群試設計，距離正則圖，圖譜理論及它們的關連

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(1)國立交通大學應用數學系博士論文群試設計，距離正則圖，圖譜理論及它們的關連 Pooling Designs, Distance-regular Graphs, Spectral Graph Theory and Their Links 博士生：黃喻培指導教授：翁志文. 教授. 中華民國一百零二年六月.

(2) 群試設計，距離正則圖，圖譜理論及它們的關連 Pooling Designs, Distance-regular Graphs, Spectral Graph Theory and Their Links. 國立交通大學應用數學系博士論文博士生：黃喻培. Student : Yu-Pei Huang. 指導教授：翁志文. Advisor : Chih-Wen Weng. A Dissertation Submitted to Department of Applied Mathematics College of Science National Chiao Tung University in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy in Applied Mathematics June 2013 Hsinchu, Taiwan, Republic of China. 中華民國一百零二年六月.

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(4) Pooling Designs, Distance-regular Graphs, Spectral Graph Theory and Their Links Student Yu-Pei Huang. Advisor Chih-Wen Weng. Department of Applied Mathematics National Chiao Tung University. Abstract This dissertation contains three quite different subjects: posets, distance-regular graphs, and spectral graph theory. Motivated by the constructions of pooling designs, we study these three subjects through interesting links among them. A pooling space is a ranked poset P such that the subposet w+ induced by the elements above w is atomic for each element w of P . Pooling spaces were introduced in [Discrete Mathematics 282:163-169, 2004] for the purpose of giving a uniform way to construct pooling designs, which have applications to the screening of DNA sequences. We find that a geometric lattice, a well-studied structure in literature, is also a pooling space. This provides us many classes of pooling designs, some old and some new. Following the same concept, the poset constructed from a distance-regular graph with its distance-regular subgraphs is also a pooling space. For a special class of distance-regular graphs, we show the existence of their distance-regular subgraphs with any given diameter. The nonexistence of a class of distance-regular graphs follows from the line of study. Distance-regular graphs appear often in some extremal class of combinatorial or linear algebraic inequalities. As we can see from the inequality of arithmetic and geometric means of a sequence of positive real numbers, the equality occurs when the sequence has some regular patterns. We consider the maximum eigenvalues of the adjacency matrices of graphs and present sharp upper bounds of them. The graphs which attain the bounds also satisfy a special kind of regularity. Keywords: pooling spaces, pooling designs, ranked posets, atomic, geometric lattices, distance-regular graphs. ii.

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(6) Table of Contents Abstract (in Chinese). i. Abstract (in English). ii. Acknowledgement. iii. Table of Contents. iv. List of Figures. vi. 1 Introduction. 1. 2 Preliminaries. 3. 2.1. Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3. 2.2. Distance-regular Graphs . . . . . . . . . . . . . . . . . . . . . . . . . .. 4. 2.3. Posets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 4. 2.4. Nonnegative Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 7. 2.5. Binomial Coefficients and Their q-analogue . . . . . . . . . . . . . . . .. 8. 3 Construct Pooling Spaces from Geometric Lattices. 10. 3.1. Pooling Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 10. 3.2. The Contractions of Graphs . . . . . . . . . . . . . . . . . . . . . . . .. 12. 3.3. Geometric Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 15. 3.4. Affine Geometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 16. 4 Distance-regular Subgraphs in a Distance-regular Graph. 20. 4.1. Strongly Closed Subgraphs . . . . . . . . . . . . . . . . . . . . . . . . .. 20. 4.2. D-bounded Property and Known Results . . . . . . . . . . . . . . . . .. 21. 4.3. The Shapes of Pentagons . . . . . . . . . . . . . . . . . . . . . . . . . .. 22. iv.

(7) 4.4. D-bounded Property and Nonexistence of Parallelograms . . . . . . . .. 25. 4.5. Classical Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 35. 5 Spectral Radius and Average 2-Degree Sequence of a Graph. 40. 5.1. Average 2-degree Sequence of a Graph . . . . . . . . . . . . . . . . . .. 40. 5.2. Upper Bounds of Spectral Radii . . . . . . . . . . . . . . . . . . . . . .. 41. 5.3. Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 42. 5.4. The Shape of the Sequence ϕ1 , ϕ2 , . . . , ϕn . . . . . . . . . . . . . . . . .. 45. Bibliography. 49. v.

(8) List of Figures Figure 1. A distance-regular graph with many distance-regular . . . . . .. 2. Figure 2. A poset . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 5. Figure 3. An upper semi-modular lattice that is not lower semi-modular .. 7. Figure 4. A pooling space which is not a meet semi-lattice . . . . . . . . .. 12. Figure 5. The poset P (C4 ) of contractions of C4 . . . . . . . . . . . . . . .. 13. Figure 6. Two pentagons in the proof of Lemma 4.4.4(ii) . . . . . . . . . .. 29. Figure 7. Three pentagons in the proof of Lemma 4.4.4(ii) . . . . . . . . .. 30. Figure 8. A graph with Mi = 2 . . . . . . . . . . . . . . . . . . . . . . . .. 41. Figure 9. A graph with Mi = 3 . . . . . . . . . . . . . . . . . . . . . . . .. 41. Figure 10. Graphs with Mi = 3 . . . . . . . . . . . . . . . . . . . . . . . .. 41. Figure 11. A graph with ϕ2 > ϕ3. 47. . . . . . . . . . . . . . . . . . . . . . . .. vi.

(9) Chapter 1 Introduction Group testing is a topic about strategies of experiment arrangements. The main idea behind it is that when we want to find some relatively few abnormal items out of a large set of items, testing items gathering together should be efficient with some smart arrangements. In 1964, W. H. Kautz and R. C. Singleton [23] introduced the now so-called disjunct matrices that are useful for us to deal with the group testing problems. With the error-tolerance ability being considered, the concepts of bd -disjunct matrices, a generalization of the original disjunct matrices, was introduced by A. G. D’yachkov, V. V. Rykov, and A. M. Rachad in 1983 [11]. A binary matrix M is bd disjunct if for any b + 1 columns x, x1 , x2 , . . . , xb of M with x different to the others, there exist d + 1 rows such that x has values 1, and x1 , x2 , . . . , xb all have values 0 at these d + 1 rows. In particular, a b0 -disjunct matrix is also called a b-disjunct matrix for short. A bd -disjunct matrix can be used to construct an error-tolerable design for non-adaptive group testing, which has applications to the screening of DNA sequences, and the corresponding decoding algorithm is efficient. See [9, 18] for details. Hence a bd -disjunct matrix is also called a pooling design. The constructions of bd -disjunct matrices were given by many authors, e.g. [29, 30, 32, 10]. These constructions use some properties of a ranked poset. In [19], the name pooling spaces was given to describe these ranked posets (formal definition in Section 3.1). Fix a pooling space P and positive integers r < k. Let M denote the incidence matrix between the rank r elements and the rank k elements in P . It was shown in [19] that M is bd -disjunct for b = r and d = 0. A binary matrix is fully bd disjunct if it is bd -disjunct but neither bd+1 - nor (b+1)d -disjunct. Some fully bd -disjunct matrices are given in [10].. 1.

(10) So far we know that the incidence relation between two levels in a pooling space can help us to construct pooling designs. Roughly speaking, the supporting structure behind the pooling space must be “good” enough. In particular, the poset of distanceregular subgraphs in a given distance-regular graph (formal definition in Section 2.2), ordered by the containment relation between subgraphs, forms a pooling space [40]. r. r. r. r r. r. r. r. Figure 1. A distance-regular graph with many distance-regular subgraphs. In Figure 1, the distance-regular subgraphs of this “cube” are the “points”, “edges”, “faces”, and the “cube” itself. In general, the determination of distance-regular subgraphs may not be so obvious. With some restrictions on the intersection numbers of a distance-regular graph, we introduce a systematical way in Chapter 4 that helps us to construct distance-regular subgraphs of it. The results involved also help us to show the nonexistence of a class of distance-regular graphs. Distance-regular graphs appear often in some extremal class of combinatorial or linear algebraic inequalities. For example it is well-known that the number of edges of a graph of girth 5 and order n is at most. √ n n−1 , 2. and its maximum number of edges is. attained when the graph is distance-regular [27, Theorem 4.2]. Sometimes other graphs with certain regularity appear as extremal class. For example, the average degree of a graph is at most the maximum eigenvalue of its adjacency matrix, and a regular graph attains the maximum [5, Lemma 3.2.1]. As we can see from the inequality of arithmetic and geometric means of a sequence of positive real numbers, extremal conditions for inequalities occur when the sequence has some regular pattern. In the last chapter of this dissertation, we consider the maximum eigenvalues of the adjacency matrices of graphs, and present sharp upper bounds of them. The graphs attain the bound also satisfy a special kind of regularity.. 2.

(11) Chapter 2 Preliminaries In this chapter we review some definitions and basic concepts concerning graphs, distance-regular graphs, posets, nonnegative matrices, and binomial coefficients.. 2.1. Graphs. A graph Γ is an ordered pair (X, R) consisting of a finite vertex set X and an edge set R where each element in R is a 2-element subset of X. Two vertices x, y ∈ X are adjacent if {x, y} ∈ R and we use x ∼ y to denote that x, y are adjacent. A path between vertices x and y in Γ is a sequence x0 , x1 , · · · , xℓ of distinct vertices where x0 = x and xℓ = y, such that xi ∼ xi+1 for i = 0, 1, · · · , ℓ − 1. The length of a path is the number of edges on it. The distance between x, y ∈ X is the length of the shortest path between x and y and is denoted by ∂(x, y). The diameter D of Γ is defined as D:=max{ ∂(x, y) | x, y ∈ X}. For a vertex x ∈ X and an integer 0 ≤ i ≤ D, set Γi (x) := { z ∈ X | ∂(x, z) = i}. The valency of a vertex x ∈ X is the cardinality of Γ1 (x) and is denoted by dx . For the adjacency matrix A = (axy ) of Γ, we mean a binary square matrix of order |X| with rows and columns indexed by the vertices in X, such that for any pair x, y ∈ X, axy = 1 iff x ∼ y. A cycle of length ℓ, denoted by Cℓ , is a graph with ℓ vertices and ℓ edges whose vertices can be placed around a circle so that two vertices are adjacent if and only if they appear consequently along the cycle. The girth of a graph with a cycle is the length of its shortest cycle.. 3.

(12) 2.2 Distance-regular Graphs A graph Γ = (X, R) is called regular (with valency k) if each vertex in X has valency k. A graph Γ is said to be distance-regular whenever for all integers 0 ≤ h, i, j ≤ D, and all vertices x, y ∈ X with ∂(x, y) = h, the number phij = |Γi (x) ∩ Γj (y)| is independent of x, y. The constants phij are known as the intersection numbers of Γ. Suppose that Γ = (X, R) is a distance-regular graph with diameter D ≥ 3. For two vertices x, y ∈ X, with ∂(x, y) = i, set B(x, y) := Γ1 (x) ∩ Γi+1 (y), C(x, y) := Γ1 (x) ∩ Γi−1 (y), A(x, y) := Γ1 (x) ∩ Γi (y). Note that |B(x, y)| = pi1. i+1 ,. |C(x, y)| = pi1. i−1 ,. |A(x, y)| = pi1. i. are independent of x, y. For convenience, set ci := pi1 0 ≤ i ≤ D, bi := pi1. i+1. i−1. for 1 ≤ i ≤ D, ai := pi1 i for. for 0 ≤ i ≤ D − 1 and put bD := 0, c0 := 0. Note that k := b0. is the valency of each vertex in Γ. It is immediate from the definition of phij that bi ̸= 0 for 0 ≤ i ≤ D − 1 and ci ̸= 0 for 1 ≤ i ≤ D. Moreover for 0 ≤ i ≤ D.. k = ai + bi + ci. 2.3. (2.2.1). Posets. Let P denote a finite set. By a partial order on P, we mean a binary relation ≤ on P such that (i) x ≤ x. for x ∈ P,. (ii) x ≤ y and y ≤ z. =⇒ x ≤ z. for x, y, z ∈ P, 4.

(13) (iii) x ≤ y and y ≤ x =⇒ x = y. for x, y ∈ P.. By a partially ordered set (or poset, for short), we mean a pair (P, ≤), where P is a finite set, and where ≤ is a partial order on P. We may suppress reference to ≤, and just write P instead of (P, ≤) if no confusion occurs. Let P denote a poset with partial order ≤, and let x and y denote any two elements in P. As usual, we write x < y whenever x ≤ y and x ̸= y, and write x ̸< y whenever x < y is not true. We say that y covers x whenever x < y, and there is no z ∈ P such that x < z < y. A sequence x0 , x1 , . . . , xt of elements of P is said to be a direct chain of length t whenever xi covers xi−1 for 1 ≤ i ≤ t. A poset can be described by a diagram in the plane in which the elements are corresponding to dots, and y covers x whenever dot y is placed above dot x with an edge connecting them. See Figure 2 for the diagram of the poset with five elements {0, x, y, z, w}, and x, y cover 0; z, w cover both x and y. Note that 0, x, z is a direct chain of length 2.. z x. e e H w HH H HH e e y @ @ @e. 0 Figure 2. A poset.. Let P denote any finite poset, and let S denote any subset of P. Then there is a unique partial order on S such that for all x, y ∈ S, x ≤ y in S if and only if x ≤ y in P. This partial order is said to be induced from P. By a subposet of P, we mean a subset of P, together with the partial order induced from P. An element x ∈ S is said to be minimal (resp. maximal) in S whenever there is no y ∈ S such that y < x (resp. x < y). Let min(S) (resp. max(S)) denote the set of all minimal (resp. maximal) elements in P. Whenever min(P ) (resp. max(P )) consists of a single element, we denote it by 0 (resp. 1), and we say that P has the least element 0 (resp. the greatest element 1). Throughout the remaining of the dissertation we assume that P is a poset with the least element 0. By an atom in P, we mean an element in P that covers 0. We let AP. 5.

(14) denote the set of atoms in P. By the interval [x, y], where x, y ∈ P with x ≤ y, we mean the subposet [x, y] := {z|z ∈ P, x ≤ z ≤ y} of P. By a rank function on P, we mean a function “rank” from P to the set of nonnegative integers such that rank(0) = 0, and for all x, y ∈ P, y covers x implies rank(y) − rank(x) = 1. Observe that the rank function is unique if it exists. P is said to be ranked whenever P has a rank function. In this case, we set rank(P ) := max{rank(x) | x ∈ P }, Pi := {x | x ∈ P, rank(x) = i}, and observe that P0 = {0}, P1 = AP . Also observe that P is ranked if and only if every direct chain from 0 to x has the same length for any x ∈ P . Let P be a ranked poset of rank n and fix two integers 1 ≤ r < k ≤ n. The incidence matrix M between Pr and Pk is a |Pr | × |Pk | binary matrix with rows indexed by Pr and columns indexed by Pk such that. { Mxy :=. 1, x ≤ y;. for x ∈ Pr , y ∈ Pk .. 0, else. Let S be a subset of P. Fix z ∈ P. Then z is said to be an upper bound (resp. lower bound) of S, if z ≥ x (resp. z ≤ x) for all x ∈ S. Suppose the subposet of upper bounds (resp. lower bounds) of S has a unique minimal (resp. maximal) element. In this case we call this element the least upper bound or join (resp. the greatest lower bound or meet) of S. If S = {x1 , x2 , . . . , xt } we write x1 ∨ x2 ∨ · · · ∨ xt for the join of S and x1 ∧ x2 ∧ · · · ∧ xt for the meet of S. P is said to be atomic whenever for each nonzero element x of P, x is the join of atoms in the interval [0, x]. Suppose P is atomic and x < y are two elements in P . Observe that the atoms in the interval [0, x] is a proper subset of the atoms in the interval [0, y]. P is said to be a meet semi-lattice (resp. join semi-lattice) whenever P is nonempty, and x ∧ y (resp. x ∨ y) exists for all x, y ∈ P. A meet semi-lattice (resp. join semi-lattice) has a 0 (resp. 1). A meet and join semi-lattice is called a lattice. Note that if a nonempty set S in a meet semi-lattice has an upper bound then the join of S exists. 6.

(15) Suppose P is a lattice. Then P is said to be upper semi-modular (resp. lower semi-modular ) whenever for all x, y ∈ P,. (resp.. y covers x ∧ y. −→. x ∨ y covers x. x ∨ y covers x. −→. y covers x ∧ y).. P is said to be modular whenever P is both upper semi-modular and lower semimodular. Figure 3 is a diagram of an upper semi-modular lattice with 7 elements. This lattice is not lower semi-modular since 1 = x ∨ y covers x but y does not cover 0 = x ∧ y. 1 e. @ @. @ @ ey e JJ. JJ. J e. J e. JJ. J e. x e. 0. Figure 3. An upper semi-modular lattice that is not lower semi-modular.. 2.4. Nonnegative Matrices. Let A = (aij ) be a square n×n matrix. We say that A is positive (resp. nonnegative) if aij > 0 (resp. aij ≥ 0) for all i, j. We say that A is reducible if the indices 1, 2, · · · , n can be divided into two disjoint nonempty sets i1 , i2 , · · · , iµ and j1 , j2 , · · · , jν where µ + ν = n such that aiα jβ = 0 for α = 1, 2, · · · , µ and β = 1, 2, · · · ν. A square matrix is called irreducible if it is not reducible. Simply consider the adjacent relation of a graph and the definition of irreducible matrices, we have the following proposition. Proposition 2.4.1. The adjacency matrix of a simple graph Γ is irreducible if and only if Γ is connected. The following theorem is a fundamental result of the study on matrix theory. It is referred to as Perron-Frobenius Theorem [31, Chapter 2]. 7.

(16) Theorem 2.4.2. If B is a nonnegative irreducible n × n matrix with largest eigenvalue ρ(B) and row-sums r1 , r2 , . . . , rn , then ρ(B) ≤ max ri , 1≤i≤n. with equality if and only if the row-sums of B are all equal.. 2.5 Binomial Coefficients and Their q-analogue For all nonnegative integers k and n, we define the binomial coefficients. (n ) k. as. follows. Definition 2.5.1..     0    ( )  n := 1  k      . if k > n, if k = 0,. n! k!(n−k)!. =. n(n−1)···(n−k+1) k(k−1)···1. otherwise.. A q-analogue of a known expression is a generalization of it involving a new parameter q such that as q → 1, the generalization returns to the original expression. The equality 1 − qn =n q→1 1 − q lim. suggests the q-analogue of n, known as the q-bracket or q-number of n, to be that defined in the following definition. Definition 2.5.2. [n]q :=. 1 − qn = 1 + q + q 2 + · · · + q n−1 . 1−q. Having the q-analogue of n, we naturally define the q-factorial as follows. Definition 2.5.3. [n]q ! := [1]q · [2]q · · · [n − 1]q · [n]q =. 1 − q 1 − q2 1 − q n−1 1 − q n · ··· · 1−q 1−q 1−q 1−q. = 1 · (1 + q) · · · (1 + q + · · · + q n−2 ) · (1 + q + · · · + q n−1 ). 8.

(17) From q-factorial, we also define the following q-binomial coefficients. Definition 2.5.4. [ ] n k q. := =. In particular,. [n] 0 q. [n]q ! [k]q ![n − k]q ! (q n − 1)(q n−1 − 1) · · · (q n−k+1 − 1) (q k − 1)(q k−1 − 1) · · · (q − 1).. := 1.. The q-binomial coefficients are also called Gaussian numbers or Gaussian coeffi[ ] cients. It is well known that nk q is just the number of k-dimensional subspaces of an n-dimensional vector space over a finite field Fq [27, p. 291].. 9.

(18) Chapter 3 Construct Pooling Spaces from Geometric Lattices The name pooling space was given in [19] to describe a special class of ranked posets which are employed to construct pooling designs. In this chapter, we clarify a few things about the definition of pooling spaces. Then we find that a geometric lattice, a well-studied structure in literature, is also a pooling space. This provides us many classes of pooling designs. In particular we study the pooling designs constructed from affine geometries and then find some of them meet the optimal bounds related to a conjecture of Erdös, Frankl, and Füredi.. 3.1. Pooling Spaces. Definition 3.1.1. Let P be a ranked poset. For any w ∈ P, define w+ = {y ≥ w | y ∈ P }. P is said to be a pooling space whenever w+ is atomic for each w ∈ P. In particular a pooling space is atomic. It is immediate from the definition that if P is a pooling space, then so is w+ for any w ∈ P. The following theorem evolves the study of pooling spaces. Theorem 3.1.2. [19, Corollary 3.2] Let P be a pooling space with rank D. Fix an integer ℓ (1 ≤ ℓ ≤ D). Let M = M (D, ℓ) be the matrix over {0, 1} whose rows (resp. 10.

(19) columns) are indexed by Pℓ (resp. PD ) such that Muv = 1 iff u ≤ v. Then for each integer b (1 ≤ b ≤ ℓ), M is bd -disjunct, where d := min|. ∪. [y, x] ∩ Pℓ | − 1. with the minimum taken over all pairs (x, T ) such that x ∈ PD , T ⊆ PD , x ∈ / T, |T | ≤ b, and with the union taken over all y such that y ∈ Pb , y ≤ x, y z for all z ∈ T. Lemma 3.1.3. Let P be a pooling space. Then each interval in P is atomic. Proof. Let [x, y] be an interval in P and z ∈ [x, y] with z ̸= x. Suppose x ∈ Pi . Note that the set of atoms contained in [x, z], no matter considered in x+ or in [x, y], is the same set [x, z] ∩ Pi+1 . Since z is the join of [x, z] ∩ Pi+1 in x+ by assumption, z is also the join of [x, z] ∩ Pi+1 in [x, y]. Remark 3.1.4. The definition of pooling space was first given in [19]. However in the abstract of that paper, it was stated in an alternative way that a pooling space is a ranked poset with atomic intervals. The following example shows that this is not correct. Example 3.1.5. Let P = {0, x, y, z, w} and the partial order is defined as in Fig. 2 of Section 2.3. Then each interval in P is atomic. Since neither z nor w is the least upper bound of x and y, P is not atomic. Observe that P is not a meet semi-lattice. We now give a revised version. Proposition 3.1.6. Let P be a ranked meet semi-lattice. Then P is a pooling space if and only if each interval in P is atomic. Proof. We have just proved the necessary condition in the previous lemma. To prove the sufficient condition we fix an element w ∈ P and suppose w ∈ Ps for some integer 0 ≤ s ≤ rank(P ). We shall prove that w+ is atomic. To do this fix x ∈ w+ \ {w} and we need to prove that x is the join of [w, x] ∩ Ps+1 in w+ . By the assumption [w, x] being atomic, x is the join of [w, x] ∩ Ps+1 in [w, x]. In particular, x is an upper bound 11.

(20) of [w, x] ∩ Ps+1 . Since P is a meet semi-lattice, the upper bounds of [w, x] ∩ Ps+1 have a least element and denote it by y. Hence y ≤ x and clearly w ≤ y, so equivalently y ∈ [w, x]. This forces x ≤ y, since x is the least upper bound and y is also an upper bound of [w, x] ∩ Ps+1 . Then we obtain x = y. We give a poling space which is not a meet semi-lattice. Example 3.1.7. Let P = {0, u, x, y, v, z, w} and let the partial order be defined as in the Figure 4 below. Observe z = u ∨ x ∨ y and w = x ∨ y ∨ v. The remaining properties of a pooling space hold trivially. Hence P is a pooling space. P is not a meet semi-lattice since z ∧ w does not exist. e e H HH @w @ HH H ey @ e e e u HHx v @ HH @ HH @ e. z. 0. Figure 4. A pooling space which is not a meet semi-lattice.. 3.2. The Contractions of Graphs. Many examples of pooling spaces were given in [19]. They are related to the Hamming matroids, the attenuated spaces, and six classical polar spaces. Among these examples there is a common property: each interval is modular. In this section we will construct pooling spaces without modular intervals. The construction in this section also can be obtained as a consequence of our main theorem in the next section. We do it earlier and repeatedly here for the purpose to give the readers a concrete impression of a pooling space, and hope that one can find one’s own class of examples in the sequel.. Throughout the section let Γ denote a simple connected graph on n vertices.. 12.

(21) Definition 3.2.1. Let P = P (Γ) denote the set of partitions S of the vertex set V (Γ) such that the subgraph induced by each block of S is connected. For S, Q ∈ P , define S ≤ Q ⇐⇒ S is a refinement of Q. The poset (P (Γ), ≤) is called the poset of contractions of Γ. Example 3.2.2. Let Γ denote a graph with the vertex set {x, y, z, w} and edge set {xy, yz, zw, wx}, i.e. Γ is the 4-cycle C4 . Then the poset P (Γ) is as in Figure 5. We delete the single element blocks in the notation of a partition, e.g. the notation 0 is used to denote the partition with four blocks {x}, {y}, {z}, {w} respectively, and xy is used to denote the partition with three blocks {x, y}, {z}, {w} respectively. The poset is a lattice, but not a modular lattice. This is because the join of xy zw and yz wx is xyzw, which covers xy zw, but yz wx does not cover their meet 0. Observe the subposet induced on xy + is P (C3 ), the poset of contractions of a triangle.. e xyzw PP H PP HH H PPP HH PP HH PPP P eyz wx xy zw e xyz e yzw eQ ezwx H e wxy P Q @ @ @ Q @ @ Q @ @ @ @ Q Q @ @ @ Q @e @ e @e Qe xy QQ yz J wx zw J. Q Q. Q J. Q J Q J. Q J e. Q 0. Figure 5. The poset P (C4 ) of contractions of C4 .. Lemma 3.2.3. P (Γ) is a ranked poset of rank n − 1. The rank i elements are those elements in P (Γ) with n − i blocks for 0 ≤ i ≤ n − 1. Proof. For N ∈ P (Γ) with n−i blocks define the rank of N to be i, where 0 ≤ i ≤ n−1. We claim that this is a rank function. Suppose that Q covers S and rank(S) = i. Since 13.

(22) S is a proper refinement of Q, rank(Q) ≥ i + 1 and there are two blocks in S that are contained in the same block of Q. Let T be an element in P (Γ) that glues these two blocks of S. Then S < T ≤ Q and rank(T ) = rank(S) + 1. This shows T = Q and rank(Q) = i + 1. Proposition 3.2.4. P (Γ) is a pooling space of rank n − 1. Proof. P (Γ) is ranked by previous lemma. From previous lemma and the definition, each atom in P (Γ) contains n − 1 blocks, one block containing two adjacent vertices and each of the remaining n − 2 blocks containing a single vertex. By identifying the atoms with the edges of Γ we find that each element S ∈ P (Γ) is the join of those edges contained in the induced subgraph of Γ corresponding to each block of S. This shows that P (Γ) is atomic. More generally, for Q ∈ P (Γ), the subposet Q+ is also atomic. This is because the subposet Q+ is isomorphic to the poset P (QΓ ) of contractions of QΓ , where QΓ is the graph with the vertex set Q, and for two distinct blocks x, y ∈ Q x is adjacent to y whenever some vertex in x is adjacent to some vertex in y. Remark 3.2.5. Let Γ = Kn denote the complete graph on n vertices. Then the elements in P = P (Kn ) are all the partitions of the vertex set of Kn . S(n, k) := |Pn−k | is called the Stirling number of the second kind where k ≥ 1. It is well known that S(n, k) can be solved by the recurrence relation S(n, k) = S(n − 1, k − 1) + kS(n − 1, k). for 1 ≤ k ≤ n − 1. with initial condition S(n, 0) := 0 for n ≥ 1, S(0, 0) := 1, and S(n, n) = 1 for n ≥ 1. See [4, Section 8.2] for details. By applying Proposition 3.2.4 and Remark 3.2.5 with the result in [19, Corollary 3.2] we immediately have the following corollary. Corollary 3.2.6. Let Γ denote a simple connected graph on n vertices and P = P (Γ). Let C(Γ, k, r) denote the incidence matrix between Pr and Pk where 1 ≤ r < k ≤ n − 1. Then C(Γ, k, r) is r-disjunct. In particular if Γ = Kn , then the matrix C(Γ, k, r) has size S(n, n − r) × S(n, n − k). 14.

(23) 3.3 Geometric Lattices The concept of geometric lattices can be described in very different ways. See [27, Chapter 23] for details. For the purpose to derive our main result easily, we adopt the definition that a geometric lattice is an upper semi-modular atomic lattice [27, Page 271]. We will show that a geometric lattice is a pooling space in this section. The following lemma is immediate from the definition. Lemma 3.3.1. Let P be an upper semi-modular lattice. Then the poset induced on every interval of P is an upper semi-modular lattice. Lemma 3.3.2. Let P be a geometric lattice. Then the poset induced on every interval of P is a geometric lattice. Proof. Let [x, y] denote an interval in P where x ∈ Pi . By previous lemma, it remains to show [x, y] is atomic. Fix z ∈ [x, y] with z ̸= x. Suppose that w is the join of Pi+1 ∩ [x, z], the atoms in [x, z]. Then w ≤ z. We are done if w = z, so assume w < z. Then there exists an atom in a ∈ [0, z] \ [0, w]. Note that a ̸< x. By the upper semi-modularity, a ∨ x ∈ Pi+1 ∩ [x, z] is an atom in [x, z], a contradiction to a ∨ x ̸< w. Lemma 3.3.3. An upper semi-modular lattice is ranked. In particular, a geometric lattice is ranked. Proof. Let P be an upper semi-modular lattice and suppose that P is not ranked. Then there exists x ∈ P such that [x, 1] is not ranked, but for all atoms a of [x, 1], [a, 1] is ranked. Pick an atom a ∈ [x, 1]. Let f be a rank function on [a, 1]. We extend the function f to a function f ′ in [x, 1] by defining    f (y) + 1, if y ∈ [a, 1]; ′ f (y) :=   f (a ∨ y), else. We shall prove f ′ is a rank function in [x, 1]. Suppose u, v ∈ [x, 1] and u covers v. We need to show f ′ (u) = f ′ (v) + 1. This is clear if v ∈ [a, 1]. Assume v ̸∈ [a, 1]. Suppose u ∈ [a, 1]. Then u = a ∨ v and f ′ (u) = f (a ∨ v) + 1 = f ′ (v) + 1. Suppose 15.

(24) u ̸∈ [a, 1]. Since u covers v = (a ∨ v) ∧ u, we have a ∨ u = (a ∨ v) ∨ u covers a ∨ v. Then f ′ (u) = f (a ∨ u) = f (a ∨ v) + 1 = f ′ (v) + 1. This concludes that [x, 1] is ranked, a contradiction. Theorem 3.3.4. Let P be a geometric lattice. Then P is a pooling space. Proof. P is ranked by Lemma 3.3.3. Since each interval of P is a geometry lattice by Lemma 3.3.2, each interval is atomic. The theorem now follows from Proposition 3.1.6.. By applying Theorem 3.3.4 and Theorem 3.1.2 we immediately have the following corollary. Corollary 3.3.5. Let P be a geometric lattice with rank n. Let G(P, k, r) denote the incidence matrix between Pr and Pk where 1 ≤ r < k ≤ n. Then G(P, k, r) is r-disjunct.. Many examples of geometry lattices are listed in Chapter 23 of [27]. These are related to linear spaces, Steiner systems, affine geometries, projective geometries, and contractions of graphs. More examples are given in [17]. In some cases the corresponding results in Corollary 3.3.5 are not fully disjunct. The fully disjunct properties on projective geometries were studied in [10].. 3.4. Affine Geometries. In this section we study the fully disjunct properties of the binary matrices constructed from affine geometries. The idea is exactly the same as the study of projective geometries in [10]. In fact this idea works for any geometric lattices with each interval isomorphic to a projective geometry. For completeness of the dissertation, we still provide the proof. Also there are some small computation mistakes in [10]. We will point out these mistakes after Corollary 3.4.6. In the beginning, we give the definition of affine geometries. Definition 3.4.1. Let V denote an n-dimensional vector space over a finite field Fq , where q is the number of elements in the field. Let P = P (V ) denote the poset with 16.

(25) element set P = {u + W | u ∈ V and W ⊆ V is a subspace} ∪ {∅}, where ∅ denote the empty set. The order is defined by inclusion. Note that P is a geometric lattice of rank n + 1. P is called the affine geometry and is denoted by AGn (Fq ). The rank i elements in Pi are referred to as the affine (i − 1)-subspaces of V for 1 ≤ i ≤ n + 1. We say that the affine subspaces u + W and v + W are parallel for vectors u, v ∈ V and subspace W ⊆ V. We immediately have the following lemma. Lemma 3.4.2. Let V denote an n-dimensional vector space over a finite field Fq . Let u1 , u2 ∈ V be elements and let W1 , W2 ⊆ V be subspaces. Then u1 + W1 = u2 + W2 if and only if W1 = W2 and u1 − u2 ∈ W1 . Lemma 3.4.3. Let V denote an n-dimensional vector space over a finite field Fq , and A denote an affine k-subspace of V . Then the number of affine r-subspaces contained in A is q. k−r. [ ] k , r q. where r < k. These affine r-subspaces in A are partitioned into [ ] k r q. (3.4.1). classes, each class consisting of q k−r parallel affine subspaces. Proof. The parallel property defines an equivalent relation on the set of affine rsubspaces in A. The number of equivalent classes is as in (3.4.1) and each equivalent class consists of q k−r elements by Lemma 3.4.2. Theorem 3.4.4. Let V denote an n-dimensional vector space over a finite field Fq . Fix integers 1 ≤ r < k ≤ n and a positive integer b. Let A, A1 , A2 , . . . , Ab denote affine k-subspaces of V with A ̸= Ai for 1 ≤ i ≤ b. Then there are at least [ ] [ ] k−r k k−r−1 k − 1 q − bq r q r q affine r-subspaces contained in A and not contained in any of Ai for 1 ≤ i ≤ b. 17. (3.4.2).

(26) Proof. There are q. k−r. [ ] k r q. affine r-subspaces contained in A, some of them in some affine subspace A ∩ Ai for each 1 ≤ i ≤ b to be deducted. A ∩ Ai takes maximal coverage of these affine r-subspaces when A ∩ Ai is an affine (k − 1)-subspace, and in this situation the number of these affine r-subspaces is. [ q. (k−1)−r. ] k−1 . r q. Remark 3.4.5. For positive integers b ≤ q and k < n the number in (3.4.2) is optimal. We choose Ai to an affine k-subspace with the meet with A corresponding to each of the q parallel affine (k − 1)-subspaces in A. Then (3.4.2) is exactly the number of affine r-subspaces contained in A and not contained in any of Ai for 1 ≤ i ≤ b. From Lemma 3.4.3, Theorem 3.4.4, and Remark 3.4.5 we have the following corollary. Corollary 3.4.6. Let P = AGn (Fq ) and let Eq (n + 1, k + 1, r + 1) denote the incidence [] [ ] matrix between Pr+1 and Pk+1 where r < k. Let d = q k−r kr q − bq k−r−1 k−1 − 1. Then r q Eq (n + 1, k + 1, r + 1) is bd -disjunct with size q. n−r. [ ] [ ] n n−k n , ×q k q r q. q(q k − 1) to ensure d ≥ 0 by (3.4.2). Moreover, q k−r − 1 if k < n and b is a positive integer such that    b ≤ q, if r > 0; (3.4.3)   b ≤ q − 1, if r = 0,. where b is any positive integer less than. then Eq (n + 1, k + 1, r + 1) is not bd+1 -disjunct. The result in [10, Corollary 4.6] is similar to Corollary 3.4.6, but the former makes a mistake for not separating the case r = 0 in (3.4.3) from r > 0. This mistake inherits an earlier mistake in [10, Theorem 4.4], referring to the last line of its proof. The 18.

(27) case r = 0 of (3.4.3) will be important in our following discussing. In view of (3.4.2) b increases if and only if d decreases. We set r = 0 and b = q − 1 to be the largest possible integer in Corollary 3.4.6 to obtain the following result. Corollary 3.4.7. Eq (n+1, k+1, 1) is (q−1)q [ ] with size q n × q n−k nk q .. k−1 −1. -disjunct, but not (q−1)q. k−1. -disjunct,. We promised in the beginning of this chapter to give some matrices that meet some optimal bound. These matrices are Eq (3, 2, 1), where q is a power of a prime. We describe an optimal bound of an assumption below, and show the relation of this assumption and the conjecture of Erdös, Frankl and Füredi [12] later. Assumption: Any b-disjunct matrix of size s × t with s < t must have s ≥ (b + 1)2 . We don’t know if the above assumption is true, but Eq (3, 2, 1) attains the equality s = (b + 1)2 , since Eq (3, 2, 1) is a (q − 1)-disjunct matrix of size q 2 × (q 2 + q) by Corollary 3.4.7. In fact the above assumption is a consequence of the following conjecture of Erdös, Frankl, and Füredi in [12]: EFF Conjecture: Any b-disjunct matrix of size s × (b + 1)2 must have s ≥ (b + 1)2 . Also see [9, page 29] for the above conjecture. Suppose that EFF Conjecture is true and suppose that the above assumption fails. Let M be a b-disjunct matrix of size s × t with s < t, but s < (b + 1)2 . If t ≥ (b + 1)2 then we obtain a b-disjunct matrix of size s × (b + 1)2 by deleting any t − (b + 1)2 columns of M. This contradicts the EFF Conjecture. Suppose t < (b + 1)2 . Then we make a larger b-disjunct matrix by taking the direct sum of M and the ((b + 1)2 − t) × ((b + 1)2 − t) identity matrix to become a matrix of size ((b + 1)2 − t + s) × (b + 1)2 . We also have a contradiction to EFF Conjecture since (b + 1)2 − t + s < (b + 1)2 . Note that Eq (3, 2, 1) has more columns than rows. In the similar construction of disjunct matrices from a projective geometry of rank 3 [10], only square matrices can be obtained. The results in this chapter have been included in the following paper. “H. Huang, Y. Huang, and C. Weng, More on pooling spaces, Discrete Mathematics, 308 (2008), 6330-6338.” 19.

(28) Chapter 4 Distance-regular Subgraphs in a Distance-regular Graph Let Γ = (X, R) denote a distance-regular graph with diameter D. In this chapter, for given 0 ≤ d ≤ D we present a systematical way to construct a distance-regular subgraph of diameter d containing two given vertices of distance d in X. With some previous work, this construction also helps us to build a criterion that rules out the existence of some distance-regular graphs.. 4.1. Strongly Closed Subgraphs. A sequence x, z, y of vertices of Γ is geodetic whenever ∂(x, z) + ∂(z, y) = ∂(x, y), where ∂ is the distance function of Γ. A sequence x, z, y of vertices of Γ is weak-geodetic whenever ∂(x, z) + ∂(z, y) ≤ ∂(x, y) + 1. For a subset ∆ ⊆ X, ∆ is strongly closed if for any weak-geodetic sequence x, z, y of Γ, x, y ∈ ∆ =⇒ z ∈ ∆. A subset ∆ of X is strongly closed with respect to a vertex x ∈ ∆ if C(y, x) ⊆ ∆ and A(y, x) ⊆ ∆ 20. for all y ∈ ∆.. (4.1.1).

(29) Note that ∆ is strongly closed if and only if for any vertex x ∈ ∆, ∆ is strongly closed with respect to x [44, Lemma 2.3]. Strongly closed subgraphs are called weakgeodetically closed subgraphs in [44]. If a strongly closed subgraph ∆ of diameter d is regular then it has valency ad +cd = b0 −bd , where ad , cd , b0 , bd are intersection numbers of Γ. Furthermore ∆ is distance-regular with intersection numbers ai (∆) = ai (Γ) and ci (∆) = ci (Γ) for 1 ≤ i ≤ d [44, Theorem 4.5].. 4.2. D-bounded Property and Known Results. Definition 4.2.1. Γ is said to be d-bounded whenever for all x, y ∈ X with ∂(x, y) ≤ d, there is a regular strongly closed subgraph of diameter ∂(x, y) which contains x and y. Note that a (D − 1)-bounded distance-regular graph is clear to be D-bounded. The properties of D-bounded distance-regular graphs were studied in [43], and these properties were used in the classification of classical distance-regular graphs of negative type [45]. We list a few results which will be used later in this chapter. Theorem 4.2.2. ([44, Theorem 4.6]) Let Γ be a distance-regular graph with diameter D ≥ 3. Let Ω be a regular subgraph of Γ with valency γ and set d := min{i | γ ≤ ci +ai }. Then the following (i),(ii) are equivalent. (i) Ω is strongly closed with respect to at least one vertex x ∈ Ω. (ii) Ω is strongly closed with diameter d. In this case γ = cd + ad . The following Theorem is a combination of three previous results. Theorem 4.2.3. Let Γ denote a distance-regular graph with diameter D ≥ 3. Suppose that the intersection numbers a1 , a2 , c2 satisfy one of the following. (i) [14, Theorem 2] a2 > a1 = 0, c2 > 1; (ii) [44, Theorem 2] a1 ̸= 0, c2 > 1; or 21.

(30) (iii) [38, Theorem 1.1] a2 > a1 ≥ c2 = 1. Fix an integer 1 ≤ d ≤ D − 1 and suppose that Γ contains no parallelograms of any lengths up to d + 1. Then Γ is d-bounded. We will deal with the complemental case “a1 = 0, a2 ̸= 0, and c2 = 1” in Theorem 4.4.6.. 4.3. The Shapes of Pentagons. Throughout this section, let Γ = (X, R) denote a distance-regular graph with diameter D ≥ 3, and intersection numbers a1 = 0, a2 ̸= 0. Such graphs are also studied in [14, 26, 33, 34, 35]. By a pentagon in Γ, we mean a 5-tuple u1 u2 u3 u4 u5 consisting of distinct vertices in Γ such that ∂(ui , ui+1 ) = 1 for 1 ≤ i ≤ 4 and ∂(u5 , u1 ) = 1. Fix a vertex x ∈ X, a pentagon u1 u2 u3 u4 u5 has shape i1 , i2 , i3 , i4 , i5 with respect to x if ij = ∂(x, uj ) for 1 ≤ j ≤ 5. By a parallelogram of length d, we mean a 4-tuple xyzw consisting of vertices of Γ such that ∂(x, y) = ∂(z, w) = 1, ∂(x, w) = d, and ∂(x, z) = ∂(y, w) = ∂(y, z) = d − 1. Note that any two vertices at distance 2 are always contained in a pentagon since a2 ̸= 0, and two nonconsecutive vertices in a pentagon of Γ have distance 2 since a1 = 0. In this section we give a few lemmas which will be used in the next section. Lemma 4.3.1. Let Γ be a distance-regular graph with diameter D ≥ 3. Suppose a1 = 0, a2 ̸= 0, and Γ contains no parallelograms of lengths up to d + 1 for some integer d ≥ 2. Let x be a vertex of Γ, and let u1 u2 u3 u4 u5 be a pentagon of Γ such that ∂(x, u1 ) = i − 1 and ∂(x, u3 ) = i + 1 for 1 ≤ i ≤ d. Then the pentagon u1 u2 u3 u4 u5 has shape i − 1, i, i + 1, i + 1, i with respect to x. Proof. It suffices to prove ∂(x, u4 ) = i + 1. We prove this by induction on i. The case i = 1 holds otherwise ∂(x, u4 ) = 1 and ∂(x, u5 ) = 1 which contradicts the assumption a1 = 0. Suppose i ≥ 2. Suppose to the contrary that ∂(x, u4 ) = i. We can choose y ∈ C(x, u1 ). Thus ∂(y, u1 ) = i − 2 and ∂(y, u3 ) = i. By the induction hypothesis, the pentagon u1 u2 u3 u4 u5 has shape i − 2, i − 1, i, i, i − 1 with respect to y. In particular, 22.

(31) ∂(y, u3 ) = ∂(y, u4 ) = i. Then xyu4 u3 is a parallelogram of length i + 1, a contradiction.. Other versions of Lemma 4.3.1 can be seen in [44, Lemma 6.9] and [38, Lemma 4.1] under various assumptions on intersection numbers. The following three lemmas were formulated by A. Hiraki in [14] under an additional assumption c2 > 1, but this assumption is essentially not used in his proofs. For the sake of completeness, we still provide the proofs. Lemma 4.3.2. Fix an integer 1 ≤ d ≤ D − 1, and suppose Γ does not contain parallelograms of lengths up to d + 1. Then for any two vertices z, z ′ ∈ X such that ∂(x, z) ≤ d and z ′ ∈ A(z, x), we have B(x, z) = B(x, z ′ ). Proof. By symmetry, it suffices to show B(x, z) ⊆ B(x, z ′ ). Suppose there exists w ∈ B(x, z)\B(x, z ′ ). Then ∂(w, z ′ ) ̸= ∂(x, z)+1. Note that ∂(w, z ′ ) ≤ ∂(w, x)+∂(x, z ′ ) = 1 + ∂(x, z) and ∂(w, z ′ ) ≥ ∂(w, z) − ∂(z, z ′ ) = ∂(x, z). This implies ∂(w, z ′ ) = ∂(x, z) and wxz ′ z forms a parallelogram of length ∂(x, z) + 1, a contradiction. Lemma 4.3.3. Fix integers 1 ≤ i ≤ d ≤ D − 1, and suppose Γ does not contain parallelograms of any lengths up to d + 1. Let x be a vertex of Γ. Then there is no pentagon of shape i, i, i, i, i + 1 with respect to x. Proof. Let u1 u2 u3 u4 u5 be a pentagon of shape i, i, i, i, i+1 with respect to x. We derive a contradiction by induction on i. The case i = 1 is impossible since a1 = 0. Suppose i ≥ 2. Note that B(x, u1 ) = B(x, u2 ) = B(x, u3 ) = B(x, u4 ) by Lemma 4.3.2. We shall prove C(x, u1 ) = C(x, u2 ) = C(x, u3 ) = C(x, u4 ). First we prove C(x, u1 ) = C(x, u2 ). It suffices to show C(x, u2 ) ⊆ C(x, u1 ) since both sets have the same size ci . To the contrary suppose there exists v ∈ C(x, u2 ) − C(x, u1 ). Note that v ∈ A(x, u1 ) as B(x, u1 ) = B(x, u2 ). Then B(u1 , x) = B(u1 , v) by Lemma 4.3.2 and hence ∂(v, u5 ) = i + 1 since u5 ∈ B(u1 , x). Now u2 u1 u5 u4 u3 has shape i − 1, i, i + 1, i + 1, i with respect to v by Lemma 4.3.1, a contradiction since v ̸∈ B(x, u4 ) = B(x, u2 ). This proves C(x, u2 ) ⊆ C(x, u1 ) as desired. By symmetry, C(x, u3 ) = C(x, u4 ). 23.

(32) It remains to show C(x, u2 ) ⊆ C(x, u4 ). To the contrary suppose there exists u ∈ C(x, u2 )−C(x, u4 ). Note that u ∈ A(x, u4 ) as B(x, u2 ) = B(x, u4 ). Then B(u4 , x) = B(u4 , u) by Lemma 4.3.2 and hence ∂(u, u5 ) = i + 1 since u5 ∈ B(u4 , x). Hence u2 u1 u5 u4 u3 has shape i − 1, i, i + 1, i + 1, i with respect to u by Lemma 4.3.1, a contradiction since u ̸∈ B(x, u4 ). Pick a vertex v ∈ C(x, u1 ) = C(x, u2 ) = C(x, u3 ) = C(x, u4 ). Then u1 u2 u3 u4 u5 is a pentagon of shape i − 1, i − 1, i − 1, i − 1, i with respect to v, a contradiction to the inductive hypothesis. Lemma 4.3.4. Fix integers 1 ≤ i ≤ d ≤ D − 1, and suppose Γ does not contain parallelograms of any lengths up to d + 1. Let x be a vertex and u1 u2 u3 u4 u5 be a pentagon of shape i, i − 1, i, i − 1, i or of shape i, i − 1, i, i − 1, i − 1 with respect to x. Then B(x, u1 ) = B(x, u3 ). Proof. It suffices to show B(x, u3 ) ⊆ B(x, u1 ) since both sets have the same size bi . Pick u ∈ B(x, u3 ). Then ∂(u, u3 ) = i + 1. Since ∂(u3 , u2 ) = 1 and ∂(x, u2 ) = i − 1, then ∂(u, u2 ) = i and similarly ∂(u, u4 ) = i. Note that ∂(u, u1 ) ̸= i − 1, otherwise by Lemma 4.3.1, the pentagon u1 u2 u3 u4 u5 has shape i − 1, i, i + 1, i + 1, i with respect to u, a contradiction. Suppose ∂(u, u1 ) = i for this moment.. Then to avoid obtaining a pentagon. u5 u4 u3 u2 u1 of type i − 1, i, i + 1, i, i or a pentagon u4 u5 u1 u2 u3 of type i, i, i, i, i + 1 with respect to u we must have ∂(u, u5 ) = i + 1 by Lemma 4.3.1 and Lemma 4.3.3. Then ∂(x, u5 ) = i by construction. Now u5 u1 xu is a parallelogram of length i + 1, a contradiction. Hence ∂(u, u1 ) = i+1 or equivalently u ∈ B(x, u1 ). This proves B(x, u3 ) ⊆ B(x, u1 ) as desired. The following lemma rules out a class of pentagons of certain shapes with respect to a given vertex. Lemma 4.3.5. Fix integers 1 ≤ i ≤ d ≤ D − 1, and suppose Γ does not contain parallelograms of any lengths up to d + 1. Let x be a vertex. Then there is no pentagon of shape i, i, i, i + 1, i + 1 with respect to x. 24.

(33) Proof. Suppose that u2 u3 u4 u5 u1 is a pentagon of shape i, i, i, i + 1, i + 1 with respect to x. We derive a contradiction by induction on i. The case i = 1 is impossible since a1 = 0. Suppose i ≥ 2. Pick v ∈ C(x, u2 ) and note that ∂(v, u1 ) = i by construction. In particular v ̸∈ B(x, u2 ) and B(x, u2 ) = B(x, u3 ) = B(x, u4 ) by Lemma 4.3.2, so v ∈ C(x, u4 ) ∪ A(x, u4 ). In fact v ∈ C(x, u4 ); otherwise ∂(v, u4 ) = i. By considering the shape of the pentagon u2 u1 u5 u4 u3 with respect to v and applying Lemma 4.3.1, we have that ∂(v, u5 ) = i. Hence xvu4 u5 is a parallelogram of length i + 1, a contradiction. Thus ∂(v, u4 ) = i − 1, and by construction we now also have ∂(v, u5 ) = i. Note that ∂(v, u3 ) = i; otherwise ∂(v, u3 ) = i − 1 and u2 u3 u4 u5 u1 is a pentagon of shape i − 1, i − 1, i − 1, i, i with respect to v, a contradiction to the inductive hypothesis. Now setting x = v in Lemma 4.3.4, we have B(v, u1 ) = B(v, u3 ), a contradiction since x ∈ B(v, u1 ) − B(v, u3 ).. 4.4 D-bounded Property and Nonexistence of Parallelograms Let Γ = (X, R) denote a distance-regular graph with diameter D ≥ 3. Fix an integer 1 ≤ d ≤ D − 1. Throughout this section, we assume that Γ satisfies the following conditions. Assumption: (i) The intersection numbers satisfy a1 = 0, a2 ̸= 0, c2 = 1, and (ii) Γ contains no parallelograms of lengths up to d + 1. We shall prove the d-bounded property of Γ in this section. By the definition of strongly closed subgraphs, the following proposition is easily seen. Proposition 4.4.1. Suppose ∆ ⊆ X is a strongly closed subgraph of Γ and ux1 vx2 x3 or ux1 x2 vx3 is a pentagon in Γ. If u, v ∈ ∆, then x1 , x2 , x3 are all in ∆. Proof. Since a1 = 0, it’s easily seen that ∂(u, v) = 2 and u, xi , v is weak-geodetic for i = 1, 2, 3. 25.

(34) We then give a definition. Definition 4.4.2. For any vertex x ∈ X and any subset Π ⊆ X, define [x, Π] to be the subgraph induced by the set {v ∈ X | there exists y ′ ∈ Π, such that the sequence x, v, y ′ is geodetic }. For any x, y ∈ X with ∂(x, y) = d, set Πxy := {y ′ ∈ Γd (x) | B(x, y) = B(x, y ′ )}. (4.4.1). ∆(x, y) = [x, Πxy ].. (4.4.2). and. Note that ∆(x, y) contains x, y and Γd (x) ∩ ∆(x, y) = Πxy . We can also easily see the following proposition. Proposition 4.4.3. For x, y, z, w ∈ X and w ∈ ∆(x, y), if x, z, w is geodetic, then z ∈ ∆(x, y). Proof. Suppose ∂(x, y) = d, ∂(x, w) = i, and ∂(x, z) = j. Then ∂(z, w) = i − j. By the construction of Definition 4.4.2, there exists y ′ ∈ Π(x, y) such that x, w, y ′ is geodetic. Hence ∂(w, y ′ ) = d − i. Note that ∂(z, y ′ ) ≤ ∂(z, w) + ∂(w, y ′ ) = d − j, and ∂(z, y ′ ) ≥ ∂(x, y ′ ) − ∂(x, z) = d − j. So ∂(z, y ′ ) = d − j and thus x, z, y ′ are geodetic. Hence z ∈ ∆(x, y). For any 1 ≤ j ≤ d, we define the following three kinds of conditions: (Bj ) For any vertices x, y ∈ X with ∂(x, y) = j, ∆(x, y) is regular strongly closed with valency aj + cj . (Wj ) For any vertices x, y ∈ X with ∂(x, y) = j, ∆(x, y) is strongly closed with respect to x. (Rj ) For any vertices x, y ∈ X with ∂(x, y) = j, the subgraph induced on ∆(x, y) is regular with valency aj + cj 26.

(35) By referring to Theorem 4.2.2, the statement (Bj ) holds for all 1 ≤ j ≤ d is equivalent to the combination of conditions that (Wj ) and (Rj ) hold for all 1 ≤ j ≤ d. Our objective is to prove that (Bj ) holds for 1 ≤ j ≤ d under the assumptions in the beginning of this section. We use induction on j to achieve our objective. To adequately proceed the induction process, the following two lemmas are required. Lemma 4.4.4. Suppose (Wj ), (Rj ), and thus (Bj ) hold in X for 1 ≤ j ≤ d − 1. For any vertices x, y ∈ X with ∂(x, y) = d and for any vertex z ∈ ∆(x, y) ∩ Γi (x), where 1 ≤ i ≤ d, we have the following (i), (ii). (i) A(z, x) ⊆ ∆(x, y). (ii) For any vertex w ∈ Γi (x) ∩ Γ2 (z) with B(x, w) = B(x, z), we have w ∈ ∆(x, y). In particular (Wd ) holds. Proof. We prove (i), (ii) by induction on d − i. In the case i = d, z ∈ Π(x, y) and (i) follows by Lemma 4.3.2, and (ii) follows from the construction of ∆(x, y) in Definition 4.4.2. Suppose i < d. To prove (i) we note that if i = 1 then A(z, x) is an empty set as a1 = 0, clearly contained in ∆(x, y). Hence we suppose 2 ≤ i < d in this case. We pick a vertex v ∈ A(z, x) and show v ∈ ∆(x, y). Pick u ∈ ∆(x, y) ∩ Γi+1 (x) ∩ Γ1 (z). Note that (i), (ii) hold if we use u to replace z by the inductive hypothesis. Let uu2 u3 vz be a pentagon of Γ for some u2 , u3 ∈ X. Note that uu2 u3 vz cannot have shape i + 1, i, i − 1, i, i, shape i+1, i+2, i+1, i, i by Lemma 4.3.1, cannot have shape i+1, i, i, i, i by Lemma 4.3.3, and cannot have shape i + 1, i + 1, i, i, i by Lemma 4.3.5 with respect to x. Hence uu2 u3 vz has shape i + 1, i + 1, i + 1, i, i or i + 1, i, i + 1, i, i with respect to x. In the first case we have u2 ∈ A(u, x), u3 ∈ A(u2 , x), and this implies u2 , u3 ∈ ∆(x, y) by the inductive hypothesis of (i). Then v ∈ ∆(x, y) by Proposition 4.4.3 since x, v, u3 is geodetic. In the latter case we have B(x, u) = B(x, u3 ) by Lemma 4.3.4, and consequently u3 ∈ ∆(x, y) by inductive hypothesis of (ii). Then v ∈ ∆(x, y) by Proposition 4.4.3 since x, v, u3 is geodetic.. 27.

(36) To prove (ii) we first note that ∆(x, z) is a regular strongly closed subgraph of diameter i by Theorem 4.2.2 (ii) and since (Bi ) holds. Suppose to the contrary that there exists w ∈ Γi (x) ∩ Γ2 (z) with B(x, w) = B(x, z) such that w ∈ / ∆(x, y). Note that hence ∆(x, z) = ∆(x, w) by construction in Definition 4.4.2 since B(x, w) = B(x, z). Let v2 be the unique vertex in C(w, z). Claim 1. ∂(x, v2 ) = i − 1. Proof of Claim 1. Let v2 be the vertex between w and z. Let zv2 wv4 v5 be a pentagon for some v2 , v4 , v5 ∈ X. Since w ∈ ∆(x, z) = ∆(x, w), v2 , v4 , v5 ∈ ∆(x, z) by Proposition 4.4.1 and thus v2 , v4 , v5 ̸∈ Γi+1 (x). If v2 ∈ A(z, x) then v2 , w ∈ ∆(x, y) by (i), a contradiction. Hence ∂(x, v2 ) = i − 1. Let u be a vertex in ∆(x, y) ∩ Γi+1 (x) ∩ Γ1 (z), y3 ∈ A(u, v2 ), and y4 ∈ C(y3 , v2 ). Claim 2. The pentagon v2 zuy3 y4 has shape i − 1, i, i + 1, i + 1, i with respect to x. Moreover the pentagon is contained in ∆(x, y). Proof of Claim 2. The shape of the pentagon v2 zuy3 y4 is determined by Lemma 4.3.1. Since y3 ∈ A(u, x), y3 ∈ ∆(x, y) by the inductive hypothesis of (i) since d − ∂(x, u) < d − i. Let w3 ∈ A(y4 , w) and w4 ∈ C(w3 , w). Claim 3. The pentagon v2 y4 w3 w4 w has shape i − 1, i, i + 1, i + 1, i with respect to x and {w3 , w4 } ∩ {y3 , u} = ∅. Proof of Claim 3. Note that ∆(x, w) = ∆(x, z) is strongly closed of diameter i since Bi holds. Also note that v2 ∈ ∆(x, z). If ∂(x, w4 ) ≤ i then w4 ∈ ∆(x, w) and this forces y4 ∈ ∆(x, z) by Proposition 4.4.1. For the same reason, we then have y3 ∈ ∆(x, z) as z, y4 ∈ ∆(x, z). We have a contradiction since ∆(x, z) has diameter i and ∂(x, y3 ) = i + 1 > i = diam ∆(x, z). Hence ∂(x, w4 ) = i + 1 and v2 ww4 w3 y4 has shape i − 1, i, i + 1, i + 1, i with respect to x by Lemma 4.3.1. Hence by the inductive hypothesis of (i), if w3 ∈ ∆(x, y) then w4 ∈ ∆(x, y). Thus if {w3 , w4 } ∩ {y3 , u} ̸= ∅, then w4 ∈ ∆(x, y). Hence w ∈ ∆(x, y) by Proposition 4.4.3 since x, w, w4 is geodetic, a contradiction.. 28.

(37) The two pentagons v2 zuy3 y4 and v2 y4 w3 w4 w are shown in Figure 6. distance to x 0. p p p p p p. i−1. i. wr xr. p p p p p p. y4 r. v2 r. i+1. w4r w3 r y3. r. z. r. u. r. Figure 6. Two pentagons in the proof of Lemma 4.4.4(ii).. Claim 4. B(x, y3 ) ̸= B(x, w3 ). Proof of Claim 4. Note that B(x, u) = B(x, y3 ) and B(x, w3 ) = B(x, w4 ) by Lemma 4.3.2. If B(x, y3 ) = B(x, w3 ) then by the inductive hypothesis of (ii) we have w3 ∈ ∆(x, y). We then have w4 ∈ ∆(x, y) by the inductive hypothesis of (i). Thus w ∈ ∆(x, y) by Proposition 4.4.3, a contradiction. Let p3 ∈ A(y3 , w3 ) and p4 ∈ C(p3 , w3 ). Claim 5. The pentagon y4 y3 p3 p4 w3 has shape i, i + 1, i + 2, i + 2, i + 1 with respect to x. Proof of Claim 5. Since p3 is adjacent to y3 , ∂(x, p3 ) = i, i + 1 or i + 2. Suppose ∂(x, p3 ) = i + 1, then ∂(x, p4 ) ̸= i + 2 by Lemma 4.3.1, ∂(x, p4 ) ̸= i + 1 by Lemma 4.3.2, and ∂(x, p4 ) ̸= i by Lemma 4.3.4, a contradiction. Suppose ∂(x, p3 ) = i, then ∂(x, p4 ) ̸= i−1 by Lemma 4.3.1, ∂(x, p4 ) ̸= i by Lemma 4.3.4, and ∂(x, p4 ) ̸= i+1 by Lemma 4.3.4, also a contradiction. Thus ∂(x, p3 ) = i + 2 and the pentagon y4 y3 p3 p4 w3 has shape i, i + 1, i + 2, i + 2, i + 1 with respect to x by Lemma 4.3.1. Now we have three pentagons and their shapes with respect to x as shown in Figure 7.. 29.

(38) distance to x p p p p p p. 0. i−1. i. wr x. r. p p p p p p. v2. y4 r. r. i+1. w4r w3 r y3. r. z. p r. i+2. r. u. p4. r r. p3. r. Figure 7. Three pentagons in the proof of Lemma 4.4.4(ii). Claim 6. B(x, y4 ) ̸= B(x, z) and thus B(x, y4 ) − B(x, z) ̸= ∅. Proof of Claim 6. If B(x, y4 ) = B(x, z), then ∆(x, y4 ) = ∆(x, z), a strongly closed subgraph of diameter i. Since y4 , z ∈ ∆(x, z), we have y3 ∈ ∆(x, z) by Proposition 4.4.1 and ∂(x, y3 ) = i + 1, which is a contradiction as before. The fact B(x, y4 ) − B(x, z) ̸= ∅ is easily seen since |B(x, y4 )| = |B(x, z)| = bi . Pick p ∈ B(x, y4 ) − B(x, z). Claim 7. ∂(p, z) = i. Proof of Claim 7. Note that ∂(p, y4 ) = i + 1 under this assumption. Also note that for this moment ∂(p, z) = i − 1 or i. Suppose ∂(p, z) = i − 1. Then zv2 y4 y3 u is a pentagon of shape i − 1, i, i + 1, i + 1, i with respect to p by Lemma 4.3.1. Since p3 is adjacent to both y3 and ∂(x, p3 ) = i + 2, we have ∂(p, p3 ) = i + 2 or i + 1. Next we show that ∂(p, p3 ) = i + 2. If ∂(p, p3 ) = i + 1 then xpy3 p3 is a parallelogram of length i + 2 ≤ d + 1, a contradiction. Thus ∂(p, p3 ) = i + 2. Next we show that ∂(p, w3 ) = i + 2. We know that ∂(p, w3 ) = i, i + 1 or i + 2. Consider the shape of the pentagon y4 y3 p3 p4 w3 with respect to p. We have ∂(p, w3 ) ̸= i by Lemma 4.3.1. If ∂(p, w3 ) = i + 1, then ∂(p, p4 ) ̸= i + 1 by Lemma 4.3.3, and 30.

(39) ∂(p, p4 ) ̸= i + 2 by Lemma 4.3.5, a contradiction to the fact that p4 is adjacent to both w3 and p3 . Thus we have ∂(p, w3 ) = i + 2. We finally consider the shape of the pentagon v2 y4 w3 w4 w with respect to p and get a contradiction. Consider the relative distance among x, p, v2 , and y4 , we have ∂(p, v2 ) = i. Hence v2 y4 w3 w4 w is a pentagon of shape i, i + 1, i + 2, i + 2, i + 1 with respect to p by Lemma 4.3.1. That is p ∈ B(x, w), a contradiction to our assumptions B(x, z) = B(x, w) and p ∈ B(x, y4 ) − B(x, z). Claim 8. ∂(p, w) = i. Proof of Claim 8. We know that ∂(p, w) = i − 1, i or i + 1 since p is adjacent to x and ∂(x, w) = i. Suppose ∂(p, w) = i + 1, then p ∈ B(x, w) but p ∈ / B(x, z) which is a contradiction to our assumption that B(x, w) = B(x, z). Hence ∂(p, w) = i − 1 or i. Most of the following arguments are similar as the ones in the previous Step 7. Suppose ∂(p, w) = i − 1. First we have that the pentagon wv2 y4 w3 w4 is of shape i − 1, i, i + 1, i + 1, i with respect to p by Lemma 4.3.1. Next we show that then ∂(p, p4 ) = i + 2. To avoid xpw3 p4 to be a parallelogram of length i + 2 ≤ d + 1, we have ∂(p, p4 ) = i + 2. Then we show that ∂(p, y3 ) = i + 2. By applying Lemma 4.3.1, Lemma 4.3.3, and Lemma 4.3.5 to the pentagon y4 w3 p4 p3 y3 , we have that ∂(p, y3 ) = i + 2. We finally consider the shape of the pentagon v2 y4 y3 uz with respect to p and get a contradiction. Consequently v2 y4 y3 uz is a pentagon of shape i, i + 1, i + 2, i + 2, i + 1 with respect to p by Lemma 4.3.1, which is a contradiction to ∂(p, z) = i. Claim 9 ∂(p, u) = ∂(p, w4 ) = i + 1. Proof of Claim 9. Since ∂(p, z) = ∂(x, z) = i, we have p ∈ A(x, z) and thus B(z, x) = B(z, p) by Lemma 4.3.2, in particular ∂(p, u) = i + 1. Similarly, ∂(p, w4 ) = i + 1. Claim 10. ∂(p, y3 ) = i. Proof of Claim 10. As p ̸∈ B(x, u) = B(x, y3 ), we have ∂(p, y3 ) = i or i + 1. We shall prove ∂(p, y3 ) = i. Suppose ∂(p, y3 ) = i + 1.. We first show that ∂(p, p3 ) = i + 2. By applying. Lemma 4.3.2 we have B(y3 , x) = B(y3 , p). Then as p3 ∈ B(y3 , x) = B(y3 , p), ∂(p, p3 ) = 31.

(40) i + 2. Next we show that ∂(p, w3 ) = i + 2. Applying Lemma 4.3.3 and Lemma 4.3.5 to the pentagon w3 y4 y3 p3 p4 and considering its shape with respect to p, we find ∂(p, w3 ) ̸= i + 1. Applying Lemma 4.3.1 to the pentagon w3 p4 p3 y3 y4 , we find ∂(p, w3 ) ̸= i. Thus ∂(p, w3 ) = i + 2. We finally get a contradiction that pxw4 w3 is a parallelogram of length i+2 ≤ d+1. Claim 11. ∂(p, w3 ) = i. Proof of Claim 11. Similar as the arguments in the previous Step 10, as p ̸∈ B(x, w4 ) = B(x, w3 ), we have ∂(p, w3 ) = i or i+1. Suppose ∂(p, w3 ) = i+1. Applying Lemma 4.3.1 to the pentagon y3 p3 p4 w3 y4 , we then find ∂(p, p4 ) ̸= i + 2 and thus ∂(p, p4 ) = i + 1. Then xpw3 p4 us a parallelogram of length i + 2 ≤ d + 1, a contradiction. We finally consider the shape of the pentagon p4 w3 y4 y3 p3 with respect to p to get a final contradiction. Since ∂(x, p3 ) = i + 2 and ∂(p, y3 ) = i, we have ∂(p, p3 ) = i + 1 and similarly ∂(p, p4 ) = i + 1. To sum up, the pentagon p4 w3 y4 y3 p3 has shape i + 1, i, i + 1, i, i + 1 with respect to p. However, Lemma 4.3.4 now yields that B(p, p4 ) = B(p, y4 ), which is a contradiction since x ∈ B(p, p4 ) and x ∈ C(p, y4 ). Consequently, w ∈ ∆(x, y) and this completes the (ii) part of this lemma. By (i) we have A(z, x) ⊆ ∆(x, y) and by Proposition 4.4.3 we also have C(z, x) ⊆ ∆(x, y). Hence (Wd ) holds by (4.1.1).. The following lemma proves (Rd ) and hence completes the remaining of our goal. Lemma 4.4.5. Suppose (Wj ), (Rj ), and thus (Bj ) hold in X for 1 ≤ j ≤ d − 1. For any vertices x, y ∈ X with ∂(x, y) = d, ∆(x, y) is regular with valency ad + cd . Proof. Set ∆ = ∆(x, y). Clearly for any v ∈ ∆, the construction ensures us that ∂(x, v) ≤ d. Hence B(y ′ , x) ∩ ∆ = ∅ for any y ′ ∈ Πxy . Applying Lemma 4.4.4, we have |Γ1 (y ′ ) ∩ ∆| = ad + cd for any y ′ ∈ Πxy . Next we show |Γ1 (x) ∩ ∆| = ad + cd . Note that y ∈ ∆ ∩ Γd (x) by construction of ∆. For any z ∈ C(x, y) ∪ A(x, y), ∂(x, z) + ∂(z, y) ≤ ∂(x, y) + 1. 32.

(41) This implies z ∈ ∆ since ∆ is strongly closed with respect to x by Lemma 4.4.4. Hence C(x, y) ∪ A(x, y) ⊆ ∆. Suppose B(x, y) ∩ ∆ ̸= ∅. Choose t ∈ B(x, y) ∩ ∆. Then there exists y ′ ∈ Πxy such that t ∈ C(x, y ′ ), a contradiction to B(x, y) = B(x, y ′ ). Hence B(x, y) ∩ ∆ = ∅ and Γ1 (x) ∩ ∆ = C(x, y) ∪ A(x, y). This proves |Γ1 (x) ∩ ∆| = ad + cd . Since each vertex in ∆ appears in a sequence of vertices x = x0 , x1 , . . . , xd in ∆, where ∂(x, xℓ ) = ℓ, ∂(xℓ−1 , xℓ ) = 1 for 1 ≤ ℓ ≤ d, and xd ∈ Πxy , it suffices to show |Γ1 (xi ) ∩ ∆| = ad + cd. (4.4.3). for 1 ≤ i ≤ d − 1. For each integer 1 ≤ i ≤ d, we show |Γ1 (xi−1 ) \ ∆| ≤ |Γ1 (xi ) \ ∆|. (4.4.4). by the 2-way counting of the number of the pairs (z, s) for z ∈ Γ1 (xi−1 ) \ ∆, s ∈ Γ1 (xi ) \ ∆ and ∂(z, s) = 2. For a fixed s ∈ Γ1 (xi ) \ ∆, we have ∂(s, xi−1 ) = 2 since a1 = 0. Hence such a z must be one of the a2 vertices in A(xi−1 , s). The number of such pairs (z, s) is thus at most |Γ1 (xi ) \ ∆|a2 . On the other hand, we show this number is |Γ1 (xi−1 ) \ ∆|a2 exactly. Fix a z ∈ Γ1 (xi−1 ) \ ∆. Note that ∂(x, z) = i by Lemma 4.4.4, and ∂(xi , z) = 2 since a1 = 0. Pick any s ∈ A(xi , z). We shall prove s ̸∈ ∆. Suppose to the contrary s ∈ ∆ in the below arguments and choose any w ∈ C(s, z). Note that ∂(x, s) ≤ i, otherwise ∂(x, s) = i + 1 and the pentagon xi−1 xi swz has shape i − 1, i, i + 1, i + 1, i with respect to x by Lemma 4.3.1. Thus w ∈ A(s, x) and then w ∈ ∆ by Lemma 4.4.4(i). This forces z ∈ ∆ by Proposition 4.4.3, a contradiction. We also have ∂(x, w) ≤ i by considering the shape of the pentagon xi−1 zwsxi with respect to x and Lemma 4.3.1. If s ∈ A(xi , x), w ∈ A(s, x), and z ∈ A(w, x), then z ∈ ∆ by Lemma 4.4.4(i), a contradiction. Hence ∂(x, w) ≤ i − 1 or ∂(x, s) ≤ i − 1. Applying Lemma 4.3.4 to the pentagon xi xi−1 zws in the remaining cases we have B(x, z) = B(x, xi ) and then z ∈ ∆ by Lemma 4.4.4(ii), a contradiction. From the above counting, we have |Γ1 (xi−1 ) \ ∆|a2 ≤ |Γ1 (xi ) \ ∆|a2 33. (4.4.5).

(42) for 1 ≤ i ≤ d. Eliminating a2 from (4.4.5), we find (4.4.4) or equivalently |Γ1 (xi−1 ) ∩ ∆| ≥ |Γ1 (xi ) ∩ ∆|. (4.4.6). for 1 ≤ i ≤ d. We have shown previously |Γ1 (x0 ) ∩ ∆| = |Γ1 (xd ) ∩ ∆| = ad + cd . Hence (4.4.3) follows from (4.4.6).. Theorem 4.4.6. Let Γ = (X, R) denote a distance-regular graph with diameter D ≥ 3, and intersection numbers a1 = 0, a2 ̸= 0, and c2 = 1. Fix an integer 1 ≤ d ≤ D − 1 and suppose that Γ contains no parallelograms of any lengths up to d + 1. Then Γ is d-bounded. Proof. For 1 ≤ j ≤ d, we prove (Wj ) and (Rj ) by induction on j. Since a1 = 0, there are no edges in Γ1 (x) for any vertex x ∈ X. If d = 1 in Definition 4.4.2, then Πxy = {y} since for any other y ′ ∈ Γ1 (x), y ′ ∈ B(x, y) but y ′ ∈ / B(x, y ′ ). Consequently ∆(x, y) = {x, y} is an edge; in particular ∆(x, y) is regular with valency 1 = a1 + c1 and is strongly closed with respect to x since a1 = 0. This proves (R1 ) and (W1 ). For d ≥ 2, assume (Wj ), (Rj ) and thus (Bj ) hold for 1 ≤ j ≤ d − 1. By Lemma 4.4.4 and Lemma 4.4.5, we have that (Wd ), (Rd ), and thus (Bd ) hold. Then the proof is completed. Theorem 4.4.6 answers the problem proposed in [44, p. 299] and is a generalization of [5, Lemma 4.3.13], [34]. Recall that Theorem 4.2.3 (i) was proved by A. Hiraki [14]. Indeed for the lemmas stated independently in Section 4.3 we are inspired by some lemmas in [14]. Combining Theorem 4.2.3 and Theorem 4.4.6, the following characterization of dbounded distance-regular graphs is completed. Theorem 4.4.7. Suppose Γ is a distance-regular graph with diameter D ≥ 3 and the intersection number a2 ̸= 0. Fix an integer 2 ≤ d ≤ D − 1. Then the following two conditions (i), (ii) are equivalent: (i) Γ is d-bounded. 34.

(43) (ii) Γ contains no parallelograms of any lengths up to d + 1 and b1 > b2 . Proof. ((i) ⇒ (ii)) Suppose that Γ is d-bounded for d ≥ 2. Let Ω ⊆ ∆ be two regular strongly closed subgraphs of diameters 1, 2 respectively. Since Ω and ∆ have different valency b0 −b1 and b0 −b2 respectively by Theorem 4.2.2, we have b1 > b2 . It is also easy to see that Γ contains no parallelograms of any lengths up to d + 1 [44, Lemma 6.5]. ((ii) ⇒ (i)) Under the assumptions Theorem 4.4.7(ii) (hence b1 > b2 ) and a2 ̸= 0, consider the following four cases. (a) a1 = 0 and c2 > 1: This case follows from Theorem 4.2.3 (i). (b) a1 = 0 and c2 = 1: This case follows from Theorem 4.4.6. (c) a1 ̸= 0 and c2 > 1 : This case follows from Theorem 4.2.3 (ii). (d) a1 ̸= 0 and c2 = 1 : Note that in this case a2 > a1 ≥ c2 = 1. Then this case follows from Theorem 4.2.3 (iii).. Some applications of Theorem 4.4.7 were previously given in [14, 35]. We will give a new application as Theorem 4.5.7 in the following section.. 4.5. Classical Parameters. Let Γ = (X, R) denote a distance-regular graph with diameter D ≥ 3. Γ is said to have classical parameters (D, b, α, β) whenever the intersection numbers of Γ satisfy [ ]( [ ]) i i−1 ci = 1+α for 0 ≤ i ≤ D, (4.5.1) 1 b 1 b ([ ] [ ] )( [ ]) D i i bi = − β−α for 0 ≤ i ≤ D, . (4.5.2) 1 b 1 b 1 b Applying (2.2.1) with (4.5.1), (4.5.2), we have [ ]( [ ] [ ] [ ] ) i D i i−1 ai = β − 1 + α( − − ) 1 b 1 b 1 b 1 b [ ]( [ ] [ ] ) i i i−1 = a1 − α( + − 1) 1 b 1 b 1 b 35. (4.5.3) (4.5.4).

(44) for 1 ≤ i ≤ D. Classical parameters were introduced in [5, Chapter 6]. Graphs with such parameters yield P - and Q-polynomial association schemes. Bannai and Ito proposed the classification of such schemes in [1]. Suppose Γ has classical parameters (D, b, α, β) and D ≥ 3. Then b is an integer, b ̸= 0, and b ̸= −1 [5, p. 195]. Two known classes of distance-regular graphs with classical parameters (D, b, α, β) and b < −1 are the dual polar graphs 2 A2D−1 (−b) and the Hermitian forms graphs Her−b (D) as listed in [5, Table 6.1]. Here we use the notation in [5, page 274]. A.A. Ivanov and S.V. Shpectorov show that if Γ has the same intersection numbers as the dual polar graph 2. A2D−1 (−b) then Γ is the dual polar graph 2 A2D−1 (−b) [20]. They also show that if Γ. does not contain parallelograms of length 2 and has the same intersection numbers as the Hermitian forms graph Her−b (D) then Γ is the Hermitian forms graph Her−b (D) [21, 22]. P. Terwilliger shows the following theorem. Theorem 4.5.1. ([39, Theorem 2.12], [44, Lemma 7.3(ii)]) Let Γ denote a distanceregular graph with classical parameters (D, b, α, β), b < −1, and D ≥ 3. Then Γ contains no parallelograms of any lengths. More general versions of Theorem 4.5.1 can be found in [42, 26, 33]. The following is a by-product of Theorem 4.5.1. Lemma 4.5.2. ([39, Theorem 2.11], [44, Lemma 7.3(ii)]) Let Γ denote a distanceregular graph with classical parameters (D, b, α, β) and D ≥ 3. Suppose Γ contains no parallelograms of lengths 2. Then Γ contains no parallelograms of any lengths. By applying Theorem 4.5.1, the D-bounded property of Γ is proved by different authors according to different assumptions [38, 44, 34, 14]. Recall that if Γ has intersection numbers b1 > b2 and a2 ̸= 0 then Γ is D-bounded as stated in Theorem 4.4.7. A poset associated with a D-bounded distance-regular graph was constructed in [43] and further studied in [45]. This produces the following two useful theorems.. 36.

(45) Theorem 4.5.3. ([43, Corollary 3.7, Theorem 4.2]) Let Γ denote a distance-regular graph with classical parameters (D, b, α, β) and b < −1. Suppose that Γ is D-bounded with D ≥ 4. Then β=α. 1 + bD . 1−b. (4.5.5). Theorem 4.5.4. ([45, Lemma 10.2, Theorem 10.3]) Let Γ denote a distance-regular graph with classical parameters (D, b, α, β) and b < −1. Suppose that Γ is D-bounded with D ≥ 4, and Γ is neither the dual polar graph 2 A2D−1 (−b) nor the Hermitian forms graph Her−b (D). Then α = (b − 1)/2,. β = −(1 + bD )/2,. (4.5.6). where −b is a power of an odd prime.. Recently, J. Guo and K. Wang investigated other posets associated with a Dbounded distance-regular graph [13]. F. Vanhove shows that the existence of a (−b + 1)/2-ovoid in the dual polar graph 2 A2D−1 (−b) will imply the existence of Γ with parameters as in (4.5.6) of Theorem 4.5.4 [41]. The following two lemmas have been obtained by applying Theorem 4.5.3. Lemma 4.5.5. ([43, Corollary 6.4]) There is no distance-regular graph Γ with classical parameters (D, b, α, β), D ≥ 4, c2 = 1, and a2 > a1 > 1. Lemma 4.5.6. ([35, Theorem 2.2]) Let Γ denote a distance-regular graph with classical parameters (D, b, α, β) and D ≥ 3. Assume the intersection numbers a1 = 0, a2 ̸= 0, and c2 = 1. Then (b, α, β) = (−2, −2, ((−2)D+1 − 1)/3). Theorem 4.5.7. There is no distance-regular graph with classical parameters (D, b, α, β) = (D, −2, −2, ((−2)D+1 − 1)/3), where D ≥ 4. Proof. Let Γ denote a distance-regular graph with classical parameters (D, b, α, β) = (D, −2, −2, ((−2)D+1 − 1)/3), where D ≥ 4. Then Γ contains no parallelograms of 37.