Recognizing Bipartite Incident-Graphs of Circulant Digraphs

Recognizing Bipartite Incident-Graphsof Circulant DigraphsJohanne Cohen� Pierre Fraigniaudy Cyril GavoillezAbstractGiven a circulant digraph H = (V;A) of order n and of generators 0; : : : ; k, 0 � i � n�1,i = 0; : : : ; k, the bipartite incident-graph of H is a bipartite graph G = (V1; V2; E) of order 2nwhere V1 = V2 = V , and, for any x1 2 V1, and any x2 2 V2, fx1; x2g 2 E , (x1; x2) 2 A ,9i 2 f0; : : : ; kg j x2 = x1 + i (mod n). Kn�odel graphs and Fibonacci graphs are two typesof such graphs. They correspond to i = 2i � 1, and i = F (i + 1) � 1, respectively. Bothgraphs have been extensively studied for the purpose of fast communications in networks, andthey have deserved a lot of attention in this context. In this paper, we show that there existsa polynomial-time algorithm to recognize Kn�odel graphs, and that the same technique appliesto Fibonacci graphs. The algorithm is based on a characterization of the cycles of length six inthese graphs (bipartite incident-graphs of circulant digraphs always have cycles of length six).A consequence of our result is that none of the Kn�odel graphs are edge-transitive, apart thoseof 2k � 2 vertices.An open problem that arises in this �eld is to know whether a polynomial-time algorithmexists for any in�nite family of bipartite incident-graphs of circulant digraphs indexed by theirnumber of vertices.Keywords: graph isomorphism, circulant graphs, chordal rings, broadcasting, gossiping.�Laboratoire de Recherche en Informatique, Bat. 490, Univ. Paris-Sud, 91405 Orsay cedex, France. Additionalsupport by the DRET of the DGA.yLaboratoire de Recherche en Informatique, Bat. 490, Univ. Paris-Sud, 91405 Orsay cedex, France. Additionalsupport by the CNRS.zLaboratoire Bordelais de Recherche en Informatique, Univ. Bordeaux I, 33405 Talence cedex, France.

1 IntroductionSo-called Kn�odel graphs and Fibonacci graphs have been used by Kn�odel [11], and Even andMonien [6] (see also [4, 12, 17]), respectively, for the purpose of performing e�cient communicationsin networks. More precisely, consider a network of n nodes, and assume that communications amongthe nodes proceed by a sequence of synchronous calls between neighboring vertices. A round isde�ned as the set of calls performed at the same time. Kn�odel on one hand, and Even and Monienon the other hand, were interested in computing the minimum number of rounds necessary toperform a all-to-all broadcasting, also called gossiping, between the nodes (see [7, 8, 10] for surveyson gossiping and related problems). The communication constraints assume that a call involvesexactly two neighboring nodes, and that a node can communicate to at most one neighbor at atime. Kn�odel considered the 2-way mode (full-duplex) in which the two nodes involved in the samecall can exchange their information in one round, whereas Even and Monien considered the 1-waymode (half-duplex) in which the information can ow in one direction at a time, that is, duringthe call between x and y, only y can receive information from x, or x can receive information fromy, not both. Under these hypotheses, it was shown in [11] that, for n even, one cannot performgossiping in less that dlog2 ne rounds in the 2-way mode, and that there are graphs, called hereKn�odel graphs, that allow gossiping to be performed in dlog2 ne rounds. Even and Monien haveshown in [6] that, for n even, one cannot perform gossiping in less than 2 + dlog% n2 e rounds in the1-way mode, where % = 1+p52 , and that there are graphs, called here Fibonacci graphs, that allowgossiping to be performed in that number of rounds (up to an additive factor of one).Kn�odel graphs and Fibonacci graphs are bipartite graphs G = (V1; V2; E) of 2n vertices. Eachpartition has n vertices labeled from 0 to n � 1. In the Kn�odel graphs, there is an edge betweenx 2 V1 and y 2 V2 if and only if there exists i 2 f0; 1; : : : ; kg such that y = x + 2i � 1 (mod n),k = blog2 nc. In the Fibonacci graphs, there is an edge between x 2 V1 and y 2 V2 if and only ifthere exists i 2 f0; 1; : : : ; kg such that y = x+ F (i+ 1)� 1 (mod n), k = F�1(n)� 1, where F (i)denotes the ith Fibonacci number (F (0) = F (1) = 1, and F (i) = F (i� 1)+F (i� 2) for i � 2) andF�1(n) denotes the integer i for which F (i) � n < F (i + 1). Both graphs are Cayley graphs onthe dihedral group, and thus they are vertex-transitive. See Figure 1 for an example of a Kn�odelgraph and a Fibonacci graph. Note that graphs on Figure 1 look pretty dense but Kn�odel graphsand Fibonacci graphs are of degree O(logn).Knowing whether a graph G of n nodes allows gossiping to be performed optimally, that isin dlog2 ne rounds in the 2-way mode, and in (about) dlog% ne rounds in the 1-way mode, is NP-complete [4]. In particular there are graphs that are not isomorphic to Kn�odel graphs (resp.Fibonacci graphs), and that allow gossiping to be performed optimally in the 2-way mode (resp.1-way mode). In this paper, we want to recognize Kn�odel graphs and Fibonacci graphs. In otherwords, given a graph G, we want to know whether G is isomorphic to a Kn�odel graph of the sameorder, or to know whether G is isomorphic to a Fibonacci graph of the same order.A closely related topic deals with circulant digraphs. Recall that a digraph is circulant ifnodes can be labeled so that the adjacency matrix is circulant, that is node x has k + 1 out-neighbors x + ai (mod n) for i = 0; : : : ; k for some k and some constants ai's independent of x.Circulant digraphs are Cayley digraphs over Zn. Ponomarenko has given in [16] a polynomial-timealgorithm to decide whether a given tournament is a circulant digraph (a tournament is a digraphobtained by giving an orientation to the edges of a complete graph). More recently, Muzychuk andTinhofer [14] have shown that one can decide in polynomial-time whether a digraph of prime order2

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0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9Figure 1: A Kn�odel graph of 20 vertices (a), and a Fibonacci graph of 20 vertices (b).is circulant. Deciding whether two circulant digraphs are isomorphic is also a di�cult problem.�Ad�am [1] conjectured that two circulant digraphs are isomorphic if and only if the generators ofone digraph can be obtained from the generators of the other digraph via a product by a constant.This conjecture is wrong [5] although it holds in many cases. For instance, Alspach and Parsons [2]have proved that �Ad�am's conjecture is true for values of n such as the product of two primes (seealso [3, 13, 15]).Nevertheless, even if they are closely related to circulant digraphs, Kn�odel graphs and Fibonaccigraphs are not circulant graphs but bipartite incident-graphs of circulant digraphs, and they arethus sometimes called bi-circulant graphs. (The bipartite incident-graph of a digraph H = (V;A)is a graph G = (V1; V2; E) such that V1 = V2 = V , and for any x1 2 V1, and x2 2 V2, fx1; x2g 2E , (x1; x2) 2 A. Note that two non isomorphic digraphs H and H 0 can yield isomorphic bipartiteincident-graphs, eg, H = (fu; vg; f(u; u); (v; v)g) and H 0 = (fu; vg; f(u; v); (v; u)g).) It is unknownwhether there exists a polynomial-time algorithm to decide whether a given graph is isomorphicto a given circulant digraph or a given incident-graph of a circulant digraph. This is why we havestudied speci�c algorithms for the case of Kn�odel graphs and Fibonacci graphs.The paper is organized as follows. In Section 2, we study the form of the solutions of equationsinvolving powers of two. The characterization of these solutions allows us to recognize Kn�odelgraphs in polynomial time, as shown in Section 3. Section 4 concludes the paper with some remarkson bipartite incident-graphs of circulant digraphs de�ned by an arbitrary increasing sequence ( i)i�0of integers, including Fibonacci graphs.2 Preliminary ResultsLet H = (V;A) be a circulant digraph of n vertices and of generators (also called dimensions) 0; : : : ; k, and let G = (V1; V2; E) be the corresponding bipartite incident-graph. By 6-cycle, wemean an elementary cycle of length six.Lemma 1 There is a 6-cycle in G if and only if one can �nd a sequence of six generators( i0 ; i1; i2; i3; i4; i5)and � 2 f0; 1; 2g such that� i0 + i2 + i4 = i1 + i3 + i5 + � n ij 6= ij+1 (mod 6) for any j 2 f0; 1; 2; 3; 4; 5g (1)3

Proof. Let (u0; u1; u2; u3; u4; u5) be a 6-cycle in G (all the ui's are pairwise distinct), and assumewithout loss of generality that u0 = 0 2 V1. We have:8>>>>>><>>>>>>: u1 = u0 + i0u1 = u2 + i1 + �1 nu3 = u2 + i2 + �2 nu3 = u4 + i3 + �3 nu5 = u4 + i4 + �4 nu5 = u0 + i5where �i 2 f�1; 0g and ij 6= ij+1 (mod 6) for any j 2 f0; 1; 2; 3; 4; 5g since the cycle uses sixdi�erent edges. Therefore we have i0 + i2 + i4 = i1 + i3 + i5 + � nwhere � = (�1 � �2) + (�3 � �4), that is �2 � � � 2 by de�nition of the �i's.Conversely, let ( i0 ; i1; i2; i3; i4; i5) be such that� i0 + i2 + i4 = i1 + i3 + i5 + � n ij 6= ij+1 (mod 6) for any j 2 f0; 1; 2; 3; 4; 5gwith � 2 f0; 1; 2g. Then let u0 2 V1, and let8>>>>>><>>>>>>: u1 = u0 + i0 mod nu2 = u1 � i1 mod nu3 = u2 + i2 mod nu4 = u3 � i3 mod nu5 = u4 + i4 mod nu6 = u5 � i5 mod nWe have u6 = u0 + ( i0 + i2 + i4)� ( i1 + i3 + i5) mod n:Since i0 + i2 + i4 = i1 + i3 + i5 (mod n);we get u6 = u0, and therefore (u0; u1; u2; u3; u4; u5) is a cycle of length six in G. This cycle iselementary because ij 6= ij+1 (mod 6) for any j 2 f0; 1; 2; 3; 4; 5g.From Lemma 1, any bipartite incident-graph of a circulant digraph has 6-cycles since i0 = i3 , i1 = i4, and i2 = i5 is a solution of Equation 1. We will solve Equation 1 to characterize 6-cyclesof Kn�odel graphs and Fibonacci graphs, and to identify the possible dimensions of a candidate tobe a Kn�odel graph or a Fibonacci graph.Let us start �rst with Kn�odel graphs. Let x0; x1; x2; x3; x4; x5 be six integers in f0; : : : ; kg, andlet n be any integer such that 2k � n < 2k+1. From Lemma 1, we are interested in computing thesolutions of the equation� 2x0 + 2x2 + 2x4 = 2x1 + 2x3 + 2x5 + � nxi 6= xi+1 (mod 6) for any i 2 f0; 1; 2; 3; 4; 5g (2)where � 2 f0; 1; 2g. 4

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(b) (c) (d)(a) Figure 2: The four types of solutions of Equation 2 for � = 0.Lemma 2 For � = 0, Equation 2 has four types of solutions: (x0; x1; x2; x3; x4; x5) =a) ( ; 00; 0; ; 00; 0) ; 0; 00 2 f0; : : : ; kg 6= 0; 0 6= 00 and 00 6= b) ( ; 0; ; 0; 0+ 1; + 1) ; 0 2 f0; : : : ; k� 1g 6= 0c) ( ; + 1; ; 0; 0+ 1; 0) ; 0 2 f0; : : : ; k� 1g 6= 0d) ( ; 0; ; + 1; 0+ 1; 0) ; 0 2 f0; : : : ; k� 1g 6= 0up to cyclic permutations1 of the xi's.Proof. The case x0; x2; x4 pairwise distinct generates the �rst type of solutions. Assume x0 =x2 = and x4 = 0 6= . There is an impossibility to solve Equation 2 if = 0 � 1 because itwould imply either x0 = x1 or x0 = x5. If 6= 0 � 1, we get a solution if two x2i+1's are bothequal to 0� 1, and the third one is equal to + 1. This generates the three last types of solutionsby changing 0 � 1 into 0.The solutions of Lemma 2 induces cycles of length 6. These 6-cycles have their edges labeledby the generator as illustrated on Figure 2. However, cycle (b) and cycle (d) are isomorphic (justtravel (b) clockwise and (d) counterclockwise from the black node), that is the second and thefourth types of solutions induces the same labeled cycle. In the following, only cycles (a), (b), and(c) will be considered.Notation. The number of blocks of consecutive 1's in the binary representation of n will bedenoted by B1(n).For instance B1((1101100111010)2) = 4, B1((100)2) = 1, B1((101)2) = 2, and B1((0)2) = 0.Integers of the formn = (100 : : :00| {z } 1 00 : : :00| {z } z }| {11 : : :11 0 z }| {11 : : :11 0 z }| {11 : : :11 00 : : :00| {z })2satisfy B1(n) = 5, and there is a solution to Equation 2 for � = 1 with x1; x3; x5 equal to theunderlined bit-positions, and x0; x2; x4 equal to the over-lined bit-positions. We have the followinglemma:Lemma 3 If B1(n) � 6, then Equation 2 has no solution for � 6= 0.1A permutation � of p symbols is a cyclic permutation if �(x1; x2; : : : ; xp) = (x2; : : : ; xp; x1).5

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γγFigure 3: The 4-cycles in a Kn�odel graph of order 2 � 2k.Proof. If B1(n) � 6, then the sum of n and three powers of two cannot result in the sum of threepowers of two. Indeed, the binary representation of 2x0+2x2+2x4 has at most three 1-entries, thatis B1(2x0 + 2x2 + 2x4) � 3. On the other hand, for every n, and every y, B1(n+ 2y) � B1(n)� 1,and thus B1(n + 2x1 + 2x3 + 2x5) � B1(n) � 3. Moreover, for n such that B1(n) � 6, the binaryrepresentation of n+2x1 +2x3 +2x5 has at least four 1-entries. This completes the proof for � = 1.The result holds for � = 2 too because B1(2n) = B1(n).Lemma 3 has an important consequence that is, for most of the integers n, Kn�odel graphs oforder 2n have 6-cycles only of the form given on Figure 2. There are orders however for whichEquation 2 has solutions for � 6= 0. This is typically the case for n = 2k. Actually, this special casedeserves a particular interest motivated by the simplicity of the solution.Lemma 4 If n = 2k then every 4-cycle of the Kn�odel graph of order 2n is a labeled cycle of type(k; � 1; ; � 1) for 2 f1; : : : ; kg:Proof. By similar arguments as in the proof of Lemma 1, one can check that 4-cycles exist ifand only if there exist four generators i = 2xi � 1, 0 � i � 3, xi 6= xi+1 (mod 4), satisfying2x0 + 2x2 = 2x1 + 2x3 + �n for � 2 f0; 1g. The equation 2x0 + 2x2 = 2x1 + 2x3 has no solutionfor xi 6= xi+1 (mod 4). Therefore, 4-cycles exist only for solutions of 2x0 + 2x2 = 2x1 + 2x3 + 2k,that is for x0 = k, and x1 = x3 = x2 � 1, x2 2 f1; : : : ; kg. Thus the solution is (x0; x1; x2; x3) =(k; � 1; ; � 1), up to a square-cyclic permutation2 of the xi's.3 Recognizing Kn�odel GraphsLet us start with n = 2k . From Lemma 4, we know that there is only one type of labeled 4-cycle ina Kn�odel graph of order 2n, namely (k; � 1; ; � 1), for 2 f1; : : : ; kg. Therefore, for any edgeof dimension k, there are k 4-cycles using that edge (see Figure 3(a)). For any edge of dimension , 2 f1; : : : ; k� 1g, there are three 4-cycles using that edge (see Figure 3(b)). Finally, for any edge2A permutation � of p symbols is a square-cyclic permutation if �(x1; x2; x3 : : : ; xp) = (x3; : : : ; xp; x1; x2).6

of dimension 0, there are two 4-cycles using that edge (the cycle 1 in Figure 3(b) does not exist for = 0). This counting yields the following corollary of Lemma 4:Corollary 1 There exists an O(n log3 n)-time algorithm to recognize Kn�odel graphs of order 2n =2k+1, k > 0.Proof. Assume k large enough. Given an input graph G = (V;E), we count the number C4(e)of 4-cycles passing through any edge e 2 E. From the counting of the number of 4-cycles passingthrough an edge in a Kn�odel graph, if there is an edge e such that C4(e) 6= 2, C4(e) 6= 3, andC4(e) 6= k then G is not a Kn�odel graph. Otherwise, every edge e with C4(e) = 2 is a candidateto be an edge of dimension 0, and every edge e with C4(e) = k is a candidate to be an edge ofdimension k. Then, looking at all 4-cycles containing the sequence of dimensions (0; k; 0), one canidentify edges that are candidates to be edges of dimension 1. More generally, knowing the edges ofdimension i, one can look at all 4-cycles containing the sequence of dimensions (i; k; i) to identifyall edges of dimension i+ 1.Now, consider the path P = (u0; : : : ; u2n�1) using dimensions 0 and k, alternatively, from anarbitrary node u0 2 V . If this path is not Hamiltonian, then G is not a Kn�odel graph. Else, labelevery node v of G by (j mod 2; b j2c) where v = uj . Then G is a Kn�odel graph if and only if an edgeof dimension i links a node (0; x) to a node (1; x+ 2i � 1).This algorithm has a cost of O(n log3 n) because, for every edge e, counting the number of4-cycles using that edge takes at most a time of O(log2 n) assuming adjacency testing in constanttime.Let us carry on our study by considering integers n such that B1(n) � 6. In this case, onecan apply Lemmas 2 and 3, and we are dealing with the four types of cycles of Figure 2 (recallthat cycles (b) and (d) are isomorphic). Figure 2(a) implies that, for any edge of dimension , 2 f0; : : : ; kg, there are k(k � 1) 6-cycles of type (a) using that edge.The contribution of cycles of type (b) in Figure 2 to each dimension is more di�cult to calculate.We proceed as for the 4-cycles studied in the case n = 2k. The counting for n = 2k can be formalizedas follows. Consider again Figure 3. On Figure 3(c), there are four edges (whose one is of a �xeddimension, dimension k), and two possible senses of travel, clockwise and counterclockwise. Forany such that 1 � � k � 1, there are potentially four positions for . However, only three ofthem are valid because 6= k. Moreover, once the position of has been �xed, we have only twoways to travel around the cycle. This fact gives six possible labeled cycles for any edge to belongto. However, only three of these 4-cycles are distinct because there is a symmetry along the axisperpendicular to dimension k. This symmetry reduces the number of travels by a factor of 2: foreach travel of Figure 3(c), there is a corresponding travel in Figure 3(b) starting in the directionindicated by the arrow.Coming back to the 6-cycle of Figure 2(b), there are six edges, and two possible directions(clockwise and counterclockwise). Thus we get twelve possibilities to travel along the edges of alabeled 6-cycle. However, we actually get only six possibilities for cycle (b) of Figure 2 because ofa symmetry along the axis parallel to ; 0 (see Figure 4(a)).7

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Figure 4: (a) The six possible travels of the cycle of type (b) in Figure 2. (b) The three possibletravels of the cycle of type (c) in Figure 2.These six possibilities are:1: ( 0 0 0 + 1 + 1)2: ( + 1 0 + 1 0 0)3: ( 0 + 1 0 + 1 0)4: ( 0 0 + 1 + 1 0) 9>>=>>; ; 0 2 f0; : : : ; k� 1g; 6= 0;and 5: ( � 1 0 � 1 0 0 + 1)6: ( 0 + 1 0 � 1 0 � 1) � 2 f1; : : : ; kg; 0 2 f0; : : : ; k� 1g; 0 6= � 1:Note that solutions 1 and 3 are the same if 0 = + 1, solutions 2 and 4 are the same if 0 = +1,solutions 3 and 4 are the same if 0 = � 1, and solutions 5 and 6 are the same if 0 = � 2.We do the same analysis for 6-cycles of type (c) in Figure 2. The counting uses the fact thatFigure 2(c) is symmetric along the axis perpendicular to the edges + 1 and 0+ 1, and along theaxis parallel to + 1; 0+ 1 (see Figure 4(b)). Therefore, we get three new possibilities:7: ( + 1 0 0+ 1 0)8: ( 0 0 + 1 0 + 1) � ; 0 2 f0; : : : ; k � 1g; 6= 0;and 9: ( � 1 0 0 + 1 0 � 1) ; 2 f1; : : : ; kg; 0 2 f0; : : : ; k� 1g; 0 6= � 1:Note that solutions 1 and 7 are the same if 0 = + 1, solutions 2 and 7 are the same if 0 = � 1,and solutions 3 and 7 are the same if 0 = � 1. Similarly solutions 1 and 8 are the same if 0 = � 1, solutions 2 and 8 are the same if 0 = + 1, and solutions 4 and 8 are the same if 0 = +1. Finally, solutions 5 and 9 are the same if 0 = � 2, and solutions 6 and 9 are the sameif 0 = � 2.Now we can count the number of 6-cycles using a given edge of a Kn�odel graph.8

Lemma 5 Assume B1(n) � 6, and let e be an edge of a Kn�odel graph of order 2n with k � 4. LetC6(e) denote the number of 6-cycles using edge e. We have:C6(n) = 8>>>><>>>>: k2 + 5k � 10 if e is of dimension 0k2 + 8k � 16 if e is of dimension 1k2 + 8k � 18 if e is of dimension , 2 � � k � 2k2 + 8k � 14 if e is of dimension k � 1k2 + 2k � 5 if e is of dimension kProof. Let e be an edge of dimension , 0 � � k. Let us count the contribution to that edgeof the nine travels identi�ed in Figure 4. We get the following table: = 0 = 1 2 � � k � 2 = k � 1 = ktravel # cycle travel # cycle travel # cycle travel # cycle travel # cycle1 k � 1 1 k � 1 1 k � 1 1 k � 1 1 02 k � 1 2 k � 1 2 k � 1 2 k � 1 2 03 k � 2 3 k � 2 3 k � 2 3 k � 1 3 04 k � 2 4 k � 3 4 k � 3 4 k � 2 4 05 0 5 k � 1 5 k � 1 5 k � 1 5 k � 16 0 6 k � 1 6 k � 2 6 k � 2 6 k � 27 k � 2 7 k � 3 7 k � 3 7 k � 2 7 08 k � 2 8 k � 3 8 k � 3 8 k � 2 8 09 0 9 k � 1 9 k � 2 9 k � 2 9 k � 2Tot: 6k � 10 Tot: 9k � 16 Tot: 9k � 18 Tot: 9k � 14 Tot: 3k � 5Then add k(k � 1) cycles from the solutions of type (a) in Figure 2 to get the claimed result.Corollary 2 There exists an O(n log5 n)-time algorithm to recognize Kn�odel graphs of order 2n,for all n such that B1(n) � 6.Proof. Assume k large enough. From Lemma 5, one can identify edges of dimensions 0 and 1in O(n log5 n)-time. Using dimensions 0 and 1, alternatively, one can construct an Hamiltonianpath P in the input graph G (if this construction fails, then G is not a Kn�odel graph). LetP = (u0; : : : ; u2n�1), and label every vertex v of G by (j mod 2; b j2c) where v = uj . The graph G isa Kn�odel graph if and only if there is an edge between (0; x) and (1; x+2i�1) for all x = 0; : : : ; n�1and all i = 0; : : : ; k, and no extra edges in G.We are now ready to prove our main result.Theorem 1 There exists an O(n log5 n)-time algorithm to recognize Kn�odel graphs of any order.Due to lack of space, the proof is only sketched (the complete proof is given in Appendix A).Sketch of the proof. From Corollaries 1 and 2, the theorem holds for n power of two, or n suchthat B1(n) � 6.Thus, assume that n satis�es B1(n) < 6, n 6= 2k. Assume moreover that n 6= 2k+1 � 1 (thislatter case deserves to be treated separately because the Kn�odel graph with n = 2k+1 � 1 is edge-transitive [9]). The key argument is that, for almost all such n, if C6(e) denotes the number of9

6-cycles passing through an edge e of a Kn�odel graph of order n, then C6(e) 6= C6(e0) for any e ande0 of dimensions 0 and k respectively, and C6(e) 6= C6(e0) for any e dimension 0 or k, and any e0 ofdimension i, i 6= 0 and i 6= k.The di�culty of the proof comes from the fact that, if B1(n) < 6, then Equation 2 has solutionsfor � 6= 0, and proceeding to a precise counting of the the number of 6-cycles passing through theedges of a Kn�odel graphis tricky. Anyway, we were able to prove that dimension 0 and dimension kcan be identi�ed by such counting. From the knowledge of the edges of dimension 0 and k, we haveshown that one can determine the set of edges of dimension 1. Then a path using dimensions 0and 1, alternatively, is an Hamiltonian cycle that allows to label the nodes such that node (0; x) isconnected to all nodes (1; x+ 2i � 1), i = 0; : : : ; k. To summarize, the algorithm is the following:Algorithm Recognize.Input: a regular graph G = (V;E) of 2n vertices, and degree k = blog2 nc;Output: tell whether or not G is isomorphic to the Kn�odel graph K of order 2n.Phase 1. For every i 2 f0; : : : ; kg, compute pi = number of 6-cycles passing through any edge ofdimension i of K;Phase 2. For every e 2 E, compute C6(e) = number of 6-cycles passing through the edge e of G;Phase 3. Identify the two sets S0 � E, and S1 � E, of edges of G that are possibly edges ofdimension 0 and 1 in K, respectively;Phase 4. Label the vertices according to an Hamiltonian path using edges in S0 and S1, alter-natively; If this construction fails in any way, then G is not a Kn�odel graph; Else, G is aKn�odel graph if and only if, for every x, node (0; x) is connected to the nodes (1; x+ 2i � 1),i = 0; : : : ; k, and that there is no extra edge in G.Each of these phases has a cost at most O(n log5 n) because the degree of every vertex isO(logn).To prove the correctness of Algorithm Recognize, the di�cult part is to prove that Phase 3 isdoable. This is formaly proved in the complete version of the proof (see Appendix A).The case of Kn�odel graphs of order 2k+1 � 2 deserves a particular attention due to the edge-transitivity of such graphs. We have proved that one can identify dimensions 0 and k by countingthe number of 6-cycles traversing a path of length 3.Corollary 3 If n 6= 2k+1 � 1, then Kn�odel graphs of order 2n are non edge-transitive. (From [9],Kn�odel graphs of order 2k+2 � 2 are edge-transitive.)Proof. The proof of correctness of Algorithm Recognize, given in Appendix A, uses the fact that,for n 6= 2k+1 � 1, the number of 6-cycles using edges of di�erent dimensions is not the same.4 Conclusion and Further ResearchIn this paper, we have shown that Kn�odel graphs can be recognized in polynomial time. The sameresult holds for Fibonacci graphs (see Appendix B). The natural question arising in this �eld is toask for which sequences i the same result holds. Let us formalize this question.10

Given an in�nite and increasing sequence of integers � = ( i)i�0, consider the sequence (G�n)n�0of circulant digraphs of order n such that, for any n � 0, G�n has generators f 0; 1; : : : ; kg where kis the largest integer such that k � n�1 (in other words, k � n�1 < k+1). Then let (H�n )n�0 bethe corresponding sequence of bipartite incident-graphs of order 2n. The problem is the following:Problem 1Instance: An integer n, and a graph G of 2n vertices;Question: Is G isomorphic to H�n?Note that the sequence � is �xed in Problem 1, and thus that this problem is di�erent from theproblem of deciding whether a graph is isomorphic to the bipartite incident-graph of a circulantdigraph, or to decide whether two bipartite incident-graphs of circulant digraphs are isomorphic.These two latter problems are known to be di�cult, even if they might be simpler than the problemof deciding whether a graph is a Cayley graph.We have seen that Problem 1 can be solved polynomially if � = (2i � 1)i�0 or if � = (F (i +1)� 1)i�0. The question is: for which family the techniques used to solve the problem for Kn�odelgraphs and Fibonacci graphs can be extended, and how? Actually, as soon as we know how tosolve Equation 1, then we are able to enumerate the 6-cycles, and to use the same techniques asthe techniques used for Kn�odel graphs and Fibonacci graphs.Note that if one cannot directly identify the dimensions i's, that is if we can just color theedges such that there is a one-to-one correspondence between the colors and the dimensions butwithout knowing the values of the dimensions, then one can conclude anyway if k = O( lognlog logn)by testing the k! possibilities of labeling the colors. However, if k is too large, k = (logn) forinstance, then one needs to identify precisely every dimensions, and this might be di�cult by usingEquation 1 only. For instance, if i = 4i, then Equation 1 has only one type of solutions, namely( ; 0; 00; ; 0; 00) for 6= 0, 0 6= 00, and 6= 00. That is all edges have the same number of6-cycles using them. There is a way though to break the symmetry in the speci�c case i = 4i, bycounting 8-cycles. This should yield a polynomial algorithm for this sequence. More generally, anysequence of the form pi, p integral � 2, might be solvable by looking at 2p-cycles, and thus in timeO(n log2p�1 n) = O(n2). Sequences giving rise to sparse graphs, eg., i = 22i , or sequences givingrise to dense graphs, eg., i = ip, p integral � 2, seem to require di�erent techniques.We let as an open problem the characterization of the sequences � for which Problem 1 can besolved using the same techniques as for Kn�odel graphs and Fibonacci graphs.Acknowledgments. The authors are thankful to Lali Barri�ere, Guillaume Fertin, Bernard Mans,and Andr�e Raspaud for their valuable helps and comments, and to Jean-Paul Allouche for fruitfuldiscussions.References[1] A. �Ad�am, Research problem 2-10, J. Combin. Theory, 2 (1967), p. 393.[2] B. Alspach and T. Parsons, Isomorphism of circulant graphs and digraphs, Discrete Math-ematics, 25 (1979), pp. 97{108. 11

[3] J.-C. Bermond, F. Comellas, and F. Hsu, Distributed loop computer networks: a survey,Journal of Parallel and Distributed Computing, 24 (1995), pp. 2{10.[4] G. Cybenko, D. Krumme, and K. Venkataraman, Gossiping in minimum time, SIAMJournal on Computing, 21 (1992), pp. 111{139.[5] B. Elspas and J. Turner, Graphs with circulant adjacency matrices, J. Comb. Theory, 9(1970), pp. 229{240.[6] S. Even and B. Monien, On the number of rounds necessary to disseminate information, inFirst ACM Symposium on Parallel Algorithms and Architectures (SPAA), 1989.[7] P. Fraigniaud and E. Lazard, Methods and Problems of Communication in Usual Net-works, Discrete Applied Mathematics, 53 (1994), pp. 79{133.[8] S. Hedetniemi, S. Hedetniemi, and A. Liestman, A survey of gossiping and broadcastingin communication networks, Networks, 18 (1986), pp. 319{349.[9] M.-C. Heydemann, N. Marlin, and S. P�erennes, Cayley graphs with complete rotations,Technical Report 1155, LRI, Bat. 490, Univ. Paris-Sud, 91405 Orsay cedex, France, 1997.Submitted to the European Journal of Combinatorics.[10] J. Hromkovi�c, R. Klasing, B. Monien, and R. Peine, Dissemination of informationin interconnection networks (broadcasting and gossiping), in Combinatorial Network Theory,D.-Z. Du and D. F. Hsu, eds., Kluwer Academic, 1995, pp. 125{212.[11] W. Knodel, New gossips and telephones, Discrete Mathematics, 13 (1975), p. 95.[12] R. Labahn and I. Warnke, Quick gossiping by telegraphs, Discrete Mathematics, 126 (1994),pp. 421{424.[13] B. Mans, F. Pappalardi, and I. Shparlinski, On the �Ad�am conjecture on circulant graphs,in Fourth Annual International Computing and Combinatorics Conference (Cocoon '98), Lec-ture Notes in Computer Science, Springer-Verlag, 1998.[14] M. Muzychuk and G. Tinhoffer, Recognizing circulant graphs of prime order in polynomialtime, The electronic journal of combinatorics, 3 (1998).[15] A. Nayak, V. Accia, and P. Gissi, A note on isomorphic chordal rings, Information Pro-cessing Letters, 55 (1995), pp. 339{341.[16] I. Ponomarenko, Polynomial-time algorithms for recognizing and isomorphisn testing ofcyclic tournaments, Acta Applicandae Mathematicae, (1992), pp. 139{160.[17] V. Sunderam and P. Winkler, Fast information sharing in a distributed system, DiscreteApplied Mathematics, 42 (1993), pp. 75{86.12

A Proof of Theorem 1From Corollaries 1 and 2, the theorem holds for n power of two, or n such that B1(n) � 6. Thuslet n be none of such integers. If Equation 2 has no solution for � 6= 0, then, as far as the numberof 6-cycles is concerned, we are let with a similar case as the ones studied in Lemma 5, and thusthe theorem holds by the same arguments as in the proof of Corollary 2.The aim of the rest of the proof is to study the Kn�odel graphs of order n for which Equation 2has solutions for � 6= 0. We will use the same techniques as in the proofs of Corollaries 1, and 2,that is we will show that, given an input graph G, one can determine the edges of G that arepossibly edges of dimension 0 in a Kn�odel graph, and those that are possibly edges of dimension 1in a Kn�odel graph. Then, given these two sets, one can label the vertices of G according to theHamiltonian path using dimension 0 and 1, alternatively. (If such an Hamiltonian path does notexist, then G is not a Kn�odel graph.) Then, it just remains to check that the edges of G connectthe labeled vertices as speci�ed by the interconnection rules of a Kn�odel graph.To determine the edges that are candidates to be of dimension 0 or 1 in a Kn�odel graph, we countthe number C6(e) of 6-cycles passing through every edge e. To show that one can conclude fromthe knowledge of C6(e) for all the edges e, we will prove that, in a Kn�odel graph, C6(e) 6= C6(e0)for every two edges e in e0 in dimension 0 and k, respectively, and C6(e) 6= C6(e0) for every edgee in dimension 0 or k, and every edge e0 in dimension d, d 6= 0 and d 6= k. There are very fewexceptions, for speci�c values of n that will deserve a particular attention.Claim 1 Determining the edges of dimensions 0 and k allows to determine the edges of dimen-sions 0 and 1.Indeed, given the list of all the edges of dimension 0, and the list of all the edges of dimension kof a Kn�odel graph, one can compute the list of all the edges of dimension 1 of this Kn�odel graph,in O(n log5 n) time. Indeed, given the list of all the edges of dimension 0, and the list of all theedges of dimension k, let us consider 6-cycles of type(k; ; 0; 0; 0; 00)where ; 0 and 00 are a priori unknown, apart the fact that 6= k, 6= 0, 0 6= 0, 00 6= 0, and 00 6= k. From Equation 2, such a 6-cycle satis�es:2k + 20 + 20 = 2 + 2 0 + 2 00 + � nwhere � 2 f�2;�1; 0; 1; 2g. Now, � = 2 is impossible because 2n > 2k+1. If � = 1, then2 = 2 +2 0+2 00+n0 where n = 2k+n0, 0 < n0 < 2k . This is impossible because 2 +2 0+2 00+n0 � 4.If � � �1, then ��n + 2k + 2 > 2 + 2 0 + 2 00 because 2k+1 � 2 + 2 0 + 2 00 due to the fact that 6= k and 00 6= k. It remains the case � = 0 which implies that at least two of the 's are thesame. Thus we are let with the equation 2k + 2 = 2p+1 + 2q, and hence either p = 0 and q = k, orp = k � 1 and q = 1. Since two 's are distinct from k, the only 6-cycles of type(k; ; 0; 0; 0; 00)are those of type (k; k� 1; 0; k� 1; 0; 1)13

(b)(a)

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0 0Figure 5: The 6-cycles containing opposite edges of dimensions 0; 0, and k.0

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vFigure 6: The three 6-cycles containing opposite edges of dimensions 0; 0, and k, passing throughan edge of dimension k.up to any permutation of the odd positions. This yields the two non isomorphic labeled cycles ofFigure 5.Therefore, by looking at all 6-cycles containing opposite edges of dimensions 0; 0, and k, wecan construct a set S of edges such that e 2 S if and only if e is either of dimension 1, or ofdimension k � 1. To distinguish dimension 1 from dimension k � 1, we construct, for every edgefu; vg of dimension k, the three 6-cycles corresponding to Figure 5. We get a situation as depictedon Figure 6. The edges of dimension 1 are those incident to nodes u or v, and through which passesonly one of the three considered 6-cycles. This completes the proof of the claim.Thus determining the edges of dimensions 0, and k is �ne. It remains to show how to performthis identi�cation.First, recall that the contributions to the number of 6-cycles passing through an edge e, andsolutions of Equation 2 for � = 0, are of the formk2 + pk +O(1) where p = 8<: 5 if e is in dimension 0;2 if e is in dimension k;8 otherwise.as shown by lemma 5. Thence, if Equation 2 would not have other solutions than those for � = 0,then the edges in dimension 0, and k could be easily identi�ed. Unfortunately, � 6= 0 provides14

additional solutions which modify this counting. Fortunately, we will show that one can controlthis modi�cation without too much e�ort.Let us denote by p(n) the vector (p0; : : : ; pk) such that k2+pi k+O(1) is the number of 6-cyclespassing through an edge of dimension i, 0 � i � k (the contributions for � 6= 0 do not modifythe leading term k2 set for � = 0). We will show that p0 6= pk, and, apart few cases, for every i,1 � i � k � 1, p0 6= pi, and pk 6= pi. These inequalities allow to distinguish dimensions 0 and kfrom the others.Let us study the contribution to p(n) of solutions of Equation 2 if � = 2. The equation2x0 + 2x2 + 2x4 = 2x1 + 2x3 + 2x5 + 2k+1 + 2n0where n = 2k + n0, impliesx0 = x2 = k; and 2x4 = 2x1 + 2x3 + 2x5 + 2n0up to a permutation of the even indices. This equation has generally no solution apart for speci�cvalues of n0. What is important is that none of these possible solutions can contribute to changep(n) because the number of such solutions is bounded by a constant.The tricky case is thus � = 1, yielding the equation2x0 + 2x2 + 2x4 = 2x1 + 2x3 + 2x5 + 2k + n0:We distinguish two cases according to whether or not an xi of the left hand side is equal to k. Thisis motivated by speci�c values of n: an integer n 6= 2k is said to be block-special ifn = 2k + `2Xi=`1 2i; k � `2 � 2;or n = 2k + `3Xi=`2+2 2i + `2Xi=`1 2i; k � `3 � 2;or n = 2k + `3Xi=`2+3 2i + `2Xi=`1 2i; k � `3 � 2;or n = 2k + `4Xi=`3+2 2i + `3Xi=`2+2 2i + `2Xi=`1 2i; k � `4 � 2:We claim that:Claim 2 If n is not block-special then any solution of Equation 2 for � 6= 0 has one of the xi's ofthe left hand side equal to k. Moreover, if n is block-special then other solutions exist, but, for anyof them, two of the xi's of the left hand side are equal to k � 1, and the third xi is determined bythe value of n. 15

Indeed, for � 6= 0, Equation 2 can be rewritten as:� 2x0 + 2x2 + 2x4 = 2x1 + 2x3 + 2x5 + 2k + n0 if � = 12x0 + 2x2 + 2x4 = 2x1 + 2x3 + 2x5 + 2k+1 + 2n0 if � = 2where 0 < n0 < 2k, n = 2k + n0, and xi 6= xi+1 (mod 6). The case � = 2 forces one of the xi's of theleft hand side to be equal to k because 2k�1 + 2k�1 + 2k�1 < 2k+1. Similarly, if � = 1, and none ofthe xi's of the left hand side is equal to k, then at least two of these xi's must be equal to k � 1because 2x1 + 2x3 + 2x5 + n0 > 0. Thus we are let with the equation 2x4 = 2x1 + 2x3 + 2x5 + n0 inwhich the sum of n0 > 0 with three powers of two must result in a single power of two.Let us �rst turn our attention to block-special values of n. The four kinds of block-specialintegers correspond to B1(n0) = 1, B1(n0) = 2, and B1(n0) = 3, respectively, where the caseB1(n0) = 2 is in turn divided into two sub-cases according to the distance between the two blocks.� If B1(n0) = 1, then n = 2k+P`2i=`1 2i, k�`2 � 2, and the equation 2x4 = 2x1+2x3+2x5+P`2i=`1 2ihas three possible solutions for x4, namely x4 = `2 + 1, x4 = `2 + 2 and x4 = `2 + 3. The �rstcase corresponds to (x1; x3; x5) = (`1 � 1; `1 � 2; `1 � 2), up to any permutation of x1; x3; x5. Thesecond case corresponds to (x1; x3; x5) = (`2+1; `1� 1; `1� 1), up to any permutation of x1; x3; x5.The third case corresponds to (x1; x3; x5) = (`2+ 2; `2+ 1; `1), up to any permutation of x1; x3; x5.That is, we have the solutions:(k � 1; `1 � 2; k � 1; `1 � 2; `2 + 1; `1 � 1) or(k � 1; `1 � 1; k � 1; `1 � 1; `2 + 2; `2 + 1) or(k � 1; `1; k � 1; `2 + 1; `2 + 3; `2 + 2)up to any permutation of the x2i's, and any permutation of the x2i+1's.� For n = 2k +P`3i=`2+2 2i +P`2i=`1 2i, k � `3 � 2, we have the solutions:(k � 1; `1 � 1; k � 1; `1 � 1; `3 + 1; `2 + 1) or(k � 1; `1; k � 1; `2 + 1; `3 + 2; `3 + 1)up to any permutation of the x2i's, and any permutation of the x2i+1's.� For n = 2k +P`3i=`2+3 2i +P`2i=`1 2i, k � `3 � 2, we have the solution:(k � 1; `1; k � 1; `2 + 1; `3 + 1; `2 + 2)up to any permutation of the x2i's, and any permutation of the x2i+1's.� For n = 2k +P`4i=`3+2 2i +P`3i=`2+2 2i +P`2i=`1 2i, k � `4 � 2, we have the solution:(k � 1; `1; k � 1; `2 + 1; `4 + 1; `3 + 1)up to any permutation of the x2i's, and any permutation of the x2i+1's.Thus, for any block-special integers, there are solutions to Equation 2 for � 6= 0 such that noneof the xi's of the left hand side is equal to k. However, two xi's of the left hand side are equal tok � 1, and the third xi is �xed according to the value of n. The reader can check that, if n is notblock-special, then the equation 2x4 = 2x1 + 2x3 +2x5 +n0 has no solution, and thus at least one ofthe xi's of the left hand side is equal to k in every solution of Equation 2 for � 6= 0 (if any). Thiscompletes the proof of Claim 2. 16

As a consequence of Claim 2:� If none of the xi's of the left hand side is equal to k, then the solutions are those corresponding toblock-special values of n. Thus none of these solutions contribute to change p(n) since the numberof such solutions is bounded by a constant.� If one of the xi's of the left hand side is equal to k, say x4 = k, then we are let with2x0 + 2x2 = 2x1 + 2x3 + 2x5 + n0: (3){ If x0 = x2, then, as previously, the equation2x0+1 = 2x1 + 2x3 + 2x5 + n0has a constant number of solutions depending on the value of n0, and thus p(n) is not modi�ed.{ If x0 6= x2, then we consider three cases:Case 1. B1(n0) � 3.In this case, there is again a constant number of solutions for Equation 3, and p(n) remainsunchanged.Case 2. B1(n0) = 2.In this case, Equation 3 has again a constant number of solutions, but ifn0 = tXi=m+2 2i + mXi=` 2i:Indeed, for such an n0, Equation 3 yields the two non isomorphic labeled 6-cycles depictedon Figure 7. These cycles contribute to add 4k + O(1) 6-cycles for every edge in dimensionk. Indeed, if t+ 1 6= k, then Cycles (a) and (b) of Figure 7 contribute for 2k + O(1) each. Ift + 1 = k, then Cycles (a) and (b) are isomorphic, but Cycle (a) alone then contributes for4k+O(1). Note that at most a constant number of such cycles have been counted before, asone can check on Figure 2. For the other dimensions, consider four cases:Case 2.1. ` > 0 and t+1 < k. Then p0 = 5, pk = 2+4 = 6, and pi � 8 for any i, 0 < i < k.For instance, p` = 8 + 4 = 12.Case 2.2. ` = 0 and t+1 < k. Then p0 = 5+4 = 9, pk = 2+4 = 6, and for any i, 0 < i < k,either pi = 8 or pi � 10. For instance, pm+1 = 8+ 4 = 12, whereas pm = 8.Case 2.3. ` > 0 and t+1 = k. Then p0 = 5, pk = 2+4 = 6, and for any i, 0 < i < k, pi � 8.Case 2.4. ` = 0 and t + 1 = k. Then p0 = 5 + 2 = 7 (Cycles (a) and (b) are isomorphic),pk = 2 + 4 = 6, and for any i, 0 < i < k, pi � 8.Case 3. B1(n0) = 1. Then n0 =Pmi=` 2i, m � `. The \main"solutions of Equation 3 are then1) 8>><>>: x1 = `x5 = m+ 1x2 = m+ 2x0 = x3 = or 2) 8<: x1 = x5 = `� 1x2 = m+ 1x0 = x3 = or 3) 8>><>>: x1 = `x2 = m+ 1x0 = x3 = x5 = � 117

(b)(a)

γ

γ γ

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k t+1

m+1 m+1

t+1k

l l

2 f0; : : : ; k � 1g n f`;m+ 1; tg.Figure 7: Additional solutions provided by Equation 3 if B1(n0) = 2.up to valid permutations of the indices. There are other solutions, but they contribute for aconstant number of cycles. The non isomorphic labeled 6-cycles corresponding to the threeprevious sets of solutions are depicted on Figure 8. Solution 1 yields Cycles 1 and 2; Solution 2yields Cycle 3; And Solution 3 yields Cycles 4, 5 and 6. These 6-cycles are non isomorphic tothe 6-cycles of Figure 2. The remaining part of the proof consist to count the contribution ofthese solutions as a function of the values of ` and m. This cases study is summarized below.0 1 `� 1 ` m+ 1 m+ 2 k � 1 k1 m < k � 2 and ` > 1 5 � 12 18 20 12 8 � 142 m < k � 2 and ` = 1 9 18 � � 20 12 8 � 143 m < k � 2 and ` = 0 15 � � � 18 12 8 � 124 m = k � 2 and ` > 1 5 � 12 16 � � 8 18 145 m = k � 2 and ` = 1 9 16 � � � � 8 18 146 m = k � 2 and ` = 0 13 � � � � � 8 16 127 m = k � 1 and ` > 1 5 � 10 11 � � 8 � 108 m = k � 1 and ` = 1 7 11 � � � � 8 � 109 m = k � 1 and ` = 0 8 � � � � � 8 � 8In this table, each cell contains the coe�cient pi in front of k in the expression of the numberof 6-cycles passing through an edge of dimension i. For instance, if m < k�2 and ` > 1, thenrow 1 says that p0 = 5, p`�1 = 12, p` = 18, pm+1 = 20, pm+2 = 12, pk = 14, and, for all theother dimensions , p = 8. The sign \-" stands either to indicate that a cell is meaningless(for instance the column ` � 1 is meaningless if ` = 0), or to indicate that the value of pi isgiven elsewhere (for instance, if ` = 1, then the value of p`�1 is given in column 0).For the values of ` and m corresponding to row 1, 2, 4, 5, 6, or 8, the dimensions 0 and k canbe distinguished. On row 3, the counting show that pk = pm+2. To distinguish dimension kfrom dimension m + 2, we consider 6-cycles of type (0; ; 0; 0; d; 00) where d 2 fm + 2; kg.By setting two of the 's to be m+ 1 (note that dimension m+ 1 can be distinguished), weforce d = m + 2, and the last edge is in dimension 1. Therefore, dimension m + 2 can bedistinguished from dimension k. The equality between pk and p`�1 on row 7 can be solvedsimilarly. 18

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l

l

lFigure 8: Additional solutions provided by Equation 3 with B1(n0) = 1.Row 9 corresponds to Kn�odel graphs of order 2k+2 � 2. It is the last but not the least caseconsidered in this proof. Indeed, n = 2k+1 � 1 corresponds to an edge-transitive graph, andit is therefore not surprising to �nd the same value for all the pi's in the table. To solve thatcase, let us look at the number of 6-cycles using a path of length 3. Solutions for � = 0contribute for a constant number of cycles (see Figure 2). Equation 2 has no solution for� = 2. Indeed, on one hand2x1 + 2x3 + 2x5 + 2n = 2x1 + 2x3 + 2x5 + 2k+2 � 2 � 2k+2 + 1;and on the other hand 2x0 + 2x2 + 2x4 � 2k+1 + 2k:The case � = 1 yields the two cycles of Figure 9 where 6= 0. These solutions contribute fora constant number of 6-cycles if the path is distinct from (k; 0; k), and contribute for k to apath (k; 0; k). Therefore, edges of dimension 0 and dimension k can be identi�ed.This completes the cases study. In every case, we have proved that dimensions 0 and k can bedistinguished from the other dimensions by counting the number of 6-cycles passing through everyedge (or path of length 3 for n = 2k+1�1). This completes the proof of the theorem by applicationof Claim 1, and of the Hamiltonian path technique.19

γ -1

γ -1

γ -1

γ -1

(b)(a)

γ γk

0

k k

0

kFigure 9: Speci�c solutions for n = 2k+1 � 1.B Recognizing Fibonacci GraphsWe just sketch how techniques given for Kn�odel graphs can be extended to Fibonacci graphs. Letn > 0 be an integer. F�1(n) is de�ned as the largest index i such that F (i) � n. The Fibonaccidecomposition of n is a binary sequence x = (xF�1(n); : : : ; x1)F where xi, i = 1; : : : ; F�1(n), isde�ned by xF�1(n) = 1, and, for i < F�1(n),xi = ( 1 if i = F�1 �n �PF�1(n)j=i+1 xjF (j)�0 otherwise.Lemma 6 There is no consecutive 1's in the Fibonacci decomposition of any positive integer.Proof. Assume that there are two consecutive 1's in the Fibonacci decomposition x of an integern, and let k = F�1(n). Then, let i be the largest index for which xi = xi�1 = 1. We haven�Pkj=i+2 xjF (j) � F (i) + F (i� 1) because xi+1 = 0. Thus n�Pkj=i+2 xjF (j) � F (i+ 1), andthus, by de�nition of x, xi+1 = 1, a contradiction.Equation 1 can be rewritten for Fibonacci graphs as:� F (x0) + F (x2) + F (x4) = F (x1) + F (x3) + F (x5) + � nxi 6= xi+1 (mod 6) for any i 2 f0; 1; 2; 3; 4; 5g (4)where � 2 f0; 1; 2g.Lemma 7 For � = 0, Equation 4 has seven types of solutions. These solutions are expressed onFigure 10.Proof. Similar to the proof of Lemma 2.In the Fibonacci decomposition of an integer, two consecutive bits "10" plays the same role asa single bit in the binary decomposition of an integer since (10 : : :1010)F + (1)F = (100 : : :0000)Fin the Fibonacci decomposition, in the same way (1 : : :11)2 + (1)2 = (10 : : :00)2 in the binarydecomposition. 20

γ’

γ" γ’

γ"

(a)

γ

γ

γ’

γ’+1 γ-1

γ+1

γ’-1

γ’-1

γ+1

γ-1

γ+1

γ-1

γ’-1

γ’-1 γ+1

γ-1

γ’

γ’+1

γ’-1

γ-1

γ+1

γ+1 γ’-1

γ-1

γ’

γ’+1 γ’+1

γ’+1γ’+1

γ’

γ’

γ’

(b) (c) (d)

γ γ γ

(e) (f) (g)

γ γ γ

Figure 10: The seven types of solution of Equation 4 for � = 0. In (a), 6= 0, 0 6= 00, and 00 6= .In (b)-(g), 6= 0, and 6= 0� 1.Notation. B10(n) denotes the number of blocks of consecutive 2-bit strings "10" in the Fibonaccidecomposition of n (the rightmost 1-entry may be not followed by a 0-entry).For instance, B10((1010010)F) = 2, B10((1000101001)F) = 3, and B10((100)F ) = 1. Note thatn = (10 0 : : :0| {z } 10 0 : : :0| {z } z }| {1010 : : :1010 00 z }| {1010 : : :1010 00 z }| {1010 : : :10100 : : :0| {z })Fsatis�es B10(n) = 5, and there is a solution of Equation 2 for � = 1 with x1; x3; x5 equal to theunderlined bit-positions, and x0; x2; x4 equal to the over-lined bit-positions. However, we have:Lemma 8 If B10(n) � 6, then Equation 4 has no solution for � 6= 0.Proof. Similar to the proof of Lemma 3.Lemma 9 If n = F (k + 1) then the 4-cycles of the Fibonacci graph of order 2n are all of type(k; � 1; ; � 2), for 2 f2; : : : ; kg.Proof. Similar to the proof of Lemma 4From all the above lemmas, one can count the number of 6-cycles passing through an edge of agiven dimension (or a path of length 3), and, therefore, we get the following result:21

Theorem 2 There exists an O(n log5 n)-time algorithm to recognize Fibonacci graphs of any order.Proof. Similar to the proof of Theorem 1. The de�nition of block-special integers must be modi�edaccording to the de�nition of the Fibonacci decomposition of an integer. An integer n 6= F (k) issaid to be block-special if it is one of the following forms:n = F (k) +Ppi=0 F (` + 2i)n = F (k) +Ppi=0 F (` + 2i) +Pqi=0 F (`+ 2p+ 3 + 2i)n = F (k) +Ppi=0 F (` + 2i) +Pqi=0 F (`+ 2p+ 4 + 2i)n = F (k) +Ppi=0 F (` + 2i) +Pqi=0 F (`+ 2p+ 5 + 2i)n = F (k) +Ppi=0 F (` + 2i) +Pqi=0 F (`+ 2p+ 6 + 2i)n = F (k) +Ppi=0 F (` + 2i) +Pqi=0 F (`+ 2p+ 3 + 2i) +Pri=0 F (`+ 2p+ 3 + 2q + 3 + 2i)n = F (k) +Ppi=0 F (` + 2i) +Pqi=0 F (`+ 2p+ 4 + 2i) +Pri=0 F (`+ 2p+ 4 + 2q + 3 + 2i)n = F (k) +Ppi=0 F (` + 2i) +Pqi=0 F (`+ 2p+ 3 + 2i) +Pri=0 F (`+ 2p+ 3 + 2q + 4 + 2i)n = F (k) +Ppi=0 F (` + 2i) +Pqi=0 F (`+ 2p+ 4 + 2i) +Pri=0 F (`+ 2p+ 4 + 2q + 4 + 2i)Claim 3 Let n be an integer, F (k+ 1) < n < F (k + 2). If n is not block-special then any solutionof Equation 4 for � 6= 0 implies that one of the xi's of the left hand side is equal to k. If n isblock-special then other solutions exist, but, for any of them, one of the xi's of the left hand side isequal to k � 1, another is equal to k � 2, and the third one is determined by the value of n.The proof of this claim is similar to the proof of Claim 2. The solutions for block-special valuesof n are listed bellow:� For n = F (k) +Ppi=0 F (`+ 2i), we have the solutions:(k � 1; `� 2; k � 2; ` � 4; `+ 2p+ 1; ` � 5) or(k � 1; `� 2; k � 2; ` � 3; `+ 2p+ 3; ` + 2p+ 2) or(k � 1; `� 1; k � 2; ` + 2p+ 2; `+ 2p+ 5; ` + 2p+ 4)� For n = F (k) +Ppi=0 F (`+ 2i) +Pqi=0 F (`+ 2p+ 3+ 2i), we have the solutions:(k � 1; `� 2; k � 2; `� 3; ` + 2p+ 3 + 2q + 1; ` + 2p) or(k � 1; `� 1; k � 2; `+ 2p� 1; ` + 2p+ 3 + 2q + 1; ` + 2p� 2) or(k � 1; `� 1; k � 2; `+ 2p; ` + 2p+ 3 + 2q + 3; ` + 2p+ 3 + 2q + 2)� For n = F (k) +Ppi=0 F (`+ 2i) +Pqi=0 F (`+ 2p+ 4+ 2i), we have the solutions:(k � 1; `� 2; k � 2; `� 3; ` + 2p+ 4 + 2q + 1; ` + 2p+ 2) or(k � 1; `� 1; k � 2; `+ 2p; ` + 2p+ 4 + 2q + 1; ` + 2p+ 1) or(k � 1; `� 1; k � 2; `+ 2p+ 2; ` + 2p+ 4 + 2q + 3; ` + 2p+ 4 + 2q + 2)� For n = F (k) +Ppi=0 F (`+ 2i) +Pqi=0 F (`+ 2p+ 5+ 2i), we have the solutions:(k � 1; `� 1; k � 2; `+ 2p+ 2; `+ 2p+ 5 + 2q + 1; `+ 2p+ 2) or(k � 1; `� 1; k � 2; `+ 2p; `+ 2p+ 5 + 2q + 1; `+ 2p+ 3)� For n = F (k) +Ppi=0 F (`+ 2i) +Pqi=0 F (`+ 2p+ 6+ 2i), we have the solution:(k � 1; `� 1; k � 2; `+ 2p+ 2; `+ 2p+ 6 + 2q + 1; `+ 2p+ 4)22

� For n = F (k) +Ppi=0 F (`+ 2i) +Pqi=0 F (`+ 2p+ 3+ 2i) +Pri=0 F (`+ 2p+ 3+ 2q+ 3+ 2i), wehave the solution:(k � 1; `� 1; k � 2; `+ 2p; `+ 2p+ 3 + 2q + 3 + 2r+ 1; `+ 2p+ 3 + 2q)� For n = F (k) +Ppi=0 F (`+ 2i) +Pqi=0 F (`+ 2p+ 4+ 2i) +Pri=0 F (`+ 2p+ 4+ 2q+ 3+ 2i), wehave the solution:(k � 1; `� 1; k � 2; `+ 2p+ 2; ` + 2p+ 4 + 2q + 3+ 2r + 1; `+ 2p+ 4 + 2q)� For n = F (k) +Ppi=0 F (`+ 2i) +Pqi=0 F (`+ 2p+ 3+ 2i) +Pri=0 F (`+ 2p+ 3+ 2q+ 4+ 2i), wehave the solution:(k � 1; `� 1; k � 2; `+ 2p; `+ 2p+ 3 + 2q + 4 + 2r+ 1; `+ 2p+ 3 + 2q + 2)� For n = F (k) +Ppi=0 F (`+ 2i) +Pqi=0 F (`+ 2p+ 4+ 2i) +Pri=0 F (`+ 2p+ 4+ 2q+ 4+ 2i), wehave the solution:(k � 1; `� 1; k � 2; `+ 2p+ 2; `+ 2p+ 4 + 2q + 4 + 2r+ 1; `+ 2p+ 4 + 2q + 2)

23