Maximal cubic graphs with diameter 4

Discrete Applied Mathematics 101 (2000) 53–61

Maximal cubic graphs with diameter 4

Dominique BusetFacult�e des Sciences Appliqu�ees, Universit�e Libre de Bruxelles, Service de Math�ematiques CP165,

50 Avenue Roosevelt, B-1050 Bruxelles, Belgium

Received 22 January 1999; revised 18 March 1999; accepted 24 May 1999

Abstract

We prove that there is no cubic graph with diameter 4 on 40 vertices. This implies that themaximal number of vertices of a (3,4)-graph is 38. ? 2000 Elsevier Science B.V. All rightsreserved.

MSC: 05C35

Keywords: Maximal cubic graphs; Diameter 4; Vertices

1. Introduction

The construction of large interconnection or microprocessor networks gave rise to the(�;D)-graph problem [13]: given two positive integers � and D, construct a connected(�;D)-graph (i.e. a graph of degree � and diameter D) with a maximum number ofvertices (a graph is of degree � if all its vertices have degree 6�, at least one ofthem being of degree �). The largest integer n such that there exists a (�;D)-graphwith n vertices will be denoted by n(�;D).Since the 1960s, the (�;D)-graph problem has been studied by many authors (see

for instance [3,6,7,11]), but very little is known about the exact values of n(�;D). Anupper bound for n(�;D) was given in 1958 by Moore [8], who proved that n(�;D)6(�(�−1)D−2)(�−2)−1 for �¿3 and n(2; D)62D+1. If we denote by M (�;D) theMoore bound, a Moore graph is a (�;D)-graph with exactly M (�;D) vertices. Thesegraphs are [14,1,10]: the complete graphs on �+1 vertices (and so n(�; 1)=�+1), thecircuits of length 2D+1 (and so n(2; D) = 2D+1), the Petersen graph (n(3; 2)= 10),the Ho�man–Singleton graph (n(7; 2) = 50) and a hypothetical graph on 3250 vertices(n(57; 2)63250). In 1981, Bannai and Ito [2] proved that, if D¿2, the circuits oflength 2D are the only (�;D)-graphs with M (�;D) − 1 vertices; this implies that,

E-mail address: [email protected] (D. Buset)

0166-218X/00/$ - see front matter ? 2000 Elsevier Science B.V. All rights reserved.PII: S0166 -218X(99)00204 -8

54 D. Buset / Discrete Applied Mathematics 101 (2000) 53–61

if D¿2; �¿3 and � 6∈ {3; 7; 57}, then n(�;D)6M (�;D)− 2. Therefore, three graphsdescribed by Elspas [13] in 1964 are maximal, which gives three new values of n(�;D);namely n(3; 3) = 20; n(4; 2) = 15 and n(5; 2) = 24 (see [4,5] for a nice description ofthese graphs). In 1992, Jorgensen [15] proved that, if D¿4, there is no (3; D)-graphwith exactly M (3; D) − 2 vertices. At present, no other value of n(�;D) is known.Note that an up-to-date table is maintained at:http:==www-mat.upc.es=grup de grafs=table g.htmlIn the particular case of (3; 4)-graphs, the above bounds imply that n(3; 4)643.

Doty [12] has constructed a (3; 4)-graph with 38 vertices (see also von Conta [17] fora non-isomorphic one), and so n(3; 4)¿38. Moreover, since a non-regular (3; 4)-graphhas at most 31vertices, it follows that a maximal (3; 4)-graph must be regular, and sohas an even number of vertices. Therefore n(3; 4)642. In 1993, Jorgensen [16] provedthat there is no (3; 4)-graph on 42 vertices, which implies that n(3; 4) = 38 or 40. Weshall prove the following

Theorem. There is no (3; 4)-graph with 40 vertices.

Corollary. n(3; 4) = 38; and so the two (3; 4)-graphs constructed by Doty and vonConta are maximal.

Since our proof is too long to be given here with all details, we shall just sketchthe main steps. The complete proof can be found in [9].

2. Preliminaries

All graphs considered in this paper will be connected, �nite, undirected, withoutloops and multiple edges. Let v be a vertex of a graph �. For any integer k ¿ 0,we shall denote by: Dk(v) the subgraph induced on the set of all vertices of � atdistance k from v; e(Dk(v); Dk−1(v)) the number of edges of � having one vertex inDk(v) and the other in Dk−1(v); c2k+1(v) the number of edges of Dk(v). For k¿2; letc2k(v) = e(Dk(v); Dk−1(v)) − |Dk(v)|. A 4-path will be any path of length at most 4,and a k-circuit any circuit of length k.The following lemma is due to Jorgensen [16].

Lemma 1. If � is a regular (3; 4)-graph with 40 vertices; then for any vertex v of �:

c3(v) = c4(v) = 0 and 6 = 6c5(v) + 4c6(v) + 2c7(v) + c8(v):

3. Non existence of a (3; 4) -graph with 40 vertices

Suppose that there exists a regular (3; 4)-graph � with 40 vertices. Our goal is toshow that such a graph must have girth 8, the �nal contradiction arising rather quicklyfrom this fact.

D. Buset / Discrete Applied Mathematics 101 (2000) 53–61 55

Proposition 1. � has girth at least 6.

Detailed proof. For any vertex v of �, Lemma 1 implies that c3(v)= 0, and so � hasno triangle. It follows that there are exactly 6 edges between D1(v) and D2(v). Sincec4(v) = 0, this implies that |D2(v)|= 6. Thus, there is no 4-circuit in �.Suppose that � has girth 5 and let v be any vertex belonging to a 5-circuit of

�. By Lemma 1, c5(v) = 1 and c6(v) = c7(v) = c8(v) = 0, which means that thereis exactly one edge in D2(v), that any vertex of D3(v) (resp. D4(v)) is adjacent toonly one vertex of D2(v) (resp. D3(v)), and that there is no edge in D3(v). Moreover,|D1(v)| = 3; |D2(v)| = 6; |D3(v)| = 10 and |D4(v)| = 20: If vi (i = 1; : : : ; 40) are thevertices of � and if v1 ∼ v2 ∼ v6 ∼ v7 ∼ v3 ∼ v1, we may assume, without loss ofgenerality, that � contains the subgraph �′ de�ned by the following adjacencies: v1 ∼v2; v2i+1 ∼ vi ∼ v2i+2 (i=1; : : : ; 5); v6 ∼ v13; v7 ∼ v14; v2j−1 ∼ vj ∼ v2j (j=8; : : : ; 20).Let Ar={v7+2r ; v8+2r} (r=1; : : : ; 12); BS=

⋃1j=0 A4+2s−j (s=1; 2; 3; 4) and Ct={v33+2k+t

such that k = 0; 1; 2; 3} (for t = 0; 1). It follows that D1(v7) = {v3; v6; v14}, D2(v7) ={v1; v2; v8; v13}∪A10 with v1 ∼ v2; D3(v7)⊃{v4; v5}∪A4 ∪A9 and D4(v7)⊃A1 ∪A2 ∪B4(these vertices are not in D3(v7) because c7(v7)= 0). Therefore, there is only one wayof connecting the vertices of B1 to v7 by 4-paths, namely v7 ∼ v14 ∼ v27 or v28 ∼�i ∼ v17+i (∀i ∈ {0; 1; 2; 3}) with �i ∈ {v33+2i ; v34+2i}. By symmetry, we may assumethat �i = v33+2i for every i ∈ {0; 1; 2; 3}: Therefore D3(v7) = {v4; v5} ∪ A4 ∪ A9 ∪ C0and B1⊂D4(v7). Symmetrically, D1(v6) = {v2; v7; v13}, D2(v6) = A9 ∪ {v1; v3; v5; v14}(with v1 ∼ v3), D3(v6)⊃{v4; v8} ∪ A2 ∪ A10 and D4(v6)⊃A1 ∪ A4 ∪ B2 ∪C0. As above,the existence of 4-paths between v6 and the vertices of C implies that v25 or v26 isadjacent to �j with �j ∈ {v33+2j; v34+2j} (j = 0; 1; 2; 3). Therefore, the four vertices �jbelong to D3(v6) and, since C0⊂D4(v6), we have necessarily �j = v34+2j for everyj ∈ {0; 1; 2; 3}. It follows that D3(v6) = {v4; v8} ∪ A2 ∪ A10 ∪ C1, B1⊂D4(v6) andC1⊂D4(v7). These new adjacencies imply that, in order to connect v7−i (i = 0; 1) tothe four vertices of B2+2i by 4-paths, there must be a maximal matching betweenB2+2i and Ci (we shall say that there is a maximal matching between two disjointsubgraphs X and Y if any vertex of X is adjacent to only one vertex of Y and ifconversely any vertex of Y is adjacent to only one vertex of X ). On the other hand,D1(v3) = {v1; v7; v8}; D2(v3) = {v2; v4; v6; v14} ∪ A4; D3(v3) = {v5; v13} ∪ A1 ∪ A10 ∪ B4and D4(v3)⊃A2 ∪ A9 ∪ B1 ∪ C0 ∪ C1. Therefore, since the vertices of B2 must be at adistance at most 4 from v3, there is a maximal matching between B2 and B4. Note thatthe subgraphs of � induced on Bi and Cj (i=1; : : : ; 4; j=0; 1) have no edge. For anyk ∈ {1; 2; 3; 4}, let ck0 (resp. bk4) be the vertex of C0 (resp. of B4) adjacent to v20+k , andlet bk1 (resp. c

k1) be the vertex of B1 (resp. C1) adjacent to c

k0 (resp. b

k4). Then D1(v11)=

{v5; v21; v22}; D2(v11)={v2; v12; c10; c20; b14; b24}; D3(v11)⊃{v1; v6; b11; b21; c11; c21; a14; a110}∪A8,where a14 (resp. a

110) is the vertex of A4 (resp. of A10) adjacent to b

14 (resp. to c

10). Note

that only two vertices of D3(v11) are still to be determined: one in A4 adjacent to b24,and one in A10 adjacent to c20. However, D4(v11)⊃{v3; v4; v7; v8; v13; v14; c30; c40; b34; b44; a11}where a11 is the vertex of A1 adjacent to b

11. Thus, if a

14 ∼ b24 (resp. a

110 ∼ c20), the

second vertex a24 ∈ A4 (resp. a210 ∈ A10) is adjacent to three vertices v8; b34; b44 (resp.


v14; c30; c40) of D4(v11), which contradicts the fact that � has diameter 4. Therefore,

a14 � b24 (resp. a110 � c20), and so b

24 ∼ a24 (resp. c

20 ∼ a210). For the same reason,

a11 � b21 (otherwise the second vertex a21 ∈ A1 would be adjacent to three vertices

v4; b31; b41; all of them being at a distance at least 4 from v11), and then b21 ∼ a21. Thus,

D3(v11) = {v1; v6; b11; b21; c11; c21} ∪ A4 ∪ A8 ∪ A10 and a21 ∈ D4(v11). Since c31; c41; b31; b41; v25and v26 must belong to D4(v11), it follows that there is a maximal matching between{b11; b21} and {c31; c41}, between {c11; c21} and {b31; b41} and between {c11; c21} and A9. Theprevious adjacencies also imply that there is a maximal matching between {c30; c40} andA10, between {b31; b41} and A1, and between {b34; b44} and A4. Let us denote by b (resp. b′)the vertex of {b34; b44} adjacent to a14 (resp. a24); by a18 (resp. a28) the vertex of A8 adjacentto b (resp. b′); by 0 (resp. ′0) the vertex of {c30; c40} adjacent to a18 (resp. a28); by 1(resp. ′1) the vertex of {c31; c41} adjacent to b (resp. b′); by � (resp. �′) the vertex of{b31; b41} adjacent to 0 (resp. ′0), and �nally by a19 (resp. a29) the vertex of A9 adjacentto c11 (resp. c

21). It follows that D1(a

14) = {v8; b14; b}; D2(a14) = {v3; a24; v21; c11; a18; 1}

and D3(a14)⊃{v1; v7; b24; b′; v11; c10; a19; v12; 0}. In fact, only three vertices of D3(a14) aremissing: a vertex of {�; �′} adjacent to c11, a vertex of {b11; b21} adjacent to 1, anda vertex of {a19; a29} adjacent to 1. It follows that all the vertices of � which arenot in B1 ∪ A9 ∪ Di(a14) (i = 1; 2; 3) must necessarily belong to D4(a14). In particular,since � has diameter 4, the vertices of {c20; ′0}∪A1∪A10 must be at a distance 4 froma14. The above adjacencies already give D4(a

14)⊃{v2; v4; v6; v14; v22; c21; ′1; a28; v5; a110; v13}.

However, a210 is adjacent to the vertices v14 and c20 (both of which belong to D4(a

14))

and also to a vertex of { 0; ′0} (with 0 ∈ D3(a14) and ′0 6∈ D3(a14)). Thus, a

210 ∈

D4(a14) forces a210 ∼ 0, and so a110 ∼ ′0 (owing to the maximal matching between

A10 and { 0; ′0}= {c30; c40}). Therefore ′0 ∈ D4(a14) forces �′ ∈D3(a14), and so �′ ∼ c11.The maximal matching between {�; �′} and {c11; c21} implies that � ∼ c21. For similarreasons applied to the neighbours of c20, we have b

21 ∼ 1, and so b11 ∼ ′1. It follows

that {b21; �′}⊂D3(a14), and so a21 ∈ D4(a14) because a21 ∼ b21. Therefore a

11 ∈ D4(a14)

implies a11 ∼ �′, and so a21 ∼ �. On the other hand, a29 is adjacent to the vertices v13and c21, and also to a vertex of { 1; ′1}. In order to connect a29 to a14 by a 4-path, a29must be adjacent to 1, and so a19 ∼ ′1.All the edges of � are now known. Unfortunately, a110 is at a distance at least 5

from a21 (indeed, a21 is adjacent to the three vertices v4; b

21, and �, all of them belonging

to D4(a110)); which contradicts the hypothesis that � has diameter 4.

We conclude that � has girth at least 6. In particular, for any vertex v of �, c5(v)=0;and so 6 = 4c6(v) + 2c7(v) + c8(v):

Proposition 2. � has girth at least 7.

Proof (sketch). Suppose that � has girth 6. For any vertex v of a 6-circuit, Lemma1 implies that c6(v) = 1 and 2c7(v) + c8(v) = 2. Therefore, c7(v) = 0 or 1, and D3(v)contains one vertex adjacent to exactly two vertices of D2(v) and ten vertices adjacentto exactly one vertex of D2(v). Let vi (i=1; : : : ; 40) be the vertices of � and v1 ∼ v2 ∼


v6 ∼ v14 ∼ v7 ∼ v3 ∼ v1 be a 6-circuit. We may assume, without loss of generality,that � contains the subgraph �′ de�ned by v1 ∼ v2, v2i+1 ∼ vi ∼ v2i+2 (i = 1; : : : ; 6)and v2j ∼ vj ∼ v2j+1 (j = 7; : : : ; 10). Moreover, any vertex v of a 6-circuit C of �belongs to a subgraph �′

v isomorphic to �′. If we denote by �i(v) (for i = 0; 2; 4; 6)

the set of vertices of D3(v) at distance i from u in �′v; u being the opposite vertex of

v in C, we shall say that an edge of D3(v) is of type Eij if one of its vertices belongsto �i(v) and the other to �j(v):If c7(v1) = 1, there is a unique edge E in D3(v1) which must be of type E06, E24,

E26, E44 or E46 (because � has girth 6).If E is of type E06, we may assume, without restriction of generality, that E =

[v14; v18]. Let xi and x′i (i= 11; : : : ; 21; i 6= 14 and x18 = x′18) be the vertices of D4(v1)adjacent to vi. In order to connect v14 to v20 and v21 by 4-paths, we must have x20 orx′20 ∼ x18 ∼ x21 or x′21. Therefore D3(v18) contains at least 4 edges, which contradictsLemma 1.The other four possibilities for E can be reeled out by similar arguments, which will

not be detailed here. It follows that any vertex belonging to a 6-circuit of � does notbelong to a 7-circuit or to another 6-circuit of �.If c7(v1) = 0, Lemma 1 implies that c8(v1) = 2, and so either exactly one vertex of

D4(v1) is adjacent to three vertices of D3(v1), or two vertices of D4(v1) are adjacentto two vertices of D3(v1). In both cases, the other vertices of D4(v1) are adjacent toexactly one vertex of D3(v1). If v is a vertex of a 6-circuit of �, we shall say thatw ∈ D4(v1) is of type Vij (resp. Vijk) (i; j; k ∈ {0; 2; 4; 6}) if w has one neighbour in�i(v) and another in �j(v) (resp. one in �i(v), one in �j(v) and another in �k(v)).Note that a vertex of type Vijk is also of type Vij, Vjk and Vik . It follows that anyvertex w ∈ D4(v1), which is adjacent to at least two vertices of D3(v1), must be of typeV24; V44 or Vi6 (i = 0; 2; 4; 6). If w has three neighbours in D3(v1), w must be of typeV066; V246; V266; V446 or V466. A long case by case analysis, using various countingarguments, shows that all these possibilities lead to contradictions. In order to givesome idea of this part of the proof, we shall describe the �rst step a little bit.If there exists a vertex w of type V66 in �, we may assume that v18 ∼ w ∼ v20.

It follows that � contains at least two 6-circuits, namely v1 ∼ v2 ∼ v6 ∼ v14 ∼ v7 ∼v3 ∼ v1 and v4 ∼ v9 ∼ v18 ∼ w ∼ v20 ∼ v10 ∼ v4. Let xi and x′i be the vertices ofD4(v1) adjacent to vi (for i = 14; 18; : : : ; 21, with x14 = x′14 and x18 = w = x20). Notethat x14 might belong to the subgraph A of D4(v1) induced by all the neighbours ofthe vertices of �6(v1). Thus, A contains 7 vertices and has no edges.If x14 ∈ A, all the vertices of D4(v1)−{w; x14} are adjacent to exactly one vertex of

D3(v1) (because c8(v1) = 2), and so xi and x′i (for i = 11; : : : ; 17; i 6=14) are 12 newvertices of D4(v1). If x14 6= w, we may assume (by symmetry) that x14 ∈ {x′18; x19; x′19}=B. Since we must connect v9 to the vertices of �2(v1) ∪ �4(v1) by 4-paths, it followsthat for any i = 11; : : : ; 17 (i 6= 14) there is a vertex in {w} ∪ B adjacent to xi or x′i .However, xi � x14 � x′i for these values of i (indeed v2 and v3 cannot belong to two6-circuits), and so at least one vertex of �2(v1) ∪ �4(v1) is at a distance at least 5from v9, contradicting the fact that � has diameter 4. Therefore x14 = w, and in order


to connect w to the vertices of �4(v1) by 4-paths, the symmetries imply that we maychoose x′18 ∼ x11, and then also x′20 ∼ x17. It follows that there is a maximal matchingbetween {x′18; x′20} and {x12; x16}. Since w must be at a distance at most 4 fromx′i (i = 11; 12; 16; 17) with � of girth at least 6, there must exist maximal matchingsbetween {x11; x12} and {x′16; x′17}, and between {x16; x17} and {x′11 ; x′12}. It follows(in order to be able to connect v11 to v7 and v10 by 4-paths) that x′11, which is alreadyadjacent to v11 and to a vertex of {x16; x17}, must also be adjacent to two verticesof {x15; x′15; x21; x′21}, a contradiction because � is regular of degree 3. In particular,this also proves that � has no vertex of type V066.If x14 6∈ A, let B (resp. C) be the set of vertices of D4(v1)−A adjacent to a vertex of

{v11; v12; v13} (resp. {v15; v16; v17}). It follows that B∪C contains 11 vertices whichhave no neighbour in �0(v1)∪�6(v1). Moreover, there is no edge in the subgraphs of� induced on B∪{x14} and C ∪{x14}. Therefore x14 has two neighbours in A. On theother hand, in order to connect v4 to the vertices of B ∪ C by 4-paths, each of the 11vertices of B ∪ C must be adjacent to a vertex of A, and so there is no vertex of Aadjacent to a vertex of �2(v1) ∪ �4(v1). In particular, since w ∈ A, there is no vertexof type V266 or V466 in �.If y is the vertex of D4(v1) − {w} adjacent to two vertices of D3(v1), we deduce

that y 6∈ A ∪ {x14}, and then y ∈ B ∪ C is of type V24 or V44 (in fact, {y}= B ∩ C).Therefore y is adjacent to at least one vertex of �4(v1) and, by symmetry, we mayassume that y ∼ v11. Thus, either y ∼ v15 or y is adjacent to a vertex of {v16; v17}.Nevertheless, w ∼ � ∈ B ∪ {x14} ∼ � ∈ {v11; v12; v13; v14} and w ∼ ∈ C ∪ {x14} ∼� ∈ {v14; v15; v16; v17} (in order to connect w to v2 and v3 by 4-paths). Since w isalready adjacent to two vertices, it follows that � = , and so w is adjacent to x14 ory. However, x14 � w (otherwise, since x14 has two neighbours in A, either there is a4-circuit in � or w belongs to two 6-circuits), and so w ∼ y. It follows that a 4-pathbetween v6 and w cannot contain y, and so must contain z ∈ {v18; v20}. Therefore zmust be adjacent to a vertex of {x14; x13; x′13}, and so there are at least 3 vertices inD4(v1) adjacent to two vertices of D3(v1), which contradicts the fact that c8(v1) = 2.To complete the proof of Proposition 2, it su�ces to follow this sequence of as-

sumptions for the type of a vertex w ∈ D4(v1) adjacent to at least two vertices ofD3(v1): �rst type V44 or V24 (which contains the types V446 and V246), next V26 or V06,and �nally type V46. All these cases lead to contradictions, but the detailed proof israther long.

Proposition 3. � has girth 8.

Proof (idea). If � has girth 7, then for any vertex v of a 7-circuit of �, Lemma 1implies that 16c7(v)63. If 26c7(v)63, the various cases to be considered dependon the position of the edges of D3(v1), which could be a path, or the disjoint unionof a path and an edge, or the union of disjoint edges. The analysis of all these casesis very similar to the analysis done in the preceding proof. If c7(v) = 1, let y0 ∼ y1 ∼· · · ∼ y6 ∼ y0 be a 7-circuit of �; this circuit must contain the subgraph Hy0 whose


Fig. 1.

vertices are yi; zi; ti; t′i (i = 0; 1; : : : ; 6) and whose edges are yi+1 ∼ yi ∼ zi; ti ∼zi ∼ t′i (i = 0; 1; : : : ; 6; the indices being computed modulo 7). Let S =

⋃6i=0 Si where

Si = {ti; t′i} (i = 0; 1; : : : ; 6) and let �Hy0 be the set consisting of the 12 vertices of� −Hy0 . It follows that any vertex of Si (i = 0; 1; : : : ; 6) is adjacent to zi, to at mostone vertex of Si−3∪Si+3 (indices computed mod. 7), and to at least one vertex of �Hy0 .In particular, any vertex of S has at most one neighbour in S. Moreover, there is atleast one vertex of S which is adjacent to two vertices of �Hy0 . It follows, using thesymmetry of the situation, that the rest of the proof consists of 12 cases, all of whichlead to a contradiction. Therefore Lemma 1 implies that c8(v) = 6 for any vertex v of�. In other words, there is at least one 8-circuit in �, and so � is of girth 8.

Theorem. There is no (3; 4)-graph � having 40 vertices.

Proof (sketch). Let v be any vertex of �. A gj-vertex of D4(v) is a vertex of D4(v)adjacent to exactly j vertices of D3 (v) (j = 1; 2; 3). If nj is the number of gj-verticesof D4(v), then n1 + n2 + n3 = 18. On the other hand, the de�nition of c8(v) impliesthat n1 +2n2 +3n3 =24. Therefore n2 =6−2n3 and n1 =12+n3. Since nj¿0 (for anyj=1; 2; 3), it follows that n363. In other words, D4(v) contains at most 3 g3-vertices.A vertex v of � is of type Gi (i = 0; 1; 2; 3) if D4(v) contains exactly i g3-vertices.If v is of type Gi, then the 18 vertices of D4(v) can be partitioned into ig3-vertices,(6 − 2i) g2-vertices and (12 + i) g1-vertices. If w is a g3-vertex of D4(v), we shalldenote by Grw the 3-claw consisting of w and its neighbours. As before, we mayassume that � contains the subgraph whose vertices are vi (i = 1; : : : ; 22) and whoseedges are v1 ∼ v2; v2i+1 ∼ vi ∼ v2i+2 (i=1; : : : ; 22). Let xj and x′j be the neighbours ofvj in D4(v1) (for j=11; : : : ; 22). The rest of the proof relies on the following lemmas.

Lemma 2. If C1 and C2 are two 8-circuits of � containing a vertex v and if theopposite vertices ti of v in Ci (i = 1; 2) are g2-vertices of D4(v); then C1 ∪ C2 isisomorphic to one of the graphs in Fig. 1.

Proof. Let v ∼ wi ∼ yi ∼ zi ∼ ti ∼ z′i ∼ y′i ∼ w′i ∼ v be the circuit Ci (i= 1; 2). Since

the girth of � is 8, D1(v)⊃{w1; w′1; w2; w

′2}; D2(v)⊃{y1; y′1; y2; y′2}; D3(v)⊃{z1; z′1;

z2; z′2} and D4(v)⊃{t1; t2}. Moreover, since � is of degree 3, we may assume, withoutloss of generality, that w1 = w2. On the other hand, t1 6= t2 (otherwise, since t1 andt2 are g2-vertices of D4(v), C2 would contain at least v; w1 = w2; t1 = t2; z1 and z′1,contradicting the fact that � is of girth 8). Thus there are two cases, according to the


fact that C2 contains the edge [v; w′1] or not. The analysis of these two possibilities,

combined with the two cases (owing to the symmetry of the situation): [t2 ∼ z1 = z2and z′2 6∈ C1] or [z2 and z′2 6∈ C1], yields the announced result.

Lemma 3. For any vertex v of �; if t1 and t2 are g2-vertices of D4(v); then t1 � t2.

Proof. Since � has girth 8, there is an 8-circuit Ci (i = 1; 2) containing v and ti, inwhich ti is opposite to v. Therefore, C1 ∪ C2 is isomorphic to one of the 7 graphs inLemma 2, and so t1 � t2.

Lemma 4. For any vertex v of �; D4(v) contains exactly 15 edges.

Lemma 5. Any vertex v of type G0 in � is contained in exactly 21 9-circuits.

Proof. Any 9-circuit of � through v contains an edge of D4(v). If v is of type G0,then D4(v) has no g3-vertex, and so must contain 6 g2-vertices ti (i = 1; : : : ; 6) and12 g1-vertices �j (j = 1; : : : ; 12). By Lemma 3, there is no edge in the subgraph of �induced on {t1; t2; : : : ; t6}. Therefore, since � is cubic, for any i ∈ {1; : : : ; 6} thereis a j ∈ {1; : : : ; 12} such that ti ∼ �j. Each of the 6 edges [ti; �j] (resp. 15 − 6 = 9edges [�k ; �m]) belongs to exactly two (resp. one) 9-circuits containing v. It followsthat there are exactly 21 circuits of length 9 through v.

Lemma 6. There is at least one vertex in � which is not of type G0.

Proof. If all 40 vertices of � are of type G0, then � contains (40.21)=9 9-circuits.Since this is not an integer, we get a contradiction.

Lemma 7. There is no vertex of type G3 in �.

Proof (sketch). If v1 is of type G3, then D4(v1) contains exactly three g3-vertices�; �; , �fteen g1-vertices and no g2-vertex. Since � has girth 8, we have 3 cases toconsider, according to the positions of the claws Gr�; Gr� and Gr .If Gr� ∩Gr� and Gr� ∩Gr are non empty, then Gr� ∩Gr =�, and (by symmetry)

we may assume that � is adjacent to v11; v15 and v19, that � ∼ v15 and that ∼ v19.Thus, by symmetry, v14 ∼ � ∼ v21 and v13 ∼ ∼ v18. It follows that xi; x′i (i =12; 16; 17; 20; 22) and xj (for j = 11; 13; 14; 18; 21) are new g1-vertices of D4(v1). Theexamination of Di(v8) implies that x18 has two neighbours in {x12; x′12; x21; x22; x′22},which must be, respectively, x12 (thanks to Di(x19) and symmetry) and x21 (owing toDi(x15)). Therefore x′12 must be adjacent to x14 or x21, which contradicts Proposition 3.The other two cases, namely [Gr� ∩ Gr� 6= ∅ and Gr� ∩ Gr = Gr� ∩ Gr = ∅], and[Gr� ∩ Gr� = Gr� ∩ Gr = Gr� ∩ Gr = ∅], can be ruled out in the same way.An argument which is a little bit longer shows that there is no vertex of type G2 or

G1 in �, and so � contains only vertices of type G0. This contradicts Lemma 6 andends the proof of the theorem.


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Maximal cubic graphs with diameter 4