# The Even Case

In this section, we prove Theorem 1.1, that is, we are interested in proving that the graph Kn-I can be decomposed into cycles of length m when m and n are even and . The preliminaries have been completed for the proof except for the basic way in which we wish to view the graph Kn-I. We present this, along with a lemma, and then proceed to the proof of the result.

Definition 3.1   If A is a set of positive integers, then .

Definition 3.2   Let , where n is a positive even integer. Let G be the circulant graph of order n-2with connection set , and u and w be two independent vertices, neither of which are vertices of G. Define , that is, , where the vertices of are labeled uand w.

The preceding definitions imply that the graph Kn - I can be thought of as the join of a circulant graph of order n - 2 with , that is, Kn - I = H(n;A), where . Thus the edges of length (n-2)/2in the circulant together with the edge uw make up the edges of the 1-factor I.

Definition 3.3   Let uiuj be an edge of length j-i in a circulant graph of order n. The edge of length j-i diametrically opposed to uiujis the edge joining ui+n/2 and uj+n/2.

Lemma 3.4    Let m and n be positive even integers satisfying , n < 2m, and . If , where are positive integers satisfying , then .

PROOF.Label the vertices of the subcirculant with . We have if and only if . Let denote the permutation . If L is any subgraph of H(n;A), then is defined as .

Let P be the trail of length given by

Note that P uses precisely one edge of each length in A. Also note that alternate vertices starting with ua1 have strictly increasing subscripts, while alternating vertices starting with u0have strictly decreasing subscripts. Thus, the vertices of P are distinct so that P is a path. The last edge used to form P has length so that all the vertices of P lie in the interval containing u0 from to . This interval has length at most (n-4)/2. Therefore, the path is vertex-disjoint from P, and the edge of length ai in P is diametrically opposed to the edge of length ai in

Thus, we can form a cycle C of length m by taking

From the above remarks, it follows that is a partition of the edge set of H(n;A) into m-cycles.

We now present the proof of the first main theorem.

PROOF OF THEOREM 1.1. Let m and n be positive even integers with . As mentioned in the introduction, if n(n-2)/2 is an even multiple of m, then using [6,11] it is not difficult to show that Kn - I can be decomposed into m-cycles. To do so, we think of Kn-I as the wreath product , that is, replace every vertex of Kn/2 by a copy of and every edge of Kn/2 by a copy of K2,2. A path P of length m/2 in Kn/2 becomes in Kn-I, and this graph can be decomposed into two cycles of length m by the main result of [6]. The proof is completed by using one of the main results of [11] to observe that Kn/2 can be decomposed into paths of length m/2.

Because of the preceding paragraph, we may assume that n(n-2)/2 is an odd multiple of m. Next, by Lemma 2.3, we may also assume that n < 2m, say n=m+r for some positive integer r with . Now if r=0, we are done by a construction of Walecki given in [9]. Thus, we assume that 0 < r < m. We have seen that implies that , and since n(n-2)/2 is an odd multiple of m, Lemma 2.4 implies that r(r-2)/2 is an even multiple of m.

Let and label the vertices of the graph H(n;A) with , where u and w are the two vertices of (see Definition 3.2). Note that the edge joining ui and uj has length j-i or length i-j depending on which lies between 1 and (n-2)/2 modulo n-2. The proof of Theorem 1.1 proceeds as follows. Suppose that B is a subset of A such that , and that we can decompose the circulant into m-cycles. Then since and the graph H(n;A-B) can be decomposed into m-cycles by Lemma 3.4, it follows that we have a decomposition of H(n;A)into m-cycles. Thus, the rest of the proof consists of determining a convenient set B of r/2 lengths such that the circulant can be decomposed into m-cycles.

By Lemma 2.4, we know that r is congruent to 0 or 2 modulo 8, and thus the proof now breaks into these two cases.

CASE 1. SUPPOSE THAT #MATH97#. Let r = 2ea where a is odd and . Thus r(r-2)/2=2ea(2e-1a-1). Since m divides r(r-2)/2 evenly, we have that m=2da'b', where , , and . Then

n-2=m+r-2=2da'b'+2ea-2=b'(2da'+(2ea-2)/b').

Partition the vertices of the circulant H into b' segments, each with vertices. Each segment will contribute 2da' edges to an m-cycle. Define the path P0,0 by

Let , where is the permutation from the proof of Lemma 3.4. Since implies that , the vertices of P0,1 are distinct from the vertices of P0,0 except for , which is the last vertex of P0,0 and the first vertex of P0,1. Similarly, the paths

are vertex-disjoint except that the path P0,i begins at the last vertex of P0,i-1 for and P0,b'-1 ends at u0. Thus is an m-cycle and the lengths of the edges of C0 are .

The definition of P0,0 does not make sense when 2d-2a'+2=1, that is, d=2 and a'=1. In this case let

Let b=r/(2d+1a') = 2e-d-1a/a'. If b > 1, then obtain the path P1,0 by adding to the subscripts of the even vertices of P0,0, that is,

Next, obtain the paths by letting powers of act on P1,0 in the same way they acted on P0,0. Furthermore, the path P1,i begins at the last vertex of P1, i-1 for and the last vertex of P1,b'-1 is u0. Thus, is an m-cycle, where the lengths of the edges of C1 are .

Similarly, for each j with , obtain the path Pj,0 by adding to the subcripts of the even vertices of P0,0. Then the paths are obtained by letting powers of act on Pj,0. Note here that Pj,i begins at the last vertex of Pj,i-1 for and the last vertex of Pj,b'-1 is u0. Hence is an m-cycle and the lengths of the edges of Cj are .

For , let Bj denote the set of lengths that Cjuses, that is, . Let . Now the longest edge in B has length . Since 2ea < 2da'b' implies that , it follows that

Therefore, the lengths of B are distinct and hence |B|=r/2. Thus, the collection

is a partition of the edge set of the circulant into m-cycles.

CASE 2. SUPPOSE THAT #MATH137#. Let r=8k+2 for some nonnegative integer k. In this case we have r(r-2)/2 = 8k(4k+1) = 2ea(4k+1) where 2ea=8k, a is odd, and gcd(a,4k+1)=1. Then m=2da'b', where , , and . Let r/2=bb'. Then

n-2=2da'b'+2ea = 2da'(b'+2e-da/a').

Let and note that . The proof now breaks into two subcases depending on the congruence class of b' modulo 4.

SUBCASE 2.1. SUPPOSE THAT #MATH145#. Observe that because b'=1 implies that which is a contradiction. Let and note that s is odd. Consider the walk P0,0 defined by

Since , we have that the vertices of P0,0 are distinct so that P0,0 is a path. In the case that s=3, we have b'=5 and , and P0,0=u-3,u0,u2,u-2,u3. Observe that the lengths of the edges of P0,0 are . Obtain the path P0,1 as before, that is, . Since , the paths P0,0 and P0,1 are vertex-disjoint. Join the first vertex of P0,1 to the last vertex of P0,0 and note that this edge has length 1.

Again, let the powers of act on P0,0 to obtain the paths P0,0, P0,1, , P0, 2da'-1. Form the m-cycle C0by joining the last vertex of P0,i to the first vertex of P0,i+1for and joining the last vertex of P0,2da'-1to the first vertex of P0,0. Necessarily, these edges all have length 1. Let .

If b > 1, then obtain the path P1,0 from P0,0 by subtracting from the subscript of the first vertex and adding to the subscripts of the remaining odd vertices of P0,0, that is,

Then the lengths of the edges of P1,0 are . As before, let the powers of act on P1,0 to obtain the paths . Form the m-cycle C1 by joining the last vertex of P1,i to the first vertex of P1,i+1 for and joining the last vertex of P1,2da'-1 to the first vertex of P1,0. Note that these edges have length .

Similarly, for , obtain the path Pj,0 from P0,0by subtracting from the subscript of the first vertex and adding to the subscripts of the remaining odd vertices of P0,0. Again, letting the powers of act on Pj,0, we form the m-cycle Cj by joining the last vertex of Pj,i to the first vertex of Pj,i+1 for and joining the last vertex of Pj,2da'-1 to the first vertex of Pj,0. Observe that these edges have length and thus the cycle Cj has edges with lengths .

Let for , and let Then the longest edge of B has length either or . Now r= 2bb' and m = 2da'b', and since r < m, we have that 2b< 2da'. Thus, . Therefore, the lengths are all distinct. Next,

so that B consists of r/2 distinct lengths. Hence, the collection

is a partition of the edge set of the circulant into m-cycles.

SUBCASE 2.2. SUPPOSE THAT #MATH178#. Let P0,0 be the path defined by

Note that P0,0 uses exactly one edge of each length and uses no edge whose length is congruent to modulo . Consider the paths

and note that P0,i begins at vertex and ends at .

There are two special cases that must be considered. When b'=7, let

When b'=3, the condition m>r=2ea+2 forces a'=a and d=e-1 which in turn implies . The initial path P0,0=u0,u3,u-4 works in this case.

If b > 1, then obtain the path P1,0 by adding to the subscripts of the even vertices of P0,0, that is,

Next, obtain the paths by letting powers of act on P1,0 as before. Observe that these paths use edges of lengths .

Similarly, for each j with , obtain the path Pj,0by adding to the subscripts of the even vertices of P0,0. Then the paths are obtained by letting powers of act on Pj,0. For , let denote the collection of paths and Bj denote the set of lengths that uses, that is,

Observe that has b(b'-1) lengths, and thus we must add b lengths to B to use r/2 lengths. The longest edge used thus far has length or . Since r < m implies b < 2d-1a', it follows that . Thus, the lengths are distinct. Furthermore, since (n-2)/2 is an even multiple of , we have that these lengths are all less than (n-2)/2 as well. In Subcase 2.1, we saw that regardless of the congruence class of b' modulo 4. Thus B consists of b(b'-1)distinct lengths.

Recall that is the initial vertex of Pj,i and is the terminal vertex of Pj,i for and . Now we shall link the paths of to form an m-cycle. We shall use Hamilton cycles in an auxiliary circulant graph G to do so. Let the vertices of G be labeled . Suppose we are given a Hamilton cycle C in G. Arbitrarily orient the cycle C to obtain a Hamilton directed cycle . If there is an arc from vi to vk in (of length k-i), then insert an edge, necessarily of length , from the terminal vertex of Pj,i to the initial vertex of Pj,k. If we do this for each arc in , it is easy to see that the paths of are joined together to form an m-cycle. Furthermore, since the edge from the terminal vertex of Pj,i to the initial vertex of Pj,k has different length, namely , than the edge from the terminal vertex of Pj,k to the initial vertex of Pj,i, we can use the reverse orientation of to link the paths in another collection , , into an m-cycle. Observe that all the edges used to link together the paths have lengths congruent to modulo and thus none of them have been used in the construction of the families .

There are several points we need to consider. First, if a Hamilton cycle in G uses exactly one length for its edges, then it is easy to see that it can be used to link together the paths of some to form an m-cycle and, in doing so, we use exactly one length in H. However, if a Hamilton cycle in G uses different lengths for its edges, then using either orientation to link together paths of some , results in some edges of lengths congruent to modulo not being used. We handle this as follows.

J.-C. Bermond, O. Favaron and M. Maheo [3] proved that any connected circulant graph of degree 4 can be decomposed into two Hamilton cycles. So we choose two lengths s1 and s2 in G such that gcd (s1,s2)=1, which guarantees that the circulant graph with connection set is connected. We decompose it into two Hamilton cycles and orient each of them in the two possible ways. We then use these four oriented cycles to form four m-cycles in H. This then uses all the required edges in H of lengths congruent to modulo corresponding to the edges used in the Hamilton cycles. Second, when we are left with 1 or 3 families to link together, we use one or two Hamilton cycles from G with edges of the same length to complete the linking process.

We now must show that there are enough edge-disjoint Hamilton cycles in the auxiliary graph G to make the above scheme work. Recall that the number of vertices in G is 2da' and that 2da'>2b. We now proceed depending on the number of families to link together. If b=1, then let so that Gis a (2da')-cycle. In this case, the length is used in H to link into an m-cycle. If b=3, then G has at least eight vertices so that there are at least two lengths s1 and s2 relatively prime to 2da'. Then each of s1 and s2generates a Hamilton cycle in and thus we can easily link together the three families of paths into m-cycles. In this case, the lengths , and are used in H to link into m-cycles.

Finally, if b >3, then write b = 4q + 1 or b = 4q + 3 for some positive integer q. Suppose first that b = 4q + 1. Then can be used to link into an m- cycle using the length in H. Next take q successive pairs , all of which generate connected circulants, to obtain two Hamilton cycles in , for , to link together four families at a time. Then the lengths are used in H. If b = 4q + 3, then find a second integer relatively prime to 2da' to handle the two extra families, and the lengths and are used in H.

Upon obtaining all the m-cycles as described above, we then let act on each m-cycle C according to resulting in a decomposition of into m-cycles. This completes the proof of Theorem 1.1.

Brian &
2000-01-20