On orientation-preserving automorphisms of Hamiltonian cycles in the N-dimensional Boolean cube

ISSN 1990-4789, Journal of Applied and Industrial Mathematics, 2011, Vol. 5, No. 3, pp. 410–416. c© Pleiades Publishing, Ltd., 2011.Original Russian Text c© A.L. Perezhogin, 2010, published in Diskretnyi Analiz i Issledovanie Operatsii, 2010, Vol. 17, No. 4, pp. 32–42.

On Orientation-Preserving Automorphisms of Hamiltonian Cyclesin the N-Dimensional Boolean Cube

A. L. Perezhogin*

Sobolev Institute of Mathematics, pr. Akad. Koptyuga 4, Novosibirsk, 630090 RussiaReceived December 9, 2009; in final form, January 11, 2010

Abstract—We obtain an upper bound for the order of the group of orientation-preserving automor-phisms of a Hamiltonian cycle in the Boolean n-cube. We prove that the existence of a Hamiltoniancycle with the order of the group of orientation-preserving automorphisms attaining this upperbound is equivalent to the existence of a Hamiltonian cycle with an additional condition on the graphof orbits of a fixed automorphism of the n-cube.

DOI: 10.1134/S1990478911030148

Keywords: n-cube, Hamiltonian cycle, automorphism

Refer as the Boolean n-dimensional cube Qn to the graph whose vertices are all binary collectionsof length n, with two vertices joined by an edge whenever the corresponding collections differ in oneposition. Given v ∈ Qn, denote the weight of the collection v by ω(v). Also put

0 = (0, 0, . . . , 0), 1 = (1, 1, . . . , 1) ∈ Qn.

A simple cycle in Qn containing all vertices is called a Hamiltonian cycle in Qn. The Hamiltoniancycles in Qn are often called Gray codes [9].

Associate to a Hamiltonian cycle H = v0 = 0, v1, . . . , vt−1, vt = v0 in Qn the word X = X(H) =x0x1 . . . x2n−1 in the alphabet An = {1, 2, . . . , n}, where xi for 0 ≤ i ≤ t − 1 is the index of the positionin which the collections vi and vi+1 differ. Refer to this word as the transition word of H . Henceforth,the indices of both the vertices of cycles and the letters of their transition words are considered modulothe cycle length.

Given a word Y , denote the content tuple of Y modulo 2 by s(Y ). In other words, s(Y ) =(s1, s2, . . . , sn), where sm = 0 for 1 ≤ m ≤ n if the letter m occurs in Y an even number of times, andsm = 1 otherwise. Observe that s(X(H)) = 0.

It is known that a translation of the cube Qn by a length n binary collection v,

Qn ⊕ v = {u ⊕ v : u ∈ Qn},where ⊕ stands for the coordinatewise addition modulo 2, is an automorphism of the graph Qn.Let G(Qn) denote the subgroup of all these translations in the automorphism group Aut(Qn) of theBoolean n-dimensional cube. The group Sn(Qn) of permutations of the coordinates of the vertices of thecube is also a subgroup of Aut(Qn). Moreover, it is not difficult that

Aut(Qn) = G(Qn) � Sn(Qn),

where � stands for the semidirect product (for instance, see [3]). Therefore, every automorphismϕ ∈ Aut(Qn) is determined by a pair (π, v) with π ∈ Sn(Qn) and v ∈ Qn as ϕ(Qn) = π(Qn) ⊕ v.

Refer as an automorphism of a Hamiltonian cycle H in Qn to an automorphism of Qn taking H intoitself. Essentially all available constructions of the Gray codes enable us to construct Hamiltonian cycleswith nontrivial automorphism groups [1, 6, 8], although the share of these cycles is small. Studying theproperties of automorphism groups is helpful in classifying Hamiltonian cycles [4, 7]. While seeking

*E-mail: [email protected]

410

ON ORIENTATION-PRESERVING AUTOMORPHISMS 411

Hamiltonian cycles with additional properties, it often helps to restrict the problem to a class of cycleswith a large automorphism group [2, 10, 11].

Put

τ(n) = max |Aut(H)|,where the maximum is taken over all Hamiltonian cycles H in Qn.

Lemma 1. τ(n + 1) ≥ τ(n) for all n ≥ 2.

Proof. The following construction of Gray codes is available [5]: if x0x1 . . . x2n−1 is the transition se-quence of a Hamiltonian cycle H in Qn then the word x0(n + 1)x1(n + 1) . . . x2n−1(n + 1) is a transitionsequence of some Hamiltonian cycle H ′ in Qn+1. It remains to observe that |Aut(H ′)| = |Aut(H)|. Theproof is complete.

Fix an orientation of a Hamiltonian cycle H in Qn. If the action of an automorphism ϕ ∈ Aut(H)of the cycle preserves the orientation then refer to this automorphism as orientation-preserving,and as orientation-reversing otherwise. Let Aut+(H) denote the group of orientation-preservingautomorphisms of a cycle H . It is shown in [2] that if |Aut+(H)| = t then

Aut+(H) ∼= Zt, (1)

while if H admits an orientation-reversing automorphism then

Aut(H) ∼= Z2 � Aut+(H). (2)

Observe that (1) implies that t is always a power of 2. Put

τ+(n) = max |Aut+(H)|,where the maximum is over all Hamiltonian cycles H in Qn. Then (2) directly implies

Lemma 2. τ(n) ≤ 2τ+(n) for all n ≥ 2.Therefore, in studying Hamiltonian cycles with large automorphism groups, the subgroups of

orientation-preserving automorphisms are of particular interest. From now on we consider only those.Take a Hamiltonian cycle H in Qn with a nontrivial group of orientation-preserving automorphisms.

By (1), this group contains a generator ϕπ,v with ϕπ,v(Qn) = π(Qn)⊕ v. Therefore, Aut+(H) = 〈ϕπ,v〉.Observe that, since π is a permutation of the coordinates of the vertices of Qn, it is obviouslya permutation of the letters of the alphabet An = {1, 2, . . . , n}. The next is proved in [2]:

Lemma 3. Consider the transition sequence x0x1 . . . x2n−1 of a Hamiltonian cycle H in Qn

with a nontrivial group of orientation-preserving automorphisms. Then |Aut+(H)| = 2m andAut+(H) = 〈ϕπ,v〉 if and only if

(i) v = s(x0x1 . . . x2n−m−1); (ii) x(i+2n−m) mod 2n = π(xi)up to the orientation of the cycle.

Decompose the permutation π into independent cycles:

π = π1π2 . . . πq, (3)

where

πi =(a

(i)1 a

(i)2 . . . a(i)

ni

), 1 ≤ i ≤ q. (4)

If π fixes an element then assume that π constitutes a cycle of length 1. Therefore, this cycle decompo-sition involves all elements. Assume in addition that in the expression for every cycle the first element isthe smallest. Then (1) directly implies

Lemma 4. If H is a Hamiltonian cycle in Qn with a nontrivial group of orientation-preservingautomorphisms Aut+(H) = 〈ϕπ,v〉, and the permutation π is expressed as in (3), then the lengthni of each cycle πi of the form (4) is a power of 2.

Given u = (u1, u2, . . . , un) ∈ Qn and a cycle πi of the form (4), put

uπi =(u

a(i)1

, ua(i)2

, . . . , ua(i)ni

)∈ Qni . (5)

JOURNAL OF APPLIED AND INDUSTRIAL MATHEMATICS Vol. 5 No. 3 2011

412 PEREZHOGIN

Given v ∈ Qn and a permutation π of the form (3), define the set

Γπ,v = {u ∈ Qn : ω(uπi) ≡ ω(vπi) (mod 2), 1 ≤ i ≤ q}.Observe that

|Γπ,v| = 2n−q. (6)

Theorem 1. Take a Hamiltonian cycle H in Qn with a nontrivial group of orientation-preserving automorphisms Aut+(H) = 〈ϕπ,v〉. Represent the permutation π as in (3), (4). Then,for every u ∈ Γπ,v, there exists a Hamiltonian cycle H ′ in Qn with

Aut+(H ′) = 〈ϕπ,u〉, |Aut+(H ′)| = |Aut+(H)|.

Proof. Suppose that |Aut+(H)| = 2m for m ≥ 1 and take the transition sequence x0x1 . . . x2n−1 of H .Consider the sequence

s0, s1, . . . , s2n−1, (7)

of binary collections of the form

si = s(xixi+1 . . . xi+2n−m−1), 0 ≤ i ≤ 2n − 1. (8)

Henceforth, the indices are considered modulo 2n. By (8) and Lemma 3, for all i and j with 0 ≤ i, j ≤2n − 1 and i < j, we have

sj = si ⊕ s(xixi+1 . . . xj−1) ⊕ s(xi+2n−mxi+2n−m+1 . . . xj+2n−m−1)= si ⊕ s(xixi+1 . . . xj−1) ⊕ s(π(xi)π(xi+1) . . . π(xj−1))

= si ⊕ s(xixi+1 . . . xj−1) ⊕ π(s(xixi+1 . . . xj−1)

).

Therefore, for every cycle πt of (3), we have

(sj)πt = (si)πt ⊕(s(xixi+1 . . . xj−1)

)πt

⊕ π((

s(xixi+1 . . . xj−1))πt

). (9)

Consequently, the weights of the binary collections (sj)πt and (si)πt are of the same parity. However,s0 = v by Lemma 3; hence,

st ∈ Γπ,v, 0 ≤ t ≤ 2n − 1. (10)

From (9) we obtain

(si)πt = (sj)πt ⇔(s(xixi+1 . . . xj−1)

)πt

= π((

s(xixi+1 . . . xj−1))πt

). (11)

Since π((

s(xixi+1 . . . xj−1))πt

)is a cyclic translation of

(s(xixi+1 . . . xj−1)

)πt

, we have

(si)πt = (sj)πt ⇔(s(xixi+1 . . . xj−1)

)πt

∈ {0, 1}. (12)

Consequently,

si = sj ⇔

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

(s(xixi+1 . . . xj−1)

)π1

∈ {0, 1},

. . .(s(xixi+1 . . . xj−1)

)πs

∈ {0, 1}.

There exist exactly 2q length n binary vectors satisfying this system. Since

s(xixi+1 . . . xj−1) �= s(xixi+1 . . . xp−1)

for j �= p, every binary collection occurs in (7) at most 2q times. However, then (6) and (10) implythat every binary collection in Γπ,v occurs in the sequence of (7) exactly 2q times. Therefore, u = st

for some t with 0 ≤ t ≤ 2n − 1. Then the Hamiltonian cycle H ′ obtained by translating H by the vectors(x0x1 . . . xt−1) is the required one. The proof of Theorem 1 is complete.



Corollary 1. Take a Hamiltonian cycle H in Qn, n ≥ 3, such that Aut+(H) = 〈ϕπ,v〉 and|Aut+(H)| = 2k with k ≥ 1. Then all length ni cycles πi in the cycle decomposition (3) of π andall u ∈ Γπ,v satisfy

uπi ⊕ π(uπi) ⊕ · · · ⊕ πni−1(uπi) =

{0 for even ω(vπi),1 otherwise.

(13)

Proof. By Theorem 1, it suffices to consider the case u = v. Suppose that πi is of the form (4). Then, forevery j with 1 ≥ j ≥ ni, among the words

vπi , π(vπi), . . . , πni−1(vπi)

exactly ω(vπi) words contain the letter a(i)j an odd number of times. The required equality follows. The

proof of the corollary is complete.

Corollary 2. Take a Hamiltonian cycle H in Qn, n ≥ 3, such that Aut+(H) = 〈ϕπ,v〉 and|Aut+(H)| = 2k with k ≥ 1. Then the cycle decomposition (3) of π includes a length 2k−1 cycleπi such that the weight of the binary collection vπi is odd.

Proof. Suppose that π is of the form (3). By Theorem 1, there exists u ∈ Γπ,v such that, for every i with1 ≤ i ≤ s, if the weight of vπi is even then uπi = 0. Consequently, if the weight of vπi is even for everycycle πi then

uπi ⊕ π(uπi) ⊕ · · · ⊕ π2k−1−1(uπi) = 0. (14)

By Corollary 1, we deduce from (14) that if some length ni cycle πi satisfies

uπi ⊕ π(uπi) ⊕ · · · ⊕ π2k−1−1(uπi) �= 0

then ni = 2k−1 and the weight of vπi is odd. It remains to observe that

u ⊕ π(u) ⊕ · · · ⊕ π2k−1−1(u) �= 0

since |Aut+(H)| = 2k. The proof of the corollary is complete.

Corollary 3. τ+(n) ≤ 2�log2(n−1)� + 1 for all n ≥ 3.

Proof. Suppose that in Qn, n ≥ 3, there is a Hamiltonian cycle H such that Aut+(H) = 〈ϕπ,v〉 and|Aut+(H)| = 2k with k ≥ 1. Then, by Corollary 2, the cycle decomposition (3) of π includes a length2k−1 cycle πi such that the weight of the binary collection vπi is odd. However, by Lemma 3, the weightof v is even. Consequently, s ≥ 2 in (3). We conclude that n ≥ 2k−1 + 1. The proof of the corollary iscomplete.

Observe that Corollary 3 is proved in [2].Corollary 4. For every k ≥ 2, in Q2k−1+1 there is a Hamiltonian cycle with the group of

orientation-preserving automorphisms of order 2k if and only if in Q2k−1+1 there is a Hamiltoniancycle H such that |Aut+(H)| = 2k and Aut+(H) = 〈ϕπ,v〉, where

π = (12 . . . 2k−1)(2k−1 + 1), v = (1, 0, 0, . . . , 0, 1).

Henceforth put

π = (12 . . . 2k−1)(2k−1 + 1), e = (1, 0, 0, . . . , 0, 1) ∈ Q2k−1+1, ϕ = ϕπ,e ∈ Aut(Q2k−1+1),

π∗ = (12 . . . 2k−1), e∗ = (1, 0, 0, . . . , 0) ∈ Q2k−1 , ϕ∗ = ϕπ∗,e∗ ∈ Aut(Q2k−1).

Lemma 5. For every k ≥ 2, the set of vertices of Q2k−1 splits under the action of ϕ∗ into orbitsof length 2k .


414 PEREZHOGIN

Proof. It is easy that, for every v ∈ Q2k−1 , we have

ϕ2k

∗ (v) = v, ϕ2k−1

∗ (v) = v ⊕ 1.

Take the minimal integer t with ϕt∗(v) = v. Then t divides 2k, but not 2k−1. Consequently, t = 2k. The

proof of Lemma 5 is complete.

Lemma 6. For every k ≥ 2, in Q2k−1 with the action of ϕ∗ there are exactly 2k−2 orbits T suchthat T contains some elements v, u ∈ T adjacent in Q2k−1 .

Proof. Suppose that n = 2k−1 and T contains some elements v and u adjacent in Q2k−1 . If v and u differin the ith coordinate then ϕ∗(v) and ϕ∗(u) differ in the (i + 1)th coordinate. If u = ϕt(v) for some t thenv = ϕ2k−t(u). Therefore, for some t with 1 ≤ t ≤ n, there exists w = (w1, w2, . . . , wn) ∈ T such that

(wn−t+1 ⊕ 1, wn−t+2 ⊕ 1, . . . , wn ⊕ 1, w1, w2, . . . , wn−t) = (w1 ⊕ 1, w2, w3, . . . , wn). (15)

If t is even then the weights of the collections in (15) have different parities. Consequently, t is odd. Thenthe remaining coordinates are uniquely reconstructed from w1. The resulting two collections belong tothe same orbit. Consequently, for every t with 1 ≤ t ≤ 2k−1 − 1, there exists a unique orbit T containinga pair of collections of the form (15). It is not difficult to see that for different t we obtain different orbits.The proof of Lemma 6 is complete.

Given a binary collection v ∈ Q2k−1 and σ ∈ {0, 1}, let (v, σ) denote the binary collection in Q2k−1+1

obtained by appending the (2k−1 + 1)th coordinate σ in v.

Lemma 7. For every k ≥ 2, the set of vertices of Q2k−1+1 splits under the action ϕ into orbits oflength 2k.

Proof. It is easy that

ϕ(v, σ) = (ϕ∗(v), σ ⊕ 1).

Consequently, every orbit v1, v2, . . . , vt in Q2k−1 for the action of ϕ∗ yields two orbits

(v1, 0), (v2, 1), . . . , (vt−1, 0), (vt, 1),

(v1, 1), (v2, 0), . . . , (vt−1, 1), (vt, 0)(16)

of ϕ in Q2k−1+1. It remains to observe that Lemma 5 yields t = 2k. The proof of Lemma 7 is complete.

For every k ≥ 2, define the graph of orbits Gk as follows: the vertices are the orbits of ϕ in Q2k−1+1;two vertices T1 and T2 are adjacent in Gk whenever the orbits of T1 and T2 have representatives adjacentin Q2k−1+1.

Refer to the edges of Gk connecting a pair of orbits of the form (16) as α-edges. However, if theseorbits are formed from an orbit v1, v2, . . . , vt satisfying the hypotheses of Lemma 6 then refer to thecorresponding α-edge as a superedge. The next is obvious:

Lemma 8. Given T1T2 ∈ EGk, there exist v1 ∈ T1 and v2 ∈ T2 such that v1 and v2 differ only inthe last coordinate if and only if T1T2 is an α-edge. If T1T2 is an α-edge in Gk then there existv1 ∈ T1 and v2 ∈ T2 such that v1 and v2 differ only in the ith coordinate, meanwhile i < 2k−1 + 1if and only if T1T2 is a superedge.

Theorem 2. For every k ≥ 3, in Q2k−1+1, there is a Hamiltonian cycle with the group oforientation-preserving automorphisms of order 2k if and only if in Gk there is a Hamiltoniancycle either passing through an odd number of α-edges or containing at least one superedge.



Proof. Suppose that n = 2k−1 + 1.

Necessity. Take a Hamiltonian cycle H = v0 = 0, v1, . . . , v2n−1, v2n = v0 in Qn with the group oforientation-preserving automorphisms of order 2k. By Corollary 4, without restricting the generality,put Aut+(H) = 〈ϕ〉. Then the vertices v0, v1, . . . , v2n−k−1 occur in distinct orbits of ϕ on Qn. For every i

with 0 ≤ i < 2n−k, let Ti denote the vertex of Gk with vi ∈ Ti. Then T0, T1, . . . , T2n−k−1, T2n−k = T0 isa Hamiltonian cycle in Gk. Since v2n−k = ϕ(v0), the collections v2n−k and v0 differ in the last position.Consequently, the chain v0, v1, . . . , v2n−k−1, v2n−k of edges connecting the vertices differing in the lastposition consists of an odd number of edges. Then the claim of the theorem follows from Lemma 8.

Sufficiency. Take a Hamiltonian cycle T0, T1, . . . , T2n−k−1, T2n−k = T0 in Gk consisting of an evennumber of α-edges. Then by the hypotheses of the theorem, at least one of the α-edges is a superedge.Without restricting the generality, assume that T0T1 is a superedge. Construct a chain in Qn as follows:Take v0 ∈ T0 arbitrarily. By Lemma 8, there is u ∈ T1 such that v0 and u differ only in the ith coordinate,while i < n. Put v1 = u.

Suppose that, for some j with 1 ≤ j ≤ 2n−k − 1, the chain v0, v1, . . . , vj is constructed. If TjTj+1 isan α-edge then put

vj+1 = vj ⊕ (0, 0, . . . , 0, 1) ∈ Tj+1.

Otherwise, take as vj+1 an arbitrary vertex in the orbit Tj+1 adjacent to vj in Qn.

The chain v0, v1, . . . , v2n−k meets all orbits of ϕ in Qn, while only v0 and v2n−k lie in the orbit T0.Suppose that v2n−k = ϕt(v0) for some t. Since, by construction, v0 and v2n−k differ in the last position,it follows that t is odd, and consequently, coprime with the orbit length 2k. Then

v0, v1, . . . , v2n−k−1, ϕt(v0), ϕt(v1), . . . , ϕt(v2n−k−1),

ϕ2t(v0), ϕ2t(v1), . . . , ϕ2t(v2n−k−1), . . . ,

ϕ(2k−1)t(v0), ϕ(2k−1)t(v1), . . . , ϕ(2k−1)t(v2n−k−1)

is a Hamiltonian cycle in Qn and its group of orientation-preserving automorphisms is of order 2k.The case of a Hamiltonian cycle T0, T1, . . . , T2n−k−1, T2n−k = T0 containing an odd number of α-

edges is treated similarly, but no superedge is used. The proof of Theorem 2 is complete.

For every k ≥ 2, define the graph of orbits G∗k as follows: its vertices are the orbits of ϕ∗ in Q2k−1 ;

two vertices T1 and T2 are adjacent in G∗k whenever the orbits T1 and T2 have representatives adjacent

in Q2k−1 .

By Lemma 6, exactly 2k−2 vertices of G∗k produce loops.

Theorem 3. If G∗k includes a Hamiltonian chain with the endpoint at a vertex with a loop then

in Q2k−1+1 exists a Hamiltonian cycle with the group of orientation-preserving automorphismsof order 2k .

Proof. Put n = 2k−1. Take a Hamiltonian chain T0, T1, . . . , T2n−k−1 in G∗k, where T0 has a loop.

Inducting on the length, construct two disjoint chains T 00 , T 0

1 , . . . , T 02n−k−1

and T 10 , T 1

1 , . . . , T 12n−k−1

in

Gk. The orbit T0 of ϕ∗ in G∗k yields two orbits of ϕ of the form (16) in Gk. Denote them by T 0

0 and T 10 .

Observe that by definition T 00 T 1

0 is a superedge in Gk. For every i with 1 ≤ i ≤ 2n−k − 1, the orbit Ti ofϕ∗ in G∗

k yields two orbits of ϕ of the form (16) in Gk. One of them is adjacent to T 0i−1, and the other to

T 1i−1. Denote the first by T 0

i , and the second by T 1i .

Adding the superedge T 00 T 1

0 and the α-edge T 02n−k−1

T 12n−k−1

to the resulting chains, we obtaina Hamiltonian cycle in Gk containing the superedge. By Theorem 2, in Q2k−1+1 there exists a Hamilto-nian cycle with the group of orientation-preserving automorphisms of order 2k. The proof of Theorem 3is complete.


416 PEREZHOGIN

Corollary 4. If the graph G∗k is Hamiltonian then in Q2k−1+1 there exists a Hamiltonian cycle

with the group of orientation-preserving automorphisms of order 2k .

It is easy to verify by hand that G∗2 contains one vertex with a loop, G∗

3 contains exactly two adjacentvertices, both with loops, and G∗

4 is Hamiltonian. Using a computer, we found a Hamiltonian cycle inG∗

5. Then Theorems 2, 3 and Corollary 3 imply

Theorem 4. τ+(n) = 2�log2(n−1)� + 1 for all n with 3 ≤ n ≤ 32.

ACKNOWLEDGMENTS

The author was supported by the Russian Foundation for Basic Research (project no. 08–01–00671).

REFERENCES1. A. A. Evdokimov, “On Enumeration of Subsets of a Finite Set,” in: Methods of Discrete Analysis in

Solving Combinatorial Problems, Vol. 34 (Inst. Mat., Novosibirsk, 1980), pp. 8–26.2. A. L. Perezhogin, “On Automorphisms of Cycles in the n-Dimensional Boolean Cube,” Diskret. Anal. Issled.

Oper. Ser. 1, 14 (3), 67–79 (2007).3. M. Hall, Group Theory (Inostrannaya Literatura, Moscow, 1962) [in Russian].4. I. J. Dejter and A. A. Delgado, “Classes of Hamilton Cycles in the 5-Cube,” J. Comb. Math. Comb. Comput.

61, 81–95 (2007).5. E. N. Gilbert, “Gray Codes and Paths on the n-Cube,” Bell System Tech. J. 37 (3), 815–826 (1958).6. L. Goddyn and P. Gvozdjak, “Binary Gray Codes with Long Bit Runs,” Electron. J. Comb. 10, Research

paper R27 (2003) [J. Comb. 10 (3), R27 (2003)].7. G. Kreweras, “Some Remarks about Hamiltonian Circuits and Cycles on Hypercubes,” Bull. Inst. Comb.

Appl. 12, 19–22 (1994).8. M. Ramras, “A New Method of Generating Hamiltonian Cycles on the n-Cube,” Discrete Math. 85 (3),

329–331 (1990).9. C. D. Savage, “A Survey of Combinatorial Gray Codes,” SIAM Rev. 39 (4), 605–629 (1997).

10. C. D. Savage and I. Shields, “A Hamilton Path Heuristic with Applications to the Middle Two LevelsProblem,” Congr. Numerantium. 140,161–178 (1999).

11. I. Shields, B. J. Shields, and C. D. Savage, “An Update on the Middle Levels Problem,” Discrete Math. 309,5271–5277 (2009).


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On orientation-preserving automorphisms of Hamiltonian cycles in the N-dimensional Boolean cube