Combinatorica 9(1)(1989)91 99 A New Lower Boun for Snake-in-the-Box Coes Jerzy Wojciechowski Department of Pure Mathematics an Mathematical Statistics, University of Cambrige, 16 Mill Lane, Cambrige, CB2 1SB, Englan Abstract. In this paper we give a new lower boun on the length of Snake-in-the-Box Coes, i.e., inuce cycles in the -imensional cube. The boun is a linear function of the number of vertices of the cube an improves the boun c 2 /, where c is a constant, prove by Danzer an Klee. AMS subject classification 1980: 05 C 35, 94 B 25 1
1. Introuction A snake-in-the-box coe is an inuce cycle in the -imensional cube I[], i.e., the graph with all -tuples of binary igits as vertices, an all pairs of vertices iffering in exactly one coorinate as eges. For each N, let S() enote the length of the longest inuce cycle in I[]. The problem of etermining S() was first met by Kautz [10] in constructing a type of error-checking coe for certain analog-to-igital conversion systems. He showe that S() λ 2. This boun was later improve by Ramanujacharyulu an Menon [11], who prove that S() (3/2), Brown (unpublishe, quote from Danzer an Klee [3]) an Singleton [12] got S() λ( 4 6), Abbott [1] obtaine λ( 5/2), Vasil ev [14] showe that an that S() 2 S() (1 ε()) 2 1 when is a power of 2, with ε() 0 as, an finally Danzer an Klee [3] prove that S() 2+1 when is a power of 2, an 2 S() 7 4 1 for all 5. In this note we give a new lower boun for S(), namely S() 9 64 2. 2
2. The Main Lemma Our aim in this section is to state an prove Lemma 3, which will provie a construction of long snakes, leaing to the proof of the lower boun state in the introuction. Let us first introuce some notions an examine their properties. If F is a subgraph of I[], then let us enote by F (0) the subgraph of I[ + 1] obtaine as the image of F in the embeing ψ 0 : I[] I[ + 1] such that Analogously, let F (1) be the image of F in ψ 0 ((v 1,..., v )) = (v 1,..., v, 0). ψ 1 : I[] I[ + 1] such that ψ 1 ((v 1,..., v )) = (v 1,..., v, 1). For each 2, let R : [2 + 1, 2 +1 ] [2 ] be the orer reversing bijection, i.e., such that R (i) = 2 +1 + 1 i. Now, for each 2, we shall efine a function H : [2 ] V (I[]) such that is a Hamiltonian cycle in I[]. Set H = (H (1),..., H (2 ), H (1)) H 2 = ((0, 0), (0, 1), (1, 1), (1, 0), (0, 0)), an H +1 (i) = { (H (i)) (0) if 1 i 2, (H R (i)) (1) if 2 + 1 i 2 +1. (0) (1) In other wors, H +1 is obtaine by taking H an H, removing the eges connecting their last vertices (0) (1) with their first vertices, joining the first vertex of H with the first vertex of H an analogously the last with the last. The following lemma can be prove simply by checking all the possible cases. 3
e 1 1 e 1 2 a 1 e 2 1 a 2 e 2 2 e 1 3 a 3 e 2 3 b 1 b 2 e 1 4 a 4 e 2 4 e 1 5 a 5 e 2 5 Fig. 1. The graph G Lemma 1. For each 2, if 1 i < j 2 +1 an (H +1 (i), H +1 (j)) E(I[ + 1]) \ E(H +1 ), then exactly one of the following conitions hols: (1) 1 i < j 2, (i, j) (1, 2 ) an (H (i), H (j)) E(I[]) \ E(H ), (2) i = 1 an j = 2, (3) 2 + 1 i < j 2 +1, (i, j) (2 + 1, 2 +1 ) an (H R (i), H R (j)) E(I[]) \ E(H ), (4) i = 2 + 1 an j = 2 +1, (5) 2 i 2 1 an i = R (j). Let G be the graph shown in Figure 1. Given a permutation σ S 5, let ϕ σ an ϕ + σ be permutations of eges of G such that ϕ σ (e j i ) = ej σ(i) an ϕ + σ (e j i ) = e3 j σ(i). Furthermore, let σ 1 = (3 5), σ 2 = (1 3)(2 4 5), σ 3 = (1 2)(4 5) be permutations of the set {1,..., 5}. The following lemma can be easily verifie. Lemma 2. (6) For each e, e E(G) an σ S 5, the eges e an e have the same number of vertices in common as the eges ϕ σ (e) an ϕ σ (e ), an the same as the eges ϕ + σ (e) an ϕ + σ (e ). If e an e have one vertex in 4
common, then it belongs to {a 1,..., a 5 }, if an only if the common vertex of ϕ σ (e) an ϕ σ (e ) belongs to {a 1,..., a 5 }, an if an only if the common vertex of ϕ + σ (e) an ϕ + σ (e ) belongs to {a 1,..., a 5 }. (7) For each e E(G) the eges ϕ σ1 (e) an ϕ + σ 2 (e) are vertex isjoint, an the eges ϕ σ1 (e) an ϕ + σ 3 (e) are vertex isjoint. Now we can state our key lemma. We claim in it the existence, for each 2, of a close walk of length 2 in G which, after combining it with the Hamiltonian cycle H, will provie a construction of long snakes. The walk will start from a vertex belonging to the set {a 1,..., a 5 }, will not use any ege twice in turn an will posses a certain property with respect to the Hamiltonian cycle H. Namely, if we regar this walk an the Hamiltonian cycle H as sequences of length 2, the walk as a sequence of eges an H as a sequence of vertices, then any two eges corresponing to two nonconsecutive vertices of H being neighbours in I[] will be vertex isjoint. Lemma 3. For every 2 there is a function Φ : [2 ] E(G) such that (8) if 1 i 2 1, j = i + 1, or i = 2, j = 1, then Φ (i) an Φ (j) have exactly one vertex v i in common such that v i {a 1,..., a 5 } for i even, an v i {b 1, b 2 } for i o, an (9) if (H (i), H (j)) E(I[]) \ E(H ), then Φ (i) an Φ (j) are vertex isjoint. Proof. We shall prove the lemma by inuction on. In orer to make the inuction work, we shall efine functions for each 2 an such that each of the following conitions hols: (10) (1) = e1 k an Φk,l (2 ) = e 2 l, (11) if 1 i 2 1, then : [2 ] E(G) (k, l) I = {(1, 1), (1, 3), (1, 4), (2, 3), (2, 4)} (i) an Φk,l (i + 1) have exactly one vertex v i in common such that v i {a 1,..., a 5 } for i even an v i {b 1, b 2 } for i o, (12) if 2 i 2 1 an (k, l) (1, 1) (k, l ), then (i) =,l Φk (i), (13) if (H (i), H (j)) E(I[]) \ E(H ), then (i) an Φk,l(j) are vertex isjoint. In other wors, the function will escribe a walk in G starting from the vertex a k an the ege e 1 k, ening in the ege e 2 l an the vertex a l, an having all the properties we require for the walk escribe by the function Φ, i.e., it will not use any ege twice in turn an its eges corresponing to two nonconsecutive 5
vertices of H being neighbours in I[] will be vertex isjoint. Also, given 2, all the walks escribe by, for (k, l) I \ {(1, 1)}, will iffer only at the first an the last vertices. The construction of such functions will complete the proof of Lemma 3 because if we set Φ = Φ 1,1, then (9) will follow from (13), an (8) will follow from (10) an (11). Set If (k, l) (1, 1), then let an set ( 2 +1 (i) = Φ 1,1 (1), Φk,l 2 (2), Φk,l 2 (3), Φk,l 2 (4)) = (e1 k, e 1 5, e 2 5, e 2 l ). { ϕσ1 Φ k,3 (i) if 1 i 2, ϕ + σ 2 Φ σ 1 2 (l),4 R (i) if 2 + 1 i 2 +1, { +1 (i) = ϕσ1 Φ 1,3 (i) if 1 i 2, ϕ + σ 3 Φ 2,4 R (i) if 2 + 1 i 2 +1, where R is the orer reversing bijection efine in Section 1. In the inuction construction performe above, the walk w corresponing to +1 obtaine from the walks w 1 an w 2 escribe by Φ k,3 an Φ σ 1 2 (l),4 ((k, l) (1, 1)) is. To obtain w, we permute the eges of w 1 with ϕ σ1, an the eges of w 2 with ϕ + σ 2, getting w 1 an w 2, then we reverse the orer of eges of w 2 getting w 2, an finally we ientify the last vertex of w 1 with the first vertex of w 2. It can be checke irectly that for = 2 the conitions (10) (13) are satisfie. Given 2, let us assume that the conitions (10) (13) are satisfie for. We shall prove that they are satisfie for + 1. Proof of Conition (10). If (k, l) I \ {(1, 1)}, then +1 (1) = ϕ σ 1 Φ k,3 (1) = ϕ σ 1 (e 1 k) = e 1 k, an +1 (2+1 ) = ϕ + σ 2 Φ σ 1 2 (l),4 (1) = ϕ + σ 2 (e 1 σ 1 2 (l)) = e2 l. For (k, l) = (1, 1) the proof is analogous. Proof of Conition (11). We have to show that if 1 i 2 +1 1, then +1 (i) an Φk,l +1 (i + 1) have exactly one vertex in common, which belongs to {a 1,..., a 5 } for i even an to {b 1, b 2 } for i o. If i 2, it follows from conition (11) of the inuction hypothesis an conition (6) of Lemma 2. If i = 2 an (k, l) (1, 1), then the eges +1 (i) = ϕ σ 1 Φ k,3 (2 ) = ϕ σ1 (e 2 3) = e 2 5, an 2 (l),4 +1 (i + 1) = ϕ+ σ 2 Φ σ 1 (2 ) = ϕ + σ 2 (e 2 4) = e 1 5, 6
have the vertex a 5 in common, so (11) hols. If (k, l) = (1, 1), then the proof is analogous. Proof of Conition (12). We have to show that if 2 i 2 +1 1 an (k, l) (1, 1) (k, l ) then +1 (i) = Φk,l +1 (i). If 2 i 2 + 1, then this follows from the conition (12) of the inuction hypothesis, otherwise an +1 (2 ) = e 2 5 = Φ k,l +1 (2 ), +1 (2 + 1) = e 1 5 = Φ k,l +1 (2 + 1). Proof of Conition (13). Let us fix i < j such that (H +1 (i), H +1 (j)) E(I[ + 1]) \ E(H +1 ). We have to show that for each (k, l) I, +1 (i) an Φk,l +1 (j) are vertex isjoint. In the proof we shall assume that (k, l) (1, 1). For (k, l) = (1, 1) the proof is analogous. By Lemma 1, one of the conitions (1) (5) hols. If (1) hols, then by conition (13) of the inuction hypothesis, Φ k,3 By conition (6) of Lemma 2, +1 (i) an Φk,l +1 (j) are vertex isjoint. If (3) hols, then by conition (13) of the inuction hypothesis, Φ σ 1 (i) an Φk,3(j) are vertex isjoint. 2 (l),4 (R (i)) an Φ σ 1 are vertex isjoint. By conition (6) of Lemma 2, +1 (i) an Φk,l +1 (j) are vertex isjoint. 2 (l),4 (R (j)) If (2) hols, then +1 (i) = e1 k, Φk,l +1 (j) = e2 5, an if (4) hols, then +1 (i) = e1 5, +1 (j) = e2 l are vertex isjoint. If (5) hols, then by conition (12) of the inuction hypothesis, Φ k,3 (i) = Φσ 1 (i) = e, 2 (l),4 for some e E(G). Hence an +1 (i) = ϕ σ 1 (e), +1 (j) = ϕ+ σ 2 (e). By conition (7) of Lemma 2, the eges +1 (i) an Φk,l +1 (j) are vertex isjoint., completing the proof of Lemma 3. 7
3. The Lower Boun In this section we shall prove the main result of this note. Theorem 4. For each 0 2, the length S( 0 ) of the longest inuce cycle in I[ 0 ] satisfies the following lower boun S( 0 ) 9 64 20. Proof. It is known [3,5,6] that S(2) = 4, S(3) = 6, S(4) = 8, S(5) = 14, S(6) = 26, so we can assume that 0 7. Let us fix 0 = + 5, 2. Assume that we have a sequence of inuce paths (v 1 1, v 2 1,..., v r1 1 ), (v1 2, v 2 2,..., v r2 2 ),..., (v1 2,..., vr 2 2 ) in I[5] such that (14) if 1 i 2 1 an j = i + 1 (or i = 2 an j = 1), then the paths (v 1 i,..., vri i ) an (v1 j,..., vrj j ) have only the vertex v ri i = v 1 j in common, (15) if (H (i), H (j)) E(I[]) \ E(H ) (H an H are efine in section 1), then (v 1 i,..., vri i ) an (vj 1,..., vrj j ) are vertex isjoint, (16) if 1 i 2, then r i = 4 for i 1 or 2 (mo 4), an r i = 5 for i 3 or 0 (mo 4). Let us regar I[ + 5] as I[] I[5]. To construct an inuce cycle C +5 in I[ + 5] we consier I[ + 5] as the -imensional cube I[] with each vertex being a copy of I[5]. Let us take the jth path (v 1 j,..., vrj j ) in the copy of I[5] corresponing to the vertex H (j) in I[]; see Figure 2 for the case = 3. Then, let us join v ri i from the ith copy of I[5] with v 1 j from the jth copy of I[5] for all i {1,..., 2 1}, j = i + 1 an i = 2, j = 1. Hence we have C +5 = ((H (1), v 1 1), (H (1), v 2 1),..., (H (1), v r1 1 ), (H (2), v 1 2),..., (H (2), v r2 2 ),..., (H (2 ), v 1 2 ),..., (H (2 ), v r 2 2 ), (H (1), v 1 1)). It is clear that C +5 is a cycle in I[ + 5]. We claim that it is an inuce cycle. Assume that ((H (i), v k i ), (H (j), v l j)) E(I[ + 5]). Then H (i) = H (j) or v k i = vl j. If H (i) = H (j), then i = j an v k i, vl j are neighbours in I[5], so ((H (i), v k i ), (H (j), v l j)) E(C +5 ), 8
v1, 1..., v r1 1 v6, 1..., v r6 6 v8, 1..., v r8 8 v7, 1..., v r7 7 v2, 1..., v r2 2 v5, 1..., v r5 5 v3, 1..., v r3 3 v4, 1..., v r4 4 Fig. 2. The cycle C 8 since the path (v 1 i,..., vri i ) is an inuce path. If vk i = vl j, then H (i) an H (j) are neighbours in I[], an by (15), By (14), k = r i an l = 1, so an the claim is prove. By (16), the length of C +5 is equal to (H (i), H (j)) E(H ). ((H (i), v k i ), (H (j), v l j)) E(C +5 ), 9 2 1 = 9 64 2+5, so to complete the proof of the theorem it is enough to construct a sequence of inuce paths satisfying conitions (14) (16). Let G be the subivision of G obtaine by subiviing e 1 k with two new vertices c1 k an c2 k in such a way that we get the path (b 1, c 1 k, c2 k, a k), an subiviing e 2 k with three new vertices c4 k, c5 k an c6 k, giving rise to the path (a k, c 4 k, c5 k, c6 k, b 2), for each k 5. Let c 3 k = a k an ξ : V [G ] V (I[5]) be efine as follows. Set ξ(b 1 ) = (0, 0, 0, 0, 0), 9
ξ(c 1 1) = (1, 0, 0, 0, 0), ξ(c 2 1) = (1, 1, 0, 0, 0), ξ(c 3 1) = (1, 1, 0, 1, 0), ξ(c 4 1) = (0, 1, 0, 1, 0), ξ(c 5 1) = (0, 1, 1, 1, 0), ξ(c 6 1) = (0, 1, 1, 1, 1), ξ(b 2 ) = (1, 1, 1, 1, 1), an if ξ(c i 1) = (α 1,..., α 5 ), then let ξ(c i k) = (α k,..., α 5, α 1,..., α k 1 ). It is clear that the function ξ efines an embeing of G into I[5] such that the image of the subivision of any ege of G is an inuce path in I[5]. By Lemma 3, we have the function Φ : [2 ] E(G) satisfying conitions (8) an (9). Let us efine the path (vi 1,..., vri i ) to be the image uner ξ of the subivision of the ege Φ (i). Now (16) can be checke irectly an, since ξ is an embeing, (8) implies (14), an (9) implies (15), completing the proof of Theorem 4. 4. Remarks There is still a gap between the lower boun on the length of snake-in-the-box coes prove in this paper an the best known upper boun. The best upper boun is ue to Deimer [5], who showe that S() 2 1 2 1 ( 5) + 7 for 7. We believe that the upper boun can be further improve to the form of c2, where c is a constant smaller than 1/2. Acknowlegements. This research is intene to be a part of a Ph.D. thesis written at the University of Cambrige uner the irection of Dr Béla Bollobás. I am grateful to my supervisor for pointing out the problem an his comprehensive help uring the writing of this note. Note ae in proof. I was informe by the referees that the lower boun of the form λ2 was first obtaine by Evokimov [6]. The constant λ he got is very small (λ = 2 11 ), but he state that after certain changes in his construction λ can be increase to 2 9 S(8). 10
The iea of his proof is similar to this use in the present proof in the sense that the snake in I[] = I[ 0 ] I[ 0 ] (where 0 is a constant) is constructe in such a way that its projection on I[ 0 ] is a Hamiltonian cycle. The rest of the cycle is constructe in quite ifferent way. In [9] Glagolev an Evokimov prove a theorem about the chromatic number of a certain infinite graph an state that it can be use to further increase the constant λ so that λ ( 3 16, 1 4 ). One of the referees informe me also that in his issertation, Evokimov [7] prove that S() 0.26 2. Recently Abbot an Katchalski [2] foun completely ifferent way of proving a lower boun of the form λ2. They use inuction in a way resembling the proof of Danzer an Klee [3], but construct the so calle accessible snake, which is a snake with some aitional paths between its vertices, allowing to keep the inuction going without ecreasing the ratio of vertices use by the snake (what coul not be avoie in the proof of Danzer an Klee). The upper boun has been lately improve by Solov jeva [13], who got S() 2 1 (1 2 2 ) for 7. + 2 5. References [1] H. L. Abbott, A Note on the Snake-in-the-Box Problem, unpublishe manuscript, Some Problems in Combinatorial Analysis, Ph.D. thesis, University of Alberta, Emonton, Canaa, (1965). [2] H. L. Abbott an M. Katchalski, On the Snake in the Box Problem, to appear in J. Combinatorial Theory. [3] L. Danzer an V. Klee, Length of Snakes in Boxes, J. Combinatorial Theory 2 (1967), 258 265. [4] D. W. Davies, Longest Separate Paths an Loops in an N Cube, IEEE Trans. Electronic Computers 14 (1965), 261. [5] K. Deimer, A New Upper Boun for the Length of Snakes, Combinatorica 5 (2) (1985), 109 120. [6] A. A. Evokimov, The maximal length of a chain in the unit n-imensional cube. Mat. Zametki 6 (1969), 309 319. English translation in Math. Notes 6 (1969), 642 648. [7] A. A. Evokimov, The maximal length of a chain in n-imensional cube an some close problems. Dissertation Can. Sc. (1971), 58 p. [8] S. Even, Snake-in-the-Box Coes, IEEE Trans. Electronic Computers 12 (1963), 18. [9] V. V. Glagolev an A. A. Evokimov, The minimal coloring of a certain infinite graph, Diskret. Analiz 17 (1970), 9 17. 11
[10] W. H. Kautz, Unit-Distance Error-Checking Coes, IRE trans. Electronic Computers 3 (1958), 179 180. [11] C. Ramanujacharyulu an V. V. Menon, A Note on the Snake-in-the-Box Problem, Publ. Inst. Statist. Univ. Paris 13 (1964), 131 135. [12] R. C. Singleton, Generalize Snake-in-the-Box Coes, IEEE Trans. Electronic Computers 15 (1966), 596 602. [13] F. I. Solov jeva, Upper boun for the length of a cycle in an n-imensional unit cube, Methos of Diskrete Analiz 45 (1987). [14] Ju. I. Vasil ev, On the Length of a Cycle in an n-dimensional Unit Cube, Soviet Math. Dokl. 4 (1963), 160 163 (transl. from Dokl. Aca. Nauk SSSR 184 (1963) 753 756). 12