AbstractBy a snake in a finite graph G is meant a cycle without chords. Denoted by S(G) the length of a longest snake in G. In this paper we obtain a new lower bound for S(G) in the case where G is the product of d copies of the complete graph on n vertices

Abbott, H.L.

Katchalski, M.

Elsevier - Publisher Connector 

Discrete Mathematics 91 (1991) 111-120 
North-Holland 
111 
Further results on snakes in 
powers of complete graphs 
H.L. Abbott 
Department of Mathematics, University of Alberta, Edmonton, Alta., Canada T6G 2Gl 
M. Katchalski 
Department of Mathematics, The Technion, Haifa, Israel 
Received 26 July 1988 
Revised 30 January 1990 
Abstract 
Abbott, H.L. and M. Katchalski, Further results on snakes in powers of complete graphs, 
Discrete Mathematics 91 (1991) 111-120. 
By a snake in a finite graph G is meant a cycle without chords. Denoted by S(G) the length of 
a longest snake in G. In this paper we obtain a new lower bound for S(G) in the case where G 
is the product of d copies of the complete graph on n vertices. 
1. Introduction 
By a snake in a finite graph G is meant a simple cycle without chords. We 
denote by S(G) the length of a longest snake in G. Let K,, denote the complete 
graph with n vertices and let K,” denote the product of d copies of K,,. Then K$ is 
the graph of the d-dimensional unit cube and a large literature has evolved 
concerning the problem of estimating s(d) = S(K$. See [6-81 and references 
given in these papers. 
In [l] it was shown that 
S(K:) = 2n (I) 
and 
2nd-‘-’ 
S(K:) z {2&l 
if 2’ < d < 2’+’ 
if d = 2’. 
(2) 
(3) 
0012-365X/91/$03.50 0 1991- Elsevier Science Publishers B.V. (North-Holland) 
112 H. L. Abbott, M. Katchalski 
In the case IZ = 2, (3) was proved by Danzer and Klee [4]. A better lower bound 
in the case IZ = 2 was obtained by Evdokimov [6] who showed that for some 
constant A > 0, 
S(K$) > A2d. (4) 
A proof of (4) which is shorter and simpler than that of Evdokimov may be found 
in [3]. Another proof of (4) was recently given by Wojciechowski [lo]. 
In [2] it was shown that 
S(Ki) = $n’+ O(n) as 124 co, (5) 
S(K$ 2 2(QYP + 0(ndP2) as n --, 00, (6) 
and 
S(K$ =2 (1+ -& 
) 
rid-l.. 
In the case a= 2, results much stronger than (7) have been established. See 
Diemer [5] and Solov’jeva [9] for the best results presently known. 
Apart from the cases d = 2, 3 it is not known whether lim,_,S(K$/nd-’ 
exists. Set 
d,=liminfq. 
n-m n 
It is implied by (1) and (5) that A2 = 2 and A3 = 2. From (6) and (7) it follows that 
1 2($)d-’ < Ad C 1 + - 
d-l’ 
In [2] it was asked whether the sequence {Ad} is bounded away from 0. The 
object of this paper is to show that this is the case. 
Theorem 1. There exists a constant c > 0 such that Ad > c for all d. 
Actually, we prove a somewhat stronger result. 
Theorem 2. Let n = 0 (mod 4). Then for d 2 3 
S(KZ) 3 ( ;)d-lS(K:I). (8) 
It is clear that Theorem 1 is a consequence of (4) and (8). Thus we need only 
prove Theorem 2. The proof of Theorem 2 is fairly complicated. In Section 2 we 
give some notation and terminology. In Section 3 we describe some preliminary 
constructions and give some of their properties. In Section 4 we construct the 
snake in Kt which establishes (8). In Section 5 we discuss some additional 
problems which we have not been able to settle. 
Snakes in complete graphs 113 
2. Notation 
The vertices of Kt are the nd d-tuples a = (a,, u2, . . . , ad) where, for each i, a, 
is one of 0, 1,2, . . . , n - 1. Two vertices of Kt are joined by an edge if they 
differ in exactly one coordinate. Et is the subgraph of Ki induced by the set of 
vertices all of whose coordinates are even. To simplify the notation we frequently 
do not distinguish between a graph and its set of vertices; thus, for example, 
a E Ez means that a is a vertex of E,d. For a = (a,, u2, . . . , ad) E KE, d 3 2, P(a) 
denotes the vertex (al, u2, . . . , ad_l) of K,d-', and for Q = (a,, u2, . . . , ad-_l) E 
Kd-1 
P(u) denotes 
(a~,6,,...,u,) is a 
the vertex (ai, u2, . . . ) ad-l, 0) of Kf. If A = 
sequence of vertices of Kf then P(A)= 
(P(%), P(a,), * . . ) mJ) and ZP(A) is defined similarly. More generally, if 
u=(u,,u,,..., ad) and 1 C t S d - 1, P,(u) denotes the vertex (al, u2, . . . , a,) 
of Kk, so that P,_l = P. If A =(~~,a~,..., uk) is a sequence of vertices of K,d 
then A = (a,_ u~_~, . . . , a,) and, if 1 c u <t c k, A,,, = (a,, u,+~, . . . , a,). 
X'= {n&x;,..., xl} denotes a snake of maximal length in Kf-', so that 
s =S(K$-')- 1. Let X =ZP(X')= {xo,xl,...,xs}, so that X is a snake of 
length s in K$ Notation not explained above, or subsequently, follows that in [l] 
and [2]. We assume throughout that IZ = 0 (mod 4). 
3. Some preliminary constructions 
The sequences Bk, k = 1,2, . . . . We construct a sequence Bk of vertices of E,k as 
follows: 
Bi = ((0) (2) (4), . . . , (n - 2)) 
and for k 3 2, Bk is the sequence obtained by juxtaposing 
IP(Bk--l) 
ZP(Bk-J+ (0, 0,. . ., 0, 2) 
ZP(Bk-J+ (0, 0,. . ., 0, 4) 
ZP(Bk_J + (0, 0, . . . , 0, 6) 
ZP(B,_,) + (0, 0, . . . , 0, n - 2). 
Thus, for example, if at = 8, B3 is the following sequence: 
((ooO)(2~)(~0)(6~)(620)(420)(220)(020)(040)(240)(440)(640)(660)(460) 
(260)(060)(062)(262)(462)(662)(642)(442)(242)(042)(022)(222)(422)(622) 
(602)(402)(202)(002)(004)(204)(404)(604)(624)(424)(224)(024)(044)(244) 
(444)(644)(664)(464)(264)(064)(066)(266)(466)(666)(646)(446)(246)(046) 
(026)(226)(426)(626)(606)(406)(206)(006)). 
114 H. L. Abbott, M. Katchalski 
The sequences V,, k =2,3,. . . . If Bk_-l = (b,, 6,, . . . , b,), we define the 
sequence V, = (uO, ul, . . . , ul) of points in Ef: by the requirements: 
(a) P(G) = Bk--l; 
(b) For each i, the sum of coordinates of vi = 0 (mod n). 
When n = 8, V_ is the following sequence-we omit some brackets; for example 
(0000) is written 0000: 
(0000,2006,4004,6002, 6200,4202,2204,0206,0404,2402,4400,6406, 
6604,4606,2600,0602,0620,2626,4624,6622,6424,4426,2420,0422, 
0224,2222,4220,6226,6020,4022,2024,0026,0044,2042,4040,6046, 
6244,4246, 2240,0242,0440,2446, 4444,6442,6640,4642,2644,0646, 
0664,2662,4660,6666,6460,4462,2464,0466,0260,2266,4264,6262, 
6064,4066,2060,0062). 
We now give a number of properties of Bk and V,. Many of these are 
immediate consequences of the definitions. Those that are not may be proved by 
straightforward induction arguments. We omit most of the details. 
Property 1. IV,1 = (Bk-l( = (n/2)k-’ for k 3 2. 
Property 2. Zf Vi, uj E V, and i #j, then Ui and Uj differ in at least two coordinates 
and in exactly two coordinates when j = i + 1. 
Proof. It is clear that bi and bj differ in at least one coordinate if i #j and in 
exactly one coordinate if j = i + 1. If 6, and bj differ in at least two coordinates, so 
do vi and vi. If bi and bj differ in exactly one coordinate then vi and uj differ in this 
coordinate and the kth coordinate, but in no others. Note that here, and 
subsequently, u[+~ is understood to be u,. Cl 
Property 3. Zf Vi, Uj E V,, i Zj, and a E X + Vi, b E X + uj, then a and b differ in at 
least two coordinates. 
Proof. ui and uj differ in at least two coordinates and since the coordinates of 
points of X are zero or one (recall that X is a snake in K$ then a and b differ (at 
least) where Ui and Uj differ. 0 
Remark. Property 3 implies that in Kt we have (n/2)d-’ copies of the snake X 
such that points in different copies are separated by distance at least 2. Our 
eventual goal is to show that by deleting very few (consecutive) points from these 
snakes, the resulting open snakes may be linked up to give a long snake in K$ 
The construction will be such that (8) follows. 
Property 4. In the sequence V,, the (n/2y-’ points (aI, a2, . . . , ai_,, yj, 
_Yj+lP . . . 7 Yk-1, z), where Yj, Yj+l, . . . , Y&-l are specified, appear as a block 
Snakes in complete graphs 115 
of consecutive points, and the points (a,, a2, . . . , aj-1, z - y), where y G yj + yj+l + 
. . . + y,_, (mod n) are the points of y or I$ 
Proof. This follows easily from the definitions of Bk and V, and straightforward 
induction arguments. 0 
We define for each k 3 2 a function r, as follows: If V, = {u,-,, ul, . . . , y} 
then, for OCiSl, 
rk(i) = first coordinate in which uj differs from ui+i. 
The reader may wish to check that the sequence of values of r4 is 
1, 1, 1,2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 
1, 1, 1,2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 
1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 
1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3. 
Property 5. For 0 s i < 1, r,Ji) = t if and only if i + 1 is divisible by (n/2)‘-’ but 
not by (n/2)’ and rk(l) = k - 1. 
Proof. This follows easily from the definition of B,_, and a straightforward 
induction argument. 0 
Property 6. Zf rk(i) = t, 2 s t s k - 1, then either 
I-l 
r \ 
P,_l(Uj) = (0, 0, . . . , 0) 
or 
t-2 
\ 
P,_l(ui) =TO, 0, . . . , 0, n - 2). 
Proof. This follows easily via a simple induction argument and the fact that for 
each t, 1 s t s k - 1, the first term of B, is (O,O, . . . ,O) and the last is 
(0, 0, . . . ,O,n-2). •i 
Property 7. (a) The sequence obtained from the subsequence of V, whose first j 
coordinates are zero, (1 C j s k - l), by deleting these j coordinates, is Vk-i 
(b) The sequence obtained from the subsequence of V, whose first j - 1 
coordinates are zero and whose jth coordinate is n - 2 (1 s j s k - 1) by deleting 
these j coordinates and increasing the kth coordinate by 2 (mod&o n), is Vk_j. 
Proof. This follows easily by a straightforward induction argument. Cl 
116 H.L. Abbott. M. Katchafski 
Let vi, Vi E V,. We say that vi and Vj are l-different if, among their first d - 1 
coordinates, vi and Vj differ only in the Zth coordinate. In what follows we 
abbreviate r, to r. 
Property 8. Zf Vi, Uj E V,, r(i) = r(j) = 1, i and j have the same parity, and vi and 
Vj are l-different then 
vj+l - Vj = Vi+1 - Ui. 
Proof. In the case I = 1, it follows from Property 4 that Ui and Vj must occur in a 
block of length n/2 in which the first coordinates steadily increase or steadily 
decrease. Since r(j) = 1, Vj+i is also in this block, and the desired conclusion is 
seen to hold. In fact, the hypothesis that i and j have opposite parity is not 
needed in this case. 
In the case 2 c 1 cd - 1, one may use induction on d. It is easy to check 
directly that the conclusion holds if d = 3 (and I= 2). If d > 3, Property 4 enables 
one to reduce the problem to V,, and the induction step may be completed 
easily. Cl 
Property 9. Zf r(i) = r(j - 1) = 1, i and j have different parity, and vi and vj are 
l-different, then: 
(a) Zf 1 = 1, Ui+i - Vi = Vj - U,-1. 
(b) Zf 2 ~ I ~ d - 1, Vi+1 - q = ~j-1- Vi. 
Proof. Similar to that of Property 8. (a) holds without the assumption that i and j 
have opposite parity. 0 
Property 10. If r(i) = r(j) = t 2 2, vi and vj are l-different for some 12 t, i and j 
have the same parity and P,-,(S) = Pt-,(vj), then 
vi+l - V; = Vj+l - Vj. 
Proof. The proof follows closely that of Property 8, except that in the induction 
step one appeals to Property 7 instead of Property 4. 0 
Property 11. Zf r(i) = r(j - 1) = t 2 2, Vi and vj are l-different from some 13 t, i 
and j have opposite parity and P,_,(v,) = P~-,(Uj_l)~ then: 
(a) Vi+1 - Ui = Vj - Vj-1 if I = t. 
(b) t)i+l - U; = Vi-1 - Uj if t < I c d - 1. 
Proof. Use induction on 1. Property 7 is used to effect the induction step. 0 
Snakes in complete graphs 117 
4. The long snake in Kz 
The reader should recall the remark following Property 3. We now construct 
m = (n/2)d-’ open snakes AO, Al, . . . , A,_, in K,d as follows: 
Let 
1 
Xr(i_l),s + vi 
Ai= ~ 
if i is even, 
Xr(i_l),s + Vi if i is odd. 
By Property 3, points on different open snakes are separated by distance at 
least two. Let 
y; = X,(i_l) + 4, y+ = X, + Ui if i is even, 
yzT =r, + Vi, y’ =x,(~_~) + vi if i is odd, 
i.e., y; is the first point and y+ is the last point of Ai. Define y+’ and yf- by the 
following conditions: 
(a) P(yT+) = P(yT), P(yf-) = P(yf). 
(b) The dth coordinate of y:’ (respectively y;-) is the unique odd integer 
between the dth coordinates of y+ and yi,i (respectively y; and y,‘_i). 
Finally, let A denote the sequence of points in Kt obtained by juxtaposing 
Y,-,A~,Y,++,Y;-,Al,y:+,y;-,A*, . . -- . 9 ym--2, A-2, Y::,, Y,-1, Am-1, y;“i. 
It is clear that this defines a simple cycle in K$. It is a consequence of Property 
5 and the fact that X has length S(Kf-‘) that the length of A is 
H= ; 0 
d-l 
(S(Kf-‘) - 1) + 2 1 (2r-‘-((;)d2+(;)d-3+...+(5)). 
This may be seen as follows: There are (n/2)d-’ copies of X with the point x0 
deleted. This accounts for the term (n/2)d-‘(S(K$-‘) - 1). The points y+’ and 
Yf-, i=o, 1,2,. . .) m - 1, contribute the term 2(n/2)d-‘. The copies of X 
corresponding to values of i with r(i) = t > 1 must be further truncated by deleting 
a further t - 1 points. This, and Property 5, accounts for the term {(n/2)d-2 + 
(n/2)d-3 +. . . + (n/2)}. 
Since H 2 (n/2)d-‘S(K$-‘), to complete the proof of (8), we must show that A 
is a snake. In order to do this, keeping in mind Property 3, it will suffice to show 
that for each i, y+’ is adjacent only to yi+i, and yi is adjacent only to yl::. 
Observe first that if x is a point of Ai, then x is not adjacent to yf+ or y,:- if 
j #i and that x is adjacent to y+’ (or y;-) only if x = y+ (or y;). Thus there 
remains only the problem of showing that among the points y”, y;-, 
i = 0, 1, . . . , m - 1, y,?’ is adjacent only to yzr+; and yi is adjacent only to yi+_:. 
We need to consider various cases. 
Case 1: Suppose y+’ is adjacent to yf+, i <j. 
It is clear that i and j must have the same parity. Suppose i and j are even. 
Then i + 1 and j + 1 are odd, and thus by Property 5, r(i) = r(j) = 1. Since ui and 
118 H.L. Abbott, M. Katchalzki 
Vj differ in at least one of the first d - 1 coordinates, so do y: and yf and hence 
also y,?’ and yf+. If vi and Vj differ in- two or more of the first d - 1 places so do 
Y:’ and yf+. Thus we may suppose vi and V~ are l-different, for some 1. By 
Property 8, vi+i - vi = vj+i - Vje If I = 1, in order for this to hold we must have 
v;=(a ,...) x), vi+,=(af2 )...) xT2), 
Vj=(b,. . . ,X+U-_b), Vj+l = (b f 2, e . . ) x + u - b F 2), 
where, for 2 s k s d - 1, the kth coordinates of all four points are the same. The 
last coordinate of y:’ is thus x f 1 and the last coordinate of yf+ is 
x + a - b T 1. In order for y+’ and yt’ to be adjacent, their dth coordinates 
must coincide. This requires a = b and thus vi = Vjj a contradiction. In the case 
I> 1, we must have, again by Property 8, 
vi = (a, . . . , b, . . . , x), Vi+1 = (U f 2, s e * ) b, . . . ) x T 2), 
Vj= (U, . , . ) c,. . . )X + b -c), Vj+l=(U f2,. . . 7 ~3. a. )X + b -C FZ?). 
Here b and c denote the Ith coordinates of Vi and Vi, b # c. For 2 G k G d - 1, 
k # 1, the kth coordinates of all four points are the same. The last coordinate of 
Y++ is x F 1 and the last coordinate of y,T+ is x + b - c T 1. In order for y:’ and 
yT+ to be adjacent these must be equal, but this requires b = c, a contradiction. 
The case where i and j are both odd can be handled in a similar way. The only 
difference is that while r(i) = r(j), the common value is not necessarily one. 
Case 2: Suppose y;- is adjacent to y,:-. 
The argument of Case 1 applies with only minor modifications. 
Case 3: Suppose y+’ is adjacent to yI:-. 
It is clear that i and j must have opposite parity. We need to show that j = i + 1. 
Consider first the case i even and j odd. By Property 5, r(i) = r(j - 1) = 1. As in 
Case 1, we may suppose that Vi and Vj are I-different for some 1. If 1 = 1, then by 
Property 9(a), we must have 
vi = (a, . . . , x), vi+1 = (a f 2, . . . ) x T 2) 
vi-1 = (b T 2, . . . , x + a - b f 2), Vj = (b, . . . ) X + U - b), 
where a # b and where, for 2 s k G d - 1, the kth coordinates of all four points 
are the same. The last coordinate of y” . IS x T 1 and the last coordinate of y,:- 
is x +a- b f 1. In order for YE?’ and Yj to be adjacent, we must have 
b = a f 2. This implies that j = i + 1. 
If I > 1, then, by Property 9(b), we must have 
vi = (a, . . . , b, . . . , x), vi+1 = (a f 2, . . . , b, . . . , x T 2) 
vj_1 = (a f 2, . . . , c, . . . , x + b - c T 2), Vj= (Up.. s ) C,. . . ,X + b - C), 
where b and c are the fth coordinates of vi and Vi, b #c, and where, for 
2 G k 6 d - 1, k # 1, the kth coordinates of all four points are the same. The last 
coordinate of y” is x F 1 and the last coordinate of y,:- is x + b - c F 1. In 
Snakes in complete graphs 119 
order for y:’ and y,T- to be adjacent, these must be the same. This implies that 
b = c, a contradiction. 
We consider now the case where i is odd and j is even. We must have 
r(i) = r(j - 1) = t (although t need not be one) and we may suppose that Vi and Vj 
are l-different. If t = 1, the argument used in the case i even, j odd carries over 
with no change. We may therefore suppose that t 3 2. It is easy to see, by 
Property 6, that IS t. 
Consider first the case where 1 = t. We have, by Property 11(a) and Property 6, 
Vi = (0, 0, . . . ) 0, U, U, f . . ) X), 
uj+1= (0, 0, . . . ) 0, U, a f 2, . . . , x T 2), 
Uj-1= (0, 0, . . f ) 0, U, b ~ 2, . . . ) X + U - b f 2), 
Uj = (0, 0, . . * 7 0, U, b, . . . ) X + U - b), 
where u = 0 or n - 2, a and b are in the tth position of vi and Vj respectively, 
a #b and where, for t < k 6 d - 1, the kth coordinates of all four points are the 
same. The last coordinate of y” . IS x T 1 and the last coordinate of yj is 
x + a - b ‘F 1. In order for y:’ and y,:- to be adjacent, these must be the same 
and thus b = a f 2. This implies that j = i + 1. 
In the case where I > t, we have, by Property 11(b) and Property 6, 
vi = (0, 0, . . . , 0, u, a, . . . , b, . . . , x) 
ui+l = (0, 0, . . . , 0, u, a f 2, . . . , b, . . . , x T 2) 
Uj-1= (0, 0, . . . ) 0, U, U ~ 2, . . . ) C, 1 . . ) x + b - C ~ 2) 
uj = (0, 0, . . . , 0, u, a, . . . , c, . . . , x + b - c) 
where u = 0 or n - 2, a is the tth coordinate of vi, and Uj, b and c are the Ith 
coordinates of vi and Vj respectively, b Zc, and where, for t < k G d - 1, k # 1, 
the kth coordinates of all four points are the same. The last coordinate of y” is 
x F 1 and the last coordinate of y,F- is x + b - c T 1. In order for yt’ and y,:- to 
be adjacent, these must be the same. However, this implies that b = c, a 
contradiction. 
This completes the proof that A is a snake in Kz and thus the proof of Theorem 
2. 0 
5. Some additional remarks 
The condition n = 0 (mod 4) may be relaxed somewhat. The construction works 
in the case where n = 2 (mod 4), except that A is an open snake in K$ However, 
A may be ‘closed’ to give a snake in Kz+2 with no essential oss of length. It is not 
clear, however, what the situation is when n is odd. In particular, we cannot 
answer the following question: Is there a constant c > 0 such that S(K$) > ~3~ for 
120 H.L. Abbott, M. Katchalski 
all d 3 2? The reader should note that the remark following the statement of 
Property 3, so very important in the proof of Theorem 2, fails badly when n is 
odd. 
References 
[l] H.L. Abbott and P.F. Dierker, Snakes in powers of complete graphs, SIAM J. Appl. Math. 32 
(1977) 347-355. 
[2] H.L. Abbott and M. Katchalski, Snakes and pseudo snakes in powers of complete graphs, 
Discrete Math. 68 (1988) l-8. 
[3] H.L. Abbott and M. Katchalski, On the snake in the box problem, J. Combin. Theory 45 (1988) 
13-24. 
[4] L. Danzer and V. Klee, Lengths of snakes in boxes, J. Combin. Theory 2 (1966) 258-265. 
[5] K. Deimer, A new upper bound for lengths of snakes, Combinatorics 5 (2) (1985) 109-120. 
[6] A.A. Evdokimov, Maximal length of a circuit in a unitary n-dimensional cube, Mat. Zametki 6 
(1969) 309329. 
[7] V. Klee, What is the maximum length of a d-dimensional snake?, Amer. Math. Monthly 77 
(1970) 63-65. 
[8] V. Klee, A method for constructing circuit codes, J. Assoc. Comput. Mach. 14 (1967) 520-538. 
(91 F.I. Solov’jeva, Upper bounds for the length of a cycle in an n-dimensional unit cube, Methods 
Diskret. Analiz. 45 (1987). 
[lo] J. Wojciechowski, A new lower bound for snake in the box codes, Combinatorics 9 (1989) 
91-99. 


Further results on snakes in powers of complete graphs 

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Further results on snakes in powers of complete graphs

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