double coverings of 2-paths by hamilton cycles*

Double Coverings of 2-pathsby Hamilton Cycles

Midori Kobayashi,1* Nobuaki Mutoh,2 Kiyasu-Zen’iti,3 Gisaku Nakamura41School of Administration and Informatics, University of Shizuoka,Shizuoka 422-8526, Japan, E-mail: [email protected]

2School of Administration and Informatics, University of Shizuoka,Shizuoka 422-8526, Japan

3Semiconductor Research Institute, Sendaisi Aobaku Kawauti 980-0862, Japan

4Tokai University, Shibuyaku Tokyo 151-0063, Japan

Received May 21, 2001; accepted September 10, 2001

Abstract: A double Dudeney set in Kn is a multiset of Hamilton cycles in Kn having the

property that each 2-path in Kn lies in exactly two of the cycles. A double Dudeney set in Kn has

been constructed when n � 4 is even. In this paper, we construct a double Dudeney set in Kn

when n � 3 is odd. # 2002 Wiley Periodicals, Inc. J Combin Designs 10: 195–206, 2002; Published online in

Wiley InterScience (www.interscience.wiley.com). DOI 10.1002/jcd.10003

Keywords: round table problem; Dudeney set; 2-path

1. INTRODUCTION

In 1905, Dudeney [1, problem 273] proposed the Round Table Problem as follows:‘‘Seat the same n persons at a round table on ðn� 1Þðn� 2Þ=2 occasions so that no

person shall ever have the same two neighbours twice. This is, of course, equivalentto saying that every person must sit once, and only once, between every possiblepair.’’

The problem is equivalent to asking for a set of Hamilton cycles in the completegraph Kn with the property that every path of length two (2-path) lies on exactly oneof the cycles. (The length of a path is the number of edges in the path.) We call such a

*This research was supported in part by Grant-in Aid for Scientific Research (C) Japan.

Correspondence to: Midori Kobayashi, School of Administration and Informatics, University of Shizuoka, Shizuoka

422-8526, Japan. E-mail: [email protected]

# 2002 Wiley Periodicals, Inc.

195

set of cycles in Kn a Dudeney set in Kn. The cardinality of a Dudeney set in Kn isðn� 1Þðn� 2Þ=2. A Dudeney set in Kn has been constructed when n is even:

Theorem 1.1 [5]. There exists a Dudeney set in Kn when n � 4 is even.

In case when n is odd, a Dudeney set in Kn has been constructed only when

1. n ¼ 2e þ 1 (e is a natural number) [11],2. n ¼ pþ 2 ( p is an odd prime number and 2 or �2 is a primitive root of GFð pÞ)

[2,4],3. n ¼ pþ 2 ( p is an odd prime number, 2 is the square of a primitive root of

GFð pÞ and p � 3 (mod 4) [4],4. n ¼ pþ 2 ( p is an odd prime number, 2 is the square of a primitive root of

GFð pÞ, p � 1 ðmod 4Þ and 3 is not a quadratic residue modulo p) [8],5. n ¼ pþ 2 ( p is an odd prime number, �2 is the square of a primitive root of

GFð pÞ, and either(i) p � 1 ðmod 4Þ and 3 is not a quadratic residue modulo p, or(ii) p � 3 ðmod 4Þ [8],

and some sporadic cases [2].When n is odd, constructing a Dudeney set in Kn is a difficult problem, so we

consider a double Dudeney set in Kn which is a multiset of Hamilton cycles in Kn

having the property that each 2-path in Kn lies in exactly two of the cycles. BeforeTheorem 1.1 was proved, a double Dudeney set for even n had been constructed:

Theorem 1.2 [10]. There exists a double Dudeney set in Kn when n � 4 is even.

In the case when n is odd, a double Dudeney set in Kn (except the case n ¼ 2e þ 1)has been constructed when

1. n ¼ p1p2 � � � pt þ 2 ( p1; p2; . . . ; pt are different odd prime numbers and t is anatural number) [7],

2. n ¼ p2 þ 2 ( p is an odd prime number) [6].

In this paper, we will show:

Theorem 1.3. There exists a double Dudeney set in Kn when n � 3 is odd.

A proof of Theorem 1.3 is as follows. When n is odd with 3 � n � 9, there is aDudeney set in Kn [1,2], so there is a double Dudeney set in Kn. Let n � 11 be an oddinteger. Since there is an antipodal starter in Kn�1 (Prop. 3.1), there is an antipodalHamilton cycle double cover in Kn�1. From Prop. 2.1, there is a double Dudeney setin Kn. This will complete the proof of Theorem 1.3.

In general the double Dudeney set constructed here is not a Dudeney set of thecomplete directed graph.

Finally, we mention a covering of 2-paths by Hamilton paths in Kn. It is easy to prove:

Lemma 1.4. Let n � 3 be an integer.

(1) If there exists a Dudeney set in Knþ1, then there exists a set of Hamilton pathshaving the property that each 2-path in Kn lies in exactly one of the paths.

196 KOBAYASHI ET AL.

(2) If there exists a double Dudeney set in Knþ1, then there exists a multiset ofHamilton paths having the property that each 2-path in Kn lies in exactly twoof the paths.

The following two corollaries are immediate from Theorems 1.1 and 1.3 andLemma 1.4. Corollary 1.6 was proved by Verrall in 1997.

Corollary 1.5. There exists a set of Hamilton paths having the property that each2-path in Kn lies in exactly one of the paths when n � 3 is odd.

Corollary 1.6 [12]. There exists a multiset of Hamilton paths having the propertythat each 2-path in Kn lies in exactly two of the paths when n � 4 is even.

2. ANTIPODAL HAMILTON CYCLE DOUBLE COVERS

Let n � 4 be an even number. Let Kn ¼ ðVn;EnÞ be the complete graph onn vertices, where Vn is the vertex set and En is the edge set. Put m ¼ n� 1.

Let A ¼ fHi ¼ ðai1; ai2; . . . ; ainÞj1 � i � mg be a set of m Hamilton cycles in Kn.If A satisfies conditions (1) and (2), A is called an antipodal Hamilton cycle doublecover of Kn:

(1) A is a Hamilton cycle double cover of Kn, that is, if S is the multiset of edgesffaij; ai; jþ1gj1 � i � m; 1 � j � ng, then S has each edge of Kn exactly twice.(The second subscripts of the aij are calculated modulo n.)

(2) If T is the set of edges ffaij; ai; jþn=2gj1 � i � m; 1 � j � n=2g, then T haseach edge of Kn exactly once.

In this terminology, two vertices aij and ai;jþn=2 are called antipodal vertices in Hi

ð1 � i � m; 1 � j � n=2Þ.Proposition 2.1. Let n � 4 be an even integer. If there exists an antipodal Hamiltoncycle double cover of Kn, then there exists a double Dudeney set in Knþ1.

Proof. Let A be an antipodal Hamilton cycle double cover of Kn. Let H ¼ða1; a2; . . . ; anÞ be any Hamilton cycle in A. Consider a sequence

ða1; b1; a2; b2; a3; b3; . . . ; an; bnÞ;

where bi ð1 � i � nÞ is a variable.Let k be an integer with 1 � k � n. Substitute b1; b2; b3; . . . ; bn by ak; ak�1;

ak�2; . . . ; akþ1, respectively, and denote the new sequence by PHðkÞ:PHðkÞ ¼ ða1; b1 ¼ ak; a2; b2 ¼ ak�1; a3; b3 ¼ ak�2; . . . ; an; bn ¼ akþ1Þ:

The subscripts of the ai and the bi are calculated modulo n.For any sequence PHðkÞ ð1 � k � nÞ, we have either ai ¼ bi and aiþn=2 ¼ biþn=2,

or ai ¼ bi�1 and aiþn=2 ¼ biþn=2�1, for some i ð1 � i � nÞ. If ai ¼ bi andaiþn=2 ¼ biþn=2, we define a Hamilton path in Kn, QHðkÞ, as follows:

QHðkÞ ¼ ½bi; aiþ1; biþ1; aiþ2; . . . ; biþn=2�1; aiþn=2�¼ ½ai; bi�1; ai�1; bi�2; . . . ; ai�n=2þ1; bi�n=2�:

DOUBLE COVERINGS BY HAMILTON CYCLES 197

If ai ¼ bi�1 and aiþn=2 ¼ biþn=2�1, we define a Hamilton path in Kn, QHðkÞ, asfollows:

QHðkÞ ¼ ½ai; bi; aiþ1; biþ1; . . . ; aiþn=2�1; biþn=2�1�¼ ½bi�1; ai�1; bi�2; ai�2; . . . ; bi�n=2; ai�n=2�:

If we put

QH ¼ fQHðkÞj1 � k � ng;

then the set QH is uniquely determined, that is, it doesn’t depend on representationsof H. If we put

Q ¼ fQHðkÞj1 � k � n;H 2 Ag;

then Q is a set of Hamilton paths in Kn with cardinality nðn� 1Þ.Claim 1. Q has each 2-path in Kn exactly twice.

Proof. Let ða; b; cÞ be any 2-path in Kn. There exist two Hamilton cycles H;H0 inA such that the edge fa; cg lies on H and H0. If we write

H ¼ ða; c; x; y; z; . . . ;wÞ;

then we have

PHð1Þ ¼ ða; a; c; . . .ÞPHð2Þ ¼ ða; c; c; . . .ÞPHð3Þ ¼ ða; x; c; . . .ÞPHð4Þ ¼ ða; y; c; . . .ÞPHð5Þ ¼ ða; z; c; . . .Þ

� � �PHðnÞ ¼ ða;w; c; . . .Þ:

So, QHðkÞ has the 2-path ða; b; cÞ for some k, 3 � k � n. Similarly, QH0 ðkÞ has the 2-path ða; b; cÞ for some k, 3 � k � n.

Therefore Q has each 2-path in Kn at least twice. Since the number of 2-paths inKn is nðn� 1Þðn� 2Þ=2 and the number of 2-paths in Q is nðn� 1Þðn� 2Þ, Q haseach 2-path in Kn exactly twice. &

Claim 2. Each pair of distinct vertices of Kn is the pair of endpoints of exactly twoof the Hamilton paths in Q.

Proof. Let a; b be any distinct vertices of Kn. There is a Hamilton cycle H in Asuch that a and b are antipodal vertices in H. We can write

H ¼ ða1; . . . ; an=2þ1; . . .Þ;

where a1 ¼ a and an=2þ1 ¼ b. Then the endpoints of QHð1Þ are a and b. Theendpoints of QHðnÞ are also a and b. Therefore there are at least two Hamilton pathsin Q which have vertices a and b as the endpoints.


Since the number of pairs of endpoints in Q is nðn� 1Þ and the number of pairs ofdistinct vertices of Kn is nðn� 1Þ=2, we obtain Claim 2. &

Add a new vertex 10 between the endpoints of QHðkÞ and make the Hamiltoncycle RHðkÞ in Knþ1, where the vertex set of Knþ1 is Vn [ f10g:

RHðkÞ ¼ ð10;QHðkÞÞ

¼ð10; bi; aiþ1; biþ1; aiþ2; . . . ; biþn=2�1; aiþn=2Þ if ai ¼ bi

ð10; ai; bi; aiþ1; biþ1; . . . ; aiþn=2�1; biþn=2�1Þ if ai ¼ bi�1:

(

Put

D ¼ fRHðkÞj1 � k � n;H 2 Ag;

and we will show that D is a double Dudeney set in Knþ1.As any element of D is a Hamilton cycle in Knþ1 trivially, we need only show that

D has each 2-path in Knþ1 exactly twice.

(i) Let ða; b; cÞ be a 2-path in Knþ1 with a; b; c 6¼ 10. From Claim 1, Q has the2-path ða; b; cÞ exactly twice. So D has the 2-path ða; b; cÞ exactly twice.

(ii) Let ða;10; bÞ be a 2-path in Knþ1. From Claim 2, a; b are endpoints of exactlytwo of the Hamilton paths in Q. So, D has the 2-path ða;10; bÞ exactly twice.

(iii) Let ð10; a; bÞ be a 2-path in Knþ1. There exist exactly two Hamilton cyclesH;H 0 in A such that the edge fa; bg lies on H and H0. If we write

H ¼ ða; b; . . .Þ;

then we have

PHð1Þ ¼ ða; a; b; . . .Þ;

and

QHð1Þ ¼ ½a; b; . . .�:

So we have

RHð1Þ ¼ ð10; a; b; . . .Þ:

Similarly, RH0 ð1Þ ¼ ð10; a; b; . . .Þ: Hence D has the 2-path ð10; a; bÞ at least twice.Comparing the number of 2-paths in D with the number of 2-paths in Knþ1, we

obtain that D has each 2-path in Knþ1 exactly twice. Therefore D is a double Dudeneyset in Knþ1. This completes the proof of Prop. 2.1. &

3. CONSTRUCTION OF ANTIPODAL STARTERS

Let n � 4 be an even number and Kn ¼ ðVn;EnÞ the complete graph. Put m ¼ n� 1and r ¼ ðm� 1Þ=2. In this section, we put Vn ¼ f0; 1; 2; � � � ;m�1g [ f1g ¼Zm [ f1g.


Let � be the vertex-permutation ð1Þð0 1 2 . . . m� 1Þ, and put � ¼ f� jj0 �j � m� 1g. Clearly � induces a permutation of the edges of Kn; we will also denotethis permutation by �. When A is a set of cycles or circuits in Kn, define�A ¼ fC� jC 2 A; � 2 �g.

For any edge fa; bg in Kn, we define the length dða; bÞ:

dða; bÞ ¼ minfm� jb� aj; jb� ajg ða; b 6¼ 1Þ;1 ðotherwiseÞ:

�

A Hamilton cycle S in Kn is called an antipodal starter if �fSg is an antipodalHamilton cycle double cover of Kn.

Let

S ¼ ð1 ¼ a1; a2; a3; . . . ; anÞ

be a Hamilton cycle in Kn. Then S is an antipodal starter if and only if S satisfies (1)and (2):

(1) If L is the multiset fdða; bÞ j fa; bg 2 Sg, then we have L ¼ f1;1; 1; 1;2; 2; . . . ; r; rg, i.e., S has each length twice.

(2) If M is the set fdðai; aiþn=2Þ j 1 � i � n=2g, then we have M ¼ f1; 1;2; . . . ; rg.

There is an antipodal Hamilton cycle double cover of Kn when n ¼ 4 and there isno antipodal Hamilton cycle double cover of Kn when n ¼ 6; 8. Brendan McKay hascomputed all the nonisomorphic antipodal starters up to n ¼ 20 and found that thenumbers are increasing quickly.

Now we will construct an antipodal starter in Kn for all even n � 10.

Proposition 3.1. There exists an antipodal starter in Kn for all even n � 10.

Proof.(Case 1) r is even.

Put s ¼ r=2. We have r � 4, s � 2 as n � 10. Put

T1 ¼ ff1; 0g; f1;�2ggT2 ¼ ff0; rg; f�1;�rggT3 ¼ ffs� 1; sg; fs; sþ 2ggT4 ¼ ffsþ 1;�ðsþ 1Þg; fsþ 1;�sggT5 ¼ ff1;�1g; f2;�2g; f3;�3g; . . . ; fs� 1;�ðs� 1ÞggT6 ¼ ff1;�3g; f2;�4g; f3;�5g; . . . ; fs� 2;�sggT7 ¼ ffr � 1;�ðr � 1Þg; fr � 2;�ðr � 2Þg;fr � 3;�ðr � 3Þg; . . . ;fsþ 2;�ðsþ 2ÞggT8 ¼ ff�r;�ðr � 1Þg; fr;�ðr � 2Þg; fr � 1;�ðr � 3Þg; . . . ; fsþ 3;�ðsþ 1Þgg:


Note that T6 ¼ T7 ¼ �; T8 ¼ ff�r;�ðr � 1Þgg when s ¼ 2. If we denote the multi-set fdðeÞje 2 Tig by dðTiÞ, then we have

dðT1Þ ¼ f1;1gdðT2Þ ¼ fr; r � 1gdðT3Þ ¼ f1; 2gdðT4Þ ¼ fr � 1; rgdðT5Þ ¼ f2; 4; 6; . . . ; r � 2gdðT6Þ ¼ f4; 6; 8; . . . ; r � 2gdðT7Þ ¼ f3; 5; 7; . . . ; r � 3gdðT8Þ ¼ f1; 3; 5; . . . ; r � 3g:

Put S ¼ [8i¼1Ti, so that we have dðSÞ ¼ f1;1; 1; 1; 2; 2; . . . ; r; rg. We will show that

S is an antipodal starter in Kn (Figure 1).

(i) The case s is odd.Put t ¼ ðs� 1Þ=2. Define two paths S1; S2:

S1 ¼ ð1; 0; r;�ðr � 2Þ; r � 2;�ðr � 4Þ; r � 4;�ðr � 6Þ; r � 6; . . . ;

� ðr � ð2t � 2ÞÞ; r � ð2t � 2Þ;�ðr � 2tÞ ¼ �ðsþ 1Þ;sþ 1;�s; s� 2;�ðs� 2Þ; s� 4;�ðs� 4Þ; s� 6;

� ðs� 6Þ; . . . ; s� ð2t � 2Þ;�ðs� ð2t � 2ÞÞ; s� 2t ¼ 1Þ;S2 ¼ ð�1;�r;�ðr � 1Þ; r � 1;�ðr � 3Þ; r � 3;�ðr � 5Þ; r � 5; . . . ;

� ðr � ð2t � 1ÞÞ; r � ð2t � 1Þ ¼ sþ 2; s; s� 1;

� ðs� 1Þ; s� 3;�ðs� 3Þ; s� 5;�ðs� 5Þ; . . . ; s� ð2t � 1Þ;� ðs� ð2t � 1ÞÞ ¼ �2;1Þ:

Then S ¼ S1 [ S2 [ ff1;�1gg, so S is a Hamilton cycle in Kn. The length of the pathS1 is r and the length of the path S2 is r þ 1. The pairs of antipodal vertices in S are

f1;�1g; f0;�rg; fr;�ðr � 1Þg; f�ðr � 2Þ; r � 1g;fr � 2;�ðr � 3Þg; f�ðr � 4Þ; r � 3g; fr � 4;�ðr � 5Þg;f�ðr � 6Þ; r � 5g; . . . ; f�ðr � 2tÞ; r � ð2t � 1Þg;fsþ 1; sg; f�s; s� 1g; fs� 2;�ðs� 1Þg;f�ðs� 2Þ; s� 3Þg; fs� 4;�ðs� 3Þg; f�ðs� 4Þ; s� 5Þg;fs� 6;�ðs� 5Þg; . . . ; fs� 2t;�ðs� ð2t � 1ÞÞg;

and those lengths are

1; r; 2; 4; 6; 8; 10; 12; . . . ; r�2;

1; r�1; r�3; r�5; r�7; r�9; r�11; . . . ; 3;

respectively. Therefore S is an antipodal starter in Kn.


(ii) The case s is even.Put t ¼ s=2. Define two paths S1; S2:

S1 ¼ ð1; 0; r;�ðr � 2Þ; r � 2;�ðr � 4Þ; r � 4;�ðr � 6Þ; r � 6; . . . ;

� ðr � ð2t � 2ÞÞ; r � ð2t � 2Þ ¼ sþ 2; s; s� 1;

� ðs� 1Þ; s� 3;�ðs� 3Þ; s� 5;�ðs� 5Þ; . . . ; s� ð2t � 3Þ;� ðs� ð2t � 3ÞÞ; s� ð2t � 1Þ ¼ 1Þ;

S2 ¼ ð�1;�r;�ðr � 1Þ; r � 1;�ðr � 3Þ; r � 3;�ðr � 5Þ; r � 5; . . . ;

� ðr � ð2t � 1ÞÞ; r � ð2t � 1Þ ¼ sþ 1;

� s; s� 2;�ðs� 2Þ; s� 4;�ðs� 4Þ; s� 6;�ðs� 6Þ; . . . ; s� ð2t � 2Þ;� ðs� ð2t � 2ÞÞ ¼ �2;1Þ:


f1;�1g; f0;�rg; fr;�ðr � 1Þg; f�ðr � 2Þ; r � 1g;fr � 2;�ðr � 3Þg; f�ðr � 4Þ; r � 3g; fr � 4;�ðr � 5Þg;f�ðr � 6Þ; r � 5g; . . . ; fr � ð2t � 2Þ;�ðr � ð2t � 1ÞÞg;fs; sþ 1g; fs� 1;�sg; f�ðs� 1Þ; s� 2Þg;fs� 3;�ðs� 2Þg; f�ðs� 3Þ; s� 4g; fs� 5;�ðs� 4Þg;f�ðs� 5Þ; s� 6g; . . . ; fs� ð2t � 1Þ;�ðs� ð2t � 2ÞÞg;

FIGURE 1. An antipodal starter in K22.



1; r; 2; 4; 6; 8; 10; 12; . . . ; r�2;

1; r�1; r�3; r�5; r�7; r�9; r�11; . . . ; 3;


(Case 2) r is odd.Put s ¼ ðr � 1Þ=2. We have r � 5, s � 2 as n � 10. Put

T1 ¼ ff1; 0g; f1;�2ggT2 ¼ ff0;�ðr � 1Þg; f�1;�rggT3 ¼ ffs� 1; sg; fs; sþ 2ggT4 ¼ ffsþ 1;�ðsþ 1Þg; fsþ 1;�sggT5 ¼ ff1;�1g; f2;�2g; f3;�3g; . . . ; fs� 1;�ðs� 1ÞggT6 ¼ ff1;�3g; f2;�4g; f3;�5g; . . . ; fs� 2;�sggT7 ¼ ffr;�rg; fr � 1;�ðr � 1Þg; fr � 2;�ðr � 2Þg; . . . ; fsþ 2;�ðsþ 2ÞggT8 ¼ ffr;�ðr � 2Þg; fr � 1;�ðr � 3Þg; fr � 2;�ðr � 4Þg; . . . ; fsþ 3;�ðsþ 1Þgg:

Note that T6 ¼ � when s ¼ 2. We have

dðT1Þ ¼ f1;1gdðT2Þ ¼ fr � 1; r � 1gdðT3Þ ¼ f1; 2gdðT4Þ ¼ fr; rgdðT5Þ ¼ f2; 4; 6; . . . ; r � 3gdðT6Þ ¼ f4; 6; 8; . . . ; r � 3gdðT7Þ ¼ f1; 3; 5; . . . ; r � 2gdðT8Þ ¼ f3; 5; 7; . . . ; r � 2g:

Put S ¼ [8i¼1Ti, so that we have dðSÞ ¼ f1;1; 1; 1; 2; 2; . . . ; r; rg. We will show that

S is an antipodal starter in Kn (Figure 2).

(i) The case s is odd.Put t ¼ ðs� 1Þ=2. Define two paths S1; S2:

S1 ¼ ð1; 0;�ðr � 1Þ; r � 1;�ðr � 3Þ; r � 3;�ðr � 5Þ; r � 5; . . . ;

� ðr � ð2t þ 1ÞÞ; r � ð2t þ 1Þ ¼ sþ 1;�s; s� 2;

� ðs� 2Þ; s� 4;�ðs� 4Þ; s� 6;�ðs� 6Þ; . . . ; s� ð2t � 2Þ;� ðs� ð2t � 2ÞÞ; s� 2t ¼ 1Þ;

S2 ¼ ð�1;�r; r;�ðr � 2Þ; r � 2;�ðr � 4Þ; r � 4;�ðr � 6Þ; r � 6; . . . ;

� ðr � 2tÞ; r � 2t ¼ sþ 2; s; s� 1;



Then S ¼ S1 [ S2 [ ff1;�1gg, so S is a Hamilton cycle in Kn. The length of thepath S1 is r and the length of the path S2 is r þ 1. The pairs of antipodal vertices inS are

f1;�1g; f0;�rg; f�ðr � 1Þ; rg; fr � 1;�ðr � 2Þg;f�ðr � 3Þ; r � 2g; fr � 3;�ðr � 4Þg; f�ðr � 5Þ; r � 4g;fr � 5;�ðr � 6Þg; . . . ; f�ðr � ð2t þ 1ÞÞ; r � 2tg;fsþ 1; sg; f�s; s� 1g; fs� 2;�ðs� 1Þg;f�ðs� 2Þ; s� 3g; fs� 4;�ðs� 3Þg; f�ðs� 4Þ; s� 5Þg;fs� 6;�ðs� 5Þg; . . . ; fs� 2t;�ðs� ð2t � 1ÞÞg;


1; r; 2; 4; 6; 8; 10; 12; . . . ; r�1;

1; r�2; r�4; r�6; r�8; r�10; r�12; . . . ; 3;


(ii) The case s is even.Put t ¼ s=2. Define two paths S1; S2:

S1 ¼ ð1; 0;�ðr � 1Þ; r � 1;�ðr � 3Þ; r � 3;�ðr � 5Þ; r � 5; . . . ;

� ðr � ð2t � 1ÞÞ; r � ð2t � 1Þ ¼ sþ 2; s; s� 1;

� ðs� 1Þ; s� 3;�ðs� 3Þ; s� 5;�ðs� 5Þ; . . . ; s� ð2t � 3Þ;� ðs� ð2t � 3ÞÞ; s� ð2t � 1Þ ¼ 1Þ;

FIGURE 2. An antipodal starter in K24.


S2 ¼ ð�1;�r; r;�ðr � 2Þ; r � 2;�ðr � 4Þ; r � 4;�ðr � 6Þ; r � 6; . . . ;

� ðr � 2tÞ; r � 2t ¼ sþ 1;�s; s� 2;



f1;�1g; f0;�rg; f�ðr � 1Þ; rg; fr � 1;�ðr � 2Þg;f�ðr � 3Þ; r � 2g; fr � 3;�ðr � 4Þg; f�ðr � 5Þ; r � 4g;fr � 5;�ðr � 6Þg; . . . ; fr � ð2t � 1Þ;�ðr � 2tÞg;fs; sþ 1g; fs� 1;�sg; f�ðs� 1Þ; s� 2g;fs� 3;�ðs� 2Þg; f�ðs� 3Þ; s� 4g; fs� 5;�ðs� 4Þg;f�ðs� 5Þ; s� 6g; . . . ; fs� ð2t � 1Þ;�ðs� ð2t � 2ÞÞg;


1; r; 2; 4; 6; 8; 10; 12; . . . ; r�1;

1; r�2; r�4; r�6; r�8; r�10; r�12; . . . ; 3;

respectively. Therefore S is an antipodal starter in Kn. This completes the proof ofProp. 3.1. &

Finally, we note that the antipodal starter S was constructed using the starter of1-factorization N1Kn in [9].

ACKNOWLEDGMENT

The authors would like to express their thanks to Professor B. D. McKay for hishelpful comments.

REFERENCES

[1] H. E. Dudeney, ‘‘Amusements in Mathematics,’’ Thomas Nelson and Sons, London, 1917,Dover Reprint, New York, 1970.

[2] K. Heinrich, M. Kobayashi, and G. Nakamura, Dudeney’s Round Table Problem, AnnDiscrete Math 92 (1991), 107–125.

[3] K. Heinrich, D. Langdeau, and H. Verrall, Covering 2-paths uniformly, J Combin Des 8(2000), 100–121.

[4] M. Kobayashi, J. Akiyama, and G. Nakamura, On Dudeney’s round table problem forpþ 2, Ars Combin, to appear.

[5] M. Kobayashi, Kiyasu-Z., and G. Nakamura, A solution of Dudeney’s round tableproblem for an even number of people, J Combin Theory (A) 62 (1993), 26–42.


[6] M. Kobayashi, Kiyasu-Z., and G. Nakamura, Double Dudeney sets for p2 þ 2 vertices,manuscript.

[7] M. Kobayashi, N. Mutoh, Kiyasu-Z., and G. Nakamura, Double Dudeney sets for an oddnumber of vertices, Aust J Combin, to appear.

[8] M. Kobayashi, N. Mutoh, Kiyasu-Z., and G. Nakamura, New Series of Dudeney Sets forpþ 2 Vertices, Ars Combin, to appear.

[9] M. Kobayashi, N. Mutoh, and G. Nakamura, New infinite classes of 1-factorizations ofcomplete graphs, Aust J Combin 21 (2000), 257–269.

[10] M. Kobayashi and G. Nakamura, Exact coverings of 2-paths by Hamilton cycles,J Combin Theory (A) 60 (1992), 298–304.

[11] G. Nakamura, Kiyasu-Z., and N. Ikeno, Solution of the round table problem for the case ofpk þ 1 persons, Commentarii Mathematici Universitatis Sancti Pauli 29 (1980), 7–20.

[12] H. Verrall, Pairwise Compatible Hamilton Decompositions of Kn, J Combin Theory (A)79 (1997), 209–222.


double coverings of 2-paths by hamilton cycles*

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