Decomposing Complete Graphs Into Cycles of Length 2P*

Annals of Discrete Mathematics 9 (1980) 155-162 @ North-Holland Publishing Company

DECOMPOSING COMPLETE GRAPHS INTO CYCLES OF LENGTH 2p'" Brian ALSPACH and Badri N. VARMA Department of Mathematics, Simon Fraser University, Bumaby, B.C. V5A lS6, Canada It is shown that the complete graph K,, can be decomposed into edge-disjoint cycles of the same length 2p' if and only if n is odd, n 8 2 p e , and 2p" divides (2") where p is any prime and e is a positive integer.

Let K,, denote the complete graph with n vertices. If E(K,,),the edge-set of K,,, can be partitioned so that each partition set is a cycle of some fixed length r, then we say that we have an isomorphic factorization of K,, into cycles C,. We denote such an isomorphic factorization by C, 1 K,,. An isomorphic factorization of K,, into cycles C, is also often referred to as a decomposition of K,, into edge-disjoint cycles of length r. Since a cycle in a graph contributes either degree 0 or degree 2 to a vertex of the graph, it is clear that C, I K,, implies that n is odd. It also must be the case that n 2 r and that r divides JE(K,,)I= (2"). These three necessary conditions for C, I K,, to hold are also sufficient as far as is presently known. This is the case when r = 2' for e 2 2 which follows from work of Rosa [6] and Kotzig [4]. When r = 3 we have the well-known case of Steiner triple systems. The cases of 5, 7, and 9 have been done by Bermond and Sotteau [2]. The cases for r even and 4 s r s 1 6 have been done by Bermond et al. [ 11. A recent survey article on cycle decompositions is by Bermond and Sotteau [3]. In this paper we give another infinite class of r's for which the necessary conditions are also sufficient. We now state the main theorem and use the rest of the paper to prove it.

Theorem 1. Let p be any prime and e be any positive integer. Then C Z p1=K,, i f and only if n is odd, n 22p', and 2p" divides (2"). If p = 2 , then 2pe=2e+l and the necessary conditions imply that n = m * 2'+*+ 1 for some positive integer m. Kotzig [4] and Rosa [6] together proved k 1 K Z m k + l for any even k 2 4 . Setting k = 2p" proves the theorem in the that c case that p = 2 . *This research was supported by the Natural Sciences and Engineering Research Council of Canada.

155

B. Alspach, B.N. Varma

156

Henceforth, p will always be an odd prime. We now state a result of Sotteau that will be used several times in the rest of the proof.

Theorem 2 (Sotteau [7]). Let k = 2 t be an even integer with k a 4 . Then the complete bipartite graph K , , can be isomorphically factored into cycles C, if and only if r 3 t, s a t, and 2t divides rs. We now examine the arithmetic implications of 2p" dividing (3.There are two cases. Either p' divides n or p' divides n - 1. First we consider the case that p' divides n - 1. Since 2 also divides $(n- l), we know that n = 2m . 2p' + 1 for some positive integer m.We then have CZp= 1 K,, by the Kotzig-Rosa result stated above. This leaves us with the case that p' divides n. In this case it is easy to verify that either n = 4mp" + p' when p" = 1 (mod 4) or n = 4mp' + 3p' when p' = 3 (mod 4). When n = 4mp' + p' we write K,, in the form K n

= K4mp'+p' = K4(rn-l)p'+(5p'-1)+1

= K4(rn-I)p'+l

UK~peUK4(m-l)pe,5p'-l.

The complete graph K 4 ( m - l ) p e + l can be decomposed into cycles C2peby the Kotzig-Rosa result. The complete bipartite graph K 4 ( m - l ) p e , 5 p e - 1 can be decomposed into cycles CZpeby Theorem 2. Thus, we see that Czpe1 K,, if we can decompose Ksp=into cycles CZpe. When n = 4mp' +3p' we write K,, in the form Kn = K 4 m p e + 3 p e - K4mp'+(3p"-1)+1 -

-K4mpe+1

U K 3 p e - 1 UK4rnpe.3pc-1.

The Kotzig-Rosa result gives a decomposition of the complete graph K4mpe+linto and Theorem 2 gives a decomposition of the complete bipartite graph cycles CZp= K4mpe,3pe-1 into cycles Czp=. Thus, we see that C2pe1 K,, if we can decompose K3pe into cycles C2pe. The problem has now been reduced to finding two particular decompositions. We shall first show that C2p= 1 K3,,e when p3= 3 (mod 4). We do it by proving a more general result which we now state.

Theorem 3. If m > O and m = 3 (mod 4), then C,, I K,,,,. First, partition the vertices of K , , into rn disjoint triples, that is, let V(K,,) = V(K,,) defined by u ( u i i )= u i ,j + l for 1s i s m and j = 1, 2, and 3 where 3 + 1= 1. The cycle { u i j : 1s i s m and j = 1,2, and 3). Let u denote the permutation of

c=

~ l l ~ 2 2 u 3 1 u 4* 2 * . u m 1 u 1 2 ~ 2 1 u 3 2 '

* '

u r ~ - l ,lum2ull

Decomposing complete graphs into cycles of length 2p'

157

has length 2rn. Define a(C) by the action of (T on the vertices of C. The three cycles C, o(C), and a2(C) all have length 2rn, are edge-disjoint, and do not contain any of the edges of the form uijui+l, where the first subscripts are taken modulo rn. Now the set of edges of the form uiiui+l, form three vertex-disjoint cycles of length rn connecting vertices of successively indexed triples. The importance of this is that if we take any permutation of the rn triples and find three vertex-disjoint cycles of length rn connecting successive triples along the permutation, then the remaining edges may be partitioned into three edgedisjoint cycles of length 2 m connecting successive triples along the permutation. We now make this more precise. Let a be a permutation of the set {1,2,. . . , rn}. If the three cycles u a ( 1 ) , j ( l ) [ ~ a ( 2 )j(2)t ,

* * *

urn(,), j(rn)iUa(l), j(l)i

for i = 1, 2, and 3

are vertex-disjoint, then the remaining edges joining triples { u ~ ( ~ u) , ( ~ )2,, u , ( ~ 3)}, to triples { u ~ ( ~ 1,+ ~ ) , 2, 3} for i = 1, 2, . . . , rn (where rn + 1= 1)can be partioned into three edge-disjoint cycles of length 2rn. We now use the well-known decomposition of K,,, n odd and n 2 3 , into edge-disjoint Hamiltonian cycles [ 5 , p. 1611. An example of the construction is given in Fig. 1. The Hamiltonian cycles are obtained by rotating the given cycle through each of the first $(rn- 1) clockwise possible rotations (the identity rotation counts as one possible rotation). Let a be the permutation of { 1,2, . . . , rn} given by the first Hamiltonian cycle in the decomposition of K , into edge-disjoint Hamiltonian cycles, that is, a ( i )is the index of the ith vertex of the cycle starting at the vertex ul. Now let C be the cycle u11u12ua(2),2ua(2),

l U a ( 3 ) ,l U a ( 3 ) , 2 ' '

~ a ( ~ - l ) , 2 ~ a ( m - l ) , 3 ~ a ( m ) , 3l ~ u al l(. ~ ) ,

It has length 2rn. The cycles C, a(C>, and a2(C) all have length 2rn, are edge-disjoint, use all of the internal edges of the triangles formed by the triples, and their edges between successive triangles along the permutation a form three vertex-disjoint cycles of length rn. Hence, the edges not used between successive triangles can be partitioned into three edge-disjoint cycles of length 2rn.

I

/ u1

n

3

u6

Fig. 1.

B.Alspach, B.N. Vama

158

Since m = 3 (mod 4), there are an odd number of Hamiltonian cycles in the decomposition of K, into Hamiltonian cycles. We have used one of the Hamiltonian cycles above to use all the internal edges of the triangles along with the edges connecting the triangles in the order of the Hamiltonian cycle. This leaves an even number of Hamiltonian cycles in the decomposition of K,. We then pair them to complete the decomposition of K3,. We pair the Hamiltonian cycles of the decomposition of K, so that any pair are successive rotations of the cycle of Fig. 1. Thus, it suffices to consider the cycle of Fig. 1 together with the next one in the clockwise rotation of the cycle, that is, the two cycles we consider are

c=

u1u2u3

* * *

U,-1UmU1

and

c' =

u 1 u 3 us u 2 u 7 u 4 u g u 6

* ' *

um-2

urn-5

Urnl&-3

urn-1 u 1 .

These two cycles determine two permutations of the triangles (triples) of K3,. These two permutations tell us how to move among the triangles. Consider the cycle that is displayed in Fig. 2. A solid line in the figure corresponds to an actual

um-l,l

um2

m3

LL

0

--? lJm-1,2 I

8 0

0

Fig. 2.

I

0

9

Decomposing complete graphs into cycles of length 2p"

159

edge of K3,, a broken line corresponds to a path from uii to uki by having the first index change according to the cycle C from i to k in the increasing direction of indices, and the wavy line from ~ 7 to 1 urn+, corresponds to a path from u71 to such that the first indices change according to the index changes following where we pass through the vertex u4. On both the the cycle C' from u7 to broken line and wavy line paths the second indices stay constant. Call the cycle that we obtain from Fig. 2 by C*.First, the cycle C* passes through 1,2, . . . , m as a first index exactly twice. Thus, the length of C* is 2m. Second, if we rotate the Fig. 2 configuration so that the first block goes onto the second block, the second block onto the third block, the third block onto the first block, and keep the second indices fixed, then this rotation and the second power of the rotation give us two more cycles of length 2m. All of the cycles have no edges in common and they follow the edges according to the cycles C and C'. In fact, the two cycles of length 2 m we obtain are nothing more than a(C*) and u2(C").

Now we want to examine the edges of C*,v(C*), and a2(C*)that lie between triples given by C.Notice that the edge ulluzl is present since the edge u13U23 is in C*. Likewise, the edge ~ 2 1 ~ is 3 1present since the edge u23u33 is in C*.In fact, if one examines the diagram of Fig. 2, it will be noticed that every edge of C* that is following the cycle C does not have a change in the second index. Hence, the edges of C*,a(C*), and aZ(C*)contain the three vertex-disjoint cycles D = u11u21u31 * * * u m I u l l , a(D), and a2(D).Therefore, by earlier remarks, the remaining edges of K 3 , that join triples according to the cycle C of K , can be partitioned into three edge-disjoint cycles of length 2m. Now we examine the edges of C*,a(C*),and a2(C*)that lie between triples given by C'. In the diagram given in Fig. 2, there are some edges between triples lying along C' in which there is a shift in the second index between end-vertices of the edges. If one calculates the total shift involved in the cycle C",it totals to a shift of +6. Since we are moving with triples, after we have traversed all of C', we will be back at the same second index at which we started. Thus, the cycle

together with u(D') and a2(D')is contained in the edges of C*, a(C*), and a2iC*).It is then the case that the cycles D', a(D'), and a'(D') are vertexdisjoint so that the remaining edges of K3, that go between triangles that follow the cycle C' in K , can be partitioned into three edge-disjoint cycles of length 2m. We have now seen that it is possible to take two different Hamiltonian cycles among the triples of K3,,, and define three cycles of length 2 m using the Hamiltonian cycles so that the remaining edges between the triples along the Hamiltonian cycles can also be partitioned into cycles of length 2m. This allows us to decompose K , , into cycles of length 2m that are edge-disjoint. This completes the proof of Theorem 3.

B. Alspach. B.N. Varma

160

The only remaining result needed is the following. It is proved similarly to Theorem 3.

Theorem 4. If m > 0 and m = 1 (mod 4), then C2, I K,,. First, partition the vertices of K,, into m disjoint sets of cardinality five. So we have V ( K , ) = {uii:1 =sis m and j = 1 , 2 , 3 , 4 , 5 } . Let (T now denote the permutation of V(K,,) defined by a(z+)= q.i+lfor 1 Ci G m and j = 1 , 2 , 3 , 4 , and 5 where we take 5 + 1 to be 1. The cycles

cl=ullu22u31u42

* ' '

um1u12u21u32

' *

um1w13u21u33

' * ' um-1,1um3u11

*

K m - l , 1um2u11

and c2=u11u2?u31u43 *

* '

both have length 2m. The ten cycles C1,C2,u(C1),a(C2),a2(C1),a2(C2),a3(C1), a3(C2),a4(C1),and a4(C2)are all edge-disjoint, have length 2m, and do not contain any edges of the form uiiui+l,iwhere the first subscript is taken modulo m. Now the set of edges of the form uiiui+l,iform five vertex-disjoint cycles of length rn connecting vertices of successively indexed 5-sets. As before, the importance of this is that if we take any permutation of the 5-sets and find five vertex-disjoint cycles of length m connecting successive 5-sets along the permutation, then the remaining edges may be partitioned into ten edge-disjoint cycles of length 2m connecting successive 5-sets along the permutation. Since we have already done the case of 3m in detail, we outline the rest of the proof in the current case of 5m. Again we use the decomposition of K, into Hamiltonian cycles as given in Fig. 1. Since m = 1 (mod 4), there are an even number of Hamiltonian cycles in the decomposition of K,. Now let a be the permutation of {1,2, . . . , m} given by a Hamiltonian cycle of the decomposition where a ( i ) is the index of the ith vertex of the cycle starting at the vertex ul. Now let C be the cycle u11'12ua(2).2ua(2).

l'a(3),

l'a(3),2

* * '

Ka(m-3),2Ku(m-3),3Ka(~-2),3

~ ~ ~ r n - ~ ~ , ~ ~ a ~ m - ~ ~ 5 u,e ( ~ m )~, 1 a U l~ t. r n - ~ ~ ,

It has length 2m. The cycles C,c+(C),a2(C),a3(C), and a4(C)all have length 2m, are edge-disjoint, use all of the internal edges of the 5-sets that join successively indexed vertices within the 5-sets, and their edges between successive 5-sets along the permutation a form five vertex-disjoint cycles of length m. Hence, the remaining edges between successive 5-sets along the permutation a may be partitioned into ten edge-disjoint cycles of length 2m. We then do the same trick with a second Hamiltonian cycle of K, to use all of the internal edges of the 5-sets whose endvertices differ by two on their indexes. At this point we will have used two Hamiltonian cycles of K, to use all the internal edges of the 5-sets. This leaves an even number of Hamiltonian cycles in the decomposition of K,. As before we pair them and may assume that we are

Decomposing complete graphs into cycles of length 2p'

161

..= Fig. 3.

taking successive Hamiltonian cycles of Fig. 1. We assume the first cycle is u1u2u3 * u , - ~ u , u ~as we also did in the previous proof. Now consider the cycle of Fig. 3. The conventions are the same as for the cycle of Fig. 2. Hence, one can see that the cycle has length 2m and under the appropriate powers of (T we obtain five edge-disjoint cycles of length 2m. By counting the label changes in the second index, it is easy to see that one obtains five vertex-disjoint cycles of length rn along each of the Hamiltonian cycles. Thus, the remaining edges may be partitioned into ten edge-disjoint cycles of length 2m along each of the Hamiltonian cycles of K,. Theorems 3 and 4 together with earlier remarks about recursively constructing cycle decompositions complete the proof of Theorem 1. It is likely that the techniques employed in this paper will prove useful in attacking harder cycle decompositions of complete graphs.

-

References [l] J.-C. Bermond, C. Huang and D. Sotteau, Balanced cycle and circuit designs: even case, Ars Combinatoria 5 (1978) 293-318.

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[2] J.-C. Bermond and D. Sotteau, Cycle and circuit designs odd case, in: Graphen Theorie und deren Anwendungen, Proc. Int. Colloq. of Oberhof (1977) 11-32. [3] J.-C. Bermond and D. Sotteau, Graph decompositions and G-designs, Proc. Fifth British Combinatorial Conference, Congress. Num. 15, Utilitas Math. (1975) 53-72. [4] A. Kotzig, On the decomposition of complete graphs into 4k-gons, Mat.-Fyz. casopis Sloven. Akad. Vied. 15 (1965) 229-233 (in Russian). [5] E. Lucas, Recreations mathematiques, Vol. I1 (Gauthiers-Villars, Paris, 1883). [6] A. Rosa, On cyclic decompositions of the complete graph into (4m + 2)-gons, Mat.-Fyz. easopis Sloven. Akad. Vied. 16 (1966) 349-353. [7] D. Sotteau, Decomposition of K , , , J K Z , J into circuits of length 2k, J. Combin. Theory (B), to appear.