$\left|X_{0,B}\right|=q-1$. Then $\left|X_{B}\right|=q(q-1)$ and hence $X_{B}$
is a Frobenius group with elementary abelian kernel $N$ and a cyclic
complement $X_{0,B}$ by Lemma 2.3(2). Since $X_{0}$ contains a unique
conjugate class of cyclic subgroups of order $q-1$, we may assume that
$X_{0,B}=\left\langle\gamma\right\rangle$, where $\gamma$ is represented by
$\mathrm{Diag(}\omega,\omega^{-1}\mathrm{)}$ and $\omega$ is a primitive
element of $GF(q)^{\ast}$.

If $x=(x_{1},x_{2})$ is any non-zero vector of $V$ lying in $B$, then
$B=(x_{1},x_{2})^{\left\langle\gamma\right\rangle}\cup\left\\{(0,0)\right\\}$.
If $x_{1}=0$, then $B=\left\langle(0,1)\right\rangle$ and hence $T_{B}\neq 1$,
whereas $T$ acts block-semiregularly on $\mathcal{D}$ by our assumption. Thus
$x_{1}\neq 0$. Also $x_{2}\neq 0$ by a similar argument.

Let $\varphi\in N$, $\varphi\neq 1$, then there are $\alpha\in X_{0}$ and
$\tau\in T$ such that $\varphi=\alpha\tau$. Therefore
$N=\left\\{1,\alpha^{\gamma^{j}}\tau^{\gamma^{j}}:i=0,...,q-1\right\\}$, since
$N$ and $X_{0,B}$ are the kernel and a complement of the Frobenius group
$X_{B}$ respectively. Thus,
$\left(\alpha\tau\right)\left(\alpha^{\gamma}\tau^{\gamma}\right)=\left(\alpha^{\gamma}\tau^{\gamma}\right)\left(\alpha\tau\right)$,
since $N$ is abelian. It follows that
$\alpha\alpha^{\gamma}T=\alpha^{\gamma}\alpha T$, since $T$ is a normal
subgroup of $X$, hence
$\left(\alpha\alpha^{\gamma}\right)\left(\alpha^{\gamma}\alpha\right)^{-1}\in
T\cap X_{0}$. Thus $\alpha\alpha^{\gamma}=\alpha^{\gamma}\alpha$, since $T\cap
X_{0}=1$. Note that $o(\alpha)=p$, since $o(\varphi)=p$ and since $T$ is a
normal subgroup of $X$ such that $T\cap X_{0}=1$. Therefore $\alpha^{\gamma}$
lies in the same Sylow $p$-subgroup of $X_{0}$ containing $\alpha$. Let $S$ be
such a Sylow $p$-subgroup of $X_{0}$. Then $\gamma\in N_{X_{0}}(S)$, since the
intersection of any two distinct Sylow $2$-subgroups of $X_{0}$ is trivial by
[37, Satz II.8.2.(c)]. Therefore, $\gamma$ preserves
$\left\langle(x_{1},x_{2})\right\rangle$, since
$\mathrm{Fix}(S)=\mathrm{Fix}(\alpha)=\left\langle(x_{1},x_{2})\right\rangle$,
and hence either $x_{1}=0$ or $x_{2}=0$, a contradiction.

  1. (iii).

The blocks of $\mathcal{D}_{X}$ are affine $GF(p)$-subspaces of $V$.

The assertion is true for $p$ odd by (ii) and by Lemma 2.2, since $-1\in
X_{0}$. Hence assume that $p=2$. Since $r_{0}=\lambda_{0}(q+1)$ and
$\lambda_{0}=p^{f_{0}}\leqslant q$, it follows that $\left|X_{0,B}\right|$ is
divisible by $q-1$. Then $X_{B}/T_{B}$ is isomorphic to a subgroup of $X_{0}$
of order $\frac{q}{\left|T_{B}\right|}\left|X_{0,B}\right|$ by Lemma 2.3(2).
Also $q/\left|T_{B}\right|<q$ by (ii). Then $X_{B}/T_{B}\cong D_{2(q-1)}$ and
$\left|T_{B}\right|=q/2$ by [17, Tables 8.1–8.2]. However, this is impossible
by Corollary 2.4. Thus (iii) holds for $q$ even as well.

  1. (iv).

$\mathcal{D}_{X}$ is isomorphic to the $2$-design constructed in Example 1.3.

If $B$ is contained in a $1$-dimensional $GF(q)$-subspace of $V$, then $B$ is
a $1$-dimensional $GF(q)$-subspace of $V$, since $k=q$. Then
$\mathcal{D}_{X}\cong AG_{2}(q)$, whereas $\lambda\geqslant 2$ by our
assumption. Thus $B$ is not contained in any $1$-dimensional subspace of $V$
and hence $q\neq p$. Therefore $B$ is a $2$-dimensional $GF(q^{1/2})$-subspace
of $V$ when $q=p^{2}$ by (iii), hence $SL_{2}(q^{1/2})\trianglelefteq
X_{0,B}\leqslant SL_{2}(q^{1/2}).Z_{(2,q^{1/2}-1)}$ in this case.

Assume that $q\neq p,p^{2}$. Then either $Z_{q-1}\leqslant X_{0,B}$ or $q$ is
odd and $X_{0,B}\cong Q_{q-1}$, or $q$ is a square and
$SL_{2}(q^{1/2})\trianglelefteq X_{0,B}\leqslant
SL_{2}(q^{1/2}).Z_{(2,q^{1/2}-1)}$ by [17, Tables 8.1–8.2], since $q-1$
divides the order of $X_{0,B}$.

Suppose that the former occurs. Since $SL_{2}(q)$ has one conjugacy class of
subgroups isomorphic to $Z_{q-1}$, we may assume that the cyclic group
contained in $X_{0,B}$ is that generated by the element represented by
$\mathrm{Diag(}\omega,\omega^{-1}\mathrm{)}$, where $\omega$ is a primitive
element of $GF(q)^{\ast}$. Let $(v_{1},v_{2})$ be any non-zero vector of $B$.
Then

$B=\left\\{(\omega^{i}v_{1},\omega^{-i}v_{2}):i=1,...,q-1\right\\}\cup\left\\{(0,0)\right\\}\text{.}$

Then $(\omega
v_{1},\omega^{-1}v_{2})+(v_{1},v_{2})=(\omega^{s}v_{1},\omega^{-s}v_{2})$ for
some $s\in\left\\{1,...,q-1\right\\}$, since $B$ is a $GF(p)$-subspace of $V$
by (iii). Then $\omega+1=\omega^{s}$ and $\omega^{-1}+1=\omega^{-s}$ and hence
$\omega^{2}+\omega^{-2}+1=0$. So $\omega^{3}=1$ and $q=4$, whereas $q\neq
p^{2}$ by our assumption.

Assume that $X_{0,B}\cong Q_{q-1}$, $q$ odd. Since $SL_{2}(q)$ has two
conjugacy classes of subgroups isomorphic to $Q_{q-1}$, we may assume that

$X_{0,B}=\left\langle\left(\begin{array}[]{cc}\omega^{2}&0\\\
0&\omega^{-2}\end{array}\right),\left(\begin{array}[]{cc}0&-\varepsilon\\\
\varepsilon^{-1}&0\end{array}\right)\right\rangle\text{,}$

where $\varepsilon$ is either $1$ or $\omega$. Then

$B=\left\\{(\omega^{2i}v_{1},\omega^{-2i}v_{2}),(-\omega^{2i}\varepsilon
v_{2},\left(\omega^{2i}\varepsilon\right)^{-1}v_{1}):i=1,...,(q-1)/2\right\\}\cup\left\\{(0,0)\right\\}\text{,}$

for some vector $(v_{1},v_{2})$ with $v_{1},v_{2}\neq 0$, since $X_{0,B}$ acts
regularly on $B^{\ast}$. If

$(\omega^{2}v_{1},\omega^{-2}v_{2})+(v_{1},v_{2})=(\omega^{2s}v_{1},\omega^{-2s}v_{2})$

for some $s=1,...,(q-1)/2$, then the above argument implies $\omega^{6}=1$ and
hence $q-1\mid 6$, but this contradicts $q\neq p$, $p$ odd. Then

$(\omega^{2}v_{1},\omega^{-2}v_{2})+(v_{1},v_{2})=\left(-\omega^{2i_{0}}\varepsilon
v_{2},\left(\omega^{2i_{0}}\varepsilon\right)^{-1}v_{1}\right)$

for some $i_{0}=1,...,(q-1)/2$, hence $\omega^{4}+\omega^{2}+1=0$ as
$v_{1},v_{2}\neq 0$. So $\omega^{6}=1$ and we again reach a contradiction.

Finally, assume that $q$ is a square and that $SL_{2}(q^{1/2})\trianglelefteq
X_{0,B}\leqslant SL_{2}(q^{1/2}).Z_{(2,q^{1/2}-1)}$. Note that this case also
encloses $q=p^{2}$ with $p$ a prime. Suppose that $q$ is odd and that
$X_{0,B}\cong SL_{2}(q^{1/2}).Z_{q^{1/2}-1}$. Then
$r_{0}=\left[X_{0}:X_{0,B}\right]=q^{1/2}\frac{q+1}{q^{1/2}-1}$, hence
$r=\lambda\frac{q+1}{q^{1/2}-1}$ since $r/r_{0}\mid q$. However this is
impossible since $r=\left[G_{0}:G_{0,B}\right]=\lambda(q+1)$ with $\lambda\mid
q$. Thus $X_{0,B}\cong SL_{2}(q^{1/2})$.

The group $X_{0,B}$ preserves two $(q^{1/2}+1)$-sets
$\mathcal{R},\mathcal{R}^{\prime}$ each of which consisting of $2$-dimensional
$GF(q^{1/2})$-subspaces of $V$. Also $\mathcal{R}$ and $\mathcal{R}^{\prime}$
induce a regulus and its opposite in $PG_{3}(q^{1/2})$ respectively, which are
the two systems of generators of an hyperbolic quadric (e.g. see [35, Section
15.3.III] or [44, Section II.13]). Hence, $\mathcal{R}$ and
$\mathcal{R}^{\prime}$ cover the same subset of points of $V$ and $X_{0,B}$
acts naturally on each member of $\mathcal{R}^{\prime}$. Since the number
conjugacy of $SL_{2}(q)$-classes of subgroups isomorphic to $X_{0,B}$ is
$(2,q^{1/2}-1)$ by [17, Table 8.1], and these are fused in $\Gamma L_{2}(q)$,
we may assume that $X_{0,B}$ is the copy of $SL_{2}(q^{1/2})$ preserving

$\mathcal{R}=\left\\{\left\langle(1,c)\right\rangle_{GF(q)}:c\in
GF(q),\;c^{q^{1/2}+1}=1\right\\}\text{ and }\mathcal{R}^{\prime}=W^{Z},$

where

$W=\left\\{(x,x^{q^{1/2}}c):x,c\in GF(q),\;c^{q^{1/2}+1}=1\right\\}$

and $Z=Z(GL_{2}(q))$ (see [44, Theorem II.13.5]). Since any Sylow $p$-subgroup
of $X_{0,B}$ acts semiregularly on $V\setminus\bigcup\mathcal{R}^{\prime}$, if
$B\cap\left(\bigcup\mathcal{R}^{\prime}\right)=\left\\{(0,0)\right\\}$ then
$q^{1/2}\mid\left|B^{\ast}\right|$, whereas $B^{\ast}$ is coprime to $p$ as
$k=q$. Therefore
$B\cap\left(\bigcup\mathcal{R}^{\prime}\right)\neq\left\\{(0,0)\right\\}$ and
hence there is $z_{0}$ in $Z$ such that $B\cap
W^{z_{0}}\neq\left\\{(0,0)\right\\}$. Actually, $B=W^{z_{0}}$ since $k=q$ and
since $X_{0,B}$ acts transitively on $(W^{z_{0}})^{\ast}$.

  1. (v).

$\mathcal{D}$ is isomorphic to the $2$-design constructed in Example 1.3.

Since $X_{0,B}\cong SL_{2}(q^{1/2})$ by (iii), and the number of conjugacy
$SL_{2}(q)$-classes of subgroups isomorphic to $X_{0,B}$ is $(2,q^{1/2}-1)$,
it follows that $[G_{0}:X_{0}]_{p}=[G_{0,B}:X_{0,B}]_{p}$ and hence that
$r_{p}=[G_{0}:G_{0,B}]_{p}=[X_{0}:X_{0,B}]_{p}=(r_{0})_{p}$. Therefore
$r=r_{0}$, since $r=\frac{\lambda}{\lambda_{0}}r_{0}$ with
$\lambda/\lambda_{0}\mid q$, hence $\mathcal{D}=\mathcal{D}_{X}$ and the
assertion follows from (iii). ∎

######  Lemma 5.4.

If $X_{0}\cong Sp_{n}(q)$, then one of the following holds:

  1. (1)

$n=2$ and $\mathcal{D}$ is isomorphic to the $2$-design of Example 1.3.

  2. (2)

$n=4$ and $\mathcal{D}$ is isomorphic to the $2$-design of Example 1.5.

###### Proof.

Let $x$ be any non-zero vector. Then $C:Y\leqslant X_{0,x}$, where $C$ is a
$p$-group and $Y$ is a isomorphic to $Sp_{n-2}(q)$, by [41, Lemma 4.1.12]. Let
$B$ be any block of $\mathcal{D}$ incident with $0,x$. Then
$P(Y)\leqslant\left|B^{Y}\right|\leqslant q^{n/2}$, where $P(Y)$ denotes the
minimal primitive permutation representation of $Y$. Assume that $n\geqslant
6$. Then $P(Y)=\frac{q^{n-2}-1}{q-1}$, or $2^{n/2-2}(2^{n/2-1}-1)$ according
to whether $q>2$ or $q=2$, respectively by [41, Proposition 5.2.1 and Table
5.2.A]. Then $q=2$ then $n=6$, since $\left|B^{Y}\right|\leqslant 2^{n/2}$ .
However, this is impossible, since $\Phi_{6}^{\ast}(2)=1$ contradicts our
assumption (see remark before Lemma 2.7). Thus $Y$ fixes each block of
$\mathcal{D}$ incident with $0,x$, and hence it lies in the kernel of the
action of $C:Y$ on the set of blocks of $\mathcal{D}$ incident with $0,x$.
Therefore $C:Y$ fixes each block of $\mathcal{D}$ incident with $0,x$. So
$\frac{q^{n}-1}{q-1}\mid r$, which is impossible since $r=p^{f}(q^{n/2}+1)$
and $n>2$. Thus $n=2,4$. Case $n=2$ leads to $2$-design of Example 1.3, since
$Sp_{2}(q)\cong SL_{2}(q)$, which is the assertion (1).

Assume that $n=4$. Suppose that $T$ acts block-semiregularly on $\mathcal{D}$.
Then $q>2$ by Lemma 3.2, since $Sp_{4}(2)^{\prime}\cong A_{6}$. Also $G_{B}$
is isomorphic to a subgroup $J$ of $G_{0}$ of index $q^{2}+1$ by Lemma 2.3(2).
Then $\left[X_{0}:J\cap X_{0}\right]\mid q^{2}+1$, hence $X_{0}\trianglelefteq
J$ by [41, Proposition 5.2.1 and Table 5.2.A]. Therefore
$\left[G_{0}:J\right]\mid(q-1)\cdot(2,q-1)\cdot\log_{p}q$, whereas
$\left[G_{0}:J\right]=q^{2}+1$. Thus $T$ does not act block semiregularly on
$\mathcal{D}$, hence $B$ is a $GF(p)$-subspace of $V$ by Lemma 2.2. Moreover,
either $X_{0,B}\leqslant Sp_{2}(q)\wr Z_{2}$, or $X_{0,B}\cong SO^{+}_{4}(q)$
with $q$ even, by [17, Tables 8.12–8.14], since
$\left[X_{0}:X_{0,B}\right]\mid p^{f}(q^{2}+1)$.

Assume that $X_{0,B}\cong SO^{+}_{4}(q)$ with $q$ even. The group $X_{0,B}$
preserves two $(q+1)$-sets $\mathcal{R},\mathcal{R}^{\prime}$ each of which
consists of $2$-dimensional $GF(q)$-subspaces of $V$. Also $\mathcal{R}$ and
$\mathcal{R}^{\prime}$ induce a regulus and its opposite in $PG_{3}(q)$
respectively, which are the two systems of generators of an hyperbolic
quadric. Hence, $\mathcal{R}$ and $\mathcal{R}^{\prime}$ cover the same subset
of points of $V$ and $X_{0,B}$ induces a copy of $PSL_{2}(q)$ in its
$2$-transitive action of degree $q+1$ on $\mathcal{R}$ and another copy with
the same action on $\mathcal{R}^{\prime}$. Since any Sylow $p$-subgroup of
$X_{0,B}$ does not fix points in $V\setminus\bigcup\mathcal{R}$, but it fixes
at least a point $B^{\ast}$, and since $k=q^{2}$, it follows that
$B\subset\bigcup\mathcal{R}$. Therefore there is $W$ in $\mathcal{R}$ such
that $B\cap W\neq\left\\{(0,0,0,0)\right\\}$. Actually $B=W$, since $k=q^{2}$
and since $X_{0,B,W}$ induces a transitive group on $W^{\ast}$. So $X_{0,B}$
preserves $W$, whereas $X_{0,B}$ induces a transitive group on the $q+1$
members of $\mathcal{R}$. Hence, this case is ruled out.

Assume that $K\trianglelefteq X_{0,B}$, where $K\cong Sp_{2}(q)\times
Sp_{2}(q)$, and $\lambda=q^{2}$. Hence $K$ preserves the sum decomposition
$V=V_{1}\oplus V_{1}^{\perp}$ where $V_{1}$ is a non-degenerate with respect
to the symplectic form preserved by $X_{0}$. It follows that $K\leqslant
GL_{m}(p)$, since $B$ is a $GF(p)$-subspace of $V$ and since $q^{2}=p^{m}$.
Assume that $p^{m}-1$ has primitive prime divisor $w\,$. Then
$w\mid\left|K(B)\right|$, since $(q+1)^{2}\mid\left|K\right|$ and since
$K\leqslant GL_{m}(p)$. Then either $Sp_{2}(q)\times 1\leqslant K(B)$ or
$1\times Sp_{2}(q)\leqslant K(B)$, since $K(B)\trianglelefteq K$. Therefore
either $B=V_{1}$ or $B=V_{1}^{\perp}$ respectively, and hence $\mathcal{D}$ is
isomorphic to the $2$-design of Example 1.5, which is the assertion (2).

Assume that $p^{m}-1$ does not have primitive prime divisors. Then $m=2$ and
$q=p$ is a Mersenne prime by [44, Theorem II.6.2], since $q^{2}=p^{m}$. Then
$K^{B}\leqslant GL_{2}(p)$ and hence either $Sp_{2}(p)\times 1\leqslant K(B)$
or $1\times Sp_{2}(p)\leqslant K(B)$. Thus either $B=V_{1}$ or
$B=V_{1}^{\perp}$ respectively, and we again obtain the assertion (2). ∎

######  Lemma 5.5.

If $X_{0}\cong SU_{n}(q^{1/2})$, $q$ square, then $n=3$.

###### Proof.

Assume that $X_{0}\cong SU_{n}(q^{1/2})$, $q$ square. If $x$ is a non-zero
isotropic vector of $V$, then $C:Y\leqslant X_{0,x}$, where $C$ is a $p$-group
and $Y$ is a isomorphic to $SU_{n-2}(q^{1/2})$, by [41, Lemma 4.1.12]. Let $B$
be any block of $\mathcal{D}$ incident with $0,x$. Then
$P(Y)\leqslant\left|B^{Y}\right|\leqslant q^{n/2}$, where $P(Y)$ denotes the
minimal primitive permutation representation of $Y$. Assume that $n>7$. Then

$q^{n/2}\geqslant\lambda\geqslant\left|B^{Y}\right|>\frac{\left(q^{(n-2)/2}+1\right)\left(q^{(n-2)/2-1}-1\right)}{q-1}$

by [41, Proposition 52.1 and Table 5.2.A], since $n$ is odd, but the previous
inequality has no solutions. Thus $Y$ fixes each block of $\mathcal{D}$
incident with $0,x$, and hence it lies in the kernel of the action of $C:Y$ on
the set of blocks of $\mathcal{D}$ incident with $0,x$. Therefore, $C:Y$ fixes
each block of $\mathcal{D}$ incident with $0,x$. So $C:Y\leqslant X_{0,B}$ and
hence $\frac{\left(q^{n/2}+1\right)\left(q^{n/2-1}-1\right)}{q-1}\mid r$,
which is impossible since $r=p^{f}(q^{n/2}+1)$ and $n>7$.

Assume that $n=5,7$. Then $X_{0,B}\leqslant GU_{n-1}(q^{1/2})<X_{0}$ by [17,
Tables 8.20–8.21 and 8.37–8.38], since $\left[X_{0}:X_{0,B}\right]\mid
p^{f}(q^{n/2}+1)$. Moreover, $p^{f}\geqslant q^{(n-1)/2}$ and
$SU_{n-1}(q^{1/2})\trianglelefteq X_{0,B}$. Since the non-zero isotropic
vectors of $V$ lie in a unique $G_{0}$-orbit of length
$(q^{n/2}+1)(q^{n/2-1}-1)$, it follows that $\left|B\cap
x^{G_{0}}\right|=q^{n/2-1}-1$ by Lemma 2.1. Hence the copy of
$SU_{n-1}(q^{1/2})$ inside $X_{0,B}$ fixes $B\cap x^{G_{0}}$ pointwise, since
the minimal primitive permutation representation of $SU_{n-1}(q^{1/2})$ is
greater than $q^{n/2-1}-1$ again by [41, Proposition 52.1 and Table 5.2.A].
So, $SU_{n-1}(q^{1/2})\leqslant X_{0,x}$ with $x$ a non-zero isotropic vector
of $V$, but this contradicts [17, Tables 8.20 and 8.37]. Thus $n=3$, which is
the assertion. ∎

######  Lemma 5.6.

If $X_{0}\cong SU_{n}(q^{1/2})$, then $n=3$ and $\mathcal{D}$ is a
$2$-$(q^{3},q^{3/2},q)$ design isomorphic to that constructed in Example 1.3.

###### Proof.

Set $q=s^{2}$, then $X_{0}\cong SU_{3}(s)$, $V=V_{3}(s^{2})$,
$r=\lambda(s^{3}+1)$ and $\lambda\mid s^{3}$. The set of blocks incident with
$0$ is split into $X_{0}$-orbits of the same length as $X_{0}\trianglelefteq
G_{0}$, hence $\left[X_{0}:X_{0,B}\right]\theta=r$ where $B$ is any block
incident with $0$ and $\theta$ is a positive divisor of $r$. Thus
$\theta\mid\left(s+1\right)\cdot 2\log_{p}(s)$, since
$\theta\mid\frac{s^{2}-1}{(3,s+1)}\cdot\left|\mathrm{Out(}PSU_{3}(s)\mathrm{)}\right|$
and $\left|\mathrm{Out(}PSU_{3}(s)\mathrm{)}\right|=(3,s+1)\cdot
2\log_{p}(s)$. Thus

$\left|X_{0,B}\right|=\theta\frac{s^{3}}{\lambda}(s^{2}-1)\text{ with
}\theta\mid\left(s+1\right)\cdot 2\log_{p}(s).$ (5.1)

and by [17, Tables 8.5–8.6] one of the following holds:

  1. (1)

$X_{0,B}\leqslant X_{0,\left\langle x\right\rangle}$ for some isotropic
$1$-dimensional $GF(s^{2})$-subspace $\left\langle x\right\rangle$ of $V$;

  2. (2)

$X_{0,B}\leqslant(Z_{s+1}\times Z_{s+1}).S_{3}$, with $s=5,7$, and
$\lambda=s^{3}$;

  3. (3)

$SU_{2}(s)\trianglelefteq X_{0,B}\leqslant GU_{2}(s)$;

  4. (4)

$SO_{3}(s)\trianglelefteq X_{0,B}\leqslant Z_{(3,s+1)}\times SO_{3}(s)$ with
$s$ odd;

  5. (5)

$X_{0,B}\leqslant 3_{+}^{1+2}:Q_{8}$, with $s=5$, and $\lambda=5^{3}$;

  6. (6)

$X_{0,B}\cong PSL_{2}(7)$ for $s=13$, $\theta=1$ and $\lambda=13^{3}$.

Now, we are going to show that (1)–(3) and (5)–(6) are ruled out and that (4)
leads to a $2$-design isomorphic to that constructed in Example 1.3.

It is well known that $G$ is a rank $3$ group and that the non-zero isotropic,
and non-isotropic vectors of $V$ are the $G_{0}$-orbits partitioning
$V^{\ast}$ and have have length $(s^{3}+1)(s^{2}-1)$ and $s^{2}(s^{3}+1)(s-1)$
respectively. Let $x_{1},x_{2}$ be representatives of such orbits, then
$\left|B\cap x_{1}^{G_{0}}\right|=s^{2}-1$ and $\left|B\cap
x_{2}^{G_{0}}\right|=s^{2}(s-1)$ by Lemma 2.1.

  1. (i).

Case (1) cannot occur.

Assume that $X_{0,B}\leqslant X_{0,\left\langle x\right\rangle}$ for some
isotropic $1$-dimensional $GF(s^{2})$-subspace $\left\langle x\right\rangle$
of $V$. Then $X_{0,B}\cong Z_{s^{2}-1}$, or $X_{0,B}\cong Z(U):Z_{s^{2}-1}$ or
$X_{0,B}\cong U:Z_{s^{2}-1}$, where $U$ is a Sylow $p$-subgroup of $X_{0}$,
since $Z(U):Z_{s-1}$ and $U/Z(U):Z_{s^{2}-1}$ are Frobenius groups. In each
case $X_{0}$ contains a unique conjugate class of subgroups isomorphic to
$X_{0,B}$. Thus $\left[G_{0}:X_{0}\right]=\left[G_{0,B}:X_{0,B}\right]$ and
hence $r=\left[G_{0}:G_{0,B}\right]=\left[X_{0}:X_{0,B}\right]$. Therefore
$X=T:X_{0}$ acts flag-transitively on $\mathcal{D}$.

Assume that $X_{0,B}\cong U:Z_{s^{2}-1}$. Since $X_{0,B}$ is maximal in
$X_{0}$, it follows from Lemma 2.3(2) that $B$ is a $GF(p)$-subspace of $V$.
Since $U$ preserves an isotropic $1$-dimensional subspace of $V$ and acts
semiregularly on the set of the remaining non-zero isotropic vectors and since
$\left|B\cap x_{1}^{G_{0}}\right|=s^{2}-1$, it results that $B\cap
x_{1}^{G_{0}}=\left\langle z\right\rangle^{\ast}_{GF(s^{2})}$, where
$\left\langle z\right\rangle_{GF(s^{2})}$ is the isotropic $1$-dimensional
subspace of $V$ preserved by $U$.

Since $\left|B\cap x_{2}^{G_{0}}\right|=s^{2}(s-1)$ and
$\left|U\right|=s^{3}$, the stabilizer in $U$ of a non-isotropic vector $u$ in
$B$ is non-trivial. Then $\left\langle u\right\rangle_{GF(s^{2})}$ lies in the
line of $PG_{2}(s^{2})$ tangent in $\left\langle z\right\rangle_{GF(s^{2})}$
to the Hermitian unital preserved by $X_{0}$. Therefore
$\left|u^{X_{0,B}}\right|=s^{2}(s+1)$ by [37, Satz II.10.12], whereas
$u^{X_{0,B}}\subseteq B\cap x_{2}^{G_{0}}$ with $\left|B\cap
x_{2}^{G_{0}}\right|=s^{2}(s-1)$. So, this case is ruled out.

Assume that $X_{0,B}\cong Z(U):Z_{s^{2}-1}$. Therefore $r=s^{2}(s^{3}+1)$.
Thus $\left|T_{B}\right|\geqslant s$ by Lemma 2.3(1), since
$p^{m}=q^{3/2}=s^{3}$. Now $X_{B}/T_{B}$ is isomorphic to a subgroup of
$X_{0}$ and contains a copy of $X_{0,B}$ as subgroup of index
$s^{3}/\left|T_{B}\right|$ by Lemma 2.3(2). Thus either $\left|T_{B}\right|=s$
or $\left|T_{B}\right|=s^{3}$, since $X_{0,B}\cong Z(U):Z_{s^{2}-1}$, since
$U:Z_{s^{2}-1}$ is the unique maximal subgroup of $X_{0}$ containing $X_{0,B}$
and since $U/Z(U):Z_{s^{2}-1}$ is a Frobenius group. In both cases there is a
non-trivial subgroup of order a divisor of $\frac{s+1}{(2,s+1)}$ fixing at
least a non-zero vector in $0^{T_{B}}$, but this is impossible for $s\neq 3$
(e.g. see [37, Satz II.10.12]). So $s=3$ and $X_{0,B}\cong Z_{3}:Z_{8}$, but
there is a cyclic groups of order at least $4$ fixing a point in $0^{T_{B}}$
and this case is excluded.

Assume that $X_{0,B}\cong Z_{s^{2}-1}$. We may argue as in the previous case
and we get that $\left|T_{B}\right|$ is either $1$ or $s$ or $s^{3}$. In the
two latter cases is a non-trivial subgroup of order a divisor of
$\frac{s+1}{(2,s+1)}$ fixing at least a non-zero vector in $0^{T_{B}}$ and we
again reach a contradiction as above. Thus $\left|T_{B}\right|=1$ and hence
$X_{B}\cong X_{0,\left\langle x\right\rangle}$ by Lemma 2.3(2) and by [17,
Tables 8.5–8.6]. It follows from [39, Theorem 2.14.(i)] that
$H^{1}(X_{0},V)=0$, also $\left|H^{1}(X_{0,\left\langle
x\right\rangle},V)\right|\leqslant\left|H^{1}(X_{0},V)\right|$ by [22,
2.3.(g)], as $\left\langle x\right\rangle$ is isotropic. Then
$X_{B}=X_{0,\left\langle x\right\rangle}^{\tau}$ for some $\tau\in T$ by [10,
17.10], hence $X_{0,\left\langle x\right\rangle}$ acts transitively on
$B^{\tau^{-1}}$. Then all the vectors contained in $B^{\tau^{-1}}$ are either
non-zero isotropic or non-isotropic. However, this is impossible since $X$
acts flag-transitively on $\mathcal{D}$ and each block incident with $0$ must
contain (non-zero) isotropic vectors as well as non-isotropic ones.

  1. (ii).

Cases (2), (5) and (6) cannot occur.

Assume that (2) occurs. If $s=5$, from (5.1) we derive that either
$X_{0,B}\cong S_{4}$ or $3_{+}^{1+2}\leqslant X_{0,B}$. In the latter case
$X_{0,B}$ contains the center of $SU_{3}(5)$, which fixes at least two points
on $B$ since $k=125$ is equivalent to $2$ modulo $3$, and we reach a
contradiction. For the same reason $B$ cannot be a $GF(5)$-subspace of $V$
when $X_{0,B}\cong S_{4}$. On the other hand, $X_{B}/T_{B}$ is isomorphic to a
subgroup of $X_{0}$ containing an isomorphic copy of $X_{0,B}$ as a subgroup
of index $5^{3}/\left|T_{B}\right|$ by Lemma 2.3(2). It follows from [21] that
$\left|T_{B}\right|=5^{2}$, since $B$ cannot be a $GF(5)$-subspace of $V$.
Then $X_{0,B}$ acts on $W$, where $W=0^{T_{B}}$, which is not a
$1$-dimensional $GF(25)$-subspace of $V$, since $X_{0,B}$ does not fix points
of $PG_{2}(25)$. Hence $W$ intersects six points of $PG_{2}(25)$ each of them
in a non-zero vector of $V$. Also, since there are no subgroups of $GL_{2}(5)$
isomorphic to $S_{4}$, it follows that $X_{0,B}(W)\neq 1$. Hence $Z_{2}\times
Z_{2}\leqslant X_{0,B}(W)$, since $X_{0,B}\cong S_{4}$. Then $Z_{2}\times
Z_{2}$ fixes six points of $PG_{2}(25)$, namely those intersecting $W$ in a
non-zero vector of $V$, but this is not the case since $Z_{2}\times Z_{2}$
acts faithfully on $PG_{2}(25)$ as a groups of homologies in a triangular
configuration. Thus $s=5$ is ruled out.

Suppose that $s=7$. Hence $(Z_{4}\times Z_{4}).S_{3}\leqslant X_{0,B}$, since
$48$ divides the order of $X_{0,B}$ by (5.1). By Lemma 2.3(2) $X_{B}/T_{B}$ is
isomorphic to a subgroup of $X_{0}$ containing an isomorphic copy of $X_{0,B}$
as a subgroup of index $7^{3}/\left|T_{B}\right|$. It follows from [21] that
$\left|T_{B}\right|\geqslant 7^{2}$, since $(Z_{4}\times Z_{4}).S_{3}\leqslant
X_{0,B}$. If $\left|T_{B}\right|=7^{2}$, the admissible candidates of maximal
subgroups of $X_{0}$ containing an isomorphic copy of $X_{B}/T_{B}$ are either
$SL_{2}(7):Z_{8}$ or $PGL_{2}(7)$ by [21], however none of these contains a
subgroup isomorphic to $(Z_{4}\times Z_{4}).S_{3}$. Therefore
$\left|T_{B}\right|=7^{3}$ and hence $B$ is a $3$-dimensional $GF(7)$-subspace
of $V$. So $X_{0,B}\leqslant PGL_{2}(7)$ by [21], and we again reach a
contradiction as $PGL_{2}(7)$ does not contains subgroups isomorphic to
$(Z_{4}\times Z_{4}).S_{3}$, whereas $X_{0,B}$ does it. This completes the
exclusion of (2).

Assume that (5) occurs. From (5.1) we derive that either $X_{0,B}\cong
Z_{3}\times Q_{8}$ or that $X_{0,B}\cong 3_{+}^{1+2}:Q_{8}$. In each case
$X_{0,B}$ contains the center of $SU_{3}(5)$ and we reach a contradiction as
in case (2).

Assume that (6) occurs. By Lemma 2.3(2) $X_{B}/T_{B}$ is isomorphic to a
subgroup of $X_{0}$ containing an isomorphic copy of $X_{0,B}$ as a subgroup
of index $13^{3}/\left|T_{B}\right|$, it follows that
$\left|T_{B}\right|=13^{3}$, hence $B$ is a $3$-dimensional $GF(13)$-subspace
of $V$, since $X_{0,B}\cong PSL_{2}(7)$. Then $X_{0,B}$ lies in a
$\mathcal{C}_{5}$-member of $X_{0}$, but this is impossible by [17, Table
8.5].

  1. (iii).

Case (3) cannot occur.

Assume that $SU_{2}(s)\trianglelefteq X_{0,B}\leqslant GU_{2}(s)$. Then
$r=\left[G_{0}:G_{0,B}\right]=\left[X_{0}:X_{0,B}\right]$ by [17, Table 8.5],
hence $X=T:X_{0}$ acts flag-transitively on $\mathcal{D}$. Moreover,
$r=s^{2}(s^{3}+1)$ and therefore $X_{0,B}\cong SU_{2}(s)$. By Lemma 2.3(2)
$X_{B}/T_{B}$ is isomorphic to a subgroup of $X_{0}$ containing an isomorphic
copy of $X_{0,B}$ as a subgroup of index $s^{3}/\left|T_{B}\right|$, it
follows that $\left|T_{B}\right|=s^{3}$ and hence $B$ is a $GF(p)$-subspace of
$V$.

Each Sylow $p$-subgroup of $X_{0,B}$ fixes a unique isotropic $1$-dimensional
$GF(s^{2})$-subspace of $V$ and distinct Sylow $p$-subgroups of $X_{0,B}$ fix
distinct isotropic $1$-dimensional $GF(s^{2})$-subspaces of $V$. Hence $B$
intersects at least $s+1$ isotropic $1$-dimensional $GF(s^{2})$-subspaces of
$V$, say $\left\langle y_{i}\right\rangle_{GF(s^{2})}$, with $i=1,...,s+1$,
each of them in at least a non-zero vector, that we may assume to be $y_{i}$.
Moreover, $X_{0,B}$ permutes $2$-transitively the set $\\{\left\langle
y_{i}\right\rangle_{GF(s^{2})}:\;i=1,...,s+1\\}$. Since $X_{0,B,\left\langle
y_{i}\right\rangle_{GF(s^{2})}}$ partitions the non-zero vectors of
$\left\langle y_{i}\right\rangle_{GF(s^{2})}$ into $s+1$ orbits each of length
$s-1$, since $\left|B\cap x_{1}^{X_{0}}\right|=s^{2}-1$ and since
$B\cap\left\langle y_{i}\right\rangle_{GF(s^{2})}$ is a $GF(p)$-subspace of
$V$, it results that $B\cap\left\langle
y_{i}\right\rangle_{GF(s^{2})}=\left\langle y_{i}\right\rangle_{GF(s)}$ for
each $i=1,...,s+1$ and that $B\cap
x_{1}^{X_{0}}=\bigcup^{s+1}_{i=1}\left\langle
y_{i}\right\rangle^{\ast}_{GF(s)}$.

If $B\cap x_{1}^{X_{0}}\not\subset\left\langle
y_{1},y_{2}\right\rangle^{\ast}_{GF(s^{2})}$, then there is $3\leqslant
i_{0}\leqslant s+1$ such that $B=\left\langle
y_{1},y_{2},y_{i_{0}}\right\rangle_{GF(s)}$, as $k=s^{3}$. Then $X_{0,B}$ lies
in a $\mathcal{C}_{5}$-member of $X_{0}$, but this is impossible by [17, Table
8.5]. Thus $B\cap x_{1}^{X_{0}}\subset\left\langle
y_{1},y_{2}\right\rangle^{\ast}_{GF(s^{2})}$ and hence $B\cap
x_{1}^{X_{0}}=\left\langle y_{1},y_{2}\right\rangle^{\ast}_{GF(s)}$.

It is not difficult to see that $X_{0,B}$ preserves the decomposition
$V=\left\langle y_{1},y_{2}\right\rangle_{GF(s^{2})}\oplus\left\langle
y_{1},y_{2}\right\rangle^{\perp}_{GF(s^{2})}$ and that each $X_{0,B}$-orbit on
$V\setminus(\left\langle y_{1},y_{2}\right\rangle_{GF(s^{2})}\cup\left\langle
y_{1},y_{2}\right\rangle^{\perp}_{GF(s^{2})})$ is of length $s(s^{2}-1)$.
Hence $B\subset(\left\langle
y_{1},y_{2}\right\rangle_{GF(s^{2})}\cup\left\langle
y_{1},y_{2}\right\rangle^{\perp}_{GF(s^{2})})$, since $k=s^{3}$ and since
$\left|B\cap\left\langle y_{1},y_{2}\right\rangle_{GF(s^{2})}\right|\geqslant
s^{2}$. Actually, $B\subset\left\langle y_{1},y_{2}\right\rangle_{GF(s^{2})}$
since $B$ is a $GF(p)$-subspace of $V$ sharing at least $s^{2}$ vectors with
$\left\langle y_{1},y_{2}\right\rangle_{GF(s^{2})}$. The $X_{0,B}$-orbits in
$\left\langle y_{1},y_{2}\right\rangle_{GF(s^{2})}\setminus\left\langle
y_{1},y_{2}\right\rangle_{GF(s)}$ have length $s(s^{2}-1)$ and each of these
must be disjoint from $B$, so we reach a contradiction since $k=s^{3}$ and
since $\left\langle y_{1},y_{2}\right\rangle_{GF(s)}\subset B$. Thus, this
case is excluded.

  1. (iv).

Lemma’s statement holds.

Assume that $SO_{3}(s)\trianglelefteq X_{0,B}\leqslant Z_{(3,s+1)}\times
SO_{3}(s)$ with $s$ odd. Suppose that $s\equiv 2\pmod{3}$ and that
$X_{0,B}\cong Z\times SO_{3}(s)$, where $Z\cong Z_{3}$. Then $Z$ fixes at
least a non-zero vector of $B\cap x_{2}^{G_{0}}$, as $\left|B\cap
x_{2}^{G_{0}}\right|=s^{2}(s-1)$, and we reach a contradiction as $Z$ is the
center of $SU_{3}(s)$. Thus $X_{0,B}\cong SO_{3}(s)$.

Since $\left|G_{0,B}\right|_{p}=s\cdot\frac{\left(\log_{p}(s)\right)_{p}}{a}$
for some positive divisor $a$ of $\left(\log_{p}(s)\right)_{p}$ by (5.1),
hence $\lambda=s^{2}\cdot a$. Then $T_{B}\neq 1$ by Lemma 2.3(1), since
$p^{m}=q^{3/2}=s^{3}$ and since $a<s$. Then $\Omega_{3}(s)$ acts on
$0^{T_{B}}$. Clearly $\Omega_{3}(s)$ does not fix a non-zero vectors in
$0^{T_{B}}$ by [17, Table 8.5]. Thus $\left|0^{T_{B}}\right|\geqslant s^{3}$
by [41, Proposition 5.4.11]. Therefore $B=0^{T_{B}}$, as $k=s^{3}$ and
$0^{T_{B}}\subseteq B$, hence $B$ is a $GF(p)$-subspace of $V$.

$X_{0,B}$ acts faithfully on $PG_{2}(s^{2})$ and one $X_{0,B}$-orbit on
$PG_{2}(s^{2})$ is $\mathcal{H}\cap\mathcal{C}$, where $\mathcal{H}$ is the
Hermitian unital of order $s$ preserved by $G_{0}$ and where $\mathcal{C}$ is
a conic preserved by $X_{0,B}$. Each Sylow $p$-subgroup of $\Omega_{3}(s)$
fixes each non-zero vector of a point $\left\langle
y_{0}\right\rangle_{GF(s^{2})}$ of $\mathcal{H}\cap\mathcal{C}$ and acts
regularly on $(\mathcal{H}\cap\mathcal{C})\backslash\\{\left\langle
y_{0}\right\rangle_{GF(s^{2})}\\}$. Moreover, the $X_{0,B}$-orbits on
$\mathcal{H}\setminus\mathcal{C}$ have length $s(s^{2}-1)/2$. Therefore either
$B\cap x^{G_{0}}=\left\langle y_{0}\right\rangle_{GF(s^{2})}^{\ast}$, or
$\left|B\cap\left\langle y\right\rangle_{GF(s^{2})}\right|>0$ for each
$\left\langle y\right\rangle_{GF(s^{2})}\in\mathcal{H}\cap\mathcal{C}$, since
$\left|B\cap x_{1}^{G_{0}}\right|=s^{2}-1$. If $B\cap
x_{1}^{G_{0}}=\left\langle y_{0}\right\rangle_{GF(s^{2})}^{\ast}$ then
$SO_{3}(s)\leqslant X_{0,\left\langle y_{0}\right\rangle_{GF(s^{2})}}$, which
is impossible. Therefore, $\left|B\cap\left\langle
y\right\rangle_{GF(s^{2})}\right|>0$ for each $\left\langle
y\right\rangle_{GF(s^{2})}\in\mathcal{H}\cap\mathcal{C}$. Moreover, since
$\Omega_{3}(s)_{\left\langle y\right\rangle_{GF(s^{2})}}$ is a Frobenius group
with cyclic complement of order $(s-1)/2$ acting semiregularly on
$\left\langle y\right\rangle_{GF(s^{2})}^{\ast}$, and since $B$ is a
$GF(p)$-subspace of $V$, it follows that $B\cap\left\langle
y\right\rangle_{GF(s^{2})}$ is a $GF(s)$-subspace of $\left\langle
y\right\rangle_{GF(s^{2})}$. Thus $B$ is a $3$-dimensional $GF(s)$-subspace of
$V$, since $\mathcal{C}$ is a conic, and hence $B$ induces a Baer subplane of
$PG_{2}(s^{2})$. Moreover, $B$ is the unique $3$-dimensional $GF(s)$-subspace
of $V$ preserved by $X_{0,B}$. Thus, we assume that $B$ is as in Example 1.4,
since the $(3,s+1)$ conjugacy $X_{0}$-classes of subgroup isomorphic to
$X_{0,B}$ are fused in $\mathrm{Aut}(X_{0})\cong\Gamma U_{3}(s)$ (see [17,
Table 8.5]). Therefore, $\mathcal{D}$ is a $2$-$(q^{3},q^{3/2},q)$ design
isomorphic to that constructed in Example 1.3. This completes the proof. ∎

######  Lemma 5.7.

If $X_{0}\cong\Omega_{n}^{-}(q)$, then $n=4$.

###### Proof.

Assume that $X_{0}\cong\Omega_{n}^{-}(q)$. Then $C:Y\leqslant X_{0,x}$, where
$C$ is a $p$-group and $Y$ is a isomorphic to $\Omega_{n-2}^{-}(q)$, by [41,
Lemma 4.1.12]. Let $B$ be any block of $\mathcal{D}$ incident with $0,x$. Then
$P(Y)\leqslant\left|B^{Y}\right|$, where $P(Y)$ denotes the minimal primitive
permutation representation of $Y$. Hence

$q^{n/2}\geqslant\lambda\geqslant\left|B^{Y}\right|>\frac{\left(q^{(n-2)/2}+1\right)\left(q^{(n-2)/2-1}-1\right)}{q-1}$
(5.2)

by [41, Proposition 52.1 and Table 5.2.A], and we reach a contradiction for
$n>6$. Thus $Y$ fixes each block of $\mathcal{D}$ incident with $0,x$ for
$n>6$, and hence it lies in the kernel of the action of $C:Y$ on the set of
blocks of $\mathcal{D}$ incident with $0,x$. Therefore, $C:Y$ fixes each block
of $\mathcal{D}$ incident with $0,x$. So
$\frac{\left(q^{n/2}+1\right)\left(q^{n/2-1}-1\right)}{q-1}\mid r$, which
cannot occur since $r=p^{f}(q^{n/2}+1)$ and $n>6$. Thus $n\leqslant 6$.

Assume that $n=6$. Then $\left[X_{0}:X_{0,B}\right]\mid p^{f}(q^{3}+1)$ and
hence $p^{f}\mid q^{3}$. Then either $X_{0,B}\leqslant M$, where $M$ is either
$\Omega_{5}(q).Z_{2}$ or $Sp_{4}(q)$ according to whether $q$ is odd or even,
respectively, by [17, Tables 8.33–8.34]. Since
$\left[X_{0}:M\right]=q^{2}\frac{q^{3}+1}{(2,q-1)}$, it follows that
$\left[M:X_{0,B}\right]\mid\frac{p^{f}}{q^{2}}(2,q-1)$ and hence
$M^{\prime}\leqslant X_{0,B}\leqslant M^{\prime}.Z_{(2,q-1)}$ by [41,
Proposition 52.1 and Table 5.2.A], since $r=p^{f}(q^{3}+1)$. Thus
$\lambda=q^{2}$.

The group $G_{B}/T_{B}$ is isomorphic to a subgroup $J$ of $G_{0}$ such that
$[G_{0}:J]=\frac{p^{t}}{q}(q^{3}+1)$ and $J$ contains an isomorphic copy of
$G_{0,B}$ by Lemma 2.3(2). Then $\left[X_{0}:J\cap
X_{0}\right]\mid\frac{p^{t}}{q}(q^{3}+1)$ and arguing as above with $J\cap
X_{0}$ in the role of $X_{0,B}$ we deduce that $k=p^{t}=q^{3}$. Therefore, $B$
is a $GF(p)$-subspace of $V$ and hence $(X_{0,B})^{B}\leqslant GL_{m}(p)$.
Since the order of $X_{0,B}$ is divisible by a primitive prime divisor of
$p^{4m/3}-1$, being $q^{6}=p^{2m}$, whereas the order of $GL_{m}(p)$ is not,
it follows that $\left(X_{0,B}\right)^{\prime}$ fixes $B$ pointwise. So
$B\subseteq\left\langle z\right\rangle_{GF(q)}$ for some non-singular vector
$z$ of $V$, and we reach a contradiction since $k=q^{3}$.

Finally, $n=2$ cannot occur. Indeed, in this case
$X_{0}\cong\Omega_{2}^{-}(q)\cong D_{2\frac{q+1}{(2,q-1)}}\leqslant\Gamma
L_{1}(q^{2})$ and it contradicts the definition of $q$. ∎

######  Lemma 5.8.

$X_{0}$ is not isomorphic to $\Omega_{n}^{-}(q)$, $n\geqslant 2$.

###### Proof.

$X_{0}\cong\Omega_{4}^{-}(q)$ by Lemma 5.7. Since $r=\lambda(q^{2}+1)$, where
$\lambda\mid q^{2}$, and since $X_{0}\trianglelefteq G_{0}$, it follows that
$\left[X_{0}:X_{0,B}\right]\theta=\lambda(q^{2}+1)$ for some positive integer
$\theta$. On the other hand $\theta\mid[G_{0}:X_{0}]$ and hence
$\theta\mid(q-1)\cdot(2,q-1)\cdot m$, as $q^{2}=p^{m}$. Thus
$\theta\mid(2m,\lambda(q^{2}+1))$ and hence

$\left|X_{0,B}\right|=\frac{q^{2}}{\lambda}\frac{q^{2}-1}{(2,q-1)}\theta\text{.}$
(5.3)

Therefore, by [17, Table 8.17] one of the following holds:

  1. (1)

$X_{0,B}\leqslant X_{0,\left\langle x\right\rangle_{GF(q)}}$, where $x$ is a
non-zero singular vector of $V$;

  2. (2)

$PSL_{2}(q)\trianglelefteq X_{0,B}\leqslant$ $PGL_{2}(q)$;

  3. (3)

$X_{0,B}\cong D_{2\frac{(q^{2}-1)}{(2,q-1)}}$;

  4. (4)

$X_{0,B}\leqslant A_{5}$ and $q=3$.

We are going to prove that none of the cases (1)–(4) occurs.

  1. (i).

Case (1) and (4) cannot occur.

Assume that $X_{0,B}\leqslant X_{0,\left\langle x\right\rangle_{GF(q)}}$. Then
either $X_{0,B}=X_{0,\left\langle x\right\rangle_{GF(q)}}$ or $X_{0,B}$ is
cyclic of order $\frac{q^{2}-1}{(2,q-1)}$ by (5.3), since $X_{0,\left\langle
x\right\rangle_{GF(q)}}$ is a Frobenius group of order
$q^{2}\frac{q^{2}-1}{(2,q-1)}$ with a cyclic complement of order
$\frac{q^{2}-1}{(2,q-1)}$. In both cases $X_{0}$ contains a unique conjugacy
class of subgroups isomorphic to $X_{0,B}$. Therefore $X=TX_{0}$ acts flag-
transitively on $\mathcal{D}$. So $X_{0,B}\neq X_{0,\left\langle
x\right\rangle_{GF(q)}}$ since $\lambda\geqslant 2$. Thus $X_{0,B}$ is cyclic
of order $\frac{q^{2}-1}{(2,q-1)}$ and hence $r=q^{2}(q^{2}+1)$ and
$\lambda=q^{2}$.

Since $X_{B}/T_{B}$ is isomorphic to a subgroup of $X_{0}$ containing
$X_{0,B}$ as subgroup of index $q^{2}/\left|T_{B}\right|$ by Lemma 2.3(2), it
follows that $X_{B}/T_{B}$ is a Frobenius group with cyclic complement of
order $\frac{q^{2}-1}{(2,q+1)}$ by [17, Table 8.17]. Thus, either $T_{B}=1$ or
$\left|T_{B}\right|=q^{2}$.

The $X_{0,B}$-orbits on the set of non-zero singular vectors of $V$ have
length $\frac{q^{2}-1}{(2,q+1)}$ except for two ones of have length $q-1$,
which are $\left\langle y_{1}\right\rangle_{GF(q)}^{\ast}$ and $\left\langle
y_{2}\right\rangle_{GF(q)}^{\ast}$ for some suitable non-zero singular vectors
$y_{1},y_{2}$ of $V$. On the other hand $\left|B\cap x^{G_{0}}\right|=q-1$,
where $x$ is a non-zero singular vector of $V$, by Lemma 2.1. Thus
$\left|B\cap\left\langle y_{i}\right\rangle_{GF(q)}\right|>1$ for $i=1$ or
$2$. We may assume that $i=1$. Hence $B\cap x^{G_{0}}=\left\langle
y_{1}\right\rangle_{GF(q)}^{\ast}$, since $\left\langle
y_{1}\right\rangle_{GF(q)}^{\ast}$ is a $X_{0,B}$-orbit.

Assume that $\left|T_{B}\right|=q^{2}$. Then $B$ is a $GF(p)$-subspace of $V$.
The cyclic subgroup $Y$ of $X_{0,B}$ of order $\frac{q+1}{(2,q+1)}$ fixes
$B\cap x^{G_{0}}$ pointwise. Thus $\left|B\cap\mathrm{Fix}(Y)\right|\geqslant
q$, where $\mathrm{Fix}(Y)$ is a $2$-dimensional $GF(q)$-subspace of $V$.
Suppose $\left|B\cap\mathrm{Fix}(Y)\right|=q$. Then $q>3$, since
$X_{0}\cong\Omega_{4}^{-}(3)\cong PSL_{2}(9)\cong A_{6}$ and this case is
ruled out by Lemma 3.2. Then there is a non-trivial subgroup $Y_{1}$ of $Y$
fixing a vector in $B\setminus\mathrm{Fix}(Y)$, since
$\left|B\setminus\mathrm{Fix}(Y)\right|=q(q-1)$. However, this is impossible
as $\mathrm{Fix}(Y_{1})=\mathrm{Fix}(Y)$. Therefore
$\left|B\cap\mathrm{Fix}(Y)\right|>q$. Clearly, $X_{0,B}$ acts on
$\mathrm{Fix}(Y)$. In particular $X_{0,B}$ induces a cyclic group of order
$q-1$ acting regularly on $\mathrm{Fix}(Y)$, and $\left\langle
y_{1}\right\rangle_{GF(q)}^{\ast},\left\langle
y_{2}\right\rangle_{GF(q)}^{\ast}$ are the unique $X_{0,B}$-orbits consisting
of non-zero singular vectors lying in $\mathrm{Fix}(Y)$. Hence
$\left|B\cap\mathrm{Fix}(Y)\right|=q+a(q-1)$ for some positive integer $a$
such that $a\leqslant q$. On the other hand
$\left|B\cap\mathrm{Fix}(Y)\right|=p^{e}$, with $q<p^{e}\leqslant q^{2}$, as
both $B$ and $\mathrm{Fix}(Y)$ are $GF(p)$-subspaces of $V$. So $a=q$ and
hence $B=\mathrm{Fix}(Y)$, since $k=q^{2}$. However this is impossible since
$\left|B\cap x^{G_{0}}\right|=q-1$, whereas $\left|\mathrm{Fix}(Y)\cap
x^{G_{0}}\right|=2(q-1)$.

Assume that $T_{B}=1$ and that $q$ is odd. As above, $q>3$ by Lemma 3.2 (this
fact is independent of the assumptions on the order of $T_{B}$). Then
$H^{1}(X_{0},V)=0$ by [39, Theorem 2.14.(i)], also
$\left|H^{1}(X_{0,\left\langle
x\right\rangle},V)\right|\leqslant\left|H^{1}(X_{0},V)\right|$ by [22,
2.3.(g)], since $\left\langle x\right\rangle$ is singular. Then
$X_{B}=X_{0,\left\langle x\right\rangle}^{\tau}$ for some $\tau\in T$ by [10,
17.10], hence $X_{0,\left\langle x\right\rangle}=X_{B^{\tau^{-1}}}$. Therefore
the Sylow $p$-subgroup of $X_{0,\left\langle x\right\rangle}$ acts regularly
on $B^{\tau^{-1}}$, hence the actions of any cyclic subgroup of
$X_{0,\left\langle x\right\rangle}$ on $B^{\tau^{-1}}$ and on the Sylow
$p$-subgroup of $X_{0,\left\langle x\right\rangle}$ are equivalent by [50,
Proposition 4.2]. Thus, each cyclic subgroup of $X_{0,\left\langle
x\right\rangle}$ fixes a point in $B^{\tau^{-1}}$ and partitions the remaining
points of $B^{\tau^{-1}}$ into $2$ orbits each of length $\frac{q^{2}-1}{2}$.
A similar conclusion holds for $B$, since $X$ acts block-transitively on
$\mathcal{D}$. However, this is impossible since $\left|B\cap
x^{G_{0}}\right|=q-1$.

Assume that $T_{B}=1$ and that $q$ is even. Let $K$ be the Sylow $2$-subgroup
of $X_{0,\left\langle y_{1}\right\rangle_{GF(q)}}$. Therefore $K$ fixes
$\left\langle y_{1}\right\rangle_{GF(q)}$ pointwise, hence $B^{K}$ is the set
of the $\lambda=q^{2}$ blocks of $\mathcal{D}$ incident with $0$ and $y_{1}$.
Furthermore, each of them contains $\left\langle y_{1}\right\rangle_{GF(q)}$.
Thus $K$ centralizes $\tau$, where $\tau\in T_{\left\langle
y_{1}\right\rangle_{GF(q)}}$, and hence $\tau$ permutes the blocks in $B^{K}$.
Then there is $\gamma\in K$ such that $\gamma\tau\in X_{B}$. Let $\eta$ in
$Y$, where $Y$ is defined as above, then $(\gamma\tau)^{\eta}\in X_{B}$. Also,
$(\gamma\tau)^{\eta}=\gamma^{\eta}\tau$, with $\gamma^{\eta}\in K$ since
$\tau\in T_{\left\langle y_{1}\right\rangle_{GF(q)}}$, $Y$ fixes $\left\langle
y_{1}\right\rangle_{GF(q)}$ pointwise and $Y$ normalizes $K$. Then
$\gamma\gamma^{\eta}=\gamma\tau\gamma^{\eta}\tau\in X_{B}$, since $\tau$
centralizes $\gamma^{\eta}$ and since $o(\tau)=2$. Consequently
$\gamma\gamma^{\eta}\in K\cap X_{0,B}$, so $\gamma^{\eta}=\gamma$ since $K$ is
an elementary abelian $2$-group and since $X_{0,B}$ is cyclic of (odd) order
$q^{2}-1$. Therefore $\gamma=1$ by [50, Proposition 17.2], since $K:Y$ is
Frobenius group. So $\tau=\gamma\tau\in X_{B}$, but this contradicts
$T_{B}=1$. Thus (1) and (4) are ruled out.

  1. (ii).

Case (2) cannot occur.

Assume that $PSL_{2}(q)\trianglelefteq X_{0,B}\leqslant$ $PGL_{2}(q)$. Since
$\left|B\cap x^{G_{0}}\right|=q-1$ and since the minimal primitive permutation
representation of $PSL_{2}(q)$ is at least $q$ for $q\neq 9$ and $6$ for $q=9$
by [41, Theorem 5.2.2], it follows that $PSL_{2}(q)$ fixes at least a point in
$B\cap x^{G_{0}}$ and hence it lies in the stabilizer of a singular
$1$-dimensional subspace of $V$, a contradiction.

  1. (iii).

Case (3) cannot occur.

Assume that $X_{0,B}\cong D_{2\frac{(q^{2}-1)}{(2,q-1)}}$. Since $X_{0}$ has a
unique conjugacy class of subgroups isomorphic to $X_{0,B}$ by [17, Table
8.17], it follows that $B^{G_{0}}=B^{X_{0}}$. Hence $X=TX_{0}$ acts flag-
transitively on $\mathcal{D}$. The group $X_{B}/T_{B}$ is isomorphic to
subgroup of $X_{0}$ containing $X_{0,B}$ as subgroup of index
$q^{2}/\left|T_{B}\right|$ by Lemma 2.3(2). Since $X_{0,B}$ is maximal in
$X_{0}$, it follows that $\left|T_{B}\right|=q^{2}$. Thus $B$ is a
$GF(p)$-subspace of $V$.

The $X_{0,B}$-orbits on the set of non-zero singular vectors $GF(q)$-subspaces
$V$ are either $\left\langle y_{1}\right\rangle_{GF(q)}^{\ast}\cup\left\langle
y_{2}\right\rangle_{GF(q)}^{\ast}$, where $y_{1},y_{2}$ have the same meaning
as in (i), or they have length $q^{2}-1$. Since $\left|B\cap
x^{G_{0}}\right|=q-1$ and since $X_{0,B}$ contains involutions switching
$\left\langle y_{1}\right\rangle_{GF(q)}$ and $\left\langle
y_{2}\right\rangle_{GF(q)}$, it follows that $\left|B\cap
x^{G_{0}}\cap\left\langle y_{i}\right\rangle_{GF(q)}^{\ast}\right|=(q-1)/2$
for each $i=1,2$. Thus $\left|B\cap\left\langle
y_{i}\right\rangle_{GF(q)}\right|=(q+1)/2$, whereas $B\cap\left\langle
y_{i}\right\rangle_{GF(q)}$ is a $GF(p)$-subspace of $V$. So
$X_{0}\cong\Omega_{4}^{-}(q)$ is excluded and the assertion follows. ∎

Now, we are in position to prove Theorem 5.1.

###### Proof of Theorem 5.1..

Suppose that $X_{0}$ is one of the groups $SL_{n}(q)$, $SU_{n}(q^{1/2})$ with
$n$ odd, $Sp_{n}(q)$ or $\Omega_{n}^{-}(q)$. The Latter is ruled out in Lemma
5.8. In the remaining cases assertion (1), (2), and (1) and (3) follows from
Proposition 5.3, Lemma 5.6 and Lemma 5.4 respectively. ∎

##  6\. Completion of the proof of Theorem 1.1

This small final section is devoted to the completion of the the proof of
Theorem 1.1.

###### Proof of Theorem 1.1..

Let $\mathcal{D}$ be a $2$-$(p^{2m},p^{m},\lambda)$, with $\lambda\mid p^{m}$,
admitting a flag-transitive automorphism group $G=TG_{0}$. If $\lambda=1$,
then assertions (1)–(5) follows from Theorem 2.5. Hence, assume that
$\lambda\geqslant 2$. If $m>1$ and $(p,2m)\neq(2,6)$, then assertion (1),
(6)–(8), (8) for $q=2$ and (9)–(10), and (11) follow from Lemma 2.6 and
Theorems 5.1, 3.1 and 4.2 respectively. Thus, in order to complete the proof,
we need to tackle the remaining cases $m=1$ or $(p,2m)=(2,6)$. The former
implies that either $G_{0}\leqslant\Gamma L_{1}(p^{2})$ or
$SL_{2}(p)\trianglelefteq G_{0}$, since $G_{0}\leqslant GL_{2}(p)$ and
$r=p(p+1)$ divides $\left|G_{0}\right|$. The former yields assertion (1) by
Lemma 2.6, whereas we may use the arguments in the proof of Proposition 5.3 to
rule out the latter.

Finally, assume that $(p,2m)=(2,6)$. Then $G_{0}\leqslant SL_{6}(2)$ and
$r=2^{f}\cdot 3^{2}$ with $f=1,2,3$. We may use the same argument of Lemma 5.2
to rule out the case $G_{0}=SL_{6}(2)$. Then $G_{0}\leqslant M$, where $M$ is
a maximal subgroup of $SL_{6}(2)$. Then $M$ lies in a member
$\mathcal{C}_{i}(SL_{6}(2))$, where $i=1,2,3$ or $8$, by [17, Tables
8.24–8.25]. Since $G_{0}$ is maximal in $G$ (recall that G acts point-
primitively on $\mathcal{D}$), and $G=TG_{0}$, it follows that $G_{0}$ acts
irreducibly on $V$. Hence $i\neq 1$.

Suppose that $i=2$. Then $G_{0}$ preserves the sum decomposition
$V=V_{1}\oplus V_{2}$, where $\dim V_{1}=\dim V_{2}=3$ and $M\cong
SL_{3}(2)\wr Z_{2}$ again by [17, Table 8.24]. One $M$-orbit is
$V_{1}^{\ast}\cup V_{2}^{\ast}$, which is of length $14$, is a union of
$G_{0}$-orbits. However, this is impossible since each $G_{0}$-orbit is of
length divisible $3^{2}$ by Lemma 2.1, being $r=2^{f}\cdot 3^{2}$.

Suppose that $i=3$. Then either $M\cong\Gamma L_{2}(8)$ or $M\cong\Gamma
L_{3}(4)$. Suppose that the former occurs. Since $18$ divides the order of
$G_{0}$, and since the unique subgroup of order $3^{2}$ which do not fix
points are cyclic by [21], it follows that $G_{0}$ contains a cyclic group $C$
of order $3^{2}$.

If $G_{0}\cap Z(GL_{2}(8))\neq 1$ then $Z_{7}\cong Z(GL_{2}(8))\leqslant
G_{0}$, hence $Z(GL_{2}(8))$ preserves each block incident with $0$, since $r$
is coprime to $7$ and since $G_{0}$ acts transitively on the set of blocks
incident with $0$. Then $B\cap B^{\prime}$ is preserved by $Z_{7}$, where
$B^{\prime}$ is any block of $\mathcal{D}$ distinct from $B$ such that $0,x\in
B\cap B^{\prime}$, with $x\neq 0$, as $\lambda\geqslant 2$. However, this is
impossible since $k=2^{3}$. Thus $G_{0}\cap Z(GL_{2}(8))=1$.

The group $\Gamma L_{2}(8)$ contains two conjugacy classes of cyclic subgroups
of order $9$, say $\mathcal{K}_{1}$ and $\mathcal{K}_{2}$. If
$C\in\mathcal{K}_{1}$ then $C$ is a subgroup of a Singer cycle and hence
$N_{M}(C)\cong Z_{63}.Z_{2}.Z_{3}$, whereas if $C\in\mathcal{K}_{2}$ then
$N_{M}(C)\cong Z_{9}.Z_{3}$. If $C$ is not normal in $G_{0}$ then
$SL_{2}(8)\trianglelefteq G_{0}$ regardless $C\in\mathcal{K}_{1}$ or
$C\in\mathcal{K}_{2}$, since $G_{0}\cap Z(GL_{2}(8))=1$. However, this case
cannot occur by Proposition 5.3. Therefore, $C\trianglelefteq G_{0}$ and since
$18$ divides the order of $G_{0}$, it follows that $C\in\mathcal{K}_{1}$ and
hence $G_{0}\leqslant N_{M}(C)$. Moreover, since $G_{0}\cap Z(GL_{2}(8))=1$,
it follows that $D_{18}\trianglelefteq G_{0}\leqslant D_{18}.Z_{3}$.
Therefore, $\lambda=2$ and hence $B$ is a $3$-dimensional $GF(2)$-subspace of
$V$ by Corollary 2.4. Then $D_{18}\trianglelefteq G_{0}\leqslant D_{18}.Z_{3}$
leads to a $2$-$(2^{6},2^{3},2)$ design isomorphic to that of Line 1 in Table
1 by using GAP [64]. This corresponds to case (1) for $q=2^{3}$ and
$\lambda=2$.

Assume that $M\cong\Gamma L_{3}(4)$. Then $SL_{3}(4)\nleq G_{0}$ by
Proposition 5.3. Since $G_{0}$ acts irreducibly on $V$ and its order is
divisible by $18$, it follows that either $G_{0}$ preserves $V_{3}(2)$, or
$G_{0}\leqslant\Gamma U_{3}(2)$ or $G_{0}\leqslant Z_{3}.S_{6}$ by [17, Tables
8.3–8.4]. The first one is ruled out by Lemma 2.1, whereas if
$G_{0}\leqslant\Gamma U_{3}(2)$ then $\mathcal{D}$ is isomorphic to one of the
$2$-designs listed in Table 1 by using GAP [64], and these correspond to cases
(12)–(13).

Assume that $G_{0}\leqslant Z_{3}.S_{6}$. Let $S$ be a Sylow $3$-subgroup of
$G_{0}$. Since no proper subgroups of $S$ have all orbits on $V^{\ast}$ of
lengths divisible by $3^{2}$, it follows that $S\leqslant G_{0}$ by Lemma 2.1.
Therefore either $G_{0}\leqslant N_{M}(S)=S:D_{8}$ or
$Z_{3}.A_{6}\trianglelefteq G_{0}\leqslant Z_{3}.S_{6}$. Assume that the
former occurs. Note that $N_{M}(S)$ splits $V^{\ast}$ in one orbit of length
$27$ and two orbits of length $18$. The first one is also a $G_{0}$-orbit as
$S\leqslant G_{0}$, each of the remaining two $N_{M}(S)$-orbits is a union of
$G_{0}$-orbits. Let $B$ be any block incident with $0$. Since $r=9\lambda$
with $\lambda=2,4,8$, and since the intersection of $B$ with any of
$G_{0}$-orbit of length different from $27$ is at most $2$ by Lemma 2.1, it
follows that there is $\sigma\in G_{0}$, with $o(\sigma)=3$, such that
$\left|\mathrm{Fix(}\sigma\mathrm{)}\cap B\right|=4$. Actually,
$\mathrm{Fix(}\sigma\mathrm{)}\subset B$. Since $S\trianglelefteq G_{0}$,
$r=9\lambda$ with $\lambda=2,4,8$, and the order of $S$ is $27$, it follows
that $\left\langle\sigma\right\rangle\trianglelefteq G_{0,B}$, and hence
$G_{0,B}$ preserves $\mathrm{Fix(}\sigma\mathrm{)}$. Therefore, $B$ is a
$3$-dimensional subspace of $V$ by Corollary 2.4. However this is impossible
by [33, Theorem 3.3.1]. Thus $Z_{3}.A_{6}\trianglelefteq G_{0}\leqslant
Z_{3}.S_{6}$, hence $G_{0}$ splits $V^{\ast}$ into two orbits of length $27$
and $36$ by [56], and if $x_{1}$ and $x_{2}$ are representatives of these
orbits and $B$ is any block incident with $0$, it follows that $\left|B\cap
x_{1}^{G_{0}}\right|=3$ and $\left|B\cap x_{2}^{G_{0}}\right|=4$ by Lemma 2.1.
Thus $\left(G_{0,B}\right)^{\prime}$ fixes $B$ pointwise, since
$A_{5}\trianglelefteq G_{0,B}\leqslant S_{5}$ by [56], hence
$\left(G_{0,B}\right)^{\prime}\leqslant G_{0,y_{i}}$ for $y_{i}\in B\cap
x_{i}^{G}$, $i=1,2$, which is not the case.

Suppose that $i=8$. Then $M\cong Sp_{6}(2)$. The argument used in the proof of
Lemma 5.4 rules out the case $G_{0}=M$. Then $G_{0}$ lies in a maximal
subgroup $K$ of $Sp_{6}(2)$, hence $K$ is one of the groups listed in [17,
Tables 8.28–8.29]. Actually, either $K\cong\Gamma L_{2}(8)$ or $K\cong
SO_{6}^{\pm}(2)$ or $K\cong G_{2}(2)$ since $G_{0}$ acts irreducibly on $V$.

If $K\cong\Gamma L_{2}(8)$, then the above argument leads to
$D_{18}\trianglelefteq G_{0}\leqslant D_{18}.Z_{3}$ and to a
$2$-$(2^{6},2^{3},2)$ design isomorphic to that of Line 1 in Table 1. Hence
(1) holds in this case.

Assume that $K\cong SO_{6}^{+}(2)$. Then the $K$-orbits on $V^{\ast}$ are
union of $G_{0}$-orbits. Since each $G_{0}$-orbit is of length divisible
$3^{2}$ by Lemma 2.1, it follows that each $K$-orbit on $V^{\ast}$ is
divisible by $3^{2}$. However this is impossible, since the $K$-orbits have
length $35$ and $28$ by [21]. So, this case is ruled out.

Assume that $K\cong SO_{6}^{-}(2)$. Since $G_{0}$ is irreducible on $V$, it
follows that $\Omega_{6}^{-}(2)\trianglelefteq G_{0}$ by [17], Tables
8.33–8.34. However, the argument used in the proof of Lemma 5.8 excludes this
case as well.

Assume that $K\cong G_{2}(2)$. Then either $G_{0}\leqslant
N_{K}(U)=S:SD_{16}$, where $U$ is a Sylow $3$-subgroup of $K$, or
$G_{0}=K^{\prime}\cong PSU_{3}(3)$ or $G_{0}=K$ by [21]. In the first two
cases $\mathcal{D}$ is isomorphic to one of the $2$-designs listed in Table 1
again by using GAP [64], which correspond to case (1), (12) and (13). If
$G_{0}=K$ we may proceed as in Proposition 3.5 thus obtaining that
$\mathcal{D}$ isomorphic to the $2$-$(2^{6},2^{3},2^{3})$ design in Line 5 of
Table 1, which corresponds to (10) for $q=2$. This completes the proof ∎

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For future work, one may want to consider experimenting with the continuous Fourier transform instead of the Laplace. And using the inverse Fourier transform to derive the pdfs once the system has been solved in the Fourier domain. One reason being that there appears to be some work towards numerically inverting characteristic functions; Fourier transforms of random variables defined also on the negative-real line. The foundation of one direction of work was by Gil-Palaez in <cit.>. A unified and multivariate approach is presented in <cit.>. However, more recently the inversion can now be done with FFTs as shown by <cit.>, based on the earlier work of <cit.>.

Aside from using point-based methods, there are still other approaches left for experimentation and investigation. For example, one may attempt instead to derive a theory for sMRMs using Fourier extensions and the convolution algorithm presented by <cit.> such that it leads to convergent and unique results. Alternatively, faster function approximation and convolution algorithms relative to the Pacal library <cit.> may be attempted instead, along with proving convergence and uniqueness. Alternatively, the complexity within the Pacal library could be relaxed, i.e. the handling of operations including both discrete and continuous random variables together can be removed, and the handling of singularities also.

Lastly, we are aware of convolution quadrature techniques <cit.>. Without having delved deep into the theory, these formulas here appear unrelated to our formulas of convolution quadrature. Additionally, the trapezoid rule for convolution quadrature <cit.> is not necessarily the same one as the trapezoid rule presented here.

§.§ Drawbacks of the proposed algorithms

Firstly, we have previously mentioned in Section <ref>, that iterative methods may prematurely converge in particular scenarios. The solution by <cit.> requires the values of the solutions to be bounded in $[0,1]$. However, pmfs and not pdfs satisfy this property. Whilst the passage-reward densities are not bounded in this interval, the cdf of these densities are. Therefore one could try adapting the algorithm for the cdfs instead to determine convergence more soundly. Alternatively, rather than computing the passage-reward densities, we could instead compute their cdfs directly. Let $Pr(Rew \leq r\ \cap\ \vDash \Diamond B) = F_s(r) = \int_{0}^{r}f_s(x)dx$. Then the system of convolution equations become

$$\textbf{F} = (A \circ \textbf{G})\ \textcircled{$\ast$}\ \textbf{F} + \textbf{H}$$

where G is still a the matrix of reward pdfs, i.e. we have $G_{s,t} = f_{rew(s,t)}$. However $H_s = \sum_{u \in B}P_{s,u}F_{rew(s,u)}$ is a vector of reward (partial) cdfs. Solving this instead may be preferred as it avoids having to numerically integrate the passage-reward densities.

Secondly, <cit.>  presented a comprehensive algorithm that can be used for resolving $\mrmpdf$ over an sMRM. The algorithm involves taking the Laplace transform of the system of convolution equations (<ref>). This leads to a system identical in form to (<ref>), i.e.

x(s) =(A∘C(s))x(s) + d(s)

where $s$ is a complex number. The definitions of the terms above are now using the Laplace transform and not the FT, i.e.

\begin{align*}
\textbf{C} &\triangleq  (C_{s,t})_{s,t \in S_?} \triangleq (\mathcal{L}\{f_{(rew(s,t))}\})_{s,t \in S_?}&\\
\textbf{d} &\triangleq  (d_s)_{s \in S_?} \triangleq (\sum_{u \in B}\textbf{P}(s,u)\mathcal{L}\{f_{rew(s,u)}\})_{s \in S_?}
\end{align*}

with the Laplace operator $\mathcal{L}$ applied to a function $f$ being defined as
$$\mathcal{L}\{f\}(\sigma + i\omega) = \int_{0}^{\infty}f(t)e^{(\sigma + i\omega)t}dt$$

Then, the power method is used to resolve each linear system  in (<ref>) for a set of $N$ sampled points $(s_i)_{0<i\leq N-1}$. Once the power method converges (for each $s_i$), either the Laplace-Euler or the Laplace-Laguerre inversion algorithms is applied to $\textbf{x}(s)$ (described in <cit.>) to obtain the passage-reward density $\textbf{f}(r)$ at $M$ points, i.e. for each $r \in Q \subset [a,b]$, where $|Q| = M$. For the Laplace-Euler inversion algorithm (their general strategy), then $M=N/c$ for a choice of $c\in\{15,16,\dots,50\}$, which determines the accuracy of the inversion. It is expected that as $c$ increases, the accuracy increases. For the Laguerre inversion algorithm, which is less applicable, $N$ is fixed to 400, and $M$ is independent of $N$.

This technique has perhaps two main advantages over our algorithms. Using their algorithm, the property $\mrmpdf$ can be resolved for the set of points $r \in Q \subset [a,b]$. Our algorithm enforces that $Q$ be a set of equidistant points between $[0,b]$ only. Secondly, for strictly continuous pdfs, their algorithm appears to be quite efficient in obtaining high accuracy for the $M$ points for the pdfs, relative to $N$, the number points sampled from the Laplace transforms. However, a disadvantage of their method could be that it requires the analytical Laplace transforms of pdfs to be known, whereas ours do not. Nevertheless, these transforms may be approximated numerically or obtained via a CAS (computer algebra system). If the approximations are poor, then like our algorithm, the power method may fail to converge. Disregarding the accuracy of solutions, the worst-case time complexities for both ours and their power method are similar.

CHAPTER: CONCLUSION

We began our thesis presenting the theoretical derivations for reachability problems over a new class of models (in the model-checking literature) - stochastic Markov reward models. We were also able to find practical algorithms for model-checking reachability properties via the DFT for the univariate reward problem. This solution extends to the case where our rewards are random vectors, with mutually independent components.

We found in the discrete-reward setting that the best algorithm (relative to our experiments) turns out to just be the naive exact-power method. The direct algorithms are quite slow, and our experiments show that the power method outperforms its approximate counterpart, and also the approximate direct method (i.e. the LU approximation method). The other iterative algorithms were slower due to our deconvolution algorithm not being optimized (via the FFT). Additionally, it was harder to vectorize the computations of the Gauss-Seidel algorithm to take effective use of the $numpy$ library. Note however, we did not experiment with approximate variants of the other iterative methods. But they suffer from the same problem as their exact counterparts. Nevertheless, at this time we recommend the exact-power method for use.

We stated earlier that some optimizations possible would be: 1) That zero-padding can be removed almost entirely by the work of <cit.>, and 2) that kernel tiling algorithms can be used for improving speed <cit.>.  However, it is worth considering whether algebraic decision diagrams (ADD) <cit.> can be used in the discrete-reward setting for reducing space complexity. Current model-checking tools such as <cit.> employ decision diagrams such as the multi-terminal binary decision diagrams (MTBDD) to help circumvent state space explosion. The paper <cit.> presents an algorithm for matrix multiplication when matrices are represented as ADDs, and the exact-power method for example requires not much more than that in terms of unique operations.

In the continuous-reward setting, we showed that our algorithm via the DFT can handle discontinuous distributions. Unfortunately, many samples were required to ensure sufficient accuracy. If improving on our approach, one may consider replacing Romberg's method with another algorithm of a similar form, e.g. <cit.>.  Additionally, we have stated that there exists different equations for deriving the first-passage reward densities <cit.>. These approaches may yield better convergence properties.

Whilst our solution works for continuous pdfs and of which required fewer samples for sufficient accuracy, we have not compared our results with the algorithm of <cit.> using the Laplace transform and the Laplace-Euler inversion algorithm, with the power method yet due to lack of time. We have, however, discussed their effectiveness in the literature review. Thus, we cannot recommend our algorithm over theirs for problems with continuous-pdfs yet.

§ FUTURE WORK

Negative rewards The sMRM affords random variables being negative, or having positive measure over the negative portion of the real line. It is not clear how the DFT can be used since it leads to representation growth of the solution vector if applied naively. Alternatively if the continuous Fourier Transform is employed, no representation growth occurs. Then, future work is in determining both highly precise numerical methods for the transform itself (hence not relying on algebraic techniques) and its inversion. Numerical algorithms for the transform include those that use the FFT, for example <cit.>, based on the work of <cit.>.

Non-determinism It is possible to extend sMRMs with actions creating stochastic Markov reward Decision Processes. That is, at any state, the process makes a non-deterministic choice between a set of actions, each of which has a transition probability distribution over the set of states of the process.  The goal here would be to design efficient algorithms for problems over this class of models. Such problems may involve determining optimal policies or determining the maximum/minimum probability of reaching a set of states $B$.

CHAPTER: SUPPLEMENTARY PROOFS

§ PROOF OF THEOREM <REF>

Given a system of convolution equations $\textbf{E}\textcircled{$\ast$}\textbf{f} = \textbf{h}$, with equations $E_0,E_2,\cdots E_N$, and the unique solution $\textbf{x}$ (a hypervector), where each $E_i$ represents the equation
$$\sum_j E_{i,j} \ast f_j = h_i$$

If the system is transformed via these two updating/substitution operations:

* replacing an equation by a convolution of itself with a non-zero vector.

* replacing an equation with the equation itself added to another equation that has been convolved with a non-zero vector.

or these two pivoting operations:

* swapping one equation with another (i.e. swapping rows of E).

* changing the order of terms in each and every equation (i.e. swapping columns of E).

then the transformed system will still have the same solution set x, with the exception of the case where columns are swapped. In this latter situation, the transformed system will have a solution set that is a reordering of the terms of x.

Proof:  Let $E_i(\textbf{x})$ be the equation $E_i$ where the unknown terms f are replaced with the solution x. Then the conjunction
$E_0(\textbf{x}) \cap E_1(\textbf{x}) \cap \cdots \cap E_N(\textbf{x})$ is true. Concerning the substitution rules, note that convolving any equation $E_i$ both sides by any non-zero vector $g$ will also maintain the truth of the conjunction, e.g. $E_0(\textbf{x}) \cap \cdots \cap\ g\ast E_i(\textbf{x})\ \cap \cdots \cap E_N(\textbf{x})$ holds. And secondly, replacing any equation by adding the equation itself to another equation convolved with a non-zero vector $g$ will also maintain the conjunction, e.g. $E_0(\textbf{x}) \cap \cdots \cap\ (g\ast E_j(\textbf{x})) + E_i(\textbf{x})\ \cap \cdots \cap E_N(\textbf{x})$.

Concerning pivoting rules, we can re-arrange the order of the equations and the conjunction will remain true, e.g. swapping the first two rows, we have $E_1(\textbf{x}) \cap E_0(\textbf{x}) \cap \cdots \cap E_N(\textbf{x})$. Lastly, if we swap the order of the terms in f, e.g. the term $f_j$ with $f_k$, then let $E'_i$ denote equation $i$ after the terms $E_{i,j}$ and $E_{i,k}$ are swapped. Let $\textbf{x}$ have its indices swapped in the same fashion as f and denote this $\textbf{x}'$. Then the conjunction $E_1'(\textbf{x}') \cap E_1'(\textbf{x}') \cap \cdots \cap E_{N}'(\textbf{x}')$ holds.

§ PROOF OF THEOREM <REF>

The cumulated reward hypervector $\textbf{f} = (\textbf{f}[r])_{r \in \mathbb{N}} = (f_{s \vDash \Diamond B}[r])_{r \in \mathbb{N},s \in S_?} =  ( Pr(r \cap s \vDash \Diamond B) )_{r \in \mathbb{N},s \in S_?}$ is the (unique) fixed point of the operator $\Upsilon: [0,1]^{\mathbb{N} \times  S_?\times 1} \rightarrow [0,1]^{\mathbb{N} \times S_? \times 1}$. This operator is defined as:

\[\Upsilon(\textbf{f}) = ((A\circ\textbf{G})\ \textcircled{$\ast$} \ {\textbf{f}}) + \textbf{h} \]

Additionally, let $\textbf{f}^{(0)} = \textbf{0}$, and $\textbf{f}^{(n+1)} = \Upsilon(\textbf{f}^{(n)})$ where $n \geq 0$. Then, for any ${r \in \mathbb{N}}$ and  $s \in S_?$ the following three statements hold:

* $f^{(n)}_s[r] = Pr(\ r\  \cap [s \vDash \Diamond^{\leq n} B])$, for all $n \geq 0$.

* $ \lim\limits_{n \rightarrow \infty} f_s^{(n)}[r] = f_s[r]$.

* $f_s^{(0)}[r] \leq  f_s^{(1)}[r] \leq  f_s^{(2)}[r]  \leq \cdots \leq  f_s[r]$.

where statement 2. states that the solution converges to a fixed point, and 3. states that the convergence is monotonic.

Proof of 1.:

The proof involves induction on $n$. Firstly with the base case $n=0$, $f_s^{(0)}[r] = \textbf{0} \equiv Pr(\ r\  \cap [s \vDash \Diamond^{\leq 0} B]) = Pr(\ r\  \cap [s \vDash B]) = \textbf{0}$ for all $s \in S_?$ (since $S_? \subseteq S - B$). This proves the base case. Now to prove the inductive step. Firstly, note that the event $[s \vDash \Diamond^{(\leq n+1)} B]$ is equal to

$$[s \vDash \bigcirc B] \cup
[s \vDash \bigcirc S_?  \ \cap \ s \vDash \Diamond^{(\leq n+1)} B ]

i.e. the event of reaching $B$ under or equal to $n+1$ steps, is equal to the event of reaching $B$ in the next step OR the event of reaching $B$ under $n+1$ steps on the condition of passing through $S_?$ at the next step. Note that $[s \vDash \bigcirc B]$ is disjoint to $   	 [s \vDash \bigcirc S_? ]$, hence

$$Pr([s \vDash \Diamond^{(\leq n+1)} B]) = Pr([s \vDash \bigcirc B]) + Pr(   	 [s \vDash \bigcirc S_?  \ \cap \ s \vDash \Diamond^{(\leq n+1)} B ])$$

Now, starting the proof with a top-down approach, we have

\begin{align*}
&=     Pr( r \cap   [s \vDash \Diamond^{(\leq n+1)} B]) & \\
&= Pr(r \cap [s \vDash \bigcirc B]) +  Pr(r \cap    	 [s \vDash \bigcirc S_?  \ \cap \ s \vDash \Diamond^{(\leq n+1)} B ])
\end{align*}

where $Pr(r \cap [s \vDash \bigcirc B]) =  \sum_{u \in B}\textbf{P}(s,u)f_{rew(s,u)}[r]
= h_s[r]$.
Therefore this corresponds to the $\textbf{h}$ portion of the $\Upsilon()$ operator. Next, by letting  $\Pi.s$ denote the set of all paths beginning in $s$, satisfying $[s \vDash \bigcirc S_?  \ \cap \ s \vDash \Diamond^{(\leq n+1)} B ]$, we have

Pr(r ∩[s ◯S_?   ∩ s ^(≤n+1) B ])

= ∑_π̂ ∈Π.sPr(π̂)f_Rew(π̂)[r]

= ∑_t ∈S_?P(s,t)(f_rew(s,t) ∗(∑_ϕ̂ ∈Π.t

= ∑_t ∈S_? A_s,t(G_s,t	∗f_t^(n))[r]

where $f_t^{(n)}[k] = Pr( k \cap  [s \vDash \Diamond^{\leq n} B])$; the induction hypothesis which we have assumed correct. Note that the result corresponds to the $A\circ\textbf{G}$ portion of the $\Upsilon()$ operator. Thus we have that

= Pr( r ∩[s ^≤n+1 B])

=  Pr(r ∩[s ◯S_?   ∩ s ^(≤n+1) B ]) + Pr(r ∩[s ◯B])

= ∑_t ∈S_?((A_s,tG_s,t) ∗f_t^(n))[r] + h_s[r]

which proves the equivalence.

Proof of 2.: We know that the event $[s \vDash \Diamond B]$ is equal to the event $[\lim_{n \rightarrow \infty} s \vDash \Diamond^{\leq n} B]$. Thus

f_s[r] = Pr(r ∩s B)
= lim_n →∞ Pr(r ∩s  ^≤n B)
= lim_n →∞ f_s^(n)[r]

Proof of 3.: It is true that $[s \vDash \Diamond^{\leq n} B] \subseteq [s \vDash \Diamond^{\leq (n+1)} B]$. Then for any $r \in \mathbb{N}$,

$$[Rew = r\  \cap [s \vDash \Diamond^{\leq n} B]] \subseteq [Rew = r\  \cap [s \vDash \Diamond^{\leq n+1} B]]$$

From statement 1. we know that $f^{(n)}_s[r] = Pr(\ r\  \cap [s \vDash \Diamond^{\leq n} B])$. Thus,

$$f^{(n)}_s[r] = Pr(\ r\  \cap [s \vDash \Diamond^{\leq n} B]) \leq Pr(\ r\  \cap [s \vDash \Diamond^{\leq n+1} B]) = f^{(n+1)}_s[r]$$

CHAPTER: SUPPLEMENTARY RESULTS

We present here additional experiments that we have performed, but deemed to not be as useful as the results in the main thesis. For the empirical results in this chapter, unless stated otherwise, only the uniform Markov chain was used (see Fig. <ref>). That is, given a fixed size empty matrix, we sampled values uniformly between zero and one for each and normalized the matrix to ensure the rows would sum to one. This resulting matrix would then be a valid probability matrix. Uniform pmfs were sampled in a similar fashion.

§ EXPERIMENTS WITH EXACT AND APPROXIMATE POWER METHOD

§.§ Previous toy problem

We solve the same problem earlier from Section <ref>, this time including solutions from the power method - both exact and approximate. For the power method, we use an absolute tolerance level of $1e-16$ for convergence ((<ref>)). The property being solved for is $Pr(r\ \cap\ s \vDash \Diamond s_4)$, for all $s \in S_? = \{s_0, s_1, s_2, s_3\}$ and $r = 0,1,2,\cdots,N-1$, where $N= 150$. The Gaussian elimination algorithm, and the exact power method computes $Pr(r\ \cap\ s \vDash \Diamond s_4)$ for $r = 0,1,2,\cdots,N-1$, whilst the approximate LU method, and approximate power method computes the property for $r= 0,1,\cdots T$, where $T = N + n$, and $n$ is the zero padding length (needed for the approximate methods).

The results are shown in Fig. <ref>. We find that the approximate power methods align with the LU approximate method of equal $T$. There appears to be a discrepancy between them when $T=12000$ possibly due to numerical errors. In fact, the approximate power method could not converge under 1000 iterations - a hard-limit for the number of iterations we enforced. Increasing the limit to 10,000 iterations, the algorithm still did not converge (see Table <ref>).

The plot on the bottom of Fig. <ref> is the absolute (approximate) error relative to the exact power method, and not Gaussian elimination as previously. We find that Gaussian elimination is not as accurate as the power method. Supporting this is the fact that the approximate methods' accuracy nears that of the power method for large $T$. R This may be attributed to rounding errors, and methods to mitigate such problems (for linear systems) are presented in <cit.>. More specifically, it could be due to the numerical instability of deconvolution, which is less known. Nevertheless, if accuracy is important, in practice the results of Gaussian elimination can be used as an initial guess for an iterative method like the power method, since solutions can be made more precise via successive iterations.

Similar to Fig. <ref>. We also include results for the power method. Note in the bottom plot that the approximate (direct and iterative) methods for a given $T$ overlap one another in error. For the LU approximations, the different padding lengths can be determined by $T-N$ where $T$ stands for Total length.

The approximate power method (with T=12000) did not converge (reaching the maximum iteration limit of 10000), probably due to the really small threshold. The LU approximation (with T=12000) method is significantly faster than Gaussian elimination and also achieved greater accuracy rel. to the power method (shown in Fig. <ref>). Although, it did require significantly more space ($\times \frac{12000}{150}$).

We see in Table <ref> that the exact power method required fewer iterations relative to the approximate power method to reach convergence. For $T=12000$, whilst the approximate power method failed to converge under 10000 iterations, Fig. <ref> suggests (at least for state $s_0$) that the hypervector has converged seeing that the absolute error relative to the power method is near 1e-16.

In this one experiment, in comparison to the exact power method, the approximate power method was slower. Gaussian elimination was non-competitive for either speed or accuracy. The LU approximation would not be considered competitive if high speed and accuracy are required. However from a more general perspective, if we know the characteristics of an sMRM problem beforehand, we may select the direct algorithms if for example it is known that the iterative algorithms will fail to converge quickly, and then apply the iterative methods on top of their results to improve their accuracies. However if $|S_?|$ is large, both the LU approximation method and GE have time complexities which are cubic in $|S_?|$. Therefore, from a scalability perspective the iterative algorithms may be considered more scalable with respect to $|S_?|$, and therefore preferred. However, whilst we cannot offer a general recommendation for which algorithm to use (without a problem context), our current preferred choice would be to use the iterative algorithms as a first.

We now proceed to evaluate the usefulness of the approximate power method relative to the exact power method. The following series of tests show that it is not necessarily faster at convergence.

§.§ Convergence performance of exact vs approximate power method

We investigated the performance of the approximate power method and the exact power method over random systems, and timed how long they took to solve for the property $\mrmpdf$. These experiments should help us determine whether the exact power method's requirement for transforms and inversions of the solution hypervector (for each iteration) contributes significantly to solving time. We found three factors that determine the solving time: 1) $|S_?|$, the size of the set of states that can eventually reach $B$ our goal states. 2) $k$, such that $\mrmpdf$ is computed for $r = 0,1,\cdots,k-1$. 3) $\lambda$, which defines the points $r= 0,1,\cdots,\lfloor\lambda k\rfloor - 1$ for which $f_{rew(s,t)}[r]$ is allowed to be non-zero in, for all $s,t \in S^2$. Note that $\sum_{x=0}^{\lfloor\lambda k\rfloor - 1}f_{rew(s,t)}[x] = 1$, and $f_{rew(s,t)}[r] = 0$ for $r >\lfloor\lambda k\rfloor - 1$. Therefore each $f_{rew(s,t)}$ is still a pmf.

We set the approximate method's padding length to $k-1$, which makes it identical to the exact power method in terms of space requirements. Therefore the main difference between the two methods is that the exact power method requires the DFT and the inverse DFT to be applied in each iteration, whereas the approximate method does not.

For the exact power method, we used the following termination criteria
$$max_{s,r}|\textbf{f}^{(n+1)} - \textbf{f}^{(n)}| \leq \epsilon$$
where $\epsilon = 1e-7$. For the approximate power method (Sec.  <ref>), the criteria is the same,
$$max_{s,r}|\textbf{x}^{(n+1)} - \textbf{x}^{(n)}| \leq \epsilon$$

where it is to be noted that $\textbf{x}^{(n)}$ is a complex number.

Experiment set-up For our experiment, we varied each of the factors above, whilst keeping others fixed. Firstly, we will set as fixed $k = 1000,|S_?|=  30$, and $\lambda = 0.5$. Then we will vary each parameter separately to get a measure of their contribution to time. For any valuation of the parameters ($k,|S_?|,\lambda$), 200 experiments are performed, their average and worst times recorded and plotted. For each of these 200 experiments, since we are solving for the property $\mrmpdf$, we randomly and uniformly generated the terms in the system

$$\textbf{f} = (A\circ \textbf{G})\textcircled{$\ast$} \textbf{f} + \textbf{h}$$

i.e. $A,\textbf{G},\textbf{h}$ are randomly generated. The pmfs $(G_{s,t}[r])_{s,t \in S_?}$ and $(h_s[r])_{s \in S_?}$ are generated to satisfy $\lambda$.

For each experiment, we only recorded the solving time, i.e. time until convergence (solving phase). We did not record the time required to build the relevant system of equations (preparation phase) (e.g. $\textbf{G}, \textbf{h}$) since it is the same for both methods.

Our first experiment (Fig. <ref>) is to vary $\lambda$. We find generally that for $\lambda \in [0,1]$, the exact power method is faster at terminating. However for $\lambda > 1$, the approximate method increases in competitiveness.

Experiments performed with $ k = 1000, |S_?| = 30$ and $\lambda$ was varied. The top graph was achieved using $7\times200$ experiments, whilst the bottom took $6\times200$ experiments. Each experiment consisted of a different randomly generated system. For the top graph, the values of lambda were $[0.1, 0.5, 0.9, 1.3, 10, 25, 40]$. For the bottom, the values were $[0.1, 0.2, 0.4, 0.6, 0.8, 1.0]$.

Next, we experimented with $k$. We see in Figure <ref> that when $\lambda = 20$, the time taken for each method was generally small, which corresponds with our previous results shown in Fig. <ref>. With $\lambda = 0.5$, we find there to be a significant growth in time requirements for the approximate method.

Experiments performed with $|S_?| = 30$ and $k$ was varied. The top and bottom graphs were achieved with 1000 experiments each.

Finally, we investigated the complexity of the system when increasing $S_?$  (Fig. <ref>). In this setting the exact power method seems to prevail for $\lambda=20$, but we find these methods perform similarly when $\lambda=0.5$.

Experiments performed with $k = 1000$ and $S_?$ was varied. The top and bottom graphs were achieved with 1400 experiments each.

To conclude our findings, there does not currently appear to be much benefit in using the approximate power method for time savings alone. If further testing the effectiveness of the methods, the pmfs $(f_{rew(s,t)}[r])_{s,t \in S^2}$ could be sampled instead from known tailed random variables that vary in their tails' strength: light, medium and heavy. Our pmfs were randomly generated, where we sampled values between $[0,1]$ for each $f_{rew(s,t)}[r]$ and then normalized  to sum to one.

§ EXPERIMENTS WITH EXACT ITERATIVE METHODS

Implementation details For the following experiments, any conclusions derived for time should take into consideration that the different methods used do not have fully optimized implementations. For example, any deconvolutions performed was computed via an algorithm with $O(k^2)$ time complexity (where $k$ is the length of the resulting vector) whilst there are $O(klog_2k)$ algorithms. Another reason would be that the numpy package in python was used to implement a portion of the iterative methods. However, its usage was not equal between the three exact methods: power, Jacobi, and Gauss-Seidel. The package would enable us to speed up the methods significantly. Hence, the fairest indicator of performance for the iterative methods would be the number of iterations required to solve a problem.

§.§ Scalability of exact iterative methods

Our next experiment is almost identical to that described in Section <ref>. It differs in that it includes results from the exact Jacobi and Gauss-Seidel methods.

As before, $|S_?|$ and $k$ are varied independently. Now, performance is characterized by time and number of iterations. When varying $|S_?|$, $k$ was fixed to $1501$. And when varying $k$, $|S_?|$ was fixed to ten. For each valuation of the parameters ($|S_?|,k$), 200 experiments are performed. For each experiment, the system of convolution equations to be solved is sampled as before, their average and worst solving times (for each method) is recorded and plotted. Additionally, we now also record the number of iterations taken by the iterative methods to solve the problem. The convergence threshold set once more to $1e$-16.

In Figure <ref> we plot the results for when $|S_?|$ is varied. As expected, the Gauss-Seidel algorithm requires the fewest iterations on average. However, due to the time complexity of our deconvolution algorithm and the lack of an optimized implementation, the libraries used to implement these algorithms lead to the power method being quickest on average. The LU approximation algorithm was found to be the least scalable. When we varied $k$ instead (see Fig. <ref>), similar results are obtained.

The LU approximation algorithm requires no iterations, therefore it does not have a plot in the top graph. As $|S_?|$ increases, there is a slight growth in the average no. of iterations for each method. Additionally, the discrepancy between the worst and average no. of iterations decreases. In the legend, (pad = 10k) means that the LU approx. method uses a padding length (see (<ref>)) of $10k$. Each graph (top and bottom) was achieved after performing ($200 \times 6$) experiments.

The LU approx. method is the worst performer here in terms of (average and worst) time. Each graph (top and bottom) was achieved after performing ($200 \times 5$) experiments.

§ AN EXAMPLE WITH THE ROMBERG METHOD

In Figure <ref>, we apply the Romberg method of different levels to approximate the convolution integral of a Gamma distribution with parameters ($k=2, \theta=2$), and a Weibull distribution with parameters ($k=3, \lambda=1$). We find here that Romberg's method at level three shows an equivalence to Simpson's method derived using the midpoint rule (not presented in this thesis). Additionally, the trapezoid rule with 8001 points yields weaker accuracy relative to Simpson's rule with 1001 points (but using a total of around 2002 points for computing the Midpoint and trapezoid rule).

The last sub-figure of Fig. <ref> shows what happens when we interpolate our pdf approximations. We find there to be a significant loss of accuracy with Romberg at level five and less so at level 3. The trapezoid rule and midpoint rule do not appear to be affected much.

Top: The approximation of the convolution via some quadrature rules and also Pacal. Middle: The absolute error of certain rules (not specifically those from Top) with respect to the Romberg method. Bottom: The absolute error of these rules when interpolated via a cubic spline relative to Pacal. Selected rules were chosen again for each plot.

CHAPTER: MARKOV CHAIN SIMULATION

We present here the four algorithms used to simulate the respective Markov chains in the experiment section of Chapter 6. They are written in python and tested to run on python 3.8. Firstly, the necessary imports are:

import matplotlib.pyplot as plt
import numpy as np
import sparse
import seaborn as sns

The code to generate uniform MCs is directly below.

def generate_random_MC_uniform(num_states):
P_matrix = np.random.random((num_states + 1, num_states))
P_matrix += 0.01  #prevents columns summing to zero (highly improbable)
P_matrix = (P_matrix / P_matrix.sum(0)).T

b_vector = P_matrix[:, -1]
A_matrix = P_matrix[:, :-1]

return A_matrix, b_vector, P_matrix

Next is the code to generate block MCs. For our experiments, we used the default parameters.

def generate_random_MC_block(num_states, num_pass=200, block_scale=5):
P_matrix = np.zeros((num_states, num_states + 1))
full_state_idxs = np.arange(num_states)

# for each state, create the probability that it can reach B
reach_idxs = np.random.choice(full_state_idxs, int(num_states * np.random.rand()), replace=False)
reach_probs = np.random.rand(len(reach_idxs))
P_matrix[reach_idxs, -1] = reach_probs

# perform n passes
for _ in range(num_pass):
block_size = int((np.random.rand() * num_states) / (2 * block_scale))
initial_idx = np.maximum((np.random.rand(2) * num_states).astype(int) - block_size, [0, 0])
P_matrix[initial_idx[0]:initial_idx[0] + 2 * block_size,
initial_idx[1]:initial_idx[1] + 2 * block_size] += 1

row_sums = P_matrix.sum(0)  # for each col, sum up all the rows
non_zero_idxs = row_sums.nonzero()
P_matrix[:, non_zero_idxs] /= row_sums[non_zero_idxs]

b_vector = P_matrix[:, -1]
A_matrix = P_matrix[:, :-1]

return A_matrix, b_vector, P_matrix

Then, we have the algorithm to generate $N$-pass MCs. Again, the default parameter is used.

def generate_random_MC_npass(num_states, num_pass=1000):
P_matrix = np.zeros((num_states, num_states + 1))
full_state_idxs = np.arange(num_states)

# for each state, create probabilty that it can reach B
reach_idxs = np.random.choice(full_state_idxs, np.maximum(1, int(num_states * np.random.rand())), replace=False)
reach_probs = np.random.rand(len(reach_idxs))
P_matrix[reach_idxs, -1] = reach_probs

# perform n passes
for _ in range(num_pass):
sel_states = np.random.permutation(full_state_idxs)
choices = np.random.choice(reach_idxs, num_states, replace=True)
temp = np.random.uniform(0, 1 - P_matrix[sel_states, :].sum(1), len(sel_states))
P_matrix[sel_states, choices] += temp
reach_idxs = sel_states

b_vector = P_matrix[:, -1]
A_matrix = P_matrix[:, :-1]

return A_matrix, b_vector, P_matrix

Lastly, we have the algorithm to sample sparse MCs. Once more, the default param. is used.

def generate_random_MC_sparse(num_states, density=0.1):
P_matrix = sparse.random((num_states + 1, num_states), density)
P_matrix = P_matrix.todense()
full_state_idxs = np.arange(num_states)

# for each state, create probabilty that it can reach B
reach_idxs = np.random.choice(full_state_idxs, np.maximum(1, int(num_states * np.random.rand())), replace=False)
reach_probs = np.random.rand(len(reach_idxs))
P_matrix[reach_idxs, -1] = reach_probs

# normalize
row_sums = P_matrix.sum(0)
non_zero_idxs = row_sums.nonzero()
P_matrix[:, non_zero_idxs] /= row_sums[non_zero_idxs]

P_matrix = P_matrix.T
b_vector = P_matrix[:, -1]
A_matrix = P_matrix[:, :-1]

return A_matrix, b_vector, P_matrix