GRAY CODES FAULTING MATCHINGS

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1 Uniersity of Ljbljana Institte of Mathematics, Physics and Mechanics Department of Mathematics Jadranska 19, 1000 Ljbljana, Sloenia Preprint series, Vol. 45 (2007), 1036 GRAY CODES FAULTING MATCHINGS Darko Dimitro Petr Gregor Tomáš Dořák Riste Škrekoski ISSN Jly 25, 2007 Ljbljana, Jly 25, 2007

2 GRAY CODES FAULTING MATCHINGS DARKO DIMITROV, TOMÁŠ DVOŘÁK, PETR GREGOR, AND RISTE ŠKREKOVSKI Abstract. A (cyclic) n-bit Gray code is a (cyclic) ordering of all 2 n binary strings of length n sch that consectie strings differ in a single bit. Eqialently, an n-bit Gray code can be iewed as a Hamiltonian path of the n-dimensional hypercbe Q n, and a cyclic Gray code as a Hamiltonian cycle of Q n. In this paper we stdy Hamiltonian paths and cycles of Q n aoiding a gien set of falty edges that form a matching, briefly called (cyclic) Gray codes falting a gien matching. Gien a matching M and two ertices, of Q n, n 4, or main reslt proides a necessary and sfficient condition, expressed in terms of forbidden configrations for M, for the existence of a Gray code between and falting M. As a corollary, we obtain a similar characterization for a cyclic Gray code falting M. In particlar, in case that M is a perfect matching, Q n has a (cyclic) Gray code falting M if and only if Q n M is a connected graph. This complements a recent reslt of Fink, who proed that eery perfect matching of Q n can be extended to a Hamiltonian cycle. Key words. hypercbe, Gray code, Hamiltonian paths, Hamiltonian cycles, matching 1. Introdction. A (cyclic) n-bit Gray code is a (cyclic) ordering of all 2 n binary strings of length n sch that consectie strings differ in a single bit. It is named after Frank Gray, who in 1953 patented a simple scheme to generate sch a cyclic code for eery n 2 [7]. Since then, the research on Gray codes satisfying certain additional properties has receied a considerable attention, and applications hae been fond in sch dierse areas as data compression, graphics and image processing, information retrieal, signal encoding or processor allocation in hypercbic networks [12]. Alternatiely, an n-bit Gray code can be iewed as a Hamiltonian path of the n-dimensional hypercbe Q n, and a cyclic Gray code as a Hamiltonian cycle of Q n. The applications of hypercbes in parallel compting [10] inspired the inestigation of Hamiltonian paths and cycles in hypercbes aoiding a gien set of falty edges. Chan and Lee [2] proed that the problem whether Q n contains a Hamiltonian cycle aoiding a gien set F of falty edges is NP-complete. On the other hand, they showed that if F 2n 5, n 3, then sch a Hamiltonian cycle exists if and only if each ertex is incident with at least two nonfalty edges. This pper bond is sharp in the sense that for any n 3 there are 2n 4 falty edges in Q n satisfying the aboe condition, bt there is no Hamiltonian cycle aoiding them. Tsai [13] obtained a similar reslt for Hamiltonian paths with gien endertices. A related problem considers Hamiltonian cycles and paths in hypercbes passing throgh a set of prescribed edges. Caha and Kobek [1] obsered that if a set P of prescribed edges extends to a Hamiltonian cycle, then it indces a sbgraph consisting of pairwise ertex-disjoint paths. On the other hand, they showed that for any n 3 there is a set of 2n 2 edges, satisfying this necessary condition, bt there is no Hamiltonian cycle of Q n containing them. This bond is sharp in the sense that if P 2n 3, n 2, the necessary condition is also sfficient for a Hamiltonian cycle Institt für Informatik, Freie Uniersität Berlin, Takstraße 9, Berlin, Germany (Darko.Dimitro@inf.f-berlin.de). Faclty of Mathematics and Physics, Charles Uniersity, Malostranské nám. 25, Praha, Czech Repblic (Tomas.Dorak@mff.cni.cz). Faclty of Mathematics and Physics, Charles Uniersity, Malostranské nám. 25, Praha, Czech Repblic (Petr.Gregor@mff.cni.cz). Department of Mathematics, Uniersity of Ljbljana, Jadranska 19, 1111 Ljbljana, Sloenia.The work of this athor was partially spported by Sloenian ARRS Research Grant P and by bilateral project BI-CZ/ between Sloenia and Czech Repblic. 1

3 2 D. DIMITROV, T. DVOŘÁK, P. GREGOR, AND R. ŠKREKOVSKI of Q n passing throgh eery edge of P [3]. There is a similar reslt for Hamiltonian paths with prescribed endertices if P 2n 4, n 5 [4]. A special case of the aboe problem is when the prescribed edges comprise a matching of the hypercbe. Kreweras [9] conjectred that eery perfect matching of Q n, n 2, can be completed to a Hamiltonian cycle. Recently, the conjectre was proed by Fink [5]. In this paper we stdy Hamiltonian paths and cycles in Q n aoiding a gien set of falty edges that form a matching, or eqialently, (cyclic) Gray codes falting a gien matching. Gien a matching M and two ertices, of Q n, n 4, or main reslt proides a necessary and sfficient condition for the existence of a Gray code between and that falts M. As a corollary, we obtain a similar characterization for the existence of a cyclic Gray code falting M. In particlar, in case that M is a perfect matching, Q n has a (cyclic) Gray code falting M if and only if Q n M is a connected graph. Note that this complements the aboe qoted reslt of Fink. Frthermore, since or necessary and sfficient conditions are decidable in a polynomial time, or reslts imply that the problem of Hamiltonicity of Q n with falty edges, which is NP-complete in general, becomes polynomial for p to 2 n 1 edges proided they form a matching. After definitions and preliminary reslts in the next two sections, we stdy necessary conditions in Sections 4 and 5. Then we consider the initial case of n = 3 and n = 4 in Sections 6 and 7. The indction step is presented in Section 8. The last section smmarizes obtained reslts. 2. Preliminaries. In this text n is always a positie integer and [n] denotes the set {1,2,...,n}. As sal, the ertex and edge sets of a graph G are denoted by V (G) and E(G), respectiely. For a ertex V (G) and a set E E(G), let G denote the sbgraph of G indced by V (G) \ {}, and let G E denote the graph with ertices V (G) and edges E(G) \ E. Similarly, for a pair of ertices x,y V (G) let G+xy denote the graph with ertices V (G) and edges E(G) {xy}. The distance of ertices and is denoted by d(,). The distance d(,w) of a ertex and an edge w is defined as the minimm of distances d(,) and d(,w). A path with endertices a and b is denoted by P ab. In particlar, P aa denotes the path consisting of a single ertex a. If P ab and P cd are ertex disjoint paths and b and c are adjacent ertices, then P ab +P cd denotes the path with endertices a and d, obtained as a concatenation of P ab with P cd. A path P is called a sbpath of a path P if P forms a sbgraph of P. We say that paths {P i } n i=1 are spanning paths of a graph G if {V (P i )} n i=1 partitions V (G). The n-dimensional hypercbe Q n is the graph with ertex set V (Q n ) = {0,1} n and edge set E(Q n ) = { (,) = 1} where (,) = {i [n] i i }. The dimension dim() of an edge E(Q n ) is the integer d sch that (,) = {d}. For a ertex V (Q n ) let d denote the ertex of Q n sch that d is an edge of dimension d. All edges of the same dimension d form a layer (of dimension d) in Q n. Note that E(Q n ) is partitioned into n layers, each of size 2 n 1. Moreoer, for eery d (,), each path P contains an edge of dimension d. In other words, ertices and are separated by the layer of dimension d. The parity p() of a ertex V (Q n ) is defined by p() = mod 2 where = n i=1 i. Note that ertices of each parity form bipartite sets of Q n. Conseqently, p() = p() if and only if d(,) is een. We extend the definition of the parity to

4 GRAY CODES FAULTING MATCHINGS 3 edges by ptting for E(Q n ) { p(), if < ; p() = p(), otherwise. For eery d [n] let Q d L and Qd R denote the sbgraphs of Q n indced by the sets { V (Q n ) d = 0} and { V (Q n ) d = 1}, respectiely. Note that both Q d L and Q d R are isomorphic to Q n 1. Similarly, for a matching M in Q n pt M d = {e M dim(e) = d}, ML d = M E(Qd L ), and MR d = M E(Qd R ). Obsere that (Q a i )b j = (Qb j )a i and (Mi a)b j = (Mb j )a i for eery a,b [n] and i,j {L,R}. Gien a set F of falty edges of Q n, we say that a sbgraph of Q n falts F if it forms a sbgraph of Q n F. 3. Some fndamental reslts. A classical reslt of Lewinter and Widlski [11] describes Hamiltonian paths in hypercbes with one falty ertex. Lemma 3.1. For any pairwise distinct ertices,,w in Q n sch that p() = p() p(w), there exists a Hamiltonian path of Q n w between and. There is a similar characterization for hypercbes with at most one falty edge, which follows from more general reslts of Tsai et al. [14]. It shold be remarked that the special case of M = was firstly obsered by Hael [8]. Lemma 3.2. For eery n 3, ertices and of different parities, and a matching M of Q n with M 1, there exists a Hamiltonian path in Q n M between and. An extension of the aboe mentioned Hael s reslt was obtained in [3]. Lemma 3.3. For eery distinct ertices,,x,y of Q n sch that p() p() and p(x) p(y), there exist spanning paths P,P xy of Q n. We shall also employ the following lemma on the existence of a Hamiltonian path passing throgh a prescribed edge, which is a special case of more general reslts of Caha and Kobek [1]. Lemma 3.4. For eery two ertices, and an edge e of Q n sch that p() p() and e, there exists a Hamiltonian path of Q n between and passing throgh e. We conclde this section with a simple obseration. Proposition 3.5. For eery matching M in Q n with M = 2 there exists a dimension d [n] sch that ML d = Md R = 1. Proof. We arge by indction on n. Since the case n 2 is obios, assme that n 3 and choose an arbitrary d [n] \ {dim(e) e M}. If ML d = Md R = 1, we are done. Otherwise it mst be the case that M = Mi d for some i {L,R} and hence by the indction hypothesis, there exists d [n] \ {d} sch that ML d = (Md i )d L = 1 = (Mi d)d R = Md as reqired. R 4. Half-layers. A matching M in Q n, n 2, is called a half-layer in Q n if M consists exactly of all edges of dimension d and parity p for some d [n] and p {0,1}. Obsere that each layer is partitioned into two half-layers, each of size 2 n 2. Moreoer, if a perfect matching of Q n contains a half-layer, then it actally is a layer of Q n. We shall see in the next section that half-layers are obstacles that cannot be oercome. More precisely, we shall see in Lemma 5.1 that there is no Hamiltonian

5 4 D. DIMITROV, T. DVOŘÁK, P. GREGOR, AND R. ŠKREKOVSKI cycle falting a half-layer. The need to treat half-layers as a special case motiates the following definitions. Gien a matching M in Q n, we call d [n] a main dimension of M in Q n if M d contains a half-layer in Q n ; a splitting dimension for M if Mi d does not contain a half-layer in Qd i for both i {L,R}. Now we show that the main dimension is niqely determined. Proposition 4.1. Let d be a main dimension of a matching M in Q n, n 2. Then, (i) M contains only edges of dimension d. Conseqently, d is the only main dimension of M. (ii) d is the niqe main dimension of Mi a in Q a i for eery a [n] \ {d} and i {L,R} proided n 3. Proof. Since for eery edge of Q n we hae p() p(), it follows that at least one of, mst be incident with an edge of M d. Hence M and conseqently, ML d = Md R =. This, together with the fact that Md contains a half-layer and therefore mst be nonempty, erifies part (i). To see why (ii) holds, note that if n 3, then a half-layer in Q n contains a half-layer in both Q a L and Qa R by definition. It follows that d is a main dimension of Mi a in Q a i while the niqeness follows from part (i). A matching M in Q n is called aligned if there exists a dimension d [n] sch that ML d 1 and Md R 1. Otherwise, it is called naligned. Note that a matching containing a half-layer is always aligned by Proposition 4.1(i). For aligned matchings or problem is easy, as we shall see in the next section. Bt this is not the case with naligned matchings. For them, we se the following lemma. Lemma 4.2. If M is an naligned matching in Q n, n 4, then there are at least two splitting dimensions for M. Proof. First note that if eery a [n] is a splitting dimension for M, we are done. Hence we can assme that this is not the case, i. e., there exist distinct a,b [n] sch that b is the main dimension of ML a in Qa L. Case 1: b is not a splitting dimension for M. Then there exist i {L,R} and d [n] \ {b} sch that d is the main dimension of Mi b. Assme that d a. Then Proposition 4.1 (ii) implies that d is also the main dimension of (Mi b)a L in (Qb i )a L. It follows that (ML a)b i = (Mb i )a L contains a half-layer. On the other hand, as b is the main dimension of ML a, Proposition 4.1 (i) reeals that Ma L contains only edges of dimension b and therefore (ML a)b i =. This, howeer, leads to a contradiction since a half-layer is always nonempty. Hence we conclde that d = a. Now choose an arbitrary c [n] \ {a,b} and j {L,R} and sppose that Mj c contains a half-layer in Q c j. Using Proposition 4.1 (ii) again, a and b are the main dimensions of (Mi b)c j and (Ma L )c j in (Qb i )c j and (Qa L )c j, respectiely. Since (Mb i )c j = (Mj c)b i and (ML a)c j = (Mc j )a L, part (ii) of Proposition 4.1 implies that a and b are also main dimensions of Mj c in Qc j, while part (i) then reeals that a = b, which is a contradiction with or assmption. Hence c mst be a splitting dimension for M. It follows that there exist n 2 2 splitting dimensions in this case. Case 2: b is a splitting dimension for M. In this case we need to show that there exists another splitting dimension for M, different from b. Assme, by way of contradiction, that this is not the case. Then for any c [n] \ {b} there exists i {L,R} sch that Mi c contains a half-layer. Recall that b is the main dimension of ML a in Qa L. If c a, then Proposition 4.1 (ii) implies that b is the main dimension

6 GRAY CODES FAULTING MATCHINGS 5 of (ML a)c i = (Mc i )a L in (Qa L )c i and therefore b is also the main dimension of Mc i in Q c i. Hence we conclde that for any c [n] \ {b} there exists i {L,R} sch that b is the main dimension of Mi c in Q c i. Since M is an naligned matching, there mst exist i {L,R} sch that Mi b contains two distinct edges e 1,e 2. Note that both dim(e 1 ) and dim(e 2 ) are different from b. By Proposition 3.5 there mst be a dimension d [n] \ {b} sch that Mj d {e 1,e 2 } = 1 for each j {L,R}. Bt then each Mj d contains an edge of dimension distinct from b, which by Proposition 4.1 (i) means that b cannot be the main dimension of Mj d, contrary to the conclsion of the preios paragraph. 5. Admissible configrations. A nonempty matching M is called an almostlayer (of dimension d) in Q n if there exist ertices, Q d L of different parities sch that M consists of all edges of dimension d except d and d. Pairs {,} and { d, d } are called the free pairs of M. The next lemma gies necessary conditions for a Gray code falting a gien matching. Lemma 5.1. Let M be a matching and, ertices of Q n. If there exists a Hamiltonian path in Q n M between and, then (i) p() p(); (ii) for eery d (,) there exists an edge e E(Q n ) \M of dimension d sch that d(,e) is odd; (iii) for eery d [n] \ (,) there exist two edges in E(Q n ) \ M of dimension d of different parities; (i) M contains no almost-layer with a free pair {,}. Proof. Claims (i) (iii) follow from the facts that in a hypercbe the nmber of ertices of odd parity eqals the nmber of ertices of een parity, and that the parity of ertices on eery path alternates. To erify (i), obsere that if M is an almost-layer with a free pair {,} V (Q d L ), then eery path P in Q n M aoids either all ertices of V (Q d L ) \ {,}, or all ertices of V (Qd R ), and therefore cannot be Hamiltonian. We say that a triple (,,M) is a configration in Q n if M is a matching in Q n and and are ertices of different parities, i. e., condition (i) of Lemma 5.1 holds. Since we do not need to distingish between ertices and, we consider the configrations (,,M) and (,,M) to coincide. If, moreoer, conditions (ii) (i) of Lemma 5.1 hold, we say that the configration (,,M) is admissible. The conditions (ii) and (iii) of Lemma 5.1 can be nified into one condition sing a slight modification of the concept of a half-layer. Note that since a half-layer contains only edges of the same parity, the distances of a gien ertex to edges of a gien half-layer mst also hae the same parity. If the distance of a ertex to edges of a half-layer M in Q n is odd, we say that M is an odd half-layer for the ertex. Proposition 5.2. A configration (,,M) is admissible if and only if the following holds (i) M contains no odd half-layer for or ; and (ii) M contains no almost-layer with a free pair {,}. The next lemma shows that these conditions are also sfficient for the existence of a Hamiltonian path in Q n, n 4, proided that M is aligned. Lemma 5.3. Let M be an aligned matching, and let and be ertices of Q n, n 4. If (,,M) is an admissible configration, then Q n M contains a Hamiltonian path between and. Proof. Since M is an aligned matching, there exists d [n] sch that ML d 1

7 6 D. DIMITROV, T. DVOŘÁK, P. GREGOR, AND R. ŠKREKOVSKI and MR d 1. Withot a loss of generality assme that V (Qd L ) and consider the following two cases: Case 1: V (Q d R ). Since (,,M) is an admissible configration, there exists a ertex w V (Q d L ) sch that p() p(w) and wwd M. Note that then also p(w d ) p(). Now, by Lemma 3.2, there exist Hamiltonian paths P w of Q d L Md L and P wd of Q d R Md R. Then P := P w + P wd is the reqired Hamiltonian path. Case 2: V (Q d L ). Here, we consider the following sbcases. Sbcase 2.1: ML d =. Since (,,M) is an admissible configration, there are ertices x,y V (Q d L ) sch that p(x) p(y), xxd M, yy d M, and {,} {x,y}. Apply Lemma 3.2 to obtain a Hamiltonian path P xd y of d Qd R Md R. Note that as p() p() and p(x) p(y), we can assme that also p() p(x) and p() p(y), interchanging x and y if necessary. We conclde the sbcase with the following constrction. If {, } {x, y} =, then apply Lemma 3.3 to obtain spanning paths P x and P y of Q d L and pt P := P x + P x d y + P y. Otherwise it mst be the case that {,} {x,y} = 1. Assme d withot a loss of generality that = x and y. Then apply Lemma 3.1 to obtain a Hamiltonian path P y of Q d L and pt P := P y + P y d x + P xx. In both cases d we obtained the reqired Hamiltonian path. Sbcase 2.2: ML d = {e}. If e, then pt xy = e and apply the constrction of Sbcase 2.1. So, we may assme that e =. Since (,,M) is an admissible configration, there is a ertex y V (Q d L ) \ {,} sch that yyd M. Assme withot a loss of generality that p() p(y). Finally, pt x = and apply the constrction of Sbcase 2.1 in order to conclde this case. Note that the aboe lemma does not hold for n = 3, see e. g. configrations (e), (f) or (g) on Figre 6.1. The case of small dimensions is analyzed separately in the next two sections. 6. Gray codes in Q 3. A configration (,,M) in Q n is good if there is a Hamiltonian path of Q n M between and. Otherwise it is bad. In this terminology, or aim is to characterize all good configrations. A configration (,,M 1 ) contains a configration (,,M 2 ) if M 1 M 2. A bad configration is minimal if it does not contain any other bad configration. An admissible configration is maximal if it is not contained in any other admissible configration. Similarly, a matching is maximal if it is not contained in any other matching. We start with the list of all minimal bad configrations in Q 3. Note that an electronic ersion of this paper contains colored figres. Proposition 6.1. Let and be ertices of different parity and let M be a matching in Q 3. Then there is a Hamiltonian path of Q 3 M between and nless (,,M) contains one of the bad configrations depicted on Figre 6.1. (a) (b) (c) (d) (e) (f) (g) Fig The minimal bad configrations in Q 3. Proof. First obsere that each configration (,,M) on Figre 6.1 is bad. Note

8 GRAY CODES FAULTING MATCHINGS (1) (2) (3) (4) (5) x y (6) (7) (8) (9) (10) Fig All matchings in Q 3 p to isomorphism. that M is presented with dashed lines. Moreoer, obsere that there are no two configration on Figre 6.1 sch that the first one contains the other one. Now we show that the configration (,,M) is either good or contains one of the bad configrations on Figre 6.1. The hypercbe Q 3 has 10 non-isomorphic matchings, inclding the empty one, depicted on Figre 6.2(1)-(10). Note that in this figre the edges of the matchings are presented with dashed lines. So M is one of the following cases on Figre 6.2. (1), (2): In these cases M has at most one edge, ths the configration (,,M) is good by Lemma 3.2. (3), (6): Obsere that if { 1, 2 } and { 1, 2 }, then the configration (,,M) is good. Otherwise it contains a configration of type (a), (b), or (c) on Figre 6.1. (4): Obsere that the configration (,,M) is good nless it eqals ( 1, 1,M) or ( 2, 2,M) which are of type (d) on Figre 6.1. (5): Obsere that the configration (,,M) is good nless it eqals ( 1, 1,M), ( 1, 2,M), or ( 2, 2,M) which are of type (e) or (f) on Figre 6.1. (7): In this case Q 3 M xy is a Hamiltonian cycle. Ths the configration (,,M) is good if is an edge on this cycle. Frthermore, if the configration (,,M) eqals ( 1, 1,M) or ( 2, 2,M), obsere that it is also good. Otherwise it contains a configration of type (e) or (f) on Figre 6.1. (8): Obsere that if M, then the configration (,,M) is good. Frthermore, it is also good if d(,) = 3 and {,} { 1, 1 }. Otherwise it contains a configration of type (e), (f), or (g) on Figre 6.1. (9): In this case Q 3 M is disconnected so (,,M) is a bad configration. Obsere that it contains a configration of type (a), (b), or (c) on Figre 6.1. (10): In the last case Q 3 M is a Hamiltonian cycle. Ths the configration (,,M) is good if is an edge on this cycle. Otherwise it contains a configration of type (e) or (f) on Figre 6.1. This completes the analysis of all cases and establishes the lemma. The following two lemmas will be sefl for inqiry of Q 4 in the next section. The first lemma applies typically for configrations (,,M) sch that both and belong to a same sbcbe Q d L or Qd R which contains at most 2 edges of M. Lemma 6.2. Let and be ertices of different parity and let e 1 and e 2 be distinct edges in Q 3. Then there is a Hamiltonian path of Q 3 between and that (i) contains exactly one of e 1 and e 2 ; or (ii) falts both e 1,e 2 bt contains an edge xy sch that x e 1 and y e 2.

9 8 D. DIMITROV, T. DVOŘÁK, P. GREGOR, AND R. ŠKREKOVSKI e 1 e 2 e 1 (1) (2) (3) e2 Fig An illstration for Lemma 6.2. Proof. Sppose first that and are at distance 3. From Figre 6.3(1) it is obios that all Hamiltonian paths between and are isomorphic. Moreoer, each of them contains one edge of the pper leel, one edge of the lower leel, and fie edges of the middle leel. Considering few cases, regarding the leels that contain e 1 and e 2, it is easy to find a sitable path. Sppose now that is an edge of Q 3. We distingish three cases: Case 1: Edges e 1 and e 2 are adjacent at some ertex x. If x =, then assme that e 1 = y and apply Lemma 3.1 to find a Hamiltonian path of Q 3 between y and. Extending this path with the edge e 1 = y we obtain a Hamiltonian path of Q 3 between and that contains e 1 and falts e 2. If x =, then proceed similarly. Otherwise we hae x / {,}. The Hamiltonian path between and that falts e 1 from Lemma 3.2 contains e 2 since it passes throgh the ertex x. Case 2: Edges e 1 and e 2 are at distance 1. If the edge is adjacent to both e 1 and e 2, then we hae two non-isomorphic cases on Figres 6.3(2)-(3) with desired Hamiltonian paths. Otherwise, we may assme that x e 1, y e 2, and y / {,} for some edge xy. The Hamiltonian path between and that falts e 1 from Lemma 3.2 contains e 2 or xy since it passes throgh the ertex y. Case 3: Edges e 1 and e 2 are at distance 2. If e 1 =, apply Corollary 3.4 to find a Hamiltonian path between and that contains edge e 2. If e 2 =, proceed similarly. Otherwise the configration (,, {e 1,e 2 }) is bad since it is of type (a), (b), or (c) on Figre 6.1. Ths the Hamiltonian path between and that falts e 1 from Lemma 3.2 contains e 2. This completes the analysis of all cases. The second lemma applies for example when MR d = {ab,cd} as on Figre 6.4 and Q d L contains a path between ertices bd and c d. The extra edge bc represents this path. Lemma 6.3. Let a,b,c,d,e,f be ertices of Q 3 as depicted on Figre 6.4 and, V (Q 3 ) be any two ertices of different parity. Then Q 3 + bc has a Hamiltonian path between and that contains the edge bc and falts {ab,cd} nless {,} = {b,c} or {,} = {e,f}. b f a d e c Fig The configration of Lemma 6.3. Proof. By exhasting all 14 possibilities for {,}, as it is presented on Figre 6.5, one can erify the alidity of the lemma.

10 GRAY CODES FAULTING MATCHINGS 9 Fig Hamiltonian paths in Lemma 6.3 for each pair {, } of endertices. 7. Gray codes in Q 4. If M is aligned in Q 4, we can se Lemma 5.3 which holds for eery n 4. Ths, it remains to consider admissible configrations (,,M) only with an naligned matching M. We start with the list of all maximal naligned matchings in Q 4. Let m(m) denote the maximm nmber of edges of M of the same dimension, i. e., m(m) = max d [n] M d. Lemma 7.1. Let M be a maximal naligned matching in Q 4. Then m(m) {2,3,4} and M is isomorphic to one of the matchings on Figres Proof. Let d [4] be a dimension with the maximal nmber of edges of M, i. e., M d = m(m). Since M is naligned, we hae ML d 2 or Md R 2, say Md L 2. Ths among eight ertices of Q d L there are at most for ertices that are not matched by ML d. Hence m(m) 4. On the other hand, eery maximal matching in Q 4 contains at least 6 edges [6], hence at least 2 edges from M are of a same dimension. This implies that m(m) 2. So m(m) {2,3,4}. Let G L and G R be the graphs obtained from Q d L and Qd R, respectiely, by remoing all ertices incident with M d. Note that G L and G R are isomorphic, so let s write simply G when it is clear from the context for which of the graphs we speak. Since M is maximal, both ML d and Md R are maximal matchings of G. To show that M is one of the matchings on Figres , we first consider the set M d and then the combination of sets ML d and Md R. The matchings Md, ML d and MR d on the following figres are represented with dashed lines, the edges of the graph G are bold, and the remaining edges are dotted. Note that the graphs G R and G are flipped on the following figres, so the matchings MR d and the maximal matchings of G are also flipped. Case 1: m(m) = 4. Since M is naligned, there are three non-isomorphic combinations of for edges in M d ; see Figre 7.4. We consider now three possibilities. Sbcase 1.1: M d is as on Figre 7.4(1). Obiosly, there is only one maximal matching of G. Ths ML d and Md R are niqely determined and the matching M is as on Figre 7.1(1). Sbcase 1.2: M d is as on Figre 7.4(2). There are two maximal matchings of G depicted on Figre 7.5. By symmetry, we may assme that ML d is as on Figre 7.5(1). Now, if MR d is also as on Figre 7.5(1), the matching M corresponds to Figre 7.1(3). And, if MR d is as on Figre 7.5(2), then M corresponds to Figre 7.1(2). Sbcase 1.3: M d is as on Figre 7.4(3). There are two maximal matchings of G depicted on Figre 7.6. By the maximality of M, it follows that ML d and Md R are not both as on Figre 7.6(2). Now, if both of them are as on Figre 7.6(1), we infer Figre 7.1(4). Finally, if one of them is as on Figre 7.6(1), and the other one as on

11 10 D. DIMITROV, T. DVOŘÁK, P. GREGOR, AND R. ŠKREKOVSKI R M = {4, 4, 0, 0} R M = {4, 2, 2, 0} R M = {4, 4, 0, 0} (1) (2) (3) R M = {4, 2, 2, 0} R M = {4, 1, 1, 1} (4) (5) Fig All maximal naligned matchings M of Q 4 with m(m) = 4. a 1 R M = {3, 3, 1, 0} R M = {3, 2, 1, 1} R M = {3, 3, 1, 0} a 3 a 1 a 3 a 2 b 1 b 3 b 2 b 4 a 2 a 4 b 2 b 4 b 1 b 3 a 4 (1) R M = {3, 2, 1, 1} (2) R M = {3, 2, 1, 1} (3) a 1 a 3 a 2 b 2 b 1 b 3 b 4 a 4 (4) (5) Fig All maximal naligned matchings M of Q 4 with m(m) = 3. R M = {2, 2, 2, 2} R M = {2, 2, 2, 2} (1) (2) a 1 a 2a3 a 4 R M = {2, 2, 2, 0} b 1 b 2 b 3 b 4 R M = {2, 2, 2, 0} (3) (4) Fig All maximal naligned matchings M of Q 4 with m(m) = 2. Figre 7.6(2), we infer Figre 7.1(5). Case 2: m(m) = 3. Since M is naligned, there are two non-isomorphic combinations of three edges in M d which are depicted on Figre 7.7. Consider now these two cases separately. Sbcase 2.1: M d is as on Figre 7.7(1). There are three maximal matchings

12 GRAY CODES FAULTING MATCHINGS 11 (1) (2) (3) Fig All sets M d with 4 edges in a maximal naligned matching M of Q 4. (1) (2) Fig All maximal matchings of G in Sbcase 1.2. of G depicted on Figre 7.8. By symmetry, we may assme that if ML d is as on Figre 7.8(x), then MR d is as on Figre 7.8(y) where 1 x y 3. Moreoer, since M is a maximal matching, we infer x y. We hae the following possibilities for the pair (x,y): (1,3): The matching M is on Figre 7.2(1); (2,3): The matching M is on Figre 7.2(2); (1,2): This case is isomorphic to the case (1,3). Indeed, consider the isomorphism of Q 4 that interchanges ertices a i and b i on Figre 7.2(1) for eery i [4] and fixes all other ertices. Sbcase 2.2: M d is as on Figre 7.7(2). There are for maximal matchings of G depicted on Figre 7.9. By symmetry, we may assme that if ML d is as on Figre 7.9(x), then MR d is as on Figre 7.9(y) where 1 x y 4. Moreoer, since M is a maximal matching, it follows that x y. Ths, we hae the following possibilities for the pair (x,y): (1,4): The matching M is on Figre 7.2(3); (2,4): The matching M is on Figre 7.2(4); (3,4): The matching M is on Figre 7.2(5); (1,2): This case does not occr since M is a maximal matching; (1,3) and (2,3): These cases are isomorphic to the cases (2,4) and (1,4), respectiely. Indeed, consider the isomorphism of Q 4 that interchanges ertices a i and b i on Figres 7.2(4) and 7.2(3) for eery i [4] and that fixes all other ertices. Case 3: m(m) = 2. Let k(m) be the minimal distance of distinct edges of M of a same dimension, i. e., k(m) = min{d(e 1,e 2 ) e 1,e 2 M i for some i [4] and e 1 e 2 }. We assme that M d contains edges at distance k(m), otherwise choose d accordingly. For each k(m) [3] we hae M d on Figre Sbcase 3.1: k(m) = 3. Ths M d is as on Figre 7.10(1). There are fie maximal matchings of G depicted on Figre Obsere that neither ML d nor Md R is as on Figres 7.11(3)-(5) since k(m) = 3. Moreoer, Md L and MR d are not both as on Figre 7.11(1), or both as on Figre 7.11(2) since k(m) = 3. Ths, one of them is as on Figre 7.11(1), and the other one as on Figre 7.11(2). Hence M corresponds to

13 12 D. DIMITROV, T. DVOŘÁK, P. GREGOR, AND R. ŠKREKOVSKI (1) (2) Fig All maximal matchings of G in Sbcase 1.3. (1) (2) Fig All sets M d with 3 edges in a maximal naligned matching M of Q 4. Figre 7.3(1). Sbcase 3.2: k(m) = 2. Ths M d is as on Figre 7.10(2). There are seen maximal matchings of G depicted on Figre Obsere that neither ML d nor Md R is as on Figres 7.12(6)-(7) since k(m) = 2. Frthermore, neither ML d nor Md R is as on Figre 7.12(5) since all matchings of G on Figres 7.12(1)-(5) contain at least one edge of e 1 or e 2 and k(m) = 2. By symmetry, we may assme that if ML d is as on Figre 7.12(x), then MR d is as on Figre 7.12(y) where 1 x y 4. Moreoer, x y since M is a maximal matching. We hae the following possibilities for the pair (x,y): (1,2): The matching M is on Figre 7.3(4); (3,4): This case is isomorphic to the case (1,2). Indeed, consider the isomorphism of Q 4 that interchanges ertices a i and b i on Figre 7.3(4) for eery i [4] and fixes all other ertices. (1,3), (1,4), (2,3), and (2,4): These cases do not occr since M is a maximal matching. Sbcase 3.3: k(m) = 1. Ths M d is as on Figre 7.10(3). There are fie maximal matchings of G depicted on Figre Obsere that neither ML d nor Md R is as on Figre 7.13(5) since m(m) = 2. By symmetry, we may assme that if ML d is as on Figre 7.13(x), then MR d is as on Figre 7.13(y) where 1 x y 4. We hae the following possibilities for the pair (x,y): (1,2): The matching M is on Figre 7.3(2); (3,4): The matching M is on Figre 7.3(3); (1,1), (1,3), (1,4), (2,2), (2,3), and (2,4): These cases do not occr since m(m) = 2; (3,3) and (4,4): These cases do not occr since M is a maximal matching. Note that the choice of the dimension d was not deterministic. Therefore, to conclde the proof, we need to erify that the matchings on Figres are nonisomorphic. Define the mark of M to be the collection R M = { M d } d [n]. Clearly, matchings with different marks are not isomorphic. So we need to examine only matchings with different marks on Figres This can be easily erified by the reader. Lemma 7.2. Let (,,M) be a maximal admissible configration in Q n with an naligned matching M. Then M is maximal with respect to inclsion.

14 GRAY CODES FAULTING MATCHINGS 13 (1) (2) (3) Fig All maximal matchings of G in Sbcase 2.1. (1) (2) (3) (4) Fig All maximal matchings of G in Sbcase 2.2. Proof. Sppose on the contrary that M is not maximal, i. e., M {e} is a matching for some edge e E(Q n ) \ M. Let d be the dimension of e. Since M is naligned, we hae ML d 2 or Md R 2, so we may assme that Md L contains edges x 1x 2 and x 3 x 4. Let e i = x i x d i for i [4]. Since d(e 1,e 2 ) is odd, d(e 3,e 4 ) is odd, e 1,e 2,e 3,e 4 are not in M {e}, and (,,M) is admissible, it follows from the definition that (,,M {e}) is also admissible. Bt this contradicts the assmption that (,, M) is a maximal admissible configration. Now we are ready to find a Hamiltonian path of Q 4 M between and if (,, M) is admissible configration. The following lemma seres as a basis of indction in the next section. Lemma 7.3. Let M be a matching in Q 4 and, V (Q 4 ). If (,,M) is admissible, then there is a Hamiltonian path of Q 4 M between and. Proof. If M is aligned, we can se Lemma 5.3 which holds for eery n 4. Ths, it remains to consider admissible configrations (,, M) only with naligned matching M. Assme that (,, M) is a maximal admissible configration, otherwise introdce new edges in M. Notice that with this procedre M stays naligned. By Lemma 7.2, the matching M is maximal. Ths m(m) {2,3,4} and M corresponds to one of the matchings on Figres by Lemma 7.1. Let d [4] be a dimension with the maximal nmber of edges from M, i. e., M d = m(m). Now, as d is fixed, for the sake of simplicity, let s omit d in the notions of Q d L, Qd R, Md L, and Md R. Frthermore, for x V (Q R) and y V (Q L ) let x L and y R denote their corresponding ertices in the opposite sbcbe, i. e., x L = x d and y R = y d. We say that ertices and are separated if V (Q L ) and V (Q R ) or ice ersa. In what follows, we distingish three cases regarding m(m). In each case we consider good/bad configrations in Q L and Q R. Then, regarding the position of ertices and, we look separately in Q L and Q R for paths that can be combined into a Hamiltonian path of Q 4 M between and. Case 1: m(m) = 4. Applying Proposition 6.1, obsere that Figre 7.14(1)-(5) shows all bad configrations in Q L or Q R for each of the fie maximal naligned matchings M with m(m) = 4 depicted on Figre 7.1, respectiely. Vertices and that form a bad configration (,,M L ) in Q L or (,,M R ) in Q R are encompassed

15 14 D. DIMITROV, T. DVOŘÁK, P. GREGOR, AND R. ŠKREKOVSKI (1) (2) (3) Fig All sets M d with 2 edges in a maximal naligned matching M of Q 4. (1) (2) (3) (4) (5) Fig All maximal matchings of G in Sbcase 3.1. by a solid elliptic cre, the matching M is represented by dashed edges, whereas other edges of Q 4 are omitted for the sake of clarity. Ths e. g. on Figre 7.14(2) we hae two bad configrations in Q L and two bad configrations in Q R. Sbcase 1.1: Vertices and are separated. Assme that V (Q L ) and V (Q R ). We distingish the following two possibilities : M is as on Figre 7.14(1) and or is incident with M d. By symmetry, we may assme that = a. Obsere on Figre 7.15 that the statement of the lemma holds for each V (Q R ) of different parity than. The Hamiltonian paths are presented by bold edges : M is as on Figre 7.14(2)-(5) or none of and is incident with M d. Let w V (Q L ) be a ertex of different parity than sch that it is incident with M L. Then, and w R are also of different parities. Obsere on Figre 7.14(2)-(5) that the ertices w and w R are in no bad configration in Q L and Q R, respectiely. Ths the configrations (,w,m L ) and (,w R,M R ) are good. Hence there is a Hamiltonian path P w of Q L M L, and a Hamiltonian path P wr of Q R M R. The desired Hamiltonian path of Q 4 M is P w + P wr. Sbcase 1.2: Both ertices and are in Q L. Note that M L is of size 2, ths denote by e 1,e 2 these two edges. By Lemma 6.2, there is a Hamiltonian path P that either contains precisely one of e 1 and e 2, or falts both e 1,e 2 bt contains an edge adjacent to both e 1,e 2. Let xy E(Q L ) be that contained edge. Obsere on Figre 7.14(1)-(5) that there is no bad configration (w,z,m R ) in Q R sch that w L and z L are both incident with e 1 or e 2. Ths the configration (x R,y R,M R ) is good. Hence there is a Hamiltonian path P xry R of Q R M R. The desired Hamiltonian path of Q 4 M is P x +P xry R +P y where P x and P y are disjoint sbpaths of P, assming that x is closer to on P than y. Sbcase 1.3: Both ertices and are in Q R. If M is as on Figre 7.14(1)-(4), we hae symmetry to the preios sbcase. Howeer, if M is as on Figre 7.14(5), then M R contains only one edge e 1. Then jst pt e 2 = bc and proceed symmetrically as in Sbcase 1.2. Case 2: m(m) = 3. Applying Proposition 6.1 obsere that Figre 7.16(1)-(5) shows all bad configrations in Q L or Q R for each of the fie maximal naligned matchings M with m(m) = 3 depicted on Figre 7.2, respectiely. We consider the following possibilities.

16 GRAY CODES FAULTING MATCHINGS 15 e 1 (1) (2) (3) (4) (5) (6) (7) e 2 Fig All maximal matchings of G in Sbcase 3.2. (1) (2) (3) (4) (5) Fig All maximal matchings of G in Sbcase 3.3. Sbcase 2.1: Vertices and are separated. Again, assme that V (Q L ) and V (Q R ). We distingish the following two possibilities : M is as on Figre 7.16(1) and is not incident with M L. Obsere on Figre 7.17 that for each V (Q L ) that is not incident with M L, there are spanning paths P wl and P blc L of Q 4 M L. By Lemma 6.3, there is a Hamiltonian path P w of Q R + bc that contains the edge bc and falts M R. Hence a desired Hamiltonian path of Q 4 M is P wl + P wb + P blc L + P c where P wb and P c are disjoint sbpaths of P w, assming that b is closer to w on P w than c : M is as on Figre 7.16(2)-(5) or is incident with M L. Let x,y V (Q L ) be ertices of different parity than that are incident with M L. We can always choose sch x,y since M L = 2. Obsere on Figre 7.16(1) that each bad configration in Q L incldes at most one ertex incident with M L. Similarly, obsere on Figre 7.16(2)-(5) that each bad configration in Q L incldes no ertex incident with M L. Ths both configrations (,x,m L ) and (,y,m L ) are good. Moreoer, obsere on Figre 7.16(1)-(5) that for at most one z V (Q L ), if z is incident with M L, then z R is in a bad configration in Q R. Hence at most one of configrations (,x R,M R ) and (,y R,M R ) is bad. Let w {x,y} be sch that the configration (,w R,M R ) is good. Therefore, there is a Hamiltonian path P w of Q L M L, and a Hamiltonian path P wr of Q R M R. Finally, the desired Hamiltonian path is P w + P wr. Sbcase 2.2: Vertices and are both in Q L or both in Q R. We distingish the following two possibilities : M is as on Figre 7.16(1) and both and are in Q R. If {,} = {b,c} or {,} = {e,f}, obsere on Figre 7.18 that the statement of the lemma holds. Otherwise, by Lemma 6.3, there is a Hamiltonian path P of Q R +bc that contains the edge bc and falts M R. Obsere on Figre 7.16(1) that the configration (b L,c L,M L ) is good. Ths there is a Hamiltonian path P blc L of Q L M L. The desired Hamiltonian path of Q 4 M is P b + P blc L + P c where P b and P c are disjoint sbpaths of P, assming that b is closer to on P than c : M is as on Figre 7.16(1) and both and are in Q L, or M is as on Figre 7.16(2)-(5). We proceed similarly as in Sbcase 1.2. Assme that, V (Q L ), otherwise interchange the roles of Q L and Q R. Let M L = {e 1,e 2 }. By

17 16 D. DIMITROV, T. DVOŘÁK, P. GREGOR, AND R. ŠKREKOVSKI a (1) (2) (3) b c (4) (5) Fig All bad configrations in Q L or Q R for each maximal naligned matching M in Q 4 with m(m) = 4. Fig Hamiltonian paths for Sbcase Lemma 6.2, there is a Hamiltonian path P that either contains precisely one of e 1 and e 2, or falts both e 1,e 2 bt contains an edge adjacent to both e 1,e 2. Let xy be that contained edge. Obsere on Figre 7.16(1)-(5) that there is no bad configration (w,z,m R ) in Q R sch that w L and z L are both incident with e 1 or e 2. Ths the configration (x R,y R,M R ) is good. Hence there is a Hamiltonian path P xry R of Q R M R. The desired Hamiltonian path of Q 4 M is P x +P xry R +P y where P x and P y are disjoint sbpaths of P, assming that x is closer to than y on the path P. Case 3: m(m) = 2. Applying Proposition 6.1 obsere that Figre 7.19(1)-(4) shows all bad configrations in Q L or Q R for each of the for maximal naligned matchings M with m(m) = 2 depicted on Figre 7.3, respectiely. Sbcase 3.1: Vertices and are separated. Assme that V (Q L ) and V (Q R ). We distingish the following three possibilities : M is as on Figre 7.19(1) and or is incident with M d. Assme that = a, the other case is symmetric. Obsere on Figre 7.20 that the statement holds for each V (Q R ) of different parity than : M is as on Figre 7.19(4) and {,} = {d,e}. Obsere on Figre 7.21 that the statement holds : The remaining possibility; that is, M is as on Figre 7.19(1) and none of and is incident with M d, or M is as on Figre 7.19(2)-(4) and {,} {d,e}.

18 GRAY CODES FAULTING MATCHINGS 17 b f e c (1) (2) (3) (4) (5) Fig All bad configrations in Q L or Q R for each maximal naligned matching M in Q 4 with m(m) = 3. b w b w c c w w b b c c Fig Spanning paths of Q L for Sbcase A ertex w V (Q L ) is free if it has a different parity than and is not incident with M d. We claim that there is a free ertex w V (Q L ) sch that the configrations (,w,m L ) and (,w R,M R ) are good. Sppose first that M is as on Figre 7.19(1)-(3). Obsere from the same figre that the configration (,x,m L ) is bad for at most one free ertex x. Similarly, the configration (,y R,M R ) is bad for at most one free ertex y. Since there are three free ertices, the claim holds in this case. Sppose now that M is as on Figre 7.19(4). If and d hae different parity, then we hae two free ertices and both of them satisfy the claim. If and d hae the same parity, we hae for free ertices. Obsere on Figre 7.19(4) that the configrations (,x,m L ) or (,x R,M R ) are bad for at most three free ertices x, since {,} {d,e}. Hence the claim holds in all cases. Therefore, there is a Hamiltonian path P w of Q L M L, and a Hamiltonian path P wr of Q R M R. The desired Hamiltonian path of Q 4 M is P w + P wr. Sbcase 3.2: Vertices and are both in Q L or both in Q R. Assme that, V (Q L ), the case, V (Q R ) is symmetric. We distingish the following fie possibilities : M is as on Figre 7.19(1) and the configration (,,M L ) is bad. We hae three non-isomorphic cases: both, precisely one, or none of and is incident

19 18 D. DIMITROV, T. DVOŘÁK, P. GREGOR, AND R. ŠKREKOVSKI Fig Hamiltonian paths for Sbcase a x 4 b x 3 c (1) (2) e x 1 x 2 d (3) (4) Fig All bad configrations in Q L or Q R for each maximal naligned matching M in Q 4 with m(m) = 2. with M L. Obsere on Figre 7.22 that the statement holds for each of these three possibilities : M is as on Figre 7.19(1) and the configration (,,M L ) is good. We hae two non-isomorphic cases: M L and / M L. Obsere on Figre 7.23 that the statement holds for both of these cases : M is as on Figre 7.19(2) and {,} = {b,c} or {,} = {x 3,x 4 }. Obsere on Figre 7.24 that the statement holds : M is as on Figre 7.19(2)-(4), the configration (,,M L ) is good, and {,} {x 3,x 4 }. Then there is a Hamiltonian path P of Q L M L. We claim that P contains an edge xy P sch that xx R / M d, yy R / M d, and the configration (x R,y R,M R ) is good. First, if M is as on Figre 7.19(3) or 7.19(4), pt x = x 1 or x = x 2, respectiely, and choose y to be any neighbor of x on P. Obsere on Figre 7.19(3)-(4) that x is in no bad configration in Q L, also x R is in no bad configration in Q R, xx R / M d, and yy R / M d. Ths the claim holds in this sitation. Now M is as on Figre 7.19(2). Since {,} {x 3,x 4 }, we hae x 3 / {,} or x 4 / {,}, say x 3 / {,}. Ths the Hamiltonian path P contains the edge x 3 x 4. Pt x = x 3 and y = x 4 and obsere on Figre 7.19(2) that the claim is established. Hence, by the claim, there is a Hamiltonian path P xry R of Q R M R. Finally, the desired Hamiltonian path of Q 4 M is P x +P xry R +P y where P x and P y are disjoint sbpaths of P, assming that x is closer to than y on the path P : The remaining possibility; that is, M is as on Figre 7.19(2)-(4), the configration (,,M L ) is bad, and {,} {b,c}. Choose an edge xy M L sch that the configrations (,,M L \ {xy}) and (x R,y R,M R ) are good. Obsere on Figre 7.19 and Figre 6.1 that sch an edge exists. Ths there is a Hamiltonian path P xry R of Q R M R. Since the configration

20 GRAY CODES FAULTING MATCHINGS 19 Fig Hamiltonian paths for Sbcase Fig A Hamiltonian path for Sbcase (,,M L ) is bad bt the configration (,,M L \{xy}) is good, there is a Hamiltonian path P that contains edge xy and falts M L \ {xy}. The desired Hamiltonian path of Q 4 M is P x + P xry R + P y where P x and P y are disjoint sbpaths of P, assming that x is closer to than y on the path P. This completes the proof of the lemma. 8. Gray codes in hypercbes of higher dimensions. In this section we se AL(,,M,Q n ) to denote the fact that a matching M contains an almost-layer in Q n with a free pair {,}. Recall that if AL(,,M,Q n ) holds, then there exists d sch that M d consists of all edges of dimension d except d and d. The following simple obseration is sed later. Proposition 8.1. Let M be a matching which contains an almost-layer A with a free pair {,} in Q n, n 4. (i) If AL(,w,M,Q n ) holds, then = w. (ii) If at least one of ertices x,y is incident with an edge of A, then AL(x,y,M,Q n ) does not hold. Note that the assmption n 4 is for part (ii) essential, as it fails to hold in case that M is the perfect matching of Q 3 depicted on Figre 6.2(10). Now we are ready to resole the case of matchings not containing a half-layer. Lemma 8.2. Let, V (Q n ), n 4, sch that p() p() and let M be a matching in Q n containing no half-layer. Then Q n M contains a Hamiltonian path between and nless M is an almost-layer in Q n with a free pair {,}. Proof. We arge by indction on n. Since the case n = 4 is settled by Lemma 7.3, assme that n > 4. If M is aligned, the statement follows from Lemma 5.3. If M is naligned, then by Lemma 4.2, there is d [n] sch that Mi d contains no half-layer in Q d i for both i {L,R}. Withot a loss of generality assme that V (Qd L ). To obtain the desired Hamiltonian path P of Q n M, consider the following two cases: Case 1: Q d R. Since M does not contain a half-layer, there is a ertex w

21 20 D. DIMITROV, T. DVOŘÁK, P. GREGOR, AND R. ŠKREKOVSKI Fig Hamiltonian paths for Sbcase Fig Hamiltonian paths for Sbcase V (Q d L ) sch that (8.1) p() p(w) and ww d M. Notice that also p(w d ) p(). We claim that w can be chosen so that neither AL(,w,ML d,qd L ) nor AL(wd,,MR d,qd R ) holds. Indeed, if Md L contains an almostlayer A in Q d L with a free pair {,w}, then each edge of A contains one ertex that can play the role of w in (8.1). Hence there are still A = 2 n 2 2 > 2 choices for w, and MR d may contain an almost-layer in Qd R with a free pair {wd,} for no more than one of them by Proposition 8.1(i). It remains to apply the indction to obtain Hamiltonian paths P w and P wd of Q d L Md L and Qd R Md R, respectiely, and finally pt P := P w + P wd to conclde this case. Case 2: V (Q d L ). Consider the following sbcases. Sbcase 2.1: AL(,,ML d,qd L ) holds. Since Md L contains an almost-layer A with 2 n 2 2 > 3 edges, Proposition 8.1(ii) implies that there mst be an edge xy A sch that MR d contains no almost-layer in Qd R with a free pair {xd,y d }. Ths, by the indction hypothesis, there are Hamiltonian paths P and P x d y of d Qd L (Md L \{xy}) and Q d R Md R, respectiely. Note that by Lemma 5.1, the path P passes throgh the edge xy, and hence there are sbpaths P x and P y of P sch that P = P x +P y. It remains to pt P := P x + P x d y + P d y to complete this sbcase. Sbcase 2.2: AL(,,ML d,qd L ) does not hold. First apply the indction hypothesis to obtain a Hamiltonian path P L of Q d L Md L. Next, distingish the following two possibilities. Sbcase 2.2.1: {xx d,yy d } M = for some pair x,y of neighbors on P. L We claim that ertices x and y can be chosen so that AL(x d,y d,mr d,qd R ) does not hold. To erify the claim, sppose, by way of contradiction, that MR d contains an almostlayer A in Q d R with a free pair {xd,y d }. Denote by S the set of ertices z of Q d L sch that z d is incident with an edge of A. Then S = 2 A = 2 n 1 4 > 2 n 2 = V (Q d L) /2 = V (P L ) /2 and hence there exist x,y S that are neighbors on P. L Proposition 8.1(ii) then implies that AL(x d,y d,mr d,qd R ) does not hold.

22 GRAY CODES FAULTING MATCHINGS 21 Fig Hamiltonian paths for Sbcase As in Sbcase 2.1, it remains to apply the indction to obtain a Hamiltonian path P xd y d of Qd R Md R and pt P := P x + P xd y d + P y, where P x and P y are the sbpaths of P L satisfying P L = P x + P y. Sbcase 2.2.2: {xx d,yy d } M for each pair x,y of neighbors on P L. By Lemma 4.2 there exists another splitting dimension d d and we claim that Sbcase for d cannot occr. To erify the claim, first obsere that (8.2) M d V (P L ) /2 = 2 n 2. If this ineqality is strict, then M 2 n 1 implies that M d < 2 n 2, contrary to (8.2), and hence the claim holds. Ths we can assme that (8.3) M d = M d = 2 n 2. This means that M d = V (P L ) /2, which together with the assmption of this sbcase implies that {xx d,yy d } M d = 1 for each pair x,y of neighbors on P L. Since the parity of consectie ertices on P L alternates, it follows that if d M d, then xx d M d for each ertex x V (P L ) with p() = p(x). Howeer, this wold mean that M contains a half-layer, contrary to or assmption. Since the same reasons apply to the endertex, we conclde that neither nor are incident with an edge of M d. Applying the same argment to the dimension d, it follows that contrary to (8.3). M d + M d V (Q n ) \ {,} /2 = 2 n 1 1, 9. The main reslts. Now we are ready to formlate the necessary and sfficient conditions for (cyclic) Gray codes falting a gien matching. Note that the conditions are expressed in terms of half-layers and almost-layers, which were defined in Sections 4 and 5, respectiely. Theorem 9.1. Let M be a matching, and let and be ertices of Q n, n 4. Then Q n has a Gray code between and falting M if and only if M contains no odd half-layer for or and no almost layer with a free pair {,}. Proof. Recall that (,,M) is an admissible configration if and only if the left side of the eqality holds, by Proposition 5.2. The necessity is settled by Lemma 5.1. The sfficiency follows from Lemma 5.3 in case M contains a half-layer, and from Lemma 8.2 otherwise. It shold be noted that for n = 3 the problem is flly described by Proposition 6.1 in Section 6, and for n 2 it is triial. Corollary 9.2. Let M be a matching in Q n, n 4. Then Q n has a cyclic Gray code falting M if and only if M does not contain a half-layer. Proof. Obsere that the following two eqialences hold. Q n M contains a Hamiltonian cycle Q n M contains a Hamiltonian path P for some edge E(Q n );