Automorphic Decompositions of Graphs

Graphs and Combinatorics (2011) 27:149–160DOI 10.1007/s00373-010-0981-2

ORIGINAL PAPER

Automorphic Decompositions of Graphs

Robert A. Beeler · Robert E. Jamison

Received: 15 May 2009 / Revised: 11 May 2010 / Published online: 14 September 2010© Springer 2010

Abstract A decomposition D of a graph H by a graph G is a partition of the edgeset of H such that the subgraph induced by the edges in each part of the partition isisomorphic to G. The intersection graph I (D) of the decomposition D has a vertexfor each part of the partition and two parts A and B are adjacent iff they share a com-mon node in H . If I (D) ∼= H , then D is an automorphic decomposition of H . In thispaper we show how automorphic decompositions serve as a common generalization ofconfigurations from geometry and graceful labelings on graphs. We will give severalexamples of automorphic decompositions as well as necessary conditions for theirexistence.

Keywords Graph decompositions · Graph designs · Graph isomorphism ·Block design

Mathematics Subject Classification (2000) 05C51 · 05C62 · 05B05

1 Introduction

All graphs in this paper are finite, undirected, simple graphs with no isolated verti-ces. We will use notation consistent with [6,19]. In particular, for any graph G, n(G)

denotes the the number of vertices in G and e(G) denotes the number of edges in G.

R. A. Beeler (B) · R. E. JamisonDepartment of Mathematics, East Tennessee State University, Johnson City,TN 37614-1700, USAe-mail: [email protected]

R. E. Jamisone-mail: [email protected]

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150 Graphs and Combinatorics (2011) 27:149–160

As usual, Kn, Cn, and Pn will denote, respectively, the complete graph, the cycle, andthe path on n-nodes. However, the degree of a vertex v in a graph G will be denotedby degG(v).

A decomposition D of a graph H (called the host) is a partition of the edge setof H . The subgraphs of H induced by the parts of the partition are the blocks of thedecomposition. By definition, blocks are edge-disjoint but in general, they may sharecommon nodes. The intersection graph I (D) of the decomposition D has a vertexfor each block and two vertices are adjacent if and only if they share a common node.To help separate the levels of abstraction, the term “vertex” will be used for verticesof the intersection graph and the term “node” will be used for vertices of the hostgraph and its blocks. For clarity, it is also helpful to refer to I (D) as the decomposi-tion graph of D [7].

A decomposition D is automorphic iff the host H and the decomposition graphI (D) are isomorphic. D is complete iff I (D) is a complete graph. D is simple iffany two blocks share at most one common node. A decomposition is cyclic iff thehost H is (isomorphic to) a graph with node set Zn (the integers modulo n) in whichthe translations x → x + t are automorphisms of H carrying blocks into blocks.A decomposition is geometric iff all of its blocks are complete graphs. Note that ageometric decomposition must be simple.

A decomposition D is uniform iff all of its blocks are isomorphic. The commonisomorphism type of a block is the prototype for the decomposition. A uniform decom-position D with prototype G is said to be a G-decomposition of H . In this case, wesay that G is a divisor of H . If there is an automorphic decomposition of H , then His an automorphic host. Similarly, if there is an automorphic G-decomposition of H ,then G is an automorphic divisor of H . If a graph H has a geometric automorphicG-decomposition, then the prototype must be G = Kn for some n.

The goal of this paper is to gain insight into the possible structures of automorphichosts and automorphic divisors. There are two quite different classes of examples con-structed in this paper. The first class involves the decomposition of regular graphs via acomplete prototype. These examples arise from configurations in incidence geometry.The second class involves decompositions of circulants via graph labelings, such asgraceful labelings. The extent to which these two classes account for all automorphicdecompositions is not known. However, an important step is taken in the main result ofthis paper. Theorem 2.1 states that if H admits a simple automorphic decompositionover a regular prototype G, then H must be regular and G must be complete.

Several simple examples are in order. Consider the complete graph on thenodes 0, 1, 2, 3, 4. The following five ordered triples represent 3-node paths:013, 124, 230, 341, 402. These paths yield an automorphic P3-decomposition of K5.This decomposition is cyclic but not simple. It is a representative of the class of auto-morphic decompositions arising from circulants.

Figure 1 shows an automorphic K3-decomposition of the complement of the dis-joint union of C3 with C6. On the left, the C3 is labeled by a, b, c and the C6 is labeled0,1,2,3,4,5. The blocks are shown on the right; the graph is their intersection graph.This example arises from configurations. It is simple, but does not have a transitiveautomorphism group because the two cycles have different length. Thus it cannot becyclic.

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Graphs and Combinatorics (2011) 27:149–160 151

Fig. 1 (93)3 : C3 ∪ C6 and a K3-decomposition graph

The Fano plane generated cyclically from 013 modulo 7 yields an example that isboth cyclic and simple, where the lines determine K3-blocks.

To conclude this introduction, we give a number of useful observations about gen-eral automorphic decompositions.

Bosák [6] gives a list of necessary conditions for a decomposition. Not surpris-ing these are only necessary but not sufficient. We now present two necessary (butnot sufficient) conditions for the existence of automorphic decompositions. Furtherconditions are found in [3]. Here d(H) denotes the average degree of H .

Lemma 1.1 Suppose D is an automorphic G-decomposition of H. Then

(i) e(H) = n(H)e(G);(ii) The average degree of H is d(H) = 2e(G).

Proof Let D be a G-decomposition of H .

(i) Note that by definition of I (D), n(I (D)) is the number of G-blocks in D . FromBosák [6], we have that n(I (D)) = e(H)/e(G). For D to be automorphic, wemust have that n(I (D)) = n(H). It follows that e(H) = e(G)n(H).

(ii) Using (i), note that

d(H) = 1

n(H)

∑

v∈V (H)

degH (v) = 2e(H)

n(H)= 2n(H)e(G)

n(H)= 2e(G).

��

Lemma 1.1 allows us to show that several families of graphs do not have automor-phic decompositions.

Corollary 1.2 (i) No tree is an automorphic host;(ii) Any regular automorphic host must have even degree of regularity.

Proof The average degree in a tree of order n is 2 − 2n < 2. The other claim is

immediate. ��

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2 Simple Automorphic Decompositions

In this section we place the additional restriction that the decomposition is simple, i.e.,given any two G-blocks in the decomposition, these blocks share at most one commonnode in H . If D is a simple G-decomposition of H such that I (D) ∼= H , then we saythat D is a simple automorphic decomposition. If such a decomposition exists, we saythat G is a simple automorphic divisor of H and that H is a simple automorphic host.

Theorem 2.1 Let H and G be simple graphs where G is d-regular and of order p. IfG is a simple automorphic divisor of H, then it follows that H is p(p − 1)-regular.Further, G must be isomorphic to K p.

Proof For convenience of notation, let n(H) = n. Let D be an automorphicG-decomposition of H . Since G is d-regular and G is a divisor of H , by Bosák[6], we have d|degH (v) for all v ∈ V (H). Let the degree sequence of H be given bydseq(H) = [d1, . . . , dn]. Assume the degrees di are listed in nondecreasing order,and let vi be node of degree di . Since the decomposition is automorphic, let Gi be theG-block corresponding to vi under the isomorphism between H and I (D). Then di

is the degree of Gi in I (D) and dseq(H) is also the degree sequence of I (D). Thusfor each i , there is a G-block Gi ∈ D that intersects exactly di other G-blocks.

For each node vi of degree di in H , exactly did G-blocks meet at vi . Thus, every

G-block containing vi intersects exactly d∗i := di

d − 1 other G-blocks at vi . We willcall the transformed value d∗

i the weight at vi . Let T (H) be the sequence of thesetransformed values. The weight of a node in a block Gi is exactly the number of otherblocks that share that same node in H . Since the decomposition is simple, the sum ofthese values gives the degree of the corresponding vertex in I (D). Thus, the sum ofthe weights on the nodes of Gi must be di . Also, for any given node vi of degree di ,exactly d∗

i + 1 G-blocks meet at vi . Thus, each d∗i appears as a weight exactly d∗

i + 1times.

Let us explode the decomposition D . That is, we pull apart the blocks into separatepieces. Each node x will be represented by d∗

i + 1 clones because we need one clonefor each block containing x . Each clone will carry the same weight as the original.Each d∗

i appears as a node label exactly d∗i + 1 times. Thus if degree di appears with

multiplicity mi in the degree sequence, then:

there are exactly mi (d∗i + 1) clones with weight d∗

i (1)

in the exploded set of blocks. Now d1 is the smallest degree in dseq(H). Every degreeis a multiple of d, so r = d1/d is integral. Note that r − 1 is the transformed valued∗

1 = r −1. Suppose rd occurs k times as a degree. Let r be the minimum degree in H .Suppose that there are k nodes in H that are of degree rd, i.e., r −1 appears in T (H)

exactly k times. Let G1, . . . , Gk be the G-blocks representing vertices of degree rdin I (D). Note that we have exactly kr copies of r − 1 with which to label the nodesof G1, . . . , Gk and no node may have a weight smaller than r − 1.

By choice of notation, G1, . . . , Gk are the blocks representing vertices of I (D)

with the k lowest degrees—namely, all of degree d1. Let i ∈ {1, . . . , k}. Suppose that

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there are at most r − 1 nodes of Gi with weight r − 1. If sum(Gi ) is the sum of theweights of nodes in Gi , then

sum(Gi ) ≥ (r − 1)2 + (p − r + 1)r = r(p − 1) + 1 > rd.

However, the weights on Gi should sum to exactly d1 = rd, a contradiction. Thuseach Gi with i ∈ {1, . . . , k} must have at least r clone-nodes weighted r − 1. Usingthe fact (1) above, r − 1 = d∗

1 occurs as a weight exactly m1(d∗1 + 1) = kr times.

Hence each Gi with i ∈ {1, . . . , k} has exactly r clone-nodes weighted r − 1. Theremaining nodes of Gi must all be weighted at least r . This implies that

rd = sum(Gi ) ≥ r(r − 1) + (p − r)r = r(p − 1) ≥ rd.

Note that equality is forced throughout, so the remaining p − r weights on Gi mustall be r . This will be used below. Furthermore, p − 1 = d or the last inequality wouldbe strict. This implies that G ∼= K p.

Suppose that Gk+1, . . . , Gk+t are represented by the vertices of degree (r + 1)d inI (D). This means that r appears in T (H) exactly t times. Note that we have exactlyt (r + 1) copies of r with which to weight the clone-nodes of Gk+1, . . . , Gk+t . Sincethere are no weights smaller than r −1 and all of the r −1 weights have been allocatedto G1, . . . , Gk , it follows that the weights on the clone-nodes of Gk+1, . . . , Gk+t mustbe at least r .

Let j ∈ {k + 1, . . . , k + t}. Suppose that at most r nodes of G j are weightedwith r . This implies that

sum(G j ) ≥ r2 + (p − r)(r + 1) = (p − 1)(r + 1) + 1 > d(r + 1).

However, the weights on G j must sum to exactly d(r + 1), a contradiction. Thus G j

has at least r + 1 copies of r . This accounts for at least t (r + 1) copies of r on thenodes of Gk+1, . . . , Gk+t . But t (r + 1) is the number of weights r available. So eachof the t blocks Gk+1, . . . , Gk+t must have exactly r + 1 nodes of weight r . Thereforeall available weights r are used on Gk+1, . . . , Gk+t . But we saw above that each ofthe k blocks G1, . . . , Gk have r nodes weighted r − 1 and p − r nodes weighted r .This is a contradiction if r < p, so we must have r = p.

Now d1 = rd is the minimum degree of H . As seen above, the degree d of regularityof G is d = p − 1. Thus 2e(H) = ∑n

i=1 di ≥ n(H)d1 = n(H)rd = n(H)p(p − 1).

Dividing by the order of H , we see that the average degree of H satisfies d(H) =2e(H)/n(H) ≥ d1 = p(p − 1). By Lemma 1.1, d(H) = 2e(G) = 2e(K p) =p(p − 1). Combining we get equality throughout. In particular, all degrees are forcedto be d1 = p(p − 1). ��

Since any K p-decomposition must be simple, the corollary below follows directlyfrom Theorem 2.1.

Corollary 2.2 If K p is an automorphic divisor of H, then H is p(p − 1)-regular.

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Theorem 2.3 Suppose D is an automorphic G-decomposition of H. Let A be a vertexinduced subgraph of H.

(i) If any two blocks of G intersect in at most m nodes, then n(H) ≥ n(A)n(G) −2me(A);

(ii) If A is triangle-free and D is simple, then n(H) ≥ n(A)n(G) − e(A);(iii) n(G) ≤ χ(H), where χ(H) is the chromatic number of H.

Proof (i) There is a set of n(A) G-blocks that represent A in I (D). For v ∈ V (A),let Bv denote the G-block corresponding to v in I (D). Since v and Bv havethe same degree, Bv must intersect degA(v) distinct G-blocks. Each of theseblocks shares at most m nodes with Bv by hypothesis. Thus, there are at leastn(G)− m degA(v) nodes that belong to Bv but no other G-block. Let U denotethe union of all blocks Bv for v ∈ A. That is, U is the set of nodes used torepresent A. Letting U o be the set of nodes in U that belong to exactly oneBv, v ∈ A, we have

n(H) ≥ |U o| ≥∑

v∈V (A)

(n(G) − m degA(v)) = n(A)n(G) − 2me(A).

(ii) A shared node lies in just two blocks because A is triangle-free. Thus there is aone-to-one correspondence between shared nodes and edges of A, so the numberof shared nodes is e(A). Hence, there are n(A)n(G) − e(A) nodes in U . SinceU is a subgraph of H , the desired inequality follows.

(iii) H must contain an independent set A of at least n(H)/χ(H) nodes. Since noedges are induced by A, from i) we have n(H) ≥ n(A)n(G) ≥ n(H)

χ(H)n(G).

Dividing by n(H) and multiplying by χ(H) gives the desired result. ��

An automorphic divisor G of H such that χ(H) = n(G) is said to be fully auto-morphic. Fully automorphic decompositions will be investigated in [4].

Corollary 2.4 G is an automorphic divisor of a bipartite graph H iff G = K2 andH is a disjoint union of cycles.

Proof Suppose G is an automorphic divisor of a bipartite graph H . By (iii) of Theorem2.3, G must be K2. The line graph of H is thus the same as the decomposition graphand this in turn is isomorphic to H . Since this decomposition is simple, it follows fromTheorem 2.1 that H must be 2-regular. Hence H is a disjoint union of cycles. Theconverse is obvious. ��

The existence of simple automorphic K p-decompositions is guaranteed by [5].However, it may be difficult to find examples of simple automorphic divisors that arenot complete. Many of our examples are based on cyclic decompositions in which theblocks may intersect multiple times. As such, we conjecture that K p is the only simpleautomorphic divisor.

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3 Geometric Automorphic Decompositions

Our goal in this section is to establish the link between decompositions with completeprototypes and certain geometric structures. We will look first at the more generalnon-uniform situation where the blocks may be complete graphs of different sizes.The corresponding geometric notions are those of near-linear space [1,2] and config-uration [8,10,11]. As defined in [1], a near-linear space is a set X of points togetherwith a family L of subsets of X called lines such that each line contains at least twopoints and each pair of points lies on at most one line. Note that this last conditionis equivalent to the assertion that every pair of lines intersects in at most one point.If Y is a subset of X , then the restriction L |Y of L to Y consists of all sets of theform L ∩ Y where L ∈ L and |L ∩ Y | ≥ 2. (Y,L |Y ) is again a near-linear space [1].For each x ∈ X , the pencil through x consists of all lines containing x . A pencil isnontrivial iff it contains at least two lines. The dual of (X,L ) has the lines L as pointset and the nontrivial pencils as lines. The dual is again a near-linear space [1]. Thegraph �(X,L ) of a near-linear space has the points of X as nodes with two points pand q adjacent iff they are collinear—that is, lie on a common line. The complementof �(X,L ) will be denoted by �(X,L ), and the graph of the dual will be denotedby �∗(X,L ).

Two near-linear spaces are isomorphic iff there is a bijection between their pointsets which preserves collinearity. A linear space that is isomorphic to its dual is self-dual. A linear space is automorphic iff �(X,L ) is isomorphic to �∗(X,L ). Clearlya self-dual space is also automorphic. However, as we shall see, the converse fails.

Every near-linear space corresponds to a geometric decomposition of its graph.Namely, take the blocks to be the complete graphs on node sets of the lines. Sincetwo lines intersect in at most one point, these blocks are edge disjoint. By definitionthey cover the edge set of �(X,L ). Conversely, let D be any decomposition of ahost H into complete subgraphs. Declare the nodes in each block to be a line. Thenthese lines together with the node set of H form a near-linear space. Clearly, these twoconstructions are inverses of each other.

Lemma 3.1 Suppose D is a geometric decomposition of a host graph H, and (X,L )

is the associated near-linear space. Then the intersection graph for D is the graph ofthe dual of (X,L ). Further, the decomposition D is automorphic iff the associatednear-linear space is automorphic.

Proof If two blocks of D intersect at a node v of H , then their associated lines are inthe pencil at v and hence are dual “points” on the same dual “line”. Hence they areadjacent in the graph of the dual. Conversely, if two dual points are adjacent in thegraph of the dual, they correspond to two blocks in H meeting at a node in H . ��

Following Dembowski [8], Gropp [10], and Grünbaum [11], (vr )-configuration isa near-linear space with v points, v lines, r lines through each point and r points oneach line. A (vr )-configuration is often referred to simple as a (vr ).

Lemma 3.2 Suppose D is a uniform, automorphic, geometric decomposition of ahost graph H of order v. The associated near-linear space is an automorphic (vr )-configuration where Kr is the prototype. Conversely, the decomposition associated

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Table 1 Some Self-dual (vr )-configurations

(103) (93)1 3C3 (93)2 C9 (93)3 C3 ∪ C6

A : bcd O : de f 012 057 037 247 a03 c04

A′ : b′c′d D : oaa′ 345 138 268 136 b14 a15

B : ace E : obb′ 678 246 157 025 c25 b02

B′ : a′c′e F : occ′ 048 046 148 c13

C : ab f 156 358 a24

C ′ : a′b′ f 237 b35

with an automorphic (vr )-configuration is a uniform automorphic decomposition withKr as the prototype.

Proof If a decomposition has a single prototype G = Kr , then all the lines in the asso-ciated near-linear space have the same number of points, namely, r . By Corollary 2.2,the host H is regular, say, of degree d. Let x be a node of H . Each neighbor y of xlies in a line with x since the edge xy must be covered by a block. Since these blocksare edge disjoint, they partition the neighbors of x . Thus d = t (r − 1) where t is thenumber of lines through x . Since d and r are independent of x , so is t . By countingnode-block incidences, we have vt = br , where b is the number of blocks. Since Dis automorphic, the number of blocks must equal the number of points. Hence t = rfollows from vt = br . Thus the associated near-linear space is a (vr )-configuration asclaimed. The converse follows essentially from the definitions. ��

Table 1 gives several examples of automorphic decompositions. There is only one(73)-configuration, namely the Fano plane. Likewise, there is a unique (83) (K8 minusa matching). There are three types of (93) up to isomorphism indicated by the secondsubscript. The subscripts used in Table 1 to distinguish these (93)’s follows Dorwart[9]. Many more examples of (vr )-configurations can be found in Grünbaum [11]. In allfour examples, the configuration is self-dual and hence automorphic. The (103) is theDesargues configuration and the (93)1 that of Pappus. In the Desargues’ configuration,points are in small letters and lines are denoted by capitals. The self-duality is givenby a ↔ A, a′ ↔ A′, b ↔ B, etc.

Now suppose we have a graph H in Kr , the class of graphs which possess anautomorphic decomposition with a single, complete prototype Kr . By Corollary 2.2,such a graph must be regular of degree d = r(r − 1). Solving for r , we get r =(1 ± √

1 + 4d)/2. Since the discriminant is bigger than 1, only the + sign yields apositive solution and r is uniquely determined by the degree of H . Thus for any graphH there is only one possible complete automorphic divisor.

Observe that a projective plane of order n is an [n2 + n + 1]n+1-configuration andvice versa. Furthermore, a projective plane of order n is automorphic since its dualis itself a projective plane of the same order, and their associated graph is Kn2+n+1.Moreover, for H = Kn2+n+1 the unique (possible) divisor is Kn+1, then applyingLemma 3.2, the next proposition follows:

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Proposition 3.3 A uniform automorphic decomposition of Kn2+n+1 exists if and onlyif there exists a projective plane of order n.

The existence of a projective plane of order n is guaranteed whenever n = pk is apower of a prime p, but is still an open problem if n is not of this type [16]. It is alsowell-known that when k > 1, non-isomorphic planes of the same order n = pk exist[16]. Thus although the divisor is unique, the decomposition need not be. Moreover,there are projective planes that are not self-dual, such as the Sherk plane of order 27[16,18]. Such a plane will be automorphic but not self-dual.

Proposition 3.4 Let H be Kn2 with the edges of n disjoint copies of Kn removed. Anyflag in a projective plane of order n gives rise to an automorphic Kn-decompositionof H.

Proof Consider the affine plane P∗ obtained from P by deleting the line L togetherwith its points and let Lp be the class of parallel lines of P∗ corresponding to thepencil through p. Consider the incidence structure (X,L ) where the point-set X isthe point set of P∗ and the line set L is the set of lines of P∗ minus those of Lp.It is easy to show that (X,L ) is a [n2]n-configuration. The graph �(X,L ) is thecomplement of nKn in Kn2 and the graph of its dual is clearly the same. Now applyLemma 3.2 and the result follows. ��

In the non-Desarguesian case the two affine planes created in the proof may fail tobe isomorphic. But even if they do, they yield isomorphic graphs, which is all that isrequired for the decomposition to be automorphic. Note that when n = 3, this yieldsthe Pappus configuration (93)1.

Proposition 3.5 Let P be a projective plane of order n. Let H be Kn2−1 withthe edges of n + 1 disjoint copies of Kn−1 removed. Then H has an automorphicKn-decomposition.

Proof Let (L , p) be an antiflag in P. Removing the points of L from P and removingp from P leaves n2 − 1 points that can be taken as the vertices of a Kn2−1. Sincep /∈ L , there are n +1 lines through p other than L . Each of these lines contains n −1points of H = Kn2−1. Each such line together with the edges of Kn2−1 on it is a Kn−1.Thus removing these edges over all lines through p removes n + 1 disjoint copies ofKn−1. The blocks of the decomposition are determined by the remaining n2 − 1 lines.The situation with the lines is analogous, so the intersection graph is also a Kn2−1.Thus this decomposition is automorphic. ��

Note that when n = 3, Proposition 3.5 yields the unique (83)-configuration. Thetheory of conics and polarities in Desarguesian planes leads to a further construction.

Proposition 3.6 If n is a prime power, then an automorphic decomposition ofKn(n−1)/2 with prototype K(n+1)/2 exists.

Proof Let C be a conic in P = PG(2, n) where n is a prime or a prime power. Thepoints of P not on C are classified as external or internal according to whether or not

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they lie on a tangent to the conic. The non-tangent lines of the conic are classified assecant or external according as they contain two or no points of the conic. There is apolarity [15] which interchanges secants and external points as well as interchangingexternal lines with internal points.

Again from [15], there are n(n −1)/2 internal points and each internal point lies on(n + 1)/2 external lines. By the polarity, there are n(n − 1)/2 external lines and eachexternal line contains (n + 1)/2 internal points. Let (X,L ) be the linear space whosepoints are the n(n − 1)/2 internal points and in which two points are collinear iff theline through them in P is an external line. The polarity insures that this linear spaceis self-dual. Hence by Lemma 3.1, the associated decomposition is an automorphicdecomposition of Kn(n−1)/2 with prototype K(n+1)/2. ��

4 Decompositions Involving Circulants

In discussing circulants over Zn , it is customary to deal only with “positive” differ-ence, where “positive” means between 0 and �n/2�. To make this precise, it is usefulto introduce a discrete absolute value modulo n:

|x |n ={

x if 0 ≤ x ≤ n/2;n − x otherwise.

Let G be a graph with e(G) = q. A Zn-labeling of G is an injective map f : V (G) →Zn . Any Zn-labeling f of G induces an edge labeling f ∗ on E(G) by f ∗(xy) :=| f (x) − f (y)|n for all edges xy ∈ E(G). A Zn-valuation is a Zn-labeling f of G forwhich the induced edge labeling f ∗ is injective and n/2 does not appear as an edgelabel. A Zn-valuation is closed if f ∗(G) = f ∗(E), where: f ∗(E) = { f ∗(xy) : xy ∈E} and f ∗(G) = {| f (x) − f (y)|n : x, y ∈ V, x �= y} [5].

Rosa [17] introduced several types of valuations, including ρ-valuations andβ-valuations. If G is a graph of size q, then a ρ-valuation is a Z2q+1-valuation onG. A β-valuation on G is a Z2q+1-valuation in which all of the vertex labels comefrom {0, 1, . . . , q}. Rosa’s β-valuations were popularized by Golomb under the namegraceful labelings [13]. It is clear from the above definitions that a ρ-valuation isa closed Z2q+1-valuation and that a graceful labeling is a closed Zn-valuations forall n ≥ 2q + 1. Also, any valuation on a complete graph is trivially closed [5]. Anextensive list of graphs that admit graceful and related labelings can be found in [12].

Let S be a set of numbers in Zn , all lying between 1 and n/2, inclusive. The circu-lant Cn(S) has vertex set V := Zn and edge set E := {xy : x, y ∈ V, |x − y|n ∈ S}[14]. We will use valuations and circulants to construct an infinite class of automorphicdecompositions.

Theorem 4.1 G is a cyclic automorphic divisor of Cn(S) iff there exists a closedZn-valuation f on G such that S = f ∗(E).

Proof In [5], it is shown that a graph G is a cyclic divisor of a circulant H = Cn(S)

iff there is a Zn-valuation f on G such that S = f ∗(E). We need only describe theintersection graph generated by this decomposition.

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Let ba ∈ D be a G-block such that V (ba) = {a, a + a1, . . . , a + ap}. Since D iscyclic, it follows that ba intersects blocks of the form bk where k = a±(a j −ai ). Labelthe vertices of I (D) according to the index of their corresponding block. Clearly, thereare n such values. If a ∈ V (I (D)), then a intersects vertices of the form a ± k wherek ∈ S′ = f ∗(G). Thus, it follows that I (D) ∼= Cn(S′). Since f is closed, it followsthat this decomposition is automorphic. ��

It should be noted that this implies that the same automorphic host may have dif-ferent automorphic divisors. For example, the path on four vertices, the star on fourvertices, and K3 are graceful graphs of size 3. Hence they are automorphic divisors ofCn(1, 2, 3) for n ≥ 7. This contrasts with the observation in Sect. 3 that a graph canhave at most one complete automorphic divisor.

While valuations and circulants give an infinite class of automorphic decomposi-tions, not all automorphic decompositions arise this way. For example, the graph of thePappus configuration is the circulant C9(1, 2, 4). However, there is no Z9-valuationon K3 that will generate this decomposition. The automorphic decomposition in Fig. 1provides an example of a non-circulant automorphic host. Since the connected compo-nents of a circulant must have the same order, it follows that C3 ∪C6 is not isomorphicto any circulant. Since the complement of a circulant is likewise a circulant, it followsthat C3 ∪ C6 is not a circulant.

5 Open Problems

There are three quite natural problems on automorphic decompositions which so farhave eluded solution.

(i) Is every graph an automorphic divisor for some suitable host?(ii) Must an automorphic host must be regular of even degree?

(iii) Are complete graphs the only simple automorphic divisors?

Acknowledgments The authors wish to thank the anonymous referee for their thoughtful and thoroughcomments which aided in the exhibition of this paper.

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Automorphic Decompositions of Graphs