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3URAL MATHEMATICAL JOURNAL, Vol. 10, No. 1, 2024, pp. 76-83
DOI: 10.15826/umj.2024.1.007
GRAPHS r OF DIAMETER 4 FOR WHICH r3,4 IS A STRONGLY REGULAR GRAPH WITH ^ = 4,61
Alexander A. Makhneva,dt, Mikhail P. Golubyatnikova,ctt,
Konstantin S. Efimovbcttt
aKrasovskii Institute of Mathematics and Mechanics, Ural Branch of the Russian Academy of Sciences, 16 S. Kovalevskaya Str., Ekaterinburg, 620108, Russian Federation
bUral State Mining University, 30 Kuibyshev Str., Ekaterinburg, 620144, Russian Federation
cUral Federal University, 19 Mira Str., Ekaterinburg, 620002, Russian Federation
^University Hainan Province, 58 Renmin Av., Haikou 570228, Hainan, P.R. China
[email protected] [email protected] tttii0nsfianfjn_s_efim0v(agmail.com
Abstract: We consider antipodal graphs r of diameter 4 for which ri,2 is a strongly regular graph. A.A. Makhnev and D.V. Paduchikh noticed that, in this case, A = r3,4 is a strongly regular graph without triangles. It is known that in the cases p = p(A) € {2,4, 6} there are infinite series of admissible parameters of strongly regular graphs with k(A) = r + 1) + r2, where r and s = — (p + r) are nonprincipal eigenvalues of A. This paper studies graphs with p(A) = 4 and 6. In these cases, r has intersection arrays {r2 + 4r + 3, r2 + 4r, 4, 1; 1, 4, r2 + 4r, r2 + 4r + 3} and {r2 + 6r + 5, r2 + 6r, 6, 1; 1, 6, r2 + 6r, r2 + 6r + 5}, respectively. It is proved that graphs with such intersection arrays do not exist.
Keywords: Distance-regular graph, Strongly regular graph, Triple intersection numbers.
1. Introduction
We consider undirected graphs without loops or multiple edges.
Let r be a connected graph. The distance d(a, b) between two vertices a and b of r is the length of a shortest path between a and b in r. Given a vertex a in a graph r, we denote by r(a) the subgraph induced by r on the set of all vertices that are at distance i from a. The subgraph [a] = r1(a) is called the neighbourhood of the vertex a.
Let r be a graph and a, b € r. Then the number of vertices in [a] n [b] is denoted by ^(a, b) (by A(a, b)) if a and b are at distance 2 (are adjacent) in r. Further, a subgraph induced by [a] n [b] is called a subgraph (a X-subgraph). Let r be a graph of diameter d and i,j € {1,2,3,..., d}. A graph r has the same set of vertices as r and vertices u and w are adjacent in r if dr(u, w) = i. A graph pj has the same set of vertices as r and vertices u and w are adjacent in r if dr(u, w) € {i, j}.
If vertices u and w are at distance i in r, then we denote by b^(u, w) (by c(u, w)) the number of vertices in the intersection ri+1(u) (ri—1(u)) with [w]. A graph r of diameter d is called distance-regular with intersection array {b0, b1,..., bd—1; c1,..., cd} if the values b^(u, w) and c(u, w) are
xThe study was supported by National Natural Science Foundation of China (12171126) and grant Laboratory of Endgeeniring Modelling and Statistics Calculations of Hainan Province.
independent of the choice of vertices u and w at distance i in r for any i = 0, [1]. Let
a = ki — b — c. Note that, for a distance-regular graph, b0 is the degree of the graph and ci = 1.
Let r be a graph of diameter d, and let x and y be vertices of r. Denote by pj(x,y) the number of vertices in the subgraph ri(x) n r(y) if d(x,y) = l in r. In a distance-regular graph, the numbers pj(x,y) are independent of the choice of vertices x and y, are denoted by pj and are called the intersection numbers of the graph r (see [1]).
Let r be a distance-regular graph of diameter d > 3. If r is an antipodal graph of diameter 4 with antipodality index r, then, by [1, Proposition 4.2.2], r has intersection array {k,k — a1 — 1, (r — 1)c2,1; 1, c2, k — a1 — 1, k}.
Consider an antipodal distance-regular graph r of diameter 4 for which r1)2 is a strongly regular graph. Makhnev and Paduchikh noticed in [3] that, in this case, A = r3;4 is a strongly regular graph without triangles and the antipodality index of r equals 2. It is known that in the cases ^ = ^(A) € {2,4,6} there arise infinite series of admissible parameters of strongly regular graphs with k(A) = ^(r + 1) + r2, where r and s = —(^ + r) are nonprincipal eigenvalues of A.
In the present paper, we consider graphs with ^(A) = 4 and 6. In these cases, r has intersection arrays
{r2 + 4r + 3, r2 + 4r, 4,1; 1, 4, r2 + 4r, r2 + 4r + 3}
and
{r2 + 6r + 5, r2 + 6r, 6,1; 1, 6, r2 + 6r, r2 + 6r + 5},
respectively.
If ^(A) = 4, then A has parameters (v, r2 + 4r + 4,0,4), where
, , 2 , „ , ^ , (r2 + 4r + 4)(r2 + 4r + 3) v = 1 + (v + 4 r + 4) + ---1-
Further, A has nonprincipal eigenvalues r and — (r + 4), and the multiplicity of r is equal to (r + 3)(r + 2)(r2 + 5r + 8)/8.
Theorem 1. A distance-regular graph with intersection array
{r2 + 4r + 3, r2 + 4r, 4, 1; 1, 4, r2 + 4r, r2 + 4r + 3}
does not exist.
If ^(A) = 6, then A has parameters (v, r2 + 6r + 6,0,6), where
v = 1 + (r2 + 6r + 6) + (r2 + 6r + 6)(r2 + 6r + 5)/6.
Further, A has nonprincipal eigenvalues r and — (r + 6), and the multiplicity of r is equal to (r + 5)(r2 + 6r + 6)(r + 4)/12. Therefore, r is even or congruent to 3 modulo 4.
Theorem 2. A distance-regular graph with intersection array
{r2 + 6r + 5, r2 + 6r, 6,1; 1, 6, r2 + 6r, r2 + 6r + 5}
does not exist.
Corollary 1. Distance-regular graphs with intersection arrays
{32, 27, 6,1; 1, 6, 27, 32}, {45, 40, 6,1; 1, 6, 40, 45}, {77, 72, 6,1; 1, 6, 72, 77}, {96, 91, 6,1; 1, 6, 91, 96}, {117,112, 6,1; 1, 6,112,117}
do not exist.
2. Triple intersection numbers
Let r be a distance-regular graph of diameter d. If u^u2, and u3 are vertices of the graph r i r1,r2, and r; w € r such that
and r1,r2, and r3 are nonnegative integers not greater than d, then •{ ^rir^rjf r is the set of vertices
u1u2u3 fu1 U2U3 1
. r1r2r3 I r^r3 J
The numbers
rir2r3
d(w, Ui) = ri
are called triple intersection numbers. For a fixed triple u1, u2, u3 of vertices,
M1«2«3 rir2r3
we will write [r1r2r3] instead of
Unfortunately, there are no general formulas for numbers [r1r2r3]. However, [2] suggests a method for calculating some numbers [r1r2r3].
Assume that u, v, and w are vertices of the graph r, W = d(u, v), U = d(v, w), and V = d(u, w). Since there is exactly one vertex x = u such that d(x,u) = 0, then the number [0jh] is 0 or 1. Hence, [0jh] = . Similarly, [i0h] = and [ij0] = .
Another set of equations can be obtained by fixing the distance between two vertices from {u, v,w} and counting the number of vertices located at all possible distances from the third. Then, we get
d d d £[j'h] = j - [0jh], £[ilh] = - [i0h], £[ijl] = pW - [ij0]. (2.1)
1=1
1=1
1=1
At the same time, some triples disappear. If |i — j| > W or i + j < W, then pW = 0; therefore,
[ijh] = 0 for all h € {0,..., d}. Define
d
Sijh(u, v, w) = £ QriQsjQth
r,s,i=Q
uvw - rst
If Krein's parameter qij is 0, then Sijh(u, v, w) = 0.
3. A distance-regular graph with intersection array
{r2 + 4r + 3, r2 + 4r, 4,1; 1, 4, r2 + 4r, r2 + 4r + 3}
In this section, r is a distance-regular graph with intersection array
{r2 + 4r + 3, r2 + 4r, 4,1; 1, 4, r2 + 4r, r2 + 4r + 3}.
Then, r has
1 + (r2 + 4r + 3) + (r2 + 4r + 3)(r2 + 4r)/4 + (r2 + 4r + 3) + 1 vertices and the spectrum
(r + 3)(r + 1) of multiplicity 1,
r + 3 of multiplicity r - 1 of multiplicity -(r + 1) of multiplicity
(r2 + 5 r + 8) (r2 + 3 r + 4) (r + 1) 16 (r + 2) '
(r2 + 5 r + 8)(r + 4)(r + 3)(r + 1) 16 (r + 2) '
(r2 + 5 r + 8) (r2 + 3 r + 4) (r + 3)
— (r + 5) of multiplicity
16 (r + 2)
(r2 + 3r + 4)(r + 3)(r + l) 16 (r + 2)
r
The multiplicity of r + 3 is equal to
(r2 + 5 r + 8 )(r2 + 3r + 4)(r + 1) 16 (r + 2) '
Further,
(r2 + 5r + 8, r + 2) = (3r + 8, r + 2)
divides 2 and (r + 2, r2 + 3r + 4) = (r + 2, r + 4) divides 2; therefore r + 2 divides 4. Consequently, r = 2, a contradiction with the fact that the multiplicity of r + 3 is equal to
(r2 + 5r + 8)(r2 + 3r + 4)(r + 1)/(16(r + 2)) = 22 x 14 x 3/64.
Theorem 1 is proved.
4. A distance-regular graph with intersection array
{r2 + 6r + 5, r2 + 6r, 6,1; 1, 6, r2 + 6r, r2 + 6r + 5}
In this section, r is a distance-regular graph with intersection array
{r2 + 6r + 5, r2 + 6r, 6,1; 1, 6, r2 + 6r, r2 + 6r + 5}.
Then, r has
1 + (r2 + 6r + 5) + (r2 + 6r + 5)(r2 + 6r)/6 + (r2 + 6r + 5) + 1 vertices, the spectrum
(r + 5)(r + 1) of multiplicity 1, r + 5 of multiplicity f = (r + 4)(r + 3)(r + 2) (r + 1)/24,
r — 1 of multiplicity (r + 6)(r + 5) (r + 4)(r + 1)/24, —(r + 1) of multiplicity (r + 5)(r + 4)(r + 3)(r + 2)/24, —(r + 7) of multiplicity (r + 5)(r + 2)(r + 1)r/24,
and the matrix Q (see [1]) of dual eigenvalues 'i
1
f f f(r + 6)(r + 5) /(r + 5) f(r + 5)r
(r + 2)(r + 3) /(r + 6)(r-l) r + 1 / (r+ 4)(r+ 3) f (r + 7)r
r + 1 0 f (r + 2)(r + 3)(r + l) -r/2 - 2 /(r + 6)(r-l) r + 1 0 / (r + 4)(r + 3)(r+ 1) r/2 + 1 f (r + 7)r
r + 1 -f (r + 2)(r + 3)(r + l) f(r + 6)(r + 5) r + 1 /(r + 5) (r + 4)(r + 3)(r+ 1) f(r + 5)r
(r + 2)(r + 3) r + 1 (r+ 4)(r+ 3)
Lemma 1. The intersection numbers are
Pii = 4, P21 = r2 + 6r, p32 = r2 + 6r, p22 = r4/6 + 2r3 + 29r2/6 - 7r, p13 = 0, p34 = 1; Pii = 6, p12 = r2 + 6r - 7, p23 = 6, p22 = r4/6 + 2r3 + 29r2/6 - 7r + 12, p23 = r2 + 6r - 7, p24 = 1, p23 = 2; p32 = r2 + 6r, p13 = 4, p14 = 1, p22 = r4/6 + 2r3 +29r2/6 - 7r, p23 = r2 + 6r, p33 = 0; p43 = r2 + 6r + 5, p22 = r4/6 + 2r3 + 41r2/6 + 5r.
1
1
1
Proof. Direct calculations using formulas from [1, Lemma 4.1.7].
□
Fix vertices u, v, and w of the graph r and define
{h j ^ ' [ijh] =
■uvw
Let A = r2(u), and let A be a graph with vertices from A in which two vertices are adjacent if they are at distance 2 in r. Then A is a regular graph of degree
p22 = r4/6 + 2r3 + 29r2/6 — 7r + 12
on
k2 = (r2 + 6r + 5)(r2 + 6r)/6 = r4/6 + 2r3 + 41r2/6 + 5r
vertices.
Lemma 2. Let d(u, v) = d(u, w) = 2 and d(v,w) = 1. Then, the triple intersection numbers
are
[111]= r4, [112] = [121] = —r4 + 6, [122]= r3 + r4 + r2 + 6r — 19, [123] = [132] = —r3 + 6; [211] = —r3 — r4 + 4, [212] = [221] = r3 + r4 + r2 + 6r — 12, [222] = r4/6 + 2r3 + 17r2/6 — 19r + 36, [223] = [232] = r3 + r4 + r2 + 6r — 12, [233] = —r3 — r4 + 4, [234] = [243] = 1; [311]= r3, [312] = [321] = —r3 + 6, [322]= r3 + r4 + r2 + 6r — 19, [323] = [332] = —r4 + 6;
[333] = r4, [422] = 1,
where r3 + r4 < 4.
Proof. Simplification of formulas (2.1). □
By Lemma 2, we have
r4/6 + 2r3 + 17r2/6 — 19r + 28 < [222] = —2r3 — 2r4 + r4/6 + 2r3 + 17r2/6 — 19r + 36 < r4/6 + 2r3 + 17r2/6 — 19r + 36.
Lemma 3. Let d(u, v) = d(u, w) = 2 and d(v,w) = 3. Then, the triple intersection numbers
are
[112] = —rn + 6, [113]= rn, [121] = — ri2 + 6, [122]= rn + ri2 + r2 + 6r — 19, [123] = —rn +6, [132] = — ri2 + 6; [212] = [221] = rn + ri2 + r2 + 6r — 12, [213] = [231] = —rn — ri2 + 4, [214] = [241] = 1, [222] = —2r3 — 2r4 + r4/6 + 2r3 + 17r2/6 — 19r + 36, [223] = [232] = rn + ri2 + r2 + 6r — 12; [312] = — ri2 + 6, [313]= ri2, [321] = —rn +6, [322]= rn + ri2 + r2 + 6r — 19, [323] = — ri2 + 6, [331]= rn, [332] = —rn +6; [422] = 1,
where rii + ri2 < 4.
Proof. Simplification of formulas (2.1). □
By Lemma 3, we have
r4/6 + 2r3 + 17r2/6 - 19r + 28 < [222] = -2r3 - 2r4 + r4/6 + 2r3 + 17r2/6 - 19r + 36 < r4/6 + 2r3 + 17r2/6 - 19r + 36.
Lemma 4. Let d(u, v) = d(u, w) = 2 and d(v,w) = 4. Then, the triple intersection numbers
are
[113] = [131] = 6, [122] = r2 + 6r - 7; [213] = [231] = r2 + 6r - 7, [222] = r4/6 + 2r3 + 29r2/6 - 7r + 12; [313] = [331] = 6, [322] = r2 + 6r - 7; [422] = 1.
Proof. Simplification of formulas (2.1). □
By Lemma 4, we have
[222] = r4/6 + 2r3 + 29r2/6 - 7r + 12.
Recall that
p22 = r2 + 6r - 7, p22 = r4/6 + 2r3 + 29r2/6 - 7r + 12, p23 = r2 + 6r - 7, p24 = 1. Let v and w be vertices from A. Then the number d of edges between A(v) and A - ({v} U A(v)) is
d = P12
L 221
+ P32
L 223
+ P42
uvz L 224 J
where and z are vertices from {Ц2;} for i = 1,3, and 4, respectively. Now, d satisfies the inequalities
(r2 + 6r - 7)(r4/3 + 4r3 + 17r2/3 - 38r + 56) + r4/6 + 2r3 + 29r2/6 - 7r + 12 < d < (r2 + 6r - 7)(r4/3 + 4r3 + 17r2/3 - 38r + 72) + r4/6 + 2r3 + 29r2/6 - 7r + 12. On the other hand,
7 / 2 \ / \\ , / 2 SweA(v) Лл(v,w) \
d = £ (P222 - 1 - Aa(V, W)) = Лл (^22 - 1---J •
адеЛ(^)
So,
d = (r4/6 + 2r3 + 29r2/6 - 7r + 12)(r4/6 + 2r3 + 29r2/6 - 7r + 11 - A),
where A is the average value of degree of the vertex w in the graph Л. Consequently,
(r2 + 6r - 7)(r4/3 + 4r3 + 17r2/3 - 38r + 56) r4 3 29r2
-;—;-о-- + 1 — + 2 V -\--— 7r + 11 — A
r4/6 + 2r3 + 29r2/6 - 7r + 12 6 6
< (r2 + 6r - 7)(r4/3 + 4r3 + 17r2/3 - 38r + 72)
~ r4/6 + 2r3 + 29r2/6 - 7r + 12 +
И з 29r^ (r2 + 6r - 7)(r4/3 + 4r3 + 17r2/3 - 38r + 72)
"б + ' Г + ~ r4/6 + 2r3 + 29r2/6 — 7r + 12 "
< И з 29r^ _ (r2 + 6r - 7)(r4/3 + 4r3 + 17r2/3 - 38r + 56)
~ ~6 +'7 +_6 r4/6 + 2r3 + 29r2/6 - 7 r + 12 '
and
Lemma 5. Let v) = w) = d(v,w) = 2. Then, the triple intersection numbers are
[111]= rg, [112]= -r7 - rg +6, [113]= rr, [121]= -rio - rg +6, [122] = r7 + r8 + r9 + r10 + r2 + 6r - 19, [123] = -r7 - r8 + 6, [131] = rio, [132] = -rio - r8 + 6, [133] = r8; [211] = -r8 - rg + 6, [212] = [221] = rr + r8 + rg + rio + r2 + 6r - 19, [213] = [231] = -rio - rr + 6, [222] = -2rr - 2r8 - 2rg - 2rio + r4/6 + 2r3 + 17r2/6 - 19r + 48, [223] = [232] = rr + r8 + rg + rio + r2 + 6r - 19, [224] = [242] = 1, [233] = -r8 - r8 + 6; [311]= r8, [312] = -rio - r8 + 6, [313]= rio, [321] = -rr - r8 + 6, [322] = r7 + r8 + rg + rio + r2 + 6r - 19, [323] = -rio - rg + 6, [331] = rr, [332] = -rr - rg + 6, [333] = rg; [422] = 1,
where
rg + rr, rg + rio, rr + r8, rio + r8, r8 + rg, rr + rio < 6. Proof. Simplification of formulas (2.1). □
By Lemma 5, we have
r4 17r2 r4 17r2
— + 2r3 + —--19r + 24 < [222] = -2r7 - 2r8 - 2r9 - 2rw + + 2r3 + —--19r + 48
6 6 6 6
r4 17r2
<--h 2r3 H---19r + 48.
66
Let d(u, v) = 2.
Let us count the number e2 of pairs of vertices (s, t) at distance 2, where s € { UV } and t € { UV }. On the one hand, by Lemma 2, we have
r4/6 + 2r3 + 17r2/6 - 19r + 28 < [222] < r4/6 + 2r3 + 17r2/6 - 19r + 36,
so,
rr4 17r2 \ /r4 17r2 \
(r2 + 6r - 7) — + 2r3 +--19r + 28 < e2 < (r2 + 6r - 7) — + 2r3 +--19r + 36 .
6 6 6 6
On the other hand, by Lemma 5, we have
[212] = rr + r8 + rg + rio + r2 + 6r - 19
and
/r4 17r2 \
(r2 + 6r - 7) + 2r3 + — - 19r + 28J < e2
rr4 29r2 \
= - + r8 + r\ + rio) + (r2 + 6r - 19) + 2r3 + — - 7r + 12J
i
/r4 17r2 \
< (r2 + 6r - 7) — + 2r3 +--19r + 36 .
66
In this way,
, r4 29r2 \ / r4 i7r2 \
(r2 + 6r - 19) — + 2r3 +--7r + 12 - (r2 + 6r - 7) — + 2r3 +--19r + 36
V 6 6 / V 6 6 /
< (r7 + r8 + r9 + rio)
r4 29r2 r4 i7r2
< (r2 + 6r - 19) — + 2r3 +--7r + 12 - (r2 + 6r - 7) — + 2r3 +--19r + 28 .
6 6 6 6
Consequently,
(r7 + r8 + ri + ri0) < —145r3/6 - 16r2 - 96r - 12, a contradiction.
Theorem 2 is proved. □
The corollary follows from Theorems 1 and 2.
So, we have shown the nonexistence of graphs with intersection arrays {r2 + 4r + 3, r2 + 4r, 4,1; 1, 4, r2 + 4r, r2 + 4r + 3}
and
{r2 + 6r + 5, r2 + 6r, 6,1; 1, 6, r2 + 6r, r2 + 6r + 5}. In particular, distance-regular graphs with intersection arrays
{32, 27, 6,1; 1, 6, 27, 32}, {45, 40, 6,1; 1, 6, 40, 45}, {77, 72, 6,1; 1, 6, 72, 77}, {96, 91, 6,1; 1, 6, 91, 96}, {117,112, 6,1; 1, 6,112,117}
do not exist.
REFERENCES
1. Brouwer A.E., Cohen A.M., Neumaier A. Distance-Regular Graphs. Berlin, Heidelberg: SpringerVerlag, 1989. 495 p. DOI: 10.1007/978-3-642-74341-2
2. Coolsaet K., Jurisic A. Using equality in the Krein conditions to prove nonexistence of sertain distance-regular graphs. J. Combin. Theory Ser. A, 2018. Vol. 115, No. 6. P. 1086-1095. DOI: 10.1016/j.jcta.2007.12.001
3. Makhnev A. A., Paduchikh D. V. Inverse problems in the class of distance-regular graphs of diameter 4. Proc. Steklov Inst. Math., 2022. Vol. 317, No. Suppl. 1. P. S121-S129. DOI: 10.1134/S0081543822030105
