C 1 with Q, we need a Hamiltonian ccle in Q that passes through {, } or {, }. With our current assumptions, we ma appl the induction hpothesis to get a Hamiltonian ccle in Q. If there are onl two health links in Q incident to, then we are done since the given Hamiltonian ccle must pass through both {, } and {, }. If is incident to more than two health links in Q, mark all but three links incident to in Q as fault making sure to leave {, } and {, } remaining as health links. Recall that Q initiall had at most n faults and we have introduced at most n 3 faults for a current maimum of 4n 5 faults in our modified Q. However, in order to appl the induction hpothesis, we must additionall ensure ever node in the modified Q is incident to two health links in Q. Suppose some node in the modified Q is not incident to two health links in Q. Ever node other than is affected b at most b one of the temporar faults. Since each node was initiall incident to at least two health links, and we eplicitl leave three health links incident to, ever node in the modified Q must be incident to at least one health link. Suppose after adding the temporar faults, w is now incident to onl one health link. Then w must have initiall been incident to n faults which is the maimum possible faults in the unmodified Q. So, w is the onl node in Q incident to less than two health links. We now mark the third health link incident to as fault and reinstate the link {, w }. Ever fault in Q is incident to either or w. So, ever other node in Q loses at most two health links due to fault connections with and w. B our construction, and w are both incident to two health links. Q Figure 3.3: Health Links Incident To Since ever node in our modified Q is incident to at least two health links, we can appl our induction hpothesis. So, we have a Hamiltonian ccle C in Q. Due to the faults we temporaril placed in incidence to, C must contain either the link {, } or {, } to incorporate the node (see Figure 3.3). Form the ccle D b joining C 1 and C B relabeling the nodes of Q if necessar, we can assume WLOG that C contains the link {, }. We can join the ccles C 1 and C to form the ccle

17 3.. CASE I: MIN DEG IN EACH Q I 17 D as depicted in Figure.3. There eists a health 3-bridge between D and either Q k or Q 3 Note that D Q 1 and D Q have k n nodes each. As depicted in Figure 3.4, we can partition D to get kn 3 3-bridges that onl overlap in health links in D. Q Q Q Q k 1 3 Figure 3.4: Partitioning D We appl the same counting argument used to establish a 3-bridge between C 1 and Q to conclude that we have at least one of our desired 3-bridges. Relabeling Q k n+1 if necessar, we assume WLOG that we have a health 3- bridge connecting D with Q 3. Repeat the Last Two Steps The conditions that applied to Q in the argument above, namel that Q contains no more than n faults, also appl to Q 3. Thus we have a Hamiltonian ccle in Q 3 that passes through one of the required edges of the 3-bridge connecting D and Q 3 to create a ccle D 3 that passes through all of the nodes in Q 1, Q, and Q 3. We can appl this same argument repeatedl to create the ccles D 4, D k. Completing the Hamiltonian Ccle of Q k n+1 Creating the final ccle D k 1 requires a slightl modified argument as the 3-bridges that are disjoint in Figure 3.4 ma now share links in Q k. Thus a fault in Q k ma now render two 3-bridges unavailable. Recall that there are at most n faults in Q k. In the worst case, we have at least kn 3 (n ) 3-bridges that survive the faults in Q k, and we have at most n + 1 additional faults to consider. Recall that we have 4n 1 faults in our Q k n+1. The following edge case calculations show for k 4 and n or k = 3 and n 3, kn 3 (n ) > n + 1

18 18 CHAPTER 3. THE INDUCTION STEP k = 4, n = : ( ) k n 4 ( n ) = ( ) = (5 ) = n + 1 = + 1 = 5. k = 3, n = 3 : ( ) k n 3 3 ( n ) = ( 3 ) = (9 4) = n + 1 = = 7. So, we have a desired health 3-bridge. The conditions that applied to Q in constructing C again appl to Q k. So, we have a Hamiltonian ccle C k in Q k that passes through one of the required links in the 3-bridge connecting D k with Q k. Thus we can join C k to D k to get a Hamiltonian ccle of Q k n+1. This completes the subcase (i-a) in which no Q i contains 4n 4 faults. 3.. Case i-b: Some Q i contains 4n 4 faults In this case we assume ever node in each Q i is incident to two health links in Q i. We also assume that the maimum number of faults not residing in dimension 1 all lie within a single Q i. B relabeling the indices if necessar, we assume WLOG that Q 1 contains the maimum 4n 4 faults. Let 1 and 1 be nodes in Q 1 connected b a fault link. Suppose further that { 1, 1 } is a link in the -bridge { 1, 1,, } where the links { 1, }, {, }, and { 1, } are all health as depicted in Figure 3.5. We ehibit a Hamiltonian ccle in Q k n+1 given this -bridge. We then show via contradiction that such a -bridge alwas eists. Q 1 Q 1 1 Figure 3.5: The Given -Bridge Each Q i contains an isomorphic ccle C i

19 3.. CASE I: MIN DEG IN EACH Q I 19 Given the -bridge in Figure 3.5, temporaril mark the fault link { 1, 1 } as health. We can then appl the induction hpothesis to Q 1 as Q 1 now onl has 4n 5 faults and each node in Q 1 is incident to two health links in Q 1. Thus, we have a Hamiltonian ccle C 1 in Q 1. Note that C 1 ma or ma not contain the link { 1, 1 }. Either wa, since all of the faults not in dimension 1 are in Q 1, each Q i, i =,..., k contains a health ccle C i that is an isomorphic cop of C 1. There is a ccle D 1 containing all nodes in Q 1 and Q If { 1, 1 } C 1, then we have a Hamiltonian Path P 1 : 1 1 Q 1. We can join P 1 with C via the -bridge { 1, 1,, } to get the ccle D 1 that contains all of the nodes in Q 1 and Q. If { 1, 1 } / C 1, then we appl a counting argument similar to the one used in case (i-a) to argue for the eistence of a health -bridge {u 1, v 1, u, v } connecting C 1 and C. Since there are k n nodes in Q 1, we have kn candidate -bridges. The onl faults left to consider are the three faults in dimension 1. The following edge case calculations show that we have at least one such health -bridge. We use the -bridge to join C 1 and C to form the ccle D 1 containing all of the nodes in Q 1 and Q k = 4, n = : k = 3, n = : k n 4 = = 8 > 3 k n 3 3 = = 4 > 3 So, whether or not { 1, 1 } C 1, we have a ccle D 1 containing all of the nodes in Q 1 and Q. There is a health -bridge joining D 1 and C 3 Note that D 1 Q contains k n nodes. This gives us kn disjoint candidate -bridges connecting D 1 and C 3. The counting argument above that established the -bridge {u 1, v 1, u, v } applies to guarantee a health -bridge between D 1 and C 3. We can then join D 1 and C 3 to form the ccle D that contains all of the nodes in Q 1, Q, and Q 3. Repeat the previous step We can repeatedl appl the argument of the previous step to get the ccles D 3,, D k 1. B the nature of this construction, we avoid the etra considerations of case (i-a) in constructing D k 1. So, given the -bridge in Figure 3.5, we have a Hamiltonian ccle in Q k n+1. The -bridge in Figure 3.5 alwas eists

20 0 CHAPTER 3. THE INDUCTION STEP Suppose no such -bridge eists. Then it must be the case that no such -bridge eists between Q 1 and Q k as well as between Q 1 and Q or else we could simpl relabel the graph. Thus, each fault in Q 1 must be incident with at least two dimension 1 faults. That is, given a fault { 1, 1 } we must have one of the four scenarios depicted in Figure 3.6. k 1 k 1 k 1 k 1 k 1 k 1 k 1 k 1 Figure 3.6: The Four Was to Prevent the -bridge So, if two faults in Q 1 are not incident to one another, at least four dimension 1 faults are required to prevent the eistence of the -bridge in Figure 3.5. Thus, with onl three dimension 1 faults, it must be the case that an two faults in Q 1 must incident to one another. This constraint on the faults in Q 1 permits onl three possibilities: the faults form a 3-ccle, a star with 1 at the center, or a star with 1 at the center. Thus, in the worst case, we can have at most n faults (since 1 and 1 are both incident to at least health links in Q 1 ) and still prevent the eistence of the -bridge in Figure 3.5. Since 4n 4 > n for n, we have enough faults in Q 1 to guarantee the eistence of the -bridge in Figure 3.5. This completes the proof of the induction step for sub-case (i-b) and case (i). 3.3 Case ii: Some node i is incident to onl one health link in Q i In this case, we assume some Q i has a node incident to onl one health link within Q i. Note that we still have at least three faults in dimension 1, and ever node in Q k n+1 is incident to two health links. B relabeling the Q i s if necessar, we assume WLOG that { 1, 1 } is the onl health link in Q 1 incident to 1. We additionall label Q k n+1 such that { 1, } is a health link. It ma or ma not be the case that the link { 1, k } is health. With onl 4n 4 faults distributed among the Q i s, ever node in Q i i 1 must be incident to at least three health links in Q i. Otherwise, (n 1) + (n ) = 4n 3 faults are required. It is possible that some node 1 Q 1, 1 1 ma be incident to onl two health links in Q 1.

21 3.3. CASE II: SOME Q I HAS MIN DEG 1 1 Case (ii) breaks into two subcases depending upon whether or not there eists some w 1 such that { 1, w 1 } is fault and {w 1, w } is health Case ii-a: There eists w 1 such that { 1, w 1 } is fault and {w 1, w } is health There is a Hamiltonian Path P 1 : 1 1 in Q 1 Ever node in Q 1 other than 1 is incident to at least two health links in Q 1. Temporaril mark the fault link { 1, w 1 } as health. Then ever node in Q 1 is now incident to at least two health links in Q 1. Furthermore, Q 1 had at most 4n 4 faults initiall. So, our modified Q 1 has at most 4n 5 faults. Thus, we ma appl the induction hpothesis to our modified Q 1 to get a Hamiltonian ccle that must necessaril pass through the link { 1, w 1 }. Thus, we have a Hamiltonian Path P 1 : 1 w 1 in the unmodified Q 1. There is a Hamiltonian Path P : in Q If necessar, temporaril mark the link {, w } as health. As argued above, w is incident to at least 3 health links in Q. Select the link {, w } and one other health link incident to w in Q to leave as health. Temporaril mark the remaining links incident to w in Q as fault. We now argue that we can appl the induction hpothesis to Q. First note that Q initiall had n 3 faults and we have introduced at most n faults for a maimum total of 4n 5 faults. B the argument at the beginning of this section, each node in the unmodified Q was incident to at least three health links in Q. Since each introduced fault is incident to w, ever node in Q other than w is affected b at most one of the introduced faults. Thus, ever node other than w must be incident to at least two health links in Q. Since we eplicitl left two health links incident to w, we ma appl the induction hpothesis to our modified Q to get a Hamiltonian ccle C in Q. Whether or not the link {, w } is health in the original Q, we have a Hamiltonian Path P : w in the unmodified Q. There is a ccle D 1 containing all of the nodes in Q 1 and Q We eplicitl chose 1 and w 1 such that the links { 1, } and {w 1, w } are health. Thus we can join P 1 and P to get the ccle D 1 that contains all of the nodes in Q 1 and Q. There is a health -bridge joining D 1 with Q 3 There are k n disjoint -bridges connecting D1 and Q 3 (See Figure 3.7). With n 1 faults restricted to Q 1, we need onl consider the remaining n faults. The following edge case calculations demonstrate that for k and n 3 or k 4 and n =, the inequalit k n > n holds.

22 CHAPTER 3. THE INDUCTION STEP k = 4, n = : k = 3, n = 3 : k n 4 = = 8 > n k n 3 3 = = 13> n Q Q Q w w 1 Figure 3.7: Partitioning D 1 Thus a -bridge {u, v, u 3, v 3 } between D 1 and Q 3 survives the 4n 1 faults in Q k n+1. There is a ccle D containing all of the nodes in Q 1, Q, and Q 3 The argument that established P applies to establish a Hamiltonian path P 3 : u 3 v 3 in Q 3. We then join D 1 and P 3 to get a ccle D that contains all of the nodes in Q 1, Q, and Q 3. Repeat the last two steps We can repeat the last two steps to produce the ccles D 3, D 4,, D k 1 which gives us a Hamiltonian ccle of Q k n+1. This completes the induction step for the case (ii-a).

23 3.3. CASE II: SOME Q I HAS MIN DEG Case ii-b: There is no node w 1 such that { 1, w 1 } is fault and {w 1, w } is health In this case, each of the n 1 neighbors of 1 joined to 1 via a fault link in Q 1 is also incident to a dimension 1 fault between Q 1 and Q. This accounts for 4n faults which is all but one of the 4n 1 faults in Q k n+1. Now suppose the link { 1, k } is health. With n, 1 is incident to at least three fault links in Q 1. With onl one fault in Q k n+1 not accounted for, there is a node w 1 such that { 1, w 1 } is fault and {w 1, w k } is health. So, b smmetr, we are in case (ii-a), and we construct the ccles D 1, D,, D k 1 b incorporating Q i s in the opposite order. So, for the remainder of this section suppose { 1, k } is fault which gives us a precise accounting of all of the faults in Q k n+1. Ashir and Stewart offer two different such Hamiltonian ccles of Q k n+1 depending upon the parit of k. We start b establishing a bridge between Q k and Q 1 and a bridge between Q 1 and Q. There is a Hamiltonian path P 1 in Q 1 Let w 1 be a node in Q 1 such that { 1, w 1 } is fault. Temporaril mark the link { 1, w 1 } as health. Since Q 1 contained onl n 1 faults, the modified Q 1 contains n < 4n 5 faults. Recall from the discussion at the beginning of case (ii) that ever node α i other than 1 is incident to at least two health links within Q i. Temporaril marking { 1, w 1 } as health places two health links in Q 1 incident with 1 which allows us to appl the induction hpothesis to our modified Q 1. So, we have a Hamiltonian ccle C 1 in Q 1. B the nature of the faults incident to 1, C 1 must contain the link { 1, w 1 }. Thus we have a Hamiltonian path P 1 in the original unmodified Q 1 with endpoints 1 and w 1. Q and Q k each contain a health path isomorphic whose vertices differ from those in P 1 onl in the first coordinate In the current case, all of the faults in Q k n+1 lie in Q 1 or in dimension 1 links with an endpoint in Q 1. Thus, Q and Q k each contain a health path isomorphic to P 1. We denote the two paths P and P k respectivel. A collection of parallel paths in dimension 1 Consider a node α in Q. As noted above, the faults in Q k n+1 are restricted to Q 1 or dim-1 links with endpoints in Q 1. Thus the dim-1 links connecting α, α 3,..., α k are all health. Therefore, we have a collection of parallel paths S 1, S,..., S k n that collectivel contain all of the nodes in Q, Q 3,..., Q k. Knitting together S 1, S,..., S k n and P 1 We will use links in P and P k to connect the endpoints of the paths S 1, S,, S k n. We then complete the Hamiltonian ccle of Q k n+1 using the links {w 1, w k }, { 1, }, and the path P 1.

24 4 CHAPTER 3. THE INDUCTION STEP The parit of k n affects the manner in which we complete the knitting of the Hamiltonian ccle of Q k n+1. Since the parit of kn is eactl that of k, Ashir and Stewart offer two alternative stitchings of the final Hamiltonian ccle depending on the parit of k. To facilitate describing the stitching process, we use a double indeing notation for the graph Q k n+1. Let 1,1, 1,, 1,k n denote the nodes in P 1 where 1,1 = 1, 1, = 1, 1,k n = w 1. As indicated in Figure 3.8, we similarl label the corresponding nodes in Q i b i,1, i,, i,k n. w = w = w = k k,k n 1 1,k n,k n = = = k = = = k k, k, , 1,1,,1 Figure 3.8: Double Inde Labeling of Q k n+1 The Hamiltonian ccle when k is even To slightl compact the notation of the Hamiltonian ccle, let S i = S i { k,i } and P = P { 1, }. The following eplicit Hamiltonian ccle of Q k n+1 is depicted in Figure ,k n S k n,kn 1 S kn 1 k,kn, k,kn S kn S 3 k 1, S 1,1 S 1 k,1 k, k,3 1,3 P This completes sub-case (ii-b) and case (ii). The Hamiltonian ccle when k is odd The following eplicit Hamiltonian ccle is depicted in Figure ,k n S k n,kn 1 S kn 1 1,k n S k,1 S 1 P Case iii: Some node i is incident to ero health links in Q i We still have our initial assumptions that our given Q k n+1 contains 4n 1 faults with ever node incident to at least two health neighbors. We still have at least 3 faults confined to dimension 1. For case (iii) we further assume that some node i Q i is incident to ero health links in Q i. To see that ever

25 3.4. CASE III: SOME Q I HAS MIN DEG 0 5 1,k n 1, k,1 1,1 Figure 3.9: (Case ii-b) The Hamiltonian ccle when k is even. node α i other than 1 must be incident to at least three health links in Q i consider the remaining n 4 faults that ma be in the Q i s i 1. In the worst case all of the remaining n 4 faults ma be incident to a node w 1 alread incident to 1 via a fault link. Thus, in this worst case, w 1 is still incident to three health links in Q 1. Relabeling the Q k n+1 if necessar we assume WLOG that 1 Q 1 is incident to ero health links in Q 1. Thus the links { 1, } and { 1, k } must be health. The Subgraph (Figure 3.1) Eists We begin b establishing the eistence of one of the configurations in Figure The following counting argument shows that a desired configuration survives all of the faults in Q k n+1. Note that 1 is the center of a star consisting of n faults in Q 1. Partition those faults into n pairs. In order to not have one of the configurations in Figure 3.11, two dim-1 faults are required to prevent both possible configurations. Thus 4n faults in total are required. However, we have onl 4n 1 faults in our Q k n+1.

26 6 CHAPTER 3. THE INDUCTION STEP k,k n 1,k n 1,1,1 Figure 3.10: (Case ii-b) The Hamiltonian ccle when k is odd. So, at least one such configuration eists. B relabeling Q k n+1 if necessar,we assume WLOG that we have the configuration in Figure 3.1. The path P 1 : 1 1 in Q 1 with P 1 = k n 1 is health. Temporaril mark the links { 1, 1 } and { 1, 1 } as health. Originall, Q 1 contained at most 4n 4 faults. So, with the two newl instated health links, the modified Q 1 contains at most 4n 6 faults which is less than the 4n 5 faults of the induction hpothesis. B the argument at the beginning of case (iii), ever node in the original Q 1 other than 1 is incident to two health links in Q 1. The newl marked health links place two health links incident with 1. Thus ever node in the modified Q 1 is incident to two health links. Thus, we can appl the induction hpothesis to our modified Q 1. Thus, we have a Hamiltonian ccle C 1 in Q 1. Note that C 1 must pass through the links { 1, 1 } and { 1, 1 }. Thus we have a path P 1 : 1 1 that contains all of the nodes of Q 1 ecept 1. The Hamiltonian path P k : k k in Q k is health. Recall that Q k contains no more than n 4 faults. If necessar, temporaril mark the link { k, k } as health. Recall that k was initiall incident to at

27 3.4. CASE III: SOME Q I HAS MIN DEG 0 7 Q Q Q Q Q Q k 1 k 1 or 1 1 Figure 3.11: Case iii: Needed Configuration Q Q Q k 1 1 k 1 k 1 Figure 3.1: Case iii: Configuration In Hand least 3 health links in Q k. Choose the link { k, k } and a health link incident to k to preserve. Mark the remaining n links incident to k in Q k as fault. Note that our modified Q k has at most (n 4) + (n ) = 4n 6 < 4n 5 faults. We now verif the second condition of the induction hpothesis holds. Each fault we introduced was incident to k. So, each node in Q k other than k is affected b at most 1 of the introduced faults. Since ever node in the original Q k was incident to at least 3 health links, ever node other than k in Q k is health to at least health links. We eplicitl left health links incident to k. So, we ma appl the induction hpothesis to Q k to get a Hamiltonian ccle that must necessaril contain the link { k, k }. Therefore, we have the Hamiltonian path P k : k k in the unmodified Q k. There eists a health Hamiltonian path P : in Q. All of the conditions that eisted in the previous case eist for Q 1. Thus we can appl the previous argument mutatis mutandis to get a Hamiltonian path P : in Q. There is a ccle D containing all of the nodes in Q k, Q 1, and Q.

28 8 CHAPTER 3. THE INDUCTION STEP D = P k 1 P 1 P 1 k There is a health -bridge connecting D and Q 3. With D Q = k n, we have kn disjoint -bridges between D and Q 3. We have at most n 1 faults outside of Q 1 to consider. The following counting argument shows that we do not have enough faults to prevent the eistence of a required -bridge. Denote the -bridge b {u 3, v 3, u, v }. k = 4, n = : k = 3, n = 3 : ( ) k n 4 = = 16 n 1 = 1 = 3. ( ) k n 3 3 = = (4) = 8 n 1 = 3 1 = 5. There is a Hamiltonian path P 3 : u 3 v 3 in Q 3. The argument establishing P k and P 1 applied directl to give us the Hamiltonian path P 3 in Q 3. There is a ccle D 3 containing all of the nodes in Q k, Q 1, Q, and Q 3 Connect D and P 3 with our usual construction to get the ccle D 3. Repeat to construct the Hamiltonian ccle D k 1 in Q k n+1 Repeat the last four steps to successivel construct the ccles D 4,..., D k 1. This concludes the proof of the induction step.

29 Chapter 4 The Base Case Q k, k Argument Overview We assume we have 4() - 5 = 3 fault links. With 3 faults distributed among dimensions, some dimension must contain at least faults. Assume WLOG that dimension 1 contains at least faults. We now partition Q k across dimension 1 to get Q 1, Q,..., Q k. Note that each Q i is a k-ccle. We now consider if the third fault is in dimension 1 or in one of the Q i s. 4. Case i: All 3 faults lie in dimension 1 With all of the faults ling in dimension 1, each Q i is a health k-ccle which is necessaril Hamiltonian. For consistenc with the rest of this paper, we let C 1,..., C k stand for the Hamiltonian ccles in Q 1,..., Q k respectivel. We now need a -bridge between C 1 and C or between C 1 and C k. Since each C i contains k nodes, we have k disjoint -bridges to connect C 1 with either C or C k. With k 4, we have k > 3. So, a desired -bridge eists. Assume WLOG that we have a -bridge between C 1 and C which gives a ccle D 1 encompassing all of the nodes in Q 1 and Q. We now need a -bridge connecting D 1 with either C k or C 3. Again we have k disjoint -bridges with k 4 and onl 3 faults in dimension 1. So, some -bridge is left unaffected b the faults in Q k. Assume WLOG that we have a -bridge between D 1 and C 3 which ields a ccle D encompassing all of the nodes in Q 1, Q, and Q 3. This construction continues to get the ccle D k which encompasses all of the nodes in Q 1, Q,..., Q k 1. Up until now, at each stage of the construction, we had k disjoint -bridges to consider. When considering - bridges between D k 1 and Q k, we have -bridges between Q 1 and Q k as well as between Q k 1 and Q k. It ma be the case that these -bridges are not disjoint within Q k. However, this poses no problem since all of the faults are in dimension 1 b assumption. Thus, the counting argument above applies again to conclude there is a health 9

30 30 CHAPTER 4. THE BASE CASE Q K, K 4 -bridge between D k 1 and C k. Thus, we can construct the ccle D k 1 which is our desired Hamiltonian ccle of Q k. 4.3 Case ii: Onl faults lie in dimension 1 Assume WLOG that the third fault lies in Q 1. We label the fault in Q 1 as { 1, 1 }. If the -bridge { 1,, 1, } or the -bridge { 1, k, 1, k } consists of health dim-1 links, then we ma appl case (i) above b considering the Hamiltonian path P 1 : 1 1 in Q 1 as the starting point. The -bridge with health dim-1 links connects P 1 with C to create D 1. The argument then proceeds identicall to case (i). Now assume that neither -bridge has two health dim-1 links. With onl two dim-1 faults, it must the case that one of the dim-1 links in each of the - bridges { 1,, 1, } and { 1, k, 1, k } is fault. Since each node is incident to at least two health links b hpothesis, it cannot be the case that the edges { 1, } and { 1, k } are both fault. Similarl, onl one of { 1, } and { 1, k } can be fault. Thus, we have two possibilities for the two dim-1 faults. Either { 1, k } and { 1, } are both fault or { 1, } and { 1, k } are both fault. We can assume WLOG that the links { 1, } and { 1, k } are fault. In the case k is even, the Hamiltonian ccle in Figure 3.9 applies with 1,3, 1,,,3, and k, replaced with 1, 1,, and k respectivel. Figure 4.1 depicts this situation when k = 6. When k is odd, then figure 3.10 applies with 1,m, 1,1,,m, and k,1 replaced with 1, 1,, and k respectivel. Figure 4. depicts this scenario when k = 5. 1 k 1 Figure 4.1: Hamiltonian Ccle for Q k when k = 6

31 4.3. CASE II: ONLY FAULTS LIE IN DIMENSION k 1 Figure 4.: Hamiltonian Ccle for Q k when k = 5

32 3 CHAPTER 4. THE BASE CASE QK, K 4

33 Chapter 5 A Lemma on Q 3 Lemma 1 If Q 3 has three fault links such that ever node is incident to two health links, then Q 3 has a Hamiltonian ccle unless the three faults form a ccle of length 3. With three faults spread across two dimensions, some dimension must contain at least two faults. We assume WLOG that dimension 1 contains at least two faults. We partition Q 3 over dimension 1 and label the nodes of Q 3 as i, i, i, 1 i 3. We have two cases to consider depending upon whether or not the third fault is also in dimension 1 or if it is in dimension. 5.1 Case i: Both dimensions contain faults. In this case the third fault must lie in one of the three Q i s. We ma assume WLOG that Q 1 contains the third fault with the link { 1, 1 } as fault. With all faults accounted for, we have the health Hamiltonian paths P i : i i, 1 i 3. Furthermore, the health links {, } and { 3, 3 } permit us to etend P and P 3 to form the ccles C and C 3 in Q and Q 3 respectivel. The current case of having onl two faults in dimension one now breaks down into two more sub-cases. We consider whether or not one of the following two -bridges has health dim-1 links: { 1,, 1, } or { 1, 3, 1, 3 } Case i-a: A -bridge contains health dim-1 links Assume WLOG that the -bridge spans Q 1 and Q. Figure 5.1 shows the one known fault and a ccle D 1 joining Q 1 and Q along with an isomorphic redrawing of the graph. Figure 5. shows the four possible -bridges we can use to join D 1 and Q 3 to construct a Hamiltonian ccle of Q 3. If one of these four -bridges is health, 33

34 34 CHAPTER 5. A LEMMA ON Q = ~ Figure 5.1: Q 3 lemma - case 1a then we get our Hamiltonian ccle with the usual construction. The onl wa the two dimension 1 faults can destro all four -bridges is if the links {, 3 } and { 1, 3 } are both fault. In this case we have the Hamiltonian ccle depicted in Figure Figure 5.: Q 3 lemma - case 1a: the four possible -bridges 5.1. Case i-b: Neither -bridge has two health dim-1 links With onl two dim-1 faults, it must the case that one of the dim-1 links in each of the -bridges { 1,, 1, } and { 1, k, 1, k } is fault. Since each node is incident to at least two health links b hpothesis, it cannot be the case that the edges { 1, } and { 1, 3 } are both fault. Similarl, onl one of { 1, } and { 1, 3 } can be fault. Thus, we have two possibilities for the two dim-1 faults. Either { 1, 3 } and { 1, } are both fault or { 1, } and { 1, 3 } are both fault. We can assume WLOG that the links { 1, } and { 1, 3 } are fault. In this case, we have the Hamiltonian ccle of Q 3 depicted in Figure 5.4

35 5.. CASE II: ALL FAULTS LIE IN ONE DIMENSION Figure 5.3: Q 3 lemma - case 1a : the ccle with fault { 1, 3 } and { 3, } Figure 5.4: Q 3 lemma - case 1b 5. Case ii: All faults lie In one dimension In this case, we argue that up to isomorphism there are onl si was to place the three faults in dimension one in Q 3. We start b encoding the dim-1 edges in a Table T as depicted in Figure 5.5. Let C ij denote swapping columns i and j in T. Similarl, let R ij denote swapping rows i and j in the Table T. B direct inspection of the adjacenc relation in Figures 5.6 and 5.7 we see that C ij and R ij respectivel encode isomorphisms of Q 3. With these isomorphisms in hand, we consider the distinct was we ma place 3 faults within the nine entries in T. In subsequent drawings of T, the edge numbers are omitted and the fault edges are indicated with disks. The argument successivel considers how to distribute the three faults among a single column, among two columns, and then among all three columns of T. Starting with restricting the faults to lie in a single column, we get one wa to place the faults as depicted in Figure 5.8. Now, we consider how to distribute the three dim-1 faults among two columns in T. Note that one column will contain faults and the other will contain 1 fault. We start b considering how to place faults in a single column. Figure 5.9 shows that b combining C 1, C 3, R 1, and R 3, there is onl one wa up to isomorphism to place two faults in a single column. With the two faults placed as in the top middle table depicted in Figure

36 36 CHAPTER 5. A LEMMA ON Q Figure 5.5: Q 3 lemma - case : encoding the dim-1 links in Table T = ~ 1 3 = ~ 3 1 = ~ = ~ = ~ = ~ C C C Figure 5.6: C ij encodes an isomorphism of Q 3 5.9, we consider how to place the third fault. The two rows of isomorphisms in Figure 5.10 show that there are two was to place the third fault. To see that that we do indeed have two isomorphism classes, note that the upper row indicates the eistence of a node incident to two faults and that there is no such node in the lower row. All we have left to consider, is how to distribute the faults among the three columns. To anale this case, we successivel consider distributing the three faults among one row, two rows, and then all three rows. Starting with placing all three faults among a single row, Figure 5.11 shows that up to isomorphism there is onl wa to do this. Now we consider how to place all three faults in 3 columns and rows. In this case, we must have faults in one row and the third fault in a second row. Figure 5.1 shows that b combining C 1, C 3, R 1, and R 3, there is up to isomorphism onl one wa to place two faults in a single row. Starting with the table in the middle left of Figure 5.1, we now consider the was to add the third fault. Figure 5.13 shows that up to isomorphism, there are onl two was to spread the three faults across all three columns and two rows. To see that that the two rows represent distinct isomorphism classes,

37 5.. CASE II: ALL FAULTS LIE IN ONE DIMENSION = ~ 3 1 = ~ 3 1 = ~ = ~ 1 3 = ~ 1 3 = ~ R R R Figure 5.7: R ij encodes an isomorphism of Q 3 = ~ = ~ C C 1 3 Figure 5.8: The one wa to have 3 faults in a single column consider that the upper row contains a 4-ccle containing three faults and that no such 4-ccle eists in the lower row. Furthermore, Note that the upper row of Figure 5.13 contains a configuration that is also present in the upper row of Figure Finall, we consider how man was three faults can be distributed across three columns and three rows. Note that we have 3 1 = 6 was to place the three faults in the table. Figure 5.14 shows that all si such arrangements are isomorphic. From Figure 5.8, we get 1 configuration. From Figure 5.10, we get configurations. From Figure 5.11 we get 1 configuration. From Figure 5.13, we get onl 1 additional configuration, and we get the sith configuration from Figure Note that the onl configuration in which the three dim-1 faults form a ccle is in Figure Figure 5.15 displas an eample from each of the si congruence classes, and Figure 5.16 shows the Hamiltonian ccle for all of the graphs ecept for (a). The faults are indicated as usual with dotted lines. In addition to the faults, onl those health links required to ehibit the Hamiltonian ccle are drawn. Figure 5.17 show the graph of 5.16-(a) with the fault 3-ccle. The health links displaed are required for a Hamiltonian ccle to pass through the nodes ling on the fault 3-ccle. It is clear form this drawing that there is no Hamiltonian ccle for Q 3 in this case.

38 38 CHAPTER 5. A LEMMA ON Q 3 = ~ = ~ C C 1 3 R R 13 1 = ~ = ~ Figure 5.9: The one wa to have faults in a single column = ~ = ~ = ~ R C R = ~ C 13 Figure 5.10: The two was to have 3 faults in two columns = ~ = ~ R R 1 3 Figure 5.11: The one wa to have 3 faults in three columns and one row

39 5.. CASE II: ALL FAULTS LIE IN ONE DIMENSION 39 = ~ = ~ C C 3 1 R R 1 3 = ~ = ~ Figure 5.1: The one essential wa to have two faults in a single row = ~ = ~ = ~ R C R = ~ R 13 Figure 5.13: The two was to have 3 faults in two rows

40 40 CHAPTER 5. A LEMMA ON Q 3 = ~ = ~ R R 1 3 R 13 = ~ = ~ = ~ R R 1 3 Figure 5.14: The one essential wa to have 3 faults in distinct columns and rows (a) (b) (c) (d) (e) ( f) Figure 5.15: The Si Was to have 3 faults in dim 1 (b) (c) (d) (e) (f) Figure 5.16: The Five Hamiltonian Ccles

41 5.. CASE II: ALL FAULTS LIE IN ONE DIMENSION 41 (a) Figure 5.17: No Hamiltonian Ccle

42 4 CHAPTER 5. A LEMMA ON Q 3

43 Chapter 6 The Base Case Q 3 3 We assume we have Q 3 3 with 4 (3) 5 = 7 faults such that each node is incident to at least two health links. Our general strateg in this case is to recapitulate the argument of the induction step replacing the induction hpothesis with Lemma 1. In order to use Lemma 1 we must insure that our Q 3 subgraphs do not contain a 3-ccle of faults. Thus, we handle this scenario separatel. First, note that a 3-ccle of faults must lie within a single dimension. Thus, our troublesome case occurs when our Q 3 3 contains two 3-ccles in distinct dimensions. 6.1 Case i: Q 3 3 contains two 3-ccles in distinct dimensions Denote the two 3-ccles b C and D. Relabel our Q 3 3 if necessar to have C in dimension one and partition along dimension one. Label the Q i s such that Q 1 contains as man or more faults than Q which contains as man or more faults than Q 3. This labeling forces D to eist in Q 1. Furthermore, Q contains at most one fault, and Q 3 does not contain an faults. Q 1 contains a Hamiltonian path P 1 : 1 1 We now suppose that the seventh fault lies within Q 1 and consider whether or not the seventh fault is incident to D. We label the Q 3 3 depending upon how D sits within Q 1. We distinguish two nodes in D with the labels 1 and 1 such that 1 is incident to as man faults in Q 1 as either of the other nodes in D and 1 is arbitraril assigned to an unlabeled node in D. We now consider whether or not the seventh fault is incident to D. In the case that the seventh fault is incident to D, Figure 6.1 shows the possibilities along with our imposed labeling. Temporaril mark the link { 1, 1 } as health. Then we ma appl Lemma 1 to our modified Q 1 to get a Hamiltonian ccle that must necessaril contain the link { 1, 1 }. Thus we have a Hamiltonian path P 1 : 1 1 in the original Q 1. 43

44 44 CHAPTER 6. THE BASE CASE Q Figure 6.1: Q 3 3 case 1-a: Seventh Fault Is Incident to D Now suppose the seventh fault is not incident to D. Since Q 1 is the graph Q 3, Figure 5. shows the two cases for the seventh fault along with P Figure 6.: Q 3 3 case 1-a: Seventh Fault Not Incident to D If the seventh fault is not in Q 1, then we temporaril mark a link in Q 1 incident to D as fault and appl the preceding argument to get the Hamiltonian path P 1 : 1 1. The -bridge { 1,, 1, } has health dim-1 links We have determined si of the seven faults in our Q 3 3. None of the si faults affects the dim-1 links in either of the following -bridges: { 1, 3, 1, 3 } and { 1,, 1, }. With onl one fault not accounted for, one of these -bridges must contain health dim-1 links. We ma relabel the graph if necessar and assume WLOG that the -bridge { 1,, 1, } contains health dim-1 links. Q contains a Hamiltonian path P : Note that it ma be the case that Q contains the seventh fault in our Q 3 3. If the link {, } is fault, then temporaril mark two of the three health links in Q incident with as fault and the link {, } as health. Then we ma appl Lemma 1 to our modified Q to get a Hamiltonian ccle in Q that must include the link {, }. Thus we have a Hamiltonian path in the original Q with endpoints and. If the link {, } is health and there is a fault in Q incident to, then mark one of the three health links incident to as fault, making sure to leave