11institutetext:  A. Kumar 22institutetext: Applied Statistics Unit, Indian
Statistical Institute, Kolkata, India, 22email<EMAIL_ADDRESS>33institutetext: S. Maitra 44institutetext: Applied Statistics Unit, Indian
Statistical Institute, Kolkata, India, 44email<EMAIL_ADDRESS>55institutetext: S. Roy 66institutetext: Department of Statistics, University
of Calcutta, Kolkata, India, 66email<EMAIL_ADDRESS>
# Almost Perfect Mutually Unbiased Bases that are Sparse

Ajeet Kumar  Subhamoy Maitra  Somjit Roy

###### Abstract

In dimension $d$, Mutually Unbiased Bases (MUBs) are a collection of
orthonormal bases over $\mathbb{C}^{d}$ such that for any two vectors
$v_{1},v_{2}$ belonging to different bases, the dot or scalar product
$|\braket{v_{1}}{v_{2}}|=\frac{1}{\sqrt{d}}$. The upper bound on the number of
such bases is $d+1$. Construction methods to achieve this bound are known for
cases when $d$ is some power of prime. The situation is more restrictive in
other cases and also when we consider the results over real rather than
complex. Thus, certain relaxations of this model are considered in literature
and consequently Approximate MUBs (AMUB) are studied. This enables one to
construct potentially large number of such objects for $\mathbb{C}^{d}$ as
well as in $\mathbb{R}^{d}$. In this regard, we propose the concept of Almost
Perfect MUBs (APMUB), where we restrict the absolute value of inner product
$|\braket{v_{1}}{v_{2}}|$ to be two-valued, one being 0 and the other
$\leq\frac{1+\mathcal{O}(d^{-\lambda})}{\sqrt{d}}$, such that $\lambda>0$ and
the numerator $1+\mathcal{O}(d^{-\lambda})\leq 2$. Each such vector
constructed, has an important feature that large number of its components are
zero and the non-zero components are of equal magnitude. Our techniques are
based on combinatorial structures related to Resolvable Block Designs (RBDs).
We show that for several composite dimensions $d$, one can construct
$\mathcal{O}(\sqrt{d})$ many APMUBs, in which cases the number of MUBs are
significantly small. To be specific, this result works for $d$ of the form
$(q-e)(q+f),\ q,e,f\in\mathbb{N}$, with the conditions $0\leq f\leq e$ for
constant $e,f$ and $q$ some power of prime. We also show that such APMUBs
provide sets of Bi-angular vectors which are of the order of
$\mathcal{O}(d^{3/2})$ in numbers, having high angular distances among them.

###### Keywords:

Almost Perfect Mutually Unbiased Bases, Combinatorial Design, Hadamard
Matrices, Quantum Information Theory, Resolvable Block Design.

###### MSC:

81P68

##  1 Introduction

Mutually Unbiased Bases (MUBs) received serious attention in Quantum
Information Processing, as they are useful in different aspects of Quantum
Cryptology and Communications like Quantum Key Distribution (QKD),
Teleportation, Entaglement Swapping, Dense Coding, Quantum Tomography, etc.
(see Durt2010  and the references therein). For any finite dimensional Hilbert
space $\mathbb{C}^{d}$, the number of MUBs is bounded by $d+1$. However, in
spite of intense research for several decades, these could be constructed only
when $d$ is some prime power Ivonovic1981 ; Wootters1989 ; Bandyopadhyay2002 ;
Klappenecker2003 ; Sulc2007 . For a specific dimension $d$, constructing a
larger number of MUBs (upper bounded by $d+1$) is one of the most challenging
problems in Quantum Information Theory.

Various mathematical tools have been used to construct MUBs, among which
noteworthy being the use of finite fields Wootters1989 ; Klappenecker2003  and
maximal set of commuting bases Bandyopadhyay2002 . For dimensions which are
not power of primes, constructing large number of MUBs still remains elusive.
This is the reason, various kinds of Approximate MUBs have been constructed
using character sums over Galois Rings or Galois Fields AMUB-
Shparlinski20062006 ; ASIC-Klappenecker2005 ; AMUB-MixedCharacterSum ; AMUB-
CharacterSum ; AMUB-FrobeniusRing ; AMUB-GaloisRing ; ARMUB-fromComplexAMUB ,
combinatorial design AK20 ; ak22  and computational search AK22 .

When MUBs are constructed over $\mathbb{R}^{d}$, we get Real MUBs. They have
interesting connections with Quadratic Forms QuadraticForm , Association
Schemes ReMUB-AssociationSchemes ; AssociationScheme-Coding , Equi-angular
Lines, Equi-angular Tight Frames over $\mathbb{R}^{d}$ ReMUB-FusionFrames ,
Representation of Groups ReMUB-GroupRepresentation , Mutually Unbiased Real
Hadamard Matrices, Bi-angular vectors over $\mathbb{R}^{d}$ Holzmann2010 ;
Kharaghani2018 ; Best2015  and Codes Calderbank1997 . As we have elemented out
earlier, large number of Real MUBs are non-existent for most of the dimensions
boykin2005real . In fact only for $d=4^{s},s>1$, we have $d/2+1$ many MUBs,
whereas for most of the dimensions $d$, which are not perfect square, we have
at best only $2$ Real MUBs boykin2005real . In view of this, various attempts
have been made to construct Approximate Real MUBs (ARMUBs) which are available
in literature AK20 ; ak22 ; ARMUB-fromComplexAMUB .

Various efforts have been made to explore connections between MUBs and
geometrical objects such as polytopes and projective planes MUB-Polytopes ;
MUB-CompPolytopes ; MUBs-ProjectivePlanes ; MUBs-HjelmslevGeometry ;
GaloisUnitaries-MUB . Since the known methods for the construction of MUBs
provides complete sets only when $d$ is some power of prime, there are
conjectures related to the existence of complete sets of MUBs and finite
projective plane, which are also currently known to exist only for prime power
orders. If $d=p_{1}^{n_{1}}p_{2}^{n_{2}}\ldots p_{s}^{n_{s}}$, then the lower
bound on the number of MUBs is $p_{r}^{n_{r}}+1$ where
$p_{r}^{n_{r}}=\min\\{{p_{1}^{n_{1}},p_{2}^{n_{2}},\ldots,p_{s}^{n_{s}}}\\}$.
Thus, constructing a large number of MUBs for any composite dimension has
proven to be elusive even over $\mathbb{C}^{d}$. In fact, the number of such
bases is very small when we consider the problem over the real vector space
$\mathbb{R}^{d}$ (see boykin2005real ). For discussions on the basics and open
problems related to MUBs and the approximate version, one may refer to our
earlier works AK20 ; ak22  and the references therein.

Towards constructing the approximate MUBs over $\mathbb{C}^{d}$, certain
techniques were proposed in AMUB-Shparlinski20062006 ; ASIC-Klappenecker2005 ;
AMUB-CharacterSum . However, such techniques cannot be applied in construction
of Real MUBs. In this direction, we could propose a construction in AK20  for
Approximate Real MUBs using Real Hadamard matrices. It has been shown in AK20
that $\frac{\sqrt{d}}{4}+1$ ARMUBs with maximum value of the inner product as
$\frac{4}{\sqrt{d}}$ could be achieved for $d=(4q)^{2}$, where $q$ is a prime.
In ak22 , we have shown that such results can be generalized as well as
improved to a great extent using Resolvable Block Designs (RBDs). The earlier
result of AK20  could be generalised in ak22  for $d=sq^{2}$, where $q$ is a
prime power. Further, the parameters could be improved too in ak22 . It has
been shown in ak22  that for $d=q(q+1)$, where $q$ is a prime power and
$q\equiv 3\bmod 4$, it is possible to construct $\lceil{\sqrt{d}}\rceil=q+1$
many ARMUBs with maximum value of the inner product upper bounded by
$\frac{2}{\sqrt{d}}$, between vectors belonging to different bases. Therefore,
the improvement in the result of ak22  is two-fold. First, the number of MUBs
is greater and second, the maximum of the inner product values is lower
compared to AK20 . As these were achieved by several kinds of combinatorial
designs, we explore this idea further. One important feature of all our
construction techniques AK20 ; ak22  was that, all the components of the basis
vectors so constructed is either zero or of constant magnitude. Hence taking
the normalising factor outside the vectors would render the numerical value of
the components as zero or of unit magnitude, which is like Weighing matrices
weigh1  – a generalisation of Hadamard matrices. Finally, note that the
construction of ak22  provides the vectors which are sparse and this property
is inherited in this paper too.

Approximate MUBs have been defined in various manners in the state of the art
literature. The cue has been taken from the two initial papers ASIC-
Klappenecker2005 ; AMUB-Shparlinski20062006 . Although the work of ASIC-
Klappenecker2005  is related to Approximate SIC POVM, the definition of
“Approximate” has been carried over to MUBs as well. Various mathematical
meanings of approximations have been used in relaxing the condition on
absolute value of the inner product between two vectors say
$\ket{v_{1}},\ket{v_{2}}$. For example, in these two papers ASIC-
Klappenecker2005 ; AMUB-Shparlinski20062006 , we get references to
$|\braket{v_{1}}{v_{2}}|$ bounded by $\frac{1+o(1)}{\sqrt{d}}$,
$\frac{2+o(1)}{\sqrt{d}}$, $\mathcal{O}(\frac{log(d)}{\sqrt{d}})$,
$\mathcal{O}(\frac{1}{\sqrt[4]{d}})$, and $\mathcal{O}(\frac{1}{\sqrt{d}})$
etc. Subsequent researchers investigating Approximate MUBs have also adopted
similar mathematical definition AMUB-CharacterSum ; AMUB-MixedCharacterSum ;
AMUB-FrobeniusRing ; AMUB-GaloisRing ; ARMUB-fromComplexAMUB .

Let us now summarise the contribution of this paper and outline its
presentation.

###  1.1 Organization and Contribution

We begin with Section 2 to present a background of related combinatorial
objects. Then towards the constructions, in Section 3, we show bounds on the
values of certain parameters, expressed in terms of the block size $k$ and
number of elements in the RBD, i.e., $d$. In this regard, we define a
combinatorial quantity $\mathcal{A}(d,k,\mu)$, relevant to our analysis, which
can be of its own independent interest. Thereafter in Lemma 3 we describe an
interesting class of RBDs which can be constructed from MOLS(s) yielding
$\mu=1$ and $r=N(s)+2$. Here $\mu$ is the maximum number of common elements
between any pair of blocks from different parallel classes, as we will explain
in the following background section. A constructive proof to obtain the same
from MOLS(s) has also been given and in Lemma 4. We further show that the
converse of Lemma 3 is also true. The results are further explained with
illustrative examples.

Then we consider the Almost Perfect MUBs (APMUBs) and some generic ideas of
construction in Section 4. The basic motivation and its relationship with Bi-
angular vectors are presented in Section 4.1. In Section 4.2, we analyse
certain properties of the AMUBs which can be constructed using RBDs having
blocks of constant size. In this direction, we consider RBD$(X,A)$ with
$|X|=d=k\cdot s=(q-e)(q+f)$ and $A$ with resolution $r$, where each block is
of size $(q-e)$. We study the asymptotic behaviours of the parameters of AMUBs
thus generated. It is also shown that our construction can provide APMUBs only
when $\mu=1$, therefore putting strong constraints over the nature of the RBDs
required for this kind of constructions.

Section 5 contains our algorithms towards constructing RBDs that can be
consequently used for the obtaining APMUB’s with parameters that could be
achieved for the first time. We first show that whenever the dimension $d$ is
a composite number and can be expressed as $k\cdot s,k\leq s$, such that
$\beta=\sqrt{\frac{s}{k}}\leq 2$, one can construct $N(s)+1$ many APMUBs,
where $N(s)$ is the number of MOLS$(s)$. We refer to this as the MOLS Lower
Bound Construction for APMUBs. Since a composite number $d$ can be factored in
multiple ways ensuring $\beta\leq 2$, there can be more than one MOLS Lower
Bound Constructions for a dimension $d$. It is to be noted that if $s=q$, some
power of prime, then $N(q)=q-1$. Hence in such situations we get $q$ many
APMUBs. The best known asymptotic bound for $N(s)$ is given by
$N(s)\rightarrow\mathcal{O}(s^{\frac{1}{14.8}})$, which generally results into
a small number of APMUBs. We improve this significantly.

In this direction we show that when $d=(q-e)(q+f),e,f\in\mathbb{N}$ and $e\geq
f$, where $q$ is some power of prime, we can obtain $\mathcal{O}(q)$ many
APMUBs. That is, when $e,f$ are constant, then the number of such APMUBs is
$\mathcal{O}(\sqrt{d})$. Illustrative examples to describe the above
construction have also been provided. This is explained in Section 5.2.

We conclude with the following two comments in Section 6 towards future
direction. First, the dimensions $d$ for which we obtain $O(\sqrt{d})$ many
MUBs have the density of the order of the primes in the space of natural
numbers. Secondly, the constructions of APMUBs directly translates to the
construction of Bi-angular vectors. The other direction may also be explored.
It needs further disciplined effort to explore this part too.

##  2 Background

Let us start with the basic definition related to Mutually Unbiased Bases.

###### Definition 1

Consider two orthonormal bases,

$M_{l}=\left\\{\ket{\psi_{1}^{l}},\ket{\psi_{2}^{l}},\ldots,\ket{\psi_{d}^{l}}\right\\}\mbox{
and
}M_{m}=\left\\{\ket{\psi_{1}^{m}},\ket{\psi_{2}^{m}},\ldots,\ket{\psi_{d}^{m}}\right\\},$

in $d$-dimensional complex vector space, i.e., $\mathbb{C}^{d}$. These two
bases will be called Mutually Unbiased if we have

$\left|\braket{\psi_{i}^{l}}{\psi_{j}^{m}}\right|=\displaystyle\frac{1}{\sqrt{d}},\
\forall i,j\in\left\\{1,2,\ldots,d\right\\}.$ (1)

The set $\mathbb{M}=\left\\{M_{1},M_{2},\ldots,M_{r}\right\\}$ consisting of
such orthonormal bases will form MUBs of size $r$, if every pair in the set is
mutually unbiased.

When the conditions among two different bases are relaxed such that
$\left|\braket{\psi_{i}^{l}}{\psi_{j}^{m}}\right|$ can take values other than
$\frac{1}{\sqrt{d}}$, then we consider the approximate version of this
problem. This is due to the fact that, for most of the dimensions, which are
not power of some primes, obtaining a large number of MUBs reaching the upper
bound is elusive. In this context we denote $\Delta$ for the set containing
different values of $\left|\braket{\psi_{i}^{l}}{\psi_{j}^{m}}\right|$ for
$l\neq m$. In this initiative, extending the ideas of ak22 , we exploit the
well known combinatorial object, the Resolvable Block Design (RBD), towards
construction of Approximate MUBs with improved parameters. One may refer to
the book stinson2007combinatorial  for more details on RBDs, and we present
certain definitions in this regard.

###### Definition 2

A combinatorial block design is a pair $(X,A)$, where $X$ is a set of
elements, called elements, and $A$ is a collection of non-empty subsets of
$X$, called blocks. A combinatorial design is called simple, if there is no
repeated block in $A$.

Generally all the combinatorial designs are assumed to be simple, i.e., they
do not have any repeated blocks.

###### Definition 3

A combinatorial design $(X,A)$ is a $t-(d,k,\lambda)$ design if each block in
$A$ is of size $k$, and that any set of $t$ elements from $X$, appears as
subset of exactly $\lambda$ blocks in $A$. Note that here $t,d,k$ and
$\lambda$ are positive integers with $1<k<d$.

###  2.1 Resolvable Block Design (RBD)

Resolvable Block Design (RBD) is a special kind of Combinatorial design, where
the set $A$ can be partitioned into parallel classes which are called
resolutions of $A$. Initially RBD was defined by R. C. Bose in the context of
Balanced Incomplete Block Design Bose1939 ; Bose1942 ; Bose1947 . Later
various generalisations could be achieved as explained in varied literatures
Shrikhande1965 ; Kageyama1976 ; Patterson1976 ; John1999 . For the purpose of
our paper we consider the following simple definition of RBD as presented in
our previous paper ak22 .

###### Definition 4

Combinatorial design $(X,A)$, is called a Resolvable Block Design (RBD), if
$A$ can be partitioned into $r\geq 1$ parallel classes, called resolutions.
Where a parallel class in design $(X,A)$ is a subset of the disjoint blocks in
$A$ whose union is $X$.

There is a special kind of RBD called Affine Resolvable BIBD (ARBIBD) Bose1942
; Bose1947 ; Shrikhande1976  (see also (stinson2007combinatorial, , Chapter
5)). It is well known that whenever $q$ is some power of a prime, one can
construct $(q^{2},q,1)$ ARBIBD. An Affine plane of order $q$ is an example of
this. Here $|X|=q^{2}$, and $A$ consists of $q(q+1)$ blocks, which can be
resolved into $q+1$ many parallel classes. Each parallel class consists of $q$
many blocks of constant size $q$. Most importantly, any pair of blocks from
different parallel classes has exactly one element in common. Affine Planes
are known only when $q$ is power of a prime. For detail, one may refer to
(stinson2007combinatorial, , Sections 2.3, 5.2, 5.3, 6.4).

Let us define the notation that we will use in this connection. For
RBD$(X,A)$, with $|X|=d$, we will indicate the elements (also called elements)
of $X$ by simple numbering, i.e., $X=\\{1,2,3,\ldots,d\\}$. Here $r$ will
denote the number of parallel classes in RBD and parallel class will be
represented by $P_{1},P_{2},\ldots,P_{r}$. The blocks in the $l^{th}$ parallel
class will be represented by $\\{b_{1}^{l},b_{2}^{l},\ldots,b_{s}^{l}\\}$,
indicating that the $l^{th}$ parallel class has $s$ many blocks. Since in our
entire analysis we will be using RBDs with constant block size, let us denote
the block size by $k$. Further, we denote the number of blocks in a parallel
class of RBD by $s$. Since in our analysis we are making of use of RBDs with
constant block size, hence each parallel class will always have $s$ many
blocks and $|X|=d=k\cdot s$. The notation $b_{ij}^{l}$ would represent the
$j^{th}$ element of the $i^{th}$ block in the $l^{th}$ parallel class.
Further, the notation $b_{i}^{l}$ would represent $i^{th}$ block of $l^{th}$
parallel class. Note that $b_{ij}^{l}\in X$. In every block, we will arrange
the elements in increasing order, and we will follow this convention
throughout the paper, unless mentioned specifically. Thus $b_{ij}^{l}\leq
b_{i,j+1}^{l},\ \forall j$. This will be important to revisit when we convert
the parallel classes into orthonormal bases. Another important parameter for
our construction is the value of the maximum number of common elements between
any pair of blocks from different parallel classes. We denote this positive
integer by $\mu$. Note that $\mu\geq 1$ for any RBD, with $r\geq 2$. One may
further refer to Lemma 2 of Section 3 in this regard.

###  2.2 Mutually Orthogonal Latin Square (MOLS)

A Latin Square of order $s$ is an $s\times s$ array, and a cell of the array
consists of a single element from a set $Y$, such that $|Y|=s$. Every row of
the Latin Square is a permutation of the elements of set $Y$ and every column
of the Latin square is also permutation of the elements from the set $Y$. For
more details one may refer to (stinson2007combinatorial, , Definition 6.1) as
well as (HandBook, , Example 1.1). A pair of Latin Squares $(L_{1},L_{2})$ of
same order and having entries from same set $Y$ (or, a different set having
same number of elements) is called Mutually Orthogonal, if in the ordered pair
$\\{(Y,Y)\\}=\\{((L_{1})_{ij},(L_{2})_{ij})\\}$, every pair $x,y\in Y$ appears
exactly once. That is, if two of the Latin Squares are superimposed, and the
resulting entries in each cell is written as ordered pairs, then every $x,y\in
Y$ appears exactly once in the cell. Further, if there is a set of $w$ many
Latin Squares, say $\\{L_{1},L_{2},\ldots,L_{w}\\}$, each of order $s$, such
that, every pair of Latin Squares is orthogonal, then the set is called
Mutually Orthogonal Latin Square of order $s$, which we denote as
$w$-MOLS$(s)$.

Let $N(s)$ denote the maximal value of $w$ such that, there are $w$ many MOLS
of order $s$ HandBook ; Colbourn2001 , (stinson2007combinatorial, , Chapter
6). While using the numerical values of $N(s)$, in subsequent examples in this
paper, we will use the currently known values of $N(s)$ from (HandBook, ,
Table 3.87, page 176). Note that these are not always the actual values of
$N(s)$ (except when $s$ is some power prime or of small order) as the exact
value of $N(s)$ is still an open question in most of the cases. It is known
that, $N(s)\leq s-1$ $\forall\ s$. When this bound is attained, we say that
there is a complete set of Mutually Orthogonal Latin Squares of order $s$. The
construction for complete sets of MOLS$(s)$ is known when $s$ is some power of
prime (stinson2007combinatorial, , Section 6.4). When $s$ is not a power of
prime, $N(s)$ is much smaller than $s-1$. A table with the largest known
values for $w$ is presented in HandBook  for $s<10000$.

It is known that there exists a constant $n_{0}$, such that for all $s\geq
n_{0}$, we have, $N(s)\geq\frac{1}{3}s^{\frac{1}{91}}$ Chowla1960 , which was
later improved by Wilson Wilson1974  to $N(s)\geq s^{\frac{1}{17}}$. Further,
it was shown in (wocjan2004new, , Section 4) that the exponent can be lower
bounded by $\frac{1}{14.8}$. One may note that $N(s)\rightarrow\infty$ as
$s\rightarrow\infty$ in general, but for the finite cases only when $s$ is
some power of prime then $N(s)=s-1$, else it is considerably small. In this
regard, one may also note that the Affine Planes of order $q$ are equivalent
to $(q-1)$ MOLS$(q)$ (stinson2007combinatorial, , Theorem 6.32). For a brief
survey on construction of MOLS, one may refer to Colbourn2001 . Note that, in
connection of RBD, we used $s$ to indicate the number of blocks in a parallel
class, whereas in the context of Latin Square, we use $s$ to indicate the
order of Latin Square.

##  3 Some Important Technical Results

Let us now consider a counting of combinatorial objects that is relevant to
us.

###### Definition 5

Let ${\cal{T}}(d,k,\mu)$, $0\leq\mu<k<d$ be the maximum number of subsets each
of size $k$, that can be constructed from $d$ distinct objects, such that,
between any two different subsets there is a maximum of $\mu$ objects in
common.

One may immediately note that
${\cal{T}}(d,k,0)=\left\lfloor\frac{d}{k}\right\rfloor$, and
${\cal{T}}(d,k,k-1)=\binom{d}{k}$. For arbitrary $d,k,\mu\in\mathbb{N}$, the
following result provides an estimate of ${\cal{T}}(d,k,\mu)$.

###### Lemma 1

${\cal{T}}(d,k,\mu)\leq\left\lfloor\frac{\binom{d}{\mu+1}}{\binom{k}{\mu+1}}\right\rfloor=\left\lfloor\frac{d!(k-\mu-1)!}{k!(d-\mu-1)!}\right\rfloor$.
The upper bound of
$\left\lfloor\frac{d!(k-\mu-1)!}{k!(d-\mu-1)!}\right\rfloor$ is achieved
whenever $(\mu+1)-(d,k,1)$ design exists and in such cases
${\cal{T}}(d,k,\mu)$ is the same as the number of blocks in $(\mu+1)-(d,k,1)$
design as in Definition 3.

###### Proof

Given a set of $d$ distinct elements, we like to construct the maximum number
of subsets each of size $k$, such that any two subsets has $\mu$ elements in
common. Let us label these blocks as $\\{b_{1},b_{2},\ldots b_{r}\\}$, with
$r={\cal{T}}(d,k,\mu)$.

Now consider all the $(\mu+1)$-element subsets of $d$ distinct elements. They
will be $\binom{d}{\mu+1}$ in numbers. Now consider the blocks $b_{i}$ and
$b_{j}$. Since there are a maximum of $\mu$ elements in common, any
$(\mu+1)$-element subset of $d$ elements cannot exist, which occur in both the
blocks $b_{i}$ and $b_{j}$. Since each block $b_{i}$ is of size $k$, the
number of $(\mu+1)$-element subsets which can be constructed by the $k$
elements in $b_{i}$ is $\binom{k}{\mu+1}$. Let us denote this set by $S_{i}$.
Similarly for block $b_{j}$ the number of $(\mu+1)$-element subsets which can
be constructed with its $k$ elements is $\binom{k}{\mu+1}$. Let us denote this
set by $S_{j}$. We have already seen $S_{i}\cap S_{j}=\phi$, else there would
be $\mu+1$ element common between $b_{i}$ and $b_{j}$. Now since there are $r$
such blocks each of size $k$, hence
$|S_{1}|+|S_{2}|+\ldots|S_{r}|=r\binom{k}{\mu+1}$. This must be less than or
equal to $\binom{d}{\mu+1}$, which is the maximum possible $(\mu+1)$-element
subsets that can be constructed from $d$ distinct elements. This implies
$r\binom{k}{\mu+1}\leq\binom{d}{\mu+1}\Rightarrow{\cal{T}}(d,k,\mu)=r\leq\left\lfloor\frac{\binom{d}{\mu+1}}{\binom{k}{\mu+1}}\right\rfloor=\left\lfloor\frac{d!(k-\mu-1)!}{k!(d-\mu-1)!}\right\rfloor$.

To see the second part of the lemma, note that $t-(d,k,1)$ design is a design
$(X,A)$ where $A$ contains the subsets of $X$ called blocks, such that $|X|=d$
and each block contains exactly $k$ elements. Every $t$-element subset of $X$
is contained in exactly one block. Hence this implies that any two blocks of
the design has maximum $t-1$ elements in common. Thus it immediately follows
that if $(\mu+1)-(d,k,1)$ design exists, then blocks of the design satisfies
the property of ${\cal{T}}(d,k,\mu)$, and since number of blocks in
$(\mu+1)-(d,k,1)$ design is $\frac{\binom{d}{\mu+1}}{\binom{k}{\mu+1}}$
(stinson2007combinatorial, , Chapter 9, Theorem 9.4 and the following
observation), which is exactly the upper bound of ${\cal{T}}(d,k,\mu)$ as
proven above.

The upper bound of ${\cal{T}}(d,k,\mu)$ is achieved whenever a
$(\mu+1)-(d,k,1)$ design exists, which is known for many values of
$0\leq\mu<k<d$. This implies that the bound for ${\cal{T}}(d,k,\mu)$ given by
the above result is tight. However, getting an exact value/expression of
${\cal{T}}(d,k,\mu)$ appears to be an open and challenging problem.

To construct the AMUBs, our focus has been on RBDs with constant block size.
In this connection we now focus on few results that are relevant to our
constructions of APMUBs in Section 5. Following is a lemma related to an RBD
providing a bound for $\mu$ and $r$ in terms of the block size and the number
of blocks in parallel classes where, as defined previously, $\mu$ is the
maximum number of common elements between any pair of blocks from different
parallel classes and $r$ is the number of parallel classes.

###### Lemma 2

Consider an RBD$(X,A)$ with $|X|=d=k\cdot s$ where $k,s\in\mathbb{N}$,
consisting of $r>1$ parallel classes, each having blocks of size $k$. Then
$\mu\geq\lceil\frac{k}{s}\rceil$, where $\mu$ is the maximum number of common
elements between any pair of blocks from different parallel classes and
$r\leq{\cal{T}}(d-1,k-1,\mu-1)$. Further, if $\mu=1$, then
$r\leq\lfloor\frac{d-1}{k-1}\rfloor=s+\lfloor\frac{s-1}{k-1}\rfloor$.

###### Proof

Since $r>1$, consider any pair of parallel classes of RBD say $(P_{l},P_{m})$.
Denote the blocks of $P_{l}$ be as $b^{l}_{1},b^{l}_{2},\ldots,b^{l}_{s}$.
Since blocks are of constant size $\Rightarrow|b^{l}_{i}|=k$ and the blocks
belonging to the same parallel class have no element in common, $\Rightarrow
b^{l}_{i}\cap b^{l}_{j}=\phi\,\,\forall\,i,j=1,2,\ldots,s$ and
$X=b^{l}_{1}\cup b^{l}_{2}\cup\ldots\cup b^{l}_{s}$, $|X|=k\cdot s$. Similar
relations will hold for blocks of any other parallel class.

Consider any block of $P_{l}$, say $b^{l}_{i}$. Since,
$\mu=\max_{m}{|b^{l}_{i}\cap b^{m}_{j}|}$, we have $b^{l}_{i}\cap
b^{m}_{j}\leq\mu,\ \forall\ j=1,2,\ldots,s$. Since $X=b^{m}_{1}\cup
b^{m}_{2}\cup\ldots\cup b^{m}_{s}$, hence, $\sum_{j=1}^{s}|b^{l}_{i}\cap
b^{m}_{j}|=|b^{l}_{i}|=k$. We also have $\sum_{m=1}^{s}|b^{l}_{i}\cap
b^{m}_{j}|\leq\sum_{m=1}^{s}\mu=\mu s\Rightarrow k\leq\mu s$. Since
$k,s,\mu\in\mathbb{N}$, we get $\mu\geq\lceil\frac{k}{s}\rceil$. This implies
minimum value of $\mu=1$ which is possible only if $k\leq s$ and on the other
hand if $k>s$, then minimum value of $\mu=2$. Thus, for $\mu=1$, we must have
$k\leq s$, i.e., number of blocks must be greater than or equal to the block
size of the RBD.

To obtain a bound on $r$, fix an element, say $x\in X$. Since the blocks of a
parallel class are mutually disjoint and their union is $X$, each parallel
class will have exactly one block which will contain $x$. Collect all the
blocks that contain the element $x$ and we will obtain a set of $r$ such
blocks. Denote this set by $S$. Now, remove $x$ from every block in the set
$S$. Hence $S$ will consist of blocks of size $k-1$, the maximum number of
common elements between any two blocks in $S$ would be now $(\mu-1)$ and the
total number of elements contained in these $r$ blocks will be $\leq(d-1)$.
Therefore, we obtain $r\leq{\cal{T}}(d-1,k-1,\mu-1)$. Thus if $\mu=1$ we have
$r\leq{\cal{T}}(d-1,k-1,0)=\lfloor\frac{d-1}{k-1}\rfloor=\lfloor\frac{s\cdot
k-1}{k-1}\rfloor=s+\lfloor\frac{s-1}{k-1}\rfloor$.

We will see that, using our construction method for APMUBs, the necessary
condition on RBDs is $\mu=1$. Hence our effort will be to construct RBDs with
$\mu=1$. For this to happen, $k\leq s$ and $r\leq
s+\lfloor\frac{s-1}{k-1}\rfloor$.

###  3.1 Results relating to MOLS

We will now consider a class of Resolvable Block Designs $(X,A)$ such that
$|X|=s^{2}$, which can be constructed from a set of $w$-MOLS$(s)$. Here $A$
consists of blocks having constant size $s$, which can be resolved into $w+2$
many parallel classes, each having $s$ many blocks, such that blocks from two
different parallel classes have exactly one element in common. Through the
following construction, we explain a simple and direct way to convert a set of
$w$ many MOLS$(s)$ into such an RBD$(X,A)$.

###### Construction 1

To construct an RBD having $w+2$ number of parallel classes from $w$-MOLS$(s)$
using the following steps.

  1. 1.

Define $M_{ref}$, which is a $s\times s$ array where each cell consists of one
of the elements from $X=\\{1,2,\ldots,s^{2}\\}$, as follows:

$M_{\text{ref}}=\begin{bmatrix}1&2&3&\ldots&s\\\ s+1&s+2&s+3&\ldots&2s\\\
2s+1&2s+2&2s+3&\ldots&3s\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\
(s-1)s+1\,\,&\,\,(s-1)s+2\,\,&\,\,(s-1)s+3\,\,&\ldots&\,s^{2}\\\
\end{bmatrix},L_{k}=\begin{bmatrix}l_{11}^{k}&l_{12}^{k}&\ldots&l_{1s}^{k}\\\
\\\ l_{21}^{k}&l_{22}^{k}&\ldots&l_{2s}^{k}\\\ \vdots&\vdots&\ddots&\vdots\\\
l_{s1}^{k}&l_{n2}^{k}&\ldots&l_{ss}^{k}\end{bmatrix}$

  2. 2.

Consider a Latin Square $L_{k}$, from the set of $w$-$MOLS(s)$. Let
$(L_{k})_{ij}=l^{k}_{ij}$ as indicated above.

  3. 3.

Corresponding to the Latin Square $L_{k}$, construct a parallel class $P_{k}$
consisting of $s$ disjoint blocks $b^{k}_{t}$, each of size $s$ as follows,

$b_{t}^{k}=\\{(M_{ref})_{ij}:l^{k}_{ij}=t\\},\text{ where
}i,j\in\\{1,2,\ldots,s\\}.$

Each row of the the Latin Square is a permutation of $\\{1,2,\ldots,s\\}$.
Hence, there will be a pair $(i,j)$ in each row for which $l^{k}_{ij}=t$.
Thus, the blocks $P_{t}^{k}$ will have a total $s$ elements, one from each row
and column of $M_{ref}$. Since $t=\\{1,2,\ldots,s\\}$, there will be $s$
blocks. Thus we are essentially collecting all the elements of $M_{ref}$
corresponding to a particular symbol $t$ of $L_{k}$ in one block $b^{k}_{t}$,
and together the blocks $b^{k}_{i}$ where $i=\\{1,2,\ldots,s\\}$ form a
parallel class $P_{k}$.

  4. 4.

Repeat the above step for all Latin Squares in the set of $w$-$MOLS(s)$,
thereby giving $w$ many parallel classes.

  5. 5.

Construct two more parallel classes, one using the horizontal rows of
$M_{ref}$, and other using the vertical rows of $M_{ref}$ as follows:

$\displaystyle P_{0}$
$\displaystyle=\left\\{(1,2,\ldots,s),(s+1,s+2,\ldots,2s),\ldots((s-1)s+1,(s-1)s+2,\ldots,s^{2})\right\\}$
$\displaystyle P_{\infty}$
$\displaystyle=\left\\{(1,s+1,\ldots,(s-1)s+1),(2,s+2,\ldots,(s-1)s+2),\ldots,(s,2s,\ldots,s^{2})\right\\}$

  6. 6.

The $RBD(X,A)$ with $X=\\{1,2,3,\ldots,s^{2}\\}$ and
$A=\\{P_{0},P_{\infty},P_{1},P_{2},\ldots,P_{w}\\}$ is the desired outcome.

###### Lemma 3

A set of $w$ many MOLS$(s)$ can be used to construct an RBD$(X,A)$ such that
$|X|=s^{2}$, consisting of constant block sizes, each having $s$ elements,
that can be resolved into $w+2$ many parallel classes. Here, any two blocks
from different parallel classes will have exactly one element in common.

###### Proof

We claim that, any pair of blocks from different parallel classes
$\\{P_{0},P_{\infty},P_{1},P_{2},\ldots P_{w}\\}$ constructed above has
exactly one element in common. Consider the $t^{th}$ and $s^{th}$ blocks of
$k^{th}$ and $m^{th}$ parallel classes respectively. Then

$P_{t}^{k}\cap
P_{s}^{m}=\\{(M_{ref})_{ij}:l^{k}_{ij}=t\\}\cap\\{(M_{ref})_{ij}:l^{m}_{ij}=s\\}.$

Now since $L^{k}$ and $L^{m}$ are the orthogonal Latin Squares, there will be
exactly one pair $(i,j)$ such that, $(L^{k})_{ij}=t$ and $(L^{m})_{ij}=s$.
Hence exactly one element will be common between the blocks $P_{t}^{k}$ and
$P_{s}^{m}$.

From the construction of $P_{0}$ and $P_{\infty}$, it is clear that any block
has exactly one element in common. Since any block of $P_{t}^{k}$ is picking
one element from each row and each column of $M_{ref}$, each block of
$P_{t}^{k}$ will have exactly one element in common with blocks of $P_{0}$ and
$P_{\infty}$, which are collection of horizontal rows $(P_{0})$ and vertical
rows $(P_{\infty})$ of $M_{ref}$.

We will now sketch the idea of the above method with a simple example to
convert a $2$-MOLS$(5)$ into $4$ parallel classes.

###### Example 1

Let us consider the $2$-MOLS$(5)$ and $M_{ref}$ as follows.

$L_{1}=\begin{bmatrix}5&1&2&3&4\\\ 1&2&3&4&5\\\ 2&3&4&5&1\\\ 3&4&5&1&2\\\
4&5&1&2&3\\\ \end{bmatrix},L_{2}=\begin{bmatrix}5&1&2&3&4\\\ 2&3&4&5&1\\\
4&5&1&2&3\\\ 1&2&3&4&5\\\ 3&4&5&1&2\\\
\end{bmatrix},M_{ref}=\begin{bmatrix}1&2&3&4&5\\\ 6&7&8&9&10\\\
11&12&13&14&15\\\ 16&17&18&19&20\\\ 21&22&23&24&25\\\ \end{bmatrix}$.

We use a $5\times 5$ Reference Matrix $M_{ref}$ consisting of elements
indicated by $\\{1,2,\ldots,25\\}$. We put them in a simple row wise
increasing sequence, which is to ensure that each element occur only once in
the matrix $M_{ref}$. Now corresponding to each MOLS $L_{1}$ and $L_{2}$, we
construct a parallel class, as per the Construction 1, thereby forming blocks
of $P_{1}$ and $P_{2}$ by picking elements from $M_{ref}$.

$\displaystyle P_{1}$
$\displaystyle=\left\\{(2,6,15,19,23),(3,7,11,20,24),(4,8,12,16,25),(5,9,13,17,21),(10,14,18,22,1)\right\\},$
$\displaystyle P_{2}$
$\displaystyle=\left\\{(2,10,13,16,24),(3,6,14,17,25),(4,7,15,18,21),(5,8,11,19,22),(1,9,12,20,23)\right\\}.$

The remaining two parallel classes will be constructed using horizontal and
vertical elements of $M_{ref}$ as follows:

$\displaystyle P_{0}$
$\displaystyle=\left\\{(1,2,3,4,5),(6,7,8,9,10),(11,12,13,14,15),(16,17,18,19,20),(21,22,23,24,25)\right\\},$
$\displaystyle P_{\infty}$
$\displaystyle=\left\\{(1,6,11,16,21),(2,7,12,17,22),(3,8,13,18,23),(4,9,14,19,24),(5,10,15,20,25)\right\\}.$

Let us now consider the construction in the other way, i.e., the converse.

###### Construction 2

Consider an RBD$(X,A)$, where $|X|=s^{2}$ and $A$ consist of $w+2$ numbers of
parallel classes such that, each block of a parallel class is of a constant
size $s$ and any pair of blocks from different parallel classes have exactly
one element in common. Let us denote the elements of $X$ by
$\\{1,2,\ldots,s^{2}\\}$, parallel classes by
$\\{P_{0},P_{\infty},P_{1},\ldots,P_{w}\\}$, and the blocks of $P_{l}$ by
$b^{l}_{i}$. Since there are $s$ blocks in each parallel class, therefore
$P_{l}=\\{b^{l}_{1},b^{l}_{2},\ldots,b^{l}_{s}\\}$. Let $s$ distinct symbols
for construction of the Latin Squares be denoted by
$Y=\\{y_{1},y_{2},\ldots,y_{s}\\}$. Now construct $w$-MOLS$(s)$ using
RBD$(X,A)$ as follows.

  1. 1.

Use $P_{0}$ and $P_{\infty}$ to construct a reference matrix $M_{ref}$ having
elements from $\\{1,2,\ldots,s^{2}\\}$ in the following manner:

$M_{ref}=\begin{bmatrix}b^{0}_{1}\cap b^{\infty}_{1}&&&b^{0}_{1}\cap
b^{\infty}_{2}&&\ldots&&b^{0}_{1}\cap b^{\infty}_{s}\\\ b^{0}_{2}\cap
b^{\infty}_{1}&&&b^{0}_{2}\cap b^{\infty}_{2}&&\ldots&&b^{0}_{2}\cap
b^{\infty}_{s}\\\ \vdots&&&\vdots&&\ldots&&\vdots\\\ b^{0}_{s}\cap
b^{\infty}_{1}&&&b^{0}_{s}\cap b^{\infty}_{2}&&\ldots&&b^{0}_{s}\cap
b^{\infty}_{s}\\\ \end{bmatrix}.$

Here $M_{ref}$ contains all the elements of $X$ exactly once. As any two
blocks from different parallel classes have exactly one element in common, if
$(M_{ref})_{ij}=(M_{ref})_{lm}$ then $b^{0}_{i}\cap
b^{\infty}_{j}=b^{0}_{l}\cap b^{\infty}_{m}$, and that implies all the blocks
$\\{b^{0}_{i},b^{0}_{l},b^{\infty}_{j},b^{\infty}_{m}\\}$ have one element in
common. Here, $b^{0}_{i}$ and $b^{0}_{l}$ are the blocks in the Parallel class
$P_{0}$. Similarly, $b^{\infty}_{j}$ and $b^{\infty}_{m}$ are the blocks of
the parallel class $P_{\infty}$. Since blocks in a Parallel class are mutually
disjoints, this is not possible, i.e., $(M_{ref})_{ij}\neq(M_{ref})_{lm}$.

  2. 2.

Corresponding to the parallel class $P_{k}$, construct the Latin Squares
$L_{k}$ as follows

$\text{if}\ (M_{ref})_{ij}\in
b^{k}_{t}\,\,\,\text{then}\,\,(L_{k})_{ij}=y_{t}.$

That is, we are substituting $y_{t}$, wherever the element of the block
$b^{k}_{t}$ is appearing in $M_{ref}$ to construct $L_{k}$. Note that $L_{k}$
is a Latin Square. Since $X=b^{k}_{1}\cup b^{k}_{2}\ldots b^{k}_{s}$, for
every $(M_{ref})_{ij}$ there will be one $b^{k}_{t}$ such that
$(M_{ref})_{ij}\in b^{k}_{t}$ and since $b^{k}_{i}\cap b^{k}_{j}=\phi$,
$i,j\in\\{1,2,\ldots,s\\}$, for each $(M_{ref})_{ij}$, there will be a unique
$b^{k}_{t}$ such that $(M_{ref})_{ij}\in b^{k}_{t}$. Now if $L_{k}$ is not a
Latin square, then there would be at least a pair of $(i_{1},i_{2})$
corresponding to which $(L_{k})_{i_{1}j}=(L_{k})_{i_{2}j}$ or a pair of
$(j_{1},j_{2})$ corresponding to which $(L_{k})_{ij_{1}}=(L_{k})_{ij_{j}}$.
Consider $(L_{k})_{i_{1}j}=(L_{k})_{i_{2}j}$. This implies if
$x_{1}=(M_{ref})_{i_{1}j}=b^{0}_{i_{1}}\cap b^{\infty}_{j}\in b^{k}_{t}$ and
$x_{2}=(M_{ref})_{i_{2}j}=b^{0}_{i_{2}}\cap b^{\infty}_{j}\in b^{k}_{t}$. Thus
$x_{1}$ and $x_{2}\in b^{\infty}_{j}$. Hence $|b^{k}_{t}\cap
b^{\infty}_{j}|\geq|\\{x_{1},x_{2}\\}|=2$ as
$x_{1}=(M_{ref})_{i_{1}j}\neq(M_{ref})_{i_{2}j}=x_{2}$. This is a
contradiction as there is exactly one element common between the blocks of
different parallel classes, here $P^{\infty}$ and $P^{k}$. Similarly it can be
argued that $(L_{k})_{ij_{1}}\neq(L_{k})_{ij_{j}}$ for any pair of
$(j_{1},j_{2})$.

  3. 3.

Repeat the above step for each of the parallel class $P_{k},\ k=1,2,\ldots,w$,
thereby constructing the set of $w$ Latin squares viz
$\\{L_{1},L_{2},\ldots,L_{w}\\}$.

The converse of Lemma 3 is as follows.

###### Lemma 4

Given an RBD$(X,A)$, where $|X|=s^{2}$ and consisting of $w+2$ many parallel
classes such that, each block of a parallel class is of a constant size $s$
and any pair of blocks from different parallel classes have exactly one
element in common. Then RBD$(X,A)$ can be used to construct $w$-MOLS$(s)$.

###### Proof

We claim that the set of Latin squares $L_{1},L_{2},\ldots,L_{w}$ as
constructed above are Mutually Orthogonal Latin Squares of order $s$.

Recalling that a pair of Latin Squares $L_{1}$ and $L_{2}$ of same order and
constructed from the entries from the same set $Y$ is called mutually
orthogonal, if the ordered pair $((L_{1})_{ij},(L_{2})_{ij})\in\\{(Y,Y)\\}$
appears exactly once.

Consider the ordered pair, $((L_{k})_{ij},(L_{m})_{ij})$. Assume that, $L_{k}$
and $L_{m}$ are not Mutually Orthogonal Latin Squares, which implies that,
there would be at least one pair of Indices $\\{(i,j),(u,v):(i,j)\neq(u,v)\\}$
such that,
$((L_{k})_{ij},(L_{m})_{ij})=((L_{k})_{uv},(L_{m})_{uv})=(y_{p},y_{q})$. Let
$y_{p}=(L_{k})_{ij}\in b^{k}_{p}$ and $y_{q}=(L_{m})_{ij}\in b^{m}_{q}$. This
implies $(M_{ref})_{ij}\in b^{k}_{p}$ and $b^{m}_{q}$. Similarly,
$y_{p}=(L_{k})_{uv}\in b^{k}_{p}$ and $y_{q}=(L_{m})_{uv}\in b^{m}_{q}$ which
implies $(M_{ref})_{uv}\in b^{k}_{p}$ and $b^{m}_{q}$. However, then we will
have, $b^{k}_{p}\cap b^{m}_{q}=\\{(M_{ref})_{ij},(M_{ref})_{uv}\\}$, but if
$(i,j)\neq(u,v)$ then $(M_{ref})_{ij}\neq(M_{ref})_{uv}$. Thus,
$|b^{k}_{p}\cap b^{m}_{q}|\geq 2$ which contradicts the fact that
$|b^{k}_{p}\cap b^{m}_{q}|=1$. Hence we conclude that $L_{k}$ and $L_{m}$ are
Mutually Orthogonal Latin Squares.

We now sketch the above method with an example to obtain a $2$-MOLS$(5)$ from
$4$ parallel classes.

###### Example 2

If we proceed with the same set of $4$ parallel classes obtained as in Example
1, it would naturally result into the same pair of MOLS$(5)$, i.e., $L_{1}$
and $L_{2}$, with which we have started Example 1. Therefore, we consider a
different set of $4$ parallel classes as follows:

$\displaystyle P_{0}$
$\displaystyle=\left\\{(1,2,3,4,5),(6,7,8,9,10),(11,12,13,14,15),(16,17,18,19,20),(21,22,23,24,25)\right\\},$
$\displaystyle P_{\infty}$
$\displaystyle=\left\\{(1,6,11,16,21),(2,7,12,17,22),(3,8,13,18,23),(4,9,14,19,24),(5,10,15,20,25)\right\\},$
$\displaystyle P_{1}$
$\displaystyle=\left\\{(1,9,12,20,23),(2,10,13,16,24),(3,6,14,17,25),(4,7,15,18,21),(5,8,11,19,22)\right\\},$
$\displaystyle P_{2}$
$\displaystyle=\left\\{(1,7,13,19,25),(2,8,14,20,21),(3,9,15,16,22),(4,10,11,17,23),(5,6,12,18,24)\right\\}.$

Note that, $P_{0}$ and $P_{\infty}$ are taken as above for convenience,
providing the $M_{ref}$ with elements from $X=\\{1,2,\ldots,25\\}$. Observe
that any pair of blocks from different parallel classes have exactly one
element in common. Now corresponding to $s=5$, we simply use
$Y=\\{1,2,3,4,5\\}$ as five symbols to construct the Latin square. The
remaining parallel classes $P_{1}$ and $P_{2}$ are used to construct $L_{1}$
and $L_{2}$ respectively, which are the required $2$-MOLS$(5)$. Following the
Construction 2 above, we obtain Orthogonal Latin Squares $L_{1}$ and $L_{2}$
as follows:

$M_{ref}=\begin{bmatrix}1&2&3&4&5\\\ 6&7&8&9&10\\\ 11&12&13&14&15\\\
16&17&18&19&20\\\ 21&22&23&24&25\\\
\end{bmatrix},\,\,\,\,L_{1}=\begin{bmatrix}1&2&3&4&5\\\ 3&4&5&1&2\\\
5&1&2&3&4\\\ 2&3&4&5&1\\\ 4&5&1&2&3\\\
\end{bmatrix},\,\,\,\,L_{2}=\begin{bmatrix}1&2&3&4&5\\\ 5&1&2&3&4\\\
4&5&1&2&3\\\ 3&4&5&1&2\\\ 2&3&4&5&1\\\ \end{bmatrix}.$

Based on this we now proceed for our generic construction idea in the next
section.

##  4 Definition of APMUBs and general construction ideas

In this section we first present the motivation of proposing such a
combinatorial object and then proceed with the general construction ideas.

###  4.1 Motivation for defining APMUBs and its Characteristics

Consider a pair of orthonormal bases, say $M_{l}\text{ and }M_{m}$. If they
are $\beta$-AMUBs then
$|\braket{\psi_{i}^{l}}{\psi_{j}^{m}}|\leq\frac{\beta}{\sqrt{d}}$, where
$\beta$ is bounded by some constant. The definition of $\beta$-AMUBs does not
rule out $|\braket{\psi_{i}^{l}}{\psi_{j}^{m}}|=0$. Let for $\mathfrak{n}_{1}$
pairs, the value of $|\braket{\psi_{i}^{l}}{\psi_{j}^{m}}|$ be $0$, and for
the remaining pairs $\mathfrak{n}_{2}$, the value of
$|\braket{\psi_{i}^{l}}{\psi_{j}^{m}}|$ be non-zero ($\neq 0$). If
$|\braket{\psi_{i}^{l}}{\psi_{j}^{m}}|=\frac{\beta_{ij}}{\sqrt{d}}$ (say),
then $\frac{\beta_{ij}}{\sqrt{d}}\leq\frac{\beta}{\sqrt{d}}$. Since
$\sum^{d}_{ij}|\braket{\psi_{i}^{l}}{\psi_{j}^{m}}|^{2}=d$. Therefore,

$\mathfrak{n}_{1}\cdot 0+\sum_{\begin{subarray}{c}i,j=1\\\
|\braket{\psi_{i}^{l}}{\psi_{j}^{m}}|\neq
0\end{subarray}}^{d}\left(\frac{\beta_{ij}}{\sqrt{d}}\right)^{2}=d\Rightarrow\mathfrak{n}_{2}\frac{\beta_{\min}^{2}}{d}\leq
d\leq\mathfrak{n}_{2}\frac{\beta_{\max}^{2}}{d},$

where, $\beta_{\min}=\min_{ij}\beta_{ij}$ and
$\beta_{\max}=\max_{ij}\beta_{ij}$ This implies,

$\mathfrak{n}_{2}\frac{\beta_{\min}^{2}}{d^{2}}\leq
1\leq\mathfrak{n}_{2}\frac{\beta_{\max}^{2}}{d^{2}}.$

Note that,
$\mathfrak{n}_{1}+\mathfrak{n}_{2}=d^{2}\Rightarrow\frac{\mathfrak{n}_{1}}{d^{2}}=1-\frac{\mathfrak{n}_{2}}{d^{2}}$,
then we have,

$1-\frac{1}{\beta_{\min}^{2}}\leq\frac{\mathfrak{n}_{1}}{d^{2}}\leq
1-\frac{1}{\beta_{\max}^{2}}.$ (2)

Note that, $\frac{\mathfrak{n}_{1}}{d^{2}}$ is the probability of randomly
selecting two orthogonal vectors from two different bases. Therefore, if
$\beta_{\min}=\beta=\beta_{\max}$, i.e.,
$\Delta=\\{0,\frac{\beta}{\sqrt{d}}\\}$ and
$\beta=1+\mathcal{O}(d^{-\lambda}),\\\ lambda>0$, then,
$\frac{\mathfrak{n}_{1}}{d^{2}}=\mathcal{O}(d^{-\lambda}),\ \lambda>0$. Hence,
with these conditions on $\Delta$ and $\beta$, the probability of any pair
randomly selected vectors from different orthonormal bases being orthogonal,
tends to $0$. Similarly, the probability that the angle between them is
$\frac{\beta}{\sqrt{d}}$ tends to $1$ asymptotically. Since $\beta\rightarrow
1$, as $d$ increases, the bases of APMUBs would behave like MUBs in this
sense.

Further note that, the vectors from a set of MUBs form a set of Bi-angular
vectors as $\Delta=\\{0,\frac{1}{\sqrt{d}}\\}$. Now if we restrict
$\Delta=\\{0,\frac{\beta}{\sqrt{d}}\\}$, then basis vectors of AMUBs would
form set of Bi-angular vectors. Thus analysis of such AMUBs would also shed
light on the study of Bi-angular vectors that has close connections with
Weighing Matrices, Error Correcting Codes, Orthogonal spreads, Frame theory,
Association Schemes etc. Calderbank1997 ; Best2013 ; Best2015 ; Kharaghani2015
; Magsino2019 ; Cahill2018 ; Casazza2019 ; Holzmann2010 ; Kharaghani2018 .

With this motivation, let us define Almost Perfect MUBs (APMUBs), which is the
main focus of this paper.

###### Definition 6

The set $\mathbb{M}=\\{M_{1},M_{2},\ldots,M_{r}\\}$ will be called Almost
Perfect MUBs (APMUBs) if $\Delta=\left\\{0,\frac{\beta}{\sqrt{d}}\right\\}$,
i.e., the set contains just two values, such that
$\beta=1+\mathcal{O}(d^{-\lambda})\leq 2$, $\lambda>0$. When the bases are
real, we call them Almost Perfect Real MUBs (APRMUBs).

If the vectors are understood as states of a quantum system then the absolute
value of an inner product essentially indicates the overlap between these
states. Hence randomly picking two quantum states corresponding to different
basis of above set of APMUBs, the overlap will have magnitude equal to
$\frac{\beta}{\sqrt{d}}$ with probability almost 1. With negligible
probability it will be $0$ which corresponds to the quantum sates being
orthogonal.

The sets of basis vectors of APMUB are Bi-angular as they are either
orthogonal or have constant absolute value of inner product, i.e.,
$\Delta=\\{0,\frac{\beta}{\sqrt{d}}\\}$. Hence APMUBs form set of Bi-angular
vectors with 0 being one of the value of inner product. Thus, for any pair of
APMUBs in $\mathbb{C}^{d}(\text{or}~{}\mathbb{R}^{d})$ we have the following
lemma.

###### Lemma 5

Consider any pair of APMUBs in $\mathbb{C}^{d}$ with
$\Delta=\\{0,\frac{\beta}{\sqrt{d}}\\}$. Then any basis vector of an APMUB
will be orthogonal to $(1-\frac{1}{\beta^{2}})\times d$ many basis vectors of
another APMUB and will be at angle $\frac{\beta}{\sqrt{d}}$ with remaining
$\frac{d}{\beta^{2}}$ many basis vectors.

###### Proof

Let $M_{l}$ and $M_{n}$ be any pair of APMUBs over $\mathbb{C}^{d}$ and let
$\\{\ket{\psi_{i}^{l}}:i=1,2,\ldots,d\\}$ and
$\\{\ket{\psi_{j}^{m}}:j=1,2,\ldots,d\\}$ be the corresponding basis vectors.
Expressing $\ket{\psi_{i}^{l}}$ as a linear combination of
$\\{\ket{\psi_{j}^{m}}:j=1,2,\ldots,d\\}$, we get,

$\ket{\psi_{i}^{l}}=\alpha_{i1}\ket{\psi_{1}^{m}}+\alpha_{i2}\ket{\psi_{2}^{m}}+\ldots+\alpha_{id}\ket{\psi_{d}^{m}},$

where $\alpha_{ij}=\braket{\psi_{i}^{l}}{\psi_{j}^{m}}$. Since the bases
consist of unit vectors, i.e., $\braket{\psi_{i}^{l}}{\psi_{i}^{l}}=1\forall
l,i$ hence,

$\braket{\psi_{i}^{l}}{\psi_{i}^{l}}=|\alpha_{i1}|^{2}+|\alpha_{i2}|^{2}+\ldots+|\alpha_{id}|^{2}=\sum_{j=1}^{d}|\braket{\psi_{i}^{l}}{\psi_{j}^{m}}|^{2}=1.$

Since $\Delta=\\{0,\frac{\beta}{\sqrt{d}}\\}$, let us assume that
$\ket{\psi_{i}^{l}}$ is orthogonal, i.e.,
$|\braket{\psi_{i}^{l}}{\psi_{j}^{m}}|=0$ with $\mathfrak{t}_{1}$ many basis
vectors of $M^{m}$, and make an angle of $\frac{\beta}{\sqrt{d}}$ with
remaining $\mathfrak{t}_{2}=d-\mathfrak{t}_{1}$ many basis vectors of $M^{m}$.
Hence we have,

$\sum_{j=1}^{d}|\braket{\psi_{i}^{l}}{\psi_{j}^{m}}|^{2}=1\Rightarrow\mathfrak{t}_{1}\cdot
0+\mathfrak{t}_{2}\times\left(\frac{\beta}{\sqrt{d}}\right)^{2}=1\
\Rightarrow\ \mathfrak{t}_{1}=\left(1-\frac{1}{\beta^{2}}\right)\times d\text{
and }\mathfrak{t}_{2}=\frac{d}{\beta^{2}}.$

Since $\ket{\psi_{i}^{l}}$ is arbitrary basis vector of $M^{l}$, hence the
result.

The above result tells that $\beta^{2}$ can only be a rational number.
Further, we have the following corollary considering all the vectors in the
set of APMUBs.

###### Corollary 1

Consider a set of $r$ many APMUBs on dimension $d$ with
$\Delta=\\{0,\frac{\beta}{\sqrt{d}}\\}$, that will produce $r\times d$
vectors. In this set, each vector will have
$(d-1)+(r-1)(1-\frac{1}{\beta^{2}})d$ many vectors as its orthogonal and
$(r-1)\frac{d}{\beta^{2}}$ many vectors having the dot product
$\frac{\beta}{\sqrt{d}}$.

The main contribution here is to show that one can construct
$\mathcal{O}(\sqrt{d})$ many APMUBs with values of $\beta$ slightly more than
one. This says that our method can provide $\mathcal{O}(d^{\frac{3}{2}})$ many
Bi-angular vectors where the dot product values are 0 and
$\frac{\beta}{\sqrt{d}}$. While there are constructions of
$\mathcal{O}(d^{2})$ Bi-angular vectors mikhail2021biangular , but the angles
we obtain with our methods are quite high than the existing constructions.
Further we achieve large sparsity and non-zero components of equal magnitude.
This is related to coherence property of unit norm vectors. This shows that
the construction of APMUBs may produce interesting results in related domain.
Further note that any set of Orthonormal Basis vectors always form Unit Norm
Tight Frames (UNTF). For a brief introduction on Frame theory and its
application in Hilbert space one may refer to casazza2016brief ;
casazza2013introduction . Thus the basis vectors of the set of APMUBs also
constitute a Unit Norm Tight Frames apart from being Bi-angular, whereas in
general the Bi-angular vectors constructed in mikhail2021biangular  do not
constitute tight frame. To see this, note that since APMUBs are orthonormal
basis vectors, hence any arbitrary vector $\ket{u}$ can be uniquely expressed
in terms of each of the APMUBs. Thus in this context also the construction of
APMUBs may be of independent interest.

###  4.2 Our general construction ideas

Let us now refer to the construction method of orthonormal bases using RBDs as
given in (ak22, , Section 3). The construction idea of ak22  is generic in
nature, where unitary matrices can be employed to construct orthogonal bases
corresponding to each parallel class of an RBD. However, here we will confine
ourselves to the choice of Hadamard Matrices (which are special kind of
unitary matrices) for constructing orthonormal bases from parallel classes of
an RBD. This will enable us to bound $\beta$ as per (ak22, , Theorem 1), which
is required for constructing APMUBs with good parameters. The order of the
Hadamard matrices used in this construction must be same as the block size of
each parallel class. With this method in place for constructing the set of
orthonormal bases from RBD, our work here primarily focuses on constructing
suitable RBDs, and consequently analysing the parameters $\Delta,\beta$ and
$\epsilon$ of corresponding AMUBs constructed from them.

We focus on constructing RBDs, having constant block size. The constant block
size is essential if we want to use Hadamard matrices of same order (real or
complex) for all the blocks of the RBD. This is required as it renders all the
components of the basis finally constructed to be either zero or of a constant
magnitude $(\frac{1}{\sqrt{k}})$, which is the normalising factor of each
basis vector. We will examine when they can satisfy the conditions of APMUBs
(APRMUBs). Our construction method results into vectors which are vary sparse
with the non-zero components of constant magnitude. This provides large sets
of both real as well as complex AMUBs for the dimensions where it is known
that not more that two or three real MUBs exist.

We first present a generic result, which is dependent on the existence of
suitable RBDs. Thereafter, we explore the methods to construct such RBDs. We
further show that if an RBD satisfy $\mu=1$, then it will result into an
APMUB. In fact, $\mu$ is the most critical parameter which we control in the
construction of RBDs. In all the constructions of AMUBs, the number of
elements in an RBD, i.e., $|X|$ can be increased without bound whereas, the
parameter $\mu$ remains constant. All our constructions will have this
property, which justifies asymptotic analysis of the parameters for AMUBs thus
constructed.

###### Theorem 4.1

Consider an RBD$(X,A)$ with $|X|=d=(q-e)(q+f)$, with $q,e,f\in\mathbb{N}$. If
the said RBD$(X,A)$ consists of $r$ parallel classes, each having blocks of
size $(q-e)$, and if $0\leq(e+f)\leq\left(\frac{c^{2}-\mu^{2}}{\mu
c}\right)d^{\frac{1}{2}}$ where $c$ is some constant, then one can construct
$r$ many $\beta$-AMUBs in dimension $d$, where
$\beta=\mu\sqrt{\frac{q+f}{q-e}}\leq c$ and $\epsilon=1-\frac{1}{q+f}$. When
$\mu=1$ and $c=2$, we will get $r$ many APMUBs with
$\beta=1+\mathcal{O}(d^{-\frac{1}{2}})\leq 2$ and
$\Delta=\\{0,\frac{1}{q-e}\\}$. Further, if there exists a real Hadamard
matrix of order $(q-e)$, we can construct $r$ many APRMUBs with the same
parameters.

###### Proof

We have $|X|=(q-e)(q+f)$ with each parallel class having the block size
$k=(q-e)$. Here $\mu$ is the maximum number of elements that are common
between two blocks from different parallel classes. Following (ak22, , Theorem
1), this implies that,
$\beta=\frac{\mu\sqrt{d}}{q-e}=\mu\sqrt{\frac{q+f}{q-e}}$. Further, using the
relation $d=q^{2}+(f-e)q-ef$, we obtain $\beta=\mu(\sqrt{1+x^{2}}+x)$, where
$x=\frac{e+f}{2\sqrt{d}}$. From the definition of $\beta$-AMUBs, $\beta$ must
be bounded for all values of $d$. Let this bound for $\beta$ be $c$, then
$\mu(\sqrt{1+x^{2}}+x)\leq c\Rightarrow
0\leq(e+f)\leq\left(\frac{c^{2}-\mu^{2}}{\mu c}\right)d^{\frac{1}{2}}$. This
inequality can be restated in terms of $q$ as
$0\leq(c^{2}e+\mu^{2}f)\leq(c^{2}-\mu^{2})q$, which is the condition for
$\beta$ being bounded above by the constant $c$.

In order to see the asymptotic variation of $\beta$ in terms of $q$, we
consider the expansion of terms as follows:

$\beta=\mu\left(1+\frac{e+f}{2q}+\frac{(e+f)(3e-f)}{2^{3}q^{2}}+\frac{(e+f)(5e^{2}-2ef+f^{2})}{2^{4}q^{3}}+\ldots\right).$
(3)

To understand the asymptotic variation of $\beta$ in terms of $d$, we again
use the relation $d=q^{2}+(f-e)q-ef$ to express $q$ in terms of $\sqrt{d}$ and
thereafter, expanding the expression for
$\beta=\mu\sqrt{\frac{q+f}{q-e}}=\mu(\sqrt{1+x^{2}}+x)$, where
$x=\frac{f+e}{2\sqrt{d}}$, in terms of negative power of $\sqrt{d}$ and we
obtain

$\beta=\mu\left(1+\frac{e+f}{2\sqrt{d}}+\frac{(e+f)^{2}}{2^{3}d}-\frac{(e+f)^{4}}{2^{7}d^{2}}+\frac{(e+f)^{6}}{2^{10}d^{3}}-\frac{5(e+f)^{8}}{2^{15}d^{4}}+\ldots\right).$
(4)

Thus, for a given $e$ and $f$, for large $d~{}(\text{or}~{}q)$, asymptotically
$\beta=\mu+\mathcal{O}(\frac{1}{q})=\mu+\mathcal{O}(\frac{1}{\sqrt{d}})$.
Therefore, if $\mu=1$, the construction yields APMUBs, provided $\beta\leq
2\Rightarrow c=2$ as we have seen above that $\beta$ is bounded by $c$. And in
this situation we get $0\leq(e+f)\leq\frac{3}{2}d^{\frac{1}{2}}$.

To get the values of the set $\Delta$, note that when $\mu=1$, there is
maximum one element common between any pair of blocks from different parallel
classes. And since Hadamard matrices are used for constructing orthonormal
bases, thus $|\braket{u}{v}|=\frac{1}{q-e}$ corresponding to the situations
when one element is common between the pair of blocks and $|\braket{u}{v}|=0$
corresponding to situation, when no elements are common between the pair of
blocks. Thus $\Delta=\\{0,\frac{1}{q-e}\\}$.

To calculate sparsity, note that for each vector constructed from a block of
size $k$, we will have exactly $k$ many non-zero and $d-k$ many zero entries,
hence

$\epsilon=\frac{d-k}{d}=1-\frac{k}{d}=1-\frac{q-e}{(q+f)(q-e)}=1-\frac{1}{q+f}.$

If a real Hadamard matrix of order $(q-e)$ exists, we can exploit it to obtain
$r$ many real approximate MUBs in $\mathbb{R}^{d}$, with same values of the
parameters $\beta$, $\Delta$ and $\epsilon$.

If the construction in ak22  is to be used for APMUBs, then $\mu$ should be 1.
Thus $\mu$, which is the maximum number of elements common between any pair of
blocks from different parallel classes, is the most critical parameter here.
Further note that, $\mu$ is always greater than or equal to $1$, hence, a very
limited kinds of RBDs can be used to construct APMUBs. As per Lemma 2,
$\mu\geq\lceil\frac{k}{s}\rceil$. Thus, for $\mu=1$, an RBD having a constant
block size must have $k\leq s$, i.e., the block size must not be greater than
the number of blocks in the parallel class. In this connection, we have noted
that an RBD constructed using MOLS have $\mu=1$. In fact, between any pair of
blocks from different parallel classes, in such an RBD, there is exactly one
element in common.

In our above theorem, we have $|X|=d=(q-e)(q+f)$, where number of elements in
a block is $(q-e)$, i.e., $k=(q-e)$, and number of blocks in a parallel class
is $(q+f)$, hence $s=(q+f)$. The reason we are expressing it like this will be
clear in Theorem 5.2 where we demonstrate the construction of such an RBD.
Since $e$ and $f$ are bounded by a positive integer, if $e\geq f$, it will
ensure that $k\leq s$.
Since$\beta=\mu\sqrt{\frac{s}{k}}=\mu\sqrt{\frac{q+f}{q-e}}$, hence for large
$d=(q-e)(q+f)$ we obtain $\beta\rightarrow\mu$, which is also evident from the
asymptotic expansion of $\beta$ above.

For $|X|=d=(q-e)(q+f)$, we can have an RBD, where the block size is $(q+f)$,
hence having $(q-e)$ blocks in each parallel class. However, in such a
situation, $\mu>1$ Lemma 2, and hence we cannot get APMUBs. However, they can
provide AMUBs as in ak22 . The result of (ak22, , Theorem 4) is a particular
case of this situation, with $e=0,f=1$ and $\mu=2$. In this case, $q+1$ many
parallel classes are there in an RBD, each having a constant block size of
$(q+1)$. In this case,
$\beta=2\sqrt{\frac{q}{q+1}}=2-\mathcal{O}(\frac{1}{\sqrt{d}})$, i.e., though
the maximum value of the inner product was slightly less than
$\frac{2}{\sqrt{d}}$, but asymptotically $\beta$ converges to 2. Further
$\Delta$ is also not two-valued. Thus the construction did not satisfy the
conditions needed for Almost Perfect MUBs which is
$\beta=1+\mathcal{O}(d^{-\lambda})$, for some $\lambda>0$ and $\Delta$ being
the set consisting of just two elements with one being 0.

Thus, in order to obtain APMUBs, the RBDs in use must have $\mu=1$ and all the
block sizes must be same. With this understanding, we will explore more
suitable designs in the following sections, so that the upper bound on the
absolute inner product values can be improved than the results presented in
ak22  to obtain Almost Perfect MUBs.

##  5 Exact constructions of APMUBs through RBDs

As followed from previous section, an RBD having constant block size must have
$\mu=1$, in order to obtain APMUBs. In this section we explain the
constructions of such RBDs. Since our focus is to build APMUBs in composite,
we concentrate on $d=k\cdot s$, and consider two categories.

  * •

The first construction, being generic in nature, will work for any composite
$d=k\cdot s=(s-e)s$ with $0\leq e\leq\frac{3}{2}d^{\frac{1}{2}}$. Here the
number of APMUBs is at least $N(s)+1$ when $e>0$ and $N(s)+2$ when $e=0$.

  * •

The second one is considered when $d$ can be expressed as $(q-e)(q+f),\,\,0<\
f\leq e$, where $q$ is some power of prime. Here the number of APMUBs is at
least $\lfloor\frac{q-e}{f}\rfloor+1$.

These are presented in the Sections 5.1 and 5.2 respectively as below. Thus
here our approach to is to obtain large numbers of APMUBs, if the composite
dimension $d$ can be expressed in some generic form. In the first category our
starting point is $w$-MOLS$(s)$ and in the second category we initiate with
$(q^{2},q,1)$-ARBIBD. In each case, we will first demonstrate the construction
with an example, then outline the algorithm for the respective construction
and then provide the proof of correctness.

###  5.1 $d=k\cdot s=(s-e)s$, $0\leq e\leq\frac{3}{2}d^{\frac{1}{2}}$

Let us first demonstrate the method by explicitly constructing RBD$(X,A)$ with
$|X|=2\cdot 5=10$, i.e., here $s=5$ and $k=2$.

  1. 1.

To begin with, consider the following $4$-MOLS$(5)$ and the $M_{ref}$:

$LS_{1}=\begin{bmatrix}1&2&3&4&5\\\ 5&1&2&3&4\\\ 4&5&1&2&3\\\ 3&4&5&1&2\\\
2&3&4&5&1\\\ \end{bmatrix},LS_{2}=\begin{bmatrix}1&2&3&4&5\\\ 4&5&1&2&3\\\
2&3&4&5&1\\\ 5&1&2&3&4\\\ 3&4&5&1&2\\\
\end{bmatrix},LS_{3}=\begin{bmatrix}1&2&3&4&5\\\ 3&4&5&1&2\\\ 5&1&2&3&4\\\
2&3&4&5&1\\\ 4&5&1&2&3\\\ \end{bmatrix}$,

$LS_{4}=\begin{bmatrix}1&2&3&4&5\\\ 2&3&4&5&1\\\ 3&4&5&1&2\\\ 4&5&1&2&3\\\
5&1&2&3&4\\\ \end{bmatrix},M_{ref}=\begin{bmatrix}1&2&3&4&5\\\ 6&7&8&9&10\\\
11&12&13&14&15\\\ 16&17&18&19&20\\\ 21&22&23&24&25\\\ \end{bmatrix}$.

  2. 2.

Using Construction 1, we construct RBD$(\bar{X},\bar{A})$, with $|\bar{X}|=25$
and $\bar{A}$ having $6$ parallel classes. We used $M_{ref}$ as shown above to
obtain the following RBD. We collect blocks from each parallel class, in a
$5\times 5$ matrix, where each row represents one block of the parallel class,
and index the blocks as $\bar{b}^{l}_{i}$, where $l$ represents the index of
parallel class and $i$ represents the block number within the parallel class.

$\bar{P}_{1}=\begin{bmatrix}\bar{b}^{1}_{5}=\\{1&7&13&19&25\\}\\\
\bar{b}^{1}_{4}=\\{2&8&14&20&21\\}\\\ \bar{b}^{1}_{3}=\\{3&9&15&16&22\\}\\\
\bar{b}^{1}_{2}=\\{4&10&11&17&23\\}\\\ \bar{b}^{1}_{1}=\\{5&6&12&18&24\\}\\\
\end{bmatrix},\bar{P}_{2}=\begin{bmatrix}\bar{b}^{2}_{5}=\\{1&8&15&17&24\\}\\\
\bar{b}^{2}_{4}=\\{2&9&11&18&25\\}\\\ \bar{b}^{2}_{3}=\\{3&10&12&19&21\\}\\\
\bar{b}^{2}_{2}=\\{4&6&13&20&22\\}\\\ \bar{b}^{2}_{1}=\\{5&7&14&16&23\\}\\\
\end{bmatrix}$,

$\bar{P}_{3}=\begin{bmatrix}\bar{b}^{3}_{5}=\\{1&9&12&20&23\\}\\\
\bar{b}^{3}_{4}=\\{2&10&13&16&24\\}\\\ \bar{b}^{3}_{3}=\\{3&6&14&17&25\\}\\\
\bar{b}^{3}_{2}=\\{4&7&15&18&21\\}\\\ \bar{b}^{3}_{1}=\\{5&8&11&19&22\\}\\\
\end{bmatrix},\bar{P}_{4}=\begin{bmatrix}\bar{b}^{4}_{5}=\\{1&10&14&18&22\\}\\\
\bar{b}^{4}_{4}=\\{2&6&15&19&23\\}\\\ \bar{b}^{4}_{3}=\\{3&7&11&20&24\\}\\\
\bar{b}^{4}_{2}=\\{4&8&12&16&25\\}\\\ \bar{b}^{4}_{1}=\\{5&9&13&17&21\\}\\\
\end{bmatrix}$,

$\bar{P}_{\infty}=\begin{bmatrix}\bar{b}^{5}_{5}=\\{1&6&11&16&21\\}\\\
\bar{b}^{5}_{4}=\\{2&7&12&17&22\\}\\\ \bar{b}^{5}_{3}=\\{3&8&13&18&23\\}\\\
\bar{b}^{5}_{2}=\\{4&9&14&19&24\\}\\\ \bar{b}^{5}_{1}=\\{5&10&15&20&25\\}\\\
\end{bmatrix},\bar{P}_{0}=\begin{bmatrix}\bar{b}^{6}_{5}=\\{1&2&3&4&5\\}\\\
\bar{b}^{6}_{4}=\\{6&7&8&9&10\\}\\\ \bar{b}^{6}_{3}=\\{11&12&13&14&15\\}\\\
\bar{b}^{6}_{2}=\\{16&17&18&19&20\\}\\\
\bar{b}^{6}_{1}=\\{21&22&23&24&25\\}\\\ \end{bmatrix}$.

Here
$\bar{A}=\\{\bar{P}_{1}\cup\bar{P}_{2}\cup\bar{P}_{3}\cup\bar{P}_{4}\cup\bar{P}_{0}\cup\bar{P}_{\infty}\\}$.
Note that, any pair of blocks from different parallel classes has exactly one
element in common.

  3. 3.

Now we remove any 3 blocks from the parallel class $\bar{P}_{0}$, say
$\\{\bar{b}^{6}_{1},\bar{b}^{6}_{2},\bar{b}^{6}_{3}\\}$.

  4. 4.

Next we remove the elements contained in this block from the entire design
$(\bar{X},\bar{A})$.

  5. 5.

Then we discard $\bar{P}_{0}$ from the design. Here we get RBD$(X,A)$
consisting of $5$ parallel classes, each having $5$ blocks of size $2$, where
$X=\\{1,2,3,4,5,6,7,8,9,10\\}$ and $A=\\{P_{1}\cup P_{2}\cup P_{3}\cup
P_{4}\cup P_{\infty}\\}$. Explicitly, we have,

$P_{1}=\begin{bmatrix}b^{1}_{5}=\\{1&7\\}\\\ b^{1}_{4}=\\{2&8\\}\\\
b^{1}_{3}=\\{3&9\\}\\\ b^{1}_{2}=\\{4&10\\}\\\ b^{1}_{1}=\\{5&6\\}\\\
\end{bmatrix},P_{2}=\begin{bmatrix}b^{2}_{5}=\\{1&8\\}\\\
b^{2}_{4}=\\{2&9\\}\\\ b^{2}_{3}=\\{3&10\\}\\\ b^{2}_{2}=\\{4&6\\}\\\
b^{2}_{1}=\\{5&7\\}\\\
\end{bmatrix},P_{3}=\begin{bmatrix}b^{3}_{5}=\\{1&9\\}\\\
b^{3}_{4}=\\{2&10\\}\\\ b^{3}_{3}=\\{3&6\\}\\\ b^{3}_{2}=\\{4&7\\}\\\
b^{3}_{1}=\\{5&8\\}\\\ \end{bmatrix},$
$P_{4}=\begin{bmatrix}b^{4}_{5}=\\{1&10\\}\\\ b^{4}_{4}=\\{2&6\\}\\\
b^{4}_{3}=\\{3&7\\}\\\ b^{4}_{2}=\\{4&8\\}\\\ b^{4}_{1}=\\{5&9\\}\\\
\end{bmatrix},P_{\infty}=\begin{bmatrix}b^{5}_{5}=\\{1&6\\}\\\
b^{5}_{4}=\\{2&7\\}\\\ b^{5}_{3}=\\{3&8\\}\\\ b^{5}_{2}=\\{4&9\\}\\\
b^{5}_{1}=\\{5&10\\}\\\ \end{bmatrix}.$

Note that, for this particular case using $(10,2,1)$-BIBD, one can construct
an RBD with $9$ parallel classes each having $5$ many blocks of constant block
size $2$. One must note that this construction may not provide RBDs having
maximum number of parallel classes, with constant block size and $\mu=1$. Even
if we include the parallel class
$P_{0}=\bar{P}_{0}\setminus\\{\bar{b}^{6}_{1},\bar{b}^{6}_{2},\bar{b}^{6}_{3}\\}$
in the RBD$(X,A)$, it will not change the value of $\mu$ which will remain
equal to 1. However, the block size of $P_{0}$ will be 5 then and hence
RBD(X,A) will contain two different block sizes. Therefore we discard the
$P_{0}$. Nevertheless we will see that even $P_{0}$ can be used to construct
orthonormal basis which will be mutually unbiased with all the orthonormal
basis constructed using $\\{P_{1}\cup P_{2}\cup P_{3}\cup P_{4}\cup
P_{\infty}\\}$.

The technique is more formally explained for the general case in Construction
3 below.

###### Construction 3

Let $d=k\cdot s=(s-e)s$, with $0<e\leq s$.

  1. 1.

Using Construction 1, construct $RBD(\bar{X},\bar{A})$, where
$\bar{X}=\\{1,2,\ldots,s^{2}$}. It will have $r=N(s)+2$ many parallel classes,
namely
$\\{\bar{P}_{1},\bar{P}_{2},\ldots,\bar{P}_{w},\bar{P}_{0},\bar{P}_{\infty}\\}$,
each having $s$ many blocks of constant size $s$. Denoting blocks of the
parallel class $\bar{P}_{l}$ with $\bar{b}^{l}_{i}$, for $i=1,2,\ldots,s$, we
note that between any two blocks from different parallel classes, there is
exactly one element in common, i.e., $|\bar{b}^{l}_{i}\cap\bar{b}^{m}_{j}|=1,\
\forall~{}l\neq m$.

  2. 2.

Pick a parallel class, say $\bar{P}_{0}$. Remove $e$ many blocks from it and
denote as
$S=\\{\bar{b}^{0}_{1}\cup\bar{b}^{0}_{2}\cup\ldots\cup\bar{b}^{0}_{e}\\}$.

  3. 3.

Remove the elements in $S$ from $\bar{X}$ and let us denote the new set with
$X$, i.e., $X=\bar{X}\setminus S$. Further, we remove the elements in $S$ from
the parallel classes
$\\{\bar{P}_{2},\bar{P}_{3},\ldots,\bar{P}_{w},\bar{P}_{\infty}\\}$ and denote
them by $P_{l}$, for $l=2,3,\ldots,r$, i.e., $P_{l}=\bar{P}_{l}\setminus S$.
Then $A=\\{P_{2},P_{3},\ldots,P_{w},P_{\infty}\\}$ with
$P_{l}=\\{b^{l}_{1},b^{l}_{2},\ldots,b^{l}_{q}\\}$, where
$b^{l}_{i}=\bar{b}^{l}_{i}\setminus S$.

  4. 4.

Discard the parallel class $\bar{P}_{0}$. The resulting RBD$(X,A)$ is the
required design. For convenience, rename the elements from 1 to $(s-e)s$.

We claim that the above design $(X,A)$ is an RBD, such that $|X|=(s-e)s$ and
$A$ consist of $N(s)+1$ many parallel classes, i.e.,
$A=\\{P_{2},P_{3},\ldots,P_{w},P_{\infty}\\}$, each having $s$ many blocks,
i.e., $P_{l}=\\{b^{l}_{1},b^{l}_{2},\ldots,b^{l}_{s}\\},l=1,2,\ldots,s$, each
of size $(s-e)$, i.e., $|b^{l}_{i}|=(s-e)\ \forall i,l$, such that blocks from
different parallel classes have at most one element in common, i.e.,
$|b^{l}_{i}\cap b^{m}_{j}|\leq 1\ \forall\,l\neq m$. We formalize this in the
form of a lemma below.

###### Lemma 6

Let $d=(s-e)s$ for $s,e\in\mathbb{N}$ with $0<e\leq s$. Then one can construct
an RBD$(X,A)$, with $|X|=d$ having constant block size $(s-e)$ with $\mu=1$,
and having $N(s)+1$ many parallel classes, where $N(s)$ is the number of
MOLS$(s)$.

###### Proof

Refer to Construction 3 above. In RBD$(\bar{X},\bar{A})$ any pair of blocks
from different parallel classes is of size $s$ and has exactly one element in
common, i.e., $|\bar{b}^{l}_{i}\cap\bar{b}^{m}_{j}|=1\ \forall\ l\neq m$.
Hence removal of the elements
$S=\\{\bar{b}^{1}_{1}\cup\bar{b}^{1}_{2}\cup\ldots\cup\bar{b}^{1}_{e}\\}$ from
entire design will discard exactly $e$ elements from each block
$\bar{b}^{l}_{i},l\neq 1$. Hence, the blocks
$b^{l}_{i}=\bar{b}^{l}_{i}\backslash S$ will be of constant size
$|b^{l}_{i}|=s-e$ and $|b^{l}_{i}\cap b^{m}_{j}|\leq 1\ \forall\,l\neq m$.

Now we can use this RBD$(X,A)$ to construct APMUBs in dimension $d=|X|=(s-e)s$
following Theorem 4.1.

###### Theorem 5.1

Let $d=(s-e)s$ for $s,e\in\mathbb{N}$ with $0<e\leq\frac{3}{2}d^{\frac{1}{2}}$
. Then there exist $N(s)+1$ many APMUBs with $\Delta=\\{0,\frac{1}{s-e}\\}$
and $\beta=\sqrt{\frac{s}{s-e}}=1+\mathcal{O}(d^{-\lambda})\leq 2$, where
$\lambda=\frac{1}{2}$ and sparsity $\epsilon=1-\frac{1}{s}$. Further, if there
exists a real Hadamard matrix of order $(s-e)$, then we can construct $N(s)+1$
many APRMUBs with the same parameters. For the case $e=0$, there exist
$N(s)+2$ many MUBs, and if there exists a real Hadamard matrix of order $s$,
then we can construct $N(s)+2$ many Real MUBs.

###### Proof

In order to show that we can produce such number of APMUBs, let us consider an
RBD$(X,A)$ with $|X|=(s-e)s$ having $N(s)+1$ parallel classes of constant
block size $(s-e)$ such that between the blocks from different parallel
classes, there is at most one element in common, and hence $\mu=1$. This
follows from Lemma 6. Then using this RBD along with a Hadamard matrix of
order $(s-e)$, we can construct orthonormal bases following Theorem 4.1, with
the values of $\Delta,\beta,\epsilon$ by substituting $q=s$, and $f=0$.
Further, the condition that $\beta\leq 2$ for APMUB gives
$e\leq\frac{3}{2}d^{\frac{1}{2}}$. In terms of $s$ this inequality becomes
$e\leq\frac{3}{4}s$ and in terms of $k=(s-e)$, the inequality becomes $s\leq
4k$ so that $\beta\leq 2$. The parameters of APMUBs are
$\Delta=\\{0,\frac{1}{s-e}\\}$,
$\beta=\sqrt{\frac{s}{s-e}}=1+\mathcal{O}(d^{-\frac{1}{2}})\leq 2$ and
$\epsilon=1-\frac{1}{s}$. Since the number of parallel classes is one more
than number of Mutually Orthogonal Latin Squares of Order $s$, we have
$r=N(s)+1$.

In the situation when $d=s^{2}$, i.e., $e=0$, using Construction 1 above, we
obtain RBD$(X,A)$ having $N(s)+2$ parallel classes, such that any pair of
blocks have exactly one point in common. Now using this RBD$(X,A)$, along with
a Hadamard matrix of order $s$, we can construct orthonormal bases following
Theorem 4.1 which will provide $N(s)+2$ many MUBs. Hence for $e=0$, the result
follows directly. Construction 3 is applied only when $e>0$. Thus when
$d=s^{2}$, we get $N(s)+2$ MUBs, and when real Hadamard matrix of order $s$ is
available, that can be used to construct $N(s)+2$ real MUBs.

###### Remark 1

Note that we can construct an orthonormal basis corresponding to the parallel
class $P_{0}=\bar{P}_{0}\setminus S$, again having $(s-e)s$ elements. These
basis vectors will be mutually unbiased with all the orthonormal bases
constructed using the parallel classes $P_{1},P_{2},\ldots P_{w},P_{\infty}$.
However, if we include it in the set of orthonormal bases then
$\Delta=\\{0,\frac{1}{\sqrt{d}},\frac{\beta}{\sqrt{d}}\\}$ will have three
values, and the condition of APMUB will not be satisfied. Thus we ignore
$P_{0}$, even though it provides an orthonormal basis which is mutually
unbiased with all the bases constructed above.

For a composite $s$, $N(s)\rightarrow\mathcal{O}(s^{\frac{1}{14.8}})\ll s-1$,
which is an upper bound HandBook ; stinson2007combinatorial ; wocjan2004new .
Thus in case $d$ can be expressed as $d=k\cdot s=(s-e)s$, where
$s,e\in\mathbb{N}$ with $k<s\leq 4k$ (or $0<e\leq\frac{3}{4}s$), we can always
construct $N(s)+1$ many APMUBs. We refer to this as Mutually Orthogonal Latin
Square Lower Bound construction for APMUBs. For example, given $d=2^{2}\cdot
3^{2}\cdot 5\cdot 7$,

  * •

this can be factored as $d=35\cdot 36$, which will provide $N(36)+1=9$ many
APMUBs with $\beta=1.01$,

  * •

or $d=30\cdot 42$ which will give $N(42)+1=6$ many APMUBs with $\beta=1.18$,

  * •

or $d=28\cdot 45$, which will give $N(45)+1=7$ many APMUB with $\beta=1.27$.

Here, $N(36)=8,N(42)=5$ and $N(45)=6\,$ are the presently known values of the
maximum number of MOLS of these orders HandBook . Let us now explain the
significance of our construction method through this example.

  * •

The number of complex MUBs for this $d=2^{2}\cdot 3^{2}\cdot 5\cdot 7$ which
can be constructed using prime power decomposition, and then taking tensor
product, would be $2^{2}+1=5$. This is the lower bound and there is no better
known result than this in the number of MUBs in this dimension. The value of
$\beta$ is $1$ in this case as exact MUBs are referred.

  * •

Expressing $d=35\cdot 36$, we have more number of APMUBs (9 many) than MUBs,
with $\beta=1.01$.

  * •

Further, expressing $d=28\cdot 45$, we can get 7 many APRMUBs with
$\beta=1.27$. This is because, we have real Hadamard matrix on the dimension
$4\cdot 7=28$.

Our Theorem 5.1 can be compared with that of (wocjan2004new, , Theorem 3),
where the result could be achieved using $(k,s)$-nets. This, in turn, can be
constructed from Mutually Orthogonal Latin Squares of order $s$. Thus, for a
square dimension $d=s^{2}$, there would be $N(s)+2$ many MUBs. Moreover, the
result in (ak22, , Corollary 3) points out that using RBDs, one can construct
$q+1$ many MUBs of dimension $d$ when $d=q^{2}$. Note that $N(q)=q-1$ and thus
the number of MUBs from (ak22, , Corollary 3) is same as that presented in
(wocjan2004new, , Theorem 3). The construction of (ak22, , Corollary 3) had
the advantage of using different Hadamard matrices for the construction of
each MUBs, whereas a single Hadamard matrix can be used for construction of
MUBs in (wocjan2004new, , Theorem 3).

Since Theorem 5.1 is based on RBDs as in (ak22, , Corollary 3), here also
different Hadamard matrices can be used for the construction of each MUBs.
Hence, Theorem 5.1 is a generalization of (wocjan2004new, , Theorem 3) and
(ak22, , Corollary 3) as enumerated below.

  1. 1.

For $d=s^{2}$, Theorem 5.1 reproduces the results of (wocjan2004new, , Theorem
3) in terms of number of MUBs constructed.

  2. 2.

For $d=q^{2}$, Theorem 5.1 reproduces the results of (ak22, , Corollary 3), in
terms of having the advantage of using different Hadamard matrices for the
construction of each MUBs and the number of MUBs constructed.

  3. 3.

As the additional contribution, for any composite $d=k\cdot s=(s-e)\cdot s$,
with $0<e\leq\frac{3}{2}d^{\frac{1}{2}}$, Theorem 5.1 provides MOLS$(s)+1$
many APMUBs and one can also use different Hadamard matrices for the
construction of each basis.

The case $N(s)=s-1$ corresponds to affine plane of order $s$ and the
corresponding RBD is called Affine Resolvable BIBD. When $s=q$, where $q$ is a
prime power, we have well known methods to construct Affine Resolvable
$(q^{2},q,1)$-BIBDs. Hence in such a situation we will have
$q\sim\mathcal{O}(\sqrt{d})$ many APMUBs for composite dimensions which are
not square. For example, if $d=3^{4}\cdot 7=21\cdot 27$, we have 29 APMUBs
with $\beta=1.13$ or for $d=2^{4}\cdot 3=6\cdot 8$ we obtain 9 APMUBs, with
$\beta=1.15$ whereas number of MUBs is 8 and 4 respectively in these cases.

Now let us explain the consequences for Approximate Real MUBs. When $q$ is a
prime power, and real Hadamard matrix of order $(q-e)$ exists, then we will
obtain $(q+1)>\sqrt{d}$ many APRMUBs, which provides large numbers of such
objects over $\mathbb{R}^{d}$. This is presented in the following result.

###### Corollary 2

Let $d=(q-e)q$, where $q$ is a prime power and $e\in\mathbb{N}$, with
$0<e\leq\frac{3}{2}d^{\frac{1}{2}}$. Then there exist $q+1$ many APMUBs with
$\Delta=\\{0,\frac{1}{q-e}\\}$,
$\beta=\sqrt{\frac{q}{q-e}}=1+\mathcal{O}(d^{-\lambda})$, where
$\lambda=\frac{1}{2}$ and $\epsilon=1-\frac{1}{q}$. Further, if there exist
real Hadamard matrices of order $(q-e)$, then one can construct $q+1$ many
Almost Perfect Real MUBs with same parameters.

The condition $e\leq\frac{3}{2}d^{\frac{1}{2}}$ is because we require
$\beta\leq 2$ for APMUBs. If $e>\frac{3}{2}d^{\frac{1}{2}}$ then $\beta>2$,
hence the constructed Approximate MUBs will not satisfy the criteria for
APMUBs [Definition 6]. Further examining the expression for $\beta$ in
Equation 4, for this particular construction with $\mu=1,f=0$, we obtain the
best possible APRMUBs when $e=1$. That is, we have this situation when the
dimensions are of the form $d=(q-1)q$. This we formally state in the following
corollary.

###### Corollary 3

Consider $d=(q-1)q$, such that $q$ is a prime power and assume that a real
Hadamard matrix of order $(q-1)$ exists. Then one can construct $q$ many
Almost Perfect Real MUBs in dimension $d$ with
$\Delta=\left\\{0,\frac{1}{q-1}\right\\}$,
$\beta=\sqrt{\frac{q}{q-1}}=1+\mathcal{O}(d^{-\frac{1}{2}})$, and
$\epsilon=1-\frac{1}{q}$.

For example, when $d=20$ and $156$, there would be respectively $5$ and $13$
APRMUBs of the above type. One may note that we have $\lceil\sqrt{d}\rceil$
many APRMUBs in this case. If $m=\frac{q-3}{2}\equiv 1\bmod 4$ and $m$ is some
prime power, then using the Paley Construction paley1933orthogonal , one can
obtain Hadamard matrix of order $2(m+1)=q-1$. Hence for any prime power
$q\equiv 1\bmod 4$, if $\frac{q-3}{2}$ is also some prime power and is equal
to $1\bmod 4$, then the real Hadamard matrix of order $q-1$ will necessarily
exist through the Paley Construction. For example, one can consider
$q=13,29,5^{3}$ etc. For such $q$’s, the result will become independent of the
Hadamard Conjecture.

###  5.2 $d=k\cdot s=(q-e)(q+f),\ 0<f\leq e$ and
$0<(e+f)\leq\frac{3}{2}d^{\frac{1}{2}}$

As noted in Corollary 2 that if $d$ can be expresses as $k\cdot s=(q-e)q$ with
$\beta=\sqrt{\frac{s}{k}}=\sqrt{\frac{q}{q-e}}\leq 2$, then there we can
construct $q=\mathcal{O}(\sqrt{d})$ many APMUBs. However, if $d$ can not be
expressed in this form then one can construct $N(s)$ many APMUBs if $d$ can be
expressed as $k\cdot s$ with $s$ a composite such that $k\leq s\leq 4k$, i.e.,
for example in the cases $d=\\{2\cdot 3^{2},\ 2\cdot 11,\ 2\cdot 3\cdot 7,\
2^{2}\cdot 3\cdot 7,\ldots\\}$ etc. The condition $k\leq s\leq 4k$ ensures
$\beta\leq 2$ and if $d$ can be expressed as $(s-e)\cdot s$ then it is
equivalent to $0\leq e\leq\frac{3}{2}d^{\frac{1}{2}}$. The best known lower
bound for general $s$ is $N(s)\sim s^{\frac{1}{14.8}}$ which is much less than
$s$.

In order to obtain significantly larger number of APMUBs, for the dimensions
that cannot be expressed in the form $d=(q-e)q$ where $q$ is some prime power
with $0<e\leq\frac{3}{2}d^{\frac{1}{2}}$, we now consider the form of
$d=(q-e)(q+f)$, such that $q$ is a prime power with $e,f\in\mathbb{N}$. First
we show that, in such a case if $e\geq f$, then we can construct RBD$(X,A)$
with $|X|=d$ having constant block size $(q-e)$ such that $A$ can be
partitioned into at least
$r=\lfloor\frac{q-(e-f)}{f}\rfloor=\lfloor\frac{q-e}{f}\rfloor+1$ many
parallel classes. Hence such an RBD$(X,A)$ can be used to construct
$\lfloor\frac{q-e}{f}\rfloor+1$ many orthonormal bases following (ak22, ,
Theorem 1). Further, if $0<(e+f)\leq\frac{3}{2}d^{\frac{1}{2}}$, then
$\beta\leq 2$, and these orthonormal bases would be APMUBs, thus providing us
$\mathcal{O}(q)$ many APMUBs in such a situation.

We explain this construction in two parts. First we consider a $(q^{2},q,1)$
Affine Resolvable BIBD as the input. We call this RBD$(\bar{X},\bar{A})$ where
$|\bar{X}|=q^{2}$ and all the blocks of $A$ is of the same size $q$ and number
of parallel classes in $A$ is $q+1$. We use this to construct
RBD$(\widetilde{X},\widetilde{A})$, where $|\widetilde{X}|=(q-e)(q+f)$ having
same number of parallel class $(q+1)$, but the blocks are not of the same
size. The sizes of the blocks are from set
$\\{(q-e),(q-e+1),\ldots,(q-e+f),q\\}$.

In the second part we use the RBD$(\widetilde{X},\widetilde{A})$ as input and
construct RBD$(X,A)$ where $|X|=(q-e)(q+f)$ such that each block in $A$ is of
size $(q-e)$. However, now the number of parallel class reduces to
$\lfloor\frac{q-e}{f}\rfloor+1$.

To demonstrate the first part of the construction, we take
$|X|=(7-3)(7+1)=4\cdot 8=32$, where $q=7,e=3$ and $f=1$. We use an Affine
Resolvable $(7^{2},7,1)$-BIBD which we call RBD$(\bar{X},\bar{A})$. It will
consist of eight parallel classes. Each parallel class would consist of seven
blocks of constant size 7. We represent each parallel class as a $7\times 7$
matrix, where each row represent one block of the parallel class. Hence there
would be eight such matrices as below to represent the design.

$\bar{P}_{1}=\begin{bmatrix}\bar{b}^{1}_{7}=\\{1&2&3&4&5&6&7\\}\\\
\bar{b}^{1}_{6}=\\{8&9&10&11&12&13&14\\}\\\
\bar{b}^{1}_{5}=\\{15&16&17&18&19&20&21\\}\\\
\bar{b}^{1}_{4}=\\{22&23&24&25&26&27&28\\}\\\
\bar{b}^{1}_{3}=\\{29&30&31&32&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}33}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}34}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}35}\\}\\\
\bar{b}^{1}_{2}=\\{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}36}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}37}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}38}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}39}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}40}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}41}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}42}\\}\\\
\bar{b}^{1}_{1}=\\{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}43}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}44}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}45}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}46}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}47}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}48}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}49}\\}\\\
\end{bmatrix},\bar{P}_{2}=\begin{bmatrix}\bar{b}^{2}_{7}=\\{1&9&17&25&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}33}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}41}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}49}\\}\\\
\bar{b}^{2}_{6}=\\{2&10&18&26&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}34}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}42}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}43}\\}\\\
\bar{b}^{2}_{5}=\\{3&11&19&27&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}35}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}36}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}44}\\}\\\
\bar{b}^{2}_{4}=\\{4&12&20&28&29&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}37}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}45}\\}\\\
\bar{b}^{2}_{3}=\\{5&13&21&22&30&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}38}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}46}\\}\\\
\bar{b}^{2}_{2}=\\{6&14&15&23&31&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}39}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}47}\\}\\\
\bar{b}^{2}_{1}=\\{7&8&16&24&32&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}40}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}48}\\}\\\
\end{bmatrix},$

$\bar{P}_{3}=\begin{bmatrix}\bar{b}^{3}_{7}=\\{1&10&19&28&30&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}39}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}48}\\}\\\
\bar{b}^{3}_{6}=\\{2&11&20&22&31&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}40}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}49}\\}\\\
\bar{b}^{3}_{5}=\\{3&12&21&23&32&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}41}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}43}\\}\\\
\bar{b}^{3}_{4}=\\{4&13&15&24&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}33}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}42}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}44}\\}\\\
\bar{b}^{3}_{3}=\\{5&14&16&25&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}34}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}36}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}45}\\}\\\
\bar{b}^{3}_{2}=\\{6&8&17&26&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}35}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}37}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}46}\\}\\\
\bar{b}^{3}_{1}=\\{7&9&18&27&29&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}38}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}47}\\}\\\
\end{bmatrix},\bar{P}_{4}=\begin{bmatrix}\bar{b}^{4}_{7}=\\{1&11&21&24&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}34}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}37}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}47}\\}\\\
\bar{b}^{4}_{6}=\\{2&12&15&25&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}35}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}38}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}48}\\}\\\
\bar{b}^{4}_{5}=\\{3&13&16&26&29&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}39}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}49}\\}\\\
\bar{b}^{4}_{4}=\\{4&14&17&27&30&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}40}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}43}\\}\\\
\bar{b}^{4}_{3}=\\{5&8&18&28&31&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}41}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}44}\\}\\\
\bar{b}^{4}_{2}=\\{6&9&19&22&32&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}42}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}45}\\}\\\
\bar{b}^{4}_{1}=\\{7&10&20&23&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}33}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}36}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}46}\\}\\\
\end{bmatrix},$

$\bar{P}_{5}=\begin{bmatrix}\bar{b}^{5}_{7}=\\{1&12&16&27&31&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}42}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}46}\\}\\\
\bar{b}^{5}_{6}=\\{2&13&17&28&32&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}36}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}47}\\}\\\
\bar{b}^{5}_{5}=\\{3&14&18&22&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}33}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}37}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}48}\\}\\\
\bar{b}^{5}_{4}=\\{4&8&19&23&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}34}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}38}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}49}\\}\\\
\bar{b}^{5}_{3}=\\{5&9&20&24&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}35}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}39}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}43}\\}\\\
\bar{b}^{5}_{2}=\\{6&10&21&25&29&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}40}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}44}\\}\\\
\bar{b}^{5}_{1}=\\{7&11&15&26&30&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}41}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}45}\\}\\\
\end{bmatrix},\bar{P}_{6}=\begin{bmatrix}\bar{b}^{6}_{7}=\\{1&13&18&23&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}35}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}40}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}45}\\}\\\
\bar{b}^{6}_{6}=\\{2&14&19&24&29&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}41}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}46}\\}\\\
\bar{b}^{6}_{5}=\\{3&8&20&25&30&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}42}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}47}\\}\\\
\bar{b}^{6}_{4}=\\{4&9&21&26&31&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}36}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}48}\\}\\\
\bar{b}^{6}_{3}=\\{5&10&15&27&32&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}37}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}49}\\}\\\
\bar{b}^{6}_{2}=\\{6&11&16&28&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}33}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}38}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}43}\\}\\\
\bar{b}^{6}_{1}=\\{7&12&17&22&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}34}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}39}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}44}\\}\\\
\end{bmatrix},$

$\bar{P}_{7}=\begin{bmatrix}\bar{b}^{7}_{7}=\\{1&14&20&26&32&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}38}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}44}\\}\\\
\bar{b}^{7}_{6}=\\{2&8&21&27&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}33}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}39}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}45}\\}\\\
\bar{b}^{7}_{5}=\\{3&9&15&28&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}34}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}40}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}46}\\}\\\
\bar{b}^{7}_{4}=\\{4&10&16&22&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}35}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}41}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}47}\\}\\\
\bar{b}^{7}_{3}=\\{5&11&17&23&29&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}42}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}48}\\}\\\
\bar{b}^{7}_{2}=\\{6&12&18&24&30&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}36}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}49}\\}\\\
\bar{b}^{7}_{1}=\\{7&13&19&25&31&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}37}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}43}\\}\\\
\end{bmatrix},\bar{P}_{8}=\begin{bmatrix}\bar{b}^{8}_{7}=\\{1&8&15&22&29&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}36}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}43}\\}\\\
\bar{b}^{8}_{6}=\\{2&9&16&23&30&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}37}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}44}\\}\\\
\bar{b}^{8}_{5}=\\{3&10&17&24&31&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}38}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}45}\\}\\\
\bar{b}^{8}_{4}=\\{4&11&18&25&32&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}39}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}46}\\}\\\
\bar{b}^{8}_{3}=\\{5&12&19&26&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}33}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}40}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}47}\\}\\\
\bar{b}^{8}_{2}=\\{6&13&20&27&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}34}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}41}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}48}\\}\\\
\bar{b}^{8}_{1}=\\{7&14&21&28&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}35}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}42}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}49}\\}\\\
\end{bmatrix},$

In order to construct RBD$(\widetilde{X},\widetilde{A})$, where
$|\widetilde{X}|=(q-e)(q+f)=(7-3)(7+1)=32$ such that $\mu=1$, we consider the
following steps.

  1. 1.

Choose any $h=e-f=3-1=2$ blocks from $\bar{P}_{1}$. Let these blocks be
$\bar{b}^{1}_{1}$ and $\bar{b}^{1}_{2}$. Let
$S_{1}=\bar{b}^{1}_{1}\cup\bar{b}^{1}_{2}=\\{36,37,38,39,40,41,42,43,44,45,46,47,48,49\\}$,
as annotated in red in the above matrices.

  2. 2.

Choose another $f=1$ block from $\bar{P}_{1}$. Let it be $\bar{b}^{1}_{3}$.
Now choose any $e=3$ elements from it. Let these be $S_{2}=\\{33,34,35\\}$.
Set $S=S_{1}\cup S_{2}=\\{33,34,35,36,37,38,39,40$, $41,42$,
$43,44,45,46,47,48,49\\}$ (indicated in red).

  3. 3.

Remove the elements of set $S$ from RBD$(\bar{X},\bar{A})$. Call the resulting
combinatorial design RBD$(\widetilde{X},\widetilde{A})$, where
$|\widetilde{X}|=32$ and
$\widetilde{A}=\\{\widetilde{P}_{1},\widetilde{P}_{2},\widetilde{P}_{3},\widetilde{P}_{4},\widetilde{P}_{5},\widetilde{P}_{6},\widetilde{P}_{7},\widetilde{P}_{8}\\}$,
presented as below.

$\widetilde{P}_{1}=\begin{bmatrix}\widetilde{b}^{1}_{7}=\\{1&2&3&4&5&6&7\\}\\\
\widetilde{b}^{1}_{6}=\\{8&9&10&11&12&13&14\\}\\\
\widetilde{b}^{1}_{5}=\\{15&16&17&18&19&20&21\\}\\\
\widetilde{b}^{1}_{4}=\\{22&23&24&25&26&27&28\\}\\\
\widetilde{b}^{1}_{3}=\\{29&30&31&32\\}\\\
\end{bmatrix},\widetilde{P}_{2}=\begin{bmatrix}\widetilde{b}^{2}_{7}=\\{1&9&17&25\\}\\\
\widetilde{b}^{2}_{6}=\\{2&10&18&26\\}\\\
\widetilde{b}^{2}_{5}=\\{3&11&19&27\\}\\\
\widetilde{b}^{2}_{4}=\\{4&12&20&28&29\\}\\\
\widetilde{b}^{2}_{3}=\\{5&13&21&22&30\\}\\\
\widetilde{b}^{2}_{2}=\\{6&14&15&23&31\\}\\\
\widetilde{b}^{2}_{1}=\\{7&8&16&24&32\\}\\\ \end{bmatrix}$,

$\widetilde{P}_{3}=\begin{bmatrix}\widetilde{b}^{3}_{7}=\\{1&10&19&28&30\\}\\\
\widetilde{b}^{3}_{6}=\\{2&11&20&22&31\\}\\\
\widetilde{b}^{3}_{5}=\\{3&12&21&23&32\\}\\\
\widetilde{b}^{3}_{4}=\\{4&13&15&24\\}\\\
\widetilde{b}^{3}_{3}=\\{5&14&16&25\\}\\\
\widetilde{b}^{3}_{2}=\\{6&8&17&26\\}\\\
\widetilde{b}^{3}_{1}=\\{7&9&18&27&29\\}\\\
\end{bmatrix},\widetilde{P}_{4}=\begin{bmatrix}\widetilde{b}^{4}_{7}=\\{1&11&21&24\\}\\\
\widetilde{b}^{4}_{6}=\\{2&12&15&25\\}\\\
\widetilde{b}^{4}_{5}=\\{3&13&16&26&29\\}\\\
\widetilde{b}^{4}_{4}=\\{4&14&17&27&30\\}\\\
\widetilde{b}^{4}_{3}=\\{5&8&18&28&31\\}\\\
\widetilde{b}^{4}_{2}=\\{6&9&19&22&32\\}\\\
\widetilde{b}^{4}_{1}=\\{7&10&20&23\\}\\\ \end{bmatrix}$,

$\widetilde{P}_{5}=\begin{bmatrix}\widetilde{b}^{5}_{7}=\\{1&12&16&27&31\\}\\\
\widetilde{b}^{5}_{6}=\\{2&13&17&28&32\\}\\\
\widetilde{b}^{5}_{5}=\\{3&14&18&22\\}\\\
\widetilde{b}^{5}_{4}=\\{4&8&19&23\\}\\\
\widetilde{b}^{5}_{3}=\\{5&9&20&24\\}\\\
\widetilde{b}^{5}_{2}=\\{6&10&21&25&29\\}\\\
\widetilde{b}^{5}_{1}=\\{7&11&15&26&30\\}\\\
\end{bmatrix},\widetilde{P}_{6}=\begin{bmatrix}\widetilde{b}^{6}_{7}=\\{1&13&18&23\\}\\\
\widetilde{b}^{6}_{6}=\\{2&14&19&24&29\\}\\\
\widetilde{b}^{6}_{5}=\\{3&8&20&25&30\\}\\\
\widetilde{b}^{6}_{4}=\\{4&9&21&26&31\\}\\\
\widetilde{b}^{6}_{3}=\\{5&10&15&27&32\\}\\\
\widetilde{b}^{6}_{2}=\\{6&11&16&28\\}\\\
\widetilde{b}^{6}_{1}=\\{7&12&17&22\\}\\\ \end{bmatrix}$,

$\widetilde{P}_{7}=\begin{bmatrix}\widetilde{b}^{7}_{7}=\\{1&14&20&26&32\\}\\\
\widetilde{b}^{7}_{6}=\\{2&8&21&27\\}\\\
\widetilde{b}^{7}_{5}=\\{3&9&15&28\\}\\\
\widetilde{b}^{7}_{4}=\\{4&10&16&22\\}\\\
\widetilde{b}^{7}_{3}=\\{5&11&17&23&29\\}\\\
\widetilde{b}^{7}_{2}=\\{6&12&18&24&30\\}\\\
\widetilde{b}^{7}_{1}=\\{7&13&19&25&31\\}\\\
\end{bmatrix},\widetilde{P}_{8}=\begin{bmatrix}\widetilde{b}^{8}_{7}=\\{1&8&15&22&29\\}\\\
\widetilde{b}^{8}_{6}=\\{2&9&16&23&30\\}\\\
\widetilde{b}^{8}_{5}=\\{3&10&17&24&31\\}\\\
\widetilde{b}^{8}_{4}=\\{4&11&18&25&32\\}\\\
\widetilde{b}^{8}_{3}=\\{5&12&19&26\\}\\\
\widetilde{b}^{8}_{2}=\\{6&13&20&27\\}\\\ \bar{b}^{8}_{1}=\\{7&14&21&28\\}\\\
\end{bmatrix}$.

Note that here all the blocks are not of the same sizes, but any two blocks
from different parallel classes have at most 1 element in common, i.e.,
$\mu=1$. The blocks sizes are in the set $\\{(q-3),(q-2),q\\}=\\{4,5,7\\}$.
The number of parallel classes in $\widetilde{A}$ remains $q+1=8$. This
technique is now more formally explained for the general case in Construction
4 below. Let $d=k\cdot s=(q-e)(q+f)$, with $0<f\leq e\leq q$. The steps for
constructing the RBD$(X,A)$ are as follows.

###### Construction 4

Given $q$, a prime power, construct $(q^{2},q,1)$-ARBIBD. Call this design
$(\bar{X},\bar{A})$ with $\bar{X}=\\{1,2,\ldots,q^{2}\\}$ and
$|\bar{A}|=q(q+1)$ many blocks, each block is of constant size $q$. It will
have $r=q+1$ many parallel classes, call them
$\\{\bar{P}_{1},\bar{P}_{2},\ldots,\bar{P}_{q+1}\\}$, each parallel class
having $q$ many blocks of constant size $q$. Between any two blocks from
different parallel classes, exactly one element will be common, i.e.,
$|\bar{b}^{l}_{i}\cap\bar{b}^{m}_{j}|=1,\forall\ l\neq m$.

  1. 1.

Given $e\geq f$, choose $h=e-f\geq 0$ many blocks from $\bar{P}_{1}$, which
are $\\{\bar{b}^{1}_{1},\bar{b}^{1}_{2},\ldots,\bar{b}^{1}_{h}\\}$. Let
$S_{1}=\bar{b}^{1}_{1}\cup\bar{b}^{1}_{2}\cup\ldots\cup\bar{b}^{1}_{h}$.
Therefore, $|S_{1}|=h\cdot q$.

  2. 2.

From $\\{\bar{b}^{1}_{h+1},\bar{b}^{1}_{h+2},\ldots,\bar{b}^{1}_{h+f}\\}$
blocks of $\bar{P}_{1}$, choose any $e$ number of elements from each of them.
Let $S_{2}$ be the union of all these elements. Therefore, $|S_{2}|=e\cdot f$.
Let $S=S_{1}\cup S_{2}$.

  3. 3.

Remove the elements of set $S$ from RBD $(\bar{X},\bar{A})$ and call the
resulting design as RBD$(\widetilde{X},\widetilde{A})$.

We claim that the above RBD$(\widetilde{X},\widetilde{A})$ is such that
$|\widetilde{X}|=(q-e)(q+f)$ and $\widetilde{A}$ consists of $q+1$ many
parallel classes having different block sizes, such that blocks from different
parallel classes have at most one element in common, i.e., $|b^{l}_{i}\cap
b^{m}_{j}|\leq 1,\forall\ l\neq m$. Hence $\mu=1$. We formalize this in the
form of a lemma below.

###### Lemma 7

Let $d=(q-e)(q+f)$ for $f,e\in\mathbb{N}$ with $0<f\leq e\leq q$ and some
prime power $q$. Then one can construct an RBD$(\widetilde{X},\widetilde{A})$,
with $|X|=d$ having block sizes from the set of integers
$\\{(q-e),(q-e+1),\ldots,(q-e+f),q\\}$ with $\mu=1$, and having $r=q+1$ many
parallel classes.

###### Proof

Refer to Construction 4 above. Here RBD$(\bar{X},\bar{A})$ is an ARBIBD with
$|\bar{X}|=q^{2}$, having constant block size $q$. Note that any pair of
blocks from different parallel classes have exactly one element in common,
i.e., $|\bar{b}^{l}_{i}\cap\bar{b}^{m}_{j}|=1\ \forall\ l\neq m$. The number
of elements in the set $|S|=|S_{1}\cup
S_{2}|=|S_{1}|+|S_{2}|=(e-f)q+ef<q^{2}$, which is a proper subset of
$\bar{X}$. These are removed from all the parallel classes of
RBD$(\bar{X},\bar{A})$. Hence, the resulting design
RBD$(\widetilde{X},\widetilde{A})$ is such that
$|\widetilde{X}|=q^{2}-(e-f)q+ef=(q-e)(q+f)=d$, having same number of parallel
classes as in $\bar{A}$. The number of element common between any two blocks
from different parallel classes would be at most 1, i.e.,
$|\widetilde{b}^{l}_{i}\cap\widetilde{b}^{m}_{j}|\leq 1,\forall\ l\neq m$.

To obtain the sizes of the blocks in RBD$(\widetilde{X},\widetilde{A})$, note
that $S_{1}$ contains all elements from $h=(e-f)$ number of blocks of
$\bar{P_{1}}$. Hence $S_{1}$ would have at least $h$ elements in common with
all the blocks of remaining parallel classes. Thus, removal of the elements in
$S_{1}$ from the parallel classes $\bar{P}_{l},l\geq 2$ will remove at least
$h$ elements from each block of $\bar{P_{l}}$ which implies
$|\bar{b}^{l}_{i}\setminus S_{1}|=q-(e-f)$. Further $S_{2}$ contains $e$
elements from $f$ many blocks of $\bar{P}_{1}$. Thus, the blocks in
$\bar{P}_{2},\bar{P}_{3},\ldots,\bar{P}_{q+1}$ will have at most $f$ many
elements in common with $S_{2}$. Consequently, after removal of all the
elements in $S$, from the parallel class $\bar{P}_{l}$, the block size
$|\widetilde{b}^{l}_{i}|,\ l\geq 2$ will be maximum $q-(e-f)=(q-h)$ and
minimum $q-(e-f)-f=(q-e)$. Further, the blocks in $\widetilde{P}_{1}$ will be
of sizes $q,(q-e)$ and have total $(q-h)$ number of blocks.

We will now show how one can use $f$ many blocks of the parallel class
$\widetilde{P}_{1}$ in RBD$(\widetilde{X},\widetilde{A})$, as constructed
above, to reshape any one of the parallel classes $\widetilde{P}_{l},\ l\neq
1$ into a parallel class having $(q+f)$ many blocks each of size $(q-e)$,
which we denote by $P_{l}$ and the resulting combinatorial design by
RBD$(X,A)$. Since $\widetilde{P}_{1}$ have $q-h$ number of blocks, thus
$\lfloor\frac{q-h}{f}\rfloor$ many parallel classes of $\bar{A}$ can be
reshaped into a parallel class of $A$.

Let us first demonstrate the construction by using
RBD$(\widetilde{X},\widetilde{A})$, as constructed in the example earlier in
this section, where $|\widetilde{X}|=(7-3)(7+1)=4\cdot 8=32$, with $q=7,e=3$,
$f=1$ and $h=e-f=2$. Note that $\widetilde{P}_{1}$ has $q-h=5$ blocks, and all
the other parallel classes have $q=7$ blocks each. Let us denote the excess
number of elements on each block of $\widetilde{b}^{l}_{j},\ l\geq 2$ than
$(q-e)$ by $m^{l}_{j}$. Hence $m^{l}_{i}=|\widetilde{b}^{l}_{i}|-(q-e)$. Here
for each block of the parallel class $\widetilde{P}_{l},\ l\geq 2$, the value
of $m^{l}_{i}$ is either 0 or 1. If the block $|\widetilde{b}^{l}_{i}|=5$,
then $m^{l}_{i}=|\widetilde{b}^{l}_{i}|-(q-e)=5-(7-3)=1$ and similarly for
block $|\widetilde{b}^{l}_{i}|=4$, $m^{l}_{i}=0$. Note that
$\sum^{q}_{i=1}m^{l}_{i}=(q-e)\cdot f=(7-3)\cdot 1=4,\forall\ l\geq 2$. We
modify this RBD$(\widetilde{X},\widetilde{A})$ as follows.

  1. 1.

Since $f=1$, consider one block $\widetilde{b}^{1}_{3}$ of
$\widetilde{P}_{1}$, which has 4 elements in it. Remove these elements from
different blocks of $\widetilde{P}_{2}$ and add them as separate block in
$\widetilde{P}_{2}$. Denote the resulting parallel class as $P_{2}$.

  2. 2.

Consider the parallel class $\widetilde{P}_{3}$ and the next $f=1$ block of
$\widetilde{P}_{1}$, i.e., the block $\widetilde{b}^{1}_{4}$. Choose a block
from $\widetilde{P}_{3}$, say $\widetilde{b}^{3}_{1}$. Since
$|\widetilde{b}^{3}_{1}|=5$, i.e., it has one element more than $q-e=4$, hence
mark one common element between $\widetilde{b}^{1}_{4}$ and
$\widetilde{b}^{3}_{1}$ which is 27 in this case. Sequentially execute this
for all the blocks of $\widetilde{P}_{1}$. This will mark the elements
$\\{27,23,22,28\\}$ on $\widetilde{b}^{1}_{4}$.

  3. 3.

Since $m^{3}_{j}=0$ or $1$, the above step will mark exactly $(q-e)=4$
elements on $\widetilde{b}^{1}_{4}$. In a situation, if $m^{l}_{i}$ has more
than one elements, then further iterations are required to mark exactly
$(q-e)$ elements on the blocks of $\widetilde{P}_{1}$. Refer to Step 3 of
Construction 5 later for the exact strategy in this regard.

  4. 4.

Now remove the elements marked on $\widetilde{b}^{1}_{4}$, i.e.,
$\\{27,23,22,28\\}$ from $\widetilde{P}_{3}$ and add them as a separate block
of $\widetilde{P}_{3}$ and denote the resulting parallel class as $P_{3}$.

  5. 5.

Consider the next parallel class $\widetilde{P}_{4}$ and the next $f=1$ block
of $\widetilde{P}_{1}$, i.e., the block $\widetilde{b}^{1}_{5}$. Then repeat
the Steps 2, 3, 4 to obtain $P_{4}$.

  6. 6.

Since the number of blocks in $\widetilde{P}_{1}=5$, in this way
$r=\frac{5}{1}=5$ parallel classes, i.e., $\widetilde{P}_{l},\ l=2,3,4,5,6$
can be modified. Discard $\widetilde{P}_{7}$ and $\widetilde{P}_{8}$. The
resulting RBD$(X,A)$ is such that $|X|=32$ with $A$ consisting of parallel
classes $\\{P_{2},P_{3},P_{4},P_{5},P_{6}\\}$ as shown below.

$P_{2}=\begin{bmatrix}b^{2}_{8}=\\{29&30&31&32\\}\\\
b^{2}_{7}=\\{1&9&17&25\\}\\\ b^{2}_{6}=\\{2&10&18&26\\}\\\
b^{2}_{5}=\\{3&11&19&27\\}\\\ b^{2}_{4}=\\{4&12&20&28\\}\\\
b^{2}_{3}=\\{5&13&21&22\\}\\\ b^{2}_{2}=\\{6&14&15&23\\}\\\
b^{2}_{1}=\\{7&8&16&24\\}\\\
\end{bmatrix},P_{3}=\begin{bmatrix}b^{3}_{8}=\\{22&23&27&28\\}\\\
b^{3}_{7}=\\{1&10&19&30\\}\\\ b^{3}_{6}=\\{2&11&20&31\\}\\\
b^{3}_{5}=\\{3&12&21&32\\}\\\ b^{3}_{4}=\\{4&13&15&24\\}\\\
b^{3}_{3}=\\{5&14&16&25\\}\\\ b^{3}_{2}=\\{6&8&17&26\\}\\\
b^{3}_{1}=\\{7&9&18&29\\}\\\
\end{bmatrix},P_{4}=\begin{bmatrix}b^{4}_{8}=\\{16&17&18&19\\}\\\
b^{4}_{7}=\\{1&11&21&24\\}\\\ b^{4}_{6}=\\{2&12&15&25\\}\\\
b^{4}_{5}=\\{3&13&26&29\\}\\\ b^{4}_{4}=\\{4&14&27&30\\}\\\
b^{4}_{3}=\\{5&8&28&31\\}\\\ b^{4}_{2}=\\{6&9&22&32\\}\\\
b^{4}_{1}=\\{7&10&20&23\\}\\\ \end{bmatrix},$

$P_{5}=\begin{bmatrix}b^{5}_{8}=\\{10&11&12&13\\}\\\
b^{5}_{7}=\\{1&16&27&31\\}\\\ b^{5}_{6}=\\{2&17&28&32\\}\\\
b^{5}_{5}=\\{3&14&18&22\\}\\\ b^{5}_{4}=\\{4&8&19&23\\}\\\
b^{5}_{3}=\\{5&9&20&24\\}\\\ b^{5}_{2}=\\{6&21&25&29\\}\\\
b^{5}_{1}=\\{7&15&26&30\\}\\\
\end{bmatrix},P_{6}=\begin{bmatrix}b^{6}_{8}=\\{2&3&4&5\\}\\\
b^{6}_{7}=\\{1&13&18&23\\}\\\ b^{6}_{6}=\\{14&19&24&29\\}\\\
b^{6}_{5}=\\{8&20&25&30\\}\\\ b^{6}_{4}=\\{9&21&26&31\\}\\\
b^{6}_{3}=\\{10&15&27&32\\}\\\ b^{6}_{2}=\\{6&11&16&28\\}\\\
b^{6}_{1}=\\{7&12&17&22\\}\\\
\end{bmatrix},P_{7}=\begin{bmatrix}b^{7}_{8}=\\{5&12&19&26\\}\\\
b^{7}_{7}=\\{1&14&20&32\\}\\\ b^{7}_{6}=\\{2&8&21&27\\}\\\
b^{7}_{5}=\\{3&9&15&28\\}\\\ b^{7}_{4}=\\{4&10&16&22\\}\\\
b^{7}_{3}=\\{11&17&23&29\\}\\\ b^{7}_{2}=\\{6&18&24&30\\}\\\
b^{7}_{1}=\\{7&13&25&31\\}\\\ \end{bmatrix}.$

We will now discuss about $P_{7}$ as presented above. From the two discarded
parallel classes $\widetilde{P}_{7}$ and $\widetilde{P}_{8}$, we find that one
can use the block $\widetilde{b}^{8}_{3}$ and remove the elements in it from
$\widetilde{P}_{7}$, and place them as a separate block of parallel class
$\widetilde{P}_{7}$, resulting into another parallel class having $(q+f)=8$
many blocks each block having $(q-e)=4$ elements each. We denote this parallel
class as $P_{7}$. Certainly this is not unique as there are other
possibilities to obtain a parallel class using $\widetilde{P}_{7}$ and
$\widetilde{P}_{8}$. Thus here we actually obtain
$r=6>\left\lfloor\frac{q-e}{f}\right\rfloor+1=5$, indicating that
$\left\lfloor\frac{q-e}{f}\right\rfloor+1$ is not a tight lower bound in this
example.

The above construction is formally explained for the general case in
Construction 5 below. We consider RBD$(\widetilde{X},\widetilde{A})$
constructed in 4 with $|\widetilde{X}|=(q-e)(q+f)$ as the input for following
construction. To begin with, compute
$m^{l}_{j}=|\widetilde{b}^{l}_{j}|-(q-e),\ l\geq 2$ for each block of
$\widetilde{P}_{l},\ l\geq 2$, which is the count of the excess number of
elements on each block of $\widetilde{b}^{l}_{j},\ l\geq 2$ than what is
required, which is $q-e$. Note that
$\sum_{j=1}^{q}m^{l}_{j}=\sum_{j=1}^{q}\left(|\widetilde{b}^{l}_{j}|-(q-e)\right)=\sum_{j=1}^{q}|\widetilde{b}^{l}_{j}|-q(q-e)=(q-e)(q+f)-q(q-e)=(q-e)f,\
\forall\ l\geq 2$. Thus $\sum_{j=1}^{q}m^{l}_{j}=(q-e)f$ is constant for all
the parallel classes of RBD$(\widetilde{X},\widetilde{A})$ except
$\widetilde{P}_{1}$, which consists of $q-h$ blocks of sizes $\\{(q-e),q\\}$
and are being used to modify the $r$ number of other parallel classes and will
be discarded in the end.

###### Construction 5

Let $d=k\cdot s=(q-e)(q+f)$, with $0<f\leq e\leq q$ and we consider
RBD$(\widetilde{X},\widetilde{A})$ from Construction 4 with
$|\widetilde{X}|=(q-e)(q+f)$ as the input.

  1. 1.

Consider $f$ many blocks of $\widetilde{P}_{1}$, which has $(q-e)$ elements,
i.e., the blocks $\\{\widetilde{b}^{1}_{h+1},\widetilde{b}^{1}_{h+2},\ldots$,
$\widetilde{b}^{1}_{h+f}\\}$. Remove the elements in the blocks
$\\{\widetilde{b}^{1}_{h+1},\widetilde{b}^{1}_{h+2},\ldots,\widetilde{b}^{1}_{h+f}\\}$
from different blocks of $\widetilde{P}_{2}$ and add the blocks
$\\{\widetilde{b}^{1}_{h+1},\widetilde{b}^{1}_{h+2},\ldots,\widetilde{b}^{1}_{h+f}\\}$
as blocks of $\widetilde{P}_{2}$. Name the resulting parallel class as
$P_{2}$.

  2. 2.

Consider the parallel class $\widetilde{P}_{3}$ and next $f$ many blocks of
$\widetilde{P}_{1}$, i.e.,
$\\{\widetilde{b}^{1}_{(h+f+1)},\widetilde{b}^{1}_{(h+f+2)},\ldots$,
$\widetilde{b}^{1}_{(h+2f)}\\}$, each consisting of $q$ many elements. Call
this set of blocks as $S^{1}_{3}$. Select a block from $\widetilde{P}_{3}$,
say $\widetilde{b}^{3}_{1}$. Corresponding to this block, mark $m^{3}_{1}$
number of elements which are common with the blocks in set $S^{1}_{3}$. Then
move them to the next block of $\widetilde{P}_{3}$, namely
$\widetilde{b}^{3}_{2}$, and mark $m^{3}_{2}$ number of elements common with
the blocks in $S^{1}_{3}$. Sequentially continue this for all the blocks of
$\widetilde{P}_{3}$.

  3. 3.

Now consider $\widetilde{b}^{1}_{u},\ \widetilde{b}^{1}_{v}\in S^{1}_{3}$,
such that $\widetilde{b}^{1}_{u}$ has more elements marked than $(q-e)$ and
$\widetilde{b}^{1}_{v}$ has less elements marked than $(q-e)$. Identify the
blocks of $\widetilde{P}_{3}$ which have a marked element common with
$\widetilde{b}^{1}_{u}$, but has an unmarked element common with
$\widetilde{b}^{1}_{v}$, say block $\widetilde{b}^{3}_{j}$. Then unmark this
element in $\widetilde{b}^{1}_{u}$, and mark the common element between the
block $\widetilde{b}^{3}_{j}$ and $\widetilde{b}^{1}_{v}$ on the block
$\widetilde{b}^{1}_{v}$. Thus, the marked element on $\widetilde{b}^{1}_{u}$
is removed and the marked on $\widetilde{b}^{1}_{v}$ is added. Iterate this
for all such blocks in $S^{1}_{3}$ which has marked elements different from
$(q-e)$ and continue this till all the blocks in set $S^{1}_{3}$ have exactly
$(q-e)$ marked element.

Later in Lemma 8, we will show that such a block $\widetilde{b}^{3}_{j}$ will
always exist. Further this process will terminate in a finite number of steps
as there are finite number of blocks and elements, and every iteration adds
the marked element of a block having less elements marked than $(q-e)$ and
removes the mark element of a block having more elements marked than $(q-e)$.

  4. 4.

Remove the elements marked on each of the blocks in the set $S^{1}_{3}$ from
the parallel class $\widetilde{P}_{3}$ and add the elements marked on the
block, say $\widetilde{b}^{1}_{h+f+1}$ as separate blocks in
$\widetilde{P}_{3}$. Similarly add elements marked on the block
$\widetilde{b}^{1}_{h+f+2}$ as separate blocks in $\widetilde{P}_{3}$ and so
on for all the blocks in $S^{1}_{3}$. Call the resulting parallel class as
$P_{3}$.

  5. 5.

Then consider the next parallel class $\widetilde{P}_{4}$ and the next $f$
many blocks of $\widetilde{P}_{1}$ and repeat the steps 2, 3 and 4 as
mentioned above. Denote the resulting class as $P_{4}$ and continue till the
number of blocks in $\widetilde{P}_{1}$ becomes less than $f$ .

  6. 6.

Since the number of blocks in $\widetilde{P}_{1}$ is $q-h=q-(e-f)$, in this
way
$r=\left\lfloor\frac{q-(e-f)}{f}\right\rfloor=\left\lfloor\frac{q-e}{f}\right\rfloor+1$
many parallel classes $\widetilde{P}_{r}$ can be modified. Then the
RBD$(X,A)$, where $X=\\{1,2,\ldots,(q-e)(q+f)\\}$ and
$A=\\{P_{2},P_{3},\ldots,P_{r+1}\\}$ is the required design.

Note that in Step 4 above, since each block in the set $S^{1}_{3}$ has $q-e$
elements marked and total number of blocks is $f$, hence total number of
elements removed from $\widetilde{P}_{3}$ are $(q-e)\cdot f$. Thus, we are
basically using a set of $f$ blocks from the parallel classes
$\widetilde{P}_{1}$ to reshape one of the remaining parallel class, into
blocks of size $(q-e)$, having $(q+f)$ blocks. This is achieved by identifying
$(q-e)$ elements of one block from the parallel classes $\widetilde{P}_{1}$,
and in different blocks of parallel class say $\widetilde{P}_{l}$. Thereafter,
we delete these elements from different blocks $\widetilde{P}_{l}$, and add
these elements as a separate block in $\widetilde{P}_{l}$. Thus, the resulting
parallel class will consist of $(q+f)$ blocks, each having $(q-e)$ elements.

We claim that the above design $(X,A)$ is a RBD, such that $|X|=(q-e)(q+f)$
and $A$ consists of $r=\left\lfloor\frac{q-e}{f}\right\rfloor+1$ many parallel
classes each having $(q+f)$ many blocks each of size $(q-e)$, such that the
blocks from different parallel classes have at most one element in common,
i.e., $|b^{l}_{i}\cap b^{m}_{j}|\leq 1\ \forall\ i\neq j$. We formalize this
in the following lemma.

###### Lemma 8

Let $d=(q-e)(q+f)$, for $f,e\in\mathbb{N}$ with $0<f\leq e\leq q$ and $q$ some
power of prime. Then one can construct an RBD$(X,A)$, with $|X|=d$ having
constant block size $(q-e)$ with $\mu=1$, and having at least
$r=\left\lfloor\frac{q-e)}{f}\right\rfloor+1$ many parallel classes.

###### Proof

Refer to Construction 5 above. Since in RBD$(\widetilde{X},\widetilde{A})$ any
pair of blocks from different parallel classes has at most one element in
common, we have $|\bar{b}^{l}_{i}\cap\bar{b}^{m}_{j}|\leq 1\ \forall\ l\neq
m$. Since no element of RBD$(\widetilde{X},\widetilde{A})$ has been deleted or
added to it to obtain RBD$(X,A)$, hence $d=|\widetilde{X}|=|X|=(q-e)(q+f)$.
From the Step 6 of Construction 5, we obtain
$r=\left\lfloor\frac{q-e)}{f}\right\rfloor+1$.

Now we show that Step 2 can be successfully executed. That is, from the set of
$f$ many blocks of $\widetilde{P}_{1}$, it will be possible to mark
$\sum^{q}_{i=1}m^{l}_{i}=(q-e)\cdot f,\ \forall\ l\geq 2$. Note that each
block $\widetilde{b}^{1}_{j},\ j>e$ has exactly one element common with
$\widetilde{b}^{l}_{j},\ j=1,2,\ldots,q,\ l\geq 2$. As $0\leq m^{l}_{j}\leq
f$, corresponding to each block $\widetilde{b}^{l}_{j}$, there will always be
$m^{l}_{j}$ elements on different blocks, which is total $f$ in number. Hence
Step 2 can be successfully executed.

Steps 3 and 4 are related to the construction where elements are to be marked
on blocks in the set $S^{1}_{3}$ having $f$ many blocks. These steps will
finally result into $(q-e)$ elements being marked on each of these blocks. For
this, note that $\sum_{i=1}^{q}m^{l}_{i}=(q-e)f$. That means, if it is not
possible to mark $q-e$ elements on each of the blocks in the set $S^{1}_{3}$,
then on some block there will have more elements marked than $q-e$ and on some
blocks there are less element marked than $q-e$. It is not possible that all
the blocks have less than $(q-e)$ elements marked or all the blocks have more
than $(q-e)$ elements marked as in that case $\sum_{i=1}^{q}m^{l}_{i}<(q-e)f$
or $>(q-e)f$ accordingly.

In case there is a block $\widetilde{b}^{1}_{u},\ \widetilde{b}^{1}_{v}\in
S^{1}_{3}$, such the $\widetilde{b}^{1}_{u}$ has more elements marked than
$(q-e)$ and $\widetilde{b}^{1}_{v}$ has less element marked than $(q-e)$, then
there will exist a block of $\widetilde{P}_{3}$ which have marked element
common with $\widetilde{b}^{1}_{u}$, but non-marked element common with
$\widetilde{b}^{1}_{v}$. Suppose there is no such block in
$\widetilde{P}_{3}$. Then the marked elements of block $\widetilde{b}^{1}_{u}$
(which is $>q-e$) and the unmarked elements of $\widetilde{b}^{1}_{v}$ (which
is $>e$) would all lie on the different blocks of $\widetilde{P}_{3}$.
However, this would imply number of blocks in $|\widetilde{P}_{3}|>q-e+e=q$,
which is a contradiction as $|\widetilde{P}_{3}|=q$.

Finally, we show that $\mu=1$. Note that the blocks which has been added in
the parallel classes
$\widetilde{P}_{2},\widetilde{P}_{3},\ldots,\widetilde{P}_{r+1}$ to construct
the parallel classes $P_{2},P_{3},\ldots,P_{r+1}$ are respectively part of the
blocks of $\widetilde{P}_{1}$. Since any block of $\widetilde{P}_{1}$ has at
most one element common with any other block of $\widetilde{P}_{l},\ l\geq 2$,
$\mu$ will be 1 for RBD$(X,A)$.

Now we can use this RBD$(X,A)$ to construct APMUBs in dimension
$d=|X|=(q-e)(q+f)$ having parameters as given by Theorem 4.1, which we
formally state and prove in the next theorem.

###### Theorem 5.2

Let $d=(q-e)(q+f)$, for some prime power $q$, where $0<f\leq e$ and
$0<(e+f)\leq\frac{3}{2}d^{\frac{1}{2}}$, $e,f\in\mathbb{N}$. Then there exist
at least $r=\left\lfloor\frac{q-e}{f}\right\rfloor+1$ many Almost Perfect
MUBs, with $\Delta=\\{0,\frac{1}{q-e}\\}$,
$\beta=\sqrt{\frac{q+f}{q-e}}=1-\mathcal{O}(d^{-\lambda})\leq 2$, where
$\lambda=\frac{1}{2}$ and $\epsilon=1-\frac{1}{q+f}$. Further, if there exists
a Real Hadamard matrix of order $(q-e)$, then one can construct $r$ many
Almost Perfect Real MUBs with same parameters.

###### Proof

In order to show that we can produce such APMUBs, let us consider an
RBD$(X,A)$ with $|X|=(q-e)(q+f)$ having $r$ parallel classes of constant block
size $(q-e)$, such that between the blocks from different parallel classes,
there is at most one element in common, and hence $\mu=1$.Then using this RBD
along with a Hadamard matrix of order $(q-e)$, we can construct orthonormal
bases following Theorem 4.1. Further, the condition that $\beta<2$ for APMUB
gives $(e+f)\leq\frac{3}{2}d^{\frac{1}{2}}$. In terms of $q$, this inequality
becomes $4e+f\leq 3q$. The parameters of APMUBs are
$\Delta=\\{0,\frac{1}{q-e}\\}$,
$\beta=\sqrt{\frac{q+f}{q-e}}=1+\mathcal{O}(d^{-\frac{1}{2}})\leq 2$ and
sparsity $\epsilon=1-\frac{1}{q+f}$. Further when a real Hadamard matrix of
order $(q-e)$ is available, the same can be used in the construction of
Approximate Real MUBs. Since number of parallel classes $r$ is at least
$\lfloor\frac{q-e}{f}\rfloor+1$, hence we get at least these many APMUBs.

###### Remark 2

Note that $\frac{q-e}{f}$ is $\mathcal{O}(\sqrt{d})$, when $e,f$ are
considered to be constants. That is, in such cases we obtain
$\mathcal{O}(\sqrt{d})$ many APMUBs for the dimension $d$.

Following Theorem 5.2, we can get at least
$r=\left\lfloor\frac{q-e}{f}\right\rfloor+1$ many APMUBs. This can enable us
to beat the Mutually Orthogonal Latin Square (MOLS) Lower Bound construction
for APMUBs (Theorem 5.1), according to which we obtain $N(q+f)+1$ many APMUBs.
Let us present a few illustrative examples in this regard.

  * •

For $d=60=6\cdot 10$, the known value of $N(10)$ is $2$ hence MOLS Lower Bound
construction provides three APMUBs with $\beta$ value of 1.29. On the other
hand, if we use above construction method, by expressing $d=(9-3)(9+1)$, we
obtain $\frac{9-(3-1)}{1}=7$ many APMUBs with $\beta=1.29$. In this case the
number of complex MUBs, that can be constructed following prime factorization
formula, is $3+1=4$ only.

  * •

For $d=24=4\cdot 6$, with $N(6)=1$, the MOLS Lower Bound construction
generates $2$ APRMUBs with $\beta=1.22$. On the other hand, expressing
$d=24=(5-1)(5+1)$ we obtain 5 APRMUBs with $\beta=1.22$. The number of real
MUBs for $d=24$ is 2 boykin2005real  only.

Further, to illustrate the advantage of Construction 5 over Construction 3,
consider the example of $d=2^{2}\cdot 3^{2}\cdot 5\cdot 7$ and different ways
of expressing it as product of two factors, that we considered earlier
following Theorem 5.1.

  * •

Expressing $d=30\cdot 42$, the MOLS Lower Bound construction will provide
$N(42)+1=6$ many APMUBs with $\beta=1.18$, where as expressing
$d=(41-11)(41+1)$ and using Theorem 5.2 will provide 31 many APMUBs with
$\beta=1.18$.

  * •

or expressing $d=28\cdot 45$, the MOLS Lower Bound construction will provide
$N(45)+1=7$ many APRMUB with $\beta=1.27$, where as expressing
$d=(43-15)(43+2)$ and using Theorem 5.2 will provide 15 many APRMUB with
$\beta=1.27$.

Note that, $N(42)=5$ and $N(45)=6$ are the presently known values of the
maximum number of MOLS of these orders HandBook . Here expressing $d=35\cdot
36$, we cannot use Theorem 5.2 as it cannot be expressed as $(q-e)(q+f)$, with
$q$ some power of prime and $0\leq f\leq e$. Thus with this factorization of
$d$, MOLS Lower Bound construction provides $N(36)+1=9$ APMUBs with
$\beta=1.01$.

Note that, $r=\left\lfloor\frac{q-(e-f)}{f}\right\rfloor$, with condition
$0<f\leq e$, is maximum for a given $q$ when $f=e=1$. From the asymptotic
expression of $\beta$, it is clear that for $0<f\leq e$ we will obtain $\beta$
closest to 1 when $e=f=1$. Thus for $e=f=1$, we state the result of APMUB as a
corollary below.

###### Corollary 4

Let $d=q^{2}-1=(q-1)(q+1)$ where $q$ is a prime power. Then one can construct
$q$ many Almost Perfect MUBs with $\Delta=\left\\{0,\frac{1}{q-1}\right\\}$
and $\beta=\sqrt{\frac{q+1}{q-1}}$ with sparsity $\epsilon=1-\frac{1}{q+1}$.
If a real Hadamard Matrix of order $q-1$ exists, then we have $q$ many APRMUBs
with the same parameters.

Our observation made in connection with Hadamard matrix of order $(q-1)$,
constructed through Paley method paley1933orthogonal  after Corollary 3 is
applicable here as well. That is, if $m=\frac{q-3}{2}\equiv 1\bmod 4$ and $m$
is some prime power, then using the Paley Construction paley1933orthogonal ,
one can obtain Hadamard matrix of order $2(m+1)=q-1$. Hence for any prime
power $q\equiv 1\bmod 4$, if $\frac{q-3}{2}$ is also some prime power and is
equivalent to $1\bmod 4$, then the real Hadamard matrix of order $q-1$ will
necessarily exist through the Paley Construction. For all such $q$’s, there
exist $q$ APRMUBs in $\mathbb{R}^{q^{2}-1}$ and the above corollary will
become independent of the Hadamard Conjecture. Examples of such $q$ are
$13,29,5^{3}$ etc.

  * •

For $d=(13-1)(13+1)=2^{3}\cdot 3\cdot 7$, we obtain $13$ many APRMUBs with
$\beta=1.080$. In this case number of real MUBs is only 2 and complex MUBs is
4.

  * •

For $d=(29-1)(29+1)=2^{3}\cdot 3\cdot 5\cdot 7$, we obtain $29$ many APRMUBs
with $\beta=1.035$. In this case also the number of real MUBs is only 2 and
complex MUBs is 4.

The above examples clearly indicates that as $d$ increases, $\beta$ approaches
closer to 1, hence we obtain APRMUBs which are significantly close to the
MUBs.

###  5.3 Some problems that require further attention

It was pointed out in the example constructed above for RBD$(X,A)$, that for
$|X|=d=4\cdot 8=(7-3)(7+1)$, one could construct more number of parallel
classes than $r=\left\lfloor\frac{q-e}{f}\right\rfloor+1=5$ in this case,
$q=7,e=3,f=1$. From our experience of constructing RBD$(X,A)$, for the
situation when $|X|$ can be expressed as $(q-e)(q+e)=q^{2}-e^{2}$, i.e., for
the situation $e=f>0$, there appears to be always more than
$r=\left\lfloor\frac{q}{f}\right\rfloor+1$ many parallel classes. In this
situation it is possible to use other parallel classes, apart from the first
one of $(q^{2},q,1)$-ARBIBD, which enable us to obtain more parallel classes
for RBD$(X,A)$ than $r=\left\lfloor\frac{q}{f}\right\rfloor+1$. A proof of
this in the following form in a general setting might be an interesting open
problem.

Let $d=(q-e)(q+e)$, for $e\in\mathbb{N}$ with $0\leq e\leq q$ and $q$ a prime
power. Then one can construct an RBD$(X,A)$, with $|X|=d$ having constant
block size $(q-e)$ with maximum intersection number $\mu=1$, and having
$r\geq\frac{q}{2}$ many parallel classes.

Further our efforts for the following form of composite $d$ could not result
into number of APMUBs of the order of $\mathcal{O}(\sqrt{d})$, where $q$ is a
prime power.

  * •

For $d=q(q+f)$, RBD having block size $q$, with $q+f$ many blocks in each
parallel classes.

  * •

For $d=(q-e)(q+f),0<e<f$, RBD having block size $q-e$ with $q+f$ many blocks
in each parallel classes.

For the above forms of $d$, we could not construct more number of APMUBs than
what is given by Mutually Orthogonal Latin Square Lower Bound construction.
Further efforts in this direction or new ideas may be required for this. We
believe that it should be possible to improve the MOLS lower bound in such
cases as well.

Note that our construction of APMUBs are very sparse and hence the set of Bi-
angular vectors are very sparse. The sparsity of each vector inner
constructions, is $\epsilon=1-\frac{1}{s}\approx 1-\frac{1}{\sqrt{d}}$. Also
the non zero components of the vectors are all of the same absolute value,
which is $\frac{1}{s-e}\approx\frac{1}{\sqrt{d}}$. Our extensive search of
literature could not find any study on bounds on the cardinality of such kind
of sparse vectors, each having same sparsity. Nevertheless, there are bounds
on the cardinality of flat equiangular lines. Here flat signifies that all the
component of the vectors are of same magnitude. In such situation the
cardinality of set of equiangular lines in $\mathbb{C}^{d}$ is bounded by
$(d^{2}-d-1)$, Refer to (godsil2009equiangular, , Lemma 2.2) which is less
than $d^{2}$, which is cardinality when the constrain of flatness is relaxed.
We similarly believe that the cardinality of such Bi-angular set, with with
such large sparsity would be significantly less than those given in
(delsarte1991bounds, , Table I) and (Calderbank1997, , Equations 3.9,5.9)].
Hence we subsequently intend to study the bounds on the cardinality of the set
of Bi-angular vectors, with large sparsity.

##  6 Conclusion

In this paper we consider construction of APMUBs, which are significantly good
approximation of MUBs. In asymptotic sense, the APMUBs are almost as good as
the MUBs. That is, for a dimension $d$, the value of the dot product between
two vectors from different bases will be very close to $\frac{1}{\sqrt{d}}$,
and in a few cases 0. In this paper we have formalized the definition of
APMUBs and shown that for a good proportion of integers, we can construct
$O(\sqrt{d})$ many APMUBs. Such a generic result is elusive in cases of
perfect MUBs. Thus, for all practical purposes in the domain of quantum
information, or related areas, our construction ideas open up a larger
possibility of obtaining required combinatorial structures. How dense are
these values of $d$ for which we can construct such APMUBs? We note by
preliminary calculations that for $d<N$, there are approximately
$\frac{3}{4}\frac{N}{\log N}$ many such $d$’s, which can be expressed as
$(q-e)q$ with $0\leq e<\frac{3}{4}q$. One may note that this is of the order
of the density of the primes. As the main scope of this paper is understanding
the combinatorial techniques, we leave this as a future research effort.
Another important issue in this regard is that our constructions are directly
related to the concept of Bi-angular vectors. We primarily note that when two
vectors are randomly selected from the set of such Bi-angular vectors, there
exists a very large probability that they will be making an angle of
$\frac{\beta}{\sqrt{d}}$. In fact as $d$ increases, the probability converges
to certainty. This is the scenario that happens in our APMUB related
constructions. We leave this too for future investigation in a disciplined
manner.

##  7 Declarations: Funding and/or Competing interests

No funding was received to assist with the preparation of this manuscript. The
authors have no relevant financial or non-financial interests to disclose. The
authors have no conflict in any form related to this manuscript.

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