aa r X i v : . [ m a t h . C O ] F e b On graphs of defect at most 2
Ramiro Feria-Pur´on , ∗ Mirka Miller , , , , † Guillermo Pineda-Villavicencio , ‡ School of Electrical Engineering and Computer ScienceThe University of Newcastle, Australia Department of MathematicsUniversity of West Bohemia, Czech Republic Department of Computer ScienceKing’s College London, UK Department of MathematicsITB Bandung, Indonesia Centre for Informatics and Applied OptimizationUniversity of Ballarat, Australia
November 25, 2018
Abstract
In this paper we consider the degree/diameter problem , namely, given natural numbers ∆ ≥ D ≥
1, find the maximum number N(∆ , D ) of vertices in a graph of maximum degree ∆and diameter D . In this context, the Moore bound M(∆ , D ) represents an upper bound forN(∆ , D ).Graphs of maximum degree ∆, diameter D and order M(∆ , D ), called Moore graphs , turnedout to be very rare. Therefore, it is very interesting to investigate graphs of maximum degree∆ ≥
2, diameter D ≥ , D ) − ǫ with small ǫ >
0, that is, (∆ , D, − ǫ )-graphs.The parameter ǫ is called the defect .Graphs of defect 1 exist only for ∆ = 2. When ǫ >
1, (∆ , D, − ǫ )-graphs represent a wideunexplored area. This paper focuses on graphs of defect 2. Building on the approaches developedin [11] we obtain several new important results on this family of graphs.First, we prove that the girth of a (∆ , D, − ≥ D ≥ D . Second, andmost important, we prove the non-existence of (∆ , D, − ≥ D ≥ , , − , D, − ǫ )-graphs with D ≥ ≤ ǫ ≤ , D, − , D, − ǫ )-graphs with D ≥ ≤ ǫ ≤ ∗ [email protected] † [email protected] ‡ [email protected] (Corresponding author) ther results of this paper include necessary conditions for the existence of (∆ , D, − ≥ D ≥
4, and the non-existence of (∆ , D, − ≥ D ≥ ≡ , D ).Finally, we conjecture that there are no (∆ , D, − ≥ D ≥
4, and commenton some implications of our results for the upper bounds of N(∆ , D ). Keywords:
Moore bound, Moore graph, Degree/diameter problem, defect, repeat.
AMS Subject Classification:
Due to the diverse features and applications of interconnection networks, it is possible to find manyinterpretations about network “optimality” in the literature. Here we are concerned with the follow-ing; see [12, pp. 168].
An optimal network contains the maximum possible number of nodes, given a limit on thenumber of connections attached to a node and a limit on the distance between any twonodes of the network.
In graph-theoretical terms, this interpretation leads to the degree/diameter problem , which can bestated as follows:
Degree/diameter problem : Given natural numbers ∆ ≥ D ≥
1, find the largest possiblenumber N(∆ , D ) of vertices in a graph of maximum degree ∆ and diameter D .Note that N(∆ , D ) is well defined for ∆ ≥ D ≥
1. An upper bound for N(∆ , D ) is given bythe
Moore bound
M(∆ , D ),M(∆ , D ) = 1 + ∆ (cid:0) −
1) + · · · + (∆ − D − (cid:1) . Graphs of degree ∆, diameter D and order M(∆ , D ) are called Moore graphs .Only a few values of N(∆ , D ) are known at present. With the exception of N(4 ,
2) = M(4 , − ,
2) = M(5 , − ,
2) = M(6 , − ,
3) = M(3 , − ,
4) = M(3 , − , D ) are those for which thereexists a Moore graph.Moore graphs are very rare. For ∆ = 2 and D ≥ D + 1 vertices, whereasfor D = 1 and ∆ ≥ D = 2 and ∆ ≥
3, Mooregraphs exist for ∆ = 3 , ≥ D ≥
3, there are no Moore graphs ([1, 6]). 2herefore, we are interested in studying the existence or otherwise of graphs of given maximumdegree ∆, diameter D and order M(∆ , D ) − ǫ for small ǫ >
0, that is, (∆ , D, − ǫ )-graphs, where theparameter ǫ is called the defect .The family of graphs of defect ǫ = 1 has been fully characterized; see [2, 10, 15]. For ∆ = 2 and each D ≥
2, the cycle on 2 D vertices is the only (2 , D, − D there areno (∆ , D, − ǫ = 2 represent a wide unexplored area. The catalogue of (3 , D, −
2) was completedby Jørgensen in [14]. So far there have been several partial results achieved on the existence orotherwise of (∆ , D, − ≥ D ≥
2; see [3, 5, 9, 14, 16, 20] for D = 2 and [18, 21]for ∆ = 4 ,
5. While the paper [18] claimed to have proved the non-existence of (4 , D, − D ≥
3, it turns out that the proof contained a mistake, so that only structural properties of(4 , D, − , D, − ≥ D ≥ ǫ ≥
3, the onlyknown work is the complete catalogue of (3 , D, − , D, − ≥ D ≥
4, and advance considerably theaforementioned question of the existence or otherwise of such graphs. To obtain our results we relyon combinatorial approaches which are inspired by those developed in [11].Our first result is a proof that the girth of a (∆ , D, − ≥ D ≥ D − D . Subsequently, we offer a non-existence proof of (∆ , D, − ≥ D ≥
4. After ruling out the existence of (4 , , − , D, − ǫ )-graphs for ∆ ≥ D ≥ ≤ ǫ ≤
2, namely, the one of (4 , D, − ǫ )-graphs.Other results of the paper include structural properties and necessary conditions for the existence of(∆ , D, − ≥ D ≥
4, and the non-existence of (∆ , D, − ≥ D ≥ ≡ , D ).Finally, we conjecture there are no (∆ , D, − ≥ D ≥
4, and comment on theimplications of our results for the upper bounds for N(∆ , D ). (∆ , D, − -graphs When ∆ = 2 or D = 1, there are no graphs of defect 2.For D = 2 there is a unique (2 , , − , , − , , − , , − , , − ≥
6, has been conjectured but notyet proved in spite of the partial support given in [5, 16].3 a ) ( b )( c ) ( d )( e ) Figure 1: Known graphs of defect 2. ( a ) and ( b ) the two (3 , , − c ) the unique (3 , , − d ) the unique (4 , , − e ) the unique (5 , , −
2) graph (note that this graph isformed by connecting appropriately 3 copies of the graph ( b )).For ∆ = 3 and D ≥ , , − c ). This graph,together with the two aforementioned (3 , , − The terminology and notation used in this paper is standard and consistent with that used in [8], soonly those concepts that can vary from texts to texts will be defined.All graphs considered in this paper are simple. The vertex set of a graph Γ is denoted by V (Γ), andits edge set by E (Γ). For an edge e = { x, y } , we write x ∼ y . The set of neighbors of a vertex x in4 is denoted by N ( x ).A path of length k is called a k -path . A path from a vertex x to a vertex y is denoted by x − y . We usethe following notation for subpaths of a path P = x x . . . x k : x i P x j = x i . . . x j , where 0 ≤ i ≤ j ≤ k .A cycle of length k is called a k -cycle . The girth of Γ, denoted g=g(Γ), is the length of the shortestcycle in Γ.The union of three independent paths of length D with common endvertices is denoted by Θ D . In agraph Γ, a vertex of degree at least 3 is called a branch vertex of Γ. We begin this section with a known condition for the regularity of a (∆ , D, − ǫ )-graph, which can beeasily deduced by considering the existence of a vertex of degree at most ∆ − Proposition 4.1
For ǫ < −
1) + (∆ − + . . . + (∆ − D − , ∆ ≥ and D ≥ , a (∆ , D, − ǫ ) -graph is regular. By Proposition 4.1, a (∆ , D, − ≥ D ≥ d rather than ∆ to denote the degree of Γ, as is customary. We call a cycle of length atmost 2 D in Γ a short cycle . Proposition 4.2 (Lemma 2 from [14])
Let Γ be a ( d, D, − -graph with d ≥ and D ≥ . Then D − ≤ g(Γ) ≤ D . Furthermore, if x is a vertex in Γ then either ( i ) x is contained in one (2 D − -cycle and no other short cycle; or ( ii ) x is contained in one Θ D , and every short cycle containing x is contained in this Θ D ; or ( iii ) x is contained in exactly two D -cycles whose intersection is a ℓ -path with ≤ ℓ ≤ D − , andno other short cycle. Each case is considered as a type. For instance, a vertex satisfying case ( i ) is called a vertex of Type( i ).While the statements of Proposition 4.2 and [14, Lemma 2] slightly differ, both assertions are clearlyequivalent. However, the statement of Proposition 4.2 is more consistent with the presentation ofour results and allows us to make the following observation, which will be used implicitly throughoutthe paper. Observation 4.1
Let Γ be a ( d, D, − -graph with d ≥ and D ≥ , and C a short cycle in Γ .Then all vertices in C are of the same type.
5n view of Proposition 4.2, we define the following concepts:We say that the vertex x ′ is a repeat of the vertex x with multiplicity m x ( x ′ ) (1 ≤ m x ( x ′ ) ≤
2) ifthere are exactly m x ( x ′ ) + 1 different paths of length at most D from x to x ′ . For vertices x and x ′ lying on a short cycle C , we denote the vertex x ′ by rep C ( x ) if x and x ′ are repeats.A vertex x is called saturated if x cannot belong to any further short cycle. If two 2 D -cycles C and C are non-disjoint, we say that C and C are neighbor cycles .From now on, whenever we refer to paths we mean shortest paths. As in [11], we extend the conceptof repeat to paths. For a path P = x − y of length at most D − D -cycle C , wedenote by rep C ( P ) the path P ′ ⊂ C defined as rep C ( x ) − rep C ( y ). We say that P ′ is the repeat of P in C and vice versa, or simply that P and P ′ are repeats in C . Lemma 4.1 (Odd Saturating Lemma)
Let Γ be a ( d, D, − -graph with d ≥ and D ≥ , and C a (2 D − -cycle in Γ . Let α be a vertex in C with repeat vertices α ′ , α ′ in C , γ a neighbor of α not contained in C , and µ , µ , . . . , µ d − the neighbors of α ′ not contained in C .Then there is in Γ a vertex µ ∈ { µ , µ , . . . , µ d − } and a D -cycle C such that γ and µ are repeatsin C , and C ∩ C = ∅ . Proof.
Let α ′ be the neighbor of α ′ in C other than α ′ . For 1 ≤ i ≤ d −
2, consider the path P i = γ − µ i . Since all vertices in C are saturated, P i cannot go through C and must be a D -path,so P i ∩ C = ∅ . Also, it follows that V ( P i ∩ P j ) = { γ } for any 1 ≤ i < j ≤ d −
2; otherwise eitherg(Γ) < D − α ′ would belong to an additional short cycle, both contradictions toProposition 4.2. See Fig. 2 ( a ). αα ′ α ′ α ′ γ ρµ µ µ d − α ′ α ′ α γ ρµ = µ k P k P C C ( a ) ( b ) . . . C P P P d − D − D − D − D − D D − Figure 2: Auxiliary figure for Lemma 4.1Let ρ be a neighbor of γ other than α , not contained in any of the paths P , P , . . . , P d − (there isexactly one such vertex). Consider a path P = ρ − α ′ . P cannot go through α ′ ; otherwise there6ould be a second short cycle ρP α ′ C αγρ in Γ containing α . Similarly, P cannot go through α ′ and consequently, it must go through a vertex µ k ∈ { µ , µ , . . . , µ d − } . Finally note that, since allvertices in C are saturated and 2 D − ≤ g(Γ) ≤ D , P must be a D -path, V ( P ∩ P k ) = { µ k } and V ( P ∩ C ) = { α ′ } .This way, we obtain there is a vertex µ = µ k and a 2 D -cycle C = γρP µP k γ such that γ and µ arerepeats in C , and C ∩ C = ∅ (Fig. 2 ( b )). ✷ Lemma 4.2 (Even Saturating Lemma)
Let Γ be a ( d, D, − -graph with d ≥ and D ≥ , and C a D -cycle in Γ . Let α, α ′ be two vertices in C such that α ′ = rep C ( α ) , γ a neighbor of α notcontained in C , and µ , µ , . . . , µ d − the neighbors of α ′ not contained in C . Suppose there is no shortcycle in Γ containing the edge α ∼ γ and intersecting C at a path of length greater than D − .Then there is in Γ a vertex µ ∈ { µ , µ , . . . , µ d − } and a short cycle C such that γ and µ are repeatsin C , and C ∩ C = ∅ . Proof.
Let α ′ , α ′ be the neighbors of α ′ contained in C . First, consider a path P = γ − α ′ . Since thereis no short cycle in Γ containing the edge α ∼ γ and intersecting C at a path of length greater than D − P must be a D -path and cannot go through α ′ or α ′ . Therefore, the path P must go throughone of the neighbors of α not contained in C (say µ ). In addition, we have that V ( P ∩ C ) = { α ′ } .See Fig. 3 ( a ).Let ρ , ρ , . . . , ρ d − be the neighbors of γ other than α , not contained in P . For 1 ≤ i ≤ d − P i = ρ i − α ′ . Since there is no short cycle in Γ containing the edge α ∼ γ andintersecting C at a path of length greater than D − P i must have length at least D − { α ′ , α ′ , γ } . Consequently, P i must go through one of the vertices in { µ , µ , . . . , µ d − } . Note also that V ( P i ∩ C ) = { α ′ } and that V ( P i ∩ P j ) ⊆ { α ′ } ∪ { µ , µ , . . . , µ d − } ,for any 1 ≤ i < j ≤ d − j (1 ≤ j ≤ d − P j goes through µ then P j must be a D -path and there isa (2 D − C = γP µ P j ρ j γ in Γ such that γ and µ = µ are repeats in C , and C ∩ C = ∅ .This case is depicted in Fig. 3 ( b ).If, on the other hand, there is no j (1 ≤ j ≤ d −
2) such that P j goes through µ then there mustexist a vertex µ k (2 ≤ k ≤ d −
2) and paths P r , P s (1 ≤ r < s ≤ d −
2) such that both P r and P s gothrough µ k . Since g(Γ) ≥ D −
1, at most one of the paths P r , P s has length D −
1. If one of thesepaths (say P r ) has length D − D − C = γρ r P r µ k P s ρ s γ in Γ such that γ and µ = µ k are repeats in C , and C ∩ C = ∅ (as in Fig. 3 ( c )). If both P r and P s are D -pathsthen there is a 2 D -cycle C = γρ r P r µ k P s ρ s γ in Γ such that γ and µ = µ k are repeats in C , and C ∩ C = ∅ (as in Fig. 3 ( d )). ✷ αα αα ′ α ′ α ′ γµ ρ ρ ρ d − α ′ α ′ α ′ µ µ k µ k γγ γ ( a ) ( b )( c ) ( d ) ρ j ρ r ρ s ρ r ρ s C CC C C C C P P P j P r P s P r P s ... D − D − D − D − D − D − D − D − D − Figure 3: Auxiliary figure for Lemma 4.2
The extension of the concept of repeat to short cycles was introduced in [11] in the context of bipartitegraphs missing the bipartite Moore bound by 4 vertices. Here, inspired by the ideas put forwardin [11], we extend the concept of repeat to 2 D -cycles of graphs of defect 2; see the Repeat CycleLemma. Lemma 4.3 (Repeat Cycle Lemma)
Let Γ be a ( d, D, − -graph with d ≥ and D ≥ , and C a D -cycle in Γ . Let { C , C , . . . , C k } be the set of neighbor cycles of C , and I i = C i ∩ C for ≤ i ≤ k . Suppose at least one I j , for j ∈ { , . . . , k } , is a path of length smaller than D − . Thenthere is an additional D -cycle C ′ in Γ intersecting C i at I ′ i = rep C i ( I i ) , where ≤ i ≤ k . Proof.
We denote the neighbors of C by C , C , . . . C k and their corresponding intersection pathswith C by I = x − y , I = x − y , . . . , I k = x k − y k in such a way that C = x I y x I y . . . x k I k y k x .For 1 ≤ i ≤ k , we also denote the repeats of I i by I ′ i = x ′ i − y ′ i , where x ′ i = rep C i ( x i ) and y ′ i = rep C i ( y i )(see Fig. 4 ( a )). 8 y y ′ x x ′ CC C x x k y y y k y ′ y ′ y ′ k x ′ x ′ x ′ k C C k ... ... I k I I I I ′ I ′ I ′ I ′ k x y y ′ x x ′ CC C x x k y y y k y ′ y ′ y ′ k x ′ x ′ x ′ k C C k I k I I I I ′ I ′ I ′ I ′ k ... ... ( a ) ( b ) Figure 4: Auxiliary figure for Lemma 4.3For 1 ≤ i ≤ k , consider the cycles C i and C ( i mod k )+1 .Suppose that I i is a path of length smaller than D −
1. Since y i is saturated, there cannot be ashort cycle in Γ, other than C , containing the edge y i ∼ x ( i mod k )+1 . Since I i is a path of lengthsmaller than D −
1, we apply the Even Saturating Lemma (mapping C i to C , y i to α , y ′ i to α ′ and x ( i mod k )+1 to γ ) and obtain an additional short cycle C in Γ such that x ( i mod k )+1 is a repeatin C of a neighbor µ C i of y ′ i , and C ∩ C i = ∅ . Since x ( i mod k )+1 is saturated, we have thatnecessarily C = C ( i mod k )+1 , which, in turn, implies µ = x ′ ( i mod k )+1 . In other words, it follows that y ′ i ∼ x ′ ( i mod k )+1 ∈ E (Γ).If, instead, I i is a ( D − I ( i mod k )+1 must be a path of length smaller than D −
1, otherwisethere would not exist a path I j (1 ≤ j ≤ k ) of length smaller than D −
1, contrary to our assumptions.Therefore, we can apply the above reasoning and deduce that x ′ ( i mod k )+1 ∼ y ′ i ∈ E (Γ).This way we obtain a subgraph Υ = S ki =1 (cid:0) I ′ i ∪ y ′ i ∼ x ′ ( i mod k )+1 (cid:1) = x ′ I ′ y ′ x ′ I ′ y ′ . . . x ′ k I ′ k y ′ k x ′ in-tersecting C i at I ′ i for 1 ≤ i ≤ k (see Fig. 4( b ), where part of the subgraph Υ is highlighted inbold).We next show that Υ must be indeed a cycle. Claim 1.
Υ is a 2 D -cycle. Proof of Claim 1.
First note that Υ is connected and that | Υ | ≤ D . By Proposition 4.2, unlessΥ is a 2 D -cycle, Υ contains no short cycle. If the neighbors of C are pairwise disjoint then Υ is a2 D -cycle. Suppose that some neighbors of C are non-disjoint and that Υ is not a cycle, then Υ is atree.Let z ∈ C ℓ be an arbitrary leaf in Υ. If the repeat path I ′ ℓ = x ′ ℓ − y ′ ℓ had length greater than 0, then z would have at least two neighbors in Υ. Therefore, I ℓ = C ∩ C ℓ contains exactly one vertex, and9hus, x ℓ = y ℓ and z = x ′ ℓ = y ′ ℓ .Recall we do addition modulo k on the subscripts of the vertices and the superscripts of the cycles.Since x ′ ℓ ∼ y ′ ℓ − and x ′ ℓ ∼ x ′ ℓ +1 are edges in Υ, it holds that y ′ ℓ − and x ′ ℓ +1 denote the same vertex.Let u ′ ℓ − , v ′ ℓ − be the neighbors of y ′ ℓ − in C ℓ − ; u ′ ℓ +1 , v ′ ℓ +1 the neighbors of x ′ ℓ +1 in C ℓ +1 ; and u ℓ , v ℓ the neighbors of x ℓ in C ℓ . We have that V ( C ℓ − ∩ C ℓ +1 ) = { y ′ ℓ − } , otherwise there would be a thirdshort cycle in Γ containing x ℓ . In particular, the vertices in { u ′ ℓ − , v ′ ℓ − , u ′ ℓ +1 , v ′ ℓ +1 , x ′ ℓ } are pairwisedistinct and d ≥
5. See Fig. 5 ( a ) and ( b ) for two drawings of this situation.Now consider a path P = x ℓ − y ′ ℓ − . Since x ℓ cannot be contained in a further short cycle,we have that P must be a D -path and go through a neighbor w ′ ℓ − of y ′ ℓ − not contained in { u ′ ℓ − , v ′ ℓ − , u ′ ℓ +1 , v ′ ℓ +1 , x ′ ℓ } , which implies d ≥
6. By similar arguments, we obtain that P mustgo through a neighbor w ℓ of x ℓ not contained in { y ℓ − , x ℓ +1 , u ℓ , v ℓ } .Finally, let t , t , . . . , t d − denote the vertices in N ( x ℓ ) −{ y ℓ − , x ℓ +1 , u ℓ , v ℓ , w ℓ } ; see Fig. 5 ( c ). Considera path Q i = t i − y ′ ℓ − . Since x ℓ cannot be contained in a further short cycle, Q i must be a D -pathand go through a neighbor of y ′ ℓ − not contained in { u ′ ℓ − , v ′ ℓ − , u ′ ℓ +1 , v ′ ℓ +1 , x ′ ℓ , w ′ ℓ − } . Therefore, wehave that d ≥ Q r and Q s containing acommon neighbor of y ′ ℓ − . This way, x ℓ would be contained in a third short cycle, a contradiction.As a result, we conclude that the repeat graph Υ of C is indeed a 2 D -cycle C ′ as claimed. Thiscompletes the proof of Claim 1, and thus, of the lemma. ✷ We call the aforementioned cycle C ′ the repeat of the cycle C in Γ, and denote it by rep( C ). Somesimple consequences of the Repeat Cycle Lemma follow next. Corollary 4.1 (Repeat Cycle Uniqueness)
If a D -cycle C has a repeat cycle C ′ then C ′ isunique. Corollary 4.2 (Repeat Cycle Symmetry) If C ′ = rep( C ) then C = rep( C ′ ) . Corollary 4.3
Let Γ be a ( d, D, − -graph with d ≥ and D ≥ . Let C, C be two D -cycles in Γ which intersect at a path I of length smaller than D − , and set I ′ = rep C ( I ) . Then the repeatcycle of C intersects C at I ′ . Corollary 4.4 (Handy Corollary)
Let Γ be a ( d, D, − -graph with d ≥ and D ≥ , C a D -cycle in Γ , and x, x ′ repeat vertices in C . Let C and C be D -cycles other than C containing x and x ′ , respectively. Suppose that I = C ∩ C is a path of length smaller than D − . Then, setting y = rep C ( x ) and y ′ = rep C ( x ′ ) , we have that y and y ′ are repeat vertices in the repeat cycle of C . Proof.
We denote the k neighbor cycles of C by E , E , . . . E k and their respective intersection pathswith C by I = x − y , I = x − y , . . . , I k = x k − y k in such a way that C = x I y x I y . . . x k I k y k x .For 1 ≤ j ≤ k , we also denote I ′ j = x ′ j − y ′ j , where x ′ j = rep E j ( x j ) and y ′ j = rep E j ( y j ).10 C ℓ − C ℓ x ℓ +1 y ℓ = x ℓ C ℓ +1 ... ... z = x ′ ℓ = y ′ ℓ y ℓ − v ′ ℓ − x ′ ℓ +1 = y ′ ℓ − u ′ ℓ − x ′ ℓ +1 = y ′ ℓ − v ′ ℓ +1 u ′ ℓ +1 u ℓ v ℓ CC ℓ − C ℓ y ℓ = x ℓ ... ... y ℓ − v ′ ℓ − u ′ ℓ − x ′ ℓ +1 = y ′ ℓ − v ′ ℓ +1 u ′ ℓ +1 u ℓ v ℓ ( a ) ( b ) CC ℓ − C ℓ x ℓ +1 y ℓ = x ℓ C ℓ +1 ... ... z = x ′ ℓ = y ′ ℓ y ℓ − v ′ ℓ − x ′ ℓ +1 = y ′ ℓ − u ′ ℓ − x ′ ℓ +1 = y ′ ℓ − v ′ ℓ +1 u ′ ℓ +1 w ′ ℓ − w ℓ u ℓ v ℓ ... t t d − P ( c ) C ℓ +1 x ℓ +1 z Figure 5: Auxiliary figure for Claim 1 of Lemma 4.3.Obviously, for some r, s (1 ≤ r, s ≤ k ) we have that C = E r , C = E s , x ∈ I r , x ′ ∈ I s , y ∈ I ′ r , and y ′ ∈ I ′ s . We may assume r < s . By the Repeat Cycle Lemma, the vertices y and y ′ belong to the repeat cycle C ′ of C . Then the paths xI r y r x r +1 I r +1 y r +1 . . . x s − I s − y s − x s I s x ′ ⊂ C and yI ′ r y ′ r x ′ r +1 I ′ r +1 y ′ r +1 . . . x ′ s − I ′ s − y ′ s − x ′ s I ′ s y ′ ⊂ C ′ are both D -paths in Γ, and the corollary follows. ✷ Main Results ( d, D, − -graphs Proposition 5.1 A ( d, D, − -graph Γ with d ≥ and D ≥ does not contain (2 D − -cycles. Proof.
Suppose, by way of contradiction, that there is a (2 D − C in Γ.Let p , p be two repeat vertices in C , and q a neighbor of p not contained in C . According to theOdd Saturating Lemma, there are both a neighbor q of p not contained in C and a 2 D -cycle D ,such that q and q are repeats in D (see Fig. 6 ( a )). p p p p p q q q q q p p q q D D D D D ( a ) ( b ) C C D − D − D − Figure 6: Auxiliary figure for Proposition 5.1For 1 ≤ i ≤
3, denote by p i +2 the repeat of p i +1 in C other than p i . We now apply the Odd SaturatingLemma (mapping C to C , p to α , p to α ′ , q to γ ) and ascertain the existence of a 2 D -cycle D and a neighbor q of p not contained in C , such that q and q are repeats in D . For i = 3 , C to C , p i to α , p i +1 to α ′ , q i to γ ) weensure the existence of a 2 D -cycle D i and a neighbor q i +1 of p i +1 not contained in C , such that q i and q i +1 are repeats in D i . See Fig. 6 ( b ).Note that D ∩ D is a path of length at most 2 < D −
1; otherwise for some vertex t ∈ D ∩ D the cycle tD q p p q D t would have length at most 2 D −
1, a contradiction. Similarly, D ∩ D and D ∩ D are paths of length at most 2.We now apply the Handy Corollary. By mapping the cycle D to C , the vertex q to x , the vertex q to x ′ , the cycle D to C , the cycle D to C , the vertex q to y and the vertex q to y ′ , we obtainthat q and q are repeat vertices in the repeat cycle of D . Therefore, since q ∈ D , it follows that12 and D are repeat cycles and q = q . As a consequence, there is in Γ a cycle q p p p q of length4 < D −
1, a contradiction. ✷ From Propositions 4.2 and 5.1, it follows immediately that
Theorem 5.1
The girth of a ( d, D, − -graph Γ with d ≥ and D ≥ is D . Θ D Proposition 5.2 A ( d, D, − -graph Γ with d ≥ and D ≥ does not contain a subgraph isomor-phic to Θ D . Proof.
In this proof our reasoning resembles that of Proposition 5.1, and especially, of [11, Propo-sition 5.1 ].Suppose that Γ contains a subgraph Θ isomorphic to Θ D , with branch vertices a and b . Let p , p , p , p and p be as in Fig. 7 ( a ), and let q be one of the neighbors of p not contained inΘ.Since all vertices of Θ are saturated, there cannot be a short cycle in Γ containing any of the incidentedges of p , p , p , p or p which are not contained in Θ. According to this and by applying the EvenSaturating Lemma, there is an additional 2 D -cycle D in Γ such that q and one of the neighbors of p not contained in Θ (say q ) are repeats in D . Also, it follows that D ∩ Θ = ∅ . Analogously, byrepeatedly applying the Even Saturating Lemma, for 2 ≤ i ≤ D -cycle D i such that q i and one of the neighbors of p i +1 not contained in Θ (say q i +1 ) are repeatsin D i . Also, we have that D i ∩ Θ = ∅ (see Fig. 7 ( b )). D D D D a bp p q q p p q q p q a bp p p p q p ( a ) ( b )Θ Θ Figure 7: Auxiliary figure for Proposition 5.213ote that D ∩ D is a path of length at most 2 < D −
1; otherwise for some vertex t ∈ D ∩ D there would be a cycle tD q p bp q D t of length at most 2 D to which the vertex b would belong, acontradiction. For similar reasons, the intersection paths D ∩ D and D ∩ D have both length atmost 2.We now apply the Handy Corollary. By mapping the cycle D to C , the vertex q to x , the vertex q to x ′ , the cycle D to C , the cycle D to C , the vertex q to y and the vertex q to y ′ , we obtainthat q and q are repeat vertices in the repeat cycle of D . Therefore, since q ∈ D , it follows that D and D are repeat cycles and q = q ; but then there is a cycle q p bp q in Γ of length 4 < D ,a contradiction. ✷ Corollary 5.1
Every vertex in a ( d, D, − -graph Γ with d ≥ and D ≥ is of Type ( iii ) . ( d, D, − -graphs In view of Corollary 5.1, the following corollary, which was obtained in [7], follows immediately.
Corollary 5.2 (Corollary 2.3 from [7])
The feasible values of d for ( d, D, − -graphs are re-stricted according to the following conditions.When D is even, d is odd.When D is a power of an odd prime, d − is a multiple of D .When D ≥ is a power of 2, d − is a multiple of D/ . Proposition 5.3
The number N D of D -cycles in a ( d, D, − -graph Γ with d ≥ and D ≥ isgiven by the expression N D = nD = d (cid:0) d − ... +( d − D − (cid:1) − D , where n is the order of Γ . Proof.
According to Proposition 4.2 and Corollary 5.1, every vertex of Γ is contained in exactlytwo 2 D -cycles. We then count the number N D of 2 D -cycles of Γ . Since the order of Γ is n =1 + d (cid:0) d −
1) + . . . + ( d − D − (cid:1) −
2, we have that N D = × (cid:16) d (cid:0) d − ... +( d − D − (cid:1) − (cid:17) D = d (cid:0) d − ... +( d − D − (cid:1) − D ,and the proposition follows. ✷ Lemma 5.1
Every two non-disjoint D -cycles in a ( d, D, − -graph Γ with d ≥ and D ≥ intersect at a path of length smaller than D − . roof. We follow a strategy very similar to the one used in the proof of [11, Lemma 5.1].Since Γ does not contain a graph isomorphic to Θ D , it is only necessary to prove here that anytwo non-disjoint 2 D -cycles in Γ cannot intersect at a path of length D −
1. Suppose, by way ofcontradiction, that there are two 2 D -cycles C and C in Γ intersecting at a path I of length D − v be an arbitrary vertex on I , and v ′ = rep C ( v ). Let C be the other 2 D -cycle containing v ′ ,and I = C ∩ C . If I were a path of length smaller than D − C would intersect C at a proper subpath of I containing v . This is a clear contradiction tothe fact that v is already saturated. Consequently, I must be a ( D − C is intersectedby exactly two 2 D -cycles, namely C and C , at two independent ( D − C , C , C , . . . , C m of pairwise disctinct 2 D -cycles in Γ such that C i intersects C i +1 at a path I i of length D − ≤ i ≤ m − C j ∩ C k = ∅ for any j, k ∈ { , . . . , m } such that 2 ≤ | i − j | ≤ m −
2. Let us denote the paths I = x − y , . . . , I m − = x m − − y m − in such away that, for 1 ≤ i ≤ m − x i ∼ x i +1 and y i ∼ y i +1 are edges in Γ. Also, let x ∈ N ( x ) ∩ ( C − I ), y ∈ N ( y ) ∩ ( C − I ), x m ∈ N ( x m − ) ∩ ( C m − I m − ), and y m ∈ N ( y m − ) ∩ ( C m − I m − ). Figure 8 ( a )shows this configuration. Set I = x − y and I m = x m − y m . Since the sequence C , C , C , . . . , C m is maximal and all the vertices in I , . . . , I m − are saturated, it follows that I = I m , and we haveeither x = x m and y = y m (as in Fig. 8 ( b )), or x = y m and y = x m (as in Fig. 8 ( c )).If x = x m and y = y m then m > D ; otherwise the cycle x x . . . x m x would have length at most2 D , contradicting the saturation of x . If, conversely, x = y m and y = x m then m > D ; otherwisethe cycle x x . . . x m y y . . . y m x containing x would have length at most 2 D , a contradiction. Forour purposes, it is enough to state m > D ≥ ∪ mi =1 C i , and q a neighbor of y not contained in Φ (see Fig. 9 ( a )).Since y is saturated, the edge q ∼ y cannot be contained in a further short cycle. We apply theEven Saturating Lemma (by mapping C to C , y to α , x to α ′ , and q to γ ), and obtain in Γ anadditional 2 D -cycle D such that q and one of the neighbors of x not contained in Φ (say q ) arerepeats in D , and D ∩ C = ∅ . Analogously, there exists an additional 2 D -cycle D such that q and a neighbor of y not contained in Φ (say q ) are repeats in D , and D ∩ C = ∅ ; an additional2 D -cycle D such that q and a neighbor of x not contained in Φ (say q ) are repeats in D , and D ∩ C = ∅ ; and an additional 2 D -cycle D such that q and a neighbor of y not contained in Φ(say q ) are repeats in D , and D ∩ C = ∅ . See Fig. 9 ( b ).Note that D ∩ D cannot be a ( D − t ∈ D ∩ D there would bea cycle tD q y y y q D t of length at most 4 + D − D − D − ≥ y is saturated and g(Γ) = 2 D . Analogously, D i ∩ D i +1 cannot be a ( D − i = 2 ,
3. 15 a ) x x x x m − x m − x m y y y y y m − y m − y m C C C C m − C m I I I I m − I m − I m I x = x m x x x x x x m − y y y y y y m − y = y m ( b ) . . .. . .. . . x = y m x x x x x y y y y y y m − y = x m ( c ) C C C C C C C C C C ... ... D − D − D − D − D − D − D − I I I I I I = I m I I I I I I m − I = I m x m − I m − x Figure 8: Auxiliary figure for Lemma 5.1We now apply the Handy Corollary as in the proofs of the previous theorems. By mapping the cycles D to C , D to C and D to C , and the vertices q to x , q to x ′ , q to y , and q to y ′ , it followsthat the vertices q and q are repeat vertices in the repeat cycle of D . Since q ∈ D , we have that D and D are repeat cycles and that q = q . This way, we obtain a cycle q y y y y y q in Γ oflength 6 < D , a contradiction.This completes the proof of the lemma. ✷ We are now in a position to prove our second main result.
Theorem 5.2
There are no ( d, D, − -graphs with even d ≥ and D ≥ . Proof.
Suppose there is a ( d, D, − d ≥ D ≥ . .. . . . . .. . .x x x x x y y y y y C C C C . . .. . . . . .. . .x x x x x y y y y y q q q q q D D D D C C C C ( a )( b ) D − D − D − D − D − q Figure 9: Auxiliary figure for Lemma 5.1According to Lemma 5.1, any two non-disjoint 2 D -cycles in Γ intersect at a path of length smallerthan D −
1, which means that every 2 D -cycle C in Γ has a repeat cycle C ′ (by the Repeat CycleLemma). Because of the uniqueness and symmetry of repeat cycles, the number N D of 2 D -cyclesin Γ must be even.However, since d is even, the number N D = d (cid:0) d − ... +( d − D − (cid:1) − D of 2 D -cycles in Γ is odd, acontradiction. ✷ Note that Theorem 5.2 contains, as a special case, the result of the non-existence of (4 , D, − D ≥
4, which was claimed prematurely in [18].From Proposition 5.3 we easily derive the following results:
Theorem 5.3
There are no ( d, D, − -graphs with odd d ≥ , D ≥ and order n = d (cid:0) d −
1) + . . . + ( d − D − (cid:1) − D ) . Corollary 5.3
There are no ( d, D, − -graphs with odd d ≥ and D ≥ such that d ≡ , D ) . D ≥ d, D, − d ≥ d , by considering the set of all possible residuesof d in the division by D . If, for some r ∈ { , , . . . , D − } , we have d ≡ r (mod D ) implies d (cid:0) d −
1) + . . . + ( d − D − (cid:1) − D ), then there are no ( d, D, − d ≥ d ≡ r (mod D ).Accordingly, the following table shows all values of 4 ≤ D ≤
16 and odd d ≥ d, D, − D d d ≡ , d ≡ d ≡ d ≡ d ≡ , d ≡ d ≡ , d ≡ d ≡ , d ≡ d ≡ ,
13 (mod 14)15 d ≡ ,
13 (mod 30)16 d ≡ , (4 , , − -graphs In this section we prove the non-existence of (4 , , − , D, − D ≥ D in a ( d, D, − d ≥ D ≥
4. We next give an alternative proof for d = 4 that covers also the case D = 3. Proposition 6.1 A (4 , D, − -graph Γ with D ≥ does not contain a subgraph isomorphic to Θ D . Proof.
Suppose that Γ contains a subgraph Θ isomorphic to Θ D , where α and α ′ are its branchvertices. Let α ′ , α ′ , α ′ , γ and µ be as in Figure 10 ( a ).First consider a path P = γ − α ′ . As α cannot belong to any further short cycle, P must go through µ and be a D -path. Let ρ and ρ be the neighbors of γ other than α and not contained in P .Consider a path P = ρ − α ′ . As α is saturated, P cannot go through α ′ , α ′ or α ′ , so it must gothrough µ and be a D -path. This way, γ is contained in a (2 D − C = γP µP ρ γ , and γ αα ′ α ′ α ′ γµ α ′ µ γ ( a ) ( b ) ρ Θ C P P D − D − D − D − Θ D − D − ρ α ′ α ′ α ′ α ′ Figure 10: Auxiliary figure for Proposition 6.1.becomes saturated. Analogously, a path P = ρ − α ′ must go through µ , causing the formation ofanother short cycle containing µ , a contradiction to Proposition 4.2 ( ii ). See Figure 10 ( b ). ✷ Next we prove that the girth of a (4 , , − Proposition 6.2 A (4 , , − -graph Γ has girth . Proof.
We proceed by contradiction, supposing there is a 5-cycle C in Γ. In view of Proposition 4.2,the graph Γ contains the subgraph G of Fig. 11, where T i denotes the enclosed set of 6 vertices atdistance 2 from x i , for 1 ≤ i ≤ C T T T T T x x x x x G Figure 11: Auxiliary figure for Proposition 6.2.Since | Γ | = 51 and | G | = 45, there is a set X ⊂ V (Γ) such that | X | = 6 and X ∩ V ( G ) = ∅ . Anyvertex x ∈ X must be adjacent to a vertex in T i , for 1 ≤ i ≤
5, in order to reach x i in at most 3steps. However, this is clearly impossible since Γ has degree 4. ✷
19n view of Propositions 4.2, 6.1, 5.1 and 6.2, it follows that every vertex in a (4 , D, − D ≥ D -cycles. Proposition 6.3
The number N D +1 of (2 D + 1) -cycles in a (4 , D, − -graph Γ with D ≥ is givenby N D +1 = × D (2 × D − D +1 . Proof.
The number of (2 D + 1)-cycles in Γ is closely related to the number of edges involving onlyvertices at distance D from any vertex x in Γ. The number of vertices at level D is 4 × D − − F of edges involving only vertices at distance D from x is | F | = 2 × × D − − × D − , since x is contained in exactly two 2 D -cycles C and C .Denote by y and y the vertices at distance D from x on C and C , respectively. Before proceedingto count, we prove that y ∼ y E (Γ). Claim 1. y ∼ y E (Γ). Proof of Claim 1.
Suppose, by way of contradiction, that y ∼ y ∈ E (Γ). Since g(Γ) = 2 D , itholds that V ( C ∩ C ) = { x } ; see Fig. 12. By Corollary 4.3, the repeat cycle C ′ of C intersects C exactly at y ; consequently, C ′ contains the edge y ∼ y . However, this contradicts the fact that C and its repeat cycle C ′ must be disjoint cycles. ✷ C C xy y Figure 12: Auxiliary figure for Proposition 6.3.Accordingly, we partition the set F into F , F and F , where F and F are the sets of edges in F adjacent to the vertices y and y , respectively, and F contains the remaining edges in F .Each edge in F or F determines two (2 D +1)-cycles containing x , while each edge from F determinesonly one (2 D + 1)-cycle containing x . Therefore, given that | F | = | F | = 2, we have that the numberof (2 D + 1)-cycles passing through the vertex x is2 | F | + 2 | F | + | F | = 4 + 4 + 2 × D − × D . D + 1)-cycles in Γ is given by the expression N D +1 = 2 × D (2 × D − D + 1 , and the proposition follows. ✷ Now we can readily prove Theorem 6.1.
Theorem 6.1
There is no (4 , , − -graph. Proof.
By Proposition 6.3, the number of 7-cycles in a (4 , , − × (2 × − / / ✷ Theorems 5.2 and 6.1 tell us that the (4 , , − d ) is the only (4 , D, − D ≥
2. Thus, we have successfully completed the census of all (4 , D, − In this paper, by exploiting the idea of extending the concept of repeats to paths and cycles, putforward in [11], we obtained the results summarized below.First, we proved that the girth of a ( d, D, − d ≥ D ≥ D . By obtainingnecessary conditions for the existence of ( d, D, − d ≥ D ≥
4, we proved thenon-existence of ( d, D, − d ≥ D ≥
4. This outcome, together with anon-existence proof of (4 , , − , D, − ǫ )-graphs with D ≥ ≤ ǫ ≤ Catalogue of (4 , D, -graphs with D ≥ . There is no Moore graph of degree 4 and diameter D ≥ Catalogue of (4 , D, − -graphs with D ≥ . There is no (4 , D, − D ≥ Catalogue of (4 , D, − -graphs with D ≥ . There is a unique (4 , , − d ).We proved the non-existence of ( d, D, − d ≥ D ≥ d ≡ , D ). Furthermore, our new necessary conditions allow us also to rule out the existence of graphsof defect 2 for many other values of d and D using a simple approach.21 .1 Remarks on the upper bound for N(∆ , D ) Our results improve the upper bound on N (∆ , D ) for many combinations of ∆ and D . Proposition 7.1
For even ∆ ≥ and D ≥ , N(∆ , D ) ≤ M(∆ , D ) − . In the particular case of ∆ = 4, we have that N(4 ,
2) = M(4 , − , D ) ≤ M(4 , D ) − D ≥ , D, − , D, − ≥ D ≥ Proposition 7.2
For odd ∆ ≥ and D ≥ such that ∆ (cid:0) −
1) + . . . + (∆ − D − (cid:1) − D ) , N(∆ , D ) ≤ M(∆ , D ) − . Corollary 7.1
For odd ∆ ≥ and D ≥ such that ∆ ≡ , D ) , N(∆ , D ) ≤ M(∆ , D ) − . Finally, we feel that the following conjectures also hold.
Conjecture 7.1
There are no (∆ , D, − -graph with ∆ ≥ and D ≥ . Conjecture 7.2
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