# Projective Normality and Ehrhart Unimodality for Weighted Projective Space
Simplices

Benjamin Braun  Department of Mathematics
University of Kentucky
Lexington, KY 40506–0027<EMAIL_ADDRESS>,  Robert Davis  Department
of Mathematics
Colgate University
Hamilton, NY 13346<EMAIL_ADDRESS>,  Derek Hanely  Department of
Mathematics
University of Kentucky
Lexington, KY 40506–0027<EMAIL_ADDRESS>,  Morgan Lane  Martha Layne
Collins High School
801 Discovery Boulevard
Shelbyville, KY 40065<EMAIL_ADDRESS>and  Liam Solus
Matematik
KTH
SE-100, 44 Stockholm, Sweden<EMAIL_ADDRESS>
(Date: 31 March 2021)

###### Abstract.

Within the intersection of Ehrhart theory, commutative algebra, and algebraic
geometry lie lattice polytopes. Ehrhart theory is concerned with lattice point
enumeration in dilates of polytopes; lattice polytopes provide a sandbox in
which to test many conjectures in commutative algebra; and many properties of
projectively normal toric varieties in algebraic geometry are encoded through
corresponding lattice polytopes. In this article we focus on reflexive
simplices and work to identify when these have the integer decomposition
property (IDP), or equivalently, when certain weighted projective spaces are
projectively normal. We characterize the reflexive, IDP simplices whose
associated weighted projective spaces have one projective coordinate with
weight fixed to unity and for which the remaining coordinates can assume one
of three distinct weights. We show that several subfamilies of such reflexive
simplices have unimodal $h^{\ast}$-polynomials, thereby making progress
towards conjectures and questions of Stanley, Hibi-Ohsugi, and others
regarding the unimodality of their $h^{*}$-polynomials. We also provide
computational results and introduce the notion of reflexive stabilizations to
explore the (non-)ubiquity of reflexive simplices that are simultaneously IDP
and $h^{\ast}$-unimodal.

###### 2020 Mathematics Subject Classification:

Primary: 52B20, 05A15, 11A05

BB was partially supported by National Science Foundation award DMS-1953785.
RD was supported in part by NSF grant DMS-1922998. LS was partially supported
by the Wallenberg AI, Autonomous Systems and Software Program (WASP) funded by
the Knut and Alice Wallenberg Foundation as well as Starting Grant
(Etableringsbidrag) No. 2019-05195 from The Swedish Research Council
(Vetenskapsrådet).

##  1\. Introduction

A subset $P\subseteq\mathbb{R}^{n}$ is a $d$-dimensional (convex) lattice
polytope if it is the convex hull of finitely many points
$\mathbf{v}^{(1)},\ldots,\mathbf{v}^{(k)}\in\mathbb{Z}^{n}$ that span a
$d$-dimensional affine subspace of $\mathbb{R}^{n}$. Many interesting
geometric and algebraic properties of $P$ are revealed by considering the
_cone over $P$_, defined as the non-negative span of the vectors formed by
appending a $1$ to each vertex of $P$, i.e.

$\mathrm{cone}\left(P\right):=\textrm{span}_{\mathbb{R}_{\geq
0}}\\{(1,\mathbf{v}^{(i)}):i=1,\ldots,k\\}\,.$

Two properties of $P$ of particular importance are encoded in the arithmetical
structure of the cone over $P$. First, a lattice polytope $P$ is said to have
the integer decomposition property (or to be IDP) if for every positive
integer $m$ and each
$(m,\mathbf{w})\in\mathrm{cone}\left(P\right)\cap\mathbb{Z}^{n+1}$, there
exist $m$ points $\mathbf{x}_{1},\ldots,\mathbf{x}_{m}\in P\cap\mathbb{Z}^{n}$
for which $(k,\mathbf{w})=\sum_{i}(1,\mathbf{x}_{i})$. Second, letting
$K^{\circ}$ denote the topological interior of a space $K$, $P$ is said to be
_reflexive_ if there exists an integer vector $\mathbf{c}\in
P^{\circ}\cap\mathbb{Z}^{n+1}$ such that

$(1,\mathbf{c})+\left(\mathrm{cone}\left(P\right)\cap\mathbb{Z}^{n+1}\right)=\mathrm{cone}\left(P\right)^{\circ}\cap\mathbb{Z}^{n+1}\,.$

Equivalently, $P$ is reflexive if, possibly after translation by an integer
vector, the origin is contained in $P^{\circ}$ and the geometric dual (or
polar body) of $P$ is also a lattice polytope. There are many interesting
questions about polytopes that are IDP and/or reflexive. For example, it is
typically difficult to identify when a given polytope is IDP and unknown how
to efficiently determine the number of reflexive polytopes of a fixed
dimension (up to unimodular equivalence). Our interest in this paper is
polytopes that are simultaneously IDP and reflexive. Our motivation comes from
conjectures in Ehrhart theory and combinatorial commutative algebra.

Specifically, many open problems in combinatorics ask whether or not a given
combinatorial generating polynomial
$p=p_{0}+p_{1}x+\cdots+p_{d}x^{d}\in\mathbb{Z}_{\geq 0}[x]$ is unimodal; i.e.,
if there exists $t\in[d]\cup\\{0\\}:=\\{1,\ldots,d\\}\cup\\{0\\}$ such that
$p_{0}\leq\cdots\leq p_{t}\geq\cdots\geq p_{d}$. One approach to such
questions is to interpret the polynomial $p$ as the $h$-polynomial associated
to a graded ring, and then use the properties of this ring to derive the
desired inequalities. Perhaps one of the longest standing questions of this
nature is the following, which appeared first in the survey [8]:

######  Conjecture 1.1.

[8] The $h$-polynomial of a standard graded Gorenstein domain is unimodal.

In recent decades, much of the work aimed at resolving this conjecture has
centered around examples in which the integral domain of interest is the
coordinate ring of a projectively normal toric variety, since in this case,
the $h$-polynomial admits a combinatorial interpretation in terms of Ehrhart
theory for lattice polytopes. Hence, one can attempt to either prove or
disprove this conjecture in the special case of semigroup algebras arising
from lattice polytopes. The $h$-polynomials of standard Gorenstein domains
associated to lattice polytopes are precisely the Ehrhart $h^{*}$-polynomials
for IDP reflexive polytopes [9], where $h^{*}$-polynomials are defined in
Section 4.

While much work has been devoted to studying Conjecture 1.1 in the context of
$h^{*}$-polynomials of IDP reflexive polytopes, the conjecture remains open
even in many natural special cases. One challenge in addressing Conjecture 1.1
for arbitrary $P$ is that combinatorial aspects of the $h^{\ast}$-polynomial
are influenced by the topology of triangulations of $P$ [3, 16, 20]. However,
when $P$ is a simplex, the coefficients of the $h^{*}$-polynomial admit a
combinatorial interpretation strictly in terms of lattice point enumeration.
This supports the idea that lattice simplices form a simple, combinatorial
testing ground for Conjecture 1.1. Moreover, there is an algorithmic
classification of reflexive simplices [10]. This makes the family of reflexive
simplices a natural starting point for testing Conjecture 1.1. Despite this
observation, the status of Conjecture 1.1 is still undecided for many
reflexive simplices. This is, in part, due to the fact that those reflexive
simplices which are IDP are not well-understood, and in part due to a general
inability to test the IDP and $h^{*}$-unimodality conditions for general
polytopes living in high dimensions.

Recently, three of the authors [5] gave a characterization of the reflexive
IDP simplices associated to a family of weighted projective spaces, and a
formula for their $h^{\ast}$-polynomials, both based in elementary number
theory. This machinery yields an efficient method for checking the IDP and
unimodality conditions for a relatively broad family of reflexive simplices.
In this paper, we explore how a necessary condition for IDP that relaxes the
IDP characterization of [5] yields a natural parameterization of an infinite
class of reflexive simplices associated to weighted projective spaces. This
parametrization allows us to check the IDP condition for reflexive simplices
having high dimension and large volume, and to investigate $h^{*}$-unimodality
for the resulting IDP reflexives.

The remainder of this paper is structured as follows. In Section 2, we review
properties of the weighted projective spaces simplices
$\Delta_{(1,\mathbf{q})}$, for $\mathbf{q}\in\mathbb{Z}_{\geq 1}^{n}$, which
correspond to weighted projective spaces in which one projective coordinate
has weight $1$ and the others are given by the entries in $\mathbf{q}$. In
Section 3, we investigate when a reflexive $\Delta_{(1,\mathbf{q})}$ is IDP.
We prove Theorem 3.6, which gives a bound on the possible entries of an IDP
reflexive $\mathbf{q}$ with fixed multiplicies on the distinct entries. We
also prove Theorem 3.9, a classification of all IDP reflexive
$\Delta_{(1,\mathbf{q})}$ where $\mathbf{q}$ has three distinct entries. In
Section 4, we consider the question of $h^{*}$-unimodality for
$\Delta_{(1,\mathbf{q})}$. We provide experimental data that suggests that
$h^{*}$-unimodality is rare among the $\Delta_{(1,\mathbf{q})}$. We also prove
Theorem 4.7, establishing that some of the $\Delta_{(1,\mathbf{q})}$ with
three distinct entries in $\mathbf{q}$ are $h^{*}$-unimodal. In Theorem 4.9,
we provide an example of a $\Delta_{(1,\mathbf{q})}$ that is simultaneously
“almost IDP” and “almost unimodal”. In Section 5, we introduce the concept of
$m$-th reflexive stabilizations for a simplex $\Delta_{(1,\mathbf{q})}$. We
prove that for large $m$, these reflexive stabilizations produce non-IDP
simplices that have non-unimodal $h^{*}$-vectors. As most of the proofs for
these results are quite technical, we omit the majority of them to the
Appendix.

##  2\. The Weighted Projective Space Simplex $\Delta_{(1,\mathbf{q})}$

Consider an integer partition $\mathbf{q}\in\mathbb{Z}_{\geq 1}^{n}$ with the
convention $q_{1}\leq\cdots\leq q_{n}$. The lattice simplex associated with
$\mathbf{q}$ is

$\Delta_{(1,\mathbf{q})}:=\mathrm{conv}\left\\{\mathbf{e}_{1},\ldots,\mathbf{e}_{n},-\sum_{i=1}^{n}q_{i}\mathbf{e}_{i}\right\\}$

where $\mathbf{e}_{i}$ denotes the $i^{th}$ standard basis vector in
$\mathbb{R}^{n}$. Set $N(\mathbf{q}):=1+\sum_{i}q_{i}$. One can show, as for
instance in [15, Proposition 4.4], that $N(\mathbf{q})$ is the normalized
volume of $\Delta_{(1,\mathbf{q})}$. Let $\mathcal{Q}$ denote the set of all
lattice simplices of the form $\Delta_{(1,\mathbf{q})}$.

The simplices in $\mathcal{Q}$ correspond to a subset of the simplices
defining weighted projective spaces [10] for which the vector $(1,\mathbf{q})$
gives the weights of the projective coordinates of the associated weighted
projective space. The family of simplices $\mathcal{Q}$ are the focus of
active study [1, 4, 5, 6, 7, 14, 17], in particular in regards to Conjecture
1.1 and the associated IDP and unimodality conditions. This is because
$\mathcal{Q}$ is a relatively general family of simplices (one for each weakly
increasing vector $\mathbf{q}\in\mathbb{Z}_{\geq 1}^{n}$) with diverse lattice
point combinatorics, and because they simultaneously admit tractable
characterizations of the reflexive and IDP conditions. This makes
$\mathcal{Q}$ a natural testing ground for exploring the intricacies of the
relationship between the IDP and unimodality properties while maintaining some
combinatorial control over the examples at hand.

The family $\mathcal{Q}$ admits a natural parametrization (or stratification),
based on the distinct entries in the vector $\mathbf{q}$, that allows us to
test for the reflexive and IDP conditions even when $\Delta_{(1,q)}$ is high-
dimensional and/or has large volume. Given a vector of distinct positive
integers $\mathbf{r}=(r_{1},\ldots,r_{d})\in\mathbb{Z}_{>0}^{d}$, write

$(r_{1}^{x_{1}},r_{2}^{x_{2}},\ldots,r_{d}^{x_{d}}):=(\underbrace{r_{1},r_{1},\ldots,r_{1}}_{x_{1}\text{
times}},\underbrace{r_{2},r_{2},\ldots,r_{2}}_{x_{2}\text{
times}},\ldots,\underbrace{r_{d},r_{d},\ldots,r_{d}}_{x_{d}\text{ times}})\,.$

If
$\mathbf{q}=(q_{1},\ldots,q_{n})=(r_{1}^{x_{1}},r_{2}^{x_{2}},\ldots,r_{d}^{x_{d}})$,
we say that both $\mathbf{q}$ and $\Delta_{(1,\mathbf{q})}$ are _supported_ by
the vector $\mathbf{r}=(r_{1},\ldots,r_{d})$, which has distinct entries, with
_multiplicity_ $\mathbf{x}=(x_{1},\dots,x_{d})$. We write
$\mathbf{q}=(\mathbf{r},\mathbf{x})$ in this case and say that $\mathbf{q}$ is
_$d$ -supported_.

We are particularly interested in the case when the simplex
$\Delta_{(1,\mathbf{q})}$ is reflexive, and we say that $\mathbf{q}$ is
reflexive whenever $\Delta_{(1,\mathbf{q})}$ is reflexive. The following
theorem and setup provide us with a number-theoretic basis for studying
reflexive simplices in $\mathcal{Q}$.

######  Theorem 2.1.

[10] The simplex $\Delta_{(1,\mathbf{q})}\in\mathcal{Q}$ is reflexive if and
only if

(1) $q_{i}\text{ divides }1+\sum_{j=1}^{n}q_{j}\quad\text{for all $1\leq i\leq
n$ }\,.$

Equivalently, if $\mathbf{q}=(\mathbf{r},\mathbf{x})$, then
$\Delta_{(1,\mathbf{q})}$ is reflexive if and only if
$\mathrm{lcm}\left(r_{1},\dots,r_{d}\right)$ divides
$1+\sum_{i=1}^{d}x_{i}r_{i}$.

######  Setup 2.2.

Let $\mathbf{q}$ be reflexive and supported by the vector
$\mathbf{r}=(r_{1},\dots,r_{d})\in\mathbb{Z}_{\geq 1}^{d}$ with multiplicity
$\mathbf{x}=(x_{1},\dots,x_{d})\in\mathbb{Z}_{\geq 1}^{d}$. Let
$\ell=\ell(\mathbf{q})$ be the integer defined by

(2)
$1+\sum_{i=1}^{d}x_{i}r_{i}=\ell\cdot\mathrm{lcm}\left(r_{1},r_{2},\dots,r_{d}\right)\,.$

Finally, we define $\mathbf{s}:=(s_{1},\dots,s_{d})$ where

(3) $s_{i}:=\frac{\mathrm{lcm}\left(r_{1},\dots,r_{d}\right)}{r_{i}}$

for each $1\leq i\leq d$.

This setup provides useful restrictions on $\mathbf{q}$, such as the following
lemma.

######  Lemma 2.3 (Braun, Liu, [7]).

In Setup 2.2, we have that $\gcd(r_{1},\dots,r_{d})=1$ and thus

(4)
$\mathrm{lcm}\left(s_{1},\dots,s_{d}\right)=\mathrm{lcm}\left(r_{1},\dots,r_{d}\right).$

######  Remark 2.4.

The Hermite normal form for the simplex $\Delta_{(1,\mathbf{q})}-e_{1}$ is

$\left[\begin{array}[]{cccccc}0&1&0&\cdots&0&N(\mathbf{q})-q_{2}\\\
0&0&1&\cdots&0&N(\mathbf{q})-q_{3}\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&0&\cdots&1&N(\mathbf{q})-q_{d}\\\
0&0&0&\cdots&0&N(\mathbf{q})\end{array}\right]\,,$

that is, $\Delta_{(1,\mathbf{q})}$ is unimodularly equivalent to the convex
hull of the columns of this matrix. This is an example of a simplex with what
Hibi, Higashitani, and Li [13] call a one-column Hermite normal form. Their
paper provides a partial study of Ehrhart-theoretic properties of such
polytopes.

##  3\. The Integer Decomposition Property

###  3.1. Restrictions on IDP reflexive $\Delta_{(1,\mathbf{q})}$

For each $\mathbf{r}$-vector, it is known [5] that there are infinitely many
reflexive $\Delta_{(1,\mathbf{q})}$’s supported on $\mathbf{r}$. Given a pair
$\Delta_{(1,\mathbf{q})}$ and $\Delta_{(1,\mathbf{p})}$, both reflexive and
IDP, Braun and Davis [4] proved that a new reflexive IDP
$\Delta_{(1,\mathbf{y})}$ can be constructed as shown in the following
theorem. Suppose that $\Delta_{(1,\mathbf{q})}\subset\mathbb{R}^{n}$ and
$\Delta_{(1,\mathbf{p})}\subset\mathbb{R}^{m}$ are reflexive and the vertices
of $\Delta_{(1,\mathbf{p})}$ are labeled as $v_{0},v_{1},\ldots,v_{m}$. For
every $i=0,1,\ldots,m$, define the _affine free sum_

$\Delta_{(1,\mathbf{q})}\ast_{i}\Delta_{(1,\mathbf{p})}:=\mathrm{conv}\left\\{(\Delta_{(1,\mathbf{q})}\times
0^{m})\cup(0^{n}\times\Delta_{(1,\mathbf{p})}-v_{i})\right\\}\subset\mathbb{R}^{n+m}.$

The notion of an affine free sum can be generalized significantly [2], but in
this article it will not be necessary.

######  Theorem 3.1 (Braun, Davis [4]).

The simplex $\Delta_{(1,\mathbf{q})}$ is reflexive and arises as a free sum
$\Delta_{(1,\mathbf{p})}*_{0}\Delta_{(1,\mathbf{w})}$ if and only if
$\Delta_{(1,\mathbf{p})}$ and $\Delta_{(1,\mathbf{w})}$ are reflexive and
$\mathbf{q}=(\mathbf{p},(1+\sum_{i}p_{i})\mathbf{w})$. If
$\Delta_{(1,\mathbf{p})}$ is IDP reflexive and $\Delta_{(1,\mathbf{q})}$ is
IDP, then $\Delta_{(1,\mathbf{q})}*_{0}\Delta_{(1,\mathbf{p})}$ is IDP.
Further, if $\Delta_{(1,\mathbf{p})}$ and $\Delta_{(1,\mathbf{q})}$ are
reflexive, IDP, and $h^{*}$-unimodal (defined in Section 4), then so is
$\Delta_{(1,\mathbf{q})}*_{0}\Delta_{(1,\mathbf{p})}$.

Thus, there are infinitely many reflexive IDP $\Delta_{(1,\mathbf{q})}$’s that
arise as a result of the affine free sum operation, and these all satisfy
Conjecture 1.1. However, the support vector for
$\Delta_{(1,\mathbf{q})}*_{0}\Delta_{(1,\mathbf{p})}$ is distinct from that of
$\mathbf{p}$ and $\mathbf{q}$, so this operation does not respect the
stratification of $\mathcal{Q}$ given by support vectors. In fact, for many
$\mathbf{r}$-vectors, it is impossible to generate infinitely many reflexive
IDP $\Delta_{(1,\mathbf{q})}$’s supported on $\mathbf{r}$, as the following
theorem shows.

######  Theorem 3.2 (Braun, Davis, Solus [5]).

Given a support vector $\mathbf{r}\in\mathbb{Z}_{\geq 1}^{d}$, if there exists
some $j<d$ such that $r_{j}\nmid r_{d}$, then only finitely many reflexive IDP
$\Delta_{(1,\mathbf{q})}$’s are supported on $\mathbf{r}$.

Computational experiments suggest that $\Delta_{(1,\mathbf{q})}$ satisfying
the criteria in Theorem 3.2 are rare. Specifically, consider all
$\mathbf{r}$-vectors that are partitions of $M\leq 75$ with distinct entries,
such that there exist some $r_{j}$ such that $r_{j}\nmid r_{d}$. Table 1 shows
that only 509 IDP reflexives are supported on $\mathbf{r}$-vectors of this
type. While this suggests that IDP reflexive $\Delta_{(1,\mathbf{q})}$’s are
rare, it is important to keep in mind that this represents a relatively small
sample set of simplices. For example,

$\mathbf{q}=(210,211,211,211,211,\underbrace{1055,1055,\ldots,1055}_{41\text{
times}})$

is not among this sample, but it is both IDP and reflexive with $210\nmid
1055$.

# of $\mathbf{r}$-vectors with some $r_{j}\nmid r_{d}$ | # of IDP reflexives supported by these
---|---
501350 | 509
Table 1. Experimental results.

Fortunately, the following theorem provides a number-theoretic
characterization of the IDP property for reflexive $\Delta_{(1,\mathbf{q})}$.

######  Theorem 3.3 (Braun, Davis, Solus [5]).

The reflexive simplex $\Delta_{(1,\mathbf{q})}$ is IDP if and only if for
every $j=1,\ldots,n$, for all $b=1,\ldots,q_{j}-1$ satisfying

(5) $b\left(\genfrac{}{}{}{}{1+\sum_{i\neq j}q_{i}}{q_{j}}\right)-\sum_{i\neq
j}\left\lfloor\genfrac{}{}{}{}{bq_{i}}{q_{j}}\right\rfloor\geq 2$

there exists a positive integer $c<b$ satisfying the following equations,
where the first is considered for all $1\leq i\leq n$ with $i\neq j$:

(6)
$\left\lfloor\genfrac{}{}{}{}{bq_{i}}{q_{j}}\right\rfloor-\left\lfloor\genfrac{}{}{}{}{cq_{i}}{q_{j}}\right\rfloor=\left\lfloor\genfrac{}{}{}{}{(b-c)q_{i}}{q_{j}}\right\rfloor,\text{
and}$ (7) $c\left(\genfrac{}{}{}{}{1+\sum_{i\neq
j}q_{i}}{q_{j}}\right)-\sum_{i\neq
j}\left\lfloor\genfrac{}{}{}{}{cq_{i}}{q_{j}}\right\rfloor=1.$

The next corollary of Theorem 3.3 provides a necessary condition for a
reflexive $\Delta_{(1,\mathbf{q})}$ to be IDP. This condition is an essential
tool in our study of IDP reflexive elements of $\mathcal{Q}$.

######  Corollary 3.4 (Braun, Davis, Solus [5]).

If $\Delta_{(1,\mathbf{q})}$ is reflexive and IDP, then for all
$j=1,2,\ldots,n$,

$1+\sum_{i=1}^{n}(q_{i}\bmod q_{j})=q_{j}$

or equivalently

$1+\sum_{i=1}^{n}x_{i}(r_{i}\bmod r_{j})=r_{j}\,.$

######  Definition 3.5.

Any $\mathbf{q}$ satisfying one (hence both) of these equations for all
$j=1,\dots,n$ is said to satisfy _the necessary condition for IDP_.

Note that if $\mathbf{q}$ satisfies the necessary condition for IDP, then
$\Delta_{(1,\mathbf{q})}$ is reflexive.

###  3.2. Stratifying by multiplicity instead of support

The necessary condition for IDP allows us to produce the following new
refinement of Theorem 3.2. It has the added benefit of giving a geometric
interpretation of IDP reflexive simplices as lattice points contained in a
$d$-dimensional box with boundary determined by the coordinates of the support
vector $\mathbf{r}\in\mathbb{Z}_{\geq 1}^{d}$.

######  Theorem 3.6.

Let $(\mathbf{r},\mathbf{x})=\mathbf{q}\in\mathbb{Z}_{\geq 1}^{n}$, where
$\mathbf{q}$ has at least two distinct entries and $r_{1}<r_{2}<\cdots<r_{d}$.
If $\Delta_{(1,\mathbf{q})}$ is reflexive and IDP, then

$x_{i}\leq r_{i+1}/r_{i}$

for all $i\leq d-1$. Further, if there exists some $j<d$ such that $r_{j}\nmid
r_{d}$, then

$x_{d}\leq r_{j}/(r_{d}\bmod r_{j})\,.$

Thus, if there exists some $j<d$ such that $r_{j}\nmid r_{d}$, then there are
at most finitely many IDP reflexives supported on $\mathbf{r}$.

###### Proof.

Let $j<d$, and assume that $\Delta_{(1,\mathbf{q})}$ is reflexive and IDP.
Then by Corollary 3.4, we have

$x_{j}r_{j}\leq 1+\sum_{i=1}^{d}x_{i}(r_{i}\bmod r_{j+1})=r_{j+1}\,,$

from which the first inequality follows. Similarly, if $r_{j}\nmid r_{d}$,
then

$x_{d}(r_{d}\bmod r_{j})\leq 1+\sum_{i=1}^{d}x_{i}(r_{i}\bmod r_{j})=r_{j}\,,$

from which the second inequality follows. ∎

Theorem 3.6 indicates that there are important relationships between the
multiplicity vector $\mathbf{x}$ of $\mathbf{q}$ and the support vector
$\mathbf{r}$. By shifting our primary focus to the multiplicity vector, we are
able to give a complete classification of all reflexive IDP
$\Delta_{(1,\mathbf{q})}$’s that are supported on up to $3$ distinct entries.
If $\mathbf{x}$ has one entry, it is straightforward to prove the following.

######  Proposition 3.7.

For $\mathbf{q}=(r_{1}^{x_{1}})$, if $\Delta_{(1,\mathbf{q})}$ is IDP
reflexive, then $\mathbf{q}=(1,1,\ldots,1)$.

If $\mathbf{x}$ has two entries, meaning that $\mathbf{r}$ has two distinct
entries, the following theorem applies.

######  Theorem 3.8 (Braun, Davis, Solus [5]).

For the vector $\mathbf{q}=(r_{1}^{x_{1}},r_{2}^{x_{2}})$,
$\Delta_{(1,\mathbf{q})}$ is IDP reflexive if and only if it satisfies the
necessary condition. The following is a classification of all such vectors,
for $x_{1},x_{2}\geq 1$:

  1. (1)

$q=(1^{x_{1}},(1+x_{1})^{x_{2}})$

  2. (2)

$q=((1+x_{2})^{x_{1}},(1+(1+x_{2})x_{1})^{x_{2}})$

Note that in the first case $r_{1}\mid r_{2}$ while in the second case
$r_{1}\nmid r_{2}$. We next extend these known results using Theorem 3.6 and
Corollary 3.4 to the $3$-supported case.

######  Theorem 3.9.

Consider a $3$-supported vector $\mathbf{q}=(\mathbf{r},\mathbf{x})$ such that
$\Delta_{(1,\mathbf{q})}$ satisfies the necessity condition for IDP given in
Corollary 3.4. If $\mathbf{x}=(x_{1},x_{2},x_{3})$ is the multiplicity vector,
then $\mathbf{r}$ is of one of the following forms.

  1. (i)

$\mathbf{r}=(1,1+x_{1},(1+x_{1})(1+x_{2}))$.

  2. (ii)

$\mathbf{r}=(1+x_{2},1+x_{1}(1+x_{2}),(1+x_{1}(1+x_{2}))(1+x_{2}))$.

  3. (iii)

$\mathbf{r}=((1+x_{2})(1+x_{3}),1+x_{1}(1+x_{2})(1+x_{3}),(1+x_{1}(1+x_{2})(1+x_{3}))(1+x_{2}))$.

  4. (iv)

$\mathbf{r}=(1,(1+x_{1})(1+x_{3}),(1+x_{1})(1+x_{2}(1+x_{3})))$.

  5. (v)

$\mathbf{r}=(1+(1+x_{3})x_{2},(1+x_{3})(1+x_{1}(1+(1+x_{3})x_{2})),(1+(1+(1+x_{3})x_{2})x_{1})(1+(1+x_{3})x_{2}))$.

  6. (vi)

$\mathbf{r}=((1+x_{3})(1+(1+x_{3})x_{2}),(1+x_{3})(1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})),(1+(1+x_{3})(1+(1+x_{3})x_{2})x_{1})(1+(1+x_{3})x_{2}))$.

  7. (vii)

$\mathbf{r}=(1+x_{3},(1+x_{3})(1+x_{1}(1+x_{3})),(1+(1+x_{3})x_{1})(1+(1+x_{3})x_{2}))$.

  8. (viii)

There exists some $k,s\geq 1$, where

$\mathbf{r}=(1+kx_{2},(skx_{2}+s+k)(1+x_{1}(1+kx_{2})),(1+x_{1}(1+kx_{2}))(1+x_{2}(skx_{2}+s+k)))$

and

$\mathbf{x}=(x_{1},x_{2},skx_{2}+s-k+1)\,.$

Further, the first seven $\mathbf{r}$-vectors produce IDP
$\Delta_{(1,\mathbf{q})}$’s, while (8) does not.

As the necessary condition for IDP (see Definition 3.5) implies reflexivity,
one immediate consequence of Proposition 3.7 together with Theorems 3.8 and
3.9 is a complete classification of all IDP reflexive simplicies corresponding
to weighted projective spaces with one projective coordinate having weight
unity and the others limited to three options. Note that the first seven
$\mathbf{r}$-vectors in Theorem 3.9 each correspond to a unique divisibility
criteria for $\mathbf{r}=(r_{1},r_{2},r_{3})$, as follows:

  1. _(i)_

$r_{1}\mid r_{2},r_{1}\mid r_{3},r_{2}\mid r_{3}$

  2. _(ii)_

$r_{1}\nmid r_{2},r_{1}\mid r_{3},r_{2}\mid r_{3}$

  3. _(iii)_

$r_{1}\nmid r_{2},r_{1}\nmid r_{3},r_{2}\mid r_{3}$

  4. _(iv)_

$r_{1}\mid r_{2},r_{1}\mid r_{3},r_{2}\nmid r_{3}$

  5. _(v)_

$r_{1}\nmid r_{2},r_{1}\mid r_{3},r_{2}\nmid r_{3}$

  6. _(vi)_

$r_{1}\nmid r_{2},r_{1}\nmid r_{3},r_{2}\nmid r_{3}$

  7. _(vii)_

$r_{1}\mid r_{2},r_{1}\nmid r_{3},r_{2}\nmid r_{3}$

  8. _(viii)_

$r_{1}\nmid r_{2},r_{1}\mid r_{3},r_{2}\nmid r_{3}$

We see that _(v)_ and _(viii)_ share the same divisibility pattern, yet of
these two families only _(v)_ contains IDP simplices. Note that for each
positive integer vector $\mathbf{x}$ of length at most three, and for each
divisibility condition on the support vector $\mathbf{r}$, there is at most
one support vector $\mathbf{r}$ such that $\mathbf{q}=(\mathbf{r},\mathbf{x})$
is reflexive IDP. It would be of interest to determine if this is true for
$\mathbf{x}$ of arbitrary length.

##  4\. Ehrhart Unimodality

###  4.1. Properties of $h^{*}$-polynomials for $\Delta_{(1,\mathbf{q})}$

The _Ehrhart function_ of $P$ is the lattice point enumerator
$i(P;t):=|tP\cap\mathbb{Z}^{n}|$, where $tP:=\\{tp:p\in P\\}$ denotes the
$t^{th}$ dilate of the polytope $P$. It is well-known [11] that $i(P;t)$ is a
polynomial in $t$ of degree $d\leq\dim(P)$. The _Ehrhart series_ of $P$ is the
rational function

$\operatorname{Ehr}_{P}(z):=\sum_{t\geq
0}i(P;t)z^{t}=\frac{h_{0}^{*}+h_{1}^{*}z+\cdots+h_{d}^{*}z^{d}}{(1-z)^{\dim(P)+1}},$

where the coefficients $h^{\ast}_{0},h^{\ast}_{1},\ldots,h^{\ast}_{d}$ are all
nonnegative integers [18]. The polynomial
$h^{\ast}(P;z):=h_{0}^{*}+h_{1}^{*}z+\cdots+h_{d}^{*}z^{d}$ is called the
_(Ehrhart) $h^{\ast}$-polynomial_ of $P$, and it is the $h$-polynomial of
interest in Conjecture 1.1 when $\operatorname{Ehr}_{P}(z)$ is the Hilbert
series of the coordinate ring $A_{P}$ of the toric variety associated to the
lattice polytope $P$. It is well-known that $A_{P}$ is standard if and only if
$P$ is IDP. A result of Stanley [19] states that $A_{P}$ is Gorenstein if and
only if $h^{\ast}(P;z)$ is symmetric with respect to its degree $s$; i.e.,
$h_{i}^{\ast}=h_{s-i}^{\ast}$ for all $i$. Hibi [12] proved the special case
that $P$ is reflexive if and only if $h_{i}^{\ast}=h_{\dim(P)-i}^{\ast}$ for
all $i$. Further, Bruns and Römer [9] proved that if $A_{P}$ is Gorenstein,
then $h^{\ast}(P;z)$ is the $h^{*}$-polynomial of a reflexive polytope. The
Bruns and Römer result implies that the family of lattice polytopes of
interest in Conjecture 1.1 are the $d$-dimensional IDP lattice polytopes with
symmetric $h^{\ast}$-polynomials of degree $d$. Unlike the situation for
general lattice polytopes, for $\Delta_{(1,\mathbf{q})}$ there is an explicit
formula for $h^{*}(\Delta_{(1,\mathbf{q})};z)$.

######  Theorem 4.1 (Braun, Davis, Solus [5]).

The $h^{*}$-polynomial of $\Delta_{(1,\mathbf{q})}$ is

$h^{*}(\Delta_{(1,\mathbf{q})};z)=\sum_{b=0}^{q_{1}+\cdots+q_{n}}z^{w(\mathbf{q},b)}$

where

(8)
$w(\mathbf{q},b):=b-\sum_{i=1}^{n}\left\lfloor\frac{q_{i}b}{1+q_{1}+\cdots+q_{n}}\right\rfloor\,.$

######  Example 4.2.

Let $\mathbf{q}=(2,2,3)$. Then $\displaystyle
w(\mathbf{q},b):=b-2\left\lfloor\frac{b}{4}\right\rfloor-\left\lfloor\frac{3b}{8}\right\rfloor$,
and thus

$h^{*}(\Delta_{(1,\mathbf{q})};z)=1+2z+4z^{2}+z^{3}.$

Prior to discussing the case of reflexive IDP $\Delta_{(1,\mathbf{q})}$, it is
worth asking whether or not $h^{*}$-unimodality is expected for an arbitrary
$\Delta_{(1,\mathbf{q})}$. Based on experiments conducted via SageMath [21],
there appears to be a trend in the overall frequency of $h^{*}$-unimodality of
$\Delta_{(1,\mathbf{q})}$ as the normalized volume $N(\mathbf{q})$ grows.
Specifically, let $V(M):=\\{\Delta_{(1,\mathbf{q})}:N(\mathbf{q})=M\\}$ and
define $\mathrm{un}(M)$ to be the fraction of the simplices in $V(M)$ having a
unimodal $h^{*}$-vector. The values of $\mathrm{un}(M)$ for $M\leq 75$ have
been computed exactly, involving 61,537,394 simplices. A plot of approximate
values of $\mathrm{un}(M)$ using random samples is given in Figure 1.

Figure 1. Approximate values of $\mathrm{un}(M)$ for $M\leq 120$, based on
randomly sampled partitions.

The limiting behavior of $\mathrm{un}(M)$ is not known; various random
sampling of $\mathbf{q}$ with fixed values of $N(\mathbf{q})$ suggests that
$\mathrm{un}(M)$ continues to decrease as $M\to\infty$, but the rate of
decrease also appears to be approaching zero. This leads us to the following
question.

######  Question 4.3.

Is $\displaystyle\lim_{M\to\infty}\mathrm{un}(M)=0$? If not, does the limit
exist? If not, what are the values of
$\displaystyle\limsup_{M\to\infty}\mathrm{un}(M)$ and
$\displaystyle\liminf_{M\to\infty}\mathrm{un}(M)$?

Based on this empirical data, it seems reasonable to suspect that those
$\Delta_{(1,\mathbf{q})}$ having unimodal $h^{*}$-polynomials are uncommon,
and thus Conjecture 1.1 is suggesting that reflexive IDP
$\Delta_{(1,\mathbf{q})}$’s are unusual in this regard. Returning our focus to
the case where $\Delta_{(1,\mathbf{q})}$ is IDP reflexive, we will later need
to know that the $h^{*}$-polynomial of $\Delta_{(1,\mathbf{q})}$ often admits
a geometric series as a factor, as the following definition and theorem
demonstrate.

######  Definition 4.4.

Suppose $\mathbf{r},\mathbf{x}$, $\ell$ and $\mathbf{s}$ are as given in Setup
2.2. We define

$g_{\mathbf{r}}^{\mathbf{x}}(z):=\sum_{0\leq\alpha<\mathrm{lcm}\left(r_{1},\ldots,r_{d}\right)}z^{u(\alpha)}$

where

$u(\alpha)=u_{\mathbf{r}}^{\mathbf{x}}(\alpha):=\alpha\ell-\sum_{i=1}^{d}x_{i}\left\lfloor\frac{\alpha}{s_{i}}\right\rfloor\,.$

######  Theorem 4.5 (Braun, Liu [7]).

Assuming Setup 2.2, we have that

$h^{*}(\Delta_{(1,\mathbf{q})};z)=\left(\sum_{t=0}^{\ell-1}z^{t}\right)\cdot
g_{\mathbf{r}}^{\mathbf{x}}(z)\,.$

######  Example 4.6.

For $\mathbf{q}=(1^{7},3^{4},5^{5})$, we have

$(z^{2}+z+1)(x^{14}+x^{11}+2x^{10}+2x^{8}+3x^{7}+2x^{6}+2x^{4}+x^{3}+1)\,.$

Note that in this case, $\ell=3$ and a factor of $z^{2}+z+1$ appears in the
$h^{*}$-polynomial.

###  4.2. $h^{*}$-unimodality and $\Delta_{(1,\mathbf{q})}$

It has been previously shown [5] that all $1$-supported and $2$-supported IDP
reflexive $\Delta_{(1,\mathbf{q})}$ are $h^{*}$-unimodal. We consider here
several of the $3$-supported classes given in Theorem 3.9.

######  Theorem 4.7.

For $\mathbf{x}=(x_{1},x_{2},x_{3})$ a positive integer vector, if
$\mathbf{q}=(\mathbf{r},\mathbf{x})$ where $\mathbf{r}$ is one of the
following forms, then $\Delta_{(1,\mathbf{q})}$ is $h^{*}$-unimodal.

  1. (i)

$\mathbf{r}=(1,1+x_{1},(1+x_{1})(1+x_{2}))$.

  2. (ii)

$\mathbf{r}=(1+x_{2},1+x_{1}(1+x_{2}),(1+x_{1}(1+x_{2}))(1+x_{2}))$.

  3. (iv)

$\mathbf{r}=(1,(1+x_{1})(1+x_{3}),(1+x_{1})(1+x_{2}(1+x_{3})))$.

###### Proof.

As shown in the proof of Theorem 3.9, these cases all arise as affine free
sums. Thus, by Theorem 3.1, they are $h^{*}$-unimodal. ∎

######  Remark 4.8.

Based on experimental evidence using SageMath [21], we conjecture that the
remaining four cases, listed here, are also $h^{*}$-unimodal.

  1. (iii)

$\mathbf{r}=((1+x_{2})(1+x_{3}),1+x_{1}(1+x_{2})(1+x_{3}),(1+x_{1}(1+x_{2})(1+x_{3}))(1+x_{2}))$.

  2. (v)

$\mathbf{r}=(1+(1+x_{3})x_{2},(1+x_{3})(1+x_{1}(1+(1+x_{3})x_{2})),(1+(1+(1+x_{3})x_{2})x_{1})(1+(1+x_{3})x_{2}))$.

  3. (vi)

$\mathbf{r}=((1+x_{3})(1+(1+x_{3})x_{2}),(1+x_{3})(1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})),(1+(1+x_{3})(1+(1+x_{3})x_{2})x_{1})(1+(1+x_{3})x_{2}))$.

  4. (vii)

$\mathbf{r}=(1+x_{3},(1+x_{3})(1+x_{1}(1+x_{3})),(1+(1+x_{3})x_{1})(1+(1+x_{3})x_{2}))$.

However, these do not arise as affine free sums. While a direct proof of
$h^{*}$-unimodality for these cases might be possible, it is not clear how to
carry this out other than by ad hoc arguments given case-by-case, similar to
the proof of Theorem 3.9.

It is interesting that among the $\Delta_{(1,\mathbf{q})}$ simplices, some can
be found that are “on the boundary” of both $h^{*}$-unimodality and the IDP
condition. One example is the following. Recall that for a lattice polytope
$P$, the _Hilbert basis_ of $\mathrm{cone}\left(P\right)$ is the minimal
generating set of $\mathrm{cone}\left(P\right)\cap\mathbb{Z}^{n+1}$. Thus, $P$
is IDP if and only if the Hilbert basis of $\mathrm{cone}\left(P\right)$
consists of the elements at height $1$ in $\mathrm{cone}\left(P\right)$, i.e.
$(1,P)\cap\mathbb{Z}^{n+1}$.

######  Theorem 4.9.

For $n\geq 1$, define $\mathbf{r}(n)=(1,3n,10n,15n)$ and
$\mathbf{x}(n)=(2n-1,1,1,1)$. Thus,
$\mathbf{q}(n):=(\mathbf{r}(n),\mathbf{x}(n))=\mathrm{rs}((3n,10n,15n))$. For
$n\geq 2$ with $\mathbf{q}=(\mathbf{r}(n),\mathbf{x}(n))$, we have

$h^{*}(\Delta_{(1,\mathbf{q})};z)=(1+z^{2}+z^{4}+z^{6}+\cdots+z^{2n-2})\cdot(1+7z+14z^{2}+7z^{3}+z^{4})\,,$

which can be verified to be non-unimodal. For $n\geq 1$ with
$\mathbf{q}=(\mathbf{r}(n),\mathbf{x}(n))$, let
$V(n)=\\{(1,\mathbf{v}):\mathbf{v}\text{ a vertex of
}\Delta_{(1,\mathbf{q})}\\}$. The Hilbert basis for
$\mathrm{cone}\left(\Delta_{(1,\mathbf{q})}\right)$ consists of $V(n)$ and the
columns of the following matrix (where the height coordinate is the first
entry):

$\begin{bmatrix}1&1&1&1&1&1&1&2&2\\\ 0&0&0&0&0&0&0&0&0\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\\
0&0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&-1&-2&-3\\\ 0&0&-1&-1&-2&-3&-4&-7&-10\\\
0&-1&-1&-2&-3&-5&-6&-10&-15\\\ \end{bmatrix}$

Thus, there are exactly two elements in the Hilbert basis of height greater
than $1$, both of which are at height $2$.

As another example, for $n\geq 2$, let

$\mathbf{q}=((n,(2n-1)(n+1),2n(n+1)),(1,1,2(n-1)))\,.$

For $n\leq 20$, it has been verified that

$\displaystyle h^{*}(\Delta_{(1,\mathbf{q})};z)=(1,(n+1)^{2},$
$\displaystyle(2n+1)(n+1)+1,(2n+1)(n+1),(2n+1)(n+1)+1,\ldots,$
$\displaystyle(2n+1)(n+1)+1,(2n+1)(n+1),(2n+1)(n+1)+1,(n+1)^{2},1)$

and that the Hilbert basis of
$\mathrm{cone}\left(\Delta_{(1,\mathbf{q})}\right)$ consists of the points
$(1,\Delta_{(1,\mathbf{q})})\cap\mathbb{Z}^{2n+1}$ together with following the
lattice point at height two (as given by the first coordinate):

$(2,-1,-2n-1,-2(n-1),-2(n-1),-2(n-1),\ldots,-2(n-1))^{T}\,.$

Thus, this family of simplices is another example of polytopes on the boundary
of both IDP and $h^{*}$-unimodality; this family is more arithmetically
complicated than the one given in Theorem 4.9.

##  5\. Reflexive Stabilizations

It is unlikely that a random integer partition $\mathbf{q}$ is reflexive.
Further, Theorem 3.6 tells us that for many support vectors $\mathbf{r}$,
$\mathbf{r}$ only supports finitely many IDP reflexives, though it supports
infinitely many reflexives. Thus, it is interesting to consider both (i) how
to assign reflexive elements of $\mathcal{Q}$ to an arbitrary integer
partition and (ii) the behavior of $h^{*}$-vectors for reflexive elements of
$\mathcal{Q}$ where at least one entry has a large multiplicity. We can begin
investigating both of these through a process we call _reflexive
stabilization_ , which allows us to assign to each integer partition
$\mathbf{q}$ a family of reflexive simplices in $\mathcal{Q}$.

######  Definition 5.1.

Let $\mathbf{q}\in\mathbb{Z}_{\geq 1}^{n}$. The _first reflexive
stabilization_ of $\mathbf{q}$, denoted $\mathrm{rs}(\mathbf{q})$, is the
vector $(1,1,\ldots,1,\mathbf{q})$ such that
$\Delta_{(1,\mathrm{rs}(\mathbf{q}))}$ is reflexive and the number of $1$’s
prepended to $\mathbf{q}$ is the minimum necessary for this condition to hold;
if $\mathbf{q}$ is reflexive, then no $1$’s are prepended. We say the number
of $1$’s prepended to $\mathbf{q}$ in $\mathrm{rs}(\mathbf{q})$ is the
_reflexive stabilization number_ of $\mathbf{q}$, denoted
$\mathrm{rsn}(\mathbf{q})$. Thus, we can write

$\mathrm{rs}(\mathbf{q}):=(1^{\mathrm{rsn}(\mathbf{q})},\mathbf{q})\,.$

More generally, the _$m$ -th reflexive stabilization_ of $\mathbf{q}$, denoted
$\mathrm{rs}(\mathbf{q},m)$, is defined as

$\mathrm{rs}(\mathbf{q},m):=(1^{\mathrm{rsn}(\mathbf{q})+(m-1)\cdot\mathrm{lcm}\left(\mathbf{q}\right)},\mathbf{q})\,.$

Note that when prepending $\mathrm{lcm}\left(\mathbf{q}\right)$ copies of $1$
as many times as desired, the corresponding simplex will remain reflexive.

######  Example 5.2.

Let $\mathbf{q}=(2,2,3)$. Then $1+2+2+3=8$, and thus
$\mathrm{rs}(\mathbf{q})=(1,1,1,1,2,2,3)$ is reflexive with
$\mathrm{rsn}(\mathbf{q})=4$. Further,
$\mathrm{rs}(\mathbf{q},3)=(\mathbf{r},\mathbf{x})=((1,2,3),(16,2,1))$.

We can restrict ourselves to considering only $\mathbf{q}\in\mathbb{Z}_{\geq
2}^{n}$, since if $1$ is an entry of $\mathbf{q}$, then
$\mathrm{rs}(\mathbf{q})$ is $\mathrm{rs}(\mathbf{q}^{\prime},m)$ for some $m$
with $\mathbf{q}^{\prime}$ given by the entries of $\mathbf{q}$ that are not
equal to $1$. We begin by noting that for $m$ sufficiently large, reflexive
stabilizations are not $h^{*}$-unimodal.

######  Theorem 5.3.

Assume that $\mathbf{q}\in\mathbb{Z}_{\geq 2}^{n}$. For $m$ sufficiently
large, $\Delta_{(1,\mathrm{rs}(\mathbf{q},m))}$ is not $h^{*}$-unimodal.
Further, $h^{*}(\Delta_{(1,\mathrm{rs}(\mathbf{q},m))})$ contains only $1$’s
and $2$’s.

A proof is given in Section A. We next show that, as expected from Theorem
3.6, reflexive stabilizations generally eliminate the IDP property.

######  Theorem 5.4.

Assume that $\mathbf{q}\in\mathbb{Z}_{\geq 2}^{n}$. For $m\geq 2$,
$\Delta_{(1,\mathrm{rs}(\mathbf{q},m))}$ is not IDP.

A proof is given in Section A. In order to study those partitions $\mathbf{q}$
such that $\Delta_{(1,\mathrm{rs}(\mathbf{q}))}$ is IDP and/or
$h^{*}$-unimodal, it makes sense to begin by considering the case where
$\mathrm{rsn}(\mathbf{q})$ is small, potentially keeping the value of
$\ell(\mathbf{q})$ small as well. When $\mathrm{rsn}(\mathbf{q})=1$ and
$\ell(\mathrm{rs}(\mathbf{q}))=1$, we have the following curious result.

######  Theorem 5.5.

Suppose that $\mathrm{rsn}(\mathbf{q})=1$ and that
$\ell(\mathrm{rs}(\mathbf{q}))=1$, i.e., that
$1+\sum_{i=1}^{n}q_{i}=\mathrm{lcm}\left(\mathbf{q}\right)-1$. Then for all
$0\leq b\leq\sum_{i}q_{i}$, we have
$w(\mathbf{q},b)=w(\mathrm{rs}(\mathbf{q}),b)$ and hence

$h^{*}(\Delta_{(1,\mathrm{rs}(\mathbf{q}))};z)=h^{*}(\Delta_{(1,\mathbf{q})};z)+z^{n+1}\,.$

It would be interesting to determine whether or not there are similar additive
structures for the $h^{*}$-polynomial in other cases where
$\mathrm{rsn}(\mathbf{q})$ and that $\ell(\mathrm{rs}(\mathbf{q}))$ are small.

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##  Appendix A Proofs

###  A.1. Proof of derivation of $(i)$-$(viii)$ in Theorem 3.9

We first suppose that $(\mathbf{r},\mathbf{x})$ satisfies the necessity
condition, and show that the resulting $\mathbf{r}$-vectors must be of one of
the eight types listed. Since $r_{1}<r_{2}<r_{3}$, reducing modulo $r_{3}$
gives $r_{3}=1+r_{1}x_{1}+r_{2}x_{2}\,.$ If we next consider the modulo
$r_{2}$ necessary condition, then substituting the above for $r_{3}$ and
simplifying gives $r_{2}=1+x_{1}r_{1}+x_{3}((1+r_{1}x_{1})\bmod r_{2})\,.$ The
challenge here is that we would like to specify $r_{2}$ using this formula,
but it involves a remainder which could fluctuate. The key observation is to
recall that if the necessary condition for IDP holds, then Theorem 3.6 implies
$x_{1}\leq r_{2}/r_{1}$. Thus, we have

(9) $1+x_{1}r_{1}\leq 1+(r_{2}/r_{1})r_{1}=1+r_{2}\,.$

There are now three cases to consider.

Case 1: Suppose we have equality in (9). It is immediate that in this case

$(1+r_{1}x_{1})\bmod r_{2}=(1+r_{2})\bmod r_{2}=1\,,$

and thus $r_{2}=1+x_{1}r_{1}+x_{3}=r_{2}+1+x_{3}\,.$ As $x_{3}\geq 1$, this
yields a contradiction, and thus this case does not occur.

Case 2: Consider if $1+x_{1}r_{1}=r_{2}$ in (9). Then

$(1+r_{1}x_{1})\bmod r_{2}=r_{2}\bmod r_{2}=0\,.$

Thus, we have
$(\mathbf{r},\mathbf{x})=((r_{1},1+x_{1}r_{1},r_{3}),(x_{1},x_{2},x_{3}))\,.$
However, we know that

$r_{3}=1+r_{1}x_{1}+r_{2}x_{2}=(1+x_{1}r_{1})(1+x_{2})\,,$

and thus
$(\mathbf{r},\mathbf{x})=((r_{1},1+x_{1}r_{1},(1+x_{1}r_{1})(1+x_{2})),(x_{1},x_{2},x_{3}))\,.$
If $r_{1}=1$, then the result is
$(\mathbf{r},\mathbf{x})=((1,1+x_{1},(1+x_{1})(1+x_{2})),(x_{1},x_{2},x_{3}))\,.$
which corresponds to $(i)$ in our theorem statement. If $r_{1}\geq 2$, we
consider our necessary condition modulo $r_{1}$ and obtain
$r_{1}=1+x_{2}+x_{3}((1+x_{2})\bmod r_{1})\,.$ Since (by this equality) we
have $1+x_{2}\leq r_{1}$, it follows that there are two subcases.

Subcase 2.1: Suppose $1+x_{2}=r_{1}$. Then our vector is

$(\mathbf{r},\mathbf{x})=((1+x_{2},1+x_{1}(1+x_{2}),(1+x_{1}(1+x_{2}))(1+x_{2})),(x_{1},x_{2},x_{3}))\,,$

yielding $(ii)$ in our theorem statement.

Subcase 2.2: Suppose $1+x_{2}<r_{1}$. Then

$\displaystyle r_{1}$ $\displaystyle=1+x_{2}+x_{3}((1+x_{2})\bmod
r_{1})=1+x_{2}+x_{3}(1+x_{2})=(1+x_{2})(1+x_{3})\,.$

Then our vector is

$\displaystyle(\mathbf{r},\mathbf{x})=$ $\displaystyle(((1+x_{2})(1+x_{3}),$
$\displaystyle\quad 1+x_{1}(1+x_{2})(1+x_{3}),$
$\displaystyle\quad(1+x_{1}(1+x_{2})(1+x_{3}))(1+x_{2})),(x_{1},x_{2},x_{3}))\,,$

yielding $(iii)$ in our theorem statement.

Case 3: If $1+x_{1}r_{1}<r_{2}$ in (9), it is immediate that
$(1+r_{1}x_{1})\bmod r_{2}=1+r_{1}x_{1}\,,$ and thus
$r_{2}=1+x_{1}r_{1}+(1+r_{1}x_{1})x_{3}=(1+x_{3})(1+x_{1}r_{1})\,.$ Combining
this with

$\displaystyle r_{3}$
$\displaystyle=1+r_{1}x_{1}+r_{2}x_{2}=1+r_{1}x_{1}+(1+x_{3})(1+x_{1}r_{1})x_{2}=(1+r_{1}x_{1})(1+(1+x_{3})x_{2})\,,$

we obtain that

$(\mathbf{r},\mathbf{x})=((r_{1},(1+x_{3})(1+x_{1}r_{1}),(1+r_{1}x_{1})(1+(1+x_{3})x_{2})),(x_{1},x_{2},x_{3}))\,,$

which is a function of the multiplicities and the value $r_{1}$. If $r_{1}=1$,
then we obtain

$(1,(1+x_{1})(1+x_{3}),(1+x_{1})(1+x_{2}(1+x_{3})))$

which corresponds to item $(iv)$ in our theorem statement. If $r_{1}\geq 2$,
we again consider the necessary condition modulo $r_{1}$, for which we obtain

$\displaystyle r_{1}$ $\displaystyle=1+x_{2}((1+x_{3})(1+x_{1}r_{1})\bmod
r_{1})+x_{3}((1+r_{1}x_{1})(1+(1+x_{3})x_{2})\bmod r_{1})$
$\displaystyle=1+x_{2}((1+x_{3})\bmod r_{1})+x_{3}((1+(1+x_{3})x_{2})\bmod
r_{1}).$

We now have three subcases to consider.

Subcase 3.1: If $1+x_{3}<r_{1}$ then

$r_{1}=1+x_{2}(1+x_{3})+x_{3}((1+(1+x_{3})x_{2})\bmod r_{1})\,.$

This requires two subsubcases.

Subsubcase 3.1.1: Suppose $(1+(1+x_{3})x_{2})=r_{1}$. Then $x_{3}$ is
arbitrary, and we have

$\displaystyle(\mathbf{r},\mathbf{x})=$
$\displaystyle((1+(1+x_{3})x_{2},(1+x_{3})(1+x_{1}(1+(1+x_{3})x_{2})),$
$\displaystyle\quad(1+(1+(1+x_{3})x_{2})x_{1})(1+(1+x_{3})x_{2})),(x_{1},x_{2},x_{3}))\,,$

establishing item $(v)$ in the theorem statement.

Subsubcase 3.1.2: Suppose $(1+(1+x_{3})x_{2})<r_{1}$. Then

$r_{1}=1+x_{2}(1+x_{3})+x_{3}(1+(1+x_{3})x_{2})=(1+x_{3})(1+(1+x_{3})x_{2})\,,$

and thus it follows that for $\mathbf{x}=(x_{1},x_{2},x_{3})$, we have

$\displaystyle\mathbf{r}=$ $\displaystyle((1+x_{3})(1+(1+x_{3})x_{2}),$
$\displaystyle\quad(1+x_{3})(1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})),$
$\displaystyle\quad(1+(1+x_{3})(1+(1+x_{3})x_{2})x_{1})(1+(1+x_{3})x_{2}))\,,$

establishing item $(vi)$ in the theorem statement.

Subcase 3.2: Suppose $1+x_{3}=r_{1}$. Then for
$\mathbf{x}=(x_{1},x_{2},x_{3})$ we have

$\mathbf{r}=(1+x_{3},(1+x_{3})(1+x_{1}(1+x_{3})),(1+(1+x_{3})x_{1})(1+(1+x_{3})x_{2}))\,,$

establishing item $(vii)$ in the theorem statement.

Subcase 3.3: Suppose $1+x_{3}>r_{1}$. If $r_{1}\mid(1+x_{3})$, then we have

$r_{1}=1+x_{2}((1+x_{3})\bmod r_{1})+x_{3}((1+(1+x_{3})x_{2})\bmod
r_{1})=1+x_{3}\,.$

This is a contradiction, and thus it follows that $r_{1}\nmid 1+x_{3}$. If
$r_{1}\mid x_{2}$, then

$r_{1}=1+x_{2}((1+x_{3})\bmod r_{1})+x_{3}\,,$

which implies that $r_{1}\mid(1+x_{3})$, again a contradiction. Thus, we must
have that $r_{1}\nmid x_{2}$, and since $r_{1}\nmid(1+x_{3})$ we also know
$r_{1}>x_{2}$ since

$r_{1}=1+x_{2}((1+x_{3})\bmod r_{1})+x_{3}((1+(1+x_{3})x_{2})\bmod
r_{1})>x_{2}\,.$

We now consider two subsubcases.

Subsubcase 3.3.1: Suppose $r_{1}\mid(1+x_{2}(1+x_{3}))$. Then

$r_{1}=1+x_{2}((1+x_{3})\bmod r_{1})\,.$

Thus, there exists some $k\geq 1$ where $r_{1}=1+x_{2}k$, and we are forced to
have

$r_{1}=1+x_{2}k=1+x_{2}((1+x_{3})\bmod r_{1})\,$

implying that $k=(1+x_{3})\bmod 1+x_{2}k$. Thus, for some $s\geq 1$ we set

$x_{3}=skx_{2}+s-k+1$

and we have the case

$\displaystyle\mathbf{r}=$
$\displaystyle(1+kx_{2},(2+skx_{2}+s-k)(1+x_{1}(1+kx_{2})),$
$\displaystyle\quad(1+x_{1}(1+kx_{2}))(1+x_{2}(2+skx_{2}+s-k)))$

with

$\mathbf{x}=(x_{1},x_{2},skx_{2}+s-k+1)\,,$

corresponding to item $(viii)$ in our theorem statement.

Subsubcase 3.3.2: Suppose $r_{1}\nmid(1+x_{2}(1+x_{3}))$, so we have

$r_{1}=1+x_{2}((1+x_{3})\bmod r_{1})+x_{3}((1+(1+x_{3})x_{2})\bmod r_{1})\,.$

We also have that $1+x_{3}>r_{1}>x_{2}$. However,
$r_{1}\nmid(1+x_{2}(1+x_{3}))$ implies that

$r_{1}=1+x_{2}((1+x_{3})\bmod r_{1})+x_{3}((1+(1+x_{3})x_{2})\bmod r_{1})\geq
1+x_{3}\,,$

yielding a contradiction.

This completes our analysis of possible cases based on the necessary
condition. In particular, all of the types listed yield reflexive simplices.

###  A.2. Proof of IDP for types $(i)$, $(ii)$, and $(iv)$ and of non-IDP for
type $(viii)$ in Theorem 3.9

We now show that type $(viii)$ $\mathbf{r}$-vectors yield non-IDP simplices,
where we apply Theorem 3.3 and the notation therein. Let
$q_{j}=r_{2}=(skx_{2}+s+k)(1+x_{1}(1+kx_{2}))$, and set $b=skx_{2}+s+k$ which
is strictly less than $r_{2}$ as needed for Theorem 3.3. It is tedious but
straightforward to reduce the left-hand side of (5) to the form
$x_{3}+(x_{3}-1)x_{2}(skx_{2}+s+k)$. Note that by the assumption of type
$(viii)$, we have

$x_{3}=s+1+k(sx_{2}-1)\geq 2$

and thus

$x_{3}+(x_{3}-1)x_{2}(skx_{2}+s+k)\geq 2+x_{2}(skx_{2}+s+k)\geq 2\,,$

satisfying (5). Setting $q_{i}=r_{3}$, we now ask if there is a solution
$0<c<b$ satisfying (6) and (7). Using the fact that we must have
$0<c<b=skx_{2}+s+k$, we obtain that the left-hand side of 6 is:

$\displaystyle\left\lfloor\frac{(skx_{2}+s+k)(1+x_{1}(1+kx_{2}))(1+x_{2}(skx_{2}+s+k))}{(skx_{2}+s+k)(1+x_{1}(1+kx_{2}))}\right\rfloor-\left\lfloor\frac{c(1+x_{2}(skx_{2}+s+k))}{skx_{2}+s+k}\right\rfloor$
$\displaystyle=$
$\displaystyle\,1+x_{2}(skx_{2}+s+k)-cx_{2}-\left\lfloor\frac{c}{skx_{2}+s+k}\right\rfloor$
$\displaystyle=$ $\displaystyle\,1+x_{2}(skx_{2}+s+k-c)$

On the other hand, using $0<c<b=skx_{2}+s+k$, we find that the right-hand side
of 6 is:

$\displaystyle\left\lfloor\frac{(skx_{2}+s+k-c)(1+x_{2}(skx_{2}+s+k))}{(skx_{2}+s+k)}\right\rfloor$
$\displaystyle=$ $\displaystyle\,x_{2}(skx_{2}+s+k-c)+\left\lfloor
1-\frac{c}{skx_{2}+s+k}\right\rfloor$ $\displaystyle=$
$\displaystyle\,x_{2}(skx_{2}+s+k-c)$

Thus, there is no $c$ value in this range that satisfies (6), and hence we
find that simplices of type $(viii)$ are not IDP.

We next show that simplices of types $(i)$-$(vii)$ are IDP. Types $(i)$,
$(ii)$, and $(iv)$ all follow from affine free sum decompositions as follows.
For $(i)$, observe that

$\mathbf{r}=(1^{x_{1}})*_{0}(1^{x_{2}})*_{0}(1^{x_{3}})\,,$

and thus Theorem 3.1 applies. For $(ii)$, observe that

$\mathbf{r}=((1+x_{1})^{x_{2}},(1+x_{1}(1+x_{2}))^{x_{2}})*_{0}(1^{x_{3}})\,,$

and thus Theorem 3.8 and Theorem 3.1 apply to finish this case. Finally, for
$(iv)$ observe that

$\mathbf{r}=(1^{x_{3}})*_{0}((1+x_{2})^{x_{3}},(1+x_{2}(1+x_{3}))^{x_{3}})\,,$

and thus again Theorem 3.8 and Theorem 3.1 apply to finish this case.

Types $(iii)$, $(v)$, $(vi)$, and $(vii)$ do not follow from affine free sum
decompositions, and thus we must use Theorem 3.3 directly. Throughout the
remainder of this proof, we use the notation

$h(b):=b\left(\genfrac{}{}{}{}{1+\sum_{i\neq
j}q_{i}}{q_{j}}\right)-\sum_{i\neq
j}\left\lfloor\genfrac{}{}{}{}{bq_{i}}{q_{j}}\right\rfloor$

to denote the left-hand side of (5).

###  A.3. Proof of IDP for type $(iii)$ in Theorem 3.9

We next verify IDP for $\mathbf{r}$-vectors of type $(iii)$ using Theorem 3.3.
We must consider three cases corresponding to three possible values of
$q_{j}$.

Case: $q_{j}=(1+x_{2})(1+x_{3})$. It is straightforward to verify that

$h(b)=b-x_{3}\left\lfloor\frac{b}{(1+x_{3})}\right\rfloor\,,$

and using this formula one can check that

$h(k(1+x_{3}))=k\,.$

Combining these two observations, it follows that $h(b)=1$ only when $b=1$ and
$b=(1+x_{3})$, thus identifying the $b$-values we are required to check in
(5). To verify that (6) always has the desired solution, we consider three
cases. If $q_{i}=(1+x_{2})(1+x_{3})$, the result is trivial. If
$q_{i}=x_{1}(1+x_{2})(1+x_{3})+1$, then we may select $c=1$, from which it
follows that both sides of (6) are equal to $x_{1}(b-1)$. If
$q_{i}=(1+x_{2})(x_{1}(1+x_{2})(1+x_{3})+1)$, then we set $c=(1+x_{3})$, from
which it is straightforward to compute that both sides of (6) are equal to
$b(1+x_{2})x_{1}-x_{1}(1+x_{2})(1+x_{3})-1+\lfloor b/(1+x_{3})\rfloor$. This
completes our first case.

Case: $q_{j}=x_{1}(1+x_{2})(1+x_{3})+1$. It is straightforward to verify that

$h(b)=b-x_{1}\left\lfloor\frac{b(1+x_{2})(1+x_{3})}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor\,,$

where the values of $b$ range from $1$ to $x_{1}(1+x_{2})(1+x_{3})$. To verify
that (6) always has the desired solution, we consider three cases. If
$q_{i}=x_{1}(1+x_{2})(1+x_{3})+1$, the result is trivial. If
$q_{i}=(1+x_{2})(1+x_{3})$, then we write $b=\alpha x_{1}+\beta$, where
$0\leq\beta<x_{1}$ and $0\leq\alpha\leq(1+x_{2})(1+x_{3})$ for
$\alpha,\beta\in\mathbb{Z}$. Consequently, we have

$\displaystyle\left\lfloor\frac{b(1+x_{2})(1+x_{3})}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor$
$\displaystyle=\left\lfloor\frac{(\alpha
x_{1}+\beta)(1+x_{2})(1+x_{3})}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor$
$\displaystyle=\alpha+\left\lfloor\frac{\beta(1+x_{2})(1+x_{3})-\alpha}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor$
$\displaystyle=\alpha+\left.\begin{cases}\phantom{-}0,&\beta>0\\\
-1,&\beta=0\end{cases}\right\\}.$

In the case that $\beta>0$, we have that $h(b)=b-x_{1}\alpha=\beta$. Hence,
the viable candidates for $c$-values from 1 to $x_{1}(1+x_{2})(1+x_{3})$ that
satisfy equation (7) are precisely those $c$ such that $c\equiv 1\bmod x_{1}$.
Therefore, the $b$-values we are required to check in (5) are all $b=\alpha
x_{1}+\beta$, where $2\leq\beta\leq x_{1}-1$. In this case, we may choose
$c=\alpha x_{1}+1$, from which it follows that both sides of (6) are equal to
0, giving the desired result. On the other hand, if $\beta=0$, then we have
that $h(b)=h(\alpha x_{1})=\alpha x_{1}-x_{1}(\alpha-1)=x_{1}$. If $x_{1}=1$,
then $h(b)=1$. Thus, we need only consider when $x_{1}>1$. In order to satisfy
(5), it must be that $\alpha>0$. Given that $x_{1}>1$ and $\alpha>0$, it is
straightforward to verify that both sides of (6) when $c=1$ are equal to
$\alpha-1$. Finally, if $q_{i}=(1+x_{2})(x_{1}(1+x_{2})(1+x_{3})+1)$, then we
can set $c=1$ and the result is immediate. This completes our second case.

Case: $q_{j}=(1+x_{2})(x_{1}(1+x_{2})(1+x_{3})+1)$. We first identify those
values of $b$ that satisfy (5) and (7). It is straightforward to verify that

$h(b)=b-x_{1}\left\lfloor\frac{b(1+x_{3})}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor-((1+x_{2})-1)\left\lfloor\frac{b}{(1+x_{2})}\right\rfloor\,.$

Writing $b=m(1+x_{2})+t$ where $0\leq m\leq x_{1}(1+x_{2})(1+x_{3})$ and
$0\leq t\leq(1+x_{2})$, it follows that

$h(b)=h(m(1+x_{2})+t)=m+t-x_{1}\left\lfloor\frac{m(1+x_{2})(1+x_{3})+t(1+x_{3})}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor\,.$

We can now further divide into cases: either we have
$m=x_{1}(1+x_{2})(1+x_{3})$ or we have $m=kx_{1}+w$ where $0\leq
k<(1+x_{2})(1+x_{3})$ and $0\leq w<x_{1}$, which yields

$h(b)=h((kx_{1}+w)(1+x_{2})+t)=w+t-x_{1}\left\lfloor\frac{w(1+x_{2})(1+x_{3})+t(1+x_{3})-k}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor\,.$

For $m\neq x_{1}(1+x_{2})(1+x_{3})$, observe that since $0\leq
w(1+x_{2})(1+x_{3})\leq x_{1}-1$ and $0\leq t(1+x_{3})<(1+x_{2})(1+x_{3})$,
with $0\leq k<(1+x_{2})(1+x_{3})$, we have that $0\leq
w(1+x_{2})(1+x_{3})+t(1+x_{3})<x_{1}(1+x_{2})(1+x_{3})$. Thus,
$\left\lfloor\frac{w(1+x_{2})(1+x_{3})+t(1+x_{3})-k}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor$
is equal to either $0$ or $-1$.

Subcase 1 of 3: Suppose $m=x_{1}(1+x_{2})(1+x_{3})$. Since $0\leq
t(1+x_{3})<(1+x_{2})(1+x_{3})$, we have

$h((1+x_{2})(1+x_{3})x_{1}(1+x_{2})+t)=t-x_{1}\left\lfloor\frac{t(1+x_{3})-(1+x_{2})(1+x_{3})}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor=t+x_{1}.$

If this is equal to $1$, then it must be that $t=0$ and $x_{1}=1$. Thus, if
$x_{1}=1$, we have that $h((1+x_{2})(1+x_{3})(1+x_{2}))=1$.

Subcase 2 of 3: Suppose now that $m\neq x_{1}(1+x_{2})(1+x_{3})$ and that

$\left\lfloor\frac{w(1+x_{2})(1+x_{3})+t(1+x_{3})-k}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor=-1\,.$

Then since $w,t\geq 0$ and $x_{1}\geq 1$, we have $h(b)=w+t+x_{1}=1$ which
forces $w=t=0$ and $x_{1}=1$. In this case, $h(k(1+x_{2}))=1$ any time that
$k>0$. Thus, if $x_{1}=1$, we have that $h(k(1+x_{2}))=1$ when
$0<k<(1+x_{2})(1+x_{3})$.

Subcase 3 of 3: Suppose again that $m\neq x_{1}(1+x_{2})(1+x_{3})$ and that

$\left\lfloor\frac{w(1+x_{2})(1+x_{3})+t(1+x_{3})-k}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor=0\,.$

Then $0\leq k\leq w(1+x_{2})(1+x_{3})+t(1+x_{3})$, which implies that either
(A) $0<w<x_{1}$ with $0\leq t<(1+x_{2})$ or (B) $w=0$ with $k\leq t(1+x_{3})$.
If (A) holds, then $h(b)=w+t=1$ forces $w=1$ and $t=0$ since $w>0$, which
means that $h(b)=1$ when $b=(kx_{1}+1)(1+x_{2})$ for $0\leq
k<(1+x_{2})(1+x_{3})$. If (B) holds, then our same equation forces $w=0$ and
$t=1$ when $k\leq(1+x_{3})$, which means that $h(b)=1$ when
$b=kx_{1}(1+x_{2})+1$ for $0\leq k<(1+x_{2})(1+x_{3})$ and $k\leq(1+x_{3})$.

We summarize the values of $b$ for which $h(b)=1$ that were just derived:

  * •

If $x_{1}=1$ and $0<k\leq(1+x_{2})(1+x_{3})$ we have $b=k(1+x_{2})$.

  * •

If $x_{1}\geq 1$ and $0\leq k<(1+x_{2})(1+x_{3})$, we have
$b=(kx_{1}+1)(1+x_{2})$.

  * •

If $x_{1}\geq 1$ and $0\leq k<(1+x_{2})(1+x_{3})$ and $k\leq(1+x_{3})$, we
have $b=kx_{1}(1+x_{2})+1$.

Our next goal is to establish that (6) is always satisfied; recall that we are
in the case where $q_{j}=(1+x_{2})(x_{1}(1+x_{2})(1+x_{3})+1)$. If
$q_{i}=(1+x_{2})(x_{1}(1+x_{2})(1+x_{3})+1)$, then (6) is trivially satisfied.
If $q_{i}=x_{1}(1+x_{2})(1+x_{3})+1$, we write $b=m(1+x_{2})+t$ where $0\leq
m<x_{1}(1+x_{2})(1+x_{3})+1$ and $0\leq t<(1+x_{2})$. Substituting this form
of $b$ into (6) yields the equation

$-\left\lfloor\frac{c}{(1+x_{2})}\right\rfloor=\left\lfloor\frac{t-c}{(1+x_{2})}\right\rfloor\,.$

If $b>(1+x_{2})$, we set $c=(1+x_{2})$ and the equation is satisfied. If
$2<b<(1+x_{2})$, then we set $c=1$ and the equation is satisfied.

If $q_{i}=(1+x_{2})(1+x_{3})$, the analysis becomes more complicated. We write
$b=m(1+x_{2})+t$ where $0\leq m<d$ and $0\leq t<(1+x_{2})$. Our argument will
proceed by considering $x_{1}=1$ and $x_{1}>1$ separately.

Suppose $x_{1}=1$. Then the left-hand-side of (6) is reduced to

$\left\lfloor\frac{t(1+x_{3})-m}{(1+x_{2})(1+x_{3})+1}\right\rfloor-\left\lfloor\frac{c(1+x_{3})}{(1+x_{2})(1+x_{3})+1}\right\rfloor$

and the right-hand-side to

$\left\lfloor\frac{t(1+x_{3})-m-c(1+x_{3})}{(1+x_{2})(1+x_{3})+1}\right\rfloor\,.$

Since $m<(1+x_{3})$, if $t(1+x_{3})-m<0$ this forces $t=0$ and $0<m$, thus $b$
is a multiple of $(1+x_{2})$, and we found earlier that $h(m(1+x_{2}))=1$.
Thus, we need proceed no further. If $t(1+x_{3})-m\geq 0$, then since
$m<(1+x_{3})$ we must have $t\geq 1$, and we also have
$t(1+x_{3})-m<(1+x_{2})(1+x_{3})$. Thus,
$\left\lfloor\frac{t(1+x_{3})-m}{(1+x_{2})(1+x_{3})+1}\right\rfloor=0$, from
which it follows that (6) reduces to

$-\left\lfloor\frac{c(1+x_{3})}{(1+x_{2})(1+x_{3})+1}\right\rfloor=\left\lfloor\frac{t(1+x_{3})-m-c(1+x_{3})}{(1+x_{2})(1+x_{3})+1}\right\rfloor\,.$

If $m=0$, set $c=1$ and this equation is solved. If $m\geq 1$, set
$c=m(1+x_{2})+1$ which is less than $b$ in this case, and this equation is
again satisfied. This completes our proof for $x_{1}=1$.

We next consider when $x_{1}\geq 2$, maintaining our previous notation of
$b=m(1+x_{2})+t$. Write $m=fx_{1}+g$ where $0\leq
f\leq\lfloor(1+x_{3})/x_{1}\rfloor$ and $0\leq g<x_{1}$ except in the case
where $f=\lfloor(1+x_{3})/x_{1}\rfloor$ in which case $g$ is bounded above by
$(1+x_{3})-(1+x_{3})\lfloor(1+x_{3})/x_{1}\rfloor$. This leads to the left-
hand-side of (6) having the form

$f+\left\lfloor\frac{g(1+x_{2})(1+x_{3})+t(1+x_{3})-f}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor-\left\lfloor\frac{c(1+x_{3})}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor$

while the right-hand-side has the form

$f+\left\lfloor\frac{g(1+x_{2})(1+x_{3})+t(1+x_{3})-f-c(1+x_{3})}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor\,.$

We thus need to solve the equation

$\displaystyle\,\left\lfloor\frac{g(1+x_{2})(1+x_{3})+t(1+x_{3})-f}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor-\left\lfloor\frac{c(1+x_{3})}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor$
$\displaystyle=$
$\displaystyle\,\left\lfloor\frac{g(1+x_{2})(1+x_{3})+t(1+x_{3})-f-c(1+x_{3})}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor$

subject to the constraints $0\leq g<x_{1}$ (with the exception mentioned
above), $0\leq t\leq(1+x_{2})$, and $0\leq
f\leq\lfloor(1+x_{3})/x_{1}\rfloor$. Note that the first two inequalities
imply that $0\leq g(1+x_{2})(1+x_{3})+t(1+x_{3})<x_{1}(1+x_{2})(1+x_{3})$, and
also $f\leq\lfloor(1+x_{3})/x_{1}\rfloor\leq(1+x_{2})(1+x_{3})$, hence

$\left\lfloor\frac{g(1+x_{2})(1+x_{3})+t(1+x_{3})-f}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor=\left\\{\begin{array}[]{ll}0&\text{
if }g(1+x_{2})(1+x_{3})+t(1+x_{3})\geq f\\\ -1&\text{ if
}g(1+x_{2})(1+x_{3})+t(1+x_{3})<f\end{array}\right.$

Subcase 1 of 2: Suppose $g(1+x_{2})(1+x_{3})+t(1+x_{3})-f\geq 0$. Then (6)
reduces to

$-\left\lfloor\frac{c(1+x_{3})}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor=\left\lfloor\frac{g(1+x_{2})(1+x_{3})+t(1+x_{3})-f-c(1+x_{3})}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor\,.$

Note that $f\leq\lfloor(1+x_{3})/x_{1}\rfloor<(1+x_{3})$, and thus we can set
$c=fx_{1}(1+x_{2})+1$ which is less than $b$. The left-hand-side of our above
equation is given by

$\displaystyle\,-\left\lfloor\frac{(fx_{1}(1+x_{2})+1)(1+x_{3})}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor$
$\displaystyle=$
$\displaystyle\,-\left\lfloor\frac{fx_{1}(1+x_{2})(1+x_{3})+(1+x_{3})}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor$
$\displaystyle=$
$\displaystyle\,-\left\lfloor\frac{fx_{1}(1+x_{2})(1+x_{3})+f+(1+x_{3})-f}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor$
$\displaystyle=$ $\displaystyle\,-f\,.$

Similarly, the right-hand-side of our equation is given by

$\displaystyle\,-f+\left\lfloor\frac{f-(1+x_{3})+g(1+x_{2})(1+x_{3})+t(1+x_{3})-f}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor$
$\displaystyle=$
$\displaystyle\,-f+\left\lfloor\frac{-(1+x_{3})+g(1+x_{2})(1+x_{3})+t(1+x_{3})}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor\,.$

Since $h(b)$ is assumed to be at least $2$, we have that one or both of $g$
and $t$ are non-zero. Combining this observation with
$g(1+x_{2})(1+x_{3})+t(1+x_{3})-f\geq 0$ it follows that
$g(1+x_{2})(1+x_{3})+t(1+x_{3})>0$. Note that
$(1+x_{3})\mid(g(1+x_{2})(1+x_{3})+t(1+x_{3}))$, and thus
$x_{1}(1+x_{2})(1+x_{3})+1>g(1+x_{2})(1+x_{3})+t(1+x_{3})-(1+x_{3})\geq 0$,
which forces the right-hand-side of our equation to equal $-f$, satisfying
(6).

Subcase 2 of 2: Suppose $g(1+x_{2})(1+x_{3})+t(1+x_{3})-f<0$. Note that since
$g,(1+x_{2})(1+x_{3}),t,(1+x_{3})\geq 0$, it follows that $f\geq 1$ and thus
$b=(fx_{1}+g)(1+x_{2})+t\geq(1+x_{2})$. Set $c=(1+x_{2})$, which is less than
$b$. With these conditions, the left-hand-side of (6) is easily seen to equal
$-1$. The right-hand-side of (6) is given by

$\left\lfloor\frac{g(1+x_{2})(1+x_{3})+t(1+x_{3})-f-(1+x_{2})(1+x_{3})}{x_{1}(1+x_{2})(1+x_{3})+1}\right\rfloor\,.$

Since $g(1+x_{2})(1+x_{3})+t(1+x_{3})-f<0$ and $-(1+x_{2})(1+x_{3})<0$, the
numerator above is strictly negative. Also, since
$g,(1+x_{2})(1+x_{3}),t,(1+x_{3})\geq 0$, the numerator is minimized by
$-f-(1+x_{2})(1+x_{3})>-(1+x_{3})-(1+x_{2})(1+x_{3})\geq-2(1+x_{2})(1+x_{3})$.
But, since we assumed that $x_{1}\geq 2$, it follows that
$x_{1}(1+x_{2})(1+x_{3})+1>2(1+x_{2})(1+x_{3})$ and thus the floor function
above is equal to $-1$, satisfying equality for (6). This completes the proof
establishing IDP for $\mathbf{r}$-vectors of type $(iii)$.

###  A.4. Proof of IDP for type $(v)$ in Theorem 3.9

We next verify IDP for $\mathbf{r}$-vectors of type $(v)$ using Theorem 3.3.
Again, we must consider three cases corresponding to three possible values of
$q_{j}$.

Case: $q_{j}=1+(1+x_{3})x_{2}$. It is straightforward to verify that

$\displaystyle
h(b)=b-x_{2}\left\lfloor\frac{b(1+x_{3})}{1+(1+x_{3})x_{2}}\right\rfloor,$

where $1\leq b\leq(1+x_{3})x_{2}$. To verify that (6) always has the desired
solution, we consider three cases. If $q_{i}=1+(1+x_{3})x_{2}$, the result is
trivial. If $q_{i}=(1+x_{3})(1+x_{1}(1+(1+x_{3})x_{2}))$, then we write
$b=\alpha x_{2}+\beta$, where $0\leq\beta<x_{2}$ and $0\leq\alpha\leq 1+x_{3}$
for $\alpha,\beta\in\mathbb{Z}$. Consequently, observe that

$\displaystyle\left\lfloor\frac{b(1+x_{3})}{1+(1+x_{3})x_{2}}\right\rfloor=\left\lfloor\frac{(\alpha
x_{2}+\beta)(1+x_{3}}{1+(1+x_{3})x_{2}}\right\rfloor=\alpha+\left\lfloor\frac{\beta(1+x_{3})-\alpha}{1+(1+x_{3})x_{2}}\right\rfloor=\alpha+\left.\begin{cases}\phantom{-}0,&\beta>0\\\
-1,&\beta=0\end{cases}\right\\}.$

In the case that $\beta>0$, this formula implies that
$h(b)=b-x_{2}\alpha=\beta$. Hence, the viable candidates for $c$-values that
satisfy (7) are precisely those $c$ such that $c\equiv 1\bmod x_{2}$.
Therefore, the $b$-values we are required to check in (5) are all $b=\alpha
x_{2}+\beta$, where $2\leq\beta\leq x_{2}-1$. In this case, we may choose
$c=\alpha x_{2}+1$, from which it follows that both sides of (6) are equal to
$(\beta-1)(1+x_{3})x_{1}$. On the other hand, if $\beta=0$, then we have that
$h(b)=h(\alpha x_{2})=\alpha x_{2}-x_{2}(\alpha-1)=x_{2}$. If $x_{2}=1$, then
$h(b)=1$. Thus, we need only consider when $x_{2}>1$. Given that $x_{2}>1$ and
$0\leq\alpha\leq 1+x_{3}$, it is straightforward to verify that both sides of
(6) when $c=1$ are equal to $(\alpha x_{2}-1)(1+x_{3})x_{1}+\alpha-1$. If
$q_{i}=(1+(1+(1+x_{3})x_{2})x_{1})(1+(1+x_{3})x_{2})$, then we may again set
$c=1$, from which it is straightforward to compute that both sides of (6) are
equal to $(b-1)(1+(1+(1+x_{3})x_{2})x_{1})$. This completes our first case.

Case: $q_{j}=(1+x_{3})(1+x_{1}(1+(1+x_{3})x_{2}))$. We first identify those
values of $b$ that satisfy (5) and (7). It is straightforward to verify that

$h(b)=b-x_{1}\left\lfloor\frac{b(1+(1+x_{3})x_{2})}{(1+x_{3})(1+x_{1}(1+(1+x_{3})x_{2}))}\right\rfloor-
x_{3}\left\lfloor\frac{b}{1+x_{3}}\right\rfloor\,.$

Writing $b=\alpha(1+x_{3})+\beta$, where $0\leq\beta<1+x_{3}$ and
$0\leq\alpha\leq x_{1}(1+(1+x_{3})x_{2})$, it follows that

$\displaystyle h(b)$ $\displaystyle=h(\alpha(1+x_{3})+\beta)$
$\displaystyle=\alpha+\beta-
x_{1}\left\lfloor\frac{\alpha(1+x_{3})(1+(1+x_{3})x_{2})+\beta(1+(1+x_{3})x_{2})}{(1+x_{3})(1+x_{1}(1+(1+x_{3})x_{2}))}\right\rfloor\,.$

We can now further divide into cases: either we have
$\alpha=x_{1}(1+(1+x_{3})x_{2})$ or we have $\alpha=\gamma x_{1}+\delta$ where
$0\leq\gamma<1+(1+x_{3})x_{2}$ and $0\leq\delta<x_{1}$, which yields

$\displaystyle h(b)$ $\displaystyle=h((\gamma x_{1}+\delta)(1+x_{3})+\beta)$
$\displaystyle=\delta+\beta-
x_{1}\left\lfloor\frac{\delta(1+x_{3})(1+(1+x_{3})x_{2})+\beta(1+(1+x_{3})x_{2})-\gamma(1+x_{3})}{(1+x_{3})(1+x_{1}(1+(1+x_{3})x_{2}))}\right\rfloor\,.$

For $\alpha\neq x_{1}(1+(1+x_{3})x_{2})$, observe that since
$0\leq\delta(1+x_{3})(1+(1+x_{3})x_{2})\leq(x_{1}-1)(1+x_{3})(1+(1+x_{3})x_{2})$
and $0\leq\beta(1+(1+x_{3})x_{2})<(1+x_{3})(1+(1+x_{3})x_{2})$ with
$0\leq\gamma(1+x_{3})<(1+x_{3})(1+(1+x_{3})x_{2})$, we have that
$0\leq\delta(1+x_{3})(1+(1+x_{3})x_{2})+\beta(1+(1+x_{3})x_{2})<x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$.
Thus,
$\left\lfloor\frac{\delta(1+x_{3})(1+(1+x_{3})x_{2})+\beta(1+(1+x_{3})x_{2}-\gamma(1+x_{3})}{(1+x_{3})(1+x_{1}(1+(1+x_{3})x_{2}))}\right\rfloor$
is equal to either $0$ or $-1$.

Subcase 1 of 3: Suppose $\alpha=x_{1}(1+(1+x_{3})x_{2})$. Since
$0\leq\beta(1+(1+x_{3})x_{2})<(1+x_{3})(1+(1+x_{3})x_{2})$, we have

$\displaystyle h((1+x_{3})x_{1}(1+(1+x_{3})x_{2})+\beta)$
$\displaystyle=\beta-
x_{1}\left\lfloor\frac{\beta(1+(1+x_{3})x_{2})-(1+x_{3})(1+(1+x_{3})x_{2})}{(1+x_{3})(1+x_{1}(1+(1+x_{3})x_{2}))}\right\rfloor$
$\displaystyle=\beta+x_{1}.$

If this is equal to 1, then it must be that $\beta=0$ and $x_{1}=1$. Thus, if
$x_{1}=1$, we have that $h((1+x_{3})(1+(1+x_{3})x_{2}))=1$.

Subcase 2 of 3: Suppose now that $\alpha\neq x_{1}(1+(1+x_{3})x_{2})$ and that

$\displaystyle\left\lfloor\frac{\delta(1+x_{3})(1+(1+x_{3})x_{2})+\beta(1+(1+x_{3})x_{2}-\gamma(1+x_{3})}{(1+x_{3})(1+x_{1}(1+(1+x_{3})x_{2}))}\right\rfloor=-1.$

Then, writing $\alpha=\gamma x_{1}+\delta$ where
$0\leq\gamma<1+(1+x_{3})x_{2}$ and $0\leq\delta<x_{1}$, we have
$h(b)=\delta+\beta+x_{1}$. Since $\delta,\beta\geq 0$ and $x_{1}\geq 1$, for
this to equal 1, we must have $\delta=\beta=0$ and $x_{1}=1$. In this case,
$h(\gamma(1+x_{3}))=1$ whenever $\gamma>0$. Thus, if $x_{1}=1$, we have that
$h(\gamma(1+x_{3}))=1$ when $0<\gamma<1+(1+x_{3})x_{2}$.

Subcase 3 of 3: Suppose now that $\alpha\neq x_{1}(1+(1+x_{3})x_{2})$ and that

$\displaystyle\left\lfloor\frac{\delta(1+x_{3})(1+(1+x_{3})x_{2})+\beta(1+(1+x_{3})x_{2}-\gamma(1+x_{3})}{(1+x_{3})(1+x_{1}(1+(1+x_{3})x_{2}))}\right\rfloor=0.$

Then, it follows that
$0\leq\gamma(1+x_{3})\leq\delta(1+x_{3})(1+(1+x_{3})x_{2})+\beta(1+(1+x_{3})x_{2})$,
which given the bounds on $\gamma$, $\delta$, and $\beta$, implies that either
(A) $0<\delta<x_{1}$ or (B) $\delta=0$ with
$\gamma(1+x_{3})\leq\beta(1+(1+x_{3})x_{2})$. If (A) holds, then
$h(b)=\delta+\beta=1$ forces $\delta=1$ and $\beta=0$ since $\delta>0$.
Therefore, $h(b)=1$ when $b=(\gamma x_{1}+1)(1+x_{3})$ for
$0\leq\gamma<1+(1+x_{3})x_{2}$. If (B) holds, then our same equation forces
$\delta=0$ and $\beta=1$ when $\gamma(1+x_{3})\leq 1+1(1+x_{3})x_{2}$, which
further implies $0\leq\gamma\leq x_{2}$. This means $h(b)=1$ when $b=\gamma
x_{1}(1+x_{3})+1$ for $0\leq\gamma\leq x_{2}$.

We summarize the values of $b$ for which $h(b)=1$ that were just derived:

  * •

If $x_{1}=1$ and $0<\gamma\leq 1+(1+x_{3})x_{2}$, we have $b=\gamma(1+x_{3})$.

  * •

If $x_{1}\geq 1$ and $0\leq\gamma<1+(1+x_{3})x_{2}$, we have $b=(\gamma
x_{1}+1)(1+x_{3})$.

  * •

If $x_{1}\geq 1$ and $0\leq\gamma<1+(1+x_{3})x_{2}$ and $\gamma<x_{2}$, we
have $b=\gamma x_{1}(1+x_{3})+1$.

Next, we establish that (6) is always satisfied; recall that we are in the
case where $q_{j}=(1+x_{3})(1+x_{1}(1+(1+x_{3})x_{2}))$. If
$q_{i}=(1+x_{3})(1+x_{1}(1+(1+x_{3})x_{2}))$, then (6) is trivially satisfied.
Now suppose $q_{i}=(1+(1+(1+x_{3})x_{2})x_{1})(1+(1+x_{3})x_{2})$. Note that
$b\neq 1+x_{3}$ since $h(1+x_{3})=1$. If $b>1+x_{3}$, we set $c=1+x_{3}$ from
which it is straightforward to compute that both sides of (6) are equal to
$bx_{2}-(1+(1+x_{3})x_{2})+\left\lfloor\frac{b}{1+x_{3}}\right\rfloor$.
Otherwise, if $b<1+x_{3}$, we may choose $c=1$ from which it follows that both
sides of (6) are equal to $(b-1)x_{2}$ since $1\leq b<1+x_{3}$ implies $0\leq
b-1<1+x_{3}$. Finally, if $q_{i}=1+(1+x_{3})x_{2}$, we write
$b=\alpha(1+x_{3})+\beta$, where $0\leq\beta<1+x_{3}$ and
$0\leq\alpha<1+x_{1}(1+(1+x_{3})x_{2})$. Moreover, we write $\alpha=\gamma
x_{1}+\delta$, where $0\leq\delta<x_{1}$ and $0\leq\gamma\leq
1+(1+x_{3})x_{2}$. We consider the following possible cases:

Subcase 1 of 4: Suppose $\beta>0$ and $\delta>0$. Then, choosing $c=(\gamma
x_{1}+1)(1+x_{3})<(\gamma x_{1}+\delta)(1+x_{3})+\beta=b$, it follows that
both sides of (6) are equal to 0.

Subcase 2 of 4: Suppose $\beta=0$ and $\delta>0$. Note that $\delta\neq 1$
since $h((\gamma x_{1}+1)(1+x_{3}))=1$. Therefore, $2\leq\delta<x_{1}$, so we
may consider $c=(\gamma x_{1}+1)(1+x_{3})<(\gamma x_{1}+\delta)(1+x_{3})=b$.
Since $2\leq\delta<x_{1}$ and $0\leq\gamma<1+(1+x_{3})x_{2}$, both sides of
(6) are equal to 0.

Subcase 3 of 4: Suppose $\beta>0$ and $\delta=0$. If $0\leq\gamma\leq x_{2}$,
then $\beta\neq 1$ since $h(\gamma x_{1}(1+x_{3})+1)=1$. Thus, it must be that
$\beta>1$, thereby allowing us to consider $c=\gamma x_{1}(1+x_{3})+1<\gamma
x_{1}(1+x_{3})+\beta=b$. With this choice of $c$, it is straightforward to
verify both sides of (6) will again be equal to 0. Otherwise, if
$x_{2}<\gamma\leq 1+(1+x_{3})x_{2}$, consider $c=x_{2}x_{1}(1+x_{3})+1<\gamma
x_{1}(1+x_{3})+\beta=b$. Then, the left-hand side of (6) simplifies to

$\displaystyle\gamma-
x_{2}+\left\lfloor\frac{\beta(1+(1+x_{3})x_{2})-\gamma(1+x_{3})}{(1+x_{3})(1+x_{1}(1+(1+x_{3})x_{2}))}\right\rfloor,$

and the right-hand side of (6) simplifies to

$\displaystyle\gamma-
x_{2}+\left\lfloor\frac{\beta(1+(1+x_{3})x_{2})-\gamma(1+x_{3})-1}{(1+x_{3})(1+x_{1}(1+(1+x_{3})x_{2}))}\right\rfloor.$

Indeed, these two quantities are equivalent because
$\beta(1+(1+x_{3})x_{2})\neq\gamma(1+x_{3})$. To see this, assume otherwise.
Then, it would follow that $\beta=(\gamma-\beta x_{2})(1+x_{3})$. However,
this is a contradiction as we assumed $0<\beta<1+x_{3}$. Therefore, we have
that (6) is satisfied.

Subcase 4 of 4: Suppose $\beta=\delta=0$. In this case, $b=\gamma
x_{1}(1+x_{3})$. Moreover, note that $\gamma>0$ since otherwise, $b=0$
contradicting our bounds on $b$. If $x_{1}=1$, then
$h(b)=h(\gamma(1+x_{3}))=1$. Hence, we need only consider when $x_{1}>1$.
Since $\gamma>0$ and $x_{1}>1$, we may consider $c=1+x_{3}<\gamma
x_{1}(1+x_{3})=b$, from which it is straightforward to find that both sides of
(6) are equal to $\gamma-1$.

As these cases cover all possible values for $\beta$ and $\delta$, this
completes our second case.

Case: $q_{j}=(1+(1+(1+x_{3})x_{2})x_{1})(1+(1+x_{3})x_{2})$. Thus, we consider
$1\leq b\leq(1+(1+(1+x_{3})x_{2})x_{1})(1+(1+x_{3})x_{2})-1$. Again, we will
start by identifying those values of $b$ that satisfy (5) and (7). It is
straightforward to verify that

$\displaystyle
h(b)=b-x_{1}\left\lfloor\frac{b}{1+(1+(1+x_{3})x_{2})x_{1}}\right\rfloor-
x_{2}\left\lfloor\frac{b(1+x_{3})}{1+(1+x_{3})x_{2}}\right\rfloor.$

Writing $b=\alpha(1+(1+x_{3})x_{2})+\beta$, where
$0\leq\beta<1+(1+x_{3})x_{2}$ and $0\leq\alpha<1+(1+(1+x_{3})x_{2})x_{1}$, it
follows that

$\displaystyle h(b)$ $\displaystyle=h(\alpha(1+(1+x_{3})x_{2})+\beta)$
$\displaystyle=\alpha+\beta-
x_{1}\left\lfloor\frac{\alpha(1+(1+x_{3})x_{2})+\beta}{1+(1+(1+x_{3})x_{2})x_{1}}\right\rfloor-
x_{2}\left\lfloor\frac{\beta(1+x_{3})}{1+(1+x_{3})x_{2}}\right\rfloor.$

Now, writing $\beta=\gamma x_{2}+\delta$ , where $0\leq\delta<x_{2}$ and
$0\leq\gamma\leq 1+x_{3}$, and $\alpha=\varepsilon x_{1}+\eta$, where
$0\leq\eta<x_{1}$ and $0\leq\varepsilon\leq 1+(1+x_{3})x_{2}$, it follows that

$\displaystyle h(b)=\eta+\delta-
x_{1}\left\lfloor\frac{\eta(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta-\varepsilon}{1+(1+(1+x_{3})x_{2})x_{1}}\right\rfloor-
x_{2}\left\lfloor\frac{\delta(1+x_{3})-\gamma}{1+(1+x_{3})x_{2}}\right\rfloor.$

Since $0\leq\delta<x_{2}$ and $0\leq\gamma\leq 1+x_{3}$, observe that

$\displaystyle\left\lfloor\frac{\delta(1+x_{3})-\gamma}{1+(1+x_{3})x_{2}}\right\rfloor=\begin{cases}-1,&\delta=0,\gamma>0\\\
\phantom{-}0,&\text{otherwise}\end{cases}.$

Given the bounds on $\varepsilon$, $\eta$, $\gamma$, and $\delta$, note that
$-(1+(1+x_{3})x_{2})\leq\eta(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta-\varepsilon<1+(1+(1+x_{3})x_{2})x_{1}$. Consequently, it follows
that

$\displaystyle\left\lvert\eta(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta-\varepsilon\right\rvert<1+(1+(1+x_{3})x_{2})x_{1},$

and this implies that $\left\lfloor\frac{\eta(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta-\varepsilon}{1+(1+(1+x_{3})x_{2})x_{1}}\right\rfloor$ is equal to
either $0$ or $-1$. To resolve this floor function, we consider the following
subcases which analyze the sign of the numerator of its argument. Let $n$
denote that numerator, i.e., $n=\eta(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta-\varepsilon$.

Subcase 1 of 3: Suppose $\eta=0$ and $\gamma x_{2}+\delta\geq\varepsilon$.
Then, we have that $n\geq 0$, implying

$\displaystyle\left\lfloor\frac{\eta(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta-\varepsilon}{1+(1+(1+x_{3})x_{2})x_{1}}\right\rfloor=0.$

Therefore, our equation for $h(b)$ simplifies to

$\displaystyle h(b)=\delta-
x_{2}\cdot\left.\begin{cases}-1,&\delta=0,\gamma>0\\\
\phantom{-}0,&\text{otherwise}\end{cases}\right\\}.$

If $\delta=0$ and $\gamma>0$, then $h(b)=h(\varepsilon
x_{1}(1+(1+x_{3})x_{2})+\gamma x_{2})=x_{2}$. Thus, if $x_{2}=1$, we have that
$h(b)=h(\varepsilon x_{1}(2+x_{3})+\gamma)=1$ whenever $\gamma\geq\varepsilon$
with $\gamma>0$. Otherwise, $h(b)=\delta$, which forces $\delta=1$, i.e.,
$h(b)=h(\varepsilon x_{1}(1+(1+x_{3})x_{2})+\gamma x_{2}+\delta)=1$ whenever
$\gamma x_{2}+\delta\geq\varepsilon$.

Subcase 2 of 3: Suppose $\eta=0$ and $\gamma x_{2}+\delta<\varepsilon$. Then,
we have that $n<0$, implying

$\displaystyle\left\lfloor\frac{\eta(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta-\varepsilon}{1+(1+(1+x_{3})x_{2})x_{1}}\right\rfloor=-1.$

Therefore, our equation for $h(b)$ simplifies to

$\displaystyle
h(b)=\delta+x_{1}-x_{2}\cdot\left.\begin{cases}-1,&\delta=0,\gamma>0\\\
\phantom{-}0,&\text{otherwise}\end{cases}\right\\}.$

If $\delta=0$ and $\gamma>0$, then $h(b)=h(\varepsilon
x_{1}(1+(1+x_{3})x_{2})+\gamma x_{2})=x_{1}+x_{2}>1$. Otherwise,
$h(b)=\delta+x_{1}$, which forces $\delta=0$ and $x_{1}=1$ since $x_{1}\geq
1$, i.e., $h(b)=h(\varepsilon(1+(1+x_{3})x_{2})+\gamma x_{2})=1$ whenever
$\gamma x_{2}<\varepsilon$.

Subcase 3 of 3: Suppose $\eta\geq 1$. Then, we have that $n\geq 0$, implying

$\displaystyle\left\lfloor\frac{\eta(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta-\varepsilon}{1+(1+(1+x_{3})x_{2})x_{1}}\right\rfloor=0.$

Therefore, our equation for $h(b)$ simplifies to

$\displaystyle h(b)=\eta+\delta-
x_{2}\cdot\left.\begin{cases}-1,&\delta=0,\gamma>0\\\
\phantom{-}0,&\text{otherwise}\end{cases}\right\\}.$

If $\delta=0$ and $\gamma>0$, then $h(b)=h((\varepsilon
x_{1}+\eta)(1+(1+x_{3})x_{2})+\gamma x_{2})=\eta+x_{2}\geq 1+x_{2}>1$.
Otherwise, $h(b)=\eta+\delta$, which forces $\delta=0$ and $\eta=1$ since we
assume $\eta\geq 1$, i.e., $h(b)=h((\varepsilon
x_{1}+1)(1+(1+x_{3})x_{2})+\gamma x_{2})=1$.

We summarize the values of $b$ for which $h(b)=1$ that were just derived:

  * •

If $x_{1}=1$, $\eta=\delta=0$, and $0\leq\gamma x_{2}<\varepsilon\leq
1+(1+x_{3})x_{2}$, we have $b=\varepsilon(1+(1+x_{3})x_{2})+\gamma x_{2}$.

  * •

If $x_{2}=1$, $\eta=\delta=0$, and $0\leq\varepsilon\leq\gamma\leq 1+x_{3}$
with $\gamma>0$, we have $b=\varepsilon x_{1}(2+x_{3})+\gamma$.

  * •

If $x_{1},x_{2}\geq 1$, $\eta=0$, $\delta=1$, and $0\leq\varepsilon\leq\gamma
x_{2}+1\leq 1+(1+x_{3})x_{2}$, we have $b=\varepsilon
x_{1}(1+(1+x_{3})x_{2})+\gamma x_{2}+1$.

  * •

If $x_{1},x_{2}\geq 1$, $\eta=1$, $\delta=0$, $0\leq\gamma\leq 1+x_{3}$, and
$0\leq\varepsilon\leq 1+(1+x_{3})x_{2}$, we have $b=(\varepsilon
x_{1}+1)(1+(1+x_{3})x_{2})+\gamma x_{2}$.

Our next goal is to establish that (6) is always satisfied; recall that we are
in the case where $q_{j}=(1+(1+(1+x+3)x_{2})x_{1})(1+(1+x_{3})x_{2})$. If
$q_{i}=(1+(1+(1+x+3)x_{2})x_{1})(1+(1+x_{3})x_{2})$, the result is trivial. If
$q_{i}=1+(1+x_{3})x_{2}$, we write $b=\alpha(1+(1+x_{3})x_{2})+\beta$, where
$0\leq\beta<1+(1+x_{3})x_{2}$ and $0\leq\alpha<1+(1+(1+x_{3})x_{2})x_{1}$.
Furthermore, we write $\beta=\gamma x_{2}+\delta$, where $0\leq\delta<x_{2}$
and $0\leq\gamma\leq 1+x_{3}$, and we write $\alpha=\varepsilon x_{1}+\eta$,
where $0\leq\eta<x_{1}$ and $0\leq\varepsilon\leq 1+(1+x_{3})x_{2}$. We
consider the following possible cases:

Subcase 1 of 4: Suppose $\eta>0$ and $\delta>0$. We consider $c=(\varepsilon
x_{1}+1)(1+(1+x_{3})x_{2})+\gamma x_{2}<(\varepsilon
x_{1}+\eta)(1+(1+x_{3})x_{2})+\gamma x_{2}+\delta=b$. Since $0<\eta<x_{1}$ and
$0\leq\delta<x_{2}$, it follows that
$0<(\eta-1)(1+(1+x_{3})x_{2})+\delta<x_{1}(1+(1+x_{3})x_{2})$. Therefore, it
is straightforward to verify that both sides of (6) are equal to $0$.

Subcase 2 of 4: Suppose $\eta=0$ and $\delta>0$. If $\gamma
x_{2}+1\geq\varepsilon$, note that $\delta\neq 1$ since otherwise,
$h(b)=h(\varepsilon x_{1}(1+(1+x_{3})x_{2})+\gamma x_{2}+1)=1$. Thus, we have
that $\delta>1$, and we consider $c=\varepsilon x_{1}(1+(1+x_{3})x_{2})+\gamma
x_{2}+1<\varepsilon x_{1}(1+(1+x_{3})x_{2})+\gamma x_{2}+\delta=b$. Given that
$1<\delta<x_{2}$ implies $\gamma x_{2}+\delta>\gamma x_{2}+1\geq\varepsilon$,
it is straightforward to verify that our choice of $c$ gives that both sides
of (6) are equal to $0$. Otherwise, if $\gamma x_{2}+1<\varepsilon$ (and
hence, $\varepsilon>1$), we consider
$c=1+x_{1}(1+(1+x_{3})x_{2})\leq\delta+x_{1}(1+(1+x_{3})x_{2})<\varepsilon
x_{1}(1+(1+x_{3})x_{2})+\gamma x_{2}+\delta=b$. With this choice of $c$, it is
straightforward to verify both sides of (6) are equal to

$\displaystyle\varepsilon-1+\left\lfloor\frac{\gamma
x_{2}+\delta-\varepsilon}{1+(1+(1+x_{3})x_{2})x_{1}}\right\rfloor\,.$

Subcase 3 of 4: Suppose $\eta>0$ and $\delta=0$. Note that $\eta\neq 1$ since
otherwise, $h(b)=h((\varepsilon x_{1}+1)(1+(1+x_{3})x_{2})+\gamma x_{2})=1$.
Thus, we have that $\eta>1$, and we consider $c=(\varepsilon
x_{1}+1)(1+(1+x_{3})x_{2})+\gamma x_{2}<(\varepsilon
x_{1}+\eta)(1+(1+x_{3})x_{2})+\gamma x_{2}=b$. Since $1<\eta<x_{1}$, it is
straightforward to verify that both sides of (6) are equal to $0$.

Subcase 4 of 4: Suppose $\eta=\delta=0$. Further suppose $\gamma
x_{2}<\varepsilon$ (and hence, $\varepsilon>0$). If $x_{1}=1$, then
$h(b)=h(\varepsilon(1+(1+x_{3})x_{2})+\gamma x_{2})=1$, so we may assume
$x_{1}>1$. Consider $c=1+(1+x_{3})x_{2}<\varepsilon
x_{1}(1+(1+x_{3})x_{2})+\gamma x_{2}=b$. Since $x_{1}>1$ and $\gamma
x_{2}<\varepsilon$, it follows that $-x_{1}(1+(1+x_{3})x_{2})\leq\gamma
x_{2}-\varepsilon-(1+(1+x_{3})x_{2})<0$. Given this, it is straightforward to
verify that both sides of (6) are equal to $\varepsilon-1$. Now, suppose
$\gamma x_{2}>\varepsilon$. Note that $\gamma\neq 0$ since otherwise,
$\varepsilon<0$ contradicting our initial bounds on $\varepsilon$. Thus, we
have that $\gamma>0$. Moreover, if $x_{2}=1$, it follows that
$h(b)=h(\varepsilon x_{1}(2+x_{3})+\gamma)=1$, so we may assume $x_{2}>1$.
Taking $c=1$, the inequality $\gamma x_{2}>\varepsilon$ readily implies that
both sides of (6) are equal to $\varepsilon$. Finally, suppose $\gamma
x_{2}=\varepsilon$. Note that neither $\gamma$ nor $\varepsilon$ can be equal
to 0 since otherwise, we would have $\eta=\delta=\gamma=\varepsilon=0$,
implying $b=0$. This, of course, contradicts the bounds on $b$. Moreover, we
may again assume $x_{2}>1$ (and thus, $\varepsilon>1$) since otherwise,
$h(b)=h(\varepsilon x_{1}(2+x_{3})+\gamma)=1$. Since $\gamma
x_{2}=\varepsilon$, observe that $b=\varepsilon(1+x_{1}(1+(1+x_{3})x_{2}))$.
As such, we may consider $c=1+x_{1}(1+(1+x_{3})x_{2})$ which is strictly less
than $b$ since $\varepsilon>1$. This choice of $c$ readily gives that both
sides of (6) are equal to $\varepsilon-1$.

Finally, if $q_{i}=(1+x_{3})(1+x_{1}(1+(1+x_{3})x_{2}))$, we begin again by
writing $b=\alpha(1+(1+x_{3})x_{2})+\beta$, where
$0\leq\beta<1+(1+x_{3})x_{2}$ and $0\leq\alpha<1+(1+(1+x_{3})x_{2})x_{1}$.
Furthermore, we write $\beta=\gamma x_{2}+\delta$, where $0\leq\delta<x_{2}$
and $0\leq\gamma\leq 1+x_{3}$, and we write $\alpha=\varepsilon x_{1}+\eta$,
where $0\leq\eta<x_{1}$ and $0\leq\varepsilon\leq 1+(1+x_{3})x_{2}$. We
consider the following possible cases:

Subcase 1 of 3: Suppose $\delta>1$. Combined with our bounds on $\gamma$, this
implies that $0\leq(\delta-1)(1+x_{3})-\gamma<1+(1+x_{3})x_{2}$. Taking $c=1$,
it follows that both sides of (6) are equal to $\alpha(1+x_{3})+\gamma$.

Subcase 2 of 3: Suppose $\delta=1$. If $\gamma x_{2}+1\geq\varepsilon$, note
that $\eta\neq 0$ since otherwise, $h(b)=h(\varepsilon
x_{1}(1+(1+x_{3})x_{2})+\gamma x_{2}+1)=1$. Thus, $\eta\geq 1$, and we may
consider $c=(\varepsilon x_{1}+1)(1+(1+x_{3})x_{2})<(\varepsilon
x_{1}+\eta)(1+(1+x_{3})x_{2})+\gamma x_{2}+1=b$. Since $\gamma\leq 1+x_{3}$,
it is straightforward to show both sides of (6) are equal to
$(\eta-1)(1+x_{3})+\gamma$. On the other hand, if $\gamma x_{2}+1<\varepsilon$
(and hence, $\varepsilon>1$), we may consider $c=1+(1+x_{3})x_{2}<(\varepsilon
x_{1}+\eta)(1+(1+x_{3})x_{2})+\gamma x_{2}+1=b$, from which it follows that
both sides of (6) are equal to $(\varepsilon x_{1}+\eta-1)(1+x_{3})+\gamma$.

Subcase 3 of 3: Suppose $\delta=0$. Note that $\eta\neq 1$ since otherwise,
$h(b)=1$. This lends itself to two possibilities: (A) $\eta=0$ or (B)
$\eta>1$. If (A) holds, we first suppose $\gamma x_{2}<\varepsilon$ (and
hence, $\varepsilon>0$). If $x_{1}=1$, then $h(b)=1$, so we may assume
$x_{1}>1$. Since $\varepsilon>0$ and $x_{1}>1$, we may consider
$c=1+(1+x_{3})x_{2}<\varepsilon x_{1}(1+(1+x_{3})x_{2})+\gamma x_{2}=b$. For
this choice of $c$, it readily follows that both sides of (6) are equal to

$\displaystyle(\varepsilon x_{1}-1)(1+x_{3})+\left\lfloor\frac{\gamma
x_{2}}{1+(1+x_{3})x_{2}}\right\rfloor.$

Now, suppose $\gamma x_{2}>\varepsilon$. Since $\varepsilon\geq 0$ by
construction, this inequality implies $\gamma\neq 0$. Moreover, note that if
$x_{2}=1$, then $h(b)=1$. Thus, we may assume $x_{2}>1$, and we simply
consider $c=1$. Given that $0<\gamma\leq 1+x_{3}$ and $x_{2}>1$, it follows
that $0<\gamma+(1+x_{3})<1+(1+x_{3})x_{2}$. Therefore, with $c=1$, we find
that both sides of (6) are equal to $\varepsilon x_{1}(1+x_{3})+\gamma-1$.
Finally, suppose $\gamma x_{2}=\varepsilon$. Since $b\neq 0$, this equality
implies that both $\gamma$ and $\varepsilon$ cannot be $0$. Moreover, we may
again assume $x_{2}>1$ since $x_{2}=1$ would imply $h(b)=1$. Since $\gamma
x_{2}=\varepsilon$, it follows that
$b=\varepsilon(1+x_{1}(1+(1+x_{3})x_{2}))$, and we also get that
$\varepsilon>1$ since we assume $x_{2}>1$. Combining, we may consider
$c=1+x_{1}(1+(1+x_{3})x_{2})<\varepsilon(1+x_{1}(1+(1+x_{3})x_{2}))=b$. As
before, the inequality $0<\gamma+(1+x_{3})<1+(1+x_{3})x_{2}$ still holds in
this case, from which it is straightforward to verify both sides of (6) are
equal to $(\varepsilon-1)x_{1}(1+x_{3})+\gamma-1$.

On the other hand, if (B) holds, we have that $\eta>1$. Therefore, we consider
$c=(\varepsilon x_{1}+1)(1+(1+x_{3})x_{2})+\gamma x_{2}<(\varepsilon
x_{1}+\eta)(1+(1+x_{3})x_{2})+\gamma x_{2}=b$, from which it follows that both
sides of (6) simplify to $(\eta-1)(1+x_{3})$. In any case, we find that both
sides of (6) are equivalent for each possible $q_{i}$, thereby completing our
third and final case. Thus, we have established IDP for $\mathbf{r}$-vectors
of type $(v)$.

###  A.5. Proof of IDP for type $(vi)$ in Theorem 3.9

We next verify IDP for $\mathbf{r}$-vectors of type $(vi)$ using Theorem 3.3.
Again, we must consider three cases corresponding to three possible values of
$q_{j}$.

Case: $q_{j}=(1+x_{3})(1+(1+x_{3})x_{2})$. We first identify those values of
$b$ that satisfy (5) and (7). It is straightforward to verify that

$\displaystyle
h(b)=b-x_{2}\left\lfloor\frac{b}{1+(1+x_{3})x_{2}}\right\rfloor-
x_{3}\left\lfloor\frac{b}{1+x_{3}}\right\rfloor,$

where $1\leq b\leq(1+x_{3})(1+(1+x_{3})x_{2})-1$. Writing
$b=\alpha(1+x_{3})+\beta$, where $0\leq\beta<1+x_{3}$ and
$0\leq\alpha<1+(1+x_{3})x_{2}$ for $\alpha,\beta\in\mathbb{Z}$, it follows
that

$\displaystyle h(b)$ $\displaystyle=h(\alpha(1+x_{3})+\beta)$
$\displaystyle=\alpha+\beta-
x_{2}\left\lfloor\frac{\alpha(1+x_{3})+\beta}{1+(1+x_{3})x_{2}}\right\rfloor.$

We can now further divide into cases: either we have $\alpha=(1+x_{3})x_{2}$
or we have $\alpha=\gamma x_{2}+\delta$ where $0\leq\gamma<1+x_{3}$ and
$0\leq\delta<x_{2}$, which yields

$\displaystyle h(b)$ $\displaystyle=h((\gamma x_{2}+\delta)(1+x_{3})+\beta)$
$\displaystyle=\delta+\beta-
x_{2}\left\lfloor\frac{\delta(1+x_{3})+\beta-\gamma}{1+(1+x_{3})x_{2}}\right\rfloor.$

For $\alpha\neq(1+x_{3})x_{2}$, let $n=\delta(1+x_{3})+\beta-\gamma$. Observe
that since $0\leq\delta(1+x_{3})+\beta<1+(1+x_{3})x_{2}$ and
$0\leq\gamma<1+x_{3}$, it follows that $\left\lvert
n\right\rvert<1+(1+x_{3})x_{2}$. Thus,
$\left\lfloor\frac{\delta(1+x_{3})+\beta-\gamma}{1+(1+x_{3})x_{2}}\right\rfloor$
is equal to either $0$ or $-1$.

Subcase 1 of 4: Suppose $\alpha=(1+x_{3})x_{2}$. Since $0\leq\beta<1+x_{3}$,
we have

$\displaystyle h((1+x_{3})^{2}x_{2}+\beta)$
$\displaystyle=(1+x_{3})x_{2}+\beta-(1+x_{3})x_{2}-x_{2}\left\lfloor\frac{\beta-(1+x_{3})}{1+(1+x_{3})x_{2}}\right\rfloor$
$\displaystyle=\beta+x_{2}.$

If this is equal to $1$, then it must be that $\beta=0$ and $x_{2}=1$. Thus,
if $x_{2}=1$, we have that $h((1+x_{3})^{2})=1$.

For the next three subcases, we assume $\alpha\neq(1+x_{3})x_{2}$, so we may
write $\alpha=\gamma x_{2}+\delta$ where $0\leq\delta<x_{2}$ and
$0\leq\gamma<1+x_{3}$.

Subcase 2 of 4: Suppose $\delta=0$ and $\beta\geq\gamma$. Then, we have that
$n\geq 0$, implying

$\displaystyle\left\lfloor\frac{\delta(1+x_{3})+\beta-\gamma}{1+(1+x_{3})x_{2}}\right\rfloor=0.$

Therefore, our equation for $h(b)$ simplifies to $h(b)=\beta$, which forces
$\beta=1$, i.e., $h(b)=h(\gamma x_{2}(1+x_{3})+1)=1$ whenever $0\leq\gamma\leq
1$.

Subcase 3 of 4: Suppose $\delta=0$ and $\beta<\gamma$. Then, we have that
$n<0$, implying

$\displaystyle\left\lfloor\frac{\delta(1+x_{3})+\beta-\gamma}{1+(1+x_{3})x_{2}}\right\rfloor=-1.$

Therefore, our equation for $h(b)$ simplifies to $h(b)=\beta+x_{2}$, which
forces $\beta=0$ and $x_{2}=1$ since $x_{2}\geq 1$, i.e.,
$h(b)=h(\gamma(1+x_{3}))=1$ for $0<\gamma<1+x_{3}$.

Subcase 4 of 4: Suppose $\delta\geq 1$. Then, since $0\leq\gamma<1+x_{3}$, we
have that $n\geq 0$, implying

$\displaystyle\left\lfloor\frac{\delta(1+x_{3})+\beta-\gamma}{1+(1+x_{3})x_{2}}\right\rfloor=0.$

Therefore, our equation for $h(b)$ simplifies to $h(b)=\delta+\beta$, which
forces $\beta=0$ and $\delta=1$ since we assume $\delta\geq 1$, i.e.,
$h(b)=h((\gamma x_{2}+1)(1+x_{3}))=1$ for $0\leq\gamma<1+x_{3}$.

We summarize the values of $b$ for which $h(b)=1$ that were just derived:

  * •

If $x_{2}=1$ and $0<\gamma\leq 1+x_{3}$, we have $b=\gamma(1+x_{3})$.

  * •

If $x_{2}\geq 1$, $\delta=0$, $\beta=1$, and $0\leq\gamma\leq 1$, we have
$b=\gamma x_{2}(1+x_{3})+1$.

  * •

If $x_{2}\geq 1$, $\delta=1$, $\beta=0$, and $0\leq\gamma<1+x_{3}$, we have
$b=(\gamma x_{2}+1)(1+x_{3})$.

Our next goal is to establish that (6) is always satisfied; recall that we are
in the case where $q_{j}=(1+x_{3})(1+(1+x_{3})x_{2})$. If
$q_{i}=(1+x_{3})(1+(1+x_{3})x_{2})$, the result is trivial. If
$q_{i}=(1+(1+x_{3})(1+(1+x_{3})x_{2})x_{1})(1+(1+x_{3})x_{2})$, we write
$b=\alpha(1+x_{3})+\beta$, where $0\leq\beta<1+x_{3}$ and
$0\leq\alpha<1+(1+x_{3})x_{2}$. Note that $b\neq 1+x_{3}$ since
$h(1+x_{3})=1$. If $b>1+x_{3}$, we set $c=1+x_{3}$ from which it is
straightforward to compute that both sides of (6) are equal to
$((\alpha-1)(1+x_{3})+\beta)(1+(1+x_{3})x_{2})x_{1}+\alpha-1$. Otherwise, if
$b<1+x_{3}$, note that $\alpha=0$ and hence $b=\beta$. To satisfy (5), we need
only consider $1<\beta<1+x_{3}$. Thus, we may choose $c=1$ from which it
follows that both sides of (6) are equal to $(\beta-1)(1+(1+x_{3})x_{2})x_{1}$
since $1<b<1+x_{3}$ implies $1\leq\beta-1<1+x_{3}$. Finally, if
$q_{i}=(1+x_{3})(1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2}))$, we again write
$b=\alpha(1+x_{3})+\beta$, where $0\leq\beta<1+x_{3}$ and
$0\leq\alpha<1+(1+x_{3})x_{2}$. Moreover, in the case that
$\alpha\neq(1+x_{3})x_{2}$, we write $\alpha=\gamma x_{2}+\delta$, where
$0\leq\delta<x_{2}$ and $0\leq\gamma<(1+x_{3})$. We consider the following
possible cases:

Subcase 1 of 5: Suppose $\alpha=(1+x_{3})x_{2}$. If $x_{2}=1$, then $\beta\neq
0$ since $h((1+x_{3})^{2})=1$. Thus, we may consider
$c=(1+x_{3})^{2}<(1+x_{3})^{2}+\beta=b$. With this choice of $c$, it is
straightforward to verify that both sides of (6) are equal to $\beta
x_{1}(1+x_{3})$. Otherwise, if $x_{2}>1$, we consider
$c=1+x_{3}<(1+x_{3})^{2}x_{2}+\beta=b$. Since $x_{2}>1$ implies
$-x_{2}(1+x_{3})<\beta-2(1+x_{3})<0$, it is straightforward to verify that
both sides of (6) are equal to $(b-(1+x_{3}))x_{1}(1+x_{3})+x_{3}$.

Subcase 2 of 5: Suppose $\alpha\neq(1+x_{3})x_{2}$ with $\delta>0$ and
$\beta>0$. Then, choosing $c=(\gamma x_{2}+1)(1+x_{3})<(\gamma
x_{2}+\delta)(1+x_{3})+\beta=b$, it follows that both sides of (6) are equal
to $((\delta-1)(1+x_{3})+\beta)x_{1}(1+x_{3})$.

Subcase 3 of 5: Suppose $\alpha\neq(1+x_{3})x_{2}$ with $\delta>0$ and
$\beta=0$. Note that $\delta\neq 1$ since $h((\gamma x_{2}+1)(1+x_{3}))=1$.
Therefore, $2\leq\delta<x_{2}$, so we may consider $c=(\gamma
x_{2}+1)(1+x_{3})<(\gamma x_{2}+\delta)(1+x_{3})=b$. Given this choice, we
find that both sides of (6) are equal to $(\delta-1)(1+x_{3})x_{1}(1+x_{3})$.

Subcase 4 of 5: Suppose $\alpha\neq(1+x_{3})x_{2}$ with $\delta=0$ and
$\beta>0$. If $0\leq\gamma\leq 1$, then $\beta\neq 1$ since $h(\gamma
x_{2}(1+x_{3})+1)=1$. Thus, it must be that $\beta>1$, thereby allowing us to
consider $c=\gamma x_{2}(1+x_{3})+1<\gamma x_{2}(1+x_{3})+\beta=b$. With this
choice of $c$, it is straightforward to verify both sides of (6) are equal to
$(\beta-1)x_{1}(1+x_{3})$. Otherwise, if $1<\gamma<1+x_{3}$, consider
$c=1+(1+x_{3})x_{2}<\gamma x_{2}(1+x_{3})+\beta=b$. Then, both sides of (6)
are equal to

$\displaystyle((\gamma-1)x_{2}(1+x_{3})+\beta-1)x_{1}(1+x_{3})+\gamma-1+\left\lfloor\frac{\beta-\gamma}{1+(1+x_{3})x_{2}}\right\rfloor.$

Subcase 5 of 5: Suppose $\alpha\neq(1+x_{3})x_{2}$ with $\delta=\beta=0$. In
this case, $b=\gamma x_{2}(1+x_{3})$. Moreover, note that $\gamma>0$ since
otherwise, $b=0$ contradicting our bounds on $b$. If $x_{2}=1$, then
$h(b)=h(\gamma(1+x_{3}))=1$. Hence, we need only consider when $x_{2}>1$.
Since $\gamma>0$ and $x_{2}>1$, we may take $c=1+x_{3}<\gamma
x_{2}(1+x_{3})=b$, from which it is straightforward to find that both sides of
(6) are equal to $(\gamma x_{2}-1)(1+x_{3})^{2}x_{1}+\gamma-1$.

This completes our first case.

Case: $q_{j}=(1+x_{3})(1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2}))$. We first identify
those values of $b$ that satisfy (5) and (7). It is straightforward to verify
that

$h(b)=b-x_{1}\left\lfloor\frac{b(1+(1+x_{3})x_{2})}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor-
x_{3}\left\lfloor\frac{b}{1+x_{3}}\right\rfloor\,.$

Writing $b=\alpha(1+x_{3})+\beta$, where $0\leq\beta<1+x_{3}$ and
$0\leq\alpha\leq 1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$, it follows that

$\displaystyle h(b)$ $\displaystyle=h(\alpha(1+x_{3})+\beta)$
$\displaystyle=\alpha+\beta-
x_{1}\left\lfloor\frac{(\alpha(1+x_{3})+\beta)(1+(1+x_{3})x_{2})}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor\,.$

We can now further divide into cases: either we have
$\alpha=x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$ or we have $\alpha=\gamma
x_{1}+\delta$ where $0\leq\gamma<(1+x_{3})(1+(1+x_{3})x_{2})$ and
$0\leq\delta<x_{1}$. If $\alpha=x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$, then since
$0\leq\beta<1+x_{3}$, we have

$\displaystyle h(b)$ $\displaystyle=\beta-
x_{1}\left\lfloor\frac{\beta(1+(1+x_{3})x_{2})-(1+x_{3})(1+(1+x_{3})x_{2})}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor$
$\displaystyle=\beta+x_{1}.$

If this is equal to 1, then it must be that $\beta=0$ and $x_{1}=1$. Thus, if
$x_{1}=1$, we have that $h((1+x_{3})^{2}(1+(1+x_{3})x_{2}))=1$. Otherwise, if
$\alpha\neq x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$, we write $\alpha=\gamma
x_{1}+\delta$ where $0\leq\gamma<(1+x_{3})(1+(1+x_{3})x_{2})$ and
$0\leq\delta<x_{1}$. Thus, $h(b)$ simplifies as follows

$\displaystyle h(b)$ $\displaystyle=h((\gamma x_{1}+\delta)(1+x_{3})+\beta)$
$\displaystyle=\delta+\beta-
x_{1}\left\lfloor\frac{(\delta(1+x_{3})+\beta)(1+(1+x_{3})x_{2})-\gamma}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2}}\right\rfloor\,.$

For $\alpha\neq x_{1}(1+(1+x_{3})x_{2})$, observe that since
$0\leq\delta(1+x_{3})(1+(1+x_{3})x_{2})\leq(x_{1}-1)(1+x_{3})(1+(1+x_{3})x_{2})$
and $0\leq\beta(1+(1+x_{3})x_{2})<(1+x_{3})(1+(1+x_{3})x_{2})$ with
$0\leq\gamma<(1+x_{3})(1+(1+x_{3})x_{2})$, we have that
$0\leq(\delta(1+x_{3})+\beta)(1+(1+x_{3})x_{2})<x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$.
Let $n=(\delta(1+x_{3})+\beta)(1+(1+x_{3})x_{2})-\gamma$. The inequalities
above readily imply $\left\lvert
n\right\rvert<1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$. Thus,

$\left\lfloor\frac{\delta(1+x_{3})(1+(1+x_{3})x_{2})+\beta(1+(1+x_{3})x_{2}-\gamma}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor$

is equal to either $0$ or $-1$. We further write
$\gamma=\varepsilon(1+(1+x_{3})x_{2})+\eta$, where
$0\leq\eta<1+(1+x_{3})x_{2}$ and $0\leq\varepsilon<1+x_{3}$. Then, $h(b)$
becomes

$\displaystyle h(b)=\delta+\beta-
x_{1}\left\lfloor\frac{(\delta(1+x_{3})+\beta)(1+(1+x_{3})x_{2})-(\varepsilon(1+(1+x_{3})x_{2})+\eta)}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor\,.$

Subcase 1 of 4: Suppose $\delta=0$ and $\beta>\varepsilon$. Then, we have that
$n>0$, implying

$\displaystyle\left\lfloor\frac{\delta(1+x_{3})(1+(1+x_{3})x_{2})+\beta(1+(1+x_{3})x_{2}-\gamma}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor=0.$

Therefore, our equation for $h(b)$ simplifies to $h(b)=\beta$, which forces
$\beta=1$. Note that $\beta=1$ implies $\varepsilon=0$ as $\beta>\varepsilon$.
Thus, we have $h(b)=h(\eta x_{1}(1+x_{3})+1)=1$ whenever
$0\leq\eta\leq(1+x_{3})x_{2}$.

Subcase 2 of 4: Suppose $\delta=0$ and $\beta<\varepsilon$. Note that
$\varepsilon>0$ since $\beta<\varepsilon$. Then, we have that $n<0$, implying

$\displaystyle\left\lfloor\frac{\delta(1+x_{3})(1+(1+x_{3})x_{2})+\beta(1+(1+x_{3})x_{2}-\gamma}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor=-1.$

Consequently, our equation for $h(b)$ simplifies to $h(b)=\beta+x_{1}$, which
forces $\beta=0$ and $x_{1}=1$ since $x_{1}\geq 1$, i.e.,
$h(b)=h((\varepsilon(1+(1+x_{3})x_{2})+\eta)(1+x_{3}))=1$ whenever
$0<\varepsilon<1+x_{3}$ and $0\leq\eta\leq(1+x_{3})x_{2}$.

Subcase 3 of 4: Suppose $\delta=0$ and $\beta=\varepsilon$. If $\eta>0$, then
we have that $n<0$, implying

$\displaystyle\left\lfloor\frac{\delta(1+x_{3})(1+(1+x_{3})x_{2})+\beta(1+(1+x_{3})x_{2}-\gamma}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor=-1.$

In this case, $h(b)$ simplifies to $h(b)=\beta+x_{1}$, which again forces
$\beta=0$ and $x_{1}=1$. Since $\beta=\varepsilon$, it follows that
$\varepsilon=0$, and so we have that $h(b)=h(\eta(1+x_{3})=1$ whenever
$0<\eta\leq(1+x_{3})x_{2}$. Otherwise, if $\eta=0$, then we have that $n=0$,
implying

$\displaystyle\left\lfloor\frac{\delta(1+x_{3})(1+(1+x_{3})x_{2})+\beta(1+(1+x_{3})x_{2}-\gamma}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor=0.$

Therefore, our equation for $h(b)$ simplifies to $h(b)=\beta$, which forces
$\beta=1$ (and thus, $\varepsilon=1$), i.e.,
$h(b)=h((1+(1+x_{3})x_{2})x_{1}(1+x_{3})+1)=1$.

Subcase 4 of 4: Suppose $\delta\geq 1$. Then, we have that $n\geq 0$, implying

$\displaystyle\left\lfloor\frac{\delta(1+x_{3})(1+(1+x_{3})x_{2})+\beta(1+(1+x_{3})x_{2}-\gamma}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor=0.$

Therefore, our equation for $h(b)$ simplifies to $h(b)=\delta+\beta$, which
forces $\delta=1$ and $\beta=0$ since $\delta\geq 1$. That is, we have
$h(b)=h(((\varepsilon(1+(1+x_{3})x_{2})+\eta)x_{1}+1)(1+x_{3}))=1$ whenever
$0\leq\varepsilon<1+x_{3}$ and $0\leq\eta\leq(1+x_{3})x_{2}$.

We summarize the values of $b$ for which $h(b)=1$ that were just derived:

  * •

If $x_{1}=1$, $\beta=0$, and $\alpha=(1+x_{3})(1+(1+x_{3})x_{2})$, we have
$b=(1+x_{3})^{2}(1+(1+x_{3})x_{2}$.

  * •

If $x_{1}=1$, $\beta=\delta=0$, $0<\varepsilon<1+x_{3}$, and
$0\leq\eta\leq(1+x_{3})x_{2}$, we have
$b=(\eta(1+(1+x_{3})x_{2})+\eta)(1+x_{3})$.

  * •

If $x_{1}=1$, $\beta=\delta=\varepsilon=0$, and $0<\eta\leq(1+x_{3})x_{2}$, we
have $b=\eta(1+x_{3})$.

  * •

If $x_{1}\geq 1$, $\delta=\varepsilon=0$, $\beta=1$, and
$0\leq\eta\leq(1+x_{3})x_{2}$, we have $b=\eta x_{1}(1+x_{3})+1$.

  * •

If $x_{1}\geq 1$, $\delta=\eta=0$, and $\beta=\varepsilon=1$, we have
$b=(1+(1+x_{3})x_{2})x_{1}(1+x_{3})+1$.

  * •

If $x_{1}\geq 1$, $\delta=1$, $\beta=0$, $0\leq\varepsilon<1+x_{3}$, and
$0\leq\eta\leq(1+x_{3})x_{2}$, we have
$b=((\varepsilon(1+(1+x_{3})x_{2})+\eta)x_{1}+1)(1+x_{3})$.

Our next goal is to establish that (6) is always satisfied; recall that we are
in the case where $q_{j}=(1+x_{3})(1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2}))$. If
$q_{i}=(1+x_{3})(1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2}))$, the result is trivial.
If $q_{i}=(1+(1+x_{3})(1+(1+x_{3})x_{2})x_{1})(1+(1+x_{3})x_{2})$, note that
$b\neq 1+x_{3}$ since $h(1+x_{3})=1$. If $b>1+x_{3}$, we set $c=1+x_{3}$ from
which it is straightforward to compute that both sides of (6) are equal to
$bx_{2}-(1+(1+x_{3})x_{2})+\left\lfloor\frac{b}{1+x_{3}}\right\rfloor$.
Otherwise, if $b<1+x_{3}$, we may choose $c=1$ from which it follows that both
sides of (6) are equal to $(b-1)x_{2}$ since $1<b<1+x_{3}$ implies $1\leq
b-1<1+x_{3}$. Finally, if $q_{i}=(1+x_{3})(1+(1+x_{3})x_{2})$, we again write
$b=\alpha(1+x_{3})+\beta$, where $0\leq\beta<1+x_{3}$ and
$0\leq\alpha<1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$. Moreover, in the case that
$\alpha\neq x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$, we write $\alpha=\gamma
x_{1}+\delta$ with $\gamma=(\varepsilon(1+(1+x_{3})x_{2})+\eta)$, where
$0\leq\delta<x_{1}$, $0\leq\varepsilon<(1+x_{3})$, and
$0\leq\eta\leq(1+x_{3})x_{2}$. We consider the following possible cases:

Subcase 1 of 5: Suppose $\alpha=x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$. If
$x_{1}=1$, note that $\beta\neq 0$ since
$h((1+x_{3})^{2}(1+(1+x_{3})x_{2}))=1$. Thus, we may consider
$c=(1+x_{3})^{2}(1+(1+x_{3})x_{2})<(1+x_{3})^{2}(1+(1+x_{3})x_{2})+\beta=b$.
With this choice of $c$, it is straightforward to verify that both sides of
(6) are equal to $0$. Otherwise, if $x_{1}>1$, we consider $c=1$. Since
$x_{1}>1$ implies $-x_{1}(1+x_{3})(1+(1+x_{3})x_{2})<(\beta-
x_{3}-2)(1+(1+x_{3})x_{2})<0$, it is straightforward to verify that both sides
of (6) are equal to $(1+x_{3})(1+(1+x_{3})x_{2})-1$.

Subcase 2 of 5: Suppose $\alpha\neq x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$ with
$\delta>0$ and $\beta>0$. Then, choosing $c=(\gamma x_{1}+1)(1+x_{3})<(\gamma
x_{1}+\delta)(1+x_{3})+\beta=b$, it follows that both sides of (6) are equal
to $0$ since $0\leq\delta-1<x_{1}-1$ and $0<\beta<1+x_{3}$ together imply
$0<(\delta-1)(1+x_{3})+\beta<x_{1}(1+x_{3})$.

Subcase 3 of 5: Suppose $\alpha\neq x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$ with
$\delta>0$ and $\beta=0$. Note that $\delta\neq 1$ since $h((\gamma
x_{1}+1)(1+x_{3}))=1$. Therefore, $2\leq\delta<x_{2}$, so we may consider
$c=(\gamma x_{1}+1)(1+x_{3})<(\gamma x_{1}+\delta)(1+x_{3})=b$. Given this
choice, we find that both sides of (6) are equal to $0$ since
$0<\delta<x_{1}$.

Subcase 4 of 5: Suppose $\alpha\neq x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$ with
$\delta=0$ and $\beta>0$. If $\beta>\varepsilon$, then $\beta\neq 1$ since
otherwise, $\beta=1$ would force $\varepsilon=0$ and $h(\eta
x_{1}(1+x_{3})+1)=1$. Thus, it must be that $\beta>1$, thereby allowing us to
consider $c=\eta
x_{1}(1+x_{3})<(\varepsilon(1+(1+x_{3})x_{2})+\eta)x_{1}(1+x_{3})+\beta=b$.
With this choice of $c$ and since $\beta-1\geq\varepsilon$ with $\beta>1$, it
is straightforward to verify both sides of (6) are equal to
$\varepsilon(1+(1+x_{3})x_{2})$. Now, if $\beta<\varepsilon$, we have that
$\varepsilon>1$ since we assumed $\beta>0$. Therefore, we consider
$c=(1+(1+x_{3})x_{2})x_{1}(1+x_{3})+1<(\varepsilon(1+(1+x_{3})x_{2})+\eta)x_{1}(1+x_{3})+\beta=b$.
Then, both sides of (6) are equal to
$(\varepsilon-1)(1+(1+x_{3})x_{2})+\eta-1$. Lastly, if $\beta=\varepsilon$, we
consider two possibilities. If $\eta>0$, we may again choose
$c=(1+(1+x_{3})x_{2})x_{1}(1+x_{3})+1<(\varepsilon(1+(1+x_{3})x_{2})+\eta)x_{1}(1+x_{3})+\beta=b$
and find that both sides of (6) are equal to
$(\varepsilon-1)(1+(1+x_{3})x_{2})+\eta-1$. Otherwise, if $\eta=0$, note that
$\varepsilon=\beta\neq 1$ since otherwise,
$h(b)=h((1+(1+x_{3})x_{2})x_{1}(1+x_{3})+1)=1$. Therefore, the same value of
$c$, namely $c=(1+(1+x_{3})x_{2})x_{1}(1+x_{3})+1$, will again be strictly
less than $b$, from which it follows that both sides of (6) are equal to
$(\varepsilon-1)(1+(1+x_{3})x_{2})+\eta-1$.

Subcase 5 of 5: Suppose $\alpha\neq x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$ with
$\delta=0$ and $\beta=0$. In this case, $b=\gamma x_{1}(1+x_{3})$. Moreover,
note that $\gamma>0$ since otherwise, $b=0$ contradicting our bounds on $b$.
If $x_{1}=1$, then $h(b)=h(\gamma(1+x_{3}))=1$. Hence, we need only consider
when $x_{1}>1$. Since $\gamma>0$ and $x_{2}>1$, we may take $c=1+x_{3}<\gamma
x_{1}(1+x_{3})=b$, from which it is straightforward to find that both sides of
(6) are equal to $\gamma-1$.

This completes our second case.

Case: $q_{j}=(1+(1+x_{3})(1+(1+x_{3})x_{2})x_{1})(1+(1+x_{3})x_{2})$. Thus, we
consider $1\leq
b\leq(1+(1+x_{3})(1+(1+x_{3})x_{2})x_{1})(1+(1+x_{3})x_{2})-1$. Again, we will
start by identifying those values of $b$ that satisfy (5) and (7). It is
straightforward to verify that

$\displaystyle
h(b)=b-x_{1}\left\lfloor\frac{b(1+x_{3})}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor-
x_{2}\left\lfloor\frac{b(1+x_{3})}{1+(1+x_{3})x_{2}}\right\rfloor\,.$

Writing $b=\alpha(1+(1+x_{3})x_{2})+\beta$, where $0\leq\beta<1+x_{3}$ and
$0\leq\alpha<1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$ for
$\alpha,\beta\in\mathbb{Z}$, it follows that

$\displaystyle h(b)$ $\displaystyle=h(\alpha(1+(1+x_{3})x_{2})+\beta)$
$\displaystyle=\alpha+\beta-
x_{1}\left\lfloor\frac{(\alpha(1+(1+x_{3})x_{2})+\beta)(1+x_{3})}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor-
x_{2}\left\lfloor\frac{\beta(1+x_{3})}{1+(1+x_{3})x_{2}}\right\rfloor\,.$

There are two different possibilities for both $\alpha$ and $\beta$: either
$\alpha=x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$ or $\alpha=\varepsilon x_{1}+\eta$
where $0\leq\eta<x_{1}$ and $0\leq\varepsilon<(1+x_{3})(1+(1+x_{3})x_{2})$,
and either $\beta=(1+x_{3})x_{2}$ or $\beta=\gamma x_{2}+\delta$ where
$0\leq\delta<x_{2}$ and $0\leq\gamma<1+x_{3}$. We consider the following
subcases.

Subcase 1 of 4: Suppose $\alpha=x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$ and
$\beta=(1+x_{3})x_{2}$. Then,

$\displaystyle h(b)$
$\displaystyle=x_{1}(1+x_{3})(1+(1+x_{3})x_{2})+(1+x_{3})x_{2}$
$\displaystyle\quad-
x_{1}\left\lfloor\frac{x_{1}(1+x_{3})^{2}(1+(1+x_{3})x_{2})^{2}+(1+x_{3})^{2}x_{2}}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor-
x_{2}\left\lfloor\frac{(1+x_{3})^{2}x_{2}}{1+(1+x_{3})x_{2}}\right\rfloor$
$\displaystyle=x_{1}+x_{2}$ $\displaystyle>1.$

Subcase 2 of 4: Suppose $\alpha=x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$ and
$\beta\neq(1+x_{3})x_{2}$. Writing $\beta=\gamma x_{2}+\delta$, where
$0\leq\delta<x_{2}$ and $0\leq\gamma<1+x_{3}$, we have that

$\displaystyle h(b)$ $\displaystyle=x_{1}(1+x_{3})(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta$ $\displaystyle\quad-
x_{1}\left\lfloor\frac{(x_{1}(1+x_{3})(1+(1+x_{3})x_{2})^{2}+\beta)(1+x_{3})}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor-
x_{2}\left\lfloor\frac{(\gamma
x_{2}+\delta)(1+x_{3})}{1+(1+x_{3})x_{2}}\right\rfloor$ $\displaystyle=\delta-
x_{1}\left\lfloor\frac{\beta(1+x_{3})-(1+x_{3})(1+(1+x_{3})x_{2})}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor-
x_{2}\left\lfloor\frac{\delta(1+x_{3})-\gamma}{1+(1+x_{3})x_{2}}\right\rfloor$
$\displaystyle=\delta+x_{1}-x_{2}\cdot\left.\begin{cases}-1,&\delta=0,\gamma>0\\\
\phantom{-}0,&\text{otherwise}\end{cases}\right\\}\,.$

If $\delta=0$ and $\gamma>0$, then $h(b)=x_{1}+x_{2}>1$. Otherwise, we have
that $h(b)=\delta+x_{1}$. For this to be equal to $1$, it must be the case
that $\delta=0$ and $x_{1}=1$ since $x_{1}\geq 1$. Moreover, note that
$\beta=0$ here since $\delta=0$ implies $\gamma=0$ (otherwise, we are in the
previous case that $\delta=0$ and $\gamma>0$). Therefore, when $x_{1}=1$, we
have that $h((1+x_{3})(1+(1+x_{3})x_{2})^{2})=1$.

Subcase 3 of 4: Suppose $\alpha\neq x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$ and
$\beta=(1+x_{3})x_{2}$. Writing $\alpha=\varepsilon x_{1}+\eta$, where
$0\leq\eta<x_{1}$ and $0\leq\varepsilon<(1+x_{3})(1+(1+x_{3})x_{2})$, we have
that

$\displaystyle h(b)$ $\displaystyle=\varepsilon
x_{1}+\eta+(1+x_{3})x_{2}-x_{1}\left\lfloor\frac{((\varepsilon
x_{1}+\eta)(1+(1+x_{3})x_{2})+(1+x_{3})x_{2})(1+x_{3})}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor$
$\displaystyle\qquad\qquad-
x_{2}\left\lfloor\frac{(1+x_{3})^{2}x_{2}}{1+(1+x_{3})x_{2}}\right\rfloor$
$\displaystyle=\eta-
x_{1}\left\lfloor\frac{(\eta(1+(1+x_{3})x_{2})+(1+x_{3})x_{2})(1+x_{3})-\varepsilon}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor+x_{2}\,.$

Given that $0\leq\eta<x_{1}$ and
$0\leq\varepsilon<(1+x_{3})(1+(1+x_{3})x_{2})$, observe that

$\displaystyle\left\lvert(\eta(1+(1+x_{3})x_{2})+(1+x_{3})x_{2})(1+x_{3})-\varepsilon\right\rvert<1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2}).$

Therefore,
$\left\lfloor\frac{(\eta(1+(1+x_{3})x_{2})+(1+x_{3})x_{2})(1+x_{3})-\varepsilon}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor$
is equal to either $0$ or $-1$. If

$\displaystyle\left\lfloor\frac{(\eta(1+(1+x_{3})x_{2})+(1+x_{3})x_{2})(1+x_{3})-\varepsilon}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor=-1\,,$

then $h(b)=\eta+x_{1}+x_{2}>1$. Otherwise, if

$\displaystyle\left\lfloor\frac{(\eta(1+(1+x_{3})x_{2})+(1+x_{3})x_{2})(1+x_{3})-\varepsilon}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor=0\,,$

then either (A) $0<\eta<x_{1}$ or (B) $\eta=0$ with
$0\leq\varepsilon\leq(1+x_{3})^{2}x_{2}$. If (A) holds, then
$h(b)=\eta+x_{2}>1$ since $\eta>0$. If (B) holds, then our same equation
forces $\eta=0$ and $x_{2}=1$ when $\varepsilon\leq(1+x_{3})^{2}$, which means
that $h(\varepsilon x_{1}(2+x_{3})+(1+x_{3}))=1$ whenever
$0\leq\varepsilon\leq(1+x_{3})^{2}$.

Subcase 4 of 4: Suppose $\alpha\neq x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$ and
$\beta\neq(1+x_{3})x_{2}$. Writing $\alpha=\varepsilon x_{1}+\eta$ where
$0\leq\eta<x_{1}$ and $0\leq\varepsilon<(1+x_{3})(1+(1+x_{3})x_{2})$ and
$\beta=\gamma x_{2}+\delta$ where $0\leq\delta<x_{2}$ and
$0\leq\gamma<1+x_{3}$, we have that

$\displaystyle h(b)$ $\displaystyle=\varepsilon x_{1}+\eta+\gamma
x_{2}+\delta-x_{1}\left\lfloor\frac{((\varepsilon
x_{1}+\eta)(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta)(1+x_{3})}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor$
$\displaystyle\qquad\qquad-x_{2}\left\lfloor\frac{(\gamma
x_{2}+\delta)(1+x_{3})}{1+(1+x_{3})x_{2}}\right\rfloor$
$\displaystyle=\eta+\delta-
x_{1}\left\lfloor\frac{(\eta(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta)(1+x_{3})-\varepsilon}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor-
x_{2}\left\lfloor\frac{\delta(1+x_{3})-\gamma}{1+(1+x_{3})x_{2}}\right\rfloor$
$\displaystyle=\eta+\delta-
x_{1}\left\lfloor\frac{(\eta(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta)(1+x_{3})-\varepsilon}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor-
x_{2}\cdot\left.\begin{cases}-1,&\delta=0,\gamma>0\\\
\phantom{-}0,&\text{otherwise}\end{cases}\right\\}\,.$

Given the bounds on $\eta$, $\gamma$, $\delta$, and $\varepsilon$, observe
that

$\displaystyle\left\lvert(\eta(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta)(1+x_{3})-\varepsilon\right\rvert<1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2}).$

Therefore,

$\left\lfloor\frac{(\eta(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta)(1+x_{3})-\varepsilon}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor$

is equal to either $0$ or $-1$.

  1. (i)

Suppose $\delta=0$ and $\gamma>0$. If

$\displaystyle\left\lfloor\frac{(\eta(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta)(1+x_{3})-\varepsilon}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor=-1\,,$

then $h(b)=\eta+x_{1}+x_{2}>1$. Otherwise, if

$\displaystyle\left\lfloor\frac{(\eta(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta)(1+x_{3})-\varepsilon}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor=0\,,$

then either (A) $0<\eta<x_{1}$ or (B) $\eta=0$ with
$0\leq\varepsilon\leq\gamma x_{2}(1+x_{3})$. If (A) holds, then
$h(b)=\eta+x_{2}>1$ since $\eta>0$ and $x_{2}\geq 1$. If (B) holds, then our
same equation forces $\eta=0$ and $x_{2}=1$ when
$\varepsilon\leq\gamma(1+x_{3})$, which means that $h(\varepsilon
x_{1}(2+x_{3})+\gamma)=1$ whenever $0\leq\varepsilon\leq\gamma(1+x_{3})$.

  2. (ii)

Suppose otherwise, i.e., $\delta=0$ and $\gamma>0$ does not hold. If

$\displaystyle\left\lfloor\frac{(\eta(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta)(1+x_{3})-\varepsilon}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor=-1\,,$

note that $\eta=0$. Then, $h(b)=\delta+x_{1}$. For this to equal $1$, it must
be the case that $\delta=0$ (and hence $\gamma=0$ since otherwise, we would be
in the previous case) and $x_{1}=1$. Since $\gamma=\delta=0$, we have that
$\beta=0$. Therefore, given that $\eta=0$ as well, it must be that
$\varepsilon>0$ since otherwise, $b=0$ contradicting our bounds on $b$. In
this case, we have that $h(\varepsilon(1+(1+x_{3})x_{2}))=1$ whenever
$0<\varepsilon<(1+x_{3})(1+(1+x_{3})x_{2})$. Otherwise, if

$\displaystyle\left\lfloor\frac{(\eta(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta)(1+x_{3})-\varepsilon}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor=0\,,$

then either (A) $0<\eta<x_{1}$ or (B) $\eta=0$ with
$0\leq\varepsilon\leq(\gamma x_{2}+\delta)(1+x_{3})$. If (A) holds, then
$h(b)=\eta+\delta$, which setting equal to $1$ forces $\eta=1$ and $\delta=0$
since $\eta>0$. Note that $\delta=0$ forces $\gamma=0$ since otherwise, we
would be in the previous case. Therefore, we have that $h((\varepsilon
x_{1}+1)(1+(1+x_{3})x_{2}))=1$ whenever
$0\leq\varepsilon<(1+x_{3})(1+(1+x_{3})x_{2})$. On the other hand, if (B)
holds, then our same equation forces $\eta=0$ and $\delta=1$ when
$0\leq\varepsilon\leq(\gamma x_{2}+1)(1+x_{3})$, which means that
$h(\varepsilon x_{1}(1+(1+x_{3})x_{2})+\gamma x_{2}+1)=1$ whenever
$0\leq\gamma<1+x_{3}$ and $0\leq\varepsilon\leq(\gamma x_{2}+1)(1+x_{3})$.

We summarize the values of $b$ for which $h(b)=1$ that were just derived:

  * •

If $x_{1}=1$, $\beta=0$, and $0<\varepsilon\leq(1+x_{3})(1+(1+x_{3})x_{2})$,
we have $b=\varepsilon(1+(1+x_{3})x_{2})$.

  * •

If $x_{2}=1$, $\delta=\eta=0$, $0<\gamma\leq 1+x_{3}$, and
$0\leq\varepsilon\leq\gamma(1+x_{3})$, we have $b=\varepsilon
x_{1}(2+x_{3})+\gamma$.

  * •

If $\beta=0$, $\eta=1$, and $0\leq\varepsilon<(1+x_{3})(1+(1+x_{3})x_{2})$, we
have $b=(\varepsilon x_{1}+1)(1+(1+x_{3})x_{2})$.

  * •

If $\eta=0$, $\delta=1$, $0\leq\gamma<1+x_{3}$, and
$0\leq\varepsilon\leq(\gamma x_{2}+1)(1+x_{3})$, we have $b=\varepsilon
x_{1}(1+(1+x_{3})x_{2})+\gamma x_{2}+1$.

Our next goal is to establish that (6) is always satisfied; recall that we are
in the case where
$q_{j}=(1+(1+x_{3})(1+(1+x_{3})x_{2})x_{1})(1+(1+x_{3})x_{2})$. If
$q_{i}=(1+(1+x_{3})(1+(1+x_{3})x_{2})x_{1})(1+(1+x_{3})x_{2})$, the result is
trivial. If $q_{i}=(1+x_{3})(1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2}))$, we write
$b=\alpha(1+(1+x_{3})x_{2})+\beta$, where $0\leq\beta<1+(1+x_{3})x_{2}$ and
$0\leq\alpha<1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$. Note that $b\neq
1+(1+x_{3})x_{2}$ since $h(1+(1+x_{3})x_{2})=1$. If $b>1+(1+x_{3})x_{2}$, we
set $c=1+(1+x_{3})x_{2}$ from which it is straightforward to compute that both
sides of (6) are equal to

$\displaystyle(\alpha-1)(1+x_{3})+\left\lfloor\frac{\beta(1+x_{3})}{1+(1+x_{3})x_{2}}\right\rfloor\,.$

Otherwise, if $b<1+(1+x_{3})x_{2}$, note that $\alpha=0$ and $b=\beta$.
Therefore, to ensure we satisfy (5), we consider
$2\leq\beta<1+(1+x_{3})x_{2}$. Moreover, note that $\beta\not\equiv 1\bmod
x_{2}$ since otherwise $h(b)=h(\beta)=1$. Now, suppose $\beta=(1+x_{3})x_{2}$.
If $x_{2}=1$, then $h(\beta)=h(1+x_{3})=1$, so we may assume $x_{2}>1$.
Setting $c=1$ and since $x_{2}>1$, it is straightforward to compute that both
sides of (6) are equal to $x_{3}$. Otherwise, if $\beta\neq(1+x_{3})x_{2}$, we
write $\beta=\gamma x_{2}+\delta$, where $0\leq\delta<x_{2}$ and
$0\leq\gamma<1+x_{3}$. Note that $\delta\neq 1$ since $h(\gamma x_{2}+1)=1$
for $0\leq\gamma<1+x_{3}$. Suppose $\delta>1$. Then, choosing $c=\gamma
x_{2}+1<\gamma x_{2}+\delta=b$, it is straightforward to show that both sides
of (6) are equal to $0$. On the other hand suppose $\delta=0$, so
$\beta=\gamma x_{2}$, where $\gamma>0$. If $x_{2}=1$, then $h(\beta)=1$, so we
may assume $x_{2}>1$. Taking $c=1$ and observing that $x_{2}>1$ implies
$-x_{2}(1+x_{3})<-(1+x_{3})-\gamma$, it is straightforward to compute that
both sides of (6) are equal to $\gamma-1$.

Finally, if $q_{i}=(1+x_{3})(1+(1+x_{3})x_{2})$, the analysis becomes a bit
more complicated. We start by again writing
$b=\alpha(1+(1+x_{3})x_{2})+\beta$, where $0\leq\beta<1+(1+x_{3})x_{2}$ and
$0\leq\alpha<1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$. Suppose
$\alpha=x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$. If $x_{1}=1$, then $\beta\neq 0$
since otherwise $h(b)=1$. Thus, we may consider
$c=(1+x_{3})(1+(1+x_{3})x_{2})^{2}<(1+x_{3})(1+(1+x_{3})x_{2})^{2}+\beta=b$
from which it is straightforward to compute that both sides of (6) are equal
to $0$. If $x_{1}>1$, observe that

$-x_{1}(1+x_{3})(1+(1+x_{3})x_{2})<-(1+x_{3})(2+(1+x_{3})x_{2})\leq(\beta-1)(1+x_{3})-(1+x_{3})(1+(1+x_{3})x_{2})<0\,.$

Choosing $c=1$, the previous inequality readily gives that both sides of (6)
are equal to $(1+x_{3})(1+(1+x_{3})x_{2})-1$.

Now, suppose $\alpha\neq x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$. Then, we may write
$\alpha=\varepsilon x_{1}+\eta$, where $0\leq\eta<x_{1}$ and
$0\leq\varepsilon<(1+x_{3})(1+(1+x_{3})x_{2})$, and so $b=(\varepsilon
x_{1}+\eta)(1+(1+x_{3})x_{2})+\beta$. Suppose $\beta=(1+x_{3})x_{2}$. If
$\eta\geq 1$, then we may consider $c=(\varepsilon
x_{1}+1)(1+(1+x_{3})x_{2})<(\varepsilon
x_{1}+\eta)(1+(1+x_{3})x_{2})+(1+x_{3})x_{2}=b$ from which it is
straightforward to show that both sides of (6) are equal to $0$. Otherwise, if
$\eta=0$, note that we may assume $x_{2}>1$ since $x_{2}=1$ gives that
$h(b)=h(\varepsilon x_{1}(2+x_{3})+(1+x_{3}))=1$.

We consider two possible cases, namely when $0\leq\varepsilon<1+x_{3}$ and
when $1+x_{3}\leq\varepsilon<(1+x_{3})(1+(1+x_{3})x_{2})$. If
$0\leq\varepsilon<1+x_{3}$, we consider $c=\varepsilon
x_{1}(1+(1+x_{3})x_{2})+x_{2}x_{3}+1$ which is strictly less than
$b=\varepsilon x_{1}(1+(1+x_{3})x_{2})+(1+x_{3})x_{2}$ as $x_{2}>1$. With this
choice of $c$ and since $0\leq\varepsilon<1+x_{3}$, it is straightforward to
show that both sides of (6) are equal to $0$. On the other hand, if
$1+x_{3}\leq\varepsilon<(1+x_{3})(1+(1+x_{3})x_{2})$, we may consider
$c=1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})<\varepsilon
x_{1}(1+(1+x_{3})x_{2})+(1+x_{3})x_{2}=b$ from which it is straightforward to
compute that both sides of (6) are equal to

$\displaystyle\varepsilon-(1+x_{3})+\left\lfloor\frac{(1+x_{3})^{2}x_{2}-\varepsilon}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor\,.$

Now, suppose $\beta\neq(1+x_{3})x_{2}$. Then, we may write $\beta=\gamma
x_{2}+\delta$, where $0\leq\delta<x_{2}$ and $0\leq\gamma<1+x_{3}$, and so $b$
can be written as $b=(\varepsilon x_{1}+\eta)(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta$. We consider the following possible subcases.

Subcase 1 of 4: Suppose $\eta>0$ and $\delta>0$. We consider $c=(\varepsilon
x_{1}+1)(1+(1+x_{3})x_{2})<(\varepsilon x_{1}+1)(1+(1+x_{3})x_{2})+\gamma
x_{2}+\delta=b$. Since $0<\eta<x_{1}$ and $0<\gamma
x_{2}+\delta<(1+x_{3})x_{2}$, it follows that
$0<(\eta-1)(1+(1+x_{3})x_{2})+\gamma x_{2}+\delta<x_{1}(1+(1+x_{3})x_{2})$.
Therefore, it is straightforward to verify that both sides of (6) are equal to
$0$.

Subcase 2 of 4: Suppose $\eta=0$ and $\delta>0$. If $\varepsilon\leq(\gamma
x_{2}+1)(1+x_{3})$, note that $\delta\neq 1$ since otherwise,
$h(b)=h(\varepsilon x_{1}(1+(1+x_{3})x_{2})+\gamma x_{2}+1)=1$. Thus, we have
that $\delta>1$, and we consider $c=\varepsilon x_{1}(1+(1+x_{3})x_{2})+\gamma
x_{2}+1<\varepsilon x_{1}(1+(1+x_{3})x_{2})+\gamma x_{2}+\delta=b$. Given that
$\varepsilon\leq(\gamma x_{2}+1)(1+x_{3})$ and $1<\delta<x_{2}$, it is
straightforward to verify that our choice of $c$ gives that both sides of (6)
are equal to $0$. Otherwise, if $(\gamma x_{2}+1)(1+x_{3})<\varepsilon$ (and
hence, $\varepsilon>1+x_{3}$), we consider
$c=1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})<\varepsilon
x_{1}(1+(1+x_{3})x_{2})+\gamma x_{2}+\delta=b$. With this choice of $c$, it is
straightforward to verify that both sides of (6) are equal to

$\displaystyle\varepsilon-(1+x_{3})+\left\lfloor\frac{(\gamma
x_{2}+\delta)(1+x_{3})-\varepsilon}{1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})}\right\rfloor\,.$

Subcase 3 of 4: Suppose $\eta>0$ and $\delta=0$. If $\gamma>0$, we may
consider $c=(\varepsilon x_{1}+1)(1+(1+x_{3})x_{2})<(\varepsilon
x_{1}+\eta)(1+(1+x_{3})x_{2})+\gamma x_{2}=b$ from which it is straightforward
to compute that both sides of (6) are equal to $0$. On the other hand, if
$\gamma=0$, note that $\eta\neq 1$ since otherwise, $h(b)=h((\varepsilon
x_{1}+\eta)(1+(1+x_{3})x_{2}))=1$. Therefore, we may again choose
$c=(\varepsilon x_{1}+1)(1+(1+x_{3})x_{2})<(\varepsilon
x_{1}+\eta)(1+(1+x_{3})x_{2})=b$. Since $1<\eta<x_{1}$, it is straightforward
to verify that both sides of (6) are equal to $0$.

Subcase 4 of 4: Suppose $\eta=\delta=0$. Further suppose $\gamma
x_{2}(1+x_{3})<\varepsilon$. If $\gamma>0$, then it follows that $\gamma
x_{2}\geq 1$. Therefore, our assumed inequality implies $\varepsilon>1+x_{3}$,
so we may consider $c=1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})<\varepsilon
x_{1}(1+(1+x_{3})x_{2})+\gamma x_{2}=b$. Since $\gamma
x_{2}(1+x_{3})<\varepsilon$, our choice of $c$ readily gives that both sides
of (6) are equal to $\varepsilon-x_{3}-2$. Otherwise, if $\gamma=0$ (and
hence, $\varepsilon>0$ since we assumed $\gamma x_{2}(1+x_{3})<\varepsilon$),
we may assume $x_{1}>1$ since otherwise, we would have that
$h(b)=h(\varepsilon(1+(1+x_{3})x_{2}))=1$. Thus, since $x_{1}>1$, we may take
$c=1+(1+x_{3})x_{2}<\varepsilon x_{1}(1+(1+x_{3})x_{2})=b$. Observe that the
bounds on $\varepsilon$ and $x_{1}>1$ imply
$-x_{1}(1+x_{3})(1+(1+x_{3})x_{2})\leq-2(1+x_{3})(1+(1+x_{3})x_{2})<-(1+x_{3})(1+(1+x_{3})x_{2})-\varepsilon<0$.
Consequently, it is straightforward to verify that both sides of (6) are equal
to $\varepsilon-1$. Now, suppose $\gamma x_{2}(1+x_{3})>\varepsilon$. Note
that $\gamma\neq 0$ since otherwise, $\varepsilon<0$ contradicting our initial
bounds on $\varepsilon$. Thus, we have that $\gamma>0$. Moreover, if
$x_{2}=1$, it follows that $h(b)=h(\varepsilon x_{1}(2+x_{3})+\gamma)=1$, so
we may assume $x_{2}>1$. Given the addition restriction that
$\varepsilon\leq(\gamma x_{2}-1)(1+x_{3})$, we may choose $c=1$ from which the
inequality $\varepsilon\leq(\gamma x_{2}-1)(1+x_{3})$ readily implies both
sides of (6) are equal to $\varepsilon$. However, for $(\gamma
x_{2}-1)(1+x_{3})<\varepsilon<\gamma x_{2}(1+x_{3})$, we consider
$c=1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})$. Note that $x_{2}>1$ and $\gamma>0$
together with our restriction on $\varepsilon$ imply that
$\varepsilon>1+x_{3}$. Therefore, we satisfy $c<\varepsilon
x_{1}(1+(1+x_{3})x_{2})+\gamma x_{2}=b$, and a straightforward computation
gives that both sides of (6) are equal to $\varepsilon-(1+x_{3})$. Finally,
suppose $\varepsilon=\gamma x_{2}(1+x_{3})$. Given this equality, note that
neither $\gamma$ nor $\varepsilon$ can be equal to $0$ since otherwise, we
would have $\eta=\delta=\gamma=\varepsilon=0$, implying $b=0$. This, of
course, contradicts the bounds on $b$. Moreover, we may again assume $x_{2}>1$
(and thus, $\varepsilon>1+x_{3}$) since otherwise, $h(b)=h(\varepsilon
x_{1}(2+x_{3})+\gamma)=1$. Since $\varepsilon>1+x_{3}$, we may consider
$c=1+x_{1}(1+x_{3})(1+(1+x_{3})x_{2})<\varepsilon
x_{1}(1+(1+x_{3})x_{2})+\gamma x_{2}=b$. This choice of $c$ readily gives that
both sides of (6) are equal to $\varepsilon-(1+x_{3})$.

In any case, we find that both sides of (6) are equivalent for each possible
$q_{i}$, thereby completing our third and final case. Thus, we have
established IDP for $\mathbf{r}$-vectors of type $(vi)$.

###  A.6. Proof of IDP for type $(vii)$ in Theorem 3.9

Here, we verify IDP for $\mathbf{r}$-vectors of type $(vii)$ using Theorem
3.3. Again, we must consider three cases corresponding to three possible
values of $q_{j}$.

Case: $q_{j}=1+x_{3}$. Since $1\leq b\leq x_{3}$ in this case, it is
straightforward to verify that $h(b)=b$. Hence, the $b$-values we are required
to check in (5) are $2\leq b\leq x_{3}$. To verify that (6) always has the
desired solution, we consider three cases. If $q_{i}=1+x_{3}$, the result is
trivial. If $q_{i}=(1+x_{3})(1+x_{1}(1+x_{3}))$, then we may select $c=1$,
from which it follows that both sides of (6) are equal to
$(b-1)(1+x_{1}(1+x_{3}))$. If $q_{i}=(1+(1+x_{3})x_{1})(1+(1+x_{3})x_{2})$,
then we may again set $c=1$, from which it is straightforward to compute that
both sides of (6) are equal to $(b-1)(x_{1}+x_{2}+x_{1}x_{2}(1+x_{3}))$. This
completes our first case.

Case: $q_{j}=(1+x_{3})(1+x_{1}(1+x_{3}))$. It is straightforward to verify
that

$h(b)=b-x_{1}\left\lfloor\frac{b}{1+x_{1}(1+x_{3})}\right\rfloor-
x_{3}\left\lfloor\frac{b}{1+x_{3}}\right\rfloor\,,$

where the values of $b$ range from $1$ to $(1+x_{3})(1+x_{1}(1+x_{3}))-1$. To
verify that (6) always has the desired solution, we consider three cases. If
$q_{i}=(1+x_{3})(1+x_{1}(1+x_{3}))$, the result is trivial. If
$q_{i}=1+x_{3}$, then we write $b=\alpha(1+x_{1}(1+x_{3}))+\beta$, where
$0\leq\beta<1+x_{1}(1+x_{3})$ and $0\leq\alpha<1+x_{3}$ for
$\alpha,\beta\in\mathbb{Z}$. Consequently, we have that

$\displaystyle h(b)$ $\displaystyle=h(\alpha(1+x_{1}(1+x_{3}))+\beta)$
$\displaystyle=\alpha(1+x_{1}(1+x_{3}))+\beta-\alpha x_{1}-\alpha
x_{1}x_{3}-x_{3}\left\lfloor\frac{\alpha+\beta}{1+x_{3}}\right\rfloor$
$\displaystyle=\alpha+\beta-
x_{3}\left\lfloor\frac{\alpha+\beta}{1+x_{3}}\right\rfloor\,.$

If $\beta>0$, we may select $c=1$, from which it follows that both sides of
(6) are equal to $\alpha$. If $\beta=0$, then our formula for $h(b)$ reduces
to

$\displaystyle h(b)=h(\alpha(1+x_{1}(1+x_{3})))=\alpha$

since $0\leq\alpha<1+x_{3}$. Thus, to satisfy (5), it must be that $\alpha>1$,
implying $b=\alpha(1+x_{1}(1+x_{3}))>1+x_{1}(1+x_{3})$. In this case, taking
$c=1+x_{1}(1+x_{3})$, it is straightforward to verify that both sides of (6)
are equal to $\alpha-1$, and (7) is satisfied as $h(1+x_{1}(1+x_{3}))=1$.
Finally, if $q_{i}=(1+(1+x_{3})x_{1})(1+(1+x_{3})x_{2})$, then we write
$b=\alpha(1+x_{3})+\beta$, where $0\leq\beta<1+x_{3}$ and
$0\leq\alpha<1+x_{1}(1+x_{3})$ for $\alpha,\beta\in\mathbb{Z}$. Consequently,
since $0\leq\beta<1+x_{3}$, we have that

$\displaystyle h(b)$ $\displaystyle=h(\alpha(1+x_{3})+\beta)$
$\displaystyle=\alpha(1+x_{3})+\beta-
x_{1}\left\lfloor\frac{\alpha(1+x_{3})+\beta}{1+x_{1}(1+x_{3})}\right\rfloor-\alpha
x_{3}-x_{3}\left\lfloor\frac{\beta}{1+x_{3}}\right\rfloor$
$\displaystyle=\alpha+\beta-
x_{1}\left\lfloor\frac{\alpha(1+x_{3})+\beta}{1+x_{1}(1+x_{3})}\right\rfloor\,.$

If $\beta>0$, we may select $c=1$, from which it is straightforward to verify
that both sides of (6) are equal to $\alpha(1+(1+x_{3})x_{2})+(\beta-1)x_{2}$
(since $0\leq\beta-1<x_{3}$). On the other hand, if $\beta=0$, then our
formula for $h(b)$ reduces to

$\displaystyle
h(b)=h(\alpha(1+x_{3}))=\alpha-\left\lfloor\frac{\alpha(1+x_{3})}{1+x_{1}(1+x_{3})}\right\rfloor.$

In order to satisfy (5), it must be that $\alpha>1$, which implies
$b=\alpha(1+x+3)>1+x_{3}$. Thus, in this case, we consider $c=1+x_{3}$.
Clearly, $h(1+x_{3})=1$, giving (7), and moreover, it is straightforward to
verify that both sides of (6) when $c=1+x_{3}$ are equal to
$(\alpha-1)(1+(1+x_{3})x_{2})$. This completes our second case.

Case: $q_{j}=(1+(1+x_{3})x_{1})(1+(1+x_{3})x_{2})$. We first identify those
values of $b$ that satisfy (5) and (7). It is straightforward to verify that

$h(b)=b-x_{1}\left\lfloor\frac{b(1+x_{3})}{(1+(1+x_{3})x_{1})(1+(1+x_{3})x_{2})}\right\rfloor-
x_{2}\left\lfloor\frac{b(1+x_{3})}{(1+(1+x_{3})x_{2})}\right\rfloor\,.$

Writing $b=\alpha(1+(1+x_{3})x_{2})+\beta$, where
$0\leq\beta\leq(1+x_{3})x_{2}$ and $0\leq\alpha\leq(1+x_{3})x_{1}$, it follows
that

$\displaystyle h(b)$ $\displaystyle=h(\alpha(1+(1+x_{3})x_{2})+\beta)$
$\displaystyle=\alpha+\beta-
x_{1}\left\lfloor\frac{\alpha(1+x_{3})(1+(1+x_{3})x_{2})+\beta(1+x_{3})}{(1+(1+x_{3})x_{1})(1+(1+x_{3})x_{2})}\right\rfloor-
x_{2}\left\lfloor\frac{\beta(1+x_{3})}{1+(1+x_{3})x_{2}}\right\rfloor\,.$

Now, writing $\beta=\gamma x_{2}+\delta$, where $0\leq\delta<x_{2}$ and
$0\leq\gamma\leq 1+x_{3}$, it follows that

$\displaystyle h(b)=\alpha+\delta-
x_{1}\left\lfloor\frac{\alpha(1+x_{3})(1+(1+x_{3})x_{2})+(\gamma
x_{2}+\delta)(1+x_{3})}{(1+(1+x_{3})x_{1})(1+(1+x_{3})x_{2})}\right\rfloor-
x_{2}\left\lfloor\frac{\delta(1+x_{3})-\gamma}{1+(1+x_{3})x_{2}}\right\rfloor\,.$

Since $0\leq\delta<x_{2}$ and $0\leq\gamma\leq 1+x_{3}$, observe that

$\displaystyle\left\lfloor\frac{\delta(1+x_{3})-\gamma}{1+(1+x_{3})x_{2}}\right\rfloor=\begin{cases}-1,&\delta=0,\gamma>0\\\
\phantom{-}0,&\text{otherwise}\quad.\end{cases}$

We further write $\alpha=\varepsilon x_{1}+\eta$, where $0\leq\eta<x_{1}$ and
$0\leq\varepsilon\leq 1+x_{3}$. Then,

$\displaystyle h(b)$ $\displaystyle=\eta+\delta-
x_{1}\left\lfloor\frac{(\eta(1+x_{3})-\varepsilon)(1+(1+x_{3})x_{2})+(\gamma
x_{2}+\delta)(1+x_{3})}{(1+(1+x_{3})x_{1})(1+(1+x_{3})x_{2})}\right\rfloor-